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The impulse $\\delta_0$ is taken as a primitive signal seed and its shift orbit supplies the address system. Residue projectors split a sequence into exact polyphase channels, nested divisibility gives exact refinement, and every finite impulse-response operator becomes a matrix of packet transfers. The standard perfect-reconstruction condition for a filter bank is therefore an analysis--synthesis identity in the common operator language of the series. Haar analysis and every finite-depth wavelet-packet transform are treated on complex $\\ell^2$, together with concrete finite-signal models and a predict--update lifting step. The edge problem is separated into two statements. When a discontinuity packet is known an","e into exact polyphase channels, nested divisibility gives exact refinement, and every finite impulse-response operator becomes a matrix of packet transfers. The standard perfect-reconstruction condition for a filter bank is therefore an analysis--synthesis identity in the common operator language of the series. Haar analysis and every finite-depth wavelet-packet transform are treated on complex $\\ell^2$, together with concrete finite-signal models and a predict--update lifting step. The edge problem is separated into two statements. When a discontinuity packet is known and retained exactly, approximation error is exactly the approximation error of the remaining smooth packet. No spectral approximation of the jump is performed. When the packet is estimated, a symmetric-difference bound isolates mask error, amplitude error, and smooth-background error. Numerical studies compare global Fourier truncation, exact known-packet reconstruction, and a deliberately narrow positive-contrast estimator in one and two dimensions, including curved boundaries, corners, and multiple amplitudes. The …","Let $\\ell^2(\\bbZ;\\bbC)$, abbreviated $\\ell^2(\\bbZ)$, be the Hilbert space of square-summable complex discrete signals. The inner product is conjugate-linear in its first argument. A superscript $*$ on a bounded operator denotes its Hilbert adjoint. Write $\\delta_n$ for the unit impulse at address $n$ and let the bilateral shift be\n\\[\n(Sx)[n]=x[n-1].\n\\]\nThen\n\\[\nS^n\\delta_0=\\delta_n.\n\\]\nThe primitive signal seed is therefore $s=\\delta_0$, the address operators are the shift words $S^n$, and the completion is the $\\ell^2$ norm.","reviated $\\ell^2(\\bbZ)$, be the Hilbert space of square-summable complex discrete signals. The inner product is conjugate-linear in its first argument. A superscript $*$ on a bounded operator denotes its Hilbert adjoint. Write $\\delta_n$ for the unit impulse at address $n$ and let the bilateral shift be\n\\[\n(Sx)[n]=x[n-1].\n\\]\nThen\n\\[\nS^n\\delta_0=\\delta_n.\n\\]\nThe primitive signal seed is therefore $s=\\delta_0$, the address operators are the shift words $S^n$, and the completion is the $\\ell^2$ norm.","summable complex discrete signals. The inner product is conjugate-linear in its first argument. A superscript $*$ on a bounded operator denotes its Hilbert adjoint. Write $\\delta_n$ for the unit impulse at address $n$ and let the bilateral shift be\n\\[\n(Sx)[n]=x[n-1].\n\\]\nThen\n\\[\nS^n\\delta_0=\\delta_n.\n\\]\nThe primitive signal seed is therefore $s=\\delta_0$, the address operators are the shift words $S^n$, and the completion is the $\\ell^2$ norm.",". The inner product is conjugate-linear in its first argument. A superscript $*$ on a bounded operator denotes its Hilbert adjoint. Write $\\delta_n$ for the unit impulse at address $n$ and let the bilateral shift be\n\\[\n(Sx)[n]=x[n-1].\n\\]\nThen\n\\[\nS^n\\delta_0=\\delta_n.\n\\]\nThe primitive signal seed is therefore $s=\\delta_0$, the address operators are the shift words $S^n$, and the completion is the $\\ell^2$ norm.","\\begin{theorem}[Impulse seed theorem]\\label{result:01}\nThe closed shift-generated hull of $\\delta_0$ is all of $\\ell^2(\\bbZ)$:\n\\[\n\\boxed{\n\\overline{\\operatorname{span}\\{S^n\\delta_0:n\\in\\bbZ\\}}^{\\,\\ell^2}\n=\\ell^2(\\bbZ).\n}\n\\]\nEvery finitely supported signal has the exact finite expression\n\\[\nx=\\sum_{n\\in\\bbZ}x[n]S^n\\delta_0.\n\\]\n\\end{theorem}","\\begin{proof}\nThe shift orbit is the standard orthonormal basis $\\{\\delta_n\\}_{n\\in\\bbZ}$. For $x\\in\\ell^2$, the finite truncations\n\\[\nx_F=\\sum_{n\\in F}x[n]\\delta_n,\\qquad F\\subset\\bbZ\\text{ finite},\n\\]\nsatisfy $\\|x-x_F\\|_2^2=\\sum_{n\\notin F}|x[n]|^2\\to0$ as the finite sets exhaust $\\bbZ$. Thus the unordered Hilbert-space sum $\\sum_nx[n]\\delta_n$ converges to $x$, and the finite-support statement is exact. The shift is a surjective isometry because $n\\mapsto n-1$ is a bijection of $\\bbZ$, with inverse shift $n\\mapsto n+1$. This proves the closed-span assertion.\n\\end{proof}","mal basis $\\{\\delta_n\\}_{n\\in\\bbZ}$. For $x\\in\\ell^2$, the finite truncations\n\\[\nx_F=\\sum_{n\\in F}x[n]\\delta_n,\\qquad F\\subset\\bbZ\\text{ finite},\n\\]\nsatisfy $\\|x-x_F\\|_2^2=\\sum_{n\\notin F}|x[n]|^2\\to0$ as the finite sets exhaust $\\bbZ$. Thus the unordered Hilbert-space sum $\\sum_nx[n]\\delta_n$ converges to $x$, and the finite-support statement is exact. The shift is a surjective isometry because $n\\mapsto n-1$ is a bijection of $\\bbZ$, with inverse shift $n\\mapsto n+1$. This proves the closed-span assertion.\n\\end{proof}","}$. For $x\\in\\ell^2$, the finite truncations\n\\[\nx_F=\\sum_{n\\in F}x[n]\\delta_n,\\qquad F\\subset\\bbZ\\text{ finite},\n\\]\nsatisfy $\\|x-x_F\\|_2^2=\\sum_{n\\notin F}|x[n]|^2\\to0$ as the finite sets exhaust $\\bbZ$. Thus the unordered Hilbert-space sum $\\sum_nx[n]\\delta_n$ converges to $x$, and the finite-support statement is exact. The shift is a surjective isometry because $n\\mapsto n-1$ is a bijection of $\\bbZ$, with inverse shift $n\\mapsto n+1$. This proves the closed-span assertion.\n\\end{proof}","inite truncations\n\\[\nx_F=\\sum_{n\\in F}x[n]\\delta_n,\\qquad F\\subset\\bbZ\\text{ finite},\n\\]\nsatisfy $\\|x-x_F\\|_2^2=\\sum_{n\\notin F}|x[n]|^2\\to0$ as the finite sets exhaust $\\bbZ$. Thus the unordered Hilbert-space sum $\\sum_nx[n]\\delta_n$ converges to $x$, and the finite-support statement is exact. The shift is a surjective isometry because $n\\mapsto n-1$ is a bijection of $\\bbZ$, with inverse shift $n\\mapsto n+1$. This proves the closed-span assertion.\n\\end{proof}","This realization fixes the language used below:\n\\[\n\\begin{gathered}\n\\text{impulse seed}\\longrightarrow\\text{shift address}\\longrightarrow\\text{packet projection}\\longrightarrow\\text{filter transfer}\\\\\n\\longrightarrow\\text{refinement}\\longrightarrow\\text{reconstruction}.\n\\end{gathered}\n\\]","\\begin{remark}[Several completions]\\label{core:record:02}\nThe same algebraic seed description applies to finite sequences, $\\ell^1$, $\\ell^2$, periodic sequences, tempered distributions, and multidimensional lattices. The topology decides which infinite reconstructions are legitimate. Finite-dimensional experiments in this paper use ordinary Euclidean norms; theorem statements specify the ambient norm when it matters.\n\\end{remark}","Fix an integer $N\\ge1$. For $0\\le r<N$, define\n\\[\n(P_{N,r}x)[n]\n=\n\\mathbf 1_{n\\equiv r\\, (\\mathrm{mod}\\,N)}x[n].\n\\]\nThe corresponding decimated channel is\n\\[\nx_r[q]=x[Nq+r].\n\\]","\\begin{theorem}[Polyphase projector theorem]\\label{result:02}\nThe maps $P_{N,r}$ are bounded orthogonal projectors on $\\ell^2$, with operator norm at most one, and satisfy\n\\[\nP_{N,r}P_{N,s}=\\delta_{rs}P_{N,r},\n\\qquad\n\\sum_{r=0}^{N-1}P_{N,r}=I.\n\\]\nConsequently,\n\\[\n\\boxed{x=\\sum_{r=0}^{N-1}P_{N,r}x}\n\\]\nwith no approximation.\n\\end{theorem}","\\begin{proof}\nEach integer belongs to exactly one residue class modulo $N$. Multiplying the two indicator functions gives zero for distinct classes and the same indicator for equal classes. Summing all residue indicators gives one at every address. Moreover,\n\\[\n\\|P_{N,r}x\\|_2^2=\\sum_{n\\equiv r\\ (\\mathrm{mod}\\,N)}|x[n]|^2\\le\\|x\\|_2^2,\n\\qquad\n\\langle P_{N,r}x,y\\rangle=\\langle x,P_{N,r}y\\rangle.\n\\]\nThe mask is real-valued, so the inner-product identity follows term by term. Hence each projector is self-adjoint, and the ranges of different residue projectors are orthogonal.\n\\end{proof}","Define the analysis and synthesis maps\n\\[\n\\cA_Nx=(x_0,\\ldots,x_{N-1}),\n\\qquad\n\\cS_N(x_0,\\ldots,x_{N-1})[Nq+r]=x_r[q].\n\\]","\\begin{corollary}[Analysis--synthesis identity]\\label{result:03}\n\\[\n\\boxed{\\cS_N\\cA_N=I.}\n\\]\nOn the product of the channel spaces one also has $\\cA_N\\cS_N=I$. Equip this finite product with the Hilbert norm\n\\[\n\\|(x_0,\\ldots,x_{N-1})\\|_{\\mathrm{ch}}^2\n=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\nThen $\\cA_N$ is a linear isometric equivalence and\n\\[\n\\|x\\|_2^2=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\n\\end{corollary}\n\\begin{proof}\nEuclidean division gives a unique pair $(r,q)$ with $0\\le r<N$ and $n=Nq+r$, including negative $n$. The address maps $n\\mapsto(n\\bmod N,n\\mathbin{\\mathrm{div}}N)$ and $(r,q)\\mapsto Nq+r$ are inverse. Analysis and synthesis are reindexing by this bijection. Reindexing the nonnegative sum $\\sum_n|x[n]|^2$ proves the norm identity and shows that both maps preserve square summabil","\\begin{corollary}[Analysis--synthesis identity]\\label{result:03}\n\\[\n\\boxed{\\cS_N\\cA_N=I.}\n\\]\nOn the product of the channel spaces one also has $\\cA_N\\cS_N=I$. Equip this finite product with the Hilbert norm\n\\[\n\\|(x_0,\\ldots,x_{N-1})\\|_{\\mathrm{ch}}^2\n=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\nThen $\\cA_N$ is a linear isometric equivalence and\n\\[\n\\|x\\|_2^2=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\n\\end{corollary}\n\\begin{proof}\nEuclidean division gives a unique pair $(r,q)$ with $0\\le r<N$ and $n=Nq+r$, including negative $n$. The address maps $n\\mapsto(n\\bmod N,n\\mathbin{\\mathrm{div}}N)$ and $(r,q)\\mapsto Nq+r$ are inverse. Analysis and synthesis are reindexing by this bijection. Reindexing the nonnegative sum $\\sum_n|x[n]|^2$ proves the norm identity and shows that both maps preserve square summability. In Lean the target is $\\ell^2(\\{0,\\ldots,N-1\\}\\times\\bbZ)$, w","\\begin{corollary}[Analysis--synthesis identity]\\label{result:03}\n\\[\n\\boxed{\\cS_N\\cA_N=I.}\n\\]\nOn the product of the channel spaces one also has $\\cA_N\\cS_N=I$. Equip this finite product with the Hilbert norm\n\\[\n\\|(x_0,\\ldots,x_{N-1})\\|_{\\mathrm{ch}}^2\n=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\nThen $\\cA_N$ is a linear isometric equivalence and\n\\[\n\\|x\\|_2^2=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\n\\end{corollary}\n\\begin{proof}\nEuclidean division gives a unique pair $(r,q)$ with $0\\le r<N$ and $n=Nq+r$, including negative $n$. The address maps $n\\mapsto(n\\bmod N,n\\mathbin{\\mathrm{div}}N)$ and $(r,q)\\mapsto Nq+r$ are inverse. Analysis and synthesis are reindexing by this bijection. Reindexing the nonnegative sum $\\sum_n|x[n]|^2$ proves the norm identity and shows that both maps preserve square summability. In Lean the target is $\\ell^2(\\{0,\\ldots,N-1\\}\\times\\bbZ)$, which realizes exactly this Hilbert norm.\n\\end{proof}","bel{result:03}\n\\[\n\\boxed{\\cS_N\\cA_N=I.}\n\\]\nOn the product of the channel spaces one also has $\\cA_N\\cS_N=I$. Equip this finite product with the Hilbert norm\n\\[\n\\|(x_0,\\ldots,x_{N-1})\\|_{\\mathrm{ch}}^2\n=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\nThen $\\cA_N$ is a linear isometric equivalence and\n\\[\n\\|x\\|_2^2=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\n\\end{corollary}\n\\begin{proof}\nEuclidean division gives a unique pair $(r,q)$ with $0\\le r<N$ and $n=Nq+r$, including negative $n$. The address maps $n\\mapsto(n\\bmod N,n\\mathbin{\\mathrm{div}}N)$ and $(r,q)\\mapsto Nq+r$ are inverse. Analysis and synthesis are reindexing by this bijection. Reindexing the nonnegative sum $\\sum_n|x[n]|^2$ proves the norm identity and shows that both maps preserve square summability. In Lean the target is $\\ell^2(\\{0,\\ldots,N-1\\}\\times\\bbZ)$, which realizes exactly this Hilbert norm.\n\\end{proof}","}\n\\[\n\\boxed{\\cS_N\\cA_N=I.}\n\\]\nOn the product of the channel spaces one also has $\\cA_N\\cS_N=I$. Equip this finite product with the Hilbert norm\n\\[\n\\|(x_0,\\ldots,x_{N-1})\\|_{\\mathrm{ch}}^2\n=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\nThen $\\cA_N$ is a linear isometric equivalence and\n\\[\n\\|x\\|_2^2=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\n\\end{corollary}\n\\begin{proof}\nEuclidean division gives a unique pair $(r,q)$ with $0\\le r<N$ and $n=Nq+r$, including negative $n$. The address maps $n\\mapsto(n\\bmod N,n\\mathbin{\\mathrm{div}}N)$ and $(r,q)\\mapsto Nq+r$ are inverse. Analysis and synthesis are reindexing by this bijection. Reindexing the nonnegative sum $\\sum_n|x[n]|^2$ proves the norm identity and shows that both maps preserve square summability. In Lean the target is $\\ell^2(\\{0,\\ldots,N-1\\}\\times\\bbZ)$, which realizes exactly this Hilbert norm.\n\\end{proof}","_N\\cA_N=I.}\n\\]\nOn the product of the channel spaces one also has $\\cA_N\\cS_N=I$. Equip this finite product with the Hilbert norm\n\\[\n\\|(x_0,\\ldots,x_{N-1})\\|_{\\mathrm{ch}}^2\n=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\nThen $\\cA_N$ is a linear isometric equivalence and\n\\[\n\\|x\\|_2^2=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\n\\end{corollary}\n\\begin{proof}\nEuclidean division gives a unique pair $(r,q)$ with $0\\le r<N$ and $n=Nq+r$, including negative $n$. The address maps $n\\mapsto(n\\bmod N,n\\mathbin{\\mathrm{div}}N)$ and $(r,q)\\mapsto Nq+r$ are inverse. Analysis and synthesis are reindexing by this bijection. Reindexing the nonnegative sum $\\sum_n|x[n]|^2$ proves the norm identity and shows that both maps preserve square summability. In Lean the target is $\\ell^2(\\{0,\\ldots,N-1\\}\\times\\bbZ)$, which realizes exactly this Hilbert norm.\n\\end{proof}","of the channel spaces one also has $\\cA_N\\cS_N=I$. Equip this finite product with the Hilbert norm\n\\[\n\\|(x_0,\\ldots,x_{N-1})\\|_{\\mathrm{ch}}^2\n=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\nThen $\\cA_N$ is a linear isometric equivalence and\n\\[\n\\|x\\|_2^2=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\n\\end{corollary}\n\\begin{proof}\nEuclidean division gives a unique pair $(r,q)$ with $0\\le r<N$ and $n=Nq+r$, including negative $n$. The address maps $n\\mapsto(n\\bmod N,n\\mathbin{\\mathrm{div}}N)$ and $(r,q)\\mapsto Nq+r$ are inverse. Analysis and synthesis are reindexing by this bijection. Reindexing the nonnegative sum $\\sum_n|x[n]|^2$ proves the norm identity and shows that both maps preserve square summability. In Lean the target is $\\ell^2(\\{0,\\ldots,N-1\\}\\times\\bbZ)$, which realizes exactly this Hilbert norm.\n\\end{proof}","one also has $\\cA_N\\cS_N=I$. Equip this finite product with the Hilbert norm\n\\[\n\\|(x_0,\\ldots,x_{N-1})\\|_{\\mathrm{ch}}^2\n=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\nThen $\\cA_N$ is a linear isometric equivalence and\n\\[\n\\|x\\|_2^2=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\n\\end{corollary}\n\\begin{proof}\nEuclidean division gives a unique pair $(r,q)$ with $0\\le r<N$ and $n=Nq+r$, including negative $n$. The address maps $n\\mapsto(n\\bmod N,n\\mathbin{\\mathrm{div}}N)$ and $(r,q)\\mapsto Nq+r$ are inverse. Analysis and synthesis are reindexing by this bijection. Reindexing the nonnegative sum $\\sum_n|x[n]|^2$ proves the norm identity and shows that both maps preserve square summability. In Lean the target is $\\ell^2(\\{0,\\ldots,N-1\\}\\times\\bbZ)$, which realizes exactly this Hilbert norm.\n\\end{proof}","oduct with the Hilbert norm\n\\[\n\\|(x_0,\\ldots,x_{N-1})\\|_{\\mathrm{ch}}^2\n=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\nThen $\\cA_N$ is a linear isometric equivalence and\n\\[\n\\|x\\|_2^2=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\n\\end{corollary}\n\\begin{proof}\nEuclidean division gives a unique pair $(r,q)$ with $0\\le r<N$ and $n=Nq+r$, including negative $n$. The address maps $n\\mapsto(n\\bmod N,n\\mathbin{\\mathrm{div}}N)$ and $(r,q)\\mapsto Nq+r$ are inverse. Analysis and synthesis are reindexing by this bijection. Reindexing the nonnegative sum $\\sum_n|x[n]|^2$ proves the norm identity and shows that both maps preserve square summability. In Lean the target is $\\ell^2(\\{0,\\ldots,N-1\\}\\times\\bbZ)$, which realizes exactly this Hilbert norm.\n\\end{proof}","a linear isometric equivalence and\n\\[\n\\|x\\|_2^2=\\sum_{r=0}^{N-1}\\|x_r\\|_2^2.\n\\]\n\\end{corollary}\n\\begin{proof}\nEuclidean division gives a unique pair $(r,q)$ with $0\\le r<N$ and $n=Nq+r$, including negative $n$. The address maps $n\\mapsto(n\\bmod N,n\\mathbin{\\mathrm{div}}N)$ and $(r,q)\\mapsto Nq+r$ are inverse. Analysis and synthesis are reindexing by this bijection. Reindexing the nonnegative sum $\\sum_n|x[n]|^2$ proves the norm identity and shows that both maps preserve square summability. In Lean the target is $\\ell^2(\\{0,\\ldots,N-1\\}\\times\\bbZ)$, which realizes exactly this Hilbert norm.\n\\end{proof}","n division gives a unique pair $(r,q)$ with $0\\le r<N$ and $n=Nq+r$, including negative $n$. The address maps $n\\mapsto(n\\bmod N,n\\mathbin{\\mathrm{div}}N)$ and $(r,q)\\mapsto Nq+r$ are inverse. Analysis and synthesis are reindexing by this bijection. Reindexing the nonnegative sum $\\sum_n|x[n]|^2$ proves the norm identity and shows that both maps preserve square summability. In Lean the target is $\\ell^2(\\{0,\\ldots,N-1\\}\\times\\bbZ)$, which realizes exactly this Hilbert norm.\n\\end{proof}","Suppose $N,M\\ge1$ and $N\\mid M$. Every coarse phase has a disjoint family of fine descendants:\n\\[\n\\operatorname{Child}_{M}(N,r)\n=\n\\{s:0\\le s<M,\\ s\\equiv r\\pmod N\\}.\n\\]","\\begin{theorem}[Exact nested refinement]\\label{result:04}\nFor $N,M\\ge1$ and $N\\mid M$,\n\\[\n\\boxed{\nP_{N,r}\n=\n\\sum_{\\substack{0\\le s<M","\\begin{theorem}[Exact nested refinement]\\label{result:04}\nFor $N,M\\ge1$ and $N\\mid M$,\n\\[\n\\boxed{\nP_{N,r}\n=\n\\sum_{\\substack{0\\le s<M\\\\s\\equiv r\\, (\\mathrm{mod}\\,N)}}P_{M,s}.\n}\n\\]\n\\end{theorem}","\\begin{proof}\nAn integer $n$ satisfies $n\\equiv r\\pmod N$ exactly when its unique residue $s=n\\bmod M$ lies among the listed descendants. The fine indicators are disjoint, so their sum is the coarse indicator.\n\\end{proof}","Let $h$ be a finite impulse response and\n\\[\ny=h*x,\n\\qquad\ny[n]=\\sum_{q\\in\\bbZ}h[q]x[n-q].\n\\]\nResolve both input and output into $N$ phases. For output phase $k$ and input phase $s$, define the transfer kernel\n\\[\n\\boxed{\nh_{ks}[m]=h[Nm+k-s].\n}\n\\]\nOut-of-range taps are understood to be zero.","\\begin{theorem}[Exact FIR polyphase transfer law]\\label{result:05}\nThe output channels satisfy\n\\[\n\\boxed{\ny_k[n]\n=\n\\sum_{s=0}^{N-1}\\sum_{m\\in\\bbZ}\nh_{ks}[m]x_s[n-m].\n}\n\\]\nFor the numerical channel transform, take $z\\in\\bbC\\setminus\\{0\\}$ and assume\n\\[\n\\sum_{n\\in\\bbZ}|x_s[n]z^{-n}|<\\infty\\quad(0\\le s<N).\n\\]\nDefine $X_s(z)=\\sum_nx_s[n]z^{-n}$. The output series then also converge absolutely, and\n\\[\n\\bm Y(z)=\\bm H^{(N)}(z)\\bm X(z),\n\\qquad\nH^{(N)}_{ks}(z)=\\sum_m h[Nm+k-s]z^{-m}.\n\\]\n\\end{theorem}","\\begin{proof}\nEvaluate $y$ at address $Nn+k$ and partition every input index by its residue $s$:\n\\[\n\\begin{aligned}\ny[Nn+k]\n&=\\sum_qh[q]x[Nn+k-q]","\\begin{proof}\nEvaluate $y$ at address $Nn+k$ and partition every input index by its residue $s$:\n\\[\n\\begin{aligned}\ny[Nn+k]\n&=\\sum_qh[q]x[Nn+k-q]\\\\\n&=\\sum_{s=0}^{N-1}\\sum_m\nh[Nm+k-s]x[N(n-m)+s].\n\\end{aligned}\n\\]\nFor a fixed output phase $k$, the change of tap address $q=Nm+k-s$ is a bijection between $\\bbZ$ and the pairs $(s,m)$ with $0\\le s<N$: choose $s\\equiv k-q\\pmod N$ and then $m=(q-k+s)/N$. Thus this is a reindexing of the finite filter support, valid for every bilateral input sequence, without a convergence assumption on $x$.","&=\\sum_{s=0}^{N-1}\\sum_m\nh[Nm+k-s]x[N(n-m)+s].\n\\end{aligned}\n\\]\nFor a fixed output phase $k$, the change of tap address $q=Nm+k-s$ is a bijection between $\\bbZ$ and the pairs $(s,m)$ with $0\\le s<N$: choose $s\\equiv k-q\\pmod N$ and then $m=(q-k+s)/N$. Thus this is a reindexing of the finite filter support, valid for every bilateral input sequence, without a convergence assumption on $x$.","The last factor is $x_s[n-m]$, giving the time-domain formula. Under the additional transform hypothesis, each delayed channel has transform\n\\[\n\\sum_nx_s[n-m]z^{-n}\n=z^{-m}\\sum_jx_s[j]z^{-j}.\n\\]\nAbsolute summability permits this bijective reindexing, and only finitely many delayed channels are combined. Taking their finite linear combination gives the matrix formula. The time-domain theorem therefore remains valid even when no numerical bilateral transform exists.\n\\end{proof}","\\begin{remark}[Operator-block form]\\label{core:record:07}\nIf $H$ denotes convolution by $h$, the same identity is\n\\[\nH=\\sum_{k,s}P_{N,k}HP_{N,s}.\n\\]\nThus the familiar polyphase matrix is exactly the transfer-block matrix of the seed calculus.\n\\end{remark}","Work with finite impulse-response banks. Let $\\mathcal L=\\bbC[z,z^{-1}]$ be the Laurent-polynomial algebra, with the monomial $z^{-m}$ acting on a bilateral sequence by $x[n]\\mapsto x[n-m]$. Let $\\bm E\\in\\operatorname{Mat}_{N\\times N}(\\mathcal L)$ and $\\bm R\\in\\operatorname{Mat}_{N\\times N}(\\mathcal L)$ be analysis and synthesis matrices. Their composition acts on all bilateral channel sequences by finite convolutions. At a nonzero $z$ where the input transforms are absolutely convergent, this gives\n\\[\n\\widehat{\\bm X}(z)=\\bm R(z)\\bm E(z)\\bm X(z).\n\\]","ks. Let $\\mathcal L=\\bbC[z,z^{-1}]$ be the Laurent-polynomial algebra, with the monomial $z^{-m}$ acting on a bilateral sequence by $x[n]\\mapsto x[n-m]$. Let $\\bm E\\in\\operatorname{Mat}_{N\\times N}(\\mathcal L)$ and $\\bm R\\in\\operatorname{Mat}_{N\\times N}(\\mathcal L)$ be analysis and synthesis matrices. Their composition acts on all bilateral channel sequences by finite convolutions. At a nonzero $z$ where the input transforms are absolutely convergent, this gives\n\\[\n\\widehat{\\bm X}(z)=\\bm R(z)\\bm E(z)\\bm X(z).\n\\]","with the monomial $z^{-m}$ acting on a bilateral sequence by $x[n]\\mapsto x[n-m]$. Let $\\bm E\\in\\operatorname{Mat}_{N\\times N}(\\mathcal L)$ and $\\bm R\\in\\operatorname{Mat}_{N\\times N}(\\mathcal L)$ be analysis and synthesis matrices. Their composition acts on all bilateral channel sequences by finite convolutions. At a nonzero $z$ where the input transforms are absolutely convergent, this gives\n\\[\n\\widehat{\\bm X}(z)=\\bm R(z)\\bm E(z)\\bm X(z).\n\\]","\\begin{theorem}[Perfect-reconstruction criterion]\\label{result:06}\nFor $d\\in\\bbZ$, such a bank reconstructs every input up to a delay of $d$ \\emph{channel indices} exactly when the following Laurent-polynomial identity holds:\n\\[\n\\boxed{\n\\bm R(z)\\bm E(z)=z^{-d}I_N.\n}\n\\]\nIf the right side is $c z^{-d}I_N$ with $c\\ne0$, reconstruction holds up to gain $c$ and channel delay $d$. After interleaving, this is a delay of $Nd$ original samples. A negative delay denotes an advance.\n\\end{theorem}","rfect-reconstruction criterion]\\label{result:06}\nFor $d\\in\\bbZ$, such a bank reconstructs every input up to a delay of $d$ \\emph{channel indices} exactly when the following Laurent-polynomial identity holds:\n\\[\n\\boxed{\n\\bm R(z)\\bm E(z)=z^{-d}I_N.\n}\n\\]\nIf the right side is $c z^{-d}I_N$ with $c\\ne0$, reconstruction holds up to gain $c$ and channel delay $d$. After interleaving, this is a delay of $Nd$ original samples. A negative delay denotes an advance.\n\\end{theorem}","\\begin{proof}\nFinite convolution represents Laurent-polynomial multiplication: composing shifts adds their delays, and distributivity handles finite sums. Thus the matrix identity implies the asserted action. Conversely, let $\\bm B=\\bm R\\bm E-cz^{-d}I_N$ and suppose its action vanishes on every channel sequence. Insert an impulse at channel $s$, address zero, with every other channel zero. Output channel $k$ reads the coefficients of the Laurent polynomial $B_{ks}$ at all addresses. All of these coefficients vanish, so $B_{ks}=0$ for every $k,s$. The representation is faithful, proving necessity as a polynomial identity rather than merely equality at one evaluation point. The displayed unit-gain case is $c=1$.\n\\end{proof}","oof}\nFinite convolution represents Laurent-polynomial multiplication: composing shifts adds their delays, and distributivity handles finite sums. Thus the matrix identity implies the asserted action. Conversely, let $\\bm B=\\bm R\\bm E-cz^{-d}I_N$ and suppose its action vanishes on every channel sequence. Insert an impulse at channel $s$, address zero, with every other channel zero. Output channel $k$ reads the coefficients of the Laurent polynomial $B_{ks}$ at all addresses. All of these coefficients vanish, so $B_{ks}=0$ for every $k,s$. The representation is faithful, proving necessity as a polynomial identity rather than merely equality at one evaluation point. The displayed unit-gain case is $c=1$.\n\\end{proof}","multiplication: composing shifts adds their delays, and distributivity handles finite sums. Thus the matrix identity implies the asserted action. Conversely, let $\\bm B=\\bm R\\bm E-cz^{-d}I_N$ and suppose its action vanishes on every channel sequence. Insert an impulse at channel $s$, address zero, with every other channel zero. Output channel $k$ reads the coefficients of the Laurent polynomial $B_{ks}$ at all addresses. All of these coefficients vanish, so $B_{ks}=0$ for every $k,s$. The representation is faithful, proving necessity as a polynomial identity rather than merely equality at one evaluation point. The displayed unit-gain case is $c=1$.\n\\end{proof}","utivity handles finite sums. Thus the matrix identity implies the asserted action. Conversely, let $\\bm B=\\bm R\\bm E-cz^{-d}I_N$ and suppose its action vanishes on every channel sequence. Insert an impulse at channel $s$, address zero, with every other channel zero. Output channel $k$ reads the coefficients of the Laurent polynomial $B_{ks}$ at all addresses. All of these coefficients vanish, so $B_{ks}=0$ for every $k,s$. The representation is faithful, proving necessity as a polynomial identity rather than merely equality at one evaluation point. The displayed unit-gain case is $c=1$.\n\\end{proof}","e sums. Thus the matrix identity implies the asserted action. Conversely, let $\\bm B=\\bm R\\bm E-cz^{-d}I_N$ and suppose its action vanishes on every channel sequence. Insert an impulse at channel $s$, address zero, with every other channel zero. Output channel $k$ reads the coefficients of the Laurent polynomial $B_{ks}$ at all addresses. All of these coefficients vanish, so $B_{ks}=0$ for every $k,s$. The representation is faithful, proving necessity as a polynomial identity rather than merely equality at one evaluation point. The displayed unit-gain case is $c=1$.\n\\end{proof}","Insert an impulse at channel $s$, address zero, with every other channel zero. Output channel $k$ reads the coefficients of the Laurent polynomial $B_{ks}$ at all addresses. All of these coefficients vanish, so $B_{ks}=0$ for every $k,s$. The representation is faithful, proving necessity as a polynomial identity rather than merely equality at one evaluation point. The displayed unit-gain case is $c=1$.\n\\end{proof}","\\paragraph{Original-sample delay.}\nFor an arbitrary original-sample delay $D\\in\\bbZ$, the output phase $k$ has\n\\[\ny_k[n]=x_{r_k}[n+q_k],\\qquad\nr_k=(k-D)\\bmod N,\\qquad q_k=(k-D)\\mathbin{\\mathrm{div}}N.\n\\]\nIts channel matrix has the sole nonzero entry $z^{q_k}$ in row $k$, column $r_k$. In particular, a one-sample delay at $N=2$ has matrix\n\\[\n\\begin{pmatrix}0&z^{-1}","\\paragraph{Original-sample delay.}\nFor an arbitrary original-sample delay $D\\in\\bbZ$, the output phase $k$ has\n\\[\ny_k[n]=x_{r_k}[n+q_k],\\qquad\nr_k=(k-D)\\bmod N,\\qquad q_k=(k-D)\\mathbin{\\mathrm{div}}N.\n\\]\nIts channel matrix has the sole nonzero entry $z^{q_k}$ in row $k$, column $r_k$. In particular, a one-sample delay at $N=2$ has matrix\n\\[\n\\begin{pmatrix}0&z^{-1}\\\\1&0\\end{pmatrix},\n\\]\nwhereas $z^{-1}I_2$ delays the original signal by two samples. The general address identity is valid for negative sample addresses as well.","1&0\\end{pmatrix},\n\\]\nwhereas $z^{-1}I_2$ delays the original signal by two samples. The general address identity is valid for negative sample addresses as well.","The Haar analysis matrix is\n\\[\n\\bm E_{\\mathrm{H}}\n=\n\\frac1{\\sqrt2}\n\\begin{pmatrix}\n1&1\\\\\n1&-1\n\\end{pmatrix}.\n\\]\nIt is orthogonal, so $\\bm R_{\\mathrm{H}}=\\bm E_{\\mathrm{H}}^{\\mathsf T}$ and\n\\[\n\\bm R_{\\mathrm{H}}\\bm E_{\\mathrm{H}}=I_2.\n\\]\nThe channels are\n\\[\na[n]=\\frac{x[2n]+x[2n+1]}{\\sqrt2},\n\\qquad\nd[n]=\\frac{x[2n]-x[2n+1]}{\\sqrt2}.\n\\]","1&-1\n\\end{pmatrix}.\n\\]\nIt is orthogonal, so $\\bm R_{\\mathrm{H}}=\\bm E_{\\mathrm{H}}^{\\mathsf T}$ and\n\\[\n\\bm R_{\\mathrm{H}}\\bm E_{\\mathrm{H}}=I_2.\n\\]\nThe channels are\n\\[\na[n]=\\frac{x[2n]+x[2n+1]}{\\sqrt2},\n\\qquad\nd[n]=\\frac{x[2n]-x[2n+1]}{\\sqrt2}.\n\\]","\\begin{corollary}[Haar energy identity]\\label{result:07}\n\\[\n\\boxed{\n\\|x\\|_2^2=\\|a\\|_2^2+\\|d\\|_2^2.\n}\n\\]\n\\end{corollary}","\\begin{proof}\nFor any $u,v\\in\\bbC$, the parallelogram identity gives\n\\[\n\\left|\\frac{u+v}{\\sqrt2}\\right|^2+\n\\left|\\frac{u-v}{\\sqrt2}\\right|^2=|u|^2+|v|^2.\n\\]\nApply this to $u=x[2n]$, $v=x[2n+1]$ and sum over $n\\in\\bbZ$. Each summand is nonnegative; the right side is $\\|x\\|_2^2$ by the two-channel address equivalence. This also proves that the literal low and detail channels belong to $\\ell^2$. The real Haar matrix is unitary over $\\bbC$, so the finite-signal statement is the same calculation with a finite sum.\n\\end{proof}",", the parallelogram identity gives\n\\[\n\\left|\\frac{u+v}{\\sqrt2}\\right|^2+\n\\left|\\frac{u-v}{\\sqrt2}\\right|^2=|u|^2+|v|^2.\n\\]\nApply this to $u=x[2n]$, $v=x[2n+1]$ and sum over $n\\in\\bbZ$. Each summand is nonnegative; the right side is $\\|x\\|_2^2$ by the two-channel address equivalence. This also proves that the literal low and detail channels belong to $\\ell^2$. The real Haar matrix is unitary over $\\bbC$, so the finite-signal statement is the same calculation with a finite sum.\n\\end{proof}","Split $x$ into even and odd channels $e,o$. Define\n\\[\nd=o-e,\n\\qquad\ns=e+\\frac12d.\n\\]\nThe inverse is\n\\[\ne=s-\\frac12d,\n\\qquad\no=d+e.\n\\]","\\begin{theorem}[Exact lifting reconstruction]\\label{result:08}\nThe predict--update map $(e,o)\\mapsto(s,d)$ is invertible with determinant one and the displayed inverse. Consequently the associated finite transform reconstructs every even-length signal exactly.\n\\end{theorem}","\\begin{proof}\nThe forward matrix is\n\\[\n\\begin{pmatrix}s\\\\d\\end{pmatrix}\n=\n\\begin{pmatrix}1/2&1/2\\\\-1&1\\end{pmatrix}\n\\begin{pmatrix}e\\\\o\\end{pmatrix},\n\\]\nwhose determinant is one. Direct substitution gives the inverse formulas.\n\\end{proof}","On $\\ell^2(\\bbZ;\\bbC)$, apply the Haar analysis to both children at each stage. For a finite input, assume its length is divisible by $2^\\ell$ when a depth-$\\ell$ transform is used. A binary word $\\alpha\\in\\{0,1\\}^{\\ell}$ addresses a depth-$\\ell$ packet, denoted $W_{\\ell,\\alpha}x$. Appending $0$ or $1$ selects the low or high child.","\\begin{theorem}[Wavelet-packet reconstruction and energy]\\label{result:09}\nAt every finite depth $\\ell$,\n\\[\n\\boxed{\nx=\\sum_{\\alpha\\in\\{0,1\\}^{\\ell}}W_{\\ell,\\alpha}^{*}W_{\\ell,\\alpha}x,\n}\n\\]\nand\n\\[\n\\boxed{\n\\|x\\|_2^2\n=\n\\sum_{\\alpha\\in\\{0,1\\}^{\\ell}}\n\\|W_{\\ell,\\alpha}x\\|_2^2.\n}\n\\]\n\\end{theorem}","\\begin{proof}\nWrite $H_0,H_1$ for the literal low and detail operators. The preceding energy identity implies $\\|H_bx\\|_2\\le\\|x\\|_2$, so both have bounded Hilbert adjoints. At depth zero the only packet operator is the identity. If the energy identity holds at depth $\\ell$, apply the one-level identity to each $W_{\\ell,\\alpha}x$ and sum; this proves it at depth $\\ell+1$.","Set $A=\\sum_\\alpha W_{\\ell,\\alpha}^*W_{\\ell,\\alpha}$. This is a finite sum of bounded operators. The energy result says $\\langle x,Ax\\rangle=\\langle x,x\\rangle$ for every $x$. Complex polarization, applied also to $x+y$ and $x+\\ii y$, gives $\\langle y,(A-I)x\\rangle=0$ for all $x,y$, hence $A=I$. Applying the operator identity to $x$ proves synthesis. This argument uses the actual Hilbert adjoints of the decimation operators, not a separately postulated inverse symbol.\n\\end{proof}","Let\n\\[\nf=s+j\n\\]\nwhere $s$ belongs to a smooth approximation class and $j$ is a known discontinuity packet. Let $Q_m$ be any approximation operator applied only to $s$, and define\n\\[\n\\widetilde f_m=Q_ms+j.\n\\]","\\begin{theorem}[Exact known-edge theorem]\\label{result:10}\nFor every norm for which the expressions are defined,\n\\[\n\\boxed{\nf-\\widetilde f_m=s-Q_ms,\n\\qquad\n\\|f-\\widetilde f_m\\|=\\|s-Q_ms\\|.\n}\n\\]\nNo approximation of $j$ occurs.\n\\end{theorem}","\\begin{proof}\nSubtract:\n\\[\nf-\\widetilde f_m=(s+j)-(Q_ms+j)=s-Q_ms.\n\\]\nTaking the norm gives the second identity.\n\\end{proof}","\\begin{corollary}[Several exact edge packets]\\label{result:11}\nIf\n\\[\nf=s+\\sum_{q=1}^{M}j_q\n\\]\nand every $j_q$ is retained exactly, then\n\\[\n\\left\\|f-\\left(Q_ms+\\sum_qj_q\\right)\\right\\|\n=\\|s-Q_ms\\|.\n\\]\nCurved edges, corners, and different jump amplitudes do not alter the identity.\n\\end{corollary}\n\\begin{proof}\nThe finite sum $j=\\sum_{q=1}^M j_q$ is a single retained packet. Applying the preceding cancellation identity gives the result; the empty family $M=0$ is included with sum zero.\n\\end{proof}","lt:11}\nIf\n\\[\nf=s+\\sum_{q=1}^{M}j_q\n\\]\nand every $j_q$ is retained exactly, then\n\\[\n\\left\\|f-\\left(Q_ms+\\sum_qj_q\\right)\\right\\|\n=\\|s-Q_ms\\|.\n\\]\nCurved edges, corners, and different jump amplitudes do not alter the identity.\n\\end{corollary}\n\\begin{proof}\nThe finite sum $j=\\sum_{q=1}^M j_q$ is a single retained packet. Applying the preceding cancellation identity gives the result; the empty family $M=0$ is included with sum zero.\n\\end{proof}","Let $(X,\\Sigma,\\mu)$ be a measure space, with measurable masks $\\Omega,\\widehat\\Omega\\subseteq X$ and scalar amplitudes $a,\\widehat a\\in\\bbC$ (real amplitudes are a special case). Suppose a scalar jump packet is\n\\[\nj=a\\mathbf 1_{\\Omega},\n\\]\nand an estimated packet is\n\\[\n\\widehat j=\\widehat a\\mathbf 1_{\\widehat\\Omega}.\n\\]","\\begin{theorem}[Packet perturbation bound]\\label{result:12}\nFor $1\\le p<\\infty$, interpreted first as an inequality of extended $L^p$ seminorms,\n\\[\n\\boxed{\n\\|j-\\widehat j\\|_{L^p}\n\\le\n|a-\\widehat a|\\,\\mu(\\Omega)^{1/p}\n+\n|\\widehat a|\\,\\mu(\\Omega\\triangle\\widehat\\Omega)^{1/p}.\n}\n\\]\nIf $\\mu(\\Omega)$ and $\\mu(\\Omega\\triangle\\widehat\\Omega)$ are finite, all terms in this displayed jump bound are finite and it is an ordinary real-valued $L^p$ inequality. For the pointwise supremum, and consequently also the essential $L^\\infty$ seminorm,\n\\[\n\\|j-\\widehat j\\|_{\\infty}\n\\le |a-\\widehat a|+|\\widehat a|\\,\n\\mathbf 1_{\\Omega\\triangle\\widehat\\Omega\\ne\\varnothing}.\n\\]\n\\end{theorem}","bound]\\label{result:12}\nFor $1\\le p<\\infty$, interpreted first as an inequality of extended $L^p$ seminorms,\n\\[\n\\boxed{\n\\|j-\\widehat j\\|_{L^p}\n\\le\n|a-\\widehat a|\\,\\mu(\\Omega)^{1/p}\n+\n|\\widehat a|\\,\\mu(\\Omega\\triangle\\widehat\\Omega)^{1/p}.\n}\n\\]\nIf $\\mu(\\Omega)$ and $\\mu(\\Omega\\triangle\\widehat\\Omega)$ are finite, all terms in this displayed jump bound are finite and it is an ordinary real-valued $L^p$ inequality. For the pointwise supremum, and consequently also the essential $L^\\infty$ seminorm,\n\\[\n\\|j-\\widehat j\\|_{\\infty}\n\\le |a-\\widehat a|+|\\widehat a|\\,\n\\mathbf 1_{\\Omega\\triangle\\widehat\\Omega\\ne\\varnothing}.\n\\]\n\\end{theorem}","f extended $L^p$ seminorms,\n\\[\n\\boxed{\n\\|j-\\widehat j\\|_{L^p}\n\\le\n|a-\\widehat a|\\,\\mu(\\Omega)^{1/p}\n+\n|\\widehat a|\\,\\mu(\\Omega\\triangle\\widehat\\Omega)^{1/p}.\n}\n\\]\nIf $\\mu(\\Omega)$ and $\\mu(\\Omega\\triangle\\widehat\\Omega)$ are finite, all terms in this displayed jump bound are finite and it is an ordinary real-valued $L^p$ inequality. For the pointwise supremum, and consequently also the essential $L^\\infty$ seminorm,\n\\[\n\\|j-\\widehat j\\|_{\\infty}\n\\le |a-\\widehat a|+|\\widehat a|\\,\n\\mathbf 1_{\\Omega\\triangle\\widehat\\Omega\\ne\\varnothing}.\n\\]\n\\end{theorem}","orms,\n\\[\n\\boxed{\n\\|j-\\widehat j\\|_{L^p}\n\\le\n|a-\\widehat a|\\,\\mu(\\Omega)^{1/p}\n+\n|\\widehat a|\\,\\mu(\\Omega\\triangle\\widehat\\Omega)^{1/p}.\n}\n\\]\nIf $\\mu(\\Omega)$ and $\\mu(\\Omega\\triangle\\widehat\\Omega)$ are finite, all terms in this displayed jump bound are finite and it is an ordinary real-valued $L^p$ inequality. For the pointwise supremum, and consequently also the essential $L^\\infty$ seminorm,\n\\[\n\\|j-\\widehat j\\|_{\\infty}\n\\le |a-\\widehat a|+|\\widehat a|\\,\n\\mathbf 1_{\\Omega\\triangle\\widehat\\Omega\\ne\\varnothing}.\n\\]\n\\end{theorem}","\\begin{proof}\nWrite\n\\[\na\\mathbf 1_{\\Omega}-\\widehat a\\mathbf 1_{\\widehat\\Omega}\n=(a-\\widehat a)\\mathbf 1_{\\Omega}\n+\\widehat a(\\mathbf 1_{\\Omega}-\\mathbf 1_{\\widehat\\Omega}).\n\\]\nApply the triangle inequality and use\n\\[\n\\|\\mathbf 1_{\\Omega}-\\mathbf 1_{\\widehat\\Omega}\\|_{L^p}\n=\\mu(\\Omega\\triangle\\widehat\\Omega)^{1/p}.\n\\]\nFor the supremum estimate, $|\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}|\\le1$ pointwise, and this function vanishes everywhere when the symmetric difference is empty. The same pointwise bound also bounds the essential supremum. Nonempty null symmetric differences need not contribute to the essential norm; the stated bound remains valid in that case and need not be sharp. Measurability justifies the $L^p$ triangle inequality and the indicator norm formula.\n\\end{proof}","r the supremum estimate, $|\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}|\\le1$ pointwise, and this function vanishes everywhere when the symmetric difference is empty. The same pointwise bound also bounds the essential supremum. Nonempty null symmetric differences need not contribute to the essential norm; the stated bound remains valid in that case and need not be sharp. Measurability justifies the $L^p$ triangle inequality and the indicator norm formula.\n\\end{proof}","If $s,\\widehat s$ are measurable smooth-background models, the same triangle inequality gives, for finite $p$,\n\\[\n\\boxed{\n\\|f-(\\widehat s+\\widehat j)\\|_p\n\\le\n\\|s-\\widehat s\\|_p\n+|a-\\widehat a|\\mu(\\Omega)^{1/p}\n+|\\widehat a|\\mu(\\Omega\\triangle\\widehat\\Omega)^{1/p}.\n}\n\\]\nThis is the correct separation of errors. Detection error is not hidden inside a generic reconstruction score.","\\begin{certificate}[One-dimensional errors]\\label{core:record:18}\nFor $1024$ samples on $[-1,1)$ and Fourier cutoff $m=18$, the retained frequency indices are $-18,\\ldots,18$: 37 bins including the zero-frequency bin. The measured values were\n\\[\n\\begin{array}{lrr}\n\\toprule\n\\text{method}&\\RMSE&\\text{maximum absolute error}","\\begin{certificate}[One-dimensional errors]\\label{core:record:18}\nFor $1024$ samples on $[-1,1)$ and Fourier cutoff $m=18$, the retained frequency indices are $-18,\\ldots,18$: 37 bins including the zero-frequency bin. The measured values were\n\\[\n\\begin{array}{lrr}\n\\toprule\n\\text{method}&\\RMSE&\\text{maximum absolute error}\\\\\n\\midrule\n\\text{global Fourier truncation}&5.4023\\times10^{-2}&3.4968\\times10^{-1}\\\\\n\\text{known edge packet}&1.3692\\times10^{-4}&1.4503\\times10^{-3}\\\\\n\\text{estimated edge packet}&2.4029\\times10^{-2}&5.9530\\times10^{-2}\\\\\n\\bottomrule\n\\end{array}\n\\]\nThe simple estimator recovered the binary mask exactly in this deliberately separated two-class example. That result is a property of this test, not a universal guarantee.\n\\end{certificate}","\\begin{certificate}[Two-dimensional errors]\\label{core:record:19}\nFor the $192\\times192$ disk experiment, a square Fourier mask retains indices $|k_x|,|k_y|\\le16$, or $33^2=1089$ bins. The reported errors are\n\\[\n\\begin{array}{lrr}\n\\toprule\n\\text{method}&\\text{full-image RMSE}&\\text{boundary-band RMSE}","\\begin{certificate}[Two-dimensional errors]\\label{core:record:19}\nFor the $192\\times192$ disk experiment, a square Fourier mask retains indices $|k_x|,|k_y|\\le16$, or $33^2=1089$ bins. The reported errors are\n\\[\n\\begin{array}{lrr}\n\\toprule\n\\text{method}&\\text{full-image RMSE}&\\text{boundary-band RMSE}\\\\\n\\midrule\n\\text{global Fourier truncation}&3.7643\\times10^{-2}&1.8541\\times10^{-1}\\\\\n\\text{known edge packet}&6.4752\\times10^{-5}&1.2709\\times10^{-5}\\\\\n\\text{estimated edge packet}&3.4043\\times10^{-3}&8.8834\\times10^{-3}\\\\\n\\bottomrule\n\\end{array}\n\\]\nThe mask intersection-over-union was one in this positive-contrast synthetic case. The amplitude was estimated rather than inserted from the generating model.\n\\end{certificate}","\\begin{certificate}[Multiedge errors]\\label{core:record:20}\nThe global reconstruction had RMSE $3.3740\\times10^{-2}$ and maximum error $3.2807\\times10^{-1}$. The known-packet reconstruction had only the remaining smooth-background error. The experiment checks the finite implementation of the theorem for a composite packet; it is not a data-driven multiregion segmentation result.\n\\end{certificate}","\\midrule\n16&$1.7999\\times10^{-1}$&$6.6522\\times10^{-2}$","32&$3.8414\\times10^{-2}$&$4.9153\\times10^{-2}$","64&$1.6267\\times10^{-2}$&$3.8056\\times10^{-2}$","128&$1.7387\\times10^{-3}$&$2.5631\\times10^{-2}$","256&$1.5870\\times10^{-4}$&$1.7424\\times10^{-2}$","\\begin{definition}[Complete packet pruning]\\label{core:record:22}\nLet $\\mathcal T$ be a finite rooted refinement tree.  Every node $v$ carries a projector $P_v$, the root satisfies $P_{\\varnothing}=I$, and every internal node satisfies\n\\[\nP_v=\\sum_{u\\in\\operatorname{ch}(v)}P_u,\n\\qquad\nP_uP_w=0\\quad(u\\ne w\\text{ in }\\operatorname{ch}(v)).\n\\]\nA set $\\Lambda$ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets $\\Lambda$ exactly once.\n\\end{definition}","ore:record:22}\nLet $\\mathcal T$ be a finite rooted refinement tree.  Every node $v$ carries a projector $P_v$, the root satisfies $P_{\\varnothing}=I$, and every internal node satisfies\n\\[\nP_v=\\sum_{u\\in\\operatorname{ch}(v)}P_u,\n\\qquad\nP_uP_w=0\\quad(u\\ne w\\text{ in }\\operatorname{ch}(v)).\n\\]\nA set $\\Lambda$ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets $\\Lambda$ exactly once.\n\\end{definition}","\\begin{theorem}[Exact synthesis from every complete pruning]\\label{result:13}\nFor every complete pruning $\\Lambda$,\n\\[\n\\boxed{\\sum_{v\\in\\Lambda}P_v=I,}\n\\qquad\n\\boxed{x=\\sum_{v\\in\\Lambda}P_vx.}\n\\]\nThus changing the selected tree changes the coordinates but does not introduce reconstruction error.\n\\end{theorem}","\\begin{proof}\nStarting from the root identity $P_{\\varnothing}=I$, replace every unselected internal node by the exact sum of its children.  If a complete pruning contains the root it contains no other node. Otherwise its restriction to each child subtree is itself a complete pruning: every terminal path through that child must be met exactly once. Induction on the finite tree therefore justifies the substitution and its termination precisely at $\\Lambda$. Child subtrees have distinct node addresses, so their selected-node sets form a disjoint union. The sum identity uses only root identity and parent-to-child conservation; sibling orthogonality gives the additional interpretation as nonoverlapping projector packets.  Applying the resulting identity to $","varnothing}=I$, replace every unselected internal node by the exact sum of its children.  If a complete pruning contains the root it contains no other node. Otherwise its restriction to each child subtree is itself a complete pruning: every terminal path through that child must be met exactly once. Induction on the finite tree therefore justifies the substitution and its termination precisely at $\\Lambda$. Child subtrees have distinct node addresses, so their selected-node sets form a disjoint union. The sum identity uses only root identity and parent-to-child conservation; sibling orthogonality gives the additional interpretation as nonoverlapping projector packets.  Applying the resulting identity to $x$ gives the synthesis formula.\n\\end{proof}","the finite tree therefore justifies the substitution and its termination precisely at $\\Lambda$. Child subtrees have distinct node addresses, so their selected-node sets form a disjoint union. The sum identity uses only root identity and parent-to-child conservation; sibling orthogonality gives the additional interpretation as nonoverlapping projector packets.  Applying the resulting identity to $x$ gives the synthesis formula.\n\\end{proof}","Let $y$ denote the observed data.  For a complete pruning $\\Lambda$, consider any finite criterion\n\\[\nJ_y(\\Lambda)\n=\nR_y(\\Lambda)\n+\\eta L_y(\\Lambda)\n+\\tau K(\\Lambda),\n\\]\nwhere $R_y$ is a representation cost, $L_y$ is a cross-packet leakage cost, and $K$ is a tree-complexity penalty.  Here $R_y,L_y,K$ take finite real values, and $\\eta,\\tau$ are fixed real weights (usually nonnegative in applications). No additivity is needed for the first theorem.","\\begin{theorem}[Finite globally optimal stable selector]\\label{result:14}\nLet $\\mathfrak P(\\mathcal T)$ be the finite set of complete prunings of $\\mathcal T$.","\\begin{enumerate}[label=(\\roman*)]\n\\item A global minimizer\n\\[\n\\Lambda_y^*\\in\\operatorname*{argmin}_{\\Lambda\\in\\mathfrak P(\\mathcal T)}J_y(\\Lambda)\n\\]\nalways exists.\n\\item If the minimizer is unique and at least one competing pruning exists, its gap\n\\[\n\\Gamma_y\n=\n\\min_{\\Lambda\\ne\\Lambda_y^*}\n\\bigl(J_y(\\Lambda)-J_y(\\Lambda_y^*)\\bigr)\n\\]\nis attained and strictly positive. If the pruning set is a singleton, its only element minimizes every criterion; no real-valued minimum over an empty competitor set is introduced.\n\\item In the case with competitors, if new data $y'$ satisfy\n\\[\n\\sup_{\\Lambda\\in\\mathfrak P(\\mathcal T)}\n|J_{y'}(\\Lambda)-J_y(\\Lambda)|\n<\\frac{\\Gamma_y}{2},\n\\]\nthen $\\Lambda_{y'}^*=\\Lambda_y^*$.\n\\item Every selected minimizer preserves the exact synthesis identity of the preceding theorem.\n\\end{enumerate}\n\\end{theorem}","tem If the minimizer is unique and at least one competing pruning exists, its gap\n\\[\n\\Gamma_y\n=\n\\min_{\\Lambda\\ne\\Lambda_y^*}\n\\bigl(J_y(\\Lambda)-J_y(\\Lambda_y^*)\\bigr)\n\\]\nis attained and strictly positive. If the pruning set is a singleton, its only element minimizes every criterion; no real-valued minimum over an empty competitor set is introduced.\n\\item In the case with competitors, if new data $y'$ satisfy\n\\[\n\\sup_{\\Lambda\\in\\mathfrak P(\\mathcal T)}\n|J_{y'}(\\Lambda)-J_y(\\Lambda)|\n<\\frac{\\Gamma_y}{2},\n\\]\nthen $\\Lambda_{y'}^*=\\Lambda_y^*$.\n\\item Every selected minimizer preserves the exact synthesis identity of the preceding theorem.\n\\end{enumerate}\n\\end{theorem}","imizer is unique and at least one competing pruning exists, its gap\n\\[\n\\Gamma_y\n=\n\\min_{\\Lambda\\ne\\Lambda_y^*}\n\\bigl(J_y(\\Lambda)-J_y(\\Lambda_y^*)\\bigr)\n\\]\nis attained and strictly positive. If the pruning set is a singleton, its only element minimizes every criterion; no real-valued minimum over an empty competitor set is introduced.\n\\item In the case with competitors, if new data $y'$ satisfy\n\\[\n\\sup_{\\Lambda\\in\\mathfrak P(\\mathcal T)}\n|J_{y'}(\\Lambda)-J_y(\\Lambda)|\n<\\frac{\\Gamma_y}{2},\n\\]\nthen $\\Lambda_{y'}^*=\\Lambda_y^*$.\n\\item Every selected minimizer preserves the exact synthesis identity of the preceding theorem.\n\\end{enumerate}\n\\end{theorem}","da_y^*}\n\\bigl(J_y(\\Lambda)-J_y(\\Lambda_y^*)\\bigr)\n\\]\nis attained and strictly positive. If the pruning set is a singleton, its only element minimizes every criterion; no real-valued minimum over an empty competitor set is introduced.\n\\item In the case with competitors, if new data $y'$ satisfy\n\\[\n\\sup_{\\Lambda\\in\\mathfrak P(\\mathcal T)}\n|J_{y'}(\\Lambda)-J_y(\\Lambda)|\n<\\frac{\\Gamma_y}{2},\n\\]\nthen $\\Lambda_{y'}^*=\\Lambda_y^*$.\n\\item Every selected minimizer preserves the exact synthesis identity of the preceding theorem.\n\\end{enumerate}\n\\end{theorem}","subset of the finite node set, so the feasible set is nonempty and finite. Thus the minimum is attained. If competitors exist, their finitely many differences from a unique optimum are strictly positive, so their minimum is positive and is achieved by a competitor. In the singleton case there is no alternative selector under any objective change. In the remaining case, for any competing pruning $\\Lambda\\ne\\Lambda_y^*$,\n\\[\n\\begin{aligned}\nJ_{y'}(\\Lambda)-J_{y'}(\\Lambda_y^*)\n&\\ge\nJ_y(\\Lambda)-J_y(\\Lambda_y^*)\n-2\\sup_{\\Xi}|J_{y'}(\\Xi)-J_y(\\Xi)|","set, so the feasible set is nonempty and finite. Thus the minimum is attained. If competitors exist, their finitely many differences from a unique optimum are strictly positive, so their minimum is positive and is achieved by a competitor. In the singleton case there is no alternative selector under any objective change. In the remaining case, for any competing pruning $\\Lambda\\ne\\Lambda_y^*$,\n\\[\n\\begin{aligned}\nJ_{y'}(\\Lambda)-J_{y'}(\\Lambda_y^*)\n&\\ge\nJ_y(\\Lambda)-J_y(\\Lambda_y^*)\n-2\\sup_{\\Xi}|J_{y'}(\\Xi)-J_y(\\Xi)|\\\\\n&>\\Gamma_y-\\Gamma_y=0.\n\\end{aligned}\n\\]\nHence the same pruning remains uniquely optimal.  Exact synthesis depends only on completeness of the pruning, not on the cost used to select it.\n\\end{proof}","\\begin{theorem}[Dynamic-programming best packet basis]\\label{result:15}\nSuppose\n\\[\nJ_y(\\Lambda)=\\sum_{v\\in\\Lambda}c_v(y).\n\\]\nFor a node $v$, define recursively\n\\[\nV_y(v)\n=\n\\begin{cases}\nc_v(y),&v\\text{ terminal},","\\begin{theorem}[Dynamic-programming best packet basis]\\label{result:15}\nSuppose\n\\[\nJ_y(\\Lambda)=\\sum_{v\\in\\Lambda}c_v(y).\n\\]\nFor a node $v$, define recursively\n\\[\nV_y(v)\n=\n\\begin{cases}\nc_v(y),&v\\text{ terminal},\\\\[2mm]\n\\displaystyle\n\\min\\left\\{c_v(y),\\sum_{u\\in\\operatorname{ch}(v)}V_y(u)\\right\\},\n&v\\text{ internal}.\n\\end{cases}\n\\]\nStop at $v$ when $c_v(y)\\le\\sum_uV_y(u)$, including equality; refine otherwise. This deterministic rule produces a globally optimal complete pruning. With the node costs supplied and child values cached, the evaluator uses one comparison per internal node, one visit per node, and at most $|\\mathcal T|-1$ additions. Its arithmetic-and-visit count is therefore at most $3|\\mathcal T|$. This is a unit-cost statement about the specified evaluator; it excludes evaluation of the local costs, bit complexity of scalar arithmetic, and materialization or sorting of packet-address strings.\n\\end{theorem}","[2mm]\n\\displaystyle\n\\min\\left\\{c_v(y),\\sum_{u\\in\\operatorname{ch}(v)}V_y(u)\\right\\},\n&v\\text{ internal}.\n\\end{cases}\n\\]\nStop at $v$ when $c_v(y)\\le\\sum_uV_y(u)$, including equality; refine otherwise. This deterministic rule produces a globally optimal complete pruning. With the node costs supplied and child values cached, the evaluator uses one comparison per internal node, one visit per node, and at most $|\\mathcal T|-1$ additions. Its arithmetic-and-visit count is therefore at most $3|\\mathcal T|$. This is a unit-cost statement about the specified evaluator; it excludes evaluation of the local costs, bit complexity of scalar arithmetic, and materialization or sorting of packet-address strings.\n\\end{theorem}","playstyle\n\\min\\left\\{c_v(y),\\sum_{u\\in\\operatorname{ch}(v)}V_y(u)\\right\\},\n&v\\text{ internal}.\n\\end{cases}\n\\]\nStop at $v$ when $c_v(y)\\le\\sum_uV_y(u)$, including equality; refine otherwise. This deterministic rule produces a globally optimal complete pruning. With the node costs supplied and child values cached, the evaluator uses one comparison per internal node, one visit per node, and at most $|\\mathcal T|-1$ additions. Its arithmetic-and-visit count is therefore at most $3|\\mathcal T|$. This is a unit-cost statement about the specified evaluator; it excludes evaluation of the local costs, bit complexity of scalar arithmetic, and materialization or sorting of packet-address strings.\n\\end{theorem}","xt{ internal}.\n\\end{cases}\n\\]\nStop at $v$ when $c_v(y)\\le\\sum_uV_y(u)$, including equality; refine otherwise. This deterministic rule produces a globally optimal complete pruning. With the node costs supplied and child values cached, the evaluator uses one comparison per internal node, one visit per node, and at most $|\\mathcal T|-1$ additions. Its arithmetic-and-visit count is therefore at most $3|\\mathcal T|$. This is a unit-cost statement about the specified evaluator; it excludes evaluation of the local costs, bit complexity of scalar arithmetic, and materialization or sorting of packet-address strings.\n\\end{theorem}","\\begin{proof}\nProceed upward from the terminal nodes.  Every complete pruning below an internal node $v$ has exactly two possible forms: it either stops at $v$, with cost $c_v(y)$, or it refines and is the union of complete child prunings, whose minimum total cost is the sum of the already computed child minima.  The recursion therefore gives the exact minimum on every subtree, in particular at the root. This characterization also proves both directions of the correspondence between recursive cuts and complete covering antichains, for arbitrary positive finite branching, including unary branches. Summing cached child values charges at most one addition per child edge; a finite rooted tree has $|\\mathcal T|-1$ such edges. The number of internal nodes is at most $|\\mathcal T|$, which gives the","\\begin{proof}\nProceed upward from the terminal nodes.  Every complete pruning below an internal node $v$ has exactly two possible forms: it either stops at $v$, with cost $c_v(y)$, or it refines and is the union of complete child prunings, whose minimum total cost is the sum of the already computed child minima.  The recursion therefore gives the exact minimum on every subtree, in particular at the root. This characterization also proves both directions of the correspondence between recursive cuts and complete covering antichains, for arbitrary positive finite branching, including unary branches. Summing cached child values charges at most one addition per child edge; a finite rooted tree has $|\\mathcal T|-1$ such edges. The number of internal nodes is at most $|\\mathcal T|$, which gives the stated work bound. Ties do not affect optimality, but t","\\begin{proof}\nProceed upward from the terminal nodes.  Every complete pruning below an internal node $v$ has exactly two possible forms: it either stops at $v$, with cost $c_v(y)$, or it refines and is the union of complete child prunings, whose minimum total cost is the sum of the already computed child minima.  The recursion therefore gives the exact minimum on every subtree, in particular at the root. This characterization also proves both directions of the correspondence between recursive cuts and complete covering antichains, for arbitrary positive finite branching, including unary branches. Summing cached child values charges at most one addition per child edge; a finite rooted tree has $|\\mathcal T|-1$ such edges. The number of internal nodes is at most $|\\mathcal T|$, which gives the stated work bound. Ties do not affect optimality, but the stop-on-ties rule","ld minima.  The recursion therefore gives the exact minimum on every subtree, in particular at the root. This characterization also proves both directions of the correspondence between recursive cuts and complete covering antichains, for arbitrary positive finite branching, including unary branches. Summing cached child values charges at most one addition per child edge; a finite rooted tree has $|\\mathcal T|-1$ such edges. The number of internal nodes is at most $|\\mathcal T|$, which gives the stated work bound. Ties do not affect optimality, but the stop-on-ties rule makes the selected pruning unambiguous.\n\\end{proof}","subtree, in particular at the root. This characterization also proves both directions of the correspondence between recursive cuts and complete covering antichains, for arbitrary positive finite branching, including unary branches. Summing cached child values charges at most one addition per child edge; a finite rooted tree has $|\\mathcal T|-1$ such edges. The number of internal nodes is at most $|\\mathcal T|$, which gives the stated work bound. Ties do not affect optimality, but the stop-on-ties rule makes the selected pruning unambiguous.\n\\end{proof}","\\begin{corollary}[Local perturbation certificate]\\label{result:16}\nAssume\n\\[\n|c_v(y')-c_v(y)|\\le\\varepsilon_v.\n\\]\nDefine\n\\[\nE(v)\n=\n\\begin{cases}\n\\varepsilon_v,&v\\text{ terminal},","\\begin{corollary}[Local perturbation certificate]\\label{result:16}\nAssume\n\\[\n|c_v(y')-c_v(y)|\\le\\varepsilon_v.\n\\]\nDefine\n\\[\nE(v)\n=\n\\begin{cases}\n\\varepsilon_v,&v\\text{ terminal},\\\\[1mm]\n\\displaystyle\n\\max\\left\\{\\varepsilon_v,\n\\sum_{u\\in\\operatorname{ch}(v)}E(u)\\right\\},&v\\text{ internal}.\n\\end{cases}\n\\]\nFor the complete prunings $\\mathfrak P_v$ below $v$, the envelope is exactly\n\\[\nE(v)=\\max_{\\Lambda\\in\\mathfrak P_v}\\sum_{a\\in\\Lambda}\\varepsilon_a.\n\\]\nThe maximum is attained. Every complete-pruning cost changes by at most $E(\\varnothing)$, and the two globally optimal values differ by at most this quantity. Consequently, for a unique optimum with competitors,\n\\[\n\\boxed{\\Gamma_y>2E(\\varnothing)}\n\\]\nis a deterministic certificate that the globally optimal pruning is unchanged.  A singleton pruning set is stable without a gap condition. Assume the random node costs and the joint local-error event are measurable. If a noise model supplies \\emph{all} local bounds simultaneously on one event o","[1mm]\n\\displaystyle\n\\max\\left\\{\\varepsilon_v,\n\\sum_{u\\in\\operatorname{ch}(v)}E(u)\\right\\},&v\\text{ internal}.\n\\end{cases}\n\\]\nFor the complete prunings $\\mathfrak P_v$ below $v$, the envelope is exactly\n\\[\nE(v)=\\max_{\\Lambda\\in\\mathfrak P_v}\\sum_{a\\in\\Lambda}\\varepsilon_a.\n\\]\nThe maximum is attained. Every complete-pruning cost changes by at most $E(\\varnothing)$, and the two globally optimal values differ by at most this quantity. Consequently, for a unique optimum with competitors,\n\\[\n\\boxed{\\Gamma_y>2E(\\varnothing)}\n\\]\nis a deterministic certificate that the globally optimal pruning is unchanged.  A singleton pruning set is stable without a gap condition. Assume the random node costs and the joint local-error event are measurable. If a noise model supplies \\emph{all} local bounds simultaneously on one event of probability at least $1-\\delta$, and the","[1mm]\n\\displaystyle\n\\max\\left\\{\\varepsilon_v,\n\\sum_{u\\in\\operatorname{ch}(v)}E(u)\\right\\},&v\\text{ internal}.\n\\end{cases}\n\\]\nFor the complete prunings $\\mathfrak P_v$ below $v$, the envelope is exactly\n\\[\nE(v)=\\max_{\\Lambda\\in\\mathfrak P_v}\\sum_{a\\in\\Lambda}\\varepsilon_a.\n\\]\nThe maximum is attained. Every complete-pruning cost changes by at most $E(\\varnothing)$, and the two globally optimal values differ by at most this quantity. Consequently, for a unique optimum with competitors,\n\\[\n\\boxed{\\Gamma_y>2E(\\varnothing)}\n\\]\nis a deterministic certificate that the globally optimal pruning is unchanged.  A singleton pruning set is stable without a gap condition. Assume the random node costs and the joint local-error event are measurable. If a noise model supplies \\emph{all} local bounds simultaneously on one event of probability at least $1-\\delta$, and the displayed","[1mm]\n\\displaystyle\n\\max\\left\\{\\varepsilon_v,\n\\sum_{u\\in\\operatorname{ch}(v)}E(u)\\right\\},&v\\text{ internal}.\n\\end{cases}\n\\]\nFor the complete prunings $\\mathfrak P_v$ below $v$, the envelope is exactly\n\\[\nE(v)=\\max_{\\Lambda\\in\\mathfrak P_v}\\sum_{a\\in\\Lambda}\\varepsilon_a.\n\\]\nThe maximum is attained. Every complete-pruning cost changes by at most $E(\\varnothing)$, and the two globally optimal values differ by at most this quantity. Consequently, for a unique optimum with competitors,\n\\[\n\\boxed{\\Gamma_y>2E(\\varnothing)}\n\\]\nis a deterministic certificate that the globally optimal pruning is unchanged.  A singleton pruning set is stable without a gap condition. Assume the random node costs and the joint local-error event are measurable. If a noise model supplies \\emph{all} local bounds simultaneously on one event of probability at least $1-\\delta$, and the displayed strict-gap condition holds, the same selector is recovered with probability at least $1-\\delta$. I","[1mm]\n\\displaystyle\n\\max\\left\\{\\varepsilon_v,\n\\sum_{u\\in\\operatorname{ch}(v)}E(u)\\right\\},&v\\text{ internal}.\n\\end{cases}\n\\]\nFor the complete prunings $\\mathfrak P_v$ below $v$, the envelope is exactly\n\\[\nE(v)=\\max_{\\Lambda\\in\\mathfrak P_v}\\sum_{a\\in\\Lambda}\\varepsilon_a.\n\\]\nThe maximum is attained. Every complete-pruning cost changes by at most $E(\\varnothing)$, and the two globally optimal values differ by at most this quantity. Consequently, for a unique optimum with competitors,\n\\[\n\\boxed{\\Gamma_y>2E(\\varnothing)}\n\\]\nis a deterministic certificate that the globally optimal pruning is unchanged.  A singleton pruning set is stable without a gap condition. Assume the random node costs and the joint local-error event are measurable. If a noise model supplies \\emph{all} local bounds simultaneously on one event of probability at least $1-\\delta$, and the displayed strict-gap condition holds, the same selector is recovered with probability at least $1-\\delta$. Individual nodewise probabilities alone do not give this joint confidence without an additi","&v\\text{ internal}.\n\\end{cases}\n\\]\nFor the complete prunings $\\mathfrak P_v$ below $v$, the envelope is exactly\n\\[\nE(v)=\\max_{\\Lambda\\in\\mathfrak P_v}\\sum_{a\\in\\Lambda}\\varepsilon_a.\n\\]\nThe maximum is attained. Every complete-pruning cost changes by at most $E(\\varnothing)$, and the two globally optimal values differ by at most this quantity. Consequently, for a unique optimum with competitors,\n\\[\n\\boxed{\\Gamma_y>2E(\\varnothing)}\n\\]\nis a deterministic certificate that the globally optimal pruning is unchanged.  A singleton pruning set is stable without a gap condition. Assume the random node costs and the joint local-error event are measurable. If a noise model supplies \\emph{all} local bounds simultaneously on one event of probability at least $1-\\delta$, and the displayed strict-gap condition holds, the same selector is recovered with probability at least $1-\\delta$. Individual nodewise probabilities alone do not give this joint confidence without an additional argument, such as a union bound.\n\\end{corollary}","y, for a unique optimum with competitors,\n\\[\n\\boxed{\\Gamma_y>2E(\\varnothing)}\n\\]\nis a deterministic certificate that the globally optimal pruning is unchanged.  A singleton pruning set is stable without a gap condition. Assume the random node costs and the joint local-error event are measurable. If a noise model supplies \\emph{all} local bounds simultaneously on one event of probability at least $1-\\delta$, and the displayed strict-gap condition holds, the same selector is recovered with probability at least $1-\\delta$. Individual nodewise probabilities alone do not give this joint confidence without an additional argument, such as a union bound.\n\\end{corollary}","t the globally optimal pruning is unchanged.  A singleton pruning set is stable without a gap condition. Assume the random node costs and the joint local-error event are measurable. If a noise model supplies \\emph{all} local bounds simultaneously on one event of probability at least $1-\\delta$, and the displayed strict-gap condition holds, the same selector is recovered with probability at least $1-\\delta$. Individual nodewise probabilities alone do not give this joint confidence without an additional argument, such as a union bound.\n\\end{corollary}","\\begin{proof}\nFor any pruning of the subtree rooted at $v$, either $v$ itself is selected, producing perturbation at most $\\varepsilon_v$, or the pruning is the disjoint union of child prunings, producing perturbation at most the sum of the child envelopes.  The same decomposition applied to the nonnegative weights $\\varepsilon_a$ proves equality with the maximum weight: at each node choose either the stop option or a maximizing pruning in each child, whichever gives the larger total. Finiteness ensures these choices attain their maxima. Evaluating each cost at a minimizer of the other cost bounds the difference of the two optimal values by the uniform envelope. The selector statement follows from the strict-gap theorem. The finite deterministic c","\\begin{proof}\nFor any pruning of the subtree rooted at $v$, either $v$ itself is selected, producing perturbation at most $\\varepsilon_v$, or the pruning is the disjoint union of child prunings, producing perturbation at most the sum of the child envelopes.  The same decomposition applied to the nonnegative weights $\\varepsilon_a$ proves equality with the maximum weight: at each node choose either the stop option or a maximizing pruning in each child, whichever gives the larger total. Finiteness ensures these choices attain their maxima. Evaluating each cost at a minimizer of the other cost bounds the difference of the two optimal values by the uniform envelope. The selector statement follows from the strict-gap theorem. The finite deterministic comparisons d","\\begin{proof}\nFor any pruning of the subtree rooted at $v$, either $v$ itself is selected, producing perturbation at most $\\varepsilon_v$, or the pruning is the disjoint union of child prunings, producing perturbation at most the sum of the child envelopes.  The same decomposition applied to the nonnegative weights $\\varepsilon_a$ proves equality with the maximum weight: at each node choose either the stop option or a maximizing pruning in each child, whichever gives the larger total. Finiteness ensures these choices attain their maxima. Evaluating each cost at a minimizer of the other cost bounds the difference of the two optimal values by the uniform envelope. The selector statement follows from the strict-gap theorem. The finite deterministic comparisons defining the selector make its preservation event measurable when th","\\begin{proof}\nFor any pruning of the subtree rooted at $v$, either $v$ itself is selected, producing perturbation at most $\\varepsilon_v$, or the pruning is the disjoint union of child prunings, producing perturbation at most the sum of the child envelopes.  The same decomposition applied to the nonnegative weights $\\varepsilon_a$ proves equality with the maximum weight: at each node choose either the stop option or a maximizing pruning in each child, whichever gives the larger total. Finiteness ensures these choices attain their maxima. Evaluating each cost at a minimizer of the other cost bounds the difference of the two optimal values by the uniform envelope. The selector statement follows from the strict-gap theorem. The finite deterministic comparisons defining the selector make its preservation event measurable when the random node costs are measurable. The joint local-error event is contained in the selector-preservation event, so monotonicity of probability proves the probabilistic assertion.\n\\end{proof}","\\begin{certificate}[Adaptive-tree optimum]\\label{core:record:27}\nFor the depth-five example, the complete binary tree has $458{,}330$ admissible prunings.  Dynamic programming and exhaustive enumeration returned the same minimum with residual $0$.  The selected pruning was\n\\[\n\\{000,001,0100,0101,011,1\\},\n\\]\nand mixed-depth synthesis reconstructed the signal with maximum residual $8.88\\times10^{-16}$.  The optimum gap was $3.1909\\times10^{-4}$, while twice the declared perturbation envelope was $6.4\\times10^{-5}$, so the finite selector is certified stable.\n\\end{certificate}","ve-tree optimum]\\label{core:record:27}\nFor the depth-five example, the complete binary tree has $458{,}330$ admissible prunings.  Dynamic programming and exhaustive enumeration returned the same minimum with residual $0$.  The selected pruning was\n\\[\n\\{000,001,0100,0101,011,1\\},\n\\]\nand mixed-depth synthesis reconstructed the signal with maximum residual $8.88\\times10^{-16}$.  The optimum gap was $3.1909\\times10^{-4}$, while twice the declared perturbation envelope was $6.4\\times10^{-5}$, so the finite selector is certified stable.\n\\end{certificate}","plete binary tree has $458{,}330$ admissible prunings.  Dynamic programming and exhaustive enumeration returned the same minimum with residual $0$.  The selected pruning was\n\\[\n\\{000,001,0100,0101,011,1\\},\n\\]\nand mixed-depth synthesis reconstructed the signal with maximum residual $8.88\\times10^{-16}$.  The optimum gap was $3.1909\\times10^{-4}$, while twice the declared perturbation envelope was $6.4\\times10^{-5}$, so the finite selector is certified stable.\n\\end{certificate}","Linear time invariance is not required for exact packet coordinates. Let $A$ and $B$ be formal power series over a commutative coefficient ring (in particular $\\bbC$). Set\n\\[\nA(z)=\\sum_{n\\ge0}a_nz^n,\n\\qquad\nA_i=P_{N,i}A.\n\\]\nAll phase indices range from $0$ to $N-1$. For another representation $B(z)$, multiplication gives a nonlinear or bilinear interaction.","\\begin{theorem}[Cyclic product law]\\label{result:17}\n\\[\n\\boxed{\nP_{N,k}(AB)\n=\n\\sum_{i+j\\equiv k\\, (\\mathrm{mod}\\,N)}A_iB_j.\n}\n\\]\nFor $q\\in\\bbN$, the $q$th-power identity is\n\\[\nP_{N,k}(A^q)\n=\n\\sum_{i_1+\\cdots+i_q\\equiv k}\nA_{i_1}\\cdots A_{i_q}.\n\\]\n\\end{theorem}","\\begin{proof}\nEvery coefficient in $A_iB_j$ has degree congruent to $i+j$ modulo $N$. At any fixed degree $n$, the Cauchy coefficient sum is finite. Sorting these terms by the residues of their two indices gives the first identity coefficient by coefficient. For the power identity, use induction on $q$ and the finite product law. At $q=0$, the index set contains the single empty tuple, its product is $1$, and its residue sum is $0$; hence the right side is $1$ for $k=0$ and $0$ otherwise, as required. No convergence is needed for these formal identities. For analytic power series, the same identities hold wherever the series are absolutely convergent and their Cauchy products may be evaluated.\n\\end{proof}","n{proof}\nEvery coefficient in $A_iB_j$ has degree congruent to $i+j$ modulo $N$. At any fixed degree $n$, the Cauchy coefficient sum is finite. Sorting these terms by the residues of their two indices gives the first identity coefficient by coefficient. For the power identity, use induction on $q$ and the finite product law. At $q=0$, the index set contains the single empty tuple, its product is $1$, and its residue sum is $0$; hence the right side is $1$ for $k=0$ and $0$ otherwise, as required. No convergence is needed for these formal identities. For analytic power series, the same identities hold wherever the series are absolutely convergent and their Cauchy products may be evaluated.\n\\end{proof}","n $A_iB_j$ has degree congruent to $i+j$ modulo $N$. At any fixed degree $n$, the Cauchy coefficient sum is finite. Sorting these terms by the residues of their two indices gives the first identity coefficient by coefficient. For the power identity, use induction on $q$ and the finite product law. At $q=0$, the index set contains the single empty tuple, its product is $1$, and its residue sum is $0$; hence the right side is $1$ for $k=0$ and $0$ otherwise, as required. No convergence is needed for these formal identities. For analytic power series, the same identities hold wherever the series are absolutely convergent and their Cauchy products may be evaluated.\n\\end{proof}","nt to $i+j$ modulo $N$. At any fixed degree $n$, the Cauchy coefficient sum is finite. Sorting these terms by the residues of their two indices gives the first identity coefficient by coefficient. For the power identity, use induction on $q$ and the finite product law. At $q=0$, the index set contains the single empty tuple, its product is $1$, and its residue sum is $0$; hence the right side is $1$ for $k=0$ and $0$ otherwise, as required. No convergence is needed for these formal identities. For analytic power series, the same identities hold wherever the series are absolutely convergent and their Cauchy products may be evaluated.\n\\end{proof}","+j$ modulo $N$. At any fixed degree $n$, the Cauchy coefficient sum is finite. Sorting these terms by the residues of their two indices gives the first identity coefficient by coefficient. For the power identity, use induction on $q$ and the finite product law. At $q=0$, the index set contains the single empty tuple, its product is $1$, and its residue sum is $0$; hence the right side is $1$ for $k=0$ and $0$ otherwise, as required. No convergence is needed for these formal identities. For analytic power series, the same identities hold wherever the series are absolutely convergent and their Cauchy products may be evaluated.\n\\end{proof}","$N$. At any fixed degree $n$, the Cauchy coefficient sum is finite. Sorting these terms by the residues of their two indices gives the first identity coefficient by coefficient. For the power identity, use induction on $q$ and the finite product law. At $q=0$, the index set contains the single empty tuple, its product is $1$, and its residue sum is $0$; hence the right side is $1$ for $k=0$ and $0$ otherwise, as required. No convergence is needed for these formal identities. For analytic power series, the same identities hold wherever the series are absolutely convergent and their Cauchy products may be evaluated.\n\\end{proof}","\\begin{figure}[htbp]\n\\centering\n\\includegraphics[width=0.76\\textwidth]{14_nonlinear_phase_coupling.pdf}\n\\caption{Norms of pairwise product interactions for five residue phases. The label in each cell gives the output phase $(i+j)\\bmod5$.}\n\\label{fig:nonlinear-coupling}\n\\end{figure}","For a finite nonlinear shift operator\n\\[\nF(x)[n]=x[n]+\\alpha x[n]x[n-1],\n\\]\nthe exact packet coordinates are simply\n\\[\nF_r((x_s)_s)=P_{N,r}F\\!\\left(\\sum_sP_{N,s}x\\right).\n\\]\nAnalysis followed by this packet system and synthesis reproduces the direct nonlinear output exactly. The numerical residual in the companion implementation was zero to machine representation.","Let $Z_0>0$ be a reference impedance and\n\\[\nz=\\frac{Z}{Z_0}=r+\\ii x.\n\\]\nThe reflection coefficient is\n\\[\n\\boxed{\n\\Gam=\\frac{z-1}{z+1},\n\\qquad\nz=\\frac{1+\\Gam}{1-\\Gam}.\n}\n\\]","\\begin{classical}[Cayley geometry]\\label{core:record:30}\nThe map sends the finite passive half-plane $r\\ge0$ onto the closed unit disk with $\\Gam=1$ omitted. Adjoining the point at infinity in impedance adds that missing boundary point; the open right half-plane maps conformally onto the open disk. Constant resistance and reactance become the familiar Smith circles. This conformal geometry is classical and is not replaced by polyphase refinement.\n\\end{classical}","Writing $\\Gam=u+\\ii v$, the grid equations are\n\\[\n\\left(u-\\frac{r}{1+r}\\right)^2+v^2=\\frac1{(1+r)^2},\n\\]\nand, for $x\\ne0$,\n\\[\n(u-1)^2+\\left(v-\\frac1x\\right)^2=\\frac1{x^2}.\n\\]","Fix\n\\[\n\\omega_N=e^{2\\pi\\ii/N}\n\\]\nand rotate the reflection coefficient by\n\\[\nR_{N,k}(\\Gam)=\\omega_N^k\\Gam.\n\\]","\\begin{theorem}[Polyphase M\\\"obius action]\\label{result:18}\nRegard normalized impedance as the projective line $\\mathbb P^1(\\bbC)=\\bbC\\cup\\{\\infty\\}$. For $N\\ge1$, the pullback is a globally defined projective action. In the finite chart, wherever the denominator is nonzero, it is\n\\[\n\\boxed{\nT_{N,k}(z)\n=\n\\frac{(1+\\omega_N^k)z+(1-\\omega_N^k)}\n{(1-\\omega_N^k)z+(1+\\omega_N^k)}.\n}\n\\]\nAs maps on $\\mathbb P^1(\\bbC)$ it satisfies\n\\[\nT_{N,k}\\circ T_{N,\\ell}=T_{N,k+\\ell\\, (\\mathrm{mod}\\,N)}.\n\\]\n\\end{theorem}","obius action]\\label{result:18}\nRegard normalized impedance as the projective line $\\mathbb P^1(\\bbC)=\\bbC\\cup\\{\\infty\\}$. For $N\\ge1$, the pullback is a globally defined projective action. In the finite chart, wherever the denominator is nonzero, it is\n\\[\n\\boxed{\nT_{N,k}(z)\n=\n\\frac{(1+\\omega_N^k)z+(1-\\omega_N^k)}\n{(1-\\omega_N^k)z+(1+\\omega_N^k)}.\n}\n\\]\nAs maps on $\\mathbb P^1(\\bbC)$ it satisfies\n\\[\nT_{N,k}\\circ T_{N,\\ell}=T_{N,k+\\ell\\, (\\mathrm{mod}\\,N)}.\n\\]\n\\end{theorem}","\\begin{proof}\nLet $C(z)=(z-1)/(z+1)$. Then\n\\[\nT_{N,k}=C^{-1}\\circ R_{N,k}\\circ C.\n\\]\nFor a division-free construction, represent a point by a nonzero pair $[u:v]$, identifying $[cu:cv]=[u:v]$ for $c\\ne0$. With $w=\\omega_N^k$, use the matrix\n\\[\nA(w)=\\frac12\\begin{pmatrix}1+w&1-w","\\begin{proof}\nLet $C(z)=(z-1)/(z+1)$. Then\n\\[\nT_{N,k}=C^{-1}\\circ R_{N,k}\\circ C.\n\\]\nFor a division-free construction, represent a point by a nonzero pair $[u:v]$, identifying $[cu:cv]=[u:v]$ for $c\\ne0$. With $w=\\omega_N^k$, use the matrix\n\\[\nA(w)=\\frac12\\begin{pmatrix}1+w&1-w\\\\1-w&1+w\\end{pmatrix}.\n\\]\nIt has determinant $w\\ne0$, preserves nonzero pairs, and commutes with rescaling, so it induces a projective map. Direct multiplication gives $A(w)A(w')=A(ww')$ and $A(1)=I$. Since $\\omega_N^N=1$, this proves the cyclic composition law modulo $N$. The Cayley coordinate of $[u:v]$ is $[u-v:u+v]$, and applying $A(w)$ changes this pair to $[w(u-v):u+v]$. This proves the conjugacy also at poles. For $[z:1]$, taking the coordinate ratio gives the displayed finite-chart formula.\n\\end{proof}","1-w&1+w\\end{pmatrix}.\n\\]\nIt has determinant $w\\ne0$, preserves nonzero pairs, and commutes with rescaling, so it induces a projective map. Direct multiplication gives $A(w)A(w')=A(ww')$ and $A(1)=I$. Since $\\omega_N^N=1$, this proves the cyclic composition law modulo $N$. The Cayley coordinate of $[u:v]$ is $[u-v:u+v]$, and applying $A(w)$ changes this pair to $[w(u-v):u+v]$. This proves the conjugacy also at poles. For $[z:1]$, taking the coordinate ratio gives the displayed finite-chart formula.\n\\end{proof}","It has determinant $w\\ne0$, preserves nonzero pairs, and commutes with rescaling, so it induces a projective map. Direct multiplication gives $A(w)A(w')=A(ww')$ and $A(1)=I$. Since $\\omega_N^N=1$, this proves the cyclic composition law modulo $N$. The Cayley coordinate of $[u:v]$ is $[u-v:u+v]$, and applying $A(w)$ changes this pair to $[w(u-v):u+v]$. This proves the conjugacy also at poles. For $[z:1]$, taking the coordinate ratio gives the displayed finite-chart formula.\n\\end{proof}","\\begin{corollary}[Impedance--admittance is the two-phase split]\\label{result:19}\nFor $N=2$ and finite $z\\ne0$,\n\\[\n\\boxed{T_{2,1}(z)=z^{-1}.}\n\\]\nOn the full projective line it exchanges $0$ and $\\infty$. Thus a half-turn of the reflection disk exchanges normalized impedance and admittance.\n\\end{corollary}\n\\begin{proof}\n$\\omega_2=\\exp(\\pi\\ii)=-1$, so $A(\\omega_2)[u:v]=[v:u]$. The finite ratio is $1/z$ when $z\\ne0$; the coordinate swap also gives $[0:1]\\leftrightarrow[1:0]$. No ordinary division by zero is used to represent infinity.\n\\end{proof}","rollary}[Impedance--admittance is the two-phase split]\\label{result:19}\nFor $N=2$ and finite $z\\ne0$,\n\\[\n\\boxed{T_{2,1}(z)=z^{-1}.}\n\\]\nOn the full projective line it exchanges $0$ and $\\infty$. Thus a half-turn of the reflection disk exchanges normalized impedance and admittance.\n\\end{corollary}\n\\begin{proof}\n$\\omega_2=\\exp(\\pi\\ii)=-1$, so $A(\\omega_2)[u:v]=[v:u]$. The finite ratio is $1/z$ when $z\\ne0$; the coordinate swap also gives $[0:1]\\leftrightarrow[1:0]$. No ordinary division by zero is used to represent infinity.\n\\end{proof}","o-phase split]\\label{result:19}\nFor $N=2$ and finite $z\\ne0$,\n\\[\n\\boxed{T_{2,1}(z)=z^{-1}.}\n\\]\nOn the full projective line it exchanges $0$ and $\\infty$. Thus a half-turn of the reflection disk exchanges normalized impedance and admittance.\n\\end{corollary}\n\\begin{proof}\n$\\omega_2=\\exp(\\pi\\ii)=-1$, so $A(\\omega_2)[u:v]=[v:u]$. The finite ratio is $1/z$ when $z\\ne0$; the coordinate swap also gives $[0:1]\\leftrightarrow[1:0]$. No ordinary division by zero is used to represent infinity.\n\\end{proof}","For $0\\le\\rho<1$ and angle $\\theta$,\n\\[\n\\Gam=\\rho e^{\\ii\\theta}\n\\]\ngives\n\\[\nz(\\rho,\\theta)\n=\\frac{1+\\rho e^{\\ii\\theta}}{1-\\rho e^{\\ii\\theta}},\n\\]\nwith\n\\[\nr(\\rho,\\theta)=\\frac{1-\\rho^2}{1-2\\rho\\cos\\theta+\\rho^2},\n\\qquad\nx(\\rho,\\theta)=\\frac{2\\rho\\sin\\theta}{1-2\\rho\\cos\\theta+\\rho^2}.\n\\]\nA fixed $\\rho$ is a constant-mismatch circle. Refining $N$ increases angular resolution without changing $\\rho$.","\\begin{figure}[htbp]\n\\centering\n\\includegraphics[width=0.78\\textwidth]{12_smith_nested_refinement.pdf}\n\\caption{The same reflection circle at phase orders $2,4,8,16,32$. Each coarse phase ray appears in every finer grid whose order is a multiple of its order. The exact residue-projector refinement law concerns coefficient packets, not sums of impedance points.}\n\\label{fig:smith-refinement}\n\\end{figure}","For a lossless transmission line,\n\\[\n\\Gam_{\\rm in}=\\Gam_L e^{-2\\ii\\beta\\ell}.\n\\]\nOne $N$-phase step corresponds to a line-length increment $\\lambda/(2N)$, up to the chosen orientation convention. Hence the phase address has a direct electrical interpretation.","Write\n\\[\nN=2^sM,\n\\qquad M\\text{ odd}.\n\\]\nThe cyclic index splits by the Chinese remainder theorem:\n\\[\n\\bbZ/N\\bbZ\n\\cong\n\\bbZ/2^s\\bbZ\\times\\bbZ/M\\bbZ.\n\\]","\\begin{proposition}[Dyadic--odd packet factorization]\\label{result:20}\nAfter identifying the residue labels by the Chinese remainder map, the phase projector factors as\n\\[\n\\boxed{\nP_{N,j}=P_{2^s,j_2}P_{M,j_o}.\n}\n\\]\n\\end{proposition}","\\begin{proof}\nThe two congruence conditions are independent because $2^s$ and $M$ are coprime. Their indicator functions multiply to the indicator of the unique residue class modulo $N$ satisfying both.\n\\end{proof}","The first dyadic step is exceptional in Smith coordinates because it is the inversion $z\\mapsto z^{-1}$. Higher dyadic levels refine the open/short duality into quarter-, eighth-, and finer phase sectors.","Away from its pole $\\Gam=1$, the inverse Cayley map has derivative\n\\[\n\\frac{dz}{d\\Gam}=\\frac{2}{(1-\\Gam)^2},\n\\]\nwhile normalized admittance has derivative proportional to $(1+\\Gam)^{-2}$. Away from $\\Gam=\\pm1$, this motivates the symmetric sensitivity\n\\[\n\\boxed{\nw_{\\rm Smith}(\\Gam)\n=\n\\frac1{|1-\\Gam|^2}\n+\n\\frac1{|1+\\Gam|^2}.\n}\n\\]","\\begin{conjecture}[Edge-packet approximation class]\\label{core:record:35}\nThere should exist natural function spaces in which a signal is represented as\n\\[\nf=s+\\sum_q a_q\\mathbf1_{\\Omega_q},\n\\]\nwith separate norms for smooth approximation, boundary geometry, packet amplitude, and packet identification. A projector-resolved approximation theory on such spaces should yield rates that do not mix smooth truncation error with discontinuity-location error.\n\\end{conjecture}","\\begin{certificate}[Finite numerical residuals]\\label{core:record:36}\nThe largest recorded residuals were\n\\[\n\\begin{array}{lr}\n\\toprule\n\\text{identity}&\\text{residual}\\\\\n\\midrule\n\\text{polyphase reconstruction}&0\\\\\n\\text{nested refinement}&0\\\\\n\\text{polyphase convolution}&4.44\\times10^{-16}\\\\\n\\text{Haar reconstruction}&8.88\\times10^{-16}\\\\\n\\text{wavelet-packet reconstruction}&2.22\\times10^{-15}\\\\\n\\text{adaptive-tree optimum versus exhaustive search}&0\\\\\n\\text{adaptive mixed-depth synthesis}&8.88\\times10^{-16}\\\\\n\\text{lifting reconstruction}&4.44\\times10^{-16}\\\\\n\\text{Smith rotation conjugacy}&3.34\\times10^{-16}\\\\\n\\text{Smith impedance--admittance duality}&0\\\\\n\\text{nonlinear product-packet reconstruction}&1.33\\times10^{-15}\\\\\n\\text{nonlinear shifted-filter reconstruction}&0\\\\\n\\bottomrule\n\\end{array}\n\\]\nThe Haar and wavelet energy residuals were $5.68\\times10^{-14}$ and $2.27\\times10^{-13}$, respectively.\n\\end{certificate}","\\bottomrule\n\\end{array}\n\\]\nThe Haar and wavelet energy residuals were $5.68\\times10^{-14}$ and $2.27\\times10^{-13}$, respectively.\n\\end{certificate}","The signal realization makes the seed, address, packet, transfer, refinement, and synthesis operations concrete. The objects and laws can be read directly:\n\\[\n\\boxed{\n\\begin{aligned}\n\\text{seed}&=\\delta_0,\\\\\n\\text{address}&=S^n,\\\\\n\\text{packet}&=P_{N,r}x,\\\\\n\\text{transfer}&=P_{N,k}HP_{N,s},\\\\\n\\text{refinement}&=P_{N,r}=\\sum_{s\\succ r}P_{M,s},\\\\\n\\text{synthesis}&=\\sum_rP_{N,r}x,\\\\\n\\text{nonlinearity}&=P_{N,r}F\\left(\\sum_sP_{N,s}x\\right).\n\\end{aligned}\n}\n\\]\nFilter banks, wavelet packets, adaptive complete prunings, edge packets, and Smith coordinates inhabit the same operator grammar, but their claims differ. Polyphase and filter-bank identities are exact. Wavelet perfect reconstruction is exact. The known-edge theorem is exact conditional on the decomposition. Estimated-edge performance depends on a detection model. The Smith chart remains the classical Cayley geometry equipped with an exact cyclic refinement algebra. Keeping those distinctions explicit is part of the calculus, not an editorial afterthought.","Given $x$, $h$, and $N$:","\\begin{enumerate}\n\\item Form $x_s[n]=x[Nn+s]$ for $0\\le s<N$.\n\\item Form every transfer kernel $h_{ks}[m]=h[Nm+k-s]$.\n\\item Compute\n\\[\ny_k[n]=\\sum_s\\sum_mh_{ks}[m]x_s[n-m].\n\\]\n\\item Interleave the $y_k$ channels.\n\\end{enumerate}","Given a decomposition $f=s+j$:","\\begin{enumerate}\n\\item Apply the selected smooth approximation $Q_m$ to $s$ only.\n\\item Retain $j$ exactly.\n\\item Return $Q_ms+j$.\n\\end{enumerate}","The error certificate is $\\|s-Q_ms\\|$.","\\midrule\n1D rectangular packet&$5.4023\\times10^{-2}$&$1.3692\\times10^{-4}$&$2.4029\\times10^{-2}$","2D disk packet&$3.7643\\times10^{-2}$&$6.4752\\times10^{-5}$&$3.4043\\times10^{-3}$","2D multiedge packet&$3.3740\\times10^{-2}$&\\multicolumn{2}{c}{known composite packet only}","\\midrule\n\\endhead\n$s$&$\\delta_0$&primitive signal seed","$A_n$&$S^n$&address operator","$P_{N,r}$&residue mask&phase packet projector","$T_{ks}$&$P_{N,k}HP_{N,s}$&input-to-output phase transfer","$E,R$&analysis and synthesis matrices&coordinate conversion and reconstruction","$W(\\gamma)$&filter or path weight&refinement or implementation cost","$\\Delta(F)$&coarse/fine signal defect&aliasing, leakage, or estimator inconsistency","$\\operatorname{Hol}(\\gamma)$&closed phase transfer&cycle consistency in a network","$\\Gam$&reflection coefficient&Smith-disk coordinate","\\midrule\\endhead\n\\ref{result:01}&HilbertPackets&Complex $\\ell^2$ impulse expansion, finite-support identity, unitary shifts, and closed shift-generated span.","\\ref{result:06}&FilterBank; FiniteTransforms&Finite Laurent/FIR bank composition, faithful action on sequences, gain-delay reconstruction criterion, and the $Nd$ original-sample delay.","\\ref{result:07}&HaarChannels; HaarHilbert&Literal even/odd input samples, normalized complex low/detail channels, and $\\ell^2$ energy equality.","\\ref{result:12}&MaskPerturbation&Measurable-mask symmetric-difference norm, extended finite-$p$ error bound, pointwise and essential supremum bounds, and smooth-background error.","\\ref{result:19}&SmithCyclic&Two-phase coordinate swap, $0\\leftrightarrow\\infty$, and finite impedance inversion when $z\\ne0$.","\\paragraph{Finite filters and numerical transforms.}\nThe filter-bank proof is an algebraic theorem about finite Laurent-polynomial kernels acting faithfully on bilateral sequences. Equality at a single complex value of $z$ would be insufficient to prove a polynomial identity; the impulse argument avoids that error. The separate transform theorem proves the finite-filter action agrees with evaluation wherever the weighted input series converge absolutely. An arbitrary bilateral $\\ell^2$ sequence need not have an absolutely convergent numerical $z$-transform. The all-sequence time-domain theorem makes no such assumption.","gebraic theorem about finite Laurent-polynomial kernels acting faithfully on bilateral sequences. Equality at a single complex value of $z$ would be insufficient to prove a polynomial identity; the impulse argument avoids that error. The separate transform theorem proves the finite-filter action agrees with evaluation wherever the weighted input series converge absolutely. An arbitrary bilateral $\\ell^2$ sequence need not have an absolutely convergent numerical $z$-transform. The all-sequence time-domain theorem makes no such assumption.","hfully on bilateral sequences. Equality at a single complex value of $z$ would be insufficient to prove a polynomial identity; the impulse argument avoids that error. The separate transform theorem proves the finite-filter action agrees with evaluation wherever the weighted input series converge absolutely. An arbitrary bilateral $\\ell^2$ sequence need not have an absolutely convergent numerical $z$-transform. The all-sequence time-domain theorem makes no such assumption.","\\paragraph{Hilbert operators and finite experiments.}\nThe coordinate maps, literal Haar channels, and packet operators are constructed on complex $\\ell^2$. Square summability and boundedness are established before adjoints are used. The finite Python experiments separately check indexing, reconstruction, and energy at finite lengths. They illustrate these operators but are not the objects executed by the Lean kernel. Likewise, the Python Smith routine evaluates only finite charts; the Lean projective construction supplies the pole-safe universal statement.","For the one-dimensional low-pass experiment, ``cutoff 18'' means 37 signed frequency bins, not 18 retained coefficients. The disk experiment uses a square $33\\times33$ frequency mask, not a radial cutoff. The sparse comparison uses largest-coefficient budgets instead of either low-pass convention. The full-image and boundary-band RMSEs are normalized by the number of samples in the respective comparison set. The supplied inpainting routines perform a fixed finite number of periodic neighbor-averaging iterations; ``harmonic filling'' describes that iteration, not a separately proved exact continuum Dirichlet solver.","The second aim is to separate representation from recovery. If a signal has already been written as\n\\[\nf=s+j\n\\]\nwith a smooth part \\(s\\) and a singular packet \\(j\\), then\n\\[\n\\widetilde f=Qs+j\n\\]\nsatisfies\n\\[\nf-\\widetilde f=s-Qs.\n\\]\nThis identity is exact, but it does not discover \\(j\\). A useful signal theory must therefore distinguish three errors. The first is the approximation error of the smooth component. The second is the error in the estimated singular packet. The third is the amplification caused by applying a reconstruction operator to that packet error. The middle sections of the paper isolate all three terms.","The singular packets considered here range from indicator functions of regions to truncated powers\n\\[\n\\frac{(t-\\tau)_+^m}{m!}.\n\\]\nThe latter encode jumps in derivatives of order \\(m\\). They permit a piecewise smooth function to be reduced to a globally Sobolev remainder under explicit trace assumptions. This gives a precise version of the informal statement that a discontinuity should be represented directly rather than approximated by a global smooth basis.","The first result is an exact coordinate theorem. Let \\(L\\) be a finite index sublattice of \\(\\ZZ^d\\), and let \\(R\\) contain one representative from every coset. The analysis map\n\\[\nx\\longmapsto (x_r)_{r\\in R},\n\\qquad\nx_r(\\ell)=x(r+\\ell),\n\\]\nis an isometric isomorphism on every \\(\\ell^p\\) space. This statement includes the usual one dimensional polyphase decomposition, but it also applies to arbitrary finite index lattice packets in several dimensions. If \\(L'\\leq L\\), the coarse packets split exactly into their fine descendants. A convolution operator becomes a matrix of convolution operators on \\(L\\), with entries\n\\[\nh_{rs}(m)=h(r-s+m).\n\\]\nThe matrix is therefore a coordinate representation of the original operator, not a reduced model.","a finite index sublattice of \\(\\ZZ^d\\), and let \\(R\\) contain one representative from every coset. The analysis map\n\\[\nx\\longmapsto (x_r)_{r\\in R},\n\\qquad\nx_r(\\ell)=x(r+\\ell),\n\\]\nis an isometric isomorphism on every \\(\\ell^p\\) space. This statement includes the usual one dimensional polyphase decomposition, but it also applies to arbitrary finite index lattice packets in several dimensions. If \\(L'\\leq L\\), the coarse packets split exactly into their fine descendants. A convolution operator becomes a matrix of convolution operators on \\(L\\), with entries\n\\[\nh_{rs}(m)=h(r-s+m).\n\\]\nThe matrix is therefore a coordinate representation of the original operator, not a reduced model.","R},\n\\qquad\nx_r(\\ell)=x(r+\\ell),\n\\]\nis an isometric isomorphism on every \\(\\ell^p\\) space. This statement includes the usual one dimensional polyphase decomposition, but it also applies to arbitrary finite index lattice packets in several dimensions. If \\(L'\\leq L\\), the coarse packets split exactly into their fine descendants. A convolution operator becomes a matrix of convolution operators on \\(L\\), with entries\n\\[\nh_{rs}(m)=h(r-s+m).\n\\]\nThe matrix is therefore a coordinate representation of the original operator, not a reduced model.",")=x(r+\\ell),\n\\]\nis an isometric isomorphism on every \\(\\ell^p\\) space. This statement includes the usual one dimensional polyphase decomposition, but it also applies to arbitrary finite index lattice packets in several dimensions. If \\(L'\\leq L\\), the coarse packets split exactly into their fine descendants. A convolution operator becomes a matrix of convolution operators on \\(L\\), with entries\n\\[\nh_{rs}(m)=h(r-s+m).\n\\]\nThe matrix is therefore a coordinate representation of the original operator, not a reduced model.","The filter bank results distinguish exact reconstruction from quantitative near reconstruction. Exact recovery with delay \\(d\\) is the pointwise matrix identity\n\\[\nR(z)E(z)=z^{-d}I.\n\\]\nWhen this identity fails, the reconstruction error is controlled by the essential supremum of the delay compensated multiplier defect. If the filters themselves are perturbed, \\cref{thm:filter_perturbation} separates the analysis error, the synthesis error, and their product. The frame bound in \\cref{thm:frame_perturbation} shows that a lower singular value gap survives every sufficiently small perturbation. This makes the distinction between algebraic invertibility and stable numerical reconstruction explicit.","The packet tree results treat every admissible pruning, not only complete trees of a fixed depth. Unitary local splits give a unitary global transform. Local frame bounds propagate recursively in the nonunitary case. If the cost of a pruning is additive over retained nodes, then\n\\[\nF(v)\n=\n\\min\\left\\{\nc(v),\n\\sum_{w\\in\\operatorname{child}(v)}F(w)\n\\right\\}\n\\]\ngives the exact optimal tree. The cost and the selected basis are stable when the node costs change by less than the gap separating the optimal pruning from its competitors.","The singular packet results begin with trace data. If a piecewise smooth function has a jump in its derivative of order \\(m\\) at \\(\\tau_q\\), then the packet\n\\[\n\\jump{f^{(m)}}_{\\tau_q}\n\\frac{(t-\\tau_q)_+^m}{m!}\n\\]\nremoves exactly that trace jump. Subtracting all packets through order \\(r-1\\) leaves a global \\(W^{r,p}\\) remainder. The packet coefficients are uniquely determined by the one sided trace jumps. This gives a finite and explicit singular part for the class covered by \\cref{thm:higher_order_packet_decomposition}.","The reconstruction formula with an estimated packet is\n\\[\nf-\\widehat f_m\n=\n(I-Q_m)s+(I-Q_m)(j-\\widehat j).\n\\]\nThis identity identifies the two terms that must be controlled. The exact packet formulas in \\cref{sec:packet_error} then evaluate or bound \\(\\|j-\\widehat j\\|\\) from amplitude error, region overlap, boundary displacement, and the order of a derivative jump. A value jump has location modulus \\(|\\tau-\\widehat\\tau|^{1/p}\\) in finite \\(L^p\\), while every positive order truncated power packet is Lipschitz in its location.","reconstruction formula with an estimated packet is\n\\[\nf-\\widehat f_m\n=\n(I-Q_m)s+(I-Q_m)(j-\\widehat j).\n\\]\nThis identity identifies the two terms that must be controlled. The exact packet formulas in \\cref{sec:packet_error} then evaluate or bound \\(\\|j-\\widehat j\\|\\) from amplitude error, region overlap, boundary displacement, and the order of a derivative jump. A value jump has location modulus \\(|\\tau-\\widehat\\tau|^{1/p}\\) in finite \\(L^p\\), while every positive order truncated power packet is Lipschitz in its location.","packet is\n\\[\nf-\\widehat f_m\n=\n(I-Q_m)s+(I-Q_m)(j-\\widehat j).\n\\]\nThis identity identifies the two terms that must be controlled. The exact packet formulas in \\cref{sec:packet_error} then evaluate or bound \\(\\|j-\\widehat j\\|\\) from amplitude error, region overlap, boundary displacement, and the order of a derivative jump. A value jump has location modulus \\(|\\tau-\\widehat\\tau|^{1/p}\\) in finite \\(L^p\\), while every positive order truncated power packet is Lipschitz in its location.","The principal recovery theorem uses adjacent sample contrasts. Let\n\\[\nb_h=Mh+2\\varepsilon\n\\]\nbe the largest contrast that can be produced by a Lipschitz background and bounded noise in one grid cell. If\n\\[\na_{\\min}>2b_h,\n\\]\nthen every threshold between \\(b_h\\) and \\(a_{\\min}-b_h\\) detects exactly the cells containing jumps. The jump locations are recovered within \\(h/2\\), and each amplitude error is at most \\(b_h\\). The probability theorem replaces the deterministic noise term by\n\\[\n2\\sigma\\sqrt{\\log(2n/\\delta)}.\n\\]\nTwo half shifted Haar grids provide a second recovery mechanism. Every point is at least one quarter of a cell away from the boundary of one of the grids, so the Haar coefficient gap can be made independent of dyadic alignment.","recovery theorem uses adjacent sample contrasts. Let\n\\[\nb_h=Mh+2\\varepsilon\n\\]\nbe the largest contrast that can be produced by a Lipschitz background and bounded noise in one grid cell. If\n\\[\na_{\\min}>2b_h,\n\\]\nthen every threshold between \\(b_h\\) and \\(a_{\\min}-b_h\\) detects exactly the cells containing jumps. The jump locations are recovered within \\(h/2\\), and each amplitude error is at most \\(b_h\\). The probability theorem replaces the deterministic noise term by\n\\[\n2\\sigma\\sqrt{\\log(2n/\\delta)}.\n\\]\nTwo half shifted Haar grids provide a second recovery mechanism. Every point is at least one quarter of a cell away from the boundary of one of the grids, so the Haar coefficient gap can be made independent of dyadic alignment.","on\n\\]\nbe the largest contrast that can be produced by a Lipschitz background and bounded noise in one grid cell. If\n\\[\na_{\\min}>2b_h,\n\\]\nthen every threshold between \\(b_h\\) and \\(a_{\\min}-b_h\\) detects exactly the cells containing jumps. The jump locations are recovered within \\(h/2\\), and each amplitude error is at most \\(b_h\\). The probability theorem replaces the deterministic noise term by\n\\[\n2\\sigma\\sqrt{\\log(2n/\\delta)}.\n\\]\nTwo half shifted Haar grids provide a second recovery mechanism. Every point is at least one quarter of a cell away from the boundary of one of the grids, so the Haar coefficient gap can be made independent of dyadic alignment.","\\midrule\n\\(L\\leq\\ZZ^d\\) & a finite index sublattice","\\(R\\) & a set of representatives for \\(\\ZZ^d/L\\)","\\(P_r\\) & the coordinate projector onto the coset \\(r+L\\)","\\(\\Acal_L,\\Scal_L\\) & lattice analysis and interleaving synthesis","\\(E,R\\) & filter bank analysis and synthesis symbols","\\(Q_m\\) & a bounded linear approximation operator acting on the smooth part","\\(j,\\widehat j\\) & the true and estimated singular packets","\\(H_\\tau\\) & the step packet \\(\\one_{[\\tau,1]}\\)","\\(\\phi_{m,\\tau}\\) & the truncated power packet \\((t-\\tau)_+^m/m!\\)","\\(h\\) & the grid spacing in the recovery theorems","Let \\(1\\leq p\\leq\\infty\\). For a countable set \\(G\\), write \\(\\ell^p(G)\\) for the usual complex sequence space. For an integer \\(d\\geq1\\), when \\(G=\\ZZ^d\\), translation by \\(a\\in\\ZZ^d\\) is\n\\[\n(T_ax)(n)=x(n-a).\n\\]\nLet \\(\\delta_0\\) be the impulse at the origin.","\\begin{theorem}[Impulse generation]\\label{thm:impulse_generation}\nFor \\(1\\leq p<\\infty\\),\n\\[\n\\overline{\\Span\\{T_a\\delta_0:a\\in\\ZZ^d\\}}^{\\ell^p}\n=\n\\ell^p(\\ZZ^d).\n\\]\nThe same statement holds in \\(c_0(\\ZZ^d)\\) with the supremum norm. In \\(\\ell^\\infty(\\ZZ^d)\\), the closed span of the translation orbit is exactly \\(c_0(\\ZZ^d)\\).\n\\end{theorem}","\\begin{proof}\nThe translation orbit is the standard coordinate family \\((\\delta_a)_{a\\in\\ZZ^d}\\). Its finite span is the space of finitely supported sequences. Finitely supported sequences are dense in \\(\\ell^p\\) for finite \\(p\\) and in \\(c_0\\). Their closure in the supremum norm is \\(c_0\\), which is a proper closed subspace of \\(\\ell^\\infty\\).\n\\end{proof}","\\begin{remark}\\label{rem:infinity_boundary}\nThe distinction at \\(p=\\infty\\) is structural. The impulse generates every finitely supported signal algebraically, but it is not a topological cyclic seed for all bounded signals.\n\\end{remark}","\\begin{proposition}[Finite periodic signals]\\label{prop:finite_periodic}\nLet \\(G=\\ZZ^d/L\\), where \\(L\\leq\\ZZ^d\\) has finite index. The translation orbit of the impulse at the identity is a basis of \\(\\CC^G\\). In particular, one impulse generates every signal on the finite periodic lattice.\n\\end{proposition}","\\begin{proof}\nTranslation sends the impulse at the identity to the impulse at each element of \\(G\\). These impulses form the standard basis of \\(\\CC^G\\).\n\\end{proof}","Let \\(L\\leq\\ZZ^d\\) be a sublattice of finite index \\(q\\). Choose a set \\(R\\subset\\ZZ^d\\) containing one representative from every coset in \\(\\ZZ^d/L\\). Thus\n\\[\n\\ZZ^d=\\bigsqcup_{r\\in R}(r+L).\n\\]\nFor \\(x:\\ZZ^d\\to\\CC\\), its channel at \\(r\\) is the sequence on \\(L\\)\n\\[\nx_r(\\ell)=x(r+\\ell).\n\\]","\\begin{definition}[Lattice packet]\\label{def:lattice_packet}\nFor \\(r\\in R\\), define the coordinate projector\n\\[\n(P_rx)(n)=\\one_{r+L}(n)x(n).\n\\]\nThe family \\((P_r)_{r\\in R}\\) is called the packet system associated with \\(L\\) and \\(R\\).\n\\end{definition}","For finite \\(p\\), equip \\(\\bigoplus_{r\\in R}\\ell^p(L)\\) with\n\\[\n\\|(u_r)\\|^p=\\sum_{r\\in R}\\|u_r\\|_p^p.\n\\]\nFor \\(p=\\infty\\), use the maximum norm.","\\begin{theorem}[Exact lattice polyphase decomposition]\\label{thm:lattice_polyphase}\nFor every \\(1\\leq p\\leq\\infty\\), the analysis map\n\\[\n\\Acal_L:\\ell^p(\\ZZ^d)\\longrightarrow\\bigoplus_{r\\in R}\\ell^p(L),\n\\qquad\n\\Acal_Lx=(x_r)_{r\\in R},\n\\]\nis an isometric isomorphism. Its inverse is the interleaving map\n\\[\n\\Scal_L(u)(r+\\ell)=u_r(\\ell).\n\\]\nThe projectors satisfy\n\\[\nP_rP_s=\\delta_{rs}P_r,\n\\qquad\n\\sum_{r\\in R}P_r=I.\n\\]\n\\end{theorem}","\\begin{proof}\nEvery address has a unique representation \\(r+\\ell\\). For finite \\(p\\),\n\\[\n\\|x\\|_p^p\n=\n\\sum_{r\\in R}\\sum_{\\ell\\in L}|x(r+\\ell)|^p\n=\n\\sum_{r\\in R}\\|x_r\\|_p^p.\n\\]\nThe corresponding supremum identity is immediate when \\(p=\\infty\\). Unique address representation proves that analysis and interleaving are inverse maps. The projector identities follow from the disjoint coset partition.\n\\end{proof}","\\begin{corollary}[One dimensional residue coordinates]\\label{cor:one_dimensional_polyphase}\nFor \\(L=N\\ZZ\\) and \\(R=\\{0,\\ldots,N-1\\}\\),\n\\[\nx_r(k)=x(Nk+r)\n\\]\nand\n\\[\n\\ell^p(\\ZZ)\\cong\\bigoplus_{r=0}^{N-1}\\ell^p(\\ZZ)\n\\]\nisometrically.\n\\end{corollary}","\\begin{proposition}[Translation covariance]\\label{prop:translation_covariance}\nFix \\(a\\in\\ZZ^d\\). For every \\(r\\in R\\), write uniquely\n\\[\nr-a=r_a+\\lambda_{a,r},\n\\qquad\nr_a\\in R,\n\\quad\n\\lambda_{a,r}\\in L.\n\\]\nThen\n\\[\n(\\Acal_LT_ax)_r(\\ell)\n=\nx_{r_a}(\\ell+\\lambda_{a,r}).\n\\]\nThus a fine translation permutes the lattice packets and applies an internal translation on \\(L\\).\n\\end{proposition}","\\begin{proof}\nEvaluate at \\(r+\\ell\\):\n\\[\n(T_ax)(r+\\ell)\n=x(r+\\ell-a)\n=x(r_a+\\ell+\\lambda_{a,r}).\n\\]\nThis is the stated channel formula.\n\\end{proof}","Let \\(L'\\leq L\\leq\\ZZ^d\\), with both indices finite. Choose representative systems \\(R\\) for \\(\\ZZ^d/L\\) and \\(R'\\) for \\(\\ZZ^d/L'\\). For \\(r\\in R\\), define\n\\[\n\\operatorname{Child}(r)\n=\n\\{s\\in R':s+L'\\subset r+L\\}.\n\\]","\\begin{theorem}[Exact lattice refinement]\\label{thm:lattice_refinement}\nFor every coarse packet,\n\\[\nP_r^{L}\n=\n\\sum_{s\\in\\operatorname{Child}(r)}P_s^{L'}.\n\\]\nThe fine packets in the sum are pairwise disjoint. Refinement through several nested sublattices is transitive.\n\\end{theorem}","\\begin{proof}\nThe coset \\(r+L\\) is the disjoint union of the \\(L'\\) cosets contained in it. Multiplying by the corresponding indicator functions gives the projector identity. A nested partition of a coset is independent of whether the intermediate level is inserted, which proves transitivity.\n\\end{proof}","\\begin{example}[Rectangular image phases]\\label{ex:image_phases}\nFor\n\\[\nL=M\\ZZ\\times N\\ZZ\\subset\\ZZ^2,\n\\]\none may take\n\\[\nR=\\{0,\\ldots,M-1\\}\\times\\{0,\\ldots,N-1\\}.\n\\]\nThe analysis map separates an image into \\(MN\\) exact phases. Refining \\((M,N)\\) to \\((M',N')\\), where \\(M\\mid M'\\) and \\(N\\mid N'\\), splits every coarse phase into\n\\[\n\\frac{M'}M\\frac{N'}N\n\\]\nfine phases.\n\\end{example}","Let \\(h\\in\\ell^1(\\ZZ^d)\\) and \\(x\\in\\ell^p(\\ZZ^d)\\), with \\(1\\leq p\\leq\\infty\\), and define\n\\[\n(Hx)(n)\n=(h*x)(n)\n=\\sum_{q\\in\\ZZ^d}h(q)x(n-q).\n\\]\nYoung's inequality gives\n\\[\n\\|Hx\\|_p\\leq\\|h\\|_1\\|x\\|_p\n\\]\nfor every \\(1\\leq p\\leq\\infty\\).","\\begin{definition}[Packet transfer kernel]\\label{def:packet_transfer_kernel}\nFor output residue \\(r\\in R\\) and input residue \\(s\\in R\\), define a kernel on \\(L\\) by\n\\[\nh_{rs}(m)=h(r-s+m),\n\\qquad\nm\\in L.\n\\]\n\\end{definition}","\\begin{theorem}[Exact convolution transfer law]\\label{thm:convolution_transfer}\nLet \\(y=h*x\\). Then\n\\[\ny_r(\\ell)\n=\n\\sum_{s\\in R}\\sum_{m\\in L}h_{rs}(m)x_s(\\ell-m).\n\\]\nEquivalently,\n\\[\n\\Acal_LH\\Acal_L^{-1}\n=\n(H_{rs})_{r,s\\in R},\n\\]\nwhere \\(H_{rs}\\) is convolution on \\(L\\) by \\(h_{rs}\\).\n\\end{theorem}","\\begin{proof}\nEvaluate the output at \\(r+\\ell\\). Write every input address as \\(s+\\ell'\\). The convolution offset is\n\\[\n(r+\\ell)-(s+\\ell')\n=(r-s)+(\\ell-\\ell').\n\\]\nSetting \\(m=\\ell-\\ell'\\) gives the formula.\n\\end{proof}","\\begin{proposition}[Transfer norm control]\\label{prop:transfer_norm_control}\nFor every output residue \\(r\\),\n\\[\n\\sum_{s\\in R}\\|h_{rs}\\|_{\\ell^1(L)}\n=\n\\|h\\|_{\\ell^1(\\ZZ^d)}.\n\\]\nThe packet transfer matrix acts on the polyphase direct sum with norm at most \\(\\|h\\|_1\\) on every \\(\\ell^p\\) space.\n\\end{proposition}","\\begin{proof}\nFor fixed \\(r\\), every \\(q\\in\\ZZ^d\\) has a unique representation \\(q=r-s+m\\) with \\(s\\in R\\) and \\(m\\in L\\). This proves the equality of the absolute sums. The operator norm bound also follows by conjugating Young's inequality through the isometric analysis map.\n\\end{proof}","\\begin{proposition}[Composition of transfer matrices]\\label{prop:transfer_composition}\nIf \\(G\\) and \\(H\\) are convolution operators with packet matrices \\((G_{rs})\\) and \\((H_{rs})\\), then the packet matrix of \\(GH\\) is\n\\[\n(GH)_{rt}\n=\n\\sum_{s\\in R}G_{rs}H_{st}.\n\\]\nIn one dimensional transform notation, the corresponding matrix symbols multiply pointwise.\n\\end{proposition}","\\begin{proof}\nConjugation by \\(\\Acal_L\\) preserves composition:\n\\[\n\\Acal_LGH\\Acal_L^{-1}\n=\n(\\Acal_LG\\Acal_L^{-1})(\\Acal_LH\\Acal_L^{-1}).\n\\]\nReading the \\((r,t)\\) block gives the formula.\n\\end{proof}","For \\(L=N\\ZZ\\) with \\(N\\geq1\\), first take \\(|z|=1\\). The assumption \\(h\\in\\ell^1(\\ZZ)\\) makes the following symbol series absolutely convergent. Define\n\\[\nH_{ks}(z)=\\sum_{m\\in\\ZZ}h(Nm+k-s)z^{-m}.\n\\]\nFor \\(x\\in\\ell^1(\\ZZ)\\), the channel transforms converge absolutely on the unit circle and satisfy\n\\[\n\\bm Y(z)=\\bm H_N(z)\\bm X(z).\n\\]\nFor \\(x\\in\\ell^2(\\ZZ)\\), the same identity holds almost everywhere in the \\(L^2\\) Fourier sense. For finite impulse-response kernels it also holds at nonzero \\(z\\) wherever the input channel transforms are weighted absolutely summable. This is the classical polyphase matrix. It is the exact block matrix of the original convolution operator.","ZZ\\) with \\(N\\geq1\\), first take \\(|z|=1\\). The assumption \\(h\\in\\ell^1(\\ZZ)\\) makes the following symbol series absolutely convergent. Define\n\\[\nH_{ks}(z)=\\sum_{m\\in\\ZZ}h(Nm+k-s)z^{-m}.\n\\]\nFor \\(x\\in\\ell^1(\\ZZ)\\), the channel transforms converge absolutely on the unit circle and satisfy\n\\[\n\\bm Y(z)=\\bm H_N(z)\\bm X(z).\n\\]\nFor \\(x\\in\\ell^2(\\ZZ)\\), the same identity holds almost everywhere in the \\(L^2\\) Fourier sense. For finite impulse-response kernels it also holds at nonzero \\(z\\) wherever the input channel transforms are weighted absolutely summable. This is the classical polyphase matrix. It is the exact block matrix of the original convolution operator.","the following symbol series absolutely convergent. Define\n\\[\nH_{ks}(z)=\\sum_{m\\in\\ZZ}h(Nm+k-s)z^{-m}.\n\\]\nFor \\(x\\in\\ell^1(\\ZZ)\\), the channel transforms converge absolutely on the unit circle and satisfy\n\\[\n\\bm Y(z)=\\bm H_N(z)\\bm X(z).\n\\]\nFor \\(x\\in\\ell^2(\\ZZ)\\), the same identity holds almost everywhere in the \\(L^2\\) Fourier sense. For finite impulse-response kernels it also holds at nonzero \\(z\\) wherever the input channel transforms are weighted absolutely summable. This is the classical polyphase matrix. It is the exact block matrix of the original convolution operator.","\\begin{corollary}[Finite impulse response closure]\\label{cor:fir_closure}\nIf \\(h\\) has finite support, every packet kernel \\(h_{ks}\\) has finite support. Matrix multiplication of polyphase symbols agrees with composition of the original finite impulse response operators.\n\\end{corollary}","\\begin{example}[A four phase finite impulse response operator]\\label{ex:four_phase_fir}\nLet\n\\[\nh(0)=1,\n\\quad h(1)=2,\n\\quad h(2)=3,\n\\quad h(3)=4,\n\\quad h(4)=5,\n\\]\nwith all other taps zero. For four residue channels, the transfer symbol is\n\\[\n\\bm H_4(z)\n=\n\\begin{pmatrix}\n1+5z^{-1}&4z^{-1}&3z^{-1}&2z^{-1}","\\begin{example}[A four phase finite impulse response operator]\\label{ex:four_phase_fir}\nLet\n\\[\nh(0)=1,\n\\quad h(1)=2,\n\\quad h(2)=3,\n\\quad h(3)=4,\n\\quad h(4)=5,\n\\]\nwith all other taps zero. For four residue channels, the transfer symbol is\n\\[\n\\bm H_4(z)\n=\n\\begin{pmatrix}\n1+5z^{-1}&4z^{-1}&3z^{-1}&2z^{-1}\\\\\n2&1+5z^{-1}&4z^{-1}&3z^{-1}\\\\\n3&2&1+5z^{-1}&4z^{-1}\\\\\n4&3&2&1+5z^{-1}\n\\end{pmatrix}.\n\\]\nThe delay term records the tap that crosses a coarse cell boundary. No approximation is introduced by the matrix representation.\n\\end{example}","Throughout this section the banks act boundedly on the relevant polyphase \\(\\ell^2\\) spaces. A finite channel analysis bank acts in Fourier coordinates by a matrix \\(E(z)\\). A synthesis bank acts by a matrix \\(R(z)\\). The matrices may be square or rectangular. Their entries are Laurent polynomials for finite filters and bounded measurable functions for general stable filters.","\\begin{theorem}[Perfect reconstruction criterion]\\label{thm:perfect_reconstruction}\nLet \\(E(z)\\) be an \\(M\\) by \\(N\\) analysis matrix and let \\(R(z)\\) be an \\(N\\) by \\(M\\) synthesis matrix. The bank reconstructs every input with a delay of \\(d\\) coarse channel samples, equivalent to \\(Nd\\) original samples in the one dimensional \\(N\\)-phase convention, exactly if and only if\n\\[\nR(z)E(z)=z^{-d}I_N\n\\]\nfor almost every point on the unit circle. A nonzero scalar factor gives exact reconstruction up to gain and delay.\n\\end{theorem}","\\begin{proof}\nThe channel Fourier transform turns analysis followed by synthesis into multiplication by \\(R(z)E(z)\\). If the matrix identity holds, inverse transformation gives the delayed input. Conversely, if every input is reconstructed with that delay, the difference matrix annihilates every vector valued Fourier datum and must vanish almost everywhere.\n\\end{proof}","\\begin{theorem}[Quantitative imperfect reconstruction]\\label{thm:imperfect_reconstruction}\nDefine the delay compensated defect\n\\[\nD(\\e^{\\ii\\omega})\n=\n\\e^{\\ii d\\omega}R(\\e^{\\ii\\omega})E(\\e^{\\ii\\omega})-I_N.\n\\]\nIf\n\\[\n\\varepsilon_{\\mathrm{rec}}\n=\n\\esssup_{\\omega}\\|D(\\e^{\\ii\\omega})\\|_{\\mathrm{op}}<\\infty,\n\\]\nthen the delay compensated reconstruction \\(\\widetilde x\\) satisfies\n\\[\n\\|\\widetilde x-x\\|_2\n\\leq\n\\varepsilon_{\\mathrm{rec}}\\|x\\|_2.\n\\]\nThe constant is sharp as an operator norm.\n\\end{theorem}","\\begin{proof}\nIn polyphase Fourier coordinates, the error is \\(D(\\e^{\\ii\\omega})\\bm X(\\e^{\\ii\\omega})\\). Pointwise multiplication and Plancherel's theorem give the upper bound. The operator norm of a measurable matrix multiplier is its essential supremum, which proves sharpness.\n\\end{proof}","\\begin{example}[A reconstruction certificate]\\label{ex:reconstruction_certificate}\nSuppose a two channel bank has delay compensated defect\n\\[\nD(\\e^{\\ii\\omega})\n=\n\\begin{pmatrix}\n0.01\\cos\\omega&0\\\\\n0&0.02\\sin\\omega\n\\end{pmatrix}.\n\\]\nThen\n\\[\n\\esssup_\\omega\\|D(\\e^{\\ii\\omega})\\|_{\\mathrm{op}}\n=0.02.\n\\]\nEvery input therefore has relative reconstruction error at most two percent in \\(\\ell^2\\). The estimate is independent of the input length.\n\\end{example}","0&0.02\\sin\\omega\n\\end{pmatrix}.\n\\]\nThen\n\\[\n\\esssup_\\omega\\|D(\\e^{\\ii\\omega})\\|_{\\mathrm{op}}\n=0.02.\n\\]\nEvery input therefore has relative reconstruction error at most two percent in \\(\\ell^2\\). The estimate is independent of the input length.\n\\end{example}","\\begin{theorem}[Stability under filter perturbation]\\label{thm:filter_perturbation}\nAssume \\(RE=z^{-d}I_N\\). Let\n\\[\n\\widetilde E=E+\\Delta E,\n\\qquad\n\\widetilde R=R+\\Delta R.\n\\]\nSet\n\\[\n\\begin{aligned}\ne_0&=\\esssup\\|E\\|_{\\mathrm{op}},\n& r_0&=\\esssup\\|R\\|_{\\mathrm{op}},","\\begin{theorem}[Stability under filter perturbation]\\label{thm:filter_perturbation}\nAssume \\(RE=z^{-d}I_N\\). Let\n\\[\n\\widetilde E=E+\\Delta E,\n\\qquad\n\\widetilde R=R+\\Delta R.\n\\]\nSet\n\\[\n\\begin{aligned}\ne_0&=\\esssup\\|E\\|_{\\mathrm{op}},\n& r_0&=\\esssup\\|R\\|_{\\mathrm{op}},\\\\\n\\delta_E&=\\esssup\\|\\Delta E\\|_{\\mathrm{op}},\n& \\delta_R&=\\esssup\\|\\Delta R\\|_{\\mathrm{op}}.\n\\end{aligned}\n\\]\nThen the delay compensated reconstruction defect obeys\n\\[\n\\varepsilon_{\\mathrm{rec}}\n\\leq\nr_0\\delta_E+e_0\\delta_R+\\delta_R\\delta_E.\n\\]\n\\end{theorem}","\\delta_E&=\\esssup\\|\\Delta E\\|_{\\mathrm{op}},\n& \\delta_R&=\\esssup\\|\\Delta R\\|_{\\mathrm{op}}.\n\\end{aligned}\n\\]\nThen the delay compensated reconstruction defect obeys\n\\[\n\\varepsilon_{\\mathrm{rec}}\n\\leq\nr_0\\delta_E+e_0\\delta_R+\\delta_R\\delta_E.\n\\]\n\\end{theorem}","\\begin{proof}\nExpand\n\\[\n\\widetilde R\\widetilde E-RE\n=\nR\\Delta E+\\Delta R E+\\Delta R\\Delta E.\n\\]\nTake pointwise operator norms, then the essential supremum. Multiplication by the delay factor has unit modulus.\n\\end{proof}","\\begin{theorem}[Frame bounds from the polyphase symbol]\\label{thm:frame_bounds}\nAssume that\n\\[\nA I_N\n\\leq\nE(\\e^{\\ii\\omega})^*E(\\e^{\\ii\\omega})\n\\leq\nB I_N\n\\]\nfor almost every \\(\\omega\\), where \\(0<A\\leq B<\\infty\\). Then the analysis bank satisfies\n\\[\nA\\|x\\|_2^2\n\\leq\n\\|\\Acal_E x\\|_2^2\n\\leq\nB\\|x\\|_2^2.\n\\]\nThe pointwise matrix\n\\[\nR_{\\rm can}(\\e^{\\ii\\omega})\n=\n(E^*E)^{-1}E^*\n\\]\nis a bounded left inverse and gives the canonical dual synthesis bank.\n\\end{theorem}","\\begin{proof}\nApply the matrix inequality to the polyphase Fourier vector at every frequency and integrate. Plancherel's theorem gives the frame inequalities. The lower bound makes \\(E^*E\\) invertible almost everywhere, and direct multiplication gives \\(R_{\\rm can}E=I_N\\). The same spectral bounds show that the canonical dual is bounded.\n\\end{proof}","\\begin{corollary}[Coefficient noise under canonical synthesis]\\label{cor:canonical_noise}\nUnder the hypotheses of \\cref{thm:frame_bounds}, suppose the observed coefficient vector is\n\\[\nc=\\Acal_Ex+\\nu.\n\\]\nCanonical synthesis gives\n\\[\n\\widehat x=R_{\\mathrm{can}}c\n\\]\nand\n\\[\n\\|\\widehat x-x\\|_2\n\\leq\n\\frac1{\\sqrt A}\\|\\nu\\|_2.\n\\]\n\\end{corollary}","\\begin{proof}\nThe smallest singular value of \\(E\\) is at least \\(\\sqrt A\\), so the pointwise norm of its pseudoinverse is at most \\(1/\\sqrt A\\). Apply Plancherel's theorem to \\(R_{\\mathrm{can}}\\nu\\).\n\\end{proof}","\\begin{theorem}[Perturbation of frame bounds]\\label{thm:frame_perturbation}\nUnder the hypotheses of \\cref{thm:frame_bounds}, let \\(\\widetilde E=E+\\Delta E\\) and suppose\n\\[\n\\delta=\\esssup_\\omega\\|\\Delta E(\\e^{\\ii\\omega})\\|_{\\mathrm{op}}<\\sqrt A.\n\\]\nThen \\(\\widetilde E\\) is a frame analysis bank with bounds\n\\[\n(\\sqrt A-\\delta)^2\n\\quad\\text{and}\\quad\n(\\sqrt B+\\delta)^2.\n\\]\n\\end{theorem}","\\begin{proof}\nFor every vector \\(v\\),\n\\[\n\\|\\widetilde Ev\\|\n\\geq\n\\|Ev\\|-\\|\\Delta Ev\\|\n\\geq\n(\\sqrt A-\\delta)\\|v\\|.\n\\]\nThe upper bound follows from\n\\[\n\\|\\widetilde Ev\\|\n\\leq\n(\\sqrt B+\\delta)\\|v\\|.\n\\]\nSquare the inequalities and integrate in frequency.\n\\end{proof}","\\begin{remark}\\label{rem:canonical_dual_support}\nEven when \\(E\\) is a Laurent polynomial matrix, the canonical dual need not have finite impulse response. Finite support of both analysis and synthesis is an additional algebraic condition.\n\\end{remark}","\\begin{corollary}[Paraunitary bank]\\label{cor:paraunitary}\nIf \\(E(z)^*E(z)=I_N\\) on the unit circle, then analysis is an isometry and synthesis may be chosen as \\(R(z)=E(z)^*\\).\n\\end{corollary}","\\begin{proof}\nUse \\cref{thm:frame_bounds} with \\(A=B=1\\), then multiply the matrices.\n\\end{proof}","The Haar matrix is\n\\[\nE_H\n=\n\\frac1{\\sqrt2}\n\\begin{pmatrix}\n1&1\\\\\n1&-1\n\\end{pmatrix}.\n\\]\nFor even and odd samples \\(e(n)=x(2n)\\) and \\(o(n)=x(2n+1)\\), the low and high coordinates are\n\\[\na(n)=\\frac{e(n)+o(n)}{\\sqrt2},\n\\qquad\nd(n)=\\frac{e(n)-o(n)}{\\sqrt2}.\n\\]","1&-1\n\\end{pmatrix}.\n\\]\nFor even and odd samples \\(e(n)=x(2n)\\) and \\(o(n)=x(2n+1)\\), the low and high coordinates are\n\\[\na(n)=\\frac{e(n)+o(n)}{\\sqrt2},\n\\qquad\nd(n)=\\frac{e(n)-o(n)}{\\sqrt2}.\n\\]","\\begin{corollary}[Haar reconstruction and energy]\\label{cor:haar_reconstruction}\nThe Haar bank reconstructs exactly and\n\\[\n\\|x\\|_2^2=\\|a\\|_2^2+\\|d\\|_2^2.\n\\]\n\\end{corollary}","\\begin{proof}\nThe matrix \\(E_H\\) is unitary.\n\\end{proof}","Let \\(P(z)\\) and \\(Q(z)\\) be Laurent polynomials. Define\n\\[\nL_P(z)\n=\n\\begin{pmatrix}\n1&0","Let \\(P(z)\\) and \\(Q(z)\\) be Laurent polynomials. Define\n\\[\nL_P(z)\n=\n\\begin{pmatrix}\n1&0\\\\\nP(z)&1\n\\end{pmatrix},\n\\qquad\nU_Q(z)\n=\n\\begin{pmatrix}\n1&Q(z)\\\\\n0&1\n\\end{pmatrix}.\n\\]","\\begin{theorem}[Exact lifting update]\\label{thm:lifting_update}\nThe lifting matrices are invertible with\n\\[\nL_P^{-1}=L_{-P},\n\\qquad\nU_Q^{-1}=U_{-Q}.\n\\]\nIf \\(R_0E_0=I\\), then every finite product \\(V\\) of lifting matrices gives another exact bank by\n\\[\nE=VE_0,\n\\qquad\nR=R_0V^{-1}.\n\\]\nEvery lifting step has determinant one.\n\\end{theorem}","\\begin{proof}\nDirect multiplication proves the inverse formulas and the determinant statement. The reconstructed product is\n\\[\nRE\n=R_0V^{-1}VE_0\n=R_0E_0\n=I.\n\\]\n\\end{proof}","\\begin{example}[A two neighbor predict and update transform]\\label{ex:predict_update}\nSplit a sequence into even and odd samples. Define\n\\[\nd(n)\n=o(n)-\\frac{e(n)+e(n+1)}2\n\\]\nand\n\\[\ns(n)\n=e(n)+\\frac{d(n-1)+d(n)}4.\n\\]\nThe inverse is obtained in reverse order:\n\\[\ne(n)\n=s(n)-\\frac{d(n-1)+d(n)}4,\n\\]\nthen\n\\[\no(n)\n=d(n)+\\frac{e(n)+e(n+1)}2.\n\\]\nThus reconstruction is exact over every coefficient field in which two is invertible. This is a standard compactly supported biorthogonal lifting pattern \\cite{CohenDaubechiesFeauveau1992,Sweldens1996,DaubechiesSweldens1998}.\n\\end{example}","\\begin{definition}[Unitary packet tree]\\label{def:unitary_packet_tree}\nA finite unitary packet tree consists of Hilbert spaces \\(\\Hcal_v\\) at its vertices and, at every internal vertex \\(v\\), a unitary map\n\\[\nU_v:\\Hcal_v\\longrightarrow\\bigoplus_{w\\in\\operatorname{child}(v)}\\Hcal_w.\n\\]\nAn admissible pruning is a set of vertices that meets every root to terminal path exactly once.\n\\end{definition}","\\begin{theorem}[Reconstruction on a pruned packet tree]\\label{thm:pruned_packet_tree}\nEvery admissible pruning \\(\\Lcal\\) defines a unitary analysis map\n\\[\n\\Acal_{\\Lcal}:\\Hcal_{\\rm root}\n\\longrightarrow\n\\bigoplus_{v\\in\\Lcal}\\Hcal_v.\n\\]\nIts inverse is obtained by applying the adjoints of the local splits from the leaves toward the root. In particular,\n\\[\n\\|x\\|^2\n=\n\\sum_{v\\in\\Lcal}\\|x_v\\|^2.\n\\]\n\\end{theorem}","\\begin{proof}\nProceed by induction on the depth of the retained subtree. If the root is retained, the statement is immediate. If the root is split, apply the unitary map \\(U_{\\rm root}\\), then apply the induction hypothesis independently in each child subtree. A direct sum of unitary maps is unitary. Reversing the maps with adjoints gives synthesis.\n\\end{proof}","\\begin{corollary}[Coefficient perturbations in a unitary tree]\\label{cor:unitary_tree_noise}\nLet \\(c=\\Acal_{\\Lcal}x\\) be the leaf coefficients of an admissible unitary pruning, and let \\(\\widehat c=c+e\\). If \\(\\widehat x\\) is synthesized from \\(\\widehat c\\), then\n\\[\n\\|\\widehat x-x\\|=\\|e\\|.\n\\]\nThus coefficient quantization and coefficient noise are neither amplified nor attenuated by exact unitary synthesis.\n\\end{corollary}","\\begin{proof}\nThe synthesis map is unitary by \\cref{thm:pruned_packet_tree}. Hence\n\\[\n\\widehat x-x\n=\\Acal_{\\Lcal}^{-1}e\n\\]\nhas the same norm as \\(e\\).\n\\end{proof}","\\begin{proposition}[Bounds for nonunitary local splits]\\label{prop:nonunitary_tree_bounds}\nSuppose every internal node has an analysis map \\(U_v\\) and constants \\(0\\leq A_v\\leq B_v<\\infty\\) satisfying\n\\[\nA_v\\|x\\|^2\n\\leq\n\\sum_{w\\in\\operatorname{child}(v)}\\|(U_vx)_w\\|^2\n\\leq\nB_v\\|x\\|^2.\n\\]\nDefine recursively\n\\[\na(v)=\n\\begin{cases}\n1,&v\\text{ retained},","\\begin{proposition}[Bounds for nonunitary local splits]\\label{prop:nonunitary_tree_bounds}\nSuppose every internal node has an analysis map \\(U_v\\) and constants \\(0\\leq A_v\\leq B_v<\\infty\\) satisfying\n\\[\nA_v\\|x\\|^2\n\\leq\n\\sum_{w\\in\\operatorname{child}(v)}\\|(U_vx)_w\\|^2\n\\leq\nB_v\\|x\\|^2.\n\\]\nDefine recursively\n\\[\na(v)=\n\\begin{cases}\n1,&v\\text{ retained},\\\\\nA_v\\min_w a(w),&v\\text{ split},\n\\end{cases}\n\\]\nand\n\\[\nb(v)=\n\\begin{cases}\n1,&v\\text{ retained},","A_v\\min_w a(w),&v\\text{ split},\n\\end{cases}\n\\]\nand\n\\[\nb(v)=\n\\begin{cases}\n1,&v\\text{ retained},\\\\\nB_v\\max_w b(w),&v\\text{ split}.\n\\end{cases}\n\\]\nThen the leaf analysis of the pruned tree has bounds \\(a({\\rm root})\\) and \\(b({\\rm root})\\).\n\\end{proposition}","B_v\\max_w b(w),&v\\text{ split}.\n\\end{cases}\n\\]\nThen the leaf analysis of the pruned tree has bounds \\(a({\\rm root})\\) and \\(b({\\rm root})\\).\n\\end{proposition}","\\begin{proof}\nAssume the child subtree bounds have been proved. Then\n\\[\n\\sum_{w}\\sum_{\\ell\\in\\Lcal_w}\\|x_\\ell\\|^2\n\\geq\n\\min_w a(w)\n\\sum_w\\|(U_vx)_w\\|^2\n\\geq\nA_v\\min_w a(w)\\|x\\|^2.\n\\]\nThe upper bound follows in the same way with the maximum and \\(B_v\\). Induction completes the proof.\n\\end{proof}","Assign a real cost \\(c(v)\\) to retaining each node. For an admissible pruning \\(\\Lcal\\), let\n\\[\nC(\\Lcal)=\\sum_{v\\in\\Lcal}c(v).\n\\]\nThe cost may measure entropy, coefficient count, an approximation penalty, or any other additive local objective.","\\begin{theorem}[Exact additive best basis recursion]\\label{thm:best_basis}\nFor each vertex define\n\\[\nF(v)\n=\n\\min\\left\\{\n c(v),\n \\sum_{w\\in\\operatorname{child}(v)}F(w)\n\\right\\},\n\\]\nwith \\(F(v)=c(v)\\) at terminal vertices. Then \\(F(v)\\) is the minimum cost of every admissible pruning of the subtree rooted at \\(v\\). A minimizing pruning is recovered by retaining \\(v\\) when the first term is no larger than the second and splitting it otherwise.\n\\end{theorem}","\\begin{proof}\nEvery admissible pruning of the subtree has exactly two forms. It either retains \\(v\\), with cost \\(c(v)\\), or splits \\(v\\) and independently chooses an admissible pruning in every child subtree. The minimum in the second case is the sum of the child minima. This gives the recursion and the reconstruction rule. In an equality case both choices minimize, and the stated rule retains the node.\n\\end{proof}","\\begin{theorem}[Stability of the optimal cost]\\label{thm:best_basis_cost_stability}\nLet \\(c\\) and \\(\\widetilde c\\) be two node cost functions. For an admissible pruning \\(\\Lcal\\), set\n\\[\n\\varepsilon(\\Lcal)\n=\n\\sum_{v\\in\\Lcal}|c(v)-\\widetilde c(v)|\n\\]\nand let\n\\[\n\\varepsilon_*\n=\n\\max_{\\Lcal}\\varepsilon(\\Lcal).\n\\]\nIf \\(F_*\\) and \\(\\widetilde F_*\\) are the two optimal root costs, then\n\\[\n|F_*-\\widetilde F_*|\n\\leq\n\\varepsilon_*.\n\\]\n\\end{theorem}","\\begin{proof}\nFor every admissible pruning,\n\\[\n|C(\\Lcal)-\\widetilde C(\\Lcal)|\n\\leq\n\\varepsilon_*.\n\\]\nEvaluate \\(\\widetilde C\\) on a minimizer for \\(C\\) to obtain \\(\\widetilde F_*\\leq F_*+\\varepsilon_*\\). Reverse the roles to obtain \\(F_*\\leq\\widetilde F_*+\\varepsilon_*\\).\n\\end{proof}","\\begin{corollary}[Stability of the selected basis]\\label{cor:best_basis_selection_stability}\nAssume the optimal pruning \\(\\Lcal_*\\) for \\(c\\) is unique and at least one competing pruning exists. Define \\(g=\\min_{\\Lcal\\ne\\Lcal_*}(C(\\Lcal)-C(\\Lcal_*))>0\\). If\n\\[\n2\\varepsilon_*<g,\n\\]\nthen the optimal pruning for \\(\\widetilde c\\) is the same. If only one admissible pruning exists, it remains selected under every cost change without a gap condition.\n\\end{corollary}","\\begin{proof}\nThe cost of the original minimizer can increase by at most \\(\\varepsilon_*\\), while the cost of every competitor can decrease by at most \\(\\varepsilon_*\\). The original gap therefore remains positive.\n\\end{proof}","For \\(m\\geq0\\) and \\(0<\\tau<1\\), define\n\\[\n\\phi_{m,\\tau}(t)\n=\n\\frac{(t-\\tau)_+^m}{m!}.\n\\]\nThus \\(\\phi_{0,\\tau}=\\one_{[\\tau,1]}\\). On each side of \\(\\tau\\), the packet is a polynomial. Its derivative of order \\(m\\) has a unit jump at \\(\\tau\\).","\\begin{definition}[Sobolev step packet class]\\label{def:step_packet_class}\nFix \\(r\\geq1\\), \\(1\\leq p\\leq\\infty\\), and an integer \\(Q\\geq0\\). The class \\(\\Fcal_{r,p}(Q)\\) consists of functions\n\\[\nf=s+\\sum_{q=1}^{Q}a_q\\phi_{0,\\tau_q}\n\\]\nwith \\(s\\in W^{r,p}(0,1)\\) and\n\\[\n0<\\tau_1<\\cdots<\\tau_Q<1.\n\\]\nNo lower bound on the jump amplitudes is required for representation. Recovery results later impose such a bound.\n\\end{definition}","\\begin{theorem}[Higher order singular packet decomposition]\\label{thm:higher_order_packet_decomposition}\nLet\n\\[\n0=\\tau_0<\\tau_1<\\cdots<\\tau_Q<\\tau_{Q+1}=1.\n\\]\nSuppose the restriction of \\(f\\) to every interval \\((\\tau_q,\\tau_{q+1})\\) belongs to \\(W^{r,p}\\), and choose the one dimensional Sobolev representatives whose derivatives through order \\(r-1\\) have one sided traces at the endpoints. Define\n\\[\na_{q,m}\n=\n\\jump{f^{(m)}}_{\\tau_q}\n=\nf^{(m)}(\\tau_q+)-f^{(m)}(\\tau_q-)\n\\]\nand\n\\[\nj_f(t)\n=\n\\sum_{q=1}^{Q}\\sum_{m=0}^{r-1}a_{q,m}\\phi_{m,\\tau_q}(t).\n\\]\nThen\n\\[\ns_f=f-j_f\n\\]\nhas matching one sided derivatives through order \\(r-1\\) at every breakpoint and belongs to \\(W^{r,p}(0,1)\\). Its weak derivative of order \\(r\\) equals the piecewise classical derivative \\(f^{(r)}\\) almost everywhere.\n\\end{theorem}","egin{theorem}[Higher order singular packet decomposition]\\label{thm:higher_order_packet_decomposition}\nLet\n\\[\n0=\\tau_0<\\tau_1<\\cdots<\\tau_Q<\\tau_{Q+1}=1.\n\\]\nSuppose the restriction of \\(f\\) to every interval \\((\\tau_q,\\tau_{q+1})\\) belongs to \\(W^{r,p}\\), and choose the one dimensional Sobolev representatives whose derivatives through order \\(r-1\\) have one sided traces at the endpoints. Define\n\\[\na_{q,m}\n=\n\\jump{f^{(m)}}_{\\tau_q}\n=\nf^{(m)}(\\tau_q+)-f^{(m)}(\\tau_q-)\n\\]\nand\n\\[\nj_f(t)\n=\n\\sum_{q=1}^{Q}\\sum_{m=0}^{r-1}a_{q,m}\\phi_{m,\\tau_q}(t).\n\\]\nThen\n\\[\ns_f=f-j_f\n\\]\nhas matching one sided derivatives through order \\(r-1\\) at every breakpoint and belongs to \\(W^{r,p}(0,1)\\). Its weak derivative of order \\(r\\) equals the piecewise classical derivative \\(f^{(r)}\\) almost everywhere.\n\\end{theorem}","_packet_decomposition}\nLet\n\\[\n0=\\tau_0<\\tau_1<\\cdots<\\tau_Q<\\tau_{Q+1}=1.\n\\]\nSuppose the restriction of \\(f\\) to every interval \\((\\tau_q,\\tau_{q+1})\\) belongs to \\(W^{r,p}\\), and choose the one dimensional Sobolev representatives whose derivatives through order \\(r-1\\) have one sided traces at the endpoints. Define\n\\[\na_{q,m}\n=\n\\jump{f^{(m)}}_{\\tau_q}\n=\nf^{(m)}(\\tau_q+)-f^{(m)}(\\tau_q-)\n\\]\nand\n\\[\nj_f(t)\n=\n\\sum_{q=1}^{Q}\\sum_{m=0}^{r-1}a_{q,m}\\phi_{m,\\tau_q}(t).\n\\]\nThen\n\\[\ns_f=f-j_f\n\\]\nhas matching one sided derivatives through order \\(r-1\\) at every breakpoint and belongs to \\(W^{r,p}(0,1)\\). Its weak derivative of order \\(r\\) equals the piecewise classical derivative \\(f^{(r)}\\) almost everywhere.\n\\end{theorem}","Suppose the restriction of \\(f\\) to every interval \\((\\tau_q,\\tau_{q+1})\\) belongs to \\(W^{r,p}\\), and choose the one dimensional Sobolev representatives whose derivatives through order \\(r-1\\) have one sided traces at the endpoints. Define\n\\[\na_{q,m}\n=\n\\jump{f^{(m)}}_{\\tau_q}\n=\nf^{(m)}(\\tau_q+)-f^{(m)}(\\tau_q-)\n\\]\nand\n\\[\nj_f(t)\n=\n\\sum_{q=1}^{Q}\\sum_{m=0}^{r-1}a_{q,m}\\phi_{m,\\tau_q}(t).\n\\]\nThen\n\\[\ns_f=f-j_f\n\\]\nhas matching one sided derivatives through order \\(r-1\\) at every breakpoint and belongs to \\(W^{r,p}(0,1)\\). Its weak derivative of order \\(r\\) equals the piecewise classical derivative \\(f^{(r)}\\) almost everywhere.\n\\end{theorem}","_{q+1})\\) belongs to \\(W^{r,p}\\), and choose the one dimensional Sobolev representatives whose derivatives through order \\(r-1\\) have one sided traces at the endpoints. Define\n\\[\na_{q,m}\n=\n\\jump{f^{(m)}}_{\\tau_q}\n=\nf^{(m)}(\\tau_q+)-f^{(m)}(\\tau_q-)\n\\]\nand\n\\[\nj_f(t)\n=\n\\sum_{q=1}^{Q}\\sum_{m=0}^{r-1}a_{q,m}\\phi_{m,\\tau_q}(t).\n\\]\nThen\n\\[\ns_f=f-j_f\n\\]\nhas matching one sided derivatives through order \\(r-1\\) at every breakpoint and belongs to \\(W^{r,p}(0,1)\\). Its weak derivative of order \\(r\\) equals the piecewise classical derivative \\(f^{(r)}\\) almost everywhere.\n\\end{theorem}","e the one dimensional Sobolev representatives whose derivatives through order \\(r-1\\) have one sided traces at the endpoints. Define\n\\[\na_{q,m}\n=\n\\jump{f^{(m)}}_{\\tau_q}\n=\nf^{(m)}(\\tau_q+)-f^{(m)}(\\tau_q-)\n\\]\nand\n\\[\nj_f(t)\n=\n\\sum_{q=1}^{Q}\\sum_{m=0}^{r-1}a_{q,m}\\phi_{m,\\tau_q}(t).\n\\]\nThen\n\\[\ns_f=f-j_f\n\\]\nhas matching one sided derivatives through order \\(r-1\\) at every breakpoint and belongs to \\(W^{r,p}(0,1)\\). Its weak derivative of order \\(r\\) equals the piecewise classical derivative \\(f^{(r)}\\) almost everywhere.\n\\end{theorem}","hose derivatives through order \\(r-1\\) have one sided traces at the endpoints. Define\n\\[\na_{q,m}\n=\n\\jump{f^{(m)}}_{\\tau_q}\n=\nf^{(m)}(\\tau_q+)-f^{(m)}(\\tau_q-)\n\\]\nand\n\\[\nj_f(t)\n=\n\\sum_{q=1}^{Q}\\sum_{m=0}^{r-1}a_{q,m}\\phi_{m,\\tau_q}(t).\n\\]\nThen\n\\[\ns_f=f-j_f\n\\]\nhas matching one sided derivatives through order \\(r-1\\) at every breakpoint and belongs to \\(W^{r,p}(0,1)\\). Its weak derivative of order \\(r\\) equals the piecewise classical derivative \\(f^{(r)}\\) almost everywhere.\n\\end{theorem}","sided traces at the endpoints. Define\n\\[\na_{q,m}\n=\n\\jump{f^{(m)}}_{\\tau_q}\n=\nf^{(m)}(\\tau_q+)-f^{(m)}(\\tau_q-)\n\\]\nand\n\\[\nj_f(t)\n=\n\\sum_{q=1}^{Q}\\sum_{m=0}^{r-1}a_{q,m}\\phi_{m,\\tau_q}(t).\n\\]\nThen\n\\[\ns_f=f-j_f\n\\]\nhas matching one sided derivatives through order \\(r-1\\) at every breakpoint and belongs to \\(W^{r,p}(0,1)\\). Its weak derivative of order \\(r\\) equals the piecewise classical derivative \\(f^{(r)}\\) almost everywhere.\n\\end{theorem}","\\begin{proof}\nFor \\(0\\leq \\ell\\leq m\\), the classical derivative of \\(\\phi_{m,\\tau}\\) on the two open sides is\n\\[\n\\phi_{m,\\tau}^{(\\ell)}(t)\n=\n\\frac{(t-\\tau)_+^{m-\\ell}}{(m-\\ell)!}.\n\\]\nThis derivative has zero jump when \\(\\ell<m\\) and a unit jump when \\(\\ell=m\\). Derivatives of order larger than \\(m\\) vanish on both open sides. It follows that subtracting the term \\(a_{q,m}\\phi_{m,\\tau_q}\\) removes exactly the jump in the derivative of order \\(m\\) without changing the other classical trace jumps. Hence \\(s_f\\) has matching traces through order \\(r-1\\).","ssical derivative of \\(\\phi_{m,\\tau}\\) on the two open sides is\n\\[\n\\phi_{m,\\tau}^{(\\ell)}(t)\n=\n\\frac{(t-\\tau)_+^{m-\\ell}}{(m-\\ell)!}.\n\\]\nThis derivative has zero jump when \\(\\ell<m\\) and a unit jump when \\(\\ell=m\\). Derivatives of order larger than \\(m\\) vanish on both open sides. It follows that subtracting the term \\(a_{q,m}\\phi_{m,\\tau_q}\\) removes exactly the jump in the derivative of order \\(m\\) without changing the other classical trace jumps. Hence \\(s_f\\) has matching traces through order \\(r-1\\).","s\n\\[\n\\phi_{m,\\tau}^{(\\ell)}(t)\n=\n\\frac{(t-\\tau)_+^{m-\\ell}}{(m-\\ell)!}.\n\\]\nThis derivative has zero jump when \\(\\ell<m\\) and a unit jump when \\(\\ell=m\\). Derivatives of order larger than \\(m\\) vanish on both open sides. It follows that subtracting the term \\(a_{q,m}\\phi_{m,\\tau_q}\\) removes exactly the jump in the derivative of order \\(m\\) without changing the other classical trace jumps. Hence \\(s_f\\) has matching traces through order \\(r-1\\).","\\tau)_+^{m-\\ell}}{(m-\\ell)!}.\n\\]\nThis derivative has zero jump when \\(\\ell<m\\) and a unit jump when \\(\\ell=m\\). Derivatives of order larger than \\(m\\) vanish on both open sides. It follows that subtracting the term \\(a_{q,m}\\phi_{m,\\tau_q}\\) removes exactly the jump in the derivative of order \\(m\\) without changing the other classical trace jumps. Hence \\(s_f\\) has matching traces through order \\(r-1\\).","On each partition interval, \\(j_f\\) is a polynomial of degree at most \\(r-1\\), so its weak derivative of order \\(r\\) is zero there. A piecewise \\(W^{r,p}\\) function whose traces through order \\(r-1\\) match across every breakpoint belongs to \\(W^{r,p}(0,1)\\). Repeated integration by parts on the partition intervals shows that all boundary terms cancel. Thus the weak derivative of order \\(r\\) is the piecewise weak derivative of \\(f\\).\n\\end{proof}","f\\) is a polynomial of degree at most \\(r-1\\), so its weak derivative of order \\(r\\) is zero there. A piecewise \\(W^{r,p}\\) function whose traces through order \\(r-1\\) match across every breakpoint belongs to \\(W^{r,p}(0,1)\\). Repeated integration by parts on the partition intervals shows that all boundary terms cancel. Thus the weak derivative of order \\(r\\) is the piecewise weak derivative of \\(f\\).\n\\end{proof}","\\begin{corollary}[Uniqueness of the trace packet]\\label{cor:trace_packet_uniqueness}\nFix the breakpoints \\(\\tau_q\\). If\n\\[\nf=s+\\sum_{q=1}^{Q}\\sum_{m=0}^{r-1}b_{q,m}\\phi_{m,\\tau_q}\n\\]\nwith \\(s\\in W^{r,p}(0,1)\\), then\n\\[\nb_{q,m}=\\jump{f^{(m)}}_{\\tau_q}.\n\\]\nThe finite truncated power packet is therefore the unique packet supported at the declared breakpoints that removes all trace jumps through order \\(r-1\\).\n\\end{corollary}","gin{corollary}[Uniqueness of the trace packet]\\label{cor:trace_packet_uniqueness}\nFix the breakpoints \\(\\tau_q\\). If\n\\[\nf=s+\\sum_{q=1}^{Q}\\sum_{m=0}^{r-1}b_{q,m}\\phi_{m,\\tau_q}\n\\]\nwith \\(s\\in W^{r,p}(0,1)\\), then\n\\[\nb_{q,m}=\\jump{f^{(m)}}_{\\tau_q}.\n\\]\nThe finite truncated power packet is therefore the unique packet supported at the declared breakpoints that removes all trace jumps through order \\(r-1\\).\n\\end{corollary}","\\begin{proof}\nEvery representative in \\(W^{r,p}(0,1)\\) has derivatives through order \\(r-1\\) that admit absolutely continuous representatives on compact subintervals, with matching one sided traces at each interior point. Taking the jump of the derivative of order \\(m\\) therefore removes the contribution of \\(s\\). The packet \\(\\phi_{m,\\tau_q}\\) has unit jump in its derivative of order \\(m\\), while all other declared packets have zero jump at that derivative order and breakpoint. Hence the remaining jump is exactly \\(b_{q,m}\\).\n\\end{proof}","ous representatives on compact subintervals, with matching one sided traces at each interior point. Taking the jump of the derivative of order \\(m\\) therefore removes the contribution of \\(s\\). The packet \\(\\phi_{m,\\tau_q}\\) has unit jump in its derivative of order \\(m\\), while all other declared packets have zero jump at that derivative order and breakpoint. Hence the remaining jump is exactly \\(b_{q,m}\\).\n\\end{proof}","Let \\(X\\) be a normed linear space and let \\(Q:X\\to X\\) be bounded and linear. The next identity is elementary, but it is the central separation principle.","\\begin{theorem}[Exact retained packet identity]\\label{thm:retained_packet_identity}\nLet \\(f=s+j\\) in \\(X\\). Define\n\\[\n\\widetilde f=Qs+j.\n\\]\nThen\n\\[\n\\boxed{\nf-\\widetilde f=(I-Q)s.\n}\n\\]\nConsequently,\n\\[\n\\|f-\\widetilde f\\|=\\|s-Qs\\|.\n\\]\n\\end{theorem}","\\begin{proof}\nSubtract:\n\\[\nf-\\widetilde f\n=s+j-Qs-j\n=(I-Q)s.\n\\]\n\\end{proof}","\\begin{remark}\\label{rem:retained_packet_boundary}\nThe theorem assumes that the packet \\(j\\) is known. It does not supply a detector. It also does not say that every representation of \\(f\\) as \\(s+j\\) is useful. The point is that once a meaningful packet has been identified, its approximation error can be removed from the smooth approximation problem exactly.\n\\end{remark}","\\begin{corollary}[Sobolev rate after packet separation]\\label{cor:sobolev_rate}\nSuppose \\(Q_m\\) satisfies\n\\[\n\\|s-Q_ms\\|_{L^p}\n\\leq\nC2^{-mr}\\|s\\|_{W^{r,p}}\n\\]\nfor every \\(s\\in W^{r,p}(0,1)\\). If \\(f=s+j\\) and \\(j\\) is retained, then\n\\[\n\\|f-(Q_ms+j)\\|_{L^p}\n\\leq\nC2^{-mr}\\|s\\|_{W^{r,p}}.\n\\]\nThe same conclusion applies to the higher order decomposition in \\cref{thm:higher_order_packet_decomposition}.\n\\end{corollary}","\\begin{theorem}[Reconstruction with an estimated packet]\\label{thm:estimated_packet_reconstruction}\nLet \\(Q_m:X\\to X\\) be linear, let \\(f=s+j\\), let \\(\\widehat j\\) be an estimated packet, and define\n\\[\n\\widehat f_m\n=\nQ_m(f-\\widehat j)+\\widehat j.\n\\]\nThen\n\\[\n\\boxed{\nf-\\widehat f_m\n=\n(I-Q_m)s+(I-Q_m)(j-\\widehat j).\n}\n\\]\nIf \\(Q_m\\) is bounded, then\n\\[\n\\|f-\\widehat f_m\\|\n\\leq\n\\|s-Q_ms\\|\n+\n(1+\\|Q_m\\|)\\|j-\\widehat j\\|.\n\\]\n\\end{theorem}","\\begin{proof}\nSince \\(f-\\widehat j=s+(j-\\widehat j)\\),\n\\[\n\\widehat f_m\n=\nQ_ms+Q_m(j-\\widehat j)+\\widehat j.\n\\]\nSubtract this expression from \\(s+j\\). The norm bound follows from\n\\[\n\\|(I-Q_m)u\\|\n\\leq\n(1+\\|Q_m\\|)\\|u\\|.\n\\]\n\\end{proof}","\\begin{theorem}[Exact scalar region formula]\\label{thm:scalar_region_formula}\nLet \\(E,F\\) be measurable subsets of a measure space, let \\(a,b\\in\\CC\\), and let \\(1\\leq p<\\infty\\). Then\n\\[\n\\begin{aligned}\n\\|a\\one_E-b\\one_F\\|_p^p\n={}&\n|a-b|^p\\mu(E\\cap F)\n+|a|^p\\mu(E\\setminus F)","\\begin{theorem}[Exact scalar region formula]\\label{thm:scalar_region_formula}\nLet \\(E,F\\) be measurable subsets of a measure space, let \\(a,b\\in\\CC\\), and let \\(1\\leq p<\\infty\\). Then\n\\[\n\\begin{aligned}\n\\|a\\one_E-b\\one_F\\|_p^p\n={}&\n|a-b|^p\\mu(E\\cap F)\n+|a|^p\\mu(E\\setminus F)\\\\\n&+|b|^p\\mu(F\\setminus E).\n\\end{aligned}\n\\]\nFor equal amplitudes,\n\\[\n\\|a\\one_E-a\\one_F\\|_p\n=\n|a|\\mu(E\\mathbin\\triangle F)^{1/p}.\n\\]\n\\end{theorem}","&+|b|^p\\mu(F\\setminus E).\n\\end{aligned}\n\\]\nFor equal amplitudes,\n\\[\n\\|a\\one_E-a\\one_F\\|_p\n=\n|a|\\mu(E\\mathbin\\triangle F)^{1/p}.\n\\]\n\\end{theorem}","\\begin{proof}\nThe sets \\(E\\cap F\\), \\(E\\setminus F\\), and \\(F\\setminus E\\) are disjoint. The difference takes the values \\(a-b\\), \\(a\\), and \\(-b\\) on these three sets and is zero elsewhere. Integrating its \\(p\\)th power gives the formula.\n\\end{proof}","\\begin{corollary}[Location error for one step]\\label{cor:step_location_error}\nLet \\(0\\leq\\tau,\\widehat\\tau\\leq1\\) and \\(H_\\tau=\\one_{[\\tau,1]}\\). For \\(1\\leq p<\\infty\\),\n\\[\n\\|aH_\\tau-aH_{\\widehat\\tau}\\|_p\n=\n|a||\\tau-\\widehat\\tau|^{1/p}.\n\\]\nIn the supremum norm, the error equals \\(|a|\\) whenever \\(\\tau\\neq\\widehat\\tau\\).\n\\end{corollary}","\\begin{proof}\nThe symmetric difference of the two intervals has length \\(|\\tau-\\widehat\\tau|\\). The supremum statement follows because the two steps differ by \\(a\\) on a nonempty interval.\n\\end{proof}","\\begin{theorem}[Location stability for higher order packets]\\label{thm:higher_order_location}\nLet \\(m\\geq1\\), \\(1\\leq p<\\infty\\), and \\(0\\leq\\tau,\\widehat\\tau\\leq1\\). Then\n\\[\n\\|\\phi_{m,\\tau}-\\phi_{m,\\widehat\\tau}\\|_{L^p(0,1)}\n\\leq\n\\frac{|\\tau-\\widehat\\tau|}\n{(m-1)!((m-1)p+1)^{1/p}}.\n\\]\nFor \\(p=\\infty\\),\n\\[\n\\|\\phi_{m,\\tau}-\\phi_{m,\\widehat\\tau}\\|_\\infty\n\\leq\n\\frac{|\\tau-\\widehat\\tau|}{(m-1)!}.\n\\]\nThus location error is Lipschitz for every derivative jump packet of positive order, while a value jump has only the finite \\(L^p\\) modulus in \\cref{cor:step_location_error}.\n\\end{theorem}","y\\), and \\(0\\leq\\tau,\\widehat\\tau\\leq1\\). Then\n\\[\n\\|\\phi_{m,\\tau}-\\phi_{m,\\widehat\\tau}\\|_{L^p(0,1)}\n\\leq\n\\frac{|\\tau-\\widehat\\tau|}\n{(m-1)!((m-1)p+1)^{1/p}}.\n\\]\nFor \\(p=\\infty\\),\n\\[\n\\|\\phi_{m,\\tau}-\\phi_{m,\\widehat\\tau}\\|_\\infty\n\\leq\n\\frac{|\\tau-\\widehat\\tau|}{(m-1)!}.\n\\]\nThus location error is Lipschitz for every derivative jump packet of positive order, while a value jump has only the finite \\(L^p\\) modulus in \\cref{cor:step_location_error}.\n\\end{theorem}","\\begin{proof}\nAssume \\(\\tau<\\widehat\\tau\\). Differentiation with respect to the location parameter gives\n\\[\n\\frac{\\partial}{\\partial u}\\phi_{m,u}(t)\n=-\\phi_{m-1,u}(t)\n\\]\nfor almost every \\(t\\). Hence\n\\[\n\\phi_{m,\\tau}-\\phi_{m,\\widehat\\tau}\n=\n\\int_\\tau^{\\widehat\\tau}\\phi_{m-1,u}\\dd u.\n\\]\nMinkowski's inequality gives\n\\[\n\\|\\phi_{m,\\tau}-\\phi_{m,\\widehat\\tau}\\|_p\n\\leq\n|\\tau-\\widehat\\tau|\n\\sup_u\\|\\phi_{m-1,u}\\|_p.\n\\]\nThe supremum is bounded by the norm of \\(t^{m-1}/(m-1)!\\) on \\([0,1]\\), which is the stated constant. The supremum norm argument is identical.\n\\end{proof}","ion with respect to the location parameter gives\n\\[\n\\frac{\\partial}{\\partial u}\\phi_{m,u}(t)\n=-\\phi_{m-1,u}(t)\n\\]\nfor almost every \\(t\\). Hence\n\\[\n\\phi_{m,\\tau}-\\phi_{m,\\widehat\\tau}\n=\n\\int_\\tau^{\\widehat\\tau}\\phi_{m-1,u}\\dd u.\n\\]\nMinkowski's inequality gives\n\\[\n\\|\\phi_{m,\\tau}-\\phi_{m,\\widehat\\tau}\\|_p\n\\leq\n|\\tau-\\widehat\\tau|\n\\sup_u\\|\\phi_{m-1,u}\\|_p.\n\\]\nThe supremum is bounded by the norm of \\(t^{m-1}/(m-1)!\\) on \\([0,1]\\), which is the stated constant. The supremum norm argument is identical.\n\\end{proof}","e location parameter gives\n\\[\n\\frac{\\partial}{\\partial u}\\phi_{m,u}(t)\n=-\\phi_{m-1,u}(t)\n\\]\nfor almost every \\(t\\). Hence\n\\[\n\\phi_{m,\\tau}-\\phi_{m,\\widehat\\tau}\n=\n\\int_\\tau^{\\widehat\\tau}\\phi_{m-1,u}\\dd u.\n\\]\nMinkowski's inequality gives\n\\[\n\\|\\phi_{m,\\tau}-\\phi_{m,\\widehat\\tau}\\|_p\n\\leq\n|\\tau-\\widehat\\tau|\n\\sup_u\\|\\phi_{m-1,u}\\|_p.\n\\]\nThe supremum is bounded by the norm of \\(t^{m-1}/(m-1)!\\) on \\([0,1]\\), which is the stated constant. The supremum norm argument is identical.\n\\end{proof}","Let \\((E_i)_{i=0}^{I}\\) and \\((F_j)_{j=0}^{J}\\) be measurable partitions of the same domain, up to null sets. Let \\(a_i\\) and \\(b_j\\) be the corresponding values. The index zero may be used for a background region.","\\begin{theorem}[Overlap matrix formula]\\label{thm:overlap_matrix}\nDefine\n\\[\nM_{ij}=\\mu(E_i\\cap F_j).\n\\]\nFor \\(1\\leq p<\\infty\\),\n\\[\n\\left\\|\n\\sum_{i=0}^{I}a_i\\one_{E_i}\n-\n\\sum_{j=0}^{J}b_j\\one_{F_j}\n\\right\\|_p^p\n=\n\\sum_{i=0}^{I}\\sum_{j=0}^{J}|a_i-b_j|^pM_{ij}.\n\\]\nThe row sums of \\(M\\) are \\(\\mu(E_i)\\), and the column sums are \\(\\mu(F_j)\\).\n\\end{theorem}","\\begin{proof}\nThe intersections \\(E_i\\cap F_j\\) form a measurable partition. On such an intersection, the two piecewise constant functions have values \\(a_i\\) and \\(b_j\\). Integrate the constant error over every intersection. The marginal identities follow by summing the partition pieces in a fixed row or column.\n\\end{proof}","\\begin{corollary}[Background and packet error]\\label{cor:background_packet_error}\nSuppose the true and estimated packets use the same amplitude list \\((a_i)_{i=0}^{I}\\), but possibly different regions. Then\n\\[\n\\|j-\\widehat j\\|_p^p\n=\n\\sum_{i,k}|a_i-a_k|^p\\mu(E_i\\cap F_k).\n\\]\nOnly off diagonal overlaps contribute.\n\\end{corollary}","\\begin{proof}\nApply \\cref{thm:overlap_matrix} with \\(b_k=a_k\\). The diagonal terms vanish.\n\\end{proof}","For a closed set \\(\\Gamma\\), write\n\\[\nN_\\delta(\\Gamma)\n=\n\\{x:\\dist(x,\\Gamma)<\\delta\\}.\n\\]","\\begin{definition}[Linear tube bound]\\label{def:linear_tube_bound}\nA boundary \\(\\Gamma\\) has a linear tube bound if there are constants \\(C_\\Gamma\\) and \\(\\delta_0>0\\) such that\n\\[\n\\mu(N_\\delta(\\Gamma))\n\\leq\nC_\\Gamma\\delta\n\\]\nfor \\(0<\\delta<\\delta_0\\).\n\\end{definition}","\\begin{theorem}[Geometric packet bound]\\label{thm:geometric_packet_bound}\nLet \\(E\\) be a region with boundary \\(\\Gamma\\), and suppose\n\\[\nE\\mathbin\\triangle F\n\\subseteq\nN_\\delta(\\Gamma).\n\\]\nIf \\(\\Gamma\\) has a linear tube bound and \\(0<\\delta<\\delta_0\\) lies in its stated range, then for \\(1\\leq p<\\infty\\),\n\\[\n\\|a\\one_E-a\\one_F\\|_p\n\\leq\n|a|(C_\\Gamma\\delta)^{1/p}.\n\\]\nIf amplitudes \\(a\\) and \\(b\\) differ, then\n\\[\n\\|a\\one_E-b\\one_F\\|_p\n\\leq\n|a-b|\\mu(E\\cap F)^{1/p}\n+\n(|a|+|b|)(C_\\Gamma\\delta)^{1/p}.\n\\]\n\\end{theorem}","m}[Geometric packet bound]\\label{thm:geometric_packet_bound}\nLet \\(E\\) be a region with boundary \\(\\Gamma\\), and suppose\n\\[\nE\\mathbin\\triangle F\n\\subseteq\nN_\\delta(\\Gamma).\n\\]\nIf \\(\\Gamma\\) has a linear tube bound and \\(0<\\delta<\\delta_0\\) lies in its stated range, then for \\(1\\leq p<\\infty\\),\n\\[\n\\|a\\one_E-a\\one_F\\|_p\n\\leq\n|a|(C_\\Gamma\\delta)^{1/p}.\n\\]\nIf amplitudes \\(a\\) and \\(b\\) differ, then\n\\[\n\\|a\\one_E-b\\one_F\\|_p\n\\leq\n|a-b|\\mu(E\\cap F)^{1/p}\n+\n(|a|+|b|)(C_\\Gamma\\delta)^{1/p}.\n\\]\n\\end{theorem}","\\begin{proof}\nThe equal amplitude case follows from \\cref{thm:scalar_region_formula} and the tube inclusion. For unequal amplitudes, split the error into its value mismatch on \\(E\\cap F\\) and the two disagreement pieces. Apply the triangle inequality and the same tube estimate.\n\\end{proof}","\\begin{example}[A perturbed disk]\\label{ex:perturbed_disk}\nLet \\(E\\) and \\(F\\) be concentric disks of radii \\(r\\) and \\(r+\\delta\\) in \\(\\RR^2\\), with \\(\\delta>0\\). Then\n\\[\n\\mu(E\\mathbin\\triangle F)\n=\n\\pi\\bigl((r+\\delta)^2-r^2\\bigr)\n=\n2\\pi r\\delta+\\pi\\delta^2.\n\\]\nFor equal amplitude \\(a\\), the exact \\(L^2\\) packet error is\n\\[\n|a|\\sqrt{2\\pi r\\delta+\\pi\\delta^2}.\n\\]\nThis gives the expected square root dependence on boundary displacement.\n\\end{example}","ooth bound. The gap is a condition on the observation model, rather than an automatic consequence of packet coordinates.\n\\end{keyidea}\n\\begin{figure}[htbp]\\centering\n\\includegraphics[width=.93\\linewidth]{monograph_figures/jump_contrast_gap.pdf}\n\\caption{A sufficient threshold interval from exact rational contrast bounds. With the displayed model parameters, smooth and noise contrasts are at most $1/10$ and jump contrasts are at least $9/10$. Every threshold strictly between those values separates the cells under the sampling assumptions stated below. The figure does not assert an optimal threshold for other observation models.}\\label{monograph:fig:contrast}\n\\end{figure}","he observation model, rather than an automatic consequence of packet coordinates.\n\\end{keyidea}\n\\begin{figure}[htbp]\\centering\n\\includegraphics[width=.93\\linewidth]{monograph_figures/jump_contrast_gap.pdf}\n\\caption{A sufficient threshold interval from exact rational contrast bounds. With the displayed model parameters, smooth and noise contrasts are at most $1/10$ and jump contrasts are at least $9/10$. Every threshold strictly between those values separates the cells under the sampling assumptions stated below. The figure does not assert an optimal threshold for other observation models.}\\label{monograph:fig:contrast}\n\\end{figure}","Let\n\\[\nt_k=kh,\n\\qquad\nk=0,\\ldots,n,\n\\qquad\nh=\\frac1n.\n\\]\nAssume\n\\[\nf(t)\n=\ns(t)+\\sum_{q=1}^{Q}a_qH_{\\tau_q}(t),\n\\qquad\nH_\\tau(t)=\\one_{[\\tau,1]}(t),\n\\]\nwhere \\(s\\) is Lipschitz with constant \\(M\\), and \\(Q\\geq1\\) is finite with nonzero jump amplitudes. The zero-jump case requires no minimum-amplitude condition. Suppose no jump lies at a sample point and no grid cell contains more than one jump. Observe\n\\[\ny_k=f(t_k)+\\eta_k.\n\\]\nDefine the adjacent contrast\n\\[\nD_k=y_{k+1}-y_k.\n\\]","t_k=kh,\n\\qquad\nk=0,\\ldots,n,\n\\qquad\nh=\\frac1n.\n\\]\nAssume\n\\[\nf(t)\n=\ns(t)+\\sum_{q=1}^{Q}a_qH_{\\tau_q}(t),\n\\qquad\nH_\\tau(t)=\\one_{[\\tau,1]}(t),\n\\]\nwhere \\(s\\) is Lipschitz with constant \\(M\\), and \\(Q\\geq1\\) is finite with nonzero jump amplitudes. The zero-jump case requires no minimum-amplitude condition. Suppose no jump lies at a sample point and no grid cell contains more than one jump. Observe\n\\[\ny_k=f(t_k)+\\eta_k.\n\\]\nDefine the adjacent contrast\n\\[\nD_k=y_{k+1}-y_k.\n\\]","me\n\\[\nf(t)\n=\ns(t)+\\sum_{q=1}^{Q}a_qH_{\\tau_q}(t),\n\\qquad\nH_\\tau(t)=\\one_{[\\tau,1]}(t),\n\\]\nwhere \\(s\\) is Lipschitz with constant \\(M\\), and \\(Q\\geq1\\) is finite with nonzero jump amplitudes. The zero-jump case requires no minimum-amplitude condition. Suppose no jump lies at a sample point and no grid cell contains more than one jump. Observe\n\\[\ny_k=f(t_k)+\\eta_k.\n\\]\nDefine the adjacent contrast\n\\[\nD_k=y_{k+1}-y_k.\n\\]","\\begin{lemma}[Contrast gap]\\label{lem:contrast_gap}\nAssume \\(|\\eta_k|\\leq\\varepsilon\\) and set\n\\[\nb_h=Mh+2\\varepsilon.\n\\]\nIf the cell \\((t_k,t_{k+1})\\) contains no jump, then\n\\[\n|D_k|\\leq b_h.\n\\]\nIf it contains one jump of amplitude \\(a_q\\), then\n\\[\n|D_k-a_q|\\leq b_h\n\\]\nand therefore\n\\[\n|D_k|\\geq |a_q|-b_h.\n\\]\n\\end{lemma}","\\begin{proof}\nIn a cell without a jump,\n\\[\nD_k=s(t_{k+1})-s(t_k)+\\eta_{k+1}-\\eta_k.\n\\]\nThe Lipschitz and noise bounds give \\(|D_k|\\leq Mh+2\\varepsilon\\). In a cell with one jump, the same formula has the additional term \\(a_q\\), which proves the second estimate.\n\\end{proof}","\\begin{theorem}[Deterministic jump cell recovery]\\label{thm:deterministic_jump_recovery}\nLet\n\\[\na_{\\min}=\\min_q|a_q|\n\\]\nand assume\n\\[\na_{\\min}>2b_h.\n\\]\nChoose a threshold \\(\\theta\\) satisfying\n\\[\nb_h<\\theta<a_{\\min}-b_h.\n\\]\nThen\n\\[\n\\widehat\\Kcal\n=\n\\{k:|D_k|>\\theta\\}\n\\]\nis exactly the set of grid cells containing jumps. If \\(\\widehat\\tau_q\\) is the midpoint of the detected cell and \\(\\widehat a_q=D_k\\), then after matching by order,\n\\[\n|\\widehat\\tau_q-\\tau_q|\n\\leq\n\\frac h2\n\\]\nand\n\\[\n|\\widehat a_q-a_q|\n\\leq\nb_h.\n\\]\n\\end{theorem}","l recovery]\\label{thm:deterministic_jump_recovery}\nLet\n\\[\na_{\\min}=\\min_q|a_q|\n\\]\nand assume\n\\[\na_{\\min}>2b_h.\n\\]\nChoose a threshold \\(\\theta\\) satisfying\n\\[\nb_h<\\theta<a_{\\min}-b_h.\n\\]\nThen\n\\[\n\\widehat\\Kcal\n=\n\\{k:|D_k|>\\theta\\}\n\\]\nis exactly the set of grid cells containing jumps. If \\(\\widehat\\tau_q\\) is the midpoint of the detected cell and \\(\\widehat a_q=D_k\\), then after matching by order,\n\\[\n|\\widehat\\tau_q-\\tau_q|\n\\leq\n\\frac h2\n\\]\nand\n\\[\n|\\widehat a_q-a_q|\n\\leq\nb_h.\n\\]\n\\end{theorem}","y}\nLet\n\\[\na_{\\min}=\\min_q|a_q|\n\\]\nand assume\n\\[\na_{\\min}>2b_h.\n\\]\nChoose a threshold \\(\\theta\\) satisfying\n\\[\nb_h<\\theta<a_{\\min}-b_h.\n\\]\nThen\n\\[\n\\widehat\\Kcal\n=\n\\{k:|D_k|>\\theta\\}\n\\]\nis exactly the set of grid cells containing jumps. If \\(\\widehat\\tau_q\\) is the midpoint of the detected cell and \\(\\widehat a_q=D_k\\), then after matching by order,\n\\[\n|\\widehat\\tau_q-\\tau_q|\n\\leq\n\\frac h2\n\\]\nand\n\\[\n|\\widehat a_q-a_q|\n\\leq\nb_h.\n\\]\n\\end{theorem}","\\begin{proof}\nBy \\cref{lem:contrast_gap}, every cell without a jump has contrast at most \\(b_h<\\theta\\). Every cell with a jump has contrast at least \\(a_{\\min}-b_h>\\theta\\). Hence the detected set is exact. The midpoint of a cell is within \\(h/2\\) of every point in that cell. The amplitude estimate is the second inequality in \\cref{lem:contrast_gap}.\n\\end{proof}","\\begin{corollary}[Jump count and sign recovery]\\label{cor:jump_count_sign}\nUnder the hypotheses of \\cref{thm:deterministic_jump_recovery}, the number of detected cells equals \\(Q\\), and\n\\[\n\\sgn(\\widehat a_q)=\\sgn(a_q)\n\\]\nfor every real valued jump.\n\\end{corollary}","\\begin{proof}\nExact cell recovery gives the jump count. Since\n\\[\n|\\widehat a_q-a_q|\\leq b_h<|a_q|,\n\\]\nthe amplitude estimate cannot cross zero.\n\\end{proof}","\\begin{corollary}[Packet reconstruction after detection]\\label{cor:reconstruction_after_detection}\nLet\n\\[\nj=\\sum_{q=1}^{Q}a_qH_{\\tau_q},\n\\qquad\n\\widehat j=\\sum_{q=1}^{Q}\\widehat a_qH_{\\widehat\\tau_q}.\n\\]\nUnder the hypotheses of \\cref{thm:deterministic_jump_recovery}, for \\(1\\leq p<\\infty\\),\n\\[\n\\|j-\\widehat j\\|_p\n\\leq\nQb_h\n+\n\\left(\n\\frac h2\\sum_{q=1}^{Q}|a_q|^p\n\\right)^{1/p}.\n\\]\nIf \\(Q_m\\) is bounded and linear and\n\\[\n\\widehat f_m=Q_m(f-\\widehat j)+\\widehat j,\n\\]\nthen\n\\[\n\\begin{aligned}\n\\|f-\\widehat f_m\\|_p\n\\leq{}&\n\\|s-Q_ms\\|_p","cket reconstruction after detection]\\label{cor:reconstruction_after_detection}\nLet\n\\[\nj=\\sum_{q=1}^{Q}a_qH_{\\tau_q},\n\\qquad\n\\widehat j=\\sum_{q=1}^{Q}\\widehat a_qH_{\\widehat\\tau_q}.\n\\]\nUnder the hypotheses of \\cref{thm:deterministic_jump_recovery}, for \\(1\\leq p<\\infty\\),\n\\[\n\\|j-\\widehat j\\|_p\n\\leq\nQb_h\n+\n\\left(\n\\frac h2\\sum_{q=1}^{Q}|a_q|^p\n\\right)^{1/p}.\n\\]\nIf \\(Q_m\\) is bounded and linear and\n\\[\n\\widehat f_m=Q_m(f-\\widehat j)+\\widehat j,\n\\]\nthen\n\\[\n\\begin{aligned}\n\\|f-\\widehat f_m\\|_p\n\\leq{}&\n\\|s-Q_ms\\|_p","ruction_after_detection}\nLet\n\\[\nj=\\sum_{q=1}^{Q}a_qH_{\\tau_q},\n\\qquad\n\\widehat j=\\sum_{q=1}^{Q}\\widehat a_qH_{\\widehat\\tau_q}.\n\\]\nUnder the hypotheses of \\cref{thm:deterministic_jump_recovery}, for \\(1\\leq p<\\infty\\),\n\\[\n\\|j-\\widehat j\\|_p\n\\leq\nQb_h\n+\n\\left(\n\\frac h2\\sum_{q=1}^{Q}|a_q|^p\n\\right)^{1/p}.\n\\]\nIf \\(Q_m\\) is bounded and linear and\n\\[\n\\widehat f_m=Q_m(f-\\widehat j)+\\widehat j,\n\\]\nthen\n\\[\n\\begin{aligned}\n\\|f-\\widehat f_m\\|_p\n\\leq{}&\n\\|s-Q_ms\\|_p\\\\\n&+(1+\\|Q_m\\|)\n\\left[\nQb_h+\n\\left(\n\\frac h2\\sum_q|a_q|^p\n\\right)^{1/p}\n\\right].\n\\end{aligned}\n\\]\n\\end{corollary}","\\begin{proof}\nWrite\n\\[\nj-\\widehat j\n=\n\\sum_q(a_q-\\widehat a_q)H_{\\widehat\\tau_q}\n+\n\\sum_q a_q(H_{\\tau_q}-H_{\\widehat\\tau_q}).\n\\]\nThe first sum has norm at most \\(\\sum_q|a_q-\\widehat a_q|\\leq Qb_h\\). The location error interval for the \\(q\\)th jump is contained in its detected grid cell and has length at most \\(h/2\\). Distinct detected cells are disjoint, so the location errors have disjoint supports. Their \\(p\\)th powers therefore add exactly, which gives the displayed location term. The reconstruction estimate follows from \\cref{thm:estimated_packet_reconstruction}.\n\\end{proof}","}\nWrite\n\\[\nj-\\widehat j\n=\n\\sum_q(a_q-\\widehat a_q)H_{\\widehat\\tau_q}\n+\n\\sum_q a_q(H_{\\tau_q}-H_{\\widehat\\tau_q}).\n\\]\nThe first sum has norm at most \\(\\sum_q|a_q-\\widehat a_q|\\leq Qb_h\\). The location error interval for the \\(q\\)th jump is contained in its detected grid cell and has length at most \\(h/2\\). Distinct detected cells are disjoint, so the location errors have disjoint supports. Their \\(p\\)th powers therefore add exactly, which gives the displayed location term. The reconstruction estimate follows from \\cref{thm:estimated_packet_reconstruction}.\n\\end{proof}","\\begin{remark}[Why finite \\(L^p\\) matters]\\label{rem:finite_p_recovery}\nA shifted step does not converge in the supremum norm until the location is exact. The finite \\(L^p\\) error does converge because the disagreement is confined to an interval whose length tends to zero.\n\\end{remark}","\\begin{corollary}[Deterministic consistency]\\label{cor:deterministic_consistency}\nAssume \\(h\\to0\\) along grids satisfying the preceding sampling assumptions, \\(\\varepsilon_h\\to0\\), the number of jumps is fixed, and the nonzero jump amplitudes and positive minimum separation remain fixed. Then the detected packet converges to the true packet in every finite \\(L^p\\) norm. If \\(Q_m\\) has a vanishing smooth approximation error and uniformly bounded norm, the complete reconstruction also converges in \\(L^p\\).\n\\end{corollary}","g grids satisfying the preceding sampling assumptions, \\(\\varepsilon_h\\to0\\), the number of jumps is fixed, and the nonzero jump amplitudes and positive minimum separation remain fixed. Then the detected packet converges to the true packet in every finite \\(L^p\\) norm. If \\(Q_m\\) has a vanishing smooth approximation error and uniformly bounded norm, the complete reconstruction also converges in \\(L^p\\).\n\\end{corollary}","\\begin{proof}\nThe gap condition holds for sufficiently small \\(h\\). Then \\(b_h\\to0\\), and the bound in \\cref{cor:reconstruction_after_detection} tends to zero.\n\\end{proof}","\\begin{example}[A numerical threshold certificate]\\label{ex:jump_threshold_certificate}\nTake \\(h=1/256\\), \\(M=0.5\\), and \\(\\varepsilon=0.002\\). Then\n\\[\nb_h\n=\\frac{0.5}{256}+0.004\n<0.00596.\n\\]\nIf every jump has magnitude at least \\(0.08\\), then \\(0.08>2b_h\\). Any threshold between \\(0.00596\\) and \\(0.07404\\) recovers the exact jump cells. The location error is at most\n\\[\n\\frac1{512}\n\\]\nand every amplitude error is less than \\(0.00596\\).\n\\end{example}","hreshold certificate]\\label{ex:jump_threshold_certificate}\nTake \\(h=1/256\\), \\(M=0.5\\), and \\(\\varepsilon=0.002\\). Then\n\\[\nb_h\n=\\frac{0.5}{256}+0.004\n<0.00596.\n\\]\nIf every jump has magnitude at least \\(0.08\\), then \\(0.08>2b_h\\). Any threshold between \\(0.00596\\) and \\(0.07404\\) recovers the exact jump cells. The location error is at most\n\\[\n\\frac1{512}\n\\]\nand every amplitude error is less than \\(0.00596\\).\n\\end{example}","Assume the noise variables are independent, centered, and \\(\\sigma\\) sub Gaussian:\n\\[\n\\mathbb E\\e^{\\lambda\\eta_k}\n\\leq\n\\e^{\\sigma^2\\lambda^2/2}.\n\\]","\\begin{theorem}[Probability recovery bound]\\label{thm:probability_recovery}\nFix \\(0<\\delta<1\\) and define\n\\[\nt_{n,\\delta}\n=\n2\\sigma\n\\sqrt{\\log\\left(\\frac{2n}{\\delta}\\right)}\n\\]\nand\n\\[\nb_{h,\\delta}=Mh+t_{n,\\delta}.\n\\]\nWith probability at least \\(1-\\delta\\), every contrast satisfies the conclusions of \\cref{lem:contrast_gap} with \\(b_h\\) replaced by \\(b_{h,\\delta}\\).","If\n\\[\na_{\\min}>2b_{h,\\delta},\n\\]\nthen thresholding between \\(b_{h,\\delta}\\) and \\(a_{\\min}-b_{h,\\delta}\\) recovers every jump cell with probability at least \\(1-\\delta\\). The same location and amplitude bounds hold with \\(b_{h,\\delta}\\).\n\\end{theorem}","\\begin{proof}\nIf \\(\\sigma=0\\), every centered zero-sub-Gaussian noise variable vanishes almost surely, and the noise-free deterministic conclusion applies. For \\(\\sigma>0\\), the difference \\(\\eta_{k+1}-\\eta_k\\) is sub Gaussian with variance proxy \\(2\\sigma^2\\), so\n\\[\n\\mathbb P\\bigl(|\\eta_{k+1}-\\eta_k|>t\\bigr)\n\\leq\n2\\exp\\left(-\\frac{t^2}{4\\sigma^2}\\right).\n\\]\nA union bound over the \\(n\\) cells gives\n\\[\n\\mathbb P\\left(\n\\max_k|\\eta_{k+1}-\\eta_k|>t_{n,\\delta}\n\\right)\n\\leq\n\\delta.\n\\]\nOn the complementary event, repeat the deterministic proof.\n\\end{proof}","in{proof}\nIf \\(\\sigma=0\\), every centered zero-sub-Gaussian noise variable vanishes almost surely, and the noise-free deterministic conclusion applies. For \\(\\sigma>0\\), the difference \\(\\eta_{k+1}-\\eta_k\\) is sub Gaussian with variance proxy \\(2\\sigma^2\\), so\n\\[\n\\mathbb P\\bigl(|\\eta_{k+1}-\\eta_k|>t\\bigr)\n\\leq\n2\\exp\\left(-\\frac{t^2}{4\\sigma^2}\\right).\n\\]\nA union bound over the \\(n\\) cells gives\n\\[\n\\mathbb P\\left(\n\\max_k|\\eta_{k+1}-\\eta_k|>t_{n,\\delta}\n\\right)\n\\leq\n\\delta.\n\\]\nOn the complementary event, repeat the deterministic proof.\n\\end{proof}","\\begin{corollary}[Repeated measurements]\\label{cor:repeated_measurements}\nSuppose \\(R\\) observations are averaged at every sample location, with all noise variables independent across sites and repetitions and each centered and \\(\\sigma\\) sub Gaussian. Then \\cref{thm:probability_recovery} holds with \\(\\sigma\\) replaced by \\(\\sigma/\\sqrt R\\). In particular, a sufficient recovery condition is\n\\[\na_{\\min}\n>\n2Mh+\n\\frac{4\\sigma}{\\sqrt R}\n\\sqrt{\\log\\left(\\frac{2n}{\\delta}\\right)}.\n\\]\n\\end{corollary}","\\begin{proof}\nThe average of \\(R\\) independent \\(\\sigma\\) sub Gaussian variables is \\(\\sigma/\\sqrt R\\) sub Gaussian. Substitute this parameter in \\cref{thm:probability_recovery}.\n\\end{proof}","\\begin{remark}\\label{rem:noise_scaling}\nThe displayed uniform noise certificate scales as \\(\\sigma\\sqrt{\\log n}\\). At fixed positive \\(\\sigma\\), it does not imply consistency of this detector. Decreasing the noise scale or averaging repeated measurements gives sufficient regimes; other averaging schemes require their own analysis.\n\\end{remark}","\\begin{corollary}[Probability consistency]\\label{cor:probability_consistency}\nLet \\(\\sigma_n\\sqrt{\\log n}\\to0\\) and \\(h=1/n\\). For every fixed \\(0<\\delta<1\\), along grids satisfying the preceding sampling assumptions, all jump cells are recovered with probability at least \\(1-\\delta\\) for sufficiently large \\(n\\). For the fixed finite jump model above, along grids satisfying the preceding sampling assumptions, the packet error converges to zero in probability in every finite \\(L^p\\) norm.\n\\end{corollary}","\\(\\sigma_n\\sqrt{\\log n}\\to0\\) and \\(h=1/n\\). For every fixed \\(0<\\delta<1\\), along grids satisfying the preceding sampling assumptions, all jump cells are recovered with probability at least \\(1-\\delta\\) for sufficiently large \\(n\\). For the fixed finite jump model above, along grids satisfying the preceding sampling assumptions, the packet error converges to zero in probability in every finite \\(L^p\\) norm.\n\\end{corollary}","\\begin{proof}\nFor each fixed \\(\\delta>0\\), \\(b_{h,\\delta}\\to0\\). Apply \\cref{thm:probability_recovery,cor:reconstruction_after_detection}. For any error tolerance \\(\\epsilon>0\\), the high-probability bound is eventually smaller than \\(\\epsilon\\), so the limit superior of the probability that the packet error exceeds \\(\\epsilon\\) is at most \\(\\delta\\). Since \\(\\delta\\) is arbitrary, convergence is in probability.\n\\end{proof}","\\begin{proposition}[Vector jump recovery]\\label{prop:vector_jump_recovery}\nLet the values lie in a normed space \\(V\\). Assume\n\\[\n\\|s(t)-s(u)\\|_V\\leq M|t-u|,\n\\qquad\n\\|\\eta_k\\|_V\\leq\\varepsilon,\n\\]\nand let the jump vectors satisfy \\(\\|a_q\\|_V\\geq a_{\\min}\\). Thresholding \\(\\|D_k\\|_V\\) gives the same recovery theorem under the condition\n\\[\na_{\\min}>2(Mh+2\\varepsilon).\n\\]\n\\end{proposition}","The adjacent difference detector is local in the sample domain. Wavelet coefficients and higher differences give related multiscale packets. In the Haar subsections the signal is defined on \\(\\RR\\). Here the step convention is \\(H_\\tau(t)=\\one_{[\\tau,\\infty)}(t)\\), whose restriction to \\((0,1)\\) agrees with the preceding interval step. A chosen extension beyond the observation interval must use this half-line step convention, and the smooth extension must satisfy the stated Lipschitz hypothesis on every displayed wavelet support.","Let\n\\[\nI_{j,k}\n=\n[k2^{-j},(k+1)2^{-j})\n\\]\nand let \\(\\psi_{j,k}\\) be the normalized Haar wavelet supported on \\(I_{j,k}\\), positive on the left half and negative on the right half.","\\begin{theorem}[Smooth Haar coefficient bound]\\label{thm:smooth_haar_bound}\nIf \\(s\\) is Lipschitz with constant \\(M\\), then\n\\[\n|\\langle s,\\psi_{j,k}\\rangle|\n\\leq\n\\frac M4\\,2^{-3j/2}.\n\\]\n\\end{theorem}","\\begin{proof}\nLet \\(L=2^{-j}\\) and let \\(a\\) be the left endpoint of the support. Pair every point \\(t\\in[a,a+L/2]\\) with \\(t+L/2\\). Then\n\\[\n\\langle s,\\psi_{j,k}\\rangle\n=\nL^{-1/2}\n\\int_a^{a+L/2}\n\\overline{s(t)-s(t+L/2)}\\,\\dd t.\n\\]\nThe integrand has magnitude at most \\(ML/2\\), and the integration interval has length \\(L/2\\). The result is \\(ML^{3/2}/4\\).\n\\end{proof}","\\begin{theorem}[Exact Haar coefficient of a step]\\label{thm:haar_step_coefficient}\nLet \\(\\tau\\in I_{j,k}\\), let \\(L=2^{-j}\\), and write\n\\[\nu=\\frac{\\tau-kL}{L}\\in[0,1].\n\\]\nThen\n\\[\n\\left|\n\\left\\langle\n aH_\\tau,\\psi_{j,k}\n\\right\\rangle\n\\right|\n=\n|a|L^{1/2}\\min\\{u,1-u\\}.\n\\]\nEvery Haar coefficient whose support does not cross \\(\\tau\\) is zero.\n\\end{theorem}","\\begin{proof}\nAway from the jump, the step is constant on the support and the Haar wavelet has zero mean. When the jump lies in the support, integrate the positive and negative halves directly. If \\(u\\leq1/2\\), the magnitude is \\(|a|uL^{1/2}\\). If \\(u\\geq1/2\\), it is \\(|a|(1-u)L^{1/2}\\).\n\\end{proof}","\\begin{corollary}[Scale separation]\\label{cor:haar_scale_separation}\nSuppose \\(f=s+aH_\\tau+\\eta\\), where \\(s\\) is \\(M\\) Lipschitz and \\(\\|\\eta\\|_\\infty\\leq\\varepsilon\\). If the jump lies at relative distance at least \\(\\rho\\) from both ends of \\(I_{j,k}\\), then\n\\[\n|\\langle f,\\psi_{j,k}\\rangle|\n\\geq\n\\left(\n|a|\\rho\n-\\varepsilon\n-\\frac M4L\n\\right)L^{1/2}.\n\\]\nA wavelet support without a jump satisfies\n\\[\n|\\langle f,\\psi_{j,k}\\rangle|\n\\leq\n\\left(\n\\varepsilon+\\frac M4L\n\\right)L^{1/2}.\n\\]\nThus a threshold gap exists whenever\n\\[\n|a|\\rho\n>\n2\\varepsilon+\\frac M2L.\n\\]\n\\end{corollary}","gin{corollary}[Scale separation]\\label{cor:haar_scale_separation}\nSuppose \\(f=s+aH_\\tau+\\eta\\), where \\(s\\) is \\(M\\) Lipschitz and \\(\\|\\eta\\|_\\infty\\leq\\varepsilon\\). If the jump lies at relative distance at least \\(\\rho\\) from both ends of \\(I_{j,k}\\), then\n\\[\n|\\langle f,\\psi_{j,k}\\rangle|\n\\geq\n\\left(\n|a|\\rho\n-\\varepsilon\n-\\frac M4L\n\\right)L^{1/2}.\n\\]\nA wavelet support without a jump satisfies\n\\[\n|\\langle f,\\psi_{j,k}\\rangle|\n\\leq\n\\left(\n\\varepsilon+\\frac M4L\n\\right)L^{1/2}.\n\\]\nThus a threshold gap exists whenever\n\\[\n|a|\\rho\n>\n2\\varepsilon+\\frac M2L.\n\\]\n\\end{corollary}","itz and \\(\\|\\eta\\|_\\infty\\leq\\varepsilon\\). If the jump lies at relative distance at least \\(\\rho\\) from both ends of \\(I_{j,k}\\), then\n\\[\n|\\langle f,\\psi_{j,k}\\rangle|\n\\geq\n\\left(\n|a|\\rho\n-\\varepsilon\n-\\frac M4L\n\\right)L^{1/2}.\n\\]\nA wavelet support without a jump satisfies\n\\[\n|\\langle f,\\psi_{j,k}\\rangle|\n\\leq\n\\left(\n\\varepsilon+\\frac M4L\n\\right)L^{1/2}.\n\\]\nThus a threshold gap exists whenever\n\\[\n|a|\\rho\n>\n2\\varepsilon+\\frac M2L.\n\\]\n\\end{corollary}","\\begin{proof}\nCombine \\cref{thm:smooth_haar_bound,thm:haar_step_coefficient}. The noise coefficient is at most\n\\[\n\\|\\eta\\|_\\infty\\|\\psi_{j,k}\\|_1\n=\n\\varepsilon L^{1/2}.\n\\]\nThe final inequality makes the lower bound on a jump support larger than the upper bound on a smooth support.\n\\end{proof}","\\begin{figure}\n\\centering\n\\includegraphics[width=0.68\\textwidth]{10_haar_scale_separation.pdf}\n\\caption{The smooth contribution decays like \\(L^{3/2}\\), while a separated jump contribution decays like \\(L^{1/2}\\).}\n\\label{fig:haar_scale_separation}\n\\end{figure}","\\begin{theorem}[Half cell shift guarantee]\\label{thm:half_shift_guarantee}\nFix a cell length \\(L>0\\). Consider the grid with boundaries \\(kL\\) and the grid with boundaries \\((k+1/2)L\\). Every point \\(\\tau\\in\\RR\\) lies in a cell of at least one grid at distance at least \\(L/4\\) from both cell boundaries.\n\\end{theorem}","\\begin{proof}\nReduce \\(\\tau\\) modulo \\(L\\) and write its representative as \\(x\\in[0,L)\\). If\n\\[\nL/4\\leq x\\leq 3L/4,\n\\]\nthen the distance from \\(x\\) to both boundaries of its cell in the original grid is at least \\(L/4\\). If \\(x<L/4\\) or \\(x>3L/4\\), then \\(x\\) lies in the cell of the shifted grid bounded by \\(-L/2\\) and \\(L/2\\), or in its translate, at distance at least \\(L/4\\) from both shifted boundaries. One of the two grids therefore has the required property.\n\\end{proof}","\\begin{corollary}[Alignment free two grid Haar detection]\\label{cor:two_grid_haar_detection}\nAssume the support length \\(L\\) is smaller than the minimum jump separation. Analyze the signal with Haar wavelets on the original grid and on the half shifted grid. Every jump appears in a support with \\(\\rho\\geq1/4\\) in one of the two systems. Therefore a threshold gap exists independently of alignment whenever\n\\[\n\\frac{|a|}{4}\n>\n2\\varepsilon+\\frac M2L.\n\\]\n\\end{corollary}","ollary}[Alignment free two grid Haar detection]\\label{cor:two_grid_haar_detection}\nAssume the support length \\(L\\) is smaller than the minimum jump separation. Analyze the signal with Haar wavelets on the original grid and on the half shifted grid. Every jump appears in a support with \\(\\rho\\geq1/4\\) in one of the two systems. Therefore a threshold gap exists independently of alignment whenever\n\\[\n\\frac{|a|}{4}\n>\n2\\varepsilon+\\frac M2L.\n\\]\n\\end{corollary}","\\begin{proof}\nApply \\cref{thm:half_shift_guarantee} and then \\cref{cor:haar_scale_separation} with \\(\\rho=1/4\\). The separation assumption prevents a selected support from containing two jumps.\n\\end{proof}","Let\n\\[\n(\\Delta x)(n)=x(n+1)-x(n).\n\\]\nThen\n\\[\n\\Delta^m x(n)\n=\n\\sum_{r=0}^{m}\n(-1)^{m-r}\\binom mr x(n+r).\n\\]","\\begin{theorem}[Polynomial annihilation and boundary support]\\label{thm:polynomial_annihilation}\nLet \\(x\\) be a sequence that agrees with a polynomial of degree at most \\(d\\) on each interval of a finite partition of \\(\\ZZ\\). Then\n\\[\n\\Delta^{d+1}x(n)=0\n\\]\nwhenever the complete stencil\n\\[\n\\{n,n+1,\\ldots,n+d+1\\}\n\\]\nlies inside one polynomial piece. Hence \\(\\Delta^{d+1}x\\) is supported in a band of at most \\(d+1\\) addresses before each breakpoint.\n\\end{theorem}","eorem}[Polynomial annihilation and boundary support]\\label{thm:polynomial_annihilation}\nLet \\(x\\) be a sequence that agrees with a polynomial of degree at most \\(d\\) on each interval of a finite partition of \\(\\ZZ\\). Then\n\\[\n\\Delta^{d+1}x(n)=0\n\\]\nwhenever the complete stencil\n\\[\n\\{n,n+1,\\ldots,n+d+1\\}\n\\]\nlies inside one polynomial piece. Hence \\(\\Delta^{d+1}x\\) is supported in a band of at most \\(d+1\\) addresses before each breakpoint.\n\\end{theorem}","\\begin{proof}\nThe \\((d+1)\\)st forward difference of a polynomial of degree at most \\(d\\) is zero. The only stencils for which this argument does not apply are those crossing a breakpoint.\n\\end{proof}","\\begin{proposition}[Noise amplification by differences]\\label{prop:difference_noise}\nIf \\(\\|\\eta\\|_\\infty\\leq\\varepsilon\\), then\n\\[\n\\|\\Delta^m\\eta\\|_\\infty\n\\leq\n2^m\\varepsilon.\n\\]\nTherefore no false response above \\(2^{d+1}\\varepsilon\\) can occur on a stencil lying inside one polynomial piece.\n\\end{proposition}","\\begin{proof}\nThe absolute values of the binomial coefficients in \\(\\Delta^m\\) sum to \\(2^m\\).\n\\end{proof}","\\begin{corollary}[Robust breakpoint bands]\\label{cor:robust_breakpoint_bands}\nAssume every true breakpoint produces at least one response of magnitude \\(\\rho\\) in its \\((d+1)\\)st difference band. If\n\\[\n\\rho>2^{d+2}\\varepsilon,\n\\]\nthen thresholding at any level between \\(2^{d+1}\\varepsilon\\) and \\(\\rho-2^{d+1}\\varepsilon\\) detects a nonempty cluster near every breakpoint and creates no cluster away from the breakpoint bands.\n\\end{corollary}","\\begin{proof}\nUse \\cref{thm:polynomial_annihilation,prop:difference_noise}. Noise cannot cross the threshold away from a breakpoint. At a response of magnitude \\(\\rho\\), the perturbed magnitude remains at least \\(\\rho-2^{d+1}\\varepsilon\\), which is above the threshold.\n\\end{proof}","\\begin{standardfact}[Classical square-wave Gibbs behavior]\\label{sf:gibbs_behavior}\nLet \\(q(x)=\\tfrac12\\operatorname{sign}(\\sin x)\\), whose jump at zero has size one. For \\(n\\geq1\\), its partial sum through the first \\(n\\) positive odd frequencies is\n\\[\nQ_n(x)=\\frac{2}{\\pi}\\sum_{k=0}^{n-1}\\frac{\\sin((2k+1)x)}{2k+1}.\n\\]\nAt \\(x_n=\\pi/(2n)\\),\n\\[\nx_n\\longrightarrow0,\\qquad Q_n(x_n)\\longrightarrow\\frac{\\operatorname{Si}(\\pi)}{\\pi}\n=\\frac12+G,\\qquad G\\approx0.089489872236,\n\\]\nwhere \\(\\operatorname{Si}(z)=\\int_0^z\\sin(t)/t\\,dt\\) and \\(G=\\operatorname{Si}(\\pi)/\\pi-1/2>0\\).\nThus a nonzero overshoot persists along points approaching the jump. Scaling by a real amplitude \\(a\\) gives an overshoot or undershoot of magnitude \\(|a|G\\).","Let \\(q(x)=\\tfrac12\\operatorname{sign}(\\sin x)\\), whose jump at zero has size one. For \\(n\\geq1\\), its partial sum through the first \\(n\\) positive odd frequencies is\n\\[\nQ_n(x)=\\frac{2}{\\pi}\\sum_{k=0}^{n-1}\\frac{\\sin((2k+1)x)}{2k+1}.\n\\]\nAt \\(x_n=\\pi/(2n)\\),\n\\[\nx_n\\longrightarrow0,\\qquad Q_n(x_n)\\longrightarrow\\frac{\\operatorname{Si}(\\pi)}{\\pi}\n=\\frac12+G,\\qquad G\\approx0.089489872236,\n\\]\nwhere \\(\\operatorname{Si}(z)=\\int_0^z\\sin(t)/t\\,dt\\) and \\(G=\\operatorname{Si}(\\pi)/\\pi-1/2>0\\).\nThus a nonzero overshoot persists along points approaching the jump. Scaling by a real amplitude \\(a\\) gives an overshoot or undershoot of magnitude \\(|a|G\\).","whose jump at zero has size one. For \\(n\\geq1\\), its partial sum through the first \\(n\\) positive odd frequencies is\n\\[\nQ_n(x)=\\frac{2}{\\pi}\\sum_{k=0}^{n-1}\\frac{\\sin((2k+1)x)}{2k+1}.\n\\]\nAt \\(x_n=\\pi/(2n)\\),\n\\[\nx_n\\longrightarrow0,\\qquad Q_n(x_n)\\longrightarrow\\frac{\\operatorname{Si}(\\pi)}{\\pi}\n=\\frac12+G,\\qquad G\\approx0.089489872236,\n\\]\nwhere \\(\\operatorname{Si}(z)=\\int_0^z\\sin(t)/t\\,dt\\) and \\(G=\\operatorname{Si}(\\pi)/\\pi-1/2>0\\).\nThus a nonzero overshoot persists along points approaching the jump. Scaling by a real amplitude \\(a\\) gives an overshoot or undershoot of magnitude \\(|a|G\\).","pi}\\sum_{k=0}^{n-1}\\frac{\\sin((2k+1)x)}{2k+1}.\n\\]\nAt \\(x_n=\\pi/(2n)\\),\n\\[\nx_n\\longrightarrow0,\\qquad Q_n(x_n)\\longrightarrow\\frac{\\operatorname{Si}(\\pi)}{\\pi}\n=\\frac12+G,\\qquad G\\approx0.089489872236,\n\\]\nwhere \\(\\operatorname{Si}(z)=\\int_0^z\\sin(t)/t\\,dt\\) and \\(G=\\operatorname{Si}(\\pi)/\\pi-1/2>0\\).\nThus a nonzero overshoot persists along points approaching the jump. Scaling by a real amplitude \\(a\\) gives an overshoot or undershoot of magnitude \\(|a|G\\).","\\(x_n=\\pi/(2n)\\),\n\\[\nx_n\\longrightarrow0,\\qquad Q_n(x_n)\\longrightarrow\\frac{\\operatorname{Si}(\\pi)}{\\pi}\n=\\frac12+G,\\qquad G\\approx0.089489872236,\n\\]\nwhere \\(\\operatorname{Si}(z)=\\int_0^z\\sin(t)/t\\,dt\\) and \\(G=\\operatorname{Si}(\\pi)/\\pi-1/2>0\\).\nThus a nonzero overshoot persists along points approaching the jump. Scaling by a real amplitude \\(a\\) gives an overshoot or undershoot of magnitude \\(|a|G\\).","For the limit, put \\(u_k=(k+\\tfrac12)/n\\). The identity\n\\[\nQ_n(x_n)=\\frac{1}{\\pi n}\\sum_{k=0}^{n-1}\\frac{\\sin(\\pi u_k)}{u_k}\n\\longrightarrow\\frac1\\pi\\int_0^1\\frac{\\sin(\\pi u)}u\\,du\n=\\frac{\\operatorname{Si}(\\pi)}{\\pi}\n\\]\nis a midpoint Riemann sum, since the integrand has continuous value \\(\\pi\\) at zero. This is the normalized classical example in \\cite[Section~6.16(i), equations~6.16.1--6.16.4]{DLMFGibbs}.\n\\end{standardfact}","\\begin{theorem}[Uniform convergence with exact jump packets]\\label{thm:uniform_packet_convergence}\nLet \\(f=s+j\\) on a compact domain, where \\(j\\) is any bounded singular packet. Suppose \\(Q_ms\\to s\\) uniformly. Then\n\\[\n\\widetilde f_m=Q_ms+j\n\\]\nconverges uniformly to \\(f\\), even when \\(f\\) is discontinuous:\n\\[\n\\|f-\\widetilde f_m\\|_\\infty\n=\n\\|s-Q_ms\\|_\\infty\n\\longrightarrow0.\n\\]\nThere is no error proportional to the jump size because the jump is shared exactly by \\(f\\) and \\(\\widetilde f_m\\).\n\\end{theorem}","hm:uniform_packet_convergence}\nLet \\(f=s+j\\) on a compact domain, where \\(j\\) is any bounded singular packet. Suppose \\(Q_ms\\to s\\) uniformly. Then\n\\[\n\\widetilde f_m=Q_ms+j\n\\]\nconverges uniformly to \\(f\\), even when \\(f\\) is discontinuous:\n\\[\n\\|f-\\widetilde f_m\\|_\\infty\n=\n\\|s-Q_ms\\|_\\infty\n\\longrightarrow0.\n\\]\nThere is no error proportional to the jump size because the jump is shared exactly by \\(f\\) and \\(\\widetilde f_m\\).\n\\end{theorem}","_packet_convergence}\nLet \\(f=s+j\\) on a compact domain, where \\(j\\) is any bounded singular packet. Suppose \\(Q_ms\\to s\\) uniformly. Then\n\\[\n\\widetilde f_m=Q_ms+j\n\\]\nconverges uniformly to \\(f\\), even when \\(f\\) is discontinuous:\n\\[\n\\|f-\\widetilde f_m\\|_\\infty\n=\n\\|s-Q_ms\\|_\\infty\n\\longrightarrow0.\n\\]\nThere is no error proportional to the jump size because the jump is shared exactly by \\(f\\) and \\(\\widetilde f_m\\).\n\\end{theorem}","\\Needspace{20\\baselineskip}\n\\begin{theorem}[Estimated packet Gibbs bound]\\label{thm:estimated_packet_gibbs}\nLet \\(Q_m\\) be linear and uniformly bounded on \\(L^p\\), with\n\\[\n\\sup_m\\|Q_m\\|\\leq C_Q.\n\\]\nFor \\(f=s+j\\) and an estimated packet \\(\\widehat j_m\\), define\n\\[\n\\widehat f_m\n=Q_m(f-\\widehat j_m)+\\widehat j_m.\n\\]\nThen\n\\[\n\\|f-\\widehat f_m\\|_p\n\\leq\n\\|s-Q_ms\\|_p\n+(1+C_Q)\\|j-\\widehat j_m\\|_p.\n\\]\nConsequently, smooth approximation and packet recovery may be improved independently. No claim of Gibbs removal is justified unless both terms are controlled in the norm under consideration.\n\\end{theorem}","\\begin{corollary}[Boundary band error]\\label{cor:boundary_band_error}\nLet \\(\\Gamma\\) be the singular boundary and assume\n\\[\n\\mu(N_\\delta(\\Gamma))\\leq C_\\Gamma\\delta\n\\]\nfor small \\(\\delta\\). For finite \\(p\\),\n\\[\n\\|f-\\widetilde f_m\\|_{L^p(N_\\delta(\\Gamma))}\n\\leq\n(C_\\Gamma\\delta)^{1/p}\n\\|s-Q_ms\\|_\\infty.\n\\]\n\\end{corollary}","\\begin{proof}\nThe error equals \\(s-Q_ms\\). Bound it by its supremum and multiply by the measure of the boundary tube.\n\\end{proof}","\\begin{example}[Square wave Fourier tail]\\label{ex:square_wave_tail}\nLet\n\\[\nj(t)=\\one_{[-a,a]}(t)\n\\]\non the circle and let \\(s\\) be smooth. If \\(Q_m\\) is a Fourier partial sum, then \\(Q_m(s+j)\\) has the ordinary oscillatory boundary layer. In contrast,\n\\[\nQ_ms+j\n\\]\nhas error exactly \\(s-Q_ms\\). The visual comparison in \\cref{fig:gibbs_comparison} illustrates this change in approximation target. It is not a new Fourier summability theorem.\n\\end{example}","\\begin{theorem}[Residue law for Cauchy products]\\label{thm:cauchy_residue_law}\nLet\n\\[\nF(z)=\\sum_{n\\geq0}f_nz^n,\n\\qquad\nG(z)=\\sum_{n\\geq0}g_nz^n\n\\]\nbe formal power series. The same statement holds for power series that converge on a common disk. Decompose both series into residue packets modulo \\(N\\):\n\\[\nF=\\sum_{r=0}^{N-1}F_r,\n\\qquad\nG=\\sum_{s=0}^{N-1}G_s.\n\\]\nIf \\(H=FG\\), then its residue packet \\(H_t\\) is\n\\[\n\\boxed{\nH_t\n=\n\\sum_{r+s\\equiv t\\!\\!\\pmod N}F_rG_s.\n}\n\\]\n\\end{theorem}","rem}[Residue law for Cauchy products]\\label{thm:cauchy_residue_law}\nLet\n\\[\nF(z)=\\sum_{n\\geq0}f_nz^n,\n\\qquad\nG(z)=\\sum_{n\\geq0}g_nz^n\n\\]\nbe formal power series. The same statement holds for power series that converge on a common disk. Decompose both series into residue packets modulo \\(N\\):\n\\[\nF=\\sum_{r=0}^{N-1}F_r,\n\\qquad\nG=\\sum_{s=0}^{N-1}G_s.\n\\]\nIf \\(H=FG\\), then its residue packet \\(H_t\\) is\n\\[\n\\boxed{\nH_t\n=\n\\sum_{r+s\\equiv t\\!\\!\\pmod N}F_rG_s.\n}\n\\]\n\\end{theorem}","\\begin{proof}\nEvery monomial in \\(F_rG_s\\) has degree congruent to \\(r+s\\) modulo \\(N\\). Conversely, every coefficient of degree congruent to \\(t\\) is formed from pairs whose residues sum to \\(t\\). Grouping the Cauchy product by this residue gives the formula.\n\\end{proof}","\\begin{theorem}[Shifted pointwise monomial law]\\label{thm:shifted_monomial_law}\nLet\n\\[\ny(n)=\\prod_{j=1}^{m}x(n-a_j)\n\\]\nand write\n\\[\na_j=Nc_j+b_j,\n\\qquad\n0\\leq b_j<N.\n\\]\nFor output residue \\(r\\), define\n\\[\nr_j\\equiv r-b_j\\pmod N,\n\\qquad\n0\\leq r_j<N,\n\\]\nand let\n\\[\n\\kappa_{r,b_j}\n=\n\\begin{cases}\n1,&r<b_j,","\\begin{theorem}[Shifted pointwise monomial law]\\label{thm:shifted_monomial_law}\nLet\n\\[\ny(n)=\\prod_{j=1}^{m}x(n-a_j)\n\\]\nand write\n\\[\na_j=Nc_j+b_j,\n\\qquad\n0\\leq b_j<N.\n\\]\nFor output residue \\(r\\), define\n\\[\nr_j\\equiv r-b_j\\pmod N,\n\\qquad\n0\\leq r_j<N,\n\\]\nand let\n\\[\n\\kappa_{r,b_j}\n=\n\\begin{cases}\n1,&r<b_j,\\\\\n0,&r\\geq b_j.\n\\end{cases}\n\\]\nThen the output channel is\n\\[\n\\boxed{\ny_r(q)\n=\n\\prod_{j=1}^{m}\nx_{r_j}\n\\bigl(q-c_j-\\kappa_{r,b_j}\\bigr).\n}\n\\]\n\\end{theorem}","0,&r\\geq b_j.\n\\end{cases}\n\\]\nThen the output channel is\n\\[\n\\boxed{\ny_r(q)\n=\n\\prod_{j=1}^{m}\nx_{r_j}\n\\bigl(q-c_j-\\kappa_{r,b_j}\\bigr).\n}\n\\]\n\\end{theorem}","\\begin{proof}\nFor \\(n=Nq+r\\),\n\\[\nn-a_j\n=N(q-c_j)+r-b_j.\n\\]\nIf \\(r\\geq b_j\\), this address has residue \\(r-b_j\\) and coarse index \\(q-c_j\\). If \\(r<b_j\\), add \\(N\\) to the residue and subtract one from the coarse index. Multiplying the resulting channel values gives the formula.\n\\end{proof}","\\begin{example}[Quadratic nearest neighbor interaction]\\label{ex:quadratic_neighbor}\nFor\n\\[\ny(n)=x(n)x(n-1)\n\\]\nand modulus \\(N\\),\n\\[\ny_r(q)\n=\n\\begin{cases}\nx_r(q)x_{r-1}(q),&r\\geq1,","\\begin{example}[Quadratic nearest neighbor interaction]\\label{ex:quadratic_neighbor}\nFor\n\\[\ny(n)=x(n)x(n-1)\n\\]\nand modulus \\(N\\),\n\\[\ny_r(q)\n=\n\\begin{cases}\nx_r(q)x_{r-1}(q),&r\\geq1,\\\\\nx_0(q)x_{N-1}(q-1),&r=0.\n\\end{cases}\n\\]\nThe carry at residue zero is the nonlinear analogue of a transfer crossing a coarse cell boundary.\n\\end{example}","\\midrule\n16 & \\(1.7999\\times10^{-1}\\) & \\(6.6522\\times10^{-2}\\)","32 & \\(3.8414\\times10^{-2}\\) & \\(4.9153\\times10^{-2}\\)","64 & \\(1.6267\\times10^{-2}\\) & \\(3.8056\\times10^{-2}\\)","128 & \\(1.7387\\times10^{-3}\\) & \\(2.5631\\times10^{-2}\\)","256 & \\(1.5870\\times10^{-4}\\) & \\(1.7424\\times10^{-2}\\)","\\midrule\none dimensional step & \\(5.4023\\times10^{-2}\\) & \\(1.3692\\times10^{-4}\\) & \\(2.4029\\times10^{-2}\\)","two dimensional disk & \\(3.7643\\times10^{-2}\\) & \\(6.4752\\times10^{-5}\\) & \\(3.4043\\times10^{-3}\\)","multiple regions & \\(3.3740\\times10^{-2}\\) & \\(5.2118\\times10^{-7}\\) & not evaluated","\\begin{interpretation}[title={Keep the rotation and its chart separate}]\nFor the phase factorization below, $M$ and $N$ are positive phase orders. The reflection coordinate rotates in the unit disk. The Cayley map expresses that same motion as a fractional linear impedance action. A pole of the chart does not mean that the projective action fails. This realization is independent of the separated jump detector.\n\\end{interpretation}","Let\n\\[\n\\mathbb H\n=\n\\{z\\in\\CC:\\Re z>0\\}\n\\]\nand define\n\\[\nC(z)=\\frac{z-1}{z+1}.\n\\]\nIt maps \\(\\mathbb H\\) conformally onto the unit disk \\(\\mathbb D\\), with inverse\n\\[\nC^{-1}(w)=\\frac{1+w}{1-w}.\n\\]\nFor \\(\\theta\\in\\RR\\), let\n\\[\nR_\\theta(w)=\\e^{\\ii\\theta}w.\n\\]","\\begin{theorem}[Cyclic Mobius action]\\label{thm:cyclic_mobius}\nThe pullback\n\\[\n\\Phi_\\theta\n=C^{-1}R_\\theta C\n\\]\nis the fractional linear map\n\\[\n\\boxed{\n\\Phi_\\theta(z)\n=\n\\frac{(1+\\e^{\\ii\\theta})z+(1-\\e^{\\ii\\theta})}\n{(1-\\e^{\\ii\\theta})z+(1+\\e^{\\ii\\theta})}.\n}\n\\]\nIt preserves \\(\\mathbb H\\) and satisfies\n\\[\n\\Phi_\\theta\\Phi_\\varphi\n=\n\\Phi_{\\theta+\\varphi}.\n\\]\nFor an integer \\(N\\geq1\\) and \\(\\theta=2\\pi/N\\), the map has order \\(N\\).\n\\end{theorem}","us action]\\label{thm:cyclic_mobius}\nThe pullback\n\\[\n\\Phi_\\theta\n=C^{-1}R_\\theta C\n\\]\nis the fractional linear map\n\\[\n\\boxed{\n\\Phi_\\theta(z)\n=\n\\frac{(1+\\e^{\\ii\\theta})z+(1-\\e^{\\ii\\theta})}\n{(1-\\e^{\\ii\\theta})z+(1+\\e^{\\ii\\theta})}.\n}\n\\]\nIt preserves \\(\\mathbb H\\) and satisfies\n\\[\n\\Phi_\\theta\\Phi_\\varphi\n=\n\\Phi_{\\theta+\\varphi}.\n\\]\nFor an integer \\(N\\geq1\\) and \\(\\theta=2\\pi/N\\), the map has order \\(N\\).\n\\end{theorem}","\\begin{proof}\nSubstitute \\(\\e^{\\ii\\theta}(z-1)/(z+1)\\) into \\(C^{-1}\\) and simplify. Preservation of \\(\\mathbb H\\) follows because \\(C\\) maps it to \\(\\mathbb D\\), rotation preserves \\(\\mathbb D\\), and \\(C^{-1}\\) maps back. The group law follows by conjugating the rotation group law. When \\(\\theta=2\\pi/N\\), the \\(N\\)th power is the identity. Conversely, if its \\(k\\)th power is the identity, conjugation by \\(C\\) and evaluation at \\(w=1/2\\) give \\(\\e^{2\\pi\\ii k/N}=1\\), so \\(N\\) divides \\(k\\). Hence the order is exactly \\(N\\).\n\\end{proof}","of}\nSubstitute \\(\\e^{\\ii\\theta}(z-1)/(z+1)\\) into \\(C^{-1}\\) and simplify. Preservation of \\(\\mathbb H\\) follows because \\(C\\) maps it to \\(\\mathbb D\\), rotation preserves \\(\\mathbb D\\), and \\(C^{-1}\\) maps back. The group law follows by conjugating the rotation group law. When \\(\\theta=2\\pi/N\\), the \\(N\\)th power is the identity. Conversely, if its \\(k\\)th power is the identity, conjugation by \\(C\\) and evaluation at \\(w=1/2\\) give \\(\\e^{2\\pi\\ii k/N}=1\\), so \\(N\\) divides \\(k\\). Hence the order is exactly \\(N\\).\n\\end{proof}","\\theta}(z-1)/(z+1)\\) into \\(C^{-1}\\) and simplify. Preservation of \\(\\mathbb H\\) follows because \\(C\\) maps it to \\(\\mathbb D\\), rotation preserves \\(\\mathbb D\\), and \\(C^{-1}\\) maps back. The group law follows by conjugating the rotation group law. When \\(\\theta=2\\pi/N\\), the \\(N\\)th power is the identity. Conversely, if its \\(k\\)th power is the identity, conjugation by \\(C\\) and evaluation at \\(w=1/2\\) give \\(\\e^{2\\pi\\ii k/N}=1\\), so \\(N\\) divides \\(k\\). Hence the order is exactly \\(N\\).\n\\end{proof}","+1)\\) into \\(C^{-1}\\) and simplify. Preservation of \\(\\mathbb H\\) follows because \\(C\\) maps it to \\(\\mathbb D\\), rotation preserves \\(\\mathbb D\\), and \\(C^{-1}\\) maps back. The group law follows by conjugating the rotation group law. When \\(\\theta=2\\pi/N\\), the \\(N\\)th power is the identity. Conversely, if its \\(k\\)th power is the identity, conjugation by \\(C\\) and evaluation at \\(w=1/2\\) give \\(\\e^{2\\pi\\ii k/N}=1\\), so \\(N\\) divides \\(k\\). Hence the order is exactly \\(N\\).\n\\end{proof}","implify. Preservation of \\(\\mathbb H\\) follows because \\(C\\) maps it to \\(\\mathbb D\\), rotation preserves \\(\\mathbb D\\), and \\(C^{-1}\\) maps back. The group law follows by conjugating the rotation group law. When \\(\\theta=2\\pi/N\\), the \\(N\\)th power is the identity. Conversely, if its \\(k\\)th power is the identity, conjugation by \\(C\\) and evaluation at \\(w=1/2\\) give \\(\\e^{2\\pi\\ii k/N}=1\\), so \\(N\\) divides \\(k\\). Hence the order is exactly \\(N\\).\n\\end{proof}","rvation of \\(\\mathbb H\\) follows because \\(C\\) maps it to \\(\\mathbb D\\), rotation preserves \\(\\mathbb D\\), and \\(C^{-1}\\) maps back. The group law follows by conjugating the rotation group law. When \\(\\theta=2\\pi/N\\), the \\(N\\)th power is the identity. Conversely, if its \\(k\\)th power is the identity, conjugation by \\(C\\) and evaluation at \\(w=1/2\\) give \\(\\e^{2\\pi\\ii k/N}=1\\), so \\(N\\) divides \\(k\\). Hence the order is exactly \\(N\\).\n\\end{proof}","ecause \\(C\\) maps it to \\(\\mathbb D\\), rotation preserves \\(\\mathbb D\\), and \\(C^{-1}\\) maps back. The group law follows by conjugating the rotation group law. When \\(\\theta=2\\pi/N\\), the \\(N\\)th power is the identity. Conversely, if its \\(k\\)th power is the identity, conjugation by \\(C\\) and evaluation at \\(w=1/2\\) give \\(\\e^{2\\pi\\ii k/N}=1\\), so \\(N\\) divides \\(k\\). Hence the order is exactly \\(N\\).\n\\end{proof}","\\begin{corollary}[Impedance and admittance]\\label{cor:impedance_admittance}\nFor \\(N=2\\), \\(\\theta=\\pi\\) and\n\\[\n\\Phi_\\pi(z)=\\frac1z.\n\\]\nThus the two phase action is normalized impedance and admittance inversion.\n\\end{corollary}","\\begin{proof}\nInsert \\(\\e^{\\ii\\pi}=-1\\) in the formula of \\cref{thm:cyclic_mobius}.\n\\end{proof}","\\begin{proposition}[Radial and phase coordinates]\\label{prop:radial_phase}\nWrite\n\\[\nC(z)=\\rho\\e^{\\ii\\varphi},\n\\qquad\n0\\leq\\rho<1.\n\\]\nThen\n\\[\nC(\\Phi_\\theta(z))\n=\n\\rho\\e^{\\ii(\\varphi+\\theta)}.\n\\]\nThe radial mismatch \\(\\rho\\) is invariant and the phase coordinate advances by \\(\\theta\\).\n\\end{proposition}","\\begin{proof}\nThis is the defining conjugacy \\(C\\Phi_\\theta=R_\\theta C\\).\n\\end{proof}","\\begin{proposition}[Coprime phase factorization]\\label{prop:phase_factorization}\nLet \\(M\\) and \\(N\\) be coprime. The cyclic phase labels modulo \\(MN\\) are equivalent to pairs of labels modulo \\(M\\) and modulo \\(N\\):\n\\[\n\\ZZ/(MN)\\ZZ\n\\cong\n\\ZZ/M\\ZZ\\times\\ZZ/N\\ZZ.\n\\]\nUnder this identification, the rotation by \\(2\\pi/(MN)\\) is determined by its two projected cyclic actions.\n\\end{proposition}","\\begin{proof}\nThis is the Chinese remainder theorem applied to the additive phase label. A label modulo \\(MN\\) is uniquely determined by its two residues, and addition of labels is preserved.\n\\end{proof}","The higher order singular packet theorem removes finitely many trace jumps. A more complete approximation theory should connect packet norms to geometric measure quantities in \\(\\BV\\), piecewise Sobolev spaces, and free discontinuity models. It should also determine when a redundant packet dictionary can identify the derivative order of a singularity rather than merely its location.","The resulting chain is\n\\[\n\\begin{gathered}\n\\text{lattice coordinates}\n\\longrightarrow\n\\text{transfer blocks}\n\\longrightarrow\n\\text{stable filter banks}\n\\longrightarrow\n\\text{packet trees}\n\\\\\n\\longrightarrow\n\\text{singular packet model}\n\\longrightarrow\n\\text{local recovery}\n\\longrightarrow\n\\text{smooth approximation rate}.\n\\end{gathered}\n\\]\nEvery arrow is exact or quantitatively controlled under the hypotheses stated in the corresponding result.","Section 13 (p. 24).}\nA packet calculus can reorganize a signal exactly, choose among finitely many representations, or describe a nonlinear coordinate geometry. These are not the same task. This guide keeps one concrete object in view for each, so that the reader can compute rather than only follow arrows.\n\\vfig{three_worlds}\n\\head{A signal that can be checked by hand}\nOur main finite signal is $f=s+j\\in\\QQ^{16}$. The background repeats\n\\[\n\\tfrac14(0,1,2,1,0,-1,-2,-1),\n\\]\nand $j[n]=3$ at $6\\le n<12$, with $j[n]=0$ elsewhere. This is a discrete background model, not an assertion of continuum smoothness. Its energy is exactly $57$. We use it for residue coordinates, Haar transforms, coefficient deletion, and known-packet cancellation.\n\\head{A tree whose alternatives can all be seen}\nThe full binary tree of depth three has only $26$ complete prunings. We enumerate all of them, compute a unique mixed-depth optimum, and certify its stability with rational costs. The displayed objective is deliberately explicit, not claimed to be a universal signal-quality score.\n\\head{A coordinate orbit w…","reorganize a signal exactly, choose among finitely many representations, or describe a nonlinear coordinate geometry. These are not the same task. This guide keeps one concrete object in view for each, so that the reader can compute rather than only follow arrows.\n\\vfig{three_worlds}\n\\head{A signal that can be checked by hand}\nOur main finite signal is $f=s+j\\in\\QQ^{16}$. The background repeats\n\\[\n\\tfrac14(0,1,2,1,0,-1,-2,-1),\n\\]\nand $j[n]=3$ at $6\\le n<12$, with $j[n]=0$ elsewhere. This is a discrete background model, not an assertion of continuum smoothness. Its energy is exactly $57$. We use it for residue coordinates, Haar transforms, coefficient deletion, and known-packet cancellation.\n\\head{A tree whose alternatives can all be seen}\nThe full binary tree of depth three has only $26$ complete prunings. We enumerate all of them, compute a unique mixed-depth optimum, and certify its stability with rational costs. The displayed objective is deliberately explicit, not claimed to be a universal signal-quality score.\n\\head{A coordinate orbit with no hidden numerical poles}\nThe four ref…","ng finitely many representations, or describe a nonlinear coordinate geometry. These are not the same task. This guide keeps one concrete object in view for each, so that the reader can compute rather than only follow arrows.\n\\vfig{three_worlds}\n\\head{A signal that can be checked by hand}\nOur main finite signal is $f=s+j\\in\\QQ^{16}$. The background repeats\n\\[\n\\tfrac14(0,1,2,1,0,-1,-2,-1),\n\\]\nand $j[n]=3$ at $6\\le n<12$, with $j[n]=0$ elsewhere. This is a discrete background model, not an assertion of continuum smoothness. Its energy is exactly $57$. We use it for residue coordinates, Haar transforms, coefficient deletion, and known-packet cancellation.\n\\head{A tree whose alternatives can all be seen}\nThe full binary tree of depth three has only $26$ complete prunings. We enumerate all of them, compute a unique mixed-depth optimum, and certify its stability with rational costs. The displayed objective is deliberately explicit, not claimed to be a universal signal-quality score.\n\\head{A coordinate orbit with no hidden numerical poles}\nThe four reflection points $\\Gamma=i^k/2$ map to fo…","many representations, or describe a nonlinear coordinate geometry. These are not the same task. This guide keeps one concrete object in view for each, so that the reader can compute rather than only follow arrows.\n\\vfig{three_worlds}\n\\head{A signal that can be checked by hand}\nOur main finite signal is $f=s+j\\in\\QQ^{16}$. The background repeats\n\\[\n\\tfrac14(0,1,2,1,0,-1,-2,-1),\n\\]\nand $j[n]=3$ at $6\\le n<12$, with $j[n]=0$ elsewhere. This is a discrete background model, not an assertion of continuum smoothness. Its energy is exactly $57$. We use it for residue coordinates, Haar transforms, coefficient deletion, and known-packet cancellation.\n\\head{A tree whose alternatives can all be seen}\nThe full binary tree of depth three has only $26$ complete prunings. We enumerate all of them, compute a unique mixed-depth optimum, and certify its stability with rational costs. The displayed objective is deliberately explicit, not claimed to be a universal signal-quality score.\n\\head{A coordinate orbit with no hidden numerical poles}\nThe four reflection points $\\Gamma=i^k/2$ map to four exact imp…","ns, or describe a nonlinear coordinate geometry. These are not the same task. This guide keeps one concrete object in view for each, so that the reader can compute rather than only follow arrows.\n\\vfig{three_worlds}\n\\head{A signal that can be checked by hand}\nOur main finite signal is $f=s+j\\in\\QQ^{16}$. The background repeats\n\\[\n\\tfrac14(0,1,2,1,0,-1,-2,-1),\n\\]\nand $j[n]=3$ at $6\\le n<12$, with $j[n]=0$ elsewhere. This is a discrete background model, not an assertion of continuum smoothness. Its energy is exactly $57$. We use it for residue coordinates, Haar transforms, coefficient deletion, and known-packet cancellation.\n\\head{A tree whose alternatives can all be seen}\nThe full binary tree of depth three has only $26$ complete prunings. We enumerate all of them, compute a unique mixed-depth optimum, and certify its stability with rational costs. The displayed objective is deliberately explicit, not claimed to be a universal signal-quality score.\n\\head{A coordinate orbit with no hidden numerical poles}\nThe four reflection points $\\Gamma=i^k/2$ map to four exact impedances. A second …","w for each, so that the reader can compute rather than only follow arrows.\n\\vfig{three_worlds}\n\\head{A signal that can be checked by hand}\nOur main finite signal is $f=s+j\\in\\QQ^{16}$. The background repeats\n\\[\n\\tfrac14(0,1,2,1,0,-1,-2,-1),\n\\]\nand $j[n]=3$ at $6\\le n<12$, with $j[n]=0$ elsewhere. This is a discrete background model, not an assertion of continuum smoothness. Its energy is exactly $57$. We use it for residue coordinates, Haar transforms, coefficient deletion, and known-packet cancellation.\n\\head{A tree whose alternatives can all be seen}\nThe full binary tree of depth three has only $26$ complete prunings. We enumerate all of them, compute a unique mixed-depth optimum, and certify its stability with rational costs. The displayed objective is deliberately explicit, not claimed to be a universal signal-quality score.\n\\head{A coordinate orbit with no hidden numerical poles}\nThe four reflection points $\\Gamma=i^k/2$ map to four exact impedances. A second orbit passes through infinity; homogeneous coordinates make that point an ordinary part of the construction.\n\\takeaway{Ex…","ts\n\\[\n\\tfrac14(0,1,2,1,0,-1,-2,-1),\n\\]\nand $j[n]=3$ at $6\\le n<12$, with $j[n]=0$ elsewhere. This is a discrete background model, not an assertion of continuum smoothness. Its energy is exactly $57$. We use it for residue coordinates, Haar transforms, coefficient deletion, and known-packet cancellation.\n\\head{A tree whose alternatives can all be seen}\nThe full binary tree of depth three has only $26$ complete prunings. We enumerate all of them, compute a unique mixed-depth optimum, and certify its stability with rational costs. The displayed objective is deliberately explicit, not claimed to be a universal signal-quality score.\n\\head{A coordinate orbit with no hidden numerical poles}\nThe four reflection points $\\Gamma=i^k/2$ map to four exact impedances. A second orbit passes through infinity; homogeneous coordinates make that point an ordinary part of the construction.\n\\takeaway{Exact coordinates preserve information. An optimizer chooses coordinates. An estimator must still identify information not already supplied.}","se alternatives can all be seen}\nThe full binary tree of depth three has only $26$ complete prunings. We enumerate all of them, compute a unique mixed-depth optimum, and certify its stability with rational costs. The displayed objective is deliberately explicit, not claimed to be a universal signal-quality score.\n\\head{A coordinate orbit with no hidden numerical poles}\nThe four reflection points $\\Gamma=i^k/2$ map to four exact impedances. A second orbit passes through infinity; homogeneous coordinates make that point an ordinary part of the construction.\n\\takeaway{Exact coordinates preserve information. An optimizer chooses coordinates. An estimator must still identify information not already supplied.}","midrule\n$S,\\delta_0$ & Sample shift and unit impulse & A finite truncation of every $\\ell^2$ signal","$P_{N,r}f$ & A masked signal on the original lattice & The shorter decimated channel","$f_r[q]$ & The value $f[Nq+r]$ & An average of neighboring samples","$D$ & One channel-index delay & Necessarily one original-sample delay","$E(D),R(D)$ & Finite Laurent analysis and synthesis matrices & Values at one test frequency","$W_\\alpha$ & The addressed Haar analysis operator & An edge detector by itself","$W_\\alpha^*W_\\alpha$ & A projector on the original signal space & A scalar packet energy","$\\Lambda$ & A complete covering antichain of a finite tree & An arbitrary collection of nodes","$j,\\widehat j$ & Known and estimated discontinuity packets & Interchangeable hypotheses","$\\Gamma,z$ & Reflection coefficient and normalized impedance & Coefficients that can be linearly interchanged","bottomrule\n\\end{tabularx}\n\\head{Three arithmetic conventions}\nFIR examples use bilateral sequences with finite support and \\emph{zero extension}. Haar examples use vectors of length $16$, with lengths divisible by the required powers of two. The three-tap smoother and nonlinear neighbor example use \\emph{periodic} indexing. These boundary conventions are stated where they enter. None is silently substituted for another.\nThe polynomial variable $t$ on \\pagerefV{20} indexes coefficients. The delay symbol $D$ acts on channels. The Smith variable $z$ is impedance. Keeping them separate prevents a visual resemblance from becoming an invalid identity.\n\\takeaway{A correct formula needs the correct object, product, norm, and boundary convention. A familiar-looking matrix does not supply them automatically.}","conventions}\nFIR examples use bilateral sequences with finite support and \\emph{zero extension}. Haar examples use vectors of length $16$, with lengths divisible by the required powers of two. The three-tap smoother and nonlinear neighbor example use \\emph{periodic} indexing. These boundary conventions are stated where they enter. None is silently substituted for another.\nThe polynomial variable $t$ on \\pagerefV{20} indexes coefficients. The delay symbol $D$ acts on channels. The Smith variable $z$ is impedance. Keeping them separate prevents a visual resemblance from becoming an invalid identity.\n\\takeaway{A correct formula needs the correct object, product, norm, and boundary convention. A familiar-looking matrix does not supply them automatically.}","e support and \\emph{zero extension}. Haar examples use vectors of length $16$, with lengths divisible by the required powers of two. The three-tap smoother and nonlinear neighbor example use \\emph{periodic} indexing. These boundary conventions are stated where they enter. None is silently substituted for another.\nThe polynomial variable $t$ on \\pagerefV{20} indexes coefficients. The delay symbol $D$ acts on channels. The Smith variable $z$ is impedance. Keeping them separate prevents a visual resemblance from becoming an invalid identity.\n\\takeaway{A correct formula needs the correct object, product, norm, and boundary convention. A familiar-looking matrix does not supply them automatically.}","examples use vectors of length $16$, with lengths divisible by the required powers of two. The three-tap smoother and nonlinear neighbor example use \\emph{periodic} indexing. These boundary conventions are stated where they enter. None is silently substituted for another.\nThe polynomial variable $t$ on \\pagerefV{20} indexes coefficients. The delay symbol $D$ acts on channels. The Smith variable $z$ is impedance. Keeping them separate prevents a visual resemblance from becoming an invalid identity.\n\\takeaway{A correct formula needs the correct object, product, norm, and boundary convention. A familiar-looking matrix does not supply them automatically.}","\\vpage{3}{Decimation is a permutation, not a loss}{EXACT ADDRESSES / THE RUNNING SIGNAL}{Traditional monograph: Theorem 2.1 (p. 3), Corollary 2.2 (p. 3).}\nFor four phases, write every sample address uniquely as $n=4q+r$ with $0\\le r<4$. Analysis stores $f[n]$ at channel address $(r,q)$. Synthesis puts that exact value back at $4q+r$.\n\\vfig{addresses}\n\\platecaption{Every cell contains an original address and its rational sample value. Shading marks the known jump packet, not an estimated feature or a frequency band.}\n\\head{What has changed? Only where the values are stored}\nThe complete tuple of channels contains sixteen values, just as the original vector does. A single decimated channel is shorter, but keeping all four loses no samples. The residue projector $P_{4,r}f$ is different: it keeps the original lattice and writes zeros outside phase $r$.\n\\[\n\\mathcal S_4\\mathcal A_4f=f,\n\\qquad\n\\|f\\|_2^2=\\sum_{r=","\\vpage{3}{Decimation is a permutation, not a loss}{EXACT ADDRESSES / THE RUNNING SIGNAL}{Traditional monograph: Theorem 2.1 (p. 3), Corollary 2.2 (p. 3).}\nFor four phases, write every sample address uniquely as $n=4q+r$ with $0\\le r<4$. Analysis stores $f[n]$ at channel address $(r,q)$. Synthesis puts that exact value back at $4q+r$.\n\\vfig{addresses}\n\\platecaption{Every cell contains an original address and its rational sample value. Shading marks the known jump packet, not an estimated feature or a frequency band.}\n\\head{What has changed? Only where the values are stored}\nThe complete tuple of channels contains sixteen values, just as the original vector does. A single decimated channel is shorter, but keeping all four loses no samples. The residue projector $P_{4,r}f$ is different: it keeps the original lattice and writes zeros outside phase $r$.\n\\[\n\\mathcal S_4\\mathcal A_4f=f,\n\\qquad\n\\|f\\|_2^2=\\sum_{r=0}^3\\|f_r\\|_2^2=","\\vpage{3}{Decimation is a permutation, not a loss}{EXACT ADDRESSES / THE RUNNING SIGNAL}{Traditional monograph: Theorem 2.1 (p. 3), Corollary 2.2 (p. 3).}\nFor four phases, write every sample address uniquely as $n=4q+r$ with $0\\le r<4$. Analysis stores $f[n]$ at channel address $(r,q)$. Synthesis puts that exact value back at $4q+r$.\n\\vfig{addresses}\n\\platecaption{Every cell contains an original address and its rational sample value. Shading marks the known jump packet, not an estimated feature or a frequency band.}\n\\head{What has changed? Only where the values are stored}\nThe complete tuple of channels contains sixteen values, just as the original vector does. A single decimated channel is shorter, but keeping all four loses no samples. The residue projector $P_{4,r}f$ is different: it keeps the original lattice and writes zeros outside phase $r$.\n\\[\n\\mathcal S_4\\mathcal A_4f=f,\n\\qquad\n\\|f\\|_2^2=\\sum_{r=0}^3\\|f_r\\|_2^2=57.\n\\]\nThe norm on the c","\\vpage{3}{Decimation is a permutation, not a loss}{EXACT ADDRESSES / THE RUNNING SIGNAL}{Traditional monograph: Theorem 2.1 (p. 3), Corollary 2.2 (p. 3).}\nFor four phases, write every sample address uniquely as $n=4q+r$ with $0\\le r<4$. Analysis stores $f[n]$ at channel address $(r,q)$. Synthesis puts that exact value back at $4q+r$.\n\\vfig{addresses}\n\\platecaption{Every cell contains an original address and its rational sample value. Shading marks the known jump packet, not an estimated feature or a frequency band.}\n\\head{What has changed? Only where the values are stored}\nThe complete tuple of channels contains sixteen values, just as the original vector does. A single decimated channel is shorter, but keeping all four loses no samples. The residue projector $P_{4,r}f$ is different: it keeps the original lattice and writes zeros outside phase $r$.\n\\[\n\\mathcal S_4\\mathcal A_4f=f,\n\\qquad\n\\|f\\|_2^2=\\sum_{r=0}^3\\|f_r\\|_2^2=57.\n\\]\nThe norm on the channel tuple is the sum-of-","\\vpage{3}{Decimation is a permutation, not a loss}{EXACT ADDRESSES / THE RUNNING SIGNAL}{Traditional monograph: Theorem 2.1 (p. 3), Corollary 2.2 (p. 3).}\nFor four phases, write every sample address uniquely as $n=4q+r$ with $0\\le r<4$. Analysis stores $f[n]$ at channel address $(r,q)$. Synthesis puts that exact value back at $4q+r$.\n\\vfig{addresses}\n\\platecaption{Every cell contains an original address and its rational sample value. Shading marks the known jump packet, not an estimated feature or a frequency band.}\n\\head{What has changed? Only where the values are stored}\nThe complete tuple of channels contains sixteen values, just as the original vector does. A single decimated channel is shorter, but keeping all four loses no samples. The residue projector $P_{4,r}f$ is different: it keeps the original lattice and writes zeros outside phase $r$.\n\\[\n\\mathcal S_4\\mathcal A_4f=f,\n\\qquad\n\\|f\\|_2^2=\\sum_{r=0}^3\\|f_r\\|_2^2=57.\n\\]\nThe norm on the channel tuple is the sum-of-squares Hilbert norm, not the maximum of the cha","y cell contains an original address and its rational sample value. Shading marks the known jump packet, not an estimated feature or a frequency band.}\n\\head{What has changed? Only where the values are stored}\nThe complete tuple of channels contains sixteen values, just as the original vector does. A single decimated channel is shorter, but keeping all four loses no samples. The residue projector $P_{4,r}f$ is different: it keeps the original lattice and writes zeros outside phase $r$.\n\\[\n\\mathcal S_4\\mathcal A_4f=f,\n\\qquad\n\\|f\\|_2^2=\\sum_{r=0}^3\\|f_r\\|_2^2=57.\n\\]\nThe norm on the channel tuple is the sum-of-squares Hilbert norm, not the maximum of the channel norms. This distinction matters in the infinite-dimensional statement.\n\\head{Negative addresses are ordinary addresses}\nEuclidean division, rather than truncation toward zero, gives\n$-1=4(-1)+3$ and $-5=4(-2)+3$. The finite tests exercise both positive and negative sample addresses. The algebraic identity applies before any plotting window or finite buffer is chosen.\n\\takeaway{Downsampling one channel discards information. Splitt…","nown jump packet, not an estimated feature or a frequency band.}\n\\head{What has changed? Only where the values are stored}\nThe complete tuple of channels contains sixteen values, just as the original vector does. A single decimated channel is shorter, but keeping all four loses no samples. The residue projector $P_{4,r}f$ is different: it keeps the original lattice and writes zeros outside phase $r$.\n\\[\n\\mathcal S_4\\mathcal A_4f=f,\n\\qquad\n\\|f\\|_2^2=\\sum_{r=0}^3\\|f_r\\|_2^2=57.\n\\]\nThe norm on the channel tuple is the sum-of-squares Hilbert norm, not the maximum of the channel norms. This distinction matters in the infinite-dimensional statement.\n\\head{Negative addresses are ordinary addresses}\nEuclidean division, rather than truncation toward zero, gives\n$-1=4(-1)+3$ and $-5=4(-2)+3$. The finite tests exercise both positive and negative sample addresses. The algebraic identity applies before any plotting window or finite buffer is chosen.\n\\takeaway{Downsampling one channel discards information. Splitting into all residue channels is an invertible address change.}","ump packet, not an estimated feature or a frequency band.}\n\\head{What has changed? Only where the values are stored}\nThe complete tuple of channels contains sixteen values, just as the original vector does. A single decimated channel is shorter, but keeping all four loses no samples. The residue projector $P_{4,r}f$ is different: it keeps the original lattice and writes zeros outside phase $r$.\n\\[\n\\mathcal S_4\\mathcal A_4f=f,\n\\qquad\n\\|f\\|_2^2=\\sum_{r=0}^3\\|f_r\\|_2^2=57.\n\\]\nThe norm on the channel tuple is the sum-of-squares Hilbert norm, not the maximum of the channel norms. This distinction matters in the infinite-dimensional statement.\n\\head{Negative addresses are ordinary addresses}\nEuclidean division, rather than truncation toward zero, gives\n$-1=4(-1)+3$ and $-5=4(-2)+3$. The finite tests exercise both positive and negative sample addresses. The algebraic identity applies before any plotting window or finite buffer is chosen.\n\\takeaway{Downsampling one channel discards information. Splitting into all residue channels is an invertible address change.}","ice and writes zeros outside phase $r$.\n\\[\n\\mathcal S_4\\mathcal A_4f=f,\n\\qquad\n\\|f\\|_2^2=\\sum_{r=0}^3\\|f_r\\|_2^2=57.\n\\]\nThe norm on the channel tuple is the sum-of-squares Hilbert norm, not the maximum of the channel norms. This distinction matters in the infinite-dimensional statement.\n\\head{Negative addresses are ordinary addresses}\nEuclidean division, rather than truncation toward zero, gives\n$-1=4(-1)+3$ and $-5=4(-2)+3$. The finite tests exercise both positive and negative sample addresses. The algebraic identity applies before any plotting window or finite buffer is chosen.\n\\takeaway{Downsampling one channel discards information. Splitting into all residue channels is an invertible address change.}","ros outside phase $r$.\n\\[\n\\mathcal S_4\\mathcal A_4f=f,\n\\qquad\n\\|f\\|_2^2=\\sum_{r=0}^3\\|f_r\\|_2^2=57.\n\\]\nThe norm on the channel tuple is the sum-of-squares Hilbert norm, not the maximum of the channel norms. This distinction matters in the infinite-dimensional statement.\n\\head{Negative addresses are ordinary addresses}\nEuclidean division, rather than truncation toward zero, gives\n$-1=4(-1)+3$ and $-5=4(-2)+3$. The finite tests exercise both positive and negative sample addresses. The algebraic identity applies before any plotting window or finite buffer is chosen.\n\\takeaway{Downsampling one channel discards information. Splitting into all residue channels is an invertible address change.}","\\vpage{4}{Refine packets on a common lattice}{EXACT REFINEMENT / WHAT CAN BE ADDED}{Traditional monograph: Theorem 2.3 (p. 5).}\nAn even sample belongs to exactly one of residues $0$ and $2$ modulo four. Therefore the coarse even mask is the sum of two disjoint fine masks:\n\\[\nP_{2,0}=P_{4,0}+P_{4,2}.\n\\]\n\\vfig{refinement}\n\\platecaption{The common horizontal axis is the original sample index. Empty cells denote masked-out addresses. There are no interpolated samples between the displayed cells.}\n\\head{Why the coordinate convention matters}\nThe decimated channel $f_0[q]=f[2q]$ has its own index $q$. At modulus four the descendants are $f[4q]$ and $f[4q+2]$. These shorter arrays must be interleaved to recover the coarse channel. Adding them entrywise instead produces sums of different original samples.\nFor example, at the first descendant address the values $f[0]=0$ and $f","\\vpage{4}{Refine packets on a common lattice}{EXACT REFINEMENT / WHAT CAN BE ADDED}{Traditional monograph: Theorem 2.3 (p. 5).}\nAn even sample belongs to exactly one of residues $0$ and $2$ modulo four. Therefore the coarse even mask is the sum of two disjoint fine masks:\n\\[\nP_{2,0}=P_{4,0}+P_{4,2}.\n\\]\n\\vfig{refinement}\n\\platecaption{The common horizontal axis is the original sample index. Empty cells denote masked-out addresses. There are no interpolated samples between the displayed cells.}\n\\head{Why the coordinate convention matters}\nThe decimated channel $f_0[q]=f[2q]$ has its own index $q$. At modulus four the descendants are $f[4q]$ and $f[4q+2]$. These shorter arrays must be interleaved to recover the coarse channel. Adding them entrywise instead produces sums of different original samples.\nFor example, at the first descendant address the values $f[0]=0$ and $f[2]=1/2$","\\vpage{4}{Refine packets on a common lattice}{EXACT REFINEMENT / WHAT CAN BE ADDED}{Traditional monograph: Theorem 2.3 (p. 5).}\nAn even sample belongs to exactly one of residues $0$ and $2$ modulo four. Therefore the coarse even mask is the sum of two disjoint fine masks:\n\\[\nP_{2,0}=P_{4,0}+P_{4,2}.\n\\]\n\\vfig{refinement}\n\\platecaption{The common horizontal axis is the original sample index. Empty cells denote masked-out addresses. There are no interpolated samples between the displayed cells.}\n\\head{Why the coordinate convention matters}\nThe decimated channel $f_0[q]=f[2q]$ has its own index $q$. At modulus four the descendants are $f[4q]$ and $f[4q+2]$. These shorter arrays must be interleaved to recover the coarse channel. Adding them entrywise instead produces sums of different original samples.\nFor example, at the first descendant address the values $f[0]=0$ and $f[2]=1/2$ are two separate values. Their sum $1/2$ cannot tell us both of them. Exact refinement is a disjoint partition,","of residues $0$ and $2$ modulo four. Therefore the coarse even mask is the sum of two disjoint fine masks:\n\\[\nP_{2,0}=P_{4,0}+P_{4,2}.\n\\]\n\\vfig{refinement}\n\\platecaption{The common horizontal axis is the original sample index. Empty cells denote masked-out addresses. There are no interpolated samples between the displayed cells.}\n\\head{Why the coordinate convention matters}\nThe decimated channel $f_0[q]=f[2q]$ has its own index $q$. At modulus four the descendants are $f[4q]$ and $f[4q+2]$. These shorter arrays must be interleaved to recover the coarse channel. Adding them entrywise instead produces sums of different original samples.\nFor example, at the first descendant address the values $f[0]=0$ and $f[2]=1/2$ are two separate values. Their sum $1/2$ cannot tell us both of them. Exact refinement is a disjoint partition, not an averaging rule or a new measurement.\n\\head{The general formula}\nWhen $N$ divides $M$,\n\\[\nP_{N,r}=\\sum_{\\substack{0\\le a<M","ur. Therefore the coarse even mask is the sum of two disjoint fine masks:\n\\[\nP_{2,0}=P_{4,0}+P_{4,2}.\n\\]\n\\vfig{refinement}\n\\platecaption{The common horizontal axis is the original sample index. Empty cells denote masked-out addresses. There are no interpolated samples between the displayed cells.}\n\\head{Why the coordinate convention matters}\nThe decimated channel $f_0[q]=f[2q]$ has its own index $q$. At modulus four the descendants are $f[4q]$ and $f[4q+2]$. These shorter arrays must be interleaved to recover the coarse channel. Adding them entrywise instead produces sums of different original samples.\nFor example, at the first descendant address the values $f[0]=0$ and $f[2]=1/2$ are two separate values. Their sum $1/2$ cannot tell us both of them. Exact refinement is a disjoint partition, not an averaging rule or a new measurement.\n\\head{The general formula}\nWhen $N$ divides $M$,\n\\[\nP_{N,r}=\\sum_{\\substack{0\\le a<M","sum of two disjoint fine masks:\n\\[\nP_{2,0}=P_{4,0}+P_{4,2}.\n\\]\n\\vfig{refinement}\n\\platecaption{The common horizontal axis is the original sample index. Empty cells denote masked-out addresses. There are no interpolated samples between the displayed cells.}\n\\head{Why the coordinate convention matters}\nThe decimated channel $f_0[q]=f[2q]$ has its own index $q$. At modulus four the descendants are $f[4q]$ and $f[4q+2]$. These shorter arrays must be interleaved to recover the coarse channel. Adding them entrywise instead produces sums of different original samples.\nFor example, at the first descendant address the values $f[0]=0$ and $f[2]=1/2$ are two separate values. Their sum $1/2$ cannot tell us both of them. Exact refinement is a disjoint partition, not an averaging rule or a new measurement.\n\\head{The general formula}\nWhen $N$ divides $M$,\n\\[\nP_{N,r}=\\sum_{\\substack{0\\le a<M","disjoint fine masks:\n\\[\nP_{2,0}=P_{4,0}+P_{4,2}.\n\\]\n\\vfig{refinement}\n\\platecaption{The common horizontal axis is the original sample index. Empty cells denote masked-out addresses. There are no interpolated samples between the displayed cells.}\n\\head{Why the coordinate convention matters}\nThe decimated channel $f_0[q]=f[2q]$ has its own index $q$. At modulus four the descendants are $f[4q]$ and $f[4q+2]$. These shorter arrays must be interleaved to recover the coarse channel. Adding them entrywise instead produces sums of different original samples.\nFor example, at the first descendant address the values $f[0]=0$ and $f[2]=1/2$ are two separate values. Their sum $1/2$ cannot tell us both of them. Exact refinement is a disjoint partition, not an averaging rule or a new measurement.\n\\head{The general formula}\nWhen $N$ divides $M$,\n\\[\nP_{N,r}=\\sum_{\\substack{0\\le a<M","s between the displayed cells.}\n\\head{Why the coordinate convention matters}\nThe decimated channel $f_0[q]=f[2q]$ has its own index $q$. At modulus four the descendants are $f[4q]$ and $f[4q+2]$. These shorter arrays must be interleaved to recover the coarse channel. Adding them entrywise instead produces sums of different original samples.\nFor example, at the first descendant address the values $f[0]=0$ and $f[2]=1/2$ are two separate values. Their sum $1/2$ cannot tell us both of them. Exact refinement is a disjoint partition, not an averaging rule or a new measurement.\n\\head{The general formula}\nWhen $N$ divides $M$,\n\\[\nP_{N,r}=\\sum_{\\substack{0\\le a<M","displayed cells.}\n\\head{Why the coordinate convention matters}\nThe decimated channel $f_0[q]=f[2q]$ has its own index $q$. At modulus four the descendants are $f[4q]$ and $f[4q+2]$. These shorter arrays must be interleaved to recover the coarse channel. Adding them entrywise instead produces sums of different original samples.\nFor example, at the first descendant address the values $f[0]=0$ and $f[2]=1/2$ are two separate values. Their sum $1/2$ cannot tell us both of them. Exact refinement is a disjoint partition, not an averaging rule or a new measurement.\n\\head{The general formula}\nWhen $N$ divides $M$,\n\\[\nP_{N,r}=\\sum_{\\substack{0\\le a<M","nvention matters}\nThe decimated channel $f_0[q]=f[2q]$ has its own index $q$. At modulus four the descendants are $f[4q]$ and $f[4q+2]$. These shorter arrays must be interleaved to recover the coarse channel. Adding them entrywise instead produces sums of different original samples.\nFor example, at the first descendant address the values $f[0]=0$ and $f[2]=1/2$ are two separate values. Their sum $1/2$ cannot tell us both of them. Exact refinement is a disjoint partition, not an averaging rule or a new measurement.\n\\head{The general formula}\nWhen $N$ divides $M$,\n\\[\nP_{N,r}=\\sum_{\\substack{0\\le a<M","rays must be interleaved to recover the coarse channel. Adding them entrywise instead produces sums of different original samples.\nFor example, at the first descendant address the values $f[0]=0$ and $f[2]=1/2$ are two separate values. Their sum $1/2$ cannot tell us both of them. Exact refinement is a disjoint partition, not an averaging rule or a new measurement.\n\\head{The general formula}\nWhen $N$ divides $M$,\n\\[\nP_{N,r}=\\sum_{\\substack{0\\le a<M","interleaved to recover the coarse channel. Adding them entrywise instead produces sums of different original samples.\nFor example, at the first descendant address the values $f[0]=0$ and $f[2]=1/2$ are two separate values. Their sum $1/2$ cannot tell us both of them. Exact refinement is a disjoint partition, not an averaging rule or a new measurement.\n\\head{The general formula}\nWhen $N$ divides $M$,\n\\[\nP_{N,r}=\\sum_{\\substack{0\\le a<M","leaved to recover the coarse channel. Adding them entrywise instead produces sums of different original samples.\nFor example, at the first descendant address the values $f[0]=0$ and $f[2]=1/2$ are two separate values. Their sum $1/2$ cannot tell us both of them. Exact refinement is a disjoint partition, not an averaging rule or a new measurement.\n\\head{The general formula}\nWhen $N$ divides $M$,\n\\[\nP_{N,r}=\\sum_{\\substack{0\\le a<M\\\\a\\equiv r\\pmod N}}P_{M,a}.\n\\]\nThe divisibility hypothesis ensures that every fine residue class lies entirely inside one coarse class. Without it, this particular descendant relation is not the right refinement rule.\n\\takeaway{Finer coordinates can preserve the object exactly. Compressing their values to averages is an additional operation, with an additional information-loss question.}","\\vpage{5}{A filter becomes a matrix of actual delays}{FIR TRANSFER / A FOUR-TAP EXAMPLE}{Traditional monograph: Theorem 3.1 (p. 5).}\nTake the causal filter $h=(1,2,-1,1)$, with taps at $0,1,2,3$, acting by $y[n]=\\sum_qh[q]f[n-q]$. Resolve the input into even and odd channels. Let $D$ denote one delay in the \\emph{channel} index.\n\\vfig{fir_transfer}\n\\head{Reading one matrix entry}\nThe odd-input contribution to the even output is $(2D+D^2)f_1$. The tap at original delay one reaches the preceding odd channel sample; the tap at delay three reaches the odd channel two positions back. These are different delays despite coming from the same input phase.\nThe two channel equations are\n\\[\n\\begin{aligned}\ny_0[n]&=f_0[n]-f_0[n-1]+2f_1[n-1]+f_1[n-2],","ix of actual delays}{FIR TRANSFER / A FOUR-TAP EXAMPLE}{Traditional monograph: Theorem 3.1 (p. 5).}\nTake the causal filter $h=(1,2,-1,1)$, with taps at $0,1,2,3$, acting by $y[n]=\\sum_qh[q]f[n-q]$. Resolve the input into even and odd channels. Let $D$ denote one delay in the \\emph{channel} index.\n\\vfig{fir_transfer}\n\\head{Reading one matrix entry}\nThe odd-input contribution to the even output is $(2D+D^2)f_1$. The tap at original delay one reaches the preceding odd channel sample; the tap at delay three reaches the odd channel two positions back. These are different delays despite coming from the same input phase.\nThe two channel equations are\n\\[\n\\begin{aligned}\ny_0[n]&=f_0[n]-f_0[n-1]+2f_1[n-1]+f_1[n-2],","note one delay in the \\emph{channel} index.\n\\vfig{fir_transfer}\n\\head{Reading one matrix entry}\nThe odd-input contribution to the even output is $(2D+D^2)f_1$. The tap at original delay one reaches the preceding odd channel sample; the tap at delay three reaches the odd channel two positions back. These are different delays despite coming from the same input phase.\nThe two channel equations are\n\\[\n\\begin{aligned}\ny_0[n]&=f_0[n]-f_0[n-1]+2f_1[n-1]+f_1[n-2],\\\\\ny_1[n]&=2f_0[n]+f_0[n-1]+f_1[n]-f_1[n-1].\n\\end{aligned}\n\\]\nInterleaving gives exactly the direct convolution, including its boundary samples under zero extension.\n\\head{The general address calculation}\nFor $N$ phases, the block from input $s$ to output $k$ has kernel\n$h_{ks}[m]=h[Nm+k-s]$. This comes from the exact identity\n$Nn+k-q=N(n-m)+s$. It is a finite reindexing of taps, so no numerical transform of the input is needed. Evaluation of a bilateral transform is a separate step with the summability hypotheses stated in the manuscript.\n\\takeaway{The matrix does not approximate the filter. Its entries record which input phase is …","y_1[n]&=2f_0[n]+f_0[n-1]+f_1[n]-f_1[n-1].\n\\end{aligned}\n\\]\nInterleaving gives exactly the direct convolution, including its boundary samples under zero extension.\n\\head{The general address calculation}\nFor $N$ phases, the block from input $s$ to output $k$ has kernel\n$h_{ks}[m]=h[Nm+k-s]$. This comes from the exact identity\n$Nn+k-q=N(n-m)+s$. It is a finite reindexing of taps, so no numerical transform of the input is needed. Evaluation of a bilateral transform is a separate step with the summability hypotheses stated in the manuscript.\n\\takeaway{The matrix does not approximate the filter. Its entries record which input phase is used, how far back to read, and what coefficient to multiply.}","\\vpage{6}{One sample is not one channel step}{DELAY UNITS / AN IMPULSE MAKES THE DIFFERENCE VISIBLE}{Traditional monograph: Theorem 4.1 (p. 7).}\nA one-sample delay swaps the two phases and wraps one of them into the preceding channel index. Its matrix is $J(D)$, not $DI_2$.\n\\vfig{sample_delay}\n\\head{Trace one impulse}\nThe input $\\delta_0$ has a single nonzero value in even channel address zero. Under $J(D)$ it appears in odd channel address zero: this is the original sample $\\delta_1$. Under $DI_2$ it stays even but moves to channel address one: this is $\\delta_2$.\nThus\n\\[\nJ(D)^2=DI_2,\n\\qquad\nH(D)=I_2+2J(D)-J(D)^2+J(D)^3\n\\]\nrecovers the previous page's four-tap transfer matrix. The same delay algebra explains both the off-diagonal phase swap and the channel powers.\n\\head{Signed delays need the same care}\nFor an original-sample delay $d$ at modulus $N$,\n\\[\ny_k[n]=f_{(k-d)\\bmod N}\n\\!\\left[n+\\left\\lfloor\\frac{k-d}{N}\\right\\rfloor\\right].\n\\]\nThis inc","\\vpage{6}{One sample is not one channel step}{DELAY UNITS / AN IMPULSE MAKES THE DIFFERENCE VISIBLE}{Traditional monograph: Theorem 4.1 (p. 7).}\nA one-sample delay swaps the two phases and wraps one of them into the preceding channel index. Its matrix is $J(D)$, not $DI_2$.\n\\vfig{sample_delay}\n\\head{Trace one impulse}\nThe input $\\delta_0$ has a single nonzero value in even channel address zero. Under $J(D)$ it appears in odd channel address zero: this is the original sample $\\delta_1$. Under $DI_2$ it stays even but moves to channel address one: this is $\\delta_2$.\nThus\n\\[\nJ(D)^2=DI_2,\n\\qquad\nH(D)=I_2+2J(D)-J(D)^2+J(D)^3\n\\]\nrecovers the previous page's four-tap transfer matrix. The same delay algebra explains both the off-diagonal phase swap and the channel powers.\n\\head{Signed delays need the same care}\nFor an original-sample delay $d$ at modulus $N$,\n\\[\ny_k[n]=f_{(k-d)\\bmod N}\n\\!\\left[n+\\left\\lfloor\\frac{k-d}{N}\\right\\rfloor\\right].\n\\]\nThis includes advanc","\\vpage{6}{One sample is not one channel step}{DELAY UNITS / AN IMPULSE MAKES THE DIFFERENCE VISIBLE}{Traditional monograph: Theorem 4.1 (p. 7).}\nA one-sample delay swaps the two phases and wraps one of them into the preceding channel index. Its matrix is $J(D)$, not $DI_2$.\n\\vfig{sample_delay}\n\\head{Trace one impulse}\nThe input $\\delta_0$ has a single nonzero value in even channel address zero. Under $J(D)$ it appears in odd channel address zero: this is the original sample $\\delta_1$. Under $DI_2$ it stays even but moves to channel address one: this is $\\delta_2$.\nThus\n\\[\nJ(D)^2=DI_2,\n\\qquad\nH(D)=I_2+2J(D)-J(D)^2+J(D)^3\n\\]\nrecovers the previous page's four-tap transfer matrix. The same delay algebra explains both the off-diagonal phase swap and the channel powers.\n\\head{Signed delays need the same care}\nFor an original-sample delay $d$ at modulus $N$,\n\\[\ny_k[n]=f_{(k-d)\\bmod N}\n\\!\\left[n+\\left\\lfloor\\frac{k-d}{N}\\right\\rfloor\\right].\n\\]\nThis includes advances when $d<0$. The finite verification checks the formula on signed","e{6}{One sample is not one channel step}{DELAY UNITS / AN IMPULSE MAKES THE DIFFERENCE VISIBLE}{Traditional monograph: Theorem 4.1 (p. 7).}\nA one-sample delay swaps the two phases and wraps one of them into the preceding channel index. Its matrix is $J(D)$, not $DI_2$.\n\\vfig{sample_delay}\n\\head{Trace one impulse}\nThe input $\\delta_0$ has a single nonzero value in even channel address zero. Under $J(D)$ it appears in odd channel address zero: this is the original sample $\\delta_1$. Under $DI_2$ it stays even but moves to channel address one: this is $\\delta_2$.\nThus\n\\[\nJ(D)^2=DI_2,\n\\qquad\nH(D)=I_2+2J(D)-J(D)^2+J(D)^3\n\\]\nrecovers the previous page's four-tap transfer matrix. The same delay algebra explains both the off-diagonal phase swap and the channel powers.\n\\head{Signed delays need the same care}\nFor an original-sample delay $d$ at modulus $N$,\n\\[\ny_k[n]=f_{(k-d)\\bmod N}\n\\!\\left[n+\\left\\lfloor\\frac{k-d}{N}\\right\\rfloor\\right].\n\\]\nThis includes advances when $d<0$. The finite verification checks the formula on signed sample addresses. A bank with product $D^qI_N$ delays the origina…","DIFFERENCE VISIBLE}{Traditional monograph: Theorem 4.1 (p. 7).}\nA one-sample delay swaps the two phases and wraps one of them into the preceding channel index. Its matrix is $J(D)$, not $DI_2$.\n\\vfig{sample_delay}\n\\head{Trace one impulse}\nThe input $\\delta_0$ has a single nonzero value in even channel address zero. Under $J(D)$ it appears in odd channel address zero: this is the original sample $\\delta_1$. Under $DI_2$ it stays even but moves to channel address one: this is $\\delta_2$.\nThus\n\\[\nJ(D)^2=DI_2,\n\\qquad\nH(D)=I_2+2J(D)-J(D)^2+J(D)^3\n\\]\nrecovers the previous page's four-tap transfer matrix. The same delay algebra explains both the off-diagonal phase swap and the channel powers.\n\\head{Signed delays need the same care}\nFor an original-sample delay $d$ at modulus $N$,\n\\[\ny_k[n]=f_{(k-d)\\bmod N}\n\\!\\left[n+\\left\\lfloor\\frac{k-d}{N}\\right\\rfloor\\right].\n\\]\nThis includes advances when $d<0$. The finite verification checks the formula on signed sample addresses. A bank with product $D^qI_N$ delays the original signal by $Nq$ samples, not $q$ samples.\n\\takeaway{Always state the unit o…","E}{Traditional monograph: Theorem 4.1 (p. 7).}\nA one-sample delay swaps the two phases and wraps one of them into the preceding channel index. Its matrix is $J(D)$, not $DI_2$.\n\\vfig{sample_delay}\n\\head{Trace one impulse}\nThe input $\\delta_0$ has a single nonzero value in even channel address zero. Under $J(D)$ it appears in odd channel address zero: this is the original sample $\\delta_1$. Under $DI_2$ it stays even but moves to channel address one: this is $\\delta_2$.\nThus\n\\[\nJ(D)^2=DI_2,\n\\qquad\nH(D)=I_2+2J(D)-J(D)^2+J(D)^3\n\\]\nrecovers the previous page's four-tap transfer matrix. The same delay algebra explains both the off-diagonal phase swap and the channel powers.\n\\head{Signed delays need the same care}\nFor an original-sample delay $d$ at modulus $N$,\n\\[\ny_k[n]=f_{(k-d)\\bmod N}\n\\!\\left[n+\\left\\lfloor\\frac{k-d}{N}\\right\\rfloor\\right].\n\\]\nThis includes advances when $d<0$. The finite verification checks the formula on signed sample addresses. A bank with product $D^qI_N$ delays the original signal by $Nq$ samples, not $q$ samples.\n\\takeaway{Always state the unit of a delay. A phas…","ay swaps the two phases and wraps one of them into the preceding channel index. Its matrix is $J(D)$, not $DI_2$.\n\\vfig{sample_delay}\n\\head{Trace one impulse}\nThe input $\\delta_0$ has a single nonzero value in even channel address zero. Under $J(D)$ it appears in odd channel address zero: this is the original sample $\\delta_1$. Under $DI_2$ it stays even but moves to channel address one: this is $\\delta_2$.\nThus\n\\[\nJ(D)^2=DI_2,\n\\qquad\nH(D)=I_2+2J(D)-J(D)^2+J(D)^3\n\\]\nrecovers the previous page's four-tap transfer matrix. The same delay algebra explains both the off-diagonal phase swap and the channel powers.\n\\head{Signed delays need the same care}\nFor an original-sample delay $d$ at modulus $N$,\n\\[\ny_k[n]=f_{(k-d)\\bmod N}\n\\!\\left[n+\\left\\lfloor\\frac{k-d}{N}\\right\\rfloor\\right].\n\\]\nThis includes advances when $d<0$. The finite verification checks the formula on signed sample addresses. A bank with product $D^qI_N$ delays the original signal by $Nq$ samples, not $q$ samples.\n\\takeaway{Always state the unit of a delay. A phase permutation and a channel delay together can represent one or…","hases and wraps one of them into the preceding channel index. Its matrix is $J(D)$, not $DI_2$.\n\\vfig{sample_delay}\n\\head{Trace one impulse}\nThe input $\\delta_0$ has a single nonzero value in even channel address zero. Under $J(D)$ it appears in odd channel address zero: this is the original sample $\\delta_1$. Under $DI_2$ it stays even but moves to channel address one: this is $\\delta_2$.\nThus\n\\[\nJ(D)^2=DI_2,\n\\qquad\nH(D)=I_2+2J(D)-J(D)^2+J(D)^3\n\\]\nrecovers the previous page's four-tap transfer matrix. The same delay algebra explains both the off-diagonal phase swap and the channel powers.\n\\head{Signed delays need the same care}\nFor an original-sample delay $d$ at modulus $N$,\n\\[\ny_k[n]=f_{(k-d)\\bmod N}\n\\!\\left[n+\\left\\lfloor\\frac{k-d}{N}\\right\\rfloor\\right].\n\\]\nThis includes advances when $d<0$. The finite verification checks the formula on signed sample addresses. A bank with product $D^qI_N$ delays the original signal by $Nq$ samples, not $q$ samples.\n\\takeaway{Always state the unit of a delay. A phase permutation and a channel delay together can represent one original sample.}","ero: this is the original sample $\\delta_1$. Under $DI_2$ it stays even but moves to channel address one: this is $\\delta_2$.\nThus\n\\[\nJ(D)^2=DI_2,\n\\qquad\nH(D)=I_2+2J(D)-J(D)^2+J(D)^3\n\\]\nrecovers the previous page's four-tap transfer matrix. The same delay algebra explains both the off-diagonal phase swap and the channel powers.\n\\head{Signed delays need the same care}\nFor an original-sample delay $d$ at modulus $N$,\n\\[\ny_k[n]=f_{(k-d)\\bmod N}\n\\!\\left[n+\\left\\lfloor\\frac{k-d}{N}\\right\\rfloor\\right].\n\\]\nThis includes advances when $d<0$. The finite verification checks the formula on signed sample addresses. A bank with product $D^qI_N$ delays the original signal by $Nq$ samples, not $q$ samples.\n\\takeaway{Always state the unit of a delay. A phase permutation and a channel delay together can represent one original sample.}","e original sample $\\delta_1$. Under $DI_2$ it stays even but moves to channel address one: this is $\\delta_2$.\nThus\n\\[\nJ(D)^2=DI_2,\n\\qquad\nH(D)=I_2+2J(D)-J(D)^2+J(D)^3\n\\]\nrecovers the previous page's four-tap transfer matrix. The same delay algebra explains both the off-diagonal phase swap and the channel powers.\n\\head{Signed delays need the same care}\nFor an original-sample delay $d$ at modulus $N$,\n\\[\ny_k[n]=f_{(k-d)\\bmod N}\n\\!\\left[n+\\left\\lfloor\\frac{k-d}{N}\\right\\rfloor\\right].\n\\]\nThis includes advances when $d<0$. The finite verification checks the formula on signed sample addresses. A bank with product $D^qI_N$ delays the original signal by $Nq$ samples, not $q$ samples.\n\\takeaway{Always state the unit of a delay. A phase permutation and a channel delay together can represent one original sample.}","inal sample $\\delta_1$. Under $DI_2$ it stays even but moves to channel address one: this is $\\delta_2$.\nThus\n\\[\nJ(D)^2=DI_2,\n\\qquad\nH(D)=I_2+2J(D)-J(D)^2+J(D)^3\n\\]\nrecovers the previous page's four-tap transfer matrix. The same delay algebra explains both the off-diagonal phase swap and the channel powers.\n\\head{Signed delays need the same care}\nFor an original-sample delay $d$ at modulus $N$,\n\\[\ny_k[n]=f_{(k-d)\\bmod N}\n\\!\\left[n+\\left\\lfloor\\frac{k-d}{N}\\right\\rfloor\\right].\n\\]\nThis includes advances when $d<0$. The finite verification checks the formula on signed sample addresses. A bank with product $D^qI_N$ delays the original signal by $Nq$ samples, not $q$ samples.\n\\takeaway{Always state the unit of a delay. A phase permutation and a channel delay together can represent one original sample.}","(D)^2=DI_2,\n\\qquad\nH(D)=I_2+2J(D)-J(D)^2+J(D)^3\n\\]\nrecovers the previous page's four-tap transfer matrix. The same delay algebra explains both the off-diagonal phase swap and the channel powers.\n\\head{Signed delays need the same care}\nFor an original-sample delay $d$ at modulus $N$,\n\\[\ny_k[n]=f_{(k-d)\\bmod N}\n\\!\\left[n+\\left\\lfloor\\frac{k-d}{N}\\right\\rfloor\\right].\n\\]\nThis includes advances when $d<0$. The finite verification checks the formula on signed sample addresses. A bank with product $D^qI_N$ delays the original signal by $Nq$ samples, not $q$ samples.\n\\takeaway{Always state the unit of a delay. A phase permutation and a channel delay together can represent one original sample.}","matrix. The same delay algebra explains both the off-diagonal phase swap and the channel powers.\n\\head{Signed delays need the same care}\nFor an original-sample delay $d$ at modulus $N$,\n\\[\ny_k[n]=f_{(k-d)\\bmod N}\n\\!\\left[n+\\left\\lfloor\\frac{k-d}{N}\\right\\rfloor\\right].\n\\]\nThis includes advances when $d<0$. The finite verification checks the formula on signed sample addresses. A bank with product $D^qI_N$ delays the original signal by $Nq$ samples, not $q$ samples.\n\\takeaway{Always state the unit of a delay. A phase permutation and a channel delay together can represent one original sample.}","both the off-diagonal phase swap and the channel powers.\n\\head{Signed delays need the same care}\nFor an original-sample delay $d$ at modulus $N$,\n\\[\ny_k[n]=f_{(k-d)\\bmod N}\n\\!\\left[n+\\left\\lfloor\\frac{k-d}{N}\\right\\rfloor\\right].\n\\]\nThis includes advances when $d<0$. The finite verification checks the formula on signed sample addresses. A bank with product $D^qI_N$ delays the original signal by $Nq$ samples, not $q$ samples.\n\\takeaway{Always state the unit of a delay. A phase permutation and a channel delay together can represent one original sample.}","onal phase swap and the channel powers.\n\\head{Signed delays need the same care}\nFor an original-sample delay $d$ at modulus $N$,\n\\[\ny_k[n]=f_{(k-d)\\bmod N}\n\\!\\left[n+\\left\\lfloor\\frac{k-d}{N}\\right\\rfloor\\right].\n\\]\nThis includes advances when $d<0$. The finite verification checks the formula on signed sample addresses. A bank with product $D^qI_N$ delays the original signal by $Nq$ samples, not $q$ samples.\n\\takeaway{Always state the unit of a delay. A phase permutation and a channel delay together can represent one original sample.}","\\vpage{7}{Perfect reconstruction is an operator identity}{FILTER BANKS / WHY A FREQUENCY SPOT-CHECK IS INSUFFICIENT}{Traditional monograph: Theorem 4.1 (p. 7).}\nConsider the finite Laurent bank\n\\[\nE(D)=\\mat{1&D\\\\0&1},\\qquad\nR(D)=\\mat{1&-D\\\\0&1}.\n\\]\nIts synthesis cancels the exact delayed cross-channel contribution introduced by analysis.\n\\vfig{perfect_reconstruction}\n\\head{Two inverses, all inputs}\nDirect polynomial multiplication gives $RE=ER=I_2$. Since $D$ acts as a shift, these identities imply perfect reconstruction on every bilateral channel sequence. With finite kernels, composition requires only finitely many operations at each address.\nThe converse uses impulses. Insert an impulse in one input channel. Each output channel lists every coefficient of the corresponding transfer polynomial. If all such responses vanish for $RE-I$, every coefficient is zero; hence the polynomial matrix itself is zero.\n\\head{A deliberately mislea","0&1}.\n\\]\nIts synthesis cancels the exact delayed cross-channel contribution introduced by analysis.\n\\vfig{perfect_reconstruction}\n\\head{Two inverses, all inputs}\nDirect polynomial multiplication gives $RE=ER=I_2$. Since $D$ acts as a shift, these identities imply perfect reconstruction on every bilateral channel sequence. With finite kernels, composition requires only finitely many operations at each address.\nThe converse uses impulses. Insert an impulse in one input channel. Each output channel lists every coefficient of the corresponding transfer polynomial. If all such responses vanish for $RE-I$, every coefficient is zero; hence the polynomial matrix itself is zero.\n\\head{A deliberately misleading test}\nThe matrix $B(D)=\\operatorname{diag}(D,1)$ satisfies $B(1)=I_2$. At that one transform value it appears perfect. In the time domain, however, it delays only the first channel. An impulse in that","0&1}.\n\\]\nIts synthesis cancels the exact delayed cross-channel contribution introduced by analysis.\n\\vfig{perfect_reconstruction}\n\\head{Two inverses, all inputs}\nDirect polynomial multiplication gives $RE=ER=I_2$. Since $D$ acts as a shift, these identities imply perfect reconstruction on every bilateral channel sequence. With finite kernels, composition requires only finitely many operations at each address.\nThe converse uses impulses. Insert an impulse in one input channel. Each output channel lists every coefficient of the corresponding transfer polynomial. If all such responses vanish for $RE-I$, every coefficient is zero; hence the polynomial matrix itself is zero.\n\\head{A deliberately misleading test}\nThe matrix $B(D)=\\operatorname{diag}(D,1)$ satisfies $B(1)=I_2$. At that one transform value it appears perfect. In the time domain, however, it delays only the first channel. An impulse in that channel im","$RE=ER=I_2$. Since $D$ acts as a shift, these identities imply perfect reconstruction on every bilateral channel sequence. With finite kernels, composition requires only finitely many operations at each address.\nThe converse uses impulses. Insert an impulse in one input channel. Each output channel lists every coefficient of the corresponding transfer polynomial. If all such responses vanish for $RE-I$, every coefficient is zero; hence the polynomial matrix itself is zero.\n\\head{A deliberately misleading test}\nThe matrix $B(D)=\\operatorname{diag}(D,1)$ satisfies $B(1)=I_2$. At that one transform value it appears perfect. In the time domain, however, it delays only the first channel. An impulse in that channel immediately distinguishes it from the identity.\nThis is why a small set of successful frequency evaluations is not automatically a proof of the full Laurent identity. A separate degree bound and a sufficient interpolation argument would be needed to turn finitely many evaluations into such a certificate.\n\\takeaway{Exact reconstruction concerns every coefficient of the transfer o…","finite kernels, composition requires only finitely many operations at each address.\nThe converse uses impulses. Insert an impulse in one input channel. Each output channel lists every coefficient of the corresponding transfer polynomial. If all such responses vanish for $RE-I$, every coefficient is zero; hence the polynomial matrix itself is zero.\n\\head{A deliberately misleading test}\nThe matrix $B(D)=\\operatorname{diag}(D,1)$ satisfies $B(1)=I_2$. At that one transform value it appears perfect. In the time domain, however, it delays only the first channel. An impulse in that channel immediately distinguishes it from the identity.\nThis is why a small set of successful frequency evaluations is not automatically a proof of the full Laurent identity. A separate degree bound and a sufficient interpolation argument would be needed to turn finitely many evaluations into such a certificate.\n\\takeaway{Exact reconstruction concerns every coefficient of the transfer operator, not just an attractive response curve or one evaluation point.}","finitely many operations at each address.\nThe converse uses impulses. Insert an impulse in one input channel. Each output channel lists every coefficient of the corresponding transfer polynomial. If all such responses vanish for $RE-I$, every coefficient is zero; hence the polynomial matrix itself is zero.\n\\head{A deliberately misleading test}\nThe matrix $B(D)=\\operatorname{diag}(D,1)$ satisfies $B(1)=I_2$. At that one transform value it appears perfect. In the time domain, however, it delays only the first channel. An impulse in that channel immediately distinguishes it from the identity.\nThis is why a small set of successful frequency evaluations is not automatically a proof of the full Laurent identity. A separate degree bound and a sufficient interpolation argument would be needed to turn finitely many evaluations into such a certificate.\n\\takeaway{Exact reconstruction concerns every coefficient of the transfer operator, not just an attractive response curve or one evaluation point.}","\\vpage{8}{A jump can hide from the first Haar split}{HAAR ENERGY / EDGE ALIGNMENT}{Traditional monograph: Corollary 4.2 (p. 8).}\nFor the running signal, Haar analysis gives\n\\[\na[n]=\\frac{f[2n]+f[2n+1]}{\\sqrt2},\\qquad\nd[n]=\\frac{f[2n]-f[2n+1]}{\\sqrt2}.\n\\]\nThe plot scales both channels by $\\sqrt2$ so that all displayed ordinates are rational.\n\\vfig{haar_alignment}\n\\head{The exact energy calculation}\nThe low channel has energy $227/4$, and the detail channel has energy $1/4$. Their sum is $57$, exactly the input energy. The normalization is essential: unnormalized sums and differences would multiply the total energy by two.\n\\head{Why is the first detail channel so small?}\nThe jump starts at sample $6$ and ends just before sample $12$. Both lie \\emph{between} the pairs $(0,1),(2,3),\\ldots,(14,15)$. Inside every pair the jump contributes either $(0,0)$ or $(3,3)$, so its first-level difference is zero. The small detail values come entirely fro","\\vpage{8}{A jump can hide from the first Haar split}{HAAR ENERGY / EDGE ALIGNMENT}{Traditional monograph: Corollary 4.2 (p. 8).}\nFor the running signal, Haar analysis gives\n\\[\na[n]=\\frac{f[2n]+f[2n+1]}{\\sqrt2},\\qquad\nd[n]=\\frac{f[2n]-f[2n+1]}{\\sqrt2}.\n\\]\nThe plot scales both channels by $\\sqrt2$ so that all displayed ordinates are rational.\n\\vfig{haar_alignment}\n\\head{The exact energy calculation}\nThe low channel has energy $227/4$, and the detail channel has energy $1/4$. Their sum is $57$, exactly the input energy. The normalization is essential: unnormalized sums and differences would multiply the total energy by two.\n\\head{Why is the first detail channel so small?}\nThe jump starts at sample $6$ and ends just before sample $12$. Both lie \\emph{between} the pairs $(0,1),(2,3),\\ldots,(14,15)$. Inside every pair the jump contributes either $(0,0)$ or $(3,3)$, so its first-level difference is zero. The small detail values come entirely from the discrete background.\nThis does not me","m the first Haar split}{HAAR ENERGY / EDGE ALIGNMENT}{Traditional monograph: Corollary 4.2 (p. 8).}\nFor the running signal, Haar analysis gives\n\\[\na[n]=\\frac{f[2n]+f[2n+1]}{\\sqrt2},\\qquad\nd[n]=\\frac{f[2n]-f[2n+1]}{\\sqrt2}.\n\\]\nThe plot scales both channels by $\\sqrt2$ so that all displayed ordinates are rational.\n\\vfig{haar_alignment}\n\\head{The exact energy calculation}\nThe low channel has energy $227/4$, and the detail channel has energy $1/4$. Their sum is $57$, exactly the input energy. The normalization is essential: unnormalized sums and differences would multiply the total energy by two.\n\\head{Why is the first detail channel so small?}\nThe jump starts at sample $6$ and ends just before sample $12$. Both lie \\emph{between} the pairs $(0,1),(2,3),\\ldots,(14,15)$. Inside every pair the jump contributes either $(0,0)$ or $(3,3)$, so its first-level difference is zero. The small detail values come entirely from the discrete background.\nThis does not mean the edge is absent from the transform. The next split of the low channel compares neighboring pairs, and those comparisons do cross…","ALIGNMENT}{Traditional monograph: Corollary 4.2 (p. 8).}\nFor the running signal, Haar analysis gives\n\\[\na[n]=\\frac{f[2n]+f[2n+1]}{\\sqrt2},\\qquad\nd[n]=\\frac{f[2n]-f[2n+1]}{\\sqrt2}.\n\\]\nThe plot scales both channels by $\\sqrt2$ so that all displayed ordinates are rational.\n\\vfig{haar_alignment}\n\\head{The exact energy calculation}\nThe low channel has energy $227/4$, and the detail channel has energy $1/4$. Their sum is $57$, exactly the input energy. The normalization is essential: unnormalized sums and differences would multiply the total energy by two.\n\\head{Why is the first detail channel so small?}\nThe jump starts at sample $6$ and ends just before sample $12$. Both lie \\emph{between} the pairs $(0,1),(2,3),\\ldots,(14,15)$. Inside every pair the jump contributes either $(0,0)$ or $(3,3)$, so its first-level difference is zero. The small detail values come entirely from the discrete background.\nThis does not mean the edge is absent from the transform. The next split of the low channel compares neighboring pairs, and those comparisons do cross the jump. The next pages expose that contr…","al monograph: Corollary 4.2 (p. 8).}\nFor the running signal, Haar analysis gives\n\\[\na[n]=\\frac{f[2n]+f[2n+1]}{\\sqrt2},\\qquad\nd[n]=\\frac{f[2n]-f[2n+1]}{\\sqrt2}.\n\\]\nThe plot scales both channels by $\\sqrt2$ so that all displayed ordinates are rational.\n\\vfig{haar_alignment}\n\\head{The exact energy calculation}\nThe low channel has energy $227/4$, and the detail channel has energy $1/4$. Their sum is $57$, exactly the input energy. The normalization is essential: unnormalized sums and differences would multiply the total energy by two.\n\\head{Why is the first detail channel so small?}\nThe jump starts at sample $6$ and ends just before sample $12$. Both lie \\emph{between} the pairs $(0,1),(2,3),\\ldots,(14,15)$. Inside every pair the jump contributes either $(0,0)$ or $(3,3)$, so its first-level difference is zero. The small detail values come entirely from the discrete background.\nThis does not mean the edge is absent from the transform. The next split of the low channel compares neighboring pairs, and those comparisons do cross the jump. The next pages expose that contribution explicitly.\n…","all displayed ordinates are rational.\n\\vfig{haar_alignment}\n\\head{The exact energy calculation}\nThe low channel has energy $227/4$, and the detail channel has energy $1/4$. Their sum is $57$, exactly the input energy. The normalization is essential: unnormalized sums and differences would multiply the total energy by two.\n\\head{Why is the first detail channel so small?}\nThe jump starts at sample $6$ and ends just before sample $12$. Both lie \\emph{between} the pairs $(0,1),(2,3),\\ldots,(14,15)$. Inside every pair the jump contributes either $(0,0)$ or $(3,3)$, so its first-level difference is zero. The small detail values come entirely from the discrete background.\nThis does not mean the edge is absent from the transform. The next split of the low channel compares neighboring pairs, and those comparisons do cross the jump. The next pages expose that contribution explicitly.\n\\takeaway{Haar energy is exact, but a particular detail level is not a universal edge detector. Edge visibility depends on alignment, scale, and the chosen packets.}","onal.\n\\vfig{haar_alignment}\n\\head{The exact energy calculation}\nThe low channel has energy $227/4$, and the detail channel has energy $1/4$. Their sum is $57$, exactly the input energy. The normalization is essential: unnormalized sums and differences would multiply the total energy by two.\n\\head{Why is the first detail channel so small?}\nThe jump starts at sample $6$ and ends just before sample $12$. Both lie \\emph{between} the pairs $(0,1),(2,3),\\ldots,(14,15)$. Inside every pair the jump contributes either $(0,0)$ or $(3,3)$, so its first-level difference is zero. The small detail values come entirely from the discrete background.\nThis does not mean the edge is absent from the transform. The next split of the low channel compares neighboring pairs, and those comparisons do cross the jump. The next pages expose that contribution explicitly.\n\\takeaway{Haar energy is exact, but a particular detail level is not a universal edge detector. Edge visibility depends on alignment, scale, and the chosen packets.}","act energy calculation}\nThe low channel has energy $227/4$, and the detail channel has energy $1/4$. Their sum is $57$, exactly the input energy. The normalization is essential: unnormalized sums and differences would multiply the total energy by two.\n\\head{Why is the first detail channel so small?}\nThe jump starts at sample $6$ and ends just before sample $12$. Both lie \\emph{between} the pairs $(0,1),(2,3),\\ldots,(14,15)$. Inside every pair the jump contributes either $(0,0)$ or $(3,3)$, so its first-level difference is zero. The small detail values come entirely from the discrete background.\nThis does not mean the edge is absent from the transform. The next split of the low channel compares neighboring pairs, and those comparisons do cross the jump. The next pages expose that contribution explicitly.\n\\takeaway{Haar energy is exact, but a particular detail level is not a universal edge detector. Edge visibility depends on alignment, scale, and the chosen packets.}","hannel has energy $1/4$. Their sum is $57$, exactly the input energy. The normalization is essential: unnormalized sums and differences would multiply the total energy by two.\n\\head{Why is the first detail channel so small?}\nThe jump starts at sample $6$ and ends just before sample $12$. Both lie \\emph{between} the pairs $(0,1),(2,3),\\ldots,(14,15)$. Inside every pair the jump contributes either $(0,0)$ or $(3,3)$, so its first-level difference is zero. The small detail values come entirely from the discrete background.\nThis does not mean the edge is absent from the transform. The next split of the low channel compares neighboring pairs, and those comparisons do cross the jump. The next pages expose that contribution explicitly.\n\\takeaway{Haar energy is exact, but a particular detail level is not a universal edge detector. Edge visibility depends on alignment, scale, and the chosen packets.}","energy $1/4$. Their sum is $57$, exactly the input energy. The normalization is essential: unnormalized sums and differences would multiply the total energy by two.\n\\head{Why is the first detail channel so small?}\nThe jump starts at sample $6$ and ends just before sample $12$. Both lie \\emph{between} the pairs $(0,1),(2,3),\\ldots,(14,15)$. Inside every pair the jump contributes either $(0,0)$ or $(3,3)$, so its first-level difference is zero. The small detail values come entirely from the discrete background.\nThis does not mean the edge is absent from the transform. The next split of the low channel compares neighboring pairs, and those comparisons do cross the jump. The next pages expose that contribution explicitly.\n\\takeaway{Haar energy is exact, but a particular detail level is not a universal edge detector. Edge visibility depends on alignment, scale, and the chosen packets.}","\\vpage{9}{Lifting preserves information, not energy}{PREDICT--UPDATE / INVERTIBILITY VERSUS ISOMETRY}{Traditional monograph: Theorem 4.3 (p. 10).}\nFor an even/odd pair $(e,o)$, form the prediction residual $d=o-e$, then update $s=e+d/2$. The inverse reads $e=s-d/2$, followed by $o=d+e$.\n\\vfig{lifting_geometry}\n\\platecaption{The square $[0,4]^2$ and its exact image under the lifting matrix have the same area. The separate pair $(2,6)$ illustrates that Euclidean energy need not be preserved.}\n\\[\n\\mat{s","ormation, not energy}{PREDICT--UPDATE / INVERTIBILITY VERSUS ISOMETRY}{Traditional monograph: Theorem 4.3 (p. 10).}\nFor an even/odd pair $(e,o)$, form the prediction residual $d=o-e$, then update $s=e+d/2$. The inverse reads $e=s-d/2$, followed by $o=d+e$.\n\\vfig{lifting_geometry}\n\\platecaption{The square $[0,4]^2$ and its exact image under the lifting matrix have the same area. The separate pair $(2,6)$ illustrates that Euclidean energy need not be preserved.}\n\\[\n\\mat{s","RY}{Traditional monograph: Theorem 4.3 (p. 10).}\nFor an even/odd pair $(e,o)$, form the prediction residual $d=o-e$, then update $s=e+d/2$. The inverse reads $e=s-d/2$, followed by $o=d+e$.\n\\vfig{lifting_geometry}\n\\platecaption{The square $[0,4]^2$ and its exact image under the lifting matrix have the same area. The separate pair $(2,6)$ illustrates that Euclidean energy need not be preserved.}\n\\[\n\\mat{s\\\\d}=\\mat{1/2&1/2\\\\-1&1}\\mat{e\\\\o},\n\\qquad\n\\mat{e\\\\o}=\\mat{1&-1/2\\\\1&1/2}\\mat{s\\\\d}.\n\\]\n\\head{What determinant one does establish}\nThe matrix is invertible, and in these real coordinates it preserves oriented area. The predict and update operations can be undone in reverse order. Applied to each pair, the construction gives exact finite-signal reconstruction.\n\\head{What determinant one does not establish}\nThe pair $(2,6)$ becomes $(4,4)$. Its squared Euclidean norm changes from $40$ to $32$. The lifting matrix is not unitary. Energy accounting must therefore use its actual coordinate metric or a separately normalized transform; invertibility alone cannot replace the Haar energy identi…","d}.\n\\]\n\\head{What determinant one does establish}\nThe matrix is invertible, and in these real coordinates it preserves oriented area. The predict and update operations can be undone in reverse order. Applied to each pair, the construction gives exact finite-signal reconstruction.\n\\head{What determinant one does not establish}\nThe pair $(2,6)$ becomes $(4,4)$. Its squared Euclidean norm changes from $40$ to $32$. The lifting matrix is not unitary. Energy accounting must therefore use its actual coordinate metric or a separately normalized transform; invertibility alone cannot replace the Haar energy identity.\nAll values in this example are rational. Integer-to-integer lifting with a rounding rule is a different algorithm. Its inverse and error conventions require their own specification and are not silently inferred from the displayed rational matrix.\n\\takeaway{Perfect reconstruction, determinant one, and norm preservation are different properties. This transform has the first two, not the third.}","\\]\n\\head{What determinant one does establish}\nThe matrix is invertible, and in these real coordinates it preserves oriented area. The predict and update operations can be undone in reverse order. Applied to each pair, the construction gives exact finite-signal reconstruction.\n\\head{What determinant one does not establish}\nThe pair $(2,6)$ becomes $(4,4)$. Its squared Euclidean norm changes from $40$ to $32$. The lifting matrix is not unitary. Energy accounting must therefore use its actual coordinate metric or a separately normalized transform; invertibility alone cannot replace the Haar energy identity.\nAll values in this example are rational. Integer-to-integer lifting with a rounding rule is a different algorithm. Its inverse and error conventions require their own specification and are not silently inferred from the displayed rational matrix.\n\\takeaway{Perfect reconstruction, determinant one, and norm preservation are different properties. This transform has the first two, not the third.}","d{What determinant one does establish}\nThe matrix is invertible, and in these real coordinates it preserves oriented area. The predict and update operations can be undone in reverse order. Applied to each pair, the construction gives exact finite-signal reconstruction.\n\\head{What determinant one does not establish}\nThe pair $(2,6)$ becomes $(4,4)$. Its squared Euclidean norm changes from $40$ to $32$. The lifting matrix is not unitary. Energy accounting must therefore use its actual coordinate metric or a separately normalized transform; invertibility alone cannot replace the Haar energy identity.\nAll values in this example are rational. Integer-to-integer lifting with a rounding rule is a different algorithm. Its inverse and error conventions require their own specification and are not silently inferred from the displayed rational matrix.\n\\takeaway{Perfect reconstruction, determinant one, and norm preservation are different properties. This transform has the first two, not the third.}","\\vpage{10}{The second level resolves the hidden change}{ACTUAL WAVELET PACKETS / FOUR RATIONAL CHANNELS}{Traditional monograph: Theorem 5.1 (p. 10).}\nApply Haar analysis to \\emph{both} children. A binary address records the chronological low/detail choices, with $0$ for low and $1$ for detail. Each depth-two channel contains four coefficients.\n\\vfig{depth_two}\n\\head{The edge that first-level detail missed}\nPacket $01$ is the detail child of the first low channel. It contains $-11/4$, the large change between neighboring pair averages. Packet $10$ vanishes for this particular signal because the first-level differences alternate in cancelling pairs. Neither feature is a universal identity for all inputs.\nThe packet energies are\n\\[\n49,\\qquad 31/4,\\qquad 0,\\qquad 1/4,\n\\]\nwhich again sum to $57$. Two factors of $1/\\sqrt2$ explain why the depth-two coefficients are rational halves and quarters.\n\\head{Coefficients and projectors are different objects}\n$W_\\al","\\vpage{10}{The second level resolves the hidden change}{ACTUAL WAVELET PACKETS / FOUR RATIONAL CHANNELS}{Traditional monograph: Theorem 5.1 (p. 10).}\nApply Haar analysis to \\emph{both} children. A binary address records the chronological low/detail choices, with $0$ for low and $1$ for detail. Each depth-two channel contains four coefficients.\n\\vfig{depth_two}\n\\head{The edge that first-level detail missed}\nPacket $01$ is the detail child of the first low channel. It contains $-11/4$, the large change between neighboring pair averages. Packet $10$ vanishes for this particular signal because the first-level differences alternate in cancelling pairs. Neither feature is a universal identity for all inputs.\nThe packet energies are\n\\[\n49,\\qquad 31/4,\\qquad 0,\\qquad 1/4,\n\\]\nwhich again sum to $57$. Two factors of $1/\\sqrt2$ explain why the depth-two coefficients are rational halves and quarters.\n\\head{Coefficients and projectors are different objects}\n$W_\\alpha f$ is the sh","ond level resolves the hidden change}{ACTUAL WAVELET PACKETS / FOUR RATIONAL CHANNELS}{Traditional monograph: Theorem 5.1 (p. 10).}\nApply Haar analysis to \\emph{both} children. A binary address records the chronological low/detail choices, with $0$ for low and $1$ for detail. Each depth-two channel contains four coefficients.\n\\vfig{depth_two}\n\\head{The edge that first-level detail missed}\nPacket $01$ is the detail child of the first low channel. It contains $-11/4$, the large change between neighboring pair averages. Packet $10$ vanishes for this particular signal because the first-level differences alternate in cancelling pairs. Neither feature is a universal identity for all inputs.\nThe packet energies are\n\\[\n49,\\qquad 31/4,\\qquad 0,\\qquad 1/4,\n\\]\nwhich again sum to $57$. Two factors of $1/\\sqrt2$ explain why the depth-two coefficients are rational halves and quarters.\n\\head{Coefficients and projectors are different objects}\n$W_\\alpha f$ is the short coefficient vector. The original-space component is $W_\\alpha^*W_\\alpha f$. Summing these synthesized components, not the short array…","FOUR RATIONAL CHANNELS}{Traditional monograph: Theorem 5.1 (p. 10).}\nApply Haar analysis to \\emph{both} children. A binary address records the chronological low/detail choices, with $0$ for low and $1$ for detail. Each depth-two channel contains four coefficients.\n\\vfig{depth_two}\n\\head{The edge that first-level detail missed}\nPacket $01$ is the detail child of the first low channel. It contains $-11/4$, the large change between neighboring pair averages. Packet $10$ vanishes for this particular signal because the first-level differences alternate in cancelling pairs. Neither feature is a universal identity for all inputs.\nThe packet energies are\n\\[\n49,\\qquad 31/4,\\qquad 0,\\qquad 1/4,\n\\]\nwhich again sum to $57$. Two factors of $1/\\sqrt2$ explain why the depth-two coefficients are rational halves and quarters.\n\\head{Coefficients and projectors are different objects}\n$W_\\alpha f$ is the short coefficient vector. The original-space component is $W_\\alpha^*W_\\alpha f$. Summing these synthesized components, not the short arrays entrywise, reconstructs $f$:\n\\[\nf=\\sum_{|\\alpha|=2}W_\\alpha^*…","Apply Haar analysis to \\emph{both} children. A binary address records the chronological low/detail choices, with $0$ for low and $1$ for detail. Each depth-two channel contains four coefficients.\n\\vfig{depth_two}\n\\head{The edge that first-level detail missed}\nPacket $01$ is the detail child of the first low channel. It contains $-11/4$, the large change between neighboring pair averages. Packet $10$ vanishes for this particular signal because the first-level differences alternate in cancelling pairs. Neither feature is a universal identity for all inputs.\nThe packet energies are\n\\[\n49,\\qquad 31/4,\\qquad 0,\\qquad 1/4,\n\\]\nwhich again sum to $57$. Two factors of $1/\\sqrt2$ explain why the depth-two coefficients are rational halves and quarters.\n\\head{Coefficients and projectors are different objects}\n$W_\\alpha f$ is the short coefficient vector. The original-space component is $W_\\alpha^*W_\\alpha f$. Summing these synthesized components, not the short arrays entrywise, reconstructs $f$:\n\\[\nf=\\sum_{|\\alpha|=2}W_\\alpha^*W_\\alpha f.\n\\]\nThe finite verification constructs the actual analysis","cients.\n\\vfig{depth_two}\n\\head{The edge that first-level detail missed}\nPacket $01$ is the detail child of the first low channel. It contains $-11/4$, the large change between neighboring pair averages. Packet $10$ vanishes for this particular signal because the first-level differences alternate in cancelling pairs. Neither feature is a universal identity for all inputs.\nThe packet energies are\n\\[\n49,\\qquad 31/4,\\qquad 0,\\qquad 1/4,\n\\]\nwhich again sum to $57$. Two factors of $1/\\sqrt2$ explain why the depth-two coefficients are rational halves and quarters.\n\\head{Coefficients and projectors are different objects}\n$W_\\alpha f$ is the short coefficient vector. The original-space component is $W_\\alpha^*W_\\alpha f$. Summing these synthesized components, not the short arrays entrywise, reconstructs $f$:\n\\[\nf=\\sum_{|\\alpha|=2}W_\\alpha^*W_\\alpha f.\n\\]\nThe finite verification constructs the actual analysis matrices and checks their product with their adjoints. The paper's $\\ell^2$ statement additionally establishes boundedness and the existence of the literal Hilbert adjoints.\n\\takeaway{A p…","ail missed}\nPacket $01$ is the detail child of the first low channel. It contains $-11/4$, the large change between neighboring pair averages. Packet $10$ vanishes for this particular signal because the first-level differences alternate in cancelling pairs. Neither feature is a universal identity for all inputs.\nThe packet energies are\n\\[\n49,\\qquad 31/4,\\qquad 0,\\qquad 1/4,\n\\]\nwhich again sum to $57$. Two factors of $1/\\sqrt2$ explain why the depth-two coefficients are rational halves and quarters.\n\\head{Coefficients and projectors are different objects}\n$W_\\alpha f$ is the short coefficient vector. The original-space component is $W_\\alpha^*W_\\alpha f$. Summing these synthesized components, not the short arrays entrywise, reconstructs $f$:\n\\[\nf=\\sum_{|\\alpha|=2}W_\\alpha^*W_\\alpha f.\n\\]\nThe finite verification constructs the actual analysis matrices and checks their product with their adjoints. The paper's $\\ell^2$ statement additionally establishes boundedness and the existence of the literal Hilbert adjoints.\n\\takeaway{A packet address specifies an operator word. To add different p…","1$ is the detail child of the first low channel. It contains $-11/4$, the large change between neighboring pair averages. Packet $10$ vanishes for this particular signal because the first-level differences alternate in cancelling pairs. Neither feature is a universal identity for all inputs.\nThe packet energies are\n\\[\n49,\\qquad 31/4,\\qquad 0,\\qquad 1/4,\n\\]\nwhich again sum to $57$. Two factors of $1/\\sqrt2$ explain why the depth-two coefficients are rational halves and quarters.\n\\head{Coefficients and projectors are different objects}\n$W_\\alpha f$ is the short coefficient vector. The original-space component is $W_\\alpha^*W_\\alpha f$. Summing these synthesized components, not the short arrays entrywise, reconstructs $f$:\n\\[\nf=\\sum_{|\\alpha|=2}W_\\alpha^*W_\\alpha f.\n\\]\nThe finite verification constructs the actual analysis matrices and checks their product with their adjoints. The paper's $\\ell^2$ statement additionally establishes boundedness and the existence of the literal Hilbert adjoints.\n\\takeaway{A packet address specifies an operator word. To add different packets as signals, sy…","s for this particular signal because the first-level differences alternate in cancelling pairs. Neither feature is a universal identity for all inputs.\nThe packet energies are\n\\[\n49,\\qquad 31/4,\\qquad 0,\\qquad 1/4,\n\\]\nwhich again sum to $57$. Two factors of $1/\\sqrt2$ explain why the depth-two coefficients are rational halves and quarters.\n\\head{Coefficients and projectors are different objects}\n$W_\\alpha f$ is the short coefficient vector. The original-space component is $W_\\alpha^*W_\\alpha f$. Summing these synthesized components, not the short arrays entrywise, reconstructs $f$:\n\\[\nf=\\sum_{|\\alpha|=2}W_\\alpha^*W_\\alpha f.\n\\]\nThe finite verification constructs the actual analysis matrices and checks their product with their adjoints. The paper's $\\ell^2$ statement additionally establishes boundedness and the existence of the literal Hilbert adjoints.\n\\takeaway{A packet address specifies an operator word. To add different packets as signals, synthesize them back into the common signal space first.}","cancelling pairs. Neither feature is a universal identity for all inputs.\nThe packet energies are\n\\[\n49,\\qquad 31/4,\\qquad 0,\\qquad 1/4,\n\\]\nwhich again sum to $57$. Two factors of $1/\\sqrt2$ explain why the depth-two coefficients are rational halves and quarters.\n\\head{Coefficients and projectors are different objects}\n$W_\\alpha f$ is the short coefficient vector. The original-space component is $W_\\alpha^*W_\\alpha f$. Summing these synthesized components, not the short arrays entrywise, reconstructs $f$:\n\\[\nf=\\sum_{|\\alpha|=2}W_\\alpha^*W_\\alpha f.\n\\]\nThe finite verification constructs the actual analysis matrices and checks their product with their adjoints. The paper's $\\ell^2$ statement additionally establishes boundedness and the existence of the literal Hilbert adjoints.\n\\takeaway{A packet address specifies an operator word. To add different packets as signals, synthesize them back into the common signal space first.}","uad 31/4,\\qquad 0,\\qquad 1/4,\n\\]\nwhich again sum to $57$. Two factors of $1/\\sqrt2$ explain why the depth-two coefficients are rational halves and quarters.\n\\head{Coefficients and projectors are different objects}\n$W_\\alpha f$ is the short coefficient vector. The original-space component is $W_\\alpha^*W_\\alpha f$. Summing these synthesized components, not the short arrays entrywise, reconstructs $f$:\n\\[\nf=\\sum_{|\\alpha|=2}W_\\alpha^*W_\\alpha f.\n\\]\nThe finite verification constructs the actual analysis matrices and checks their product with their adjoints. The paper's $\\ell^2$ statement additionally establishes boundedness and the existence of the literal Hilbert adjoints.\n\\takeaway{A packet address specifies an operator word. To add different packets as signals, synthesize them back into the common signal space first.}","/4,\\qquad 0,\\qquad 1/4,\n\\]\nwhich again sum to $57$. Two factors of $1/\\sqrt2$ explain why the depth-two coefficients are rational halves and quarters.\n\\head{Coefficients and projectors are different objects}\n$W_\\alpha f$ is the short coefficient vector. The original-space component is $W_\\alpha^*W_\\alpha f$. Summing these synthesized components, not the short arrays entrywise, reconstructs $f$:\n\\[\nf=\\sum_{|\\alpha|=2}W_\\alpha^*W_\\alpha f.\n\\]\nThe finite verification constructs the actual analysis matrices and checks their product with their adjoints. The paper's $\\ell^2$ statement additionally establishes boundedness and the existence of the literal Hilbert adjoints.\n\\takeaway{A packet address specifies an operator word. To add different packets as signals, synthesize them back into the common signal space first.}","s and projectors are different objects}\n$W_\\alpha f$ is the short coefficient vector. The original-space component is $W_\\alpha^*W_\\alpha f$. Summing these synthesized components, not the short arrays entrywise, reconstructs $f$:\n\\[\nf=\\sum_{|\\alpha|=2}W_\\alpha^*W_\\alpha f.\n\\]\nThe finite verification constructs the actual analysis matrices and checks their product with their adjoints. The paper's $\\ell^2$ statement additionally establishes boundedness and the existence of the literal Hilbert adjoints.\n\\takeaway{A packet address specifies an operator word. To add different packets as signals, synthesize them back into the common signal space first.}","discard some of its sixteen coefficients. Order them by squared magnitude, with a deterministic address-and-position rule for ties.\n\\vfig{sparse_tail}\n\\head{Every discarded square has a meaning}\nIn this fixed orthonormal basis, the squared reconstruction error equals the sum of the squared discarded coefficients. For our running signal:\n\\begin{center}\n\\begin{tabular}{r|rrrrrr}\nkept coefficients $k$ & 1&2&4&8&12&16","hline\nsquared error & $59/4$&$115/16$&$11/16$&$1/4$&$0$&$0$\n\\end{tabular}\n\\end{center}\nThe four zero coefficients belong to packet $10$, so twelve coefficients suffice for exact reconstruction here. This is a property of the selected signal and basis, not a guarantee that all edge signals need only twelve coefficients.\n\\head{Do not compare incompatible budgets}\nA cutoff $m$ in a one-dimensional signed Fourier spectrum means $2m+1$ bins, including zero. A budget of $k$ largest coefficients is a different rule. Real Fourier synthesis may also require conjugate-pair closure. Neither count alone equals a bit-storage budget once amplitudes, phases, indices, or a selected tree must be encoded.\nThe manuscript's Fourier-versu","hline\nsquared error & $59/4$&$115/16$&$11/16$&$1/4$&$0$&$0$\n\\end{tabular}\n\\end{center}\nThe four zero coefficients belong to packet $10$, so twelve coefficients suffice for exact reconstruction here. This is a property of the selected signal and basis, not a guarantee that all edge signals need only twelve coefficients.\n\\head{Do not compare incompatible budgets}\nA cutoff $m$ in a one-dimensional signed Fourier spectrum means $2m+1$ bins, including zero. A budget of $k$ largest coefficients is a different rule. Real Fourier synthesis may also require conjugate-pair closure. Neither count alone equals a bit-storage budget once amplitudes, phases, indices, or a selected tree must be encoded.\nThe manuscript's Fourier-versus-Haar ex","hline\nsquared error & $59/4$&$115/16$&$11/16$&$1/4$&$0$&$0$\n\\end{tabular}\n\\end{center}\nThe four zero coefficients belong to packet $10$, so twelve coefficients suffice for exact reconstruction here. This is a property of the selected signal and basis, not a guarantee that all edge signals need only twelve coefficients.\n\\head{Do not compare incompatible budgets}\nA cutoff $m$ in a one-dimensional signed Fourier spectrum means $2m+1$ bins, including zero. A budget of $k$ largest coefficients is a different rule. Real Fourier synthesis may also require conjugate-pair closure. Neither count alone equals a bit-storage budget once amplitudes, phases, indices, or a selected tree must be encoded.\nThe manuscript's Fourier-versus-Haar experiment","hline\nsquared error & $59/4$&$115/16$&$11/16$&$1/4$&$0$&$0$\n\\end{tabular}\n\\end{center}\nThe four zero coefficients belong to packet $10$, so twelve coefficients suffice for exact reconstruction here. This is a property of the selected signal and basis, not a guarantee that all edge signals need only twelve coefficients.\n\\head{Do not compare incompatible budgets}\nA cutoff $m$ in a one-dimensional signed Fourier spectrum means $2m+1$ bins, including zero. A budget of $k$ largest coefficients is a different rule. Real Fourier synthesis may also require conjugate-pair closure. Neither count alone equals a bit-storage budget once amplitudes, phases, indices, or a selected tree must be encoded.\nThe manuscript's Fourier-versus-Haar experiment is a","hline\nsquared error & $59/4$&$115/16$&$11/16$&$1/4$&$0$&$0$\n\\end{tabular}\n\\end{center}\nThe four zero coefficients belong to packet $10$, so twelve coefficients suffice for exact reconstruction here. This is a property of the selected signal and basis, not a guarantee that all edge signals need only twelve coefficients.\n\\head{Do not compare incompatible budgets}\nA cutoff $m$ in a one-dimensional signed Fourier spectrum means $2m+1$ bins, including zero. A budget of $k$ largest coefficients is a different rule. Real Fourier synthesis may also require conjugate-pair closure. Neither count alone equals a bit-storage budget once amplitudes, phases, indices, or a selected tree must be encoded.\nThe manuscript's Fourier-versus-Haar experiment is a fini","hline\nsquared error & $59/4$&$115/16$&$11/16$&$1/4$&$0$&$0$\n\\end{tabular}\n\\end{center}\nThe four zero coefficients belong to packet $10$, so twelve coefficients suffice for exact reconstruction here. This is a property of the selected signal and basis, not a guarantee that all edge signals need only twelve coefficients.\n\\head{Do not compare incompatible budgets}\nA cutoff $m$ in a one-dimensional signed Fourier spectrum means $2m+1$ bins, including zero. A budget of $k$ largest coefficients is a different rule. Real Fourier synthesis may also require conjugate-pair closure. Neither count alone equals a bit-storage budget once amplitudes, phases, indices, or a selected tree must be encoded.\nThe manuscript's Fourier-versus-Haar experiment is a finite c","hline\nsquared error & $59/4$&$115/16$&$11/16$&$1/4$&$0$&$0$\n\\end{tabular}\n\\end{center}\nThe four zero coefficients belong to packet $10$, so twelve coefficients suffice for exact reconstruction here. This is a property of the selected signal and basis, not a guarantee that all edge signals need only twelve coefficients.\n\\head{Do not compare incompatible budgets}\nA cutoff $m$ in a one-dimensional signed Fourier spectrum means $2m+1$ bins, including zero. A budget of $k$ largest coefficients is a different rule. Real Fourier synthesis may also require conjugate-pair closure. Neither count alone equals a bit-storage budget once amplitudes, phases, indices, or a selected tree must be encoded.\nThe manuscript's Fourier-versus-Haar experiment is a finite comparison, not a proof that one basis dominates every signal class. Here we","hline\nsquared error & $59/4$&$115/16$&$11/16$&$1/4$&$0$&$0$\n\\end{tabular}\n\\end{center}\nThe four zero coefficients belong to packet $10$, so twelve coefficients suffice for exact reconstruction here. This is a property of the selected signal and basis, not a guarantee that all edge signals need only twelve coefficients.\n\\head{Do not compare incompatible budgets}\nA cutoff $m$ in a one-dimensional signed Fourier spectrum means $2m+1$ bins, including zero. A budget of $k$ largest coefficients is a different rule. Real Fourier synthesis may also require conjugate-pair closure. Neither count alone equals a bit-storage budget once amplitudes, phases, indices, or a selected tree must be encoded.\nThe manuscript's Fourier-versus-Haar experiment is a finite comparison, not a proof that one basis dominates every signal class. Here we isolate the part that can be checked exactly: reconstruction error within one fixed finite basis.\n\\takeaway{Changing coordinates creates no error. Throwing away coefficients does, and orthogonality makes that error exactly accountable.}","$115/16$&$11/16$&$1/4$&$0$&$0$\n\\end{tabular}\n\\end{center}\nThe four zero coefficients belong to packet $10$, so twelve coefficients suffice for exact reconstruction here. This is a property of the selected signal and basis, not a guarantee that all edge signals need only twelve coefficients.\n\\head{Do not compare incompatible budgets}\nA cutoff $m$ in a one-dimensional signed Fourier spectrum means $2m+1$ bins, including zero. A budget of $k$ largest coefficients is a different rule. Real Fourier synthesis may also require conjugate-pair closure. Neither count alone equals a bit-storage budget once amplitudes, phases, indices, or a selected tree must be encoded.\nThe manuscript's Fourier-versus-Haar experiment is a finite comparison, not a proof that one basis dominates every signal class. Here we isolate the part that can be checked exactly: reconstruction error within one fixed finite basis.\n\\takeaway{Changing coordinates creates no error. Throwing away coefficients does, and orthogonality makes that error exactly accountable.}","ar}\n\\end{center}\nThe four zero coefficients belong to packet $10$, so twelve coefficients suffice for exact reconstruction here. This is a property of the selected signal and basis, not a guarantee that all edge signals need only twelve coefficients.\n\\head{Do not compare incompatible budgets}\nA cutoff $m$ in a one-dimensional signed Fourier spectrum means $2m+1$ bins, including zero. A budget of $k$ largest coefficients is a different rule. Real Fourier synthesis may also require conjugate-pair closure. Neither count alone equals a bit-storage budget once amplitudes, phases, indices, or a selected tree must be encoded.\nThe manuscript's Fourier-versus-Haar experiment is a finite comparison, not a proof that one basis dominates every signal class. Here we isolate the part that can be checked exactly: reconstruction error within one fixed finite basis.\n\\takeaway{Changing coordinates creates no error. Throwing away coefficients does, and orthogonality makes that error exactly accountable.}","UNTING}{Traditional monograph: Definition 11.1 (p. 19), Theorem 11.2 (p. 20).}\nA complete pruning chooses nodes of a finite refinement tree so that every root-to-terminal path meets exactly one selected node. The horizontal positions below are \\emph{terminal packet addresses}, not time intervals in the original signal.\n\\vfig{complete_cuts}\n\\head{Three candidate selections}\nThe complete selection $\\{0,10,11\\}$ covers every terminal branch once. The selection $\\{0,00,1\\}$ covers the $00$ branch twice: it is not an antichain. The selection $\\{00,01,10\\}$ misses the $11$ branch entirely.\nFor actual Haar packets, set $P_v=W_v^*W_v$. Parent-to-child conservation gives\n\\[\nP_v=P_{v0}+P_{v1}.\n\\]\nReplacing unselected parents by their children until the complete pruning is reached proves $\\sum_{v\\in\\Lambda}P_v=I$. This is why the coefficients at different depths still form an exact coordinate system.\n\\head{An exhaustive finite check}\nAt depth three the pruning count obeys $a_0=1$ and $a_{d+1}=1+a_d^2$, giving $1,2,5,26$. The supplied suite checks all $26$ covering conditions and all $26$ actual…","11.2 (p. 20).}\nA complete pruning chooses nodes of a finite refinement tree so that every root-to-terminal path meets exactly one selected node. The horizontal positions below are \\emph{terminal packet addresses}, not time intervals in the original signal.\n\\vfig{complete_cuts}\n\\head{Three candidate selections}\nThe complete selection $\\{0,10,11\\}$ covers every terminal branch once. The selection $\\{0,00,1\\}$ covers the $00$ branch twice: it is not an antichain. The selection $\\{00,01,10\\}$ misses the $11$ branch entirely.\nFor actual Haar packets, set $P_v=W_v^*W_v$. Parent-to-child conservation gives\n\\[\nP_v=P_{v0}+P_{v1}.\n\\]\nReplacing unselected parents by their children until the complete pruning is reached proves $\\sum_{v\\in\\Lambda}P_v=I$. This is why the coefficients at different depths still form an exact coordinate system.\n\\head{An exhaustive finite check}\nAt depth three the pruning count obeys $a_0=1$ and $a_{d+1}=1+a_d^2$, giving $1,2,5,26$. The supplied suite checks all $26$ covering conditions and all $26$ actual $16\\times16$ projector-sum identities. These finite checks illu…","te pruning chooses nodes of a finite refinement tree so that every root-to-terminal path meets exactly one selected node. The horizontal positions below are \\emph{terminal packet addresses}, not time intervals in the original signal.\n\\vfig{complete_cuts}\n\\head{Three candidate selections}\nThe complete selection $\\{0,10,11\\}$ covers every terminal branch once. The selection $\\{0,00,1\\}$ covers the $00$ branch twice: it is not an antichain. The selection $\\{00,01,10\\}$ misses the $11$ branch entirely.\nFor actual Haar packets, set $P_v=W_v^*W_v$. Parent-to-child conservation gives\n\\[\nP_v=P_{v0}+P_{v1}.\n\\]\nReplacing unselected parents by their children until the complete pruning is reached proves $\\sum_{v\\in\\Lambda}P_v=I$. This is why the coefficients at different depths still form an exact coordinate system.\n\\head{An exhaustive finite check}\nAt depth three the pruning count obeys $a_0=1$ and $a_{d+1}=1+a_d^2$, giving $1,2,5,26$. The supplied suite checks all $26$ covering conditions and all $26$ actual $16\\times16$ projector-sum identities. These finite checks illustrate, but do not repl…","hat every root-to-terminal path meets exactly one selected node. The horizontal positions below are \\emph{terminal packet addresses}, not time intervals in the original signal.\n\\vfig{complete_cuts}\n\\head{Three candidate selections}\nThe complete selection $\\{0,10,11\\}$ covers every terminal branch once. The selection $\\{0,00,1\\}$ covers the $00$ branch twice: it is not an antichain. The selection $\\{00,01,10\\}$ misses the $11$ branch entirely.\nFor actual Haar packets, set $P_v=W_v^*W_v$. Parent-to-child conservation gives\n\\[\nP_v=P_{v0}+P_{v1}.\n\\]\nReplacing unselected parents by their children until the complete pruning is reached proves $\\sum_{v\\in\\Lambda}P_v=I$. This is why the coefficients at different depths still form an exact coordinate system.\n\\head{An exhaustive finite check}\nAt depth three the pruning count obeys $a_0=1$ and $a_{d+1}=1+a_d^2$, giving $1,2,5,26$. The supplied suite checks all $26$ covering conditions and all $26$ actual $16\\times16$ projector-sum identities. These finite checks illustrate, but do not replace, the universal tree theorem.\n\\takeaway{Mixed depth is…","path meets exactly one selected node. The horizontal positions below are \\emph{terminal packet addresses}, not time intervals in the original signal.\n\\vfig{complete_cuts}\n\\head{Three candidate selections}\nThe complete selection $\\{0,10,11\\}$ covers every terminal branch once. The selection $\\{0,00,1\\}$ covers the $00$ branch twice: it is not an antichain. The selection $\\{00,01,10\\}$ misses the $11$ branch entirely.\nFor actual Haar packets, set $P_v=W_v^*W_v$. Parent-to-child conservation gives\n\\[\nP_v=P_{v0}+P_{v1}.\n\\]\nReplacing unselected parents by their children until the complete pruning is reached proves $\\sum_{v\\in\\Lambda}P_v=I$. This is why the coefficients at different depths still form an exact coordinate system.\n\\head{An exhaustive finite check}\nAt depth three the pruning count obeys $a_0=1$ and $a_{d+1}=1+a_d^2$, giving $1,2,5,26$. The supplied suite checks all $26$ covering conditions and all $26$ actual $16\\times16$ projector-sum identities. These finite checks illustrate, but do not replace, the universal tree theorem.\n\\takeaway{Mixed depth is allowed. Missing a termina…","al positions below are \\emph{terminal packet addresses}, not time intervals in the original signal.\n\\vfig{complete_cuts}\n\\head{Three candidate selections}\nThe complete selection $\\{0,10,11\\}$ covers every terminal branch once. The selection $\\{0,00,1\\}$ covers the $00$ branch twice: it is not an antichain. The selection $\\{00,01,10\\}$ misses the $11$ branch entirely.\nFor actual Haar packets, set $P_v=W_v^*W_v$. Parent-to-child conservation gives\n\\[\nP_v=P_{v0}+P_{v1}.\n\\]\nReplacing unselected parents by their children until the complete pruning is reached proves $\\sum_{v\\in\\Lambda}P_v=I$. This is why the coefficients at different depths still form an exact coordinate system.\n\\head{An exhaustive finite check}\nAt depth three the pruning count obeys $a_0=1$ and $a_{d+1}=1+a_d^2$, giving $1,2,5,26$. The supplied suite checks all $26$ covering conditions and all $26$ actual $16\\times16$ projector-sum identities. These finite checks illustrate, but do not replace, the universal tree theorem.\n\\takeaway{Mixed depth is allowed. Missing a terminal path or selecting an ancestor and its descendant…","es}, not time intervals in the original signal.\n\\vfig{complete_cuts}\n\\head{Three candidate selections}\nThe complete selection $\\{0,10,11\\}$ covers every terminal branch once. The selection $\\{0,00,1\\}$ covers the $00$ branch twice: it is not an antichain. The selection $\\{00,01,10\\}$ misses the $11$ branch entirely.\nFor actual Haar packets, set $P_v=W_v^*W_v$. Parent-to-child conservation gives\n\\[\nP_v=P_{v0}+P_{v1}.\n\\]\nReplacing unselected parents by their children until the complete pruning is reached proves $\\sum_{v\\in\\Lambda}P_v=I$. This is why the coefficients at different depths still form an exact coordinate system.\n\\head{An exhaustive finite check}\nAt depth three the pruning count obeys $a_0=1$ and $a_{d+1}=1+a_d^2$, giving $1,2,5,26$. The supplied suite checks all $26$ covering conditions and all $26$ actual $16\\times16$ projector-sum identities. These finite checks illustrate, but do not replace, the universal tree theorem.\n\\takeaway{Mixed depth is allowed. Missing a terminal path or selecting an ancestor and its descendant is not.}","selection $\\{0,10,11\\}$ covers every terminal branch once. The selection $\\{0,00,1\\}$ covers the $00$ branch twice: it is not an antichain. The selection $\\{00,01,10\\}$ misses the $11$ branch entirely.\nFor actual Haar packets, set $P_v=W_v^*W_v$. Parent-to-child conservation gives\n\\[\nP_v=P_{v0}+P_{v1}.\n\\]\nReplacing unselected parents by their children until the complete pruning is reached proves $\\sum_{v\\in\\Lambda}P_v=I$. This is why the coefficients at different depths still form an exact coordinate system.\n\\head{An exhaustive finite check}\nAt depth three the pruning count obeys $a_0=1$ and $a_{d+1}=1+a_d^2$, giving $1,2,5,26$. The supplied suite checks all $26$ covering conditions and all $26$ actual $16\\times16$ projector-sum identities. These finite checks illustrate, but do not replace, the universal tree theorem.\n\\takeaway{Mixed depth is allowed. Missing a terminal path or selecting an ancestor and its descendant is not.}","nch entirely.\nFor actual Haar packets, set $P_v=W_v^*W_v$. Parent-to-child conservation gives\n\\[\nP_v=P_{v0}+P_{v1}.\n\\]\nReplacing unselected parents by their children until the complete pruning is reached proves $\\sum_{v\\in\\Lambda}P_v=I$. This is why the coefficients at different depths still form an exact coordinate system.\n\\head{An exhaustive finite check}\nAt depth three the pruning count obeys $a_0=1$ and $a_{d+1}=1+a_d^2$, giving $1,2,5,26$. The supplied suite checks all $26$ covering conditions and all $26$ actual $16\\times16$ projector-sum identities. These finite checks illustrate, but do not replace, the universal tree theorem.\n\\takeaway{Mixed depth is allowed. Missing a terminal path or selecting an ancestor and its descendant is not.}",".\nFor actual Haar packets, set $P_v=W_v^*W_v$. Parent-to-child conservation gives\n\\[\nP_v=P_{v0}+P_{v1}.\n\\]\nReplacing unselected parents by their children until the complete pruning is reached proves $\\sum_{v\\in\\Lambda}P_v=I$. This is why the coefficients at different depths still form an exact coordinate system.\n\\head{An exhaustive finite check}\nAt depth three the pruning count obeys $a_0=1$ and $a_{d+1}=1+a_d^2$, giving $1,2,5,26$. The supplied suite checks all $26$ covering conditions and all $26$ actual $16\\times16$ projector-sum identities. These finite checks illustrate, but do not replace, the universal tree theorem.\n\\takeaway{Mixed depth is allowed. Missing a terminal path or selecting an ancestor and its descendant is not.}","set $P_v=W_v^*W_v$. Parent-to-child conservation gives\n\\[\nP_v=P_{v0}+P_{v1}.\n\\]\nReplacing unselected parents by their children until the complete pruning is reached proves $\\sum_{v\\in\\Lambda}P_v=I$. This is why the coefficients at different depths still form an exact coordinate system.\n\\head{An exhaustive finite check}\nAt depth three the pruning count obeys $a_0=1$ and $a_{d+1}=1+a_d^2$, giving $1,2,5,26$. The supplied suite checks all $26$ covering conditions and all $26$ actual $16\\times16$ projector-sum identities. These finite checks illustrate, but do not replace, the universal tree theorem.\n\\takeaway{Mixed depth is allowed. Missing a terminal path or selecting an ancestor and its descendant is not.}","rvation gives\n\\[\nP_v=P_{v0}+P_{v1}.\n\\]\nReplacing unselected parents by their children until the complete pruning is reached proves $\\sum_{v\\in\\Lambda}P_v=I$. This is why the coefficients at different depths still form an exact coordinate system.\n\\head{An exhaustive finite check}\nAt depth three the pruning count obeys $a_0=1$ and $a_{d+1}=1+a_d^2$, giving $1,2,5,26$. The supplied suite checks all $26$ covering conditions and all $26$ actual $16\\times16$ projector-sum identities. These finite checks illustrate, but do not replace, the universal tree theorem.\n\\takeaway{Mixed depth is allowed. Missing a terminal path or selecting an ancestor and its descendant is not.}","}.\n\\]\nReplacing unselected parents by their children until the complete pruning is reached proves $\\sum_{v\\in\\Lambda}P_v=I$. This is why the coefficients at different depths still form an exact coordinate system.\n\\head{An exhaustive finite check}\nAt depth three the pruning count obeys $a_0=1$ and $a_{d+1}=1+a_d^2$, giving $1,2,5,26$. The supplied suite checks all $26$ covering conditions and all $26$ actual $16\\times16$ projector-sum identities. These finite checks illustrate, but do not replace, the universal tree theorem.\n\\takeaway{Mixed depth is allowed. Missing a terminal path or selecting an ancestor and its descendant is not.}","ing unselected parents by their children until the complete pruning is reached proves $\\sum_{v\\in\\Lambda}P_v=I$. This is why the coefficients at different depths still form an exact coordinate system.\n\\head{An exhaustive finite check}\nAt depth three the pruning count obeys $a_0=1$ and $a_{d+1}=1+a_d^2$, giving $1,2,5,26$. The supplied suite checks all $26$ covering conditions and all $26$ actual $16\\times16$ projector-sum identities. These finite checks illustrate, but do not replace, the universal tree theorem.\n\\takeaway{Mixed depth is allowed. Missing a terminal path or selecting an ancestor and its descendant is not.}","rnative without searching every alternative}{ADDITIVE COSTS / EXACT DYNAMIC PROGRAMMING}{Traditional monograph: Theorem 11.3 (p. 20), Theorem 11.4 (p. 21).}\nGive every node the rational stop cost shown below. A node may either stop and pay its own cost, or refine and pay the already optimized costs of its children. The shaded nodes form the selected complete pruning.\n\\vfig[.94]{dynamic_program}\n\\[\nV(v)=\\min\\{c_v,\\ V(v0)+V(v1)\\},\n\\qquad V(v)=c_v\\text{ at a terminal node}.\n\\]\n\\head{Read the calculation from the bottom upward}\nAt $10$, the children cost $1+1=2$, better than stopping for $7$. At $11$, stopping costs $4$, better than $2+3=5$. The right subtree therefore costs $6$, better than its stop cost $8$. The left subtree costs $1+2=3$, better than $5$. At the root, $3+6=9$ beats $20$.\nThe unique optimum is\n\\[\n\\Lambda_* = \\{00,01,100,101,11\\},\\qquad J(\\Lambda_*)=9.\n\\]\nAll $26$ complete prunings are separately enumerated and give the same minimum. At equality, the rule stops, making the selector deterministic. A singleton tree needs no comparison.\nThe costs here are supplied numbers.…","Theorem 11.4 (p. 21).}\nGive every node the rational stop cost shown below. A node may either stop and pay its own cost, or refine and pay the already optimized costs of its children. The shaded nodes form the selected complete pruning.\n\\vfig[.94]{dynamic_program}\n\\[\nV(v)=\\min\\{c_v,\\ V(v0)+V(v1)\\},\n\\qquad V(v)=c_v\\text{ at a terminal node}.\n\\]\n\\head{Read the calculation from the bottom upward}\nAt $10$, the children cost $1+1=2$, better than stopping for $7$. At $11$, stopping costs $4$, better than $2+3=5$. The right subtree therefore costs $6$, better than its stop cost $8$. The left subtree costs $1+2=3$, better than $5$. At the root, $3+6=9$ beats $20$.\nThe unique optimum is\n\\[\n\\Lambda_* = \\{00,01,100,101,11\\},\\qquad J(\\Lambda_*)=9.\n\\]\nAll $26$ complete prunings are separately enumerated and give the same minimum. At equality, the rule stops, making the selector deterministic. A singleton tree needs no comparison.\nThe costs here are supplied numbers. The theorem optimizes that stated additive objective; it does not prove those numbers encode the best denoising criterion. Coupled co…","ive every node the rational stop cost shown below. A node may either stop and pay its own cost, or refine and pay the already optimized costs of its children. The shaded nodes form the selected complete pruning.\n\\vfig[.94]{dynamic_program}\n\\[\nV(v)=\\min\\{c_v,\\ V(v0)+V(v1)\\},\n\\qquad V(v)=c_v\\text{ at a terminal node}.\n\\]\n\\head{Read the calculation from the bottom upward}\nAt $10$, the children cost $1+1=2$, better than stopping for $7$. At $11$, stopping costs $4$, better than $2+3=5$. The right subtree therefore costs $6$, better than its stop cost $8$. The left subtree costs $1+2=3$, better than $5$. At the root, $3+6=9$ beats $20$.\nThe unique optimum is\n\\[\n\\Lambda_* = \\{00,01,100,101,11\\},\\qquad J(\\Lambda_*)=9.\n\\]\nAll $26$ complete prunings are separately enumerated and give the same minimum. At equality, the rule stops, making the selector deterministic. A singleton tree needs no comparison.\nThe costs here are supplied numbers. The theorem optimizes that stated additive objective; it does not prove those numbers encode the best denoising criterion. Coupled costs that cannot be writt…","ost shown below. A node may either stop and pay its own cost, or refine and pay the already optimized costs of its children. The shaded nodes form the selected complete pruning.\n\\vfig[.94]{dynamic_program}\n\\[\nV(v)=\\min\\{c_v,\\ V(v0)+V(v1)\\},\n\\qquad V(v)=c_v\\text{ at a terminal node}.\n\\]\n\\head{Read the calculation from the bottom upward}\nAt $10$, the children cost $1+1=2$, better than stopping for $7$. At $11$, stopping costs $4$, better than $2+3=5$. The right subtree therefore costs $6$, better than its stop cost $8$. The left subtree costs $1+2=3$, better than $5$. At the root, $3+6=9$ beats $20$.\nThe unique optimum is\n\\[\n\\Lambda_* = \\{00,01,100,101,11\\},\\qquad J(\\Lambda_*)=9.\n\\]\nAll $26$ complete prunings are separately enumerated and give the same minimum. At equality, the rule stops, making the selector deterministic. A singleton tree needs no comparison.\nThe costs here are supplied numbers. The theorem optimizes that stated additive objective; it does not prove those numbers encode the best denoising criterion. Coupled costs that cannot be written as a sum of selected-node costs","n below. A node may either stop and pay its own cost, or refine and pay the already optimized costs of its children. The shaded nodes form the selected complete pruning.\n\\vfig[.94]{dynamic_program}\n\\[\nV(v)=\\min\\{c_v,\\ V(v0)+V(v1)\\},\n\\qquad V(v)=c_v\\text{ at a terminal node}.\n\\]\n\\head{Read the calculation from the bottom upward}\nAt $10$, the children cost $1+1=2$, better than stopping for $7$. At $11$, stopping costs $4$, better than $2+3=5$. The right subtree therefore costs $6$, better than its stop cost $8$. The left subtree costs $1+2=3$, better than $5$. At the root, $3+6=9$ beats $20$.\nThe unique optimum is\n\\[\n\\Lambda_* = \\{00,01,100,101,11\\},\\qquad J(\\Lambda_*)=9.\n\\]\nAll $26$ complete prunings are separately enumerated and give the same minimum. At equality, the rule stops, making the selector deterministic. A singleton tree needs no comparison.\nThe costs here are supplied numbers. The theorem optimizes that stated additive objective; it does not prove those numbers encode the best denoising criterion. Coupled costs that cannot be written as a sum of selected-node costs need a …","ither stop and pay its own cost, or refine and pay the already optimized costs of its children. The shaded nodes form the selected complete pruning.\n\\vfig[.94]{dynamic_program}\n\\[\nV(v)=\\min\\{c_v,\\ V(v0)+V(v1)\\},\n\\qquad V(v)=c_v\\text{ at a terminal node}.\n\\]\n\\head{Read the calculation from the bottom upward}\nAt $10$, the children cost $1+1=2$, better than stopping for $7$. At $11$, stopping costs $4$, better than $2+3=5$. The right subtree therefore costs $6$, better than its stop cost $8$. The left subtree costs $1+2=3$, better than $5$. At the root, $3+6=9$ beats $20$.\nThe unique optimum is\n\\[\n\\Lambda_* = \\{00,01,100,101,11\\},\\qquad J(\\Lambda_*)=9.\n\\]\nAll $26$ complete prunings are separately enumerated and give the same minimum. At equality, the rule stops, making the selector deterministic. A singleton tree needs no comparison.\nThe costs here are supplied numbers. The theorem optimizes that stated additive objective; it does not prove those numbers encode the best denoising criterion. Coupled costs that cannot be written as a sum of selected-node costs need a different optimizer.\n…","y its own cost, or refine and pay the already optimized costs of its children. The shaded nodes form the selected complete pruning.\n\\vfig[.94]{dynamic_program}\n\\[\nV(v)=\\min\\{c_v,\\ V(v0)+V(v1)\\},\n\\qquad V(v)=c_v\\text{ at a terminal node}.\n\\]\n\\head{Read the calculation from the bottom upward}\nAt $10$, the children cost $1+1=2$, better than stopping for $7$. At $11$, stopping costs $4$, better than $2+3=5$. The right subtree therefore costs $6$, better than its stop cost $8$. The left subtree costs $1+2=3$, better than $5$. At the root, $3+6=9$ beats $20$.\nThe unique optimum is\n\\[\n\\Lambda_* = \\{00,01,100,101,11\\},\\qquad J(\\Lambda_*)=9.\n\\]\nAll $26$ complete prunings are separately enumerated and give the same minimum. At equality, the rule stops, making the selector deterministic. A singleton tree needs no comparison.\nThe costs here are supplied numbers. The theorem optimizes that stated additive objective; it does not prove those numbers encode the best denoising criterion. Coupled costs that cannot be written as a sum of selected-node costs need a different optimizer.\n\\takeaway{Global …","dy optimized costs of its children. The shaded nodes form the selected complete pruning.\n\\vfig[.94]{dynamic_program}\n\\[\nV(v)=\\min\\{c_v,\\ V(v0)+V(v1)\\},\n\\qquad V(v)=c_v\\text{ at a terminal node}.\n\\]\n\\head{Read the calculation from the bottom upward}\nAt $10$, the children cost $1+1=2$, better than stopping for $7$. At $11$, stopping costs $4$, better than $2+3=5$. The right subtree therefore costs $6$, better than its stop cost $8$. The left subtree costs $1+2=3$, better than $5$. At the root, $3+6=9$ beats $20$.\nThe unique optimum is\n\\[\n\\Lambda_* = \\{00,01,100,101,11\\},\\qquad J(\\Lambda_*)=9.\n\\]\nAll $26$ complete prunings are separately enumerated and give the same minimum. At equality, the rule stops, making the selector deterministic. A singleton tree needs no comparison.\nThe costs here are supplied numbers. The theorem optimizes that stated additive objective; it does not prove those numbers encode the best denoising criterion. Coupled costs that cannot be written as a sum of selected-node costs need a different optimizer.\n\\takeaway{Global optimality follows from the exact stop-or-r…","ren. The shaded nodes form the selected complete pruning.\n\\vfig[.94]{dynamic_program}\n\\[\nV(v)=\\min\\{c_v,\\ V(v0)+V(v1)\\},\n\\qquad V(v)=c_v\\text{ at a terminal node}.\n\\]\n\\head{Read the calculation from the bottom upward}\nAt $10$, the children cost $1+1=2$, better than stopping for $7$. At $11$, stopping costs $4$, better than $2+3=5$. The right subtree therefore costs $6$, better than its stop cost $8$. The left subtree costs $1+2=3$, better than $5$. At the root, $3+6=9$ beats $20$.\nThe unique optimum is\n\\[\n\\Lambda_* = \\{00,01,100,101,11\\},\\qquad J(\\Lambda_*)=9.\n\\]\nAll $26$ complete prunings are separately enumerated and give the same minimum. At equality, the rule stops, making the selector deterministic. A singleton tree needs no comparison.\nThe costs here are supplied numbers. The theorem optimizes that stated additive objective; it does not prove those numbers encode the best denoising criterion. Coupled costs that cannot be written as a sum of selected-node costs need a different optimizer.\n\\takeaway{Global optimality follows from the exact stop-or-refine decomposition of a comple…","he selected complete pruning.\n\\vfig[.94]{dynamic_program}\n\\[\nV(v)=\\min\\{c_v,\\ V(v0)+V(v1)\\},\n\\qquad V(v)=c_v\\text{ at a terminal node}.\n\\]\n\\head{Read the calculation from the bottom upward}\nAt $10$, the children cost $1+1=2$, better than stopping for $7$. At $11$, stopping costs $4$, better than $2+3=5$. The right subtree therefore costs $6$, better than its stop cost $8$. The left subtree costs $1+2=3$, better than $5$. At the root, $3+6=9$ beats $20$.\nThe unique optimum is\n\\[\n\\Lambda_* = \\{00,01,100,101,11\\},\\qquad J(\\Lambda_*)=9.\n\\]\nAll $26$ complete prunings are separately enumerated and give the same minimum. At equality, the rule stops, making the selector deterministic. A singleton tree needs no comparison.\nThe costs here are supplied numbers. The theorem optimizes that stated additive objective; it does not prove those numbers encode the best denoising criterion. Coupled costs that cannot be written as a sum of selected-node costs need a different optimizer.\n\\takeaway{Global optimality follows from the exact stop-or-refine decomposition of a complete pruning, not from a greed…","pruning.\n\\vfig[.94]{dynamic_program}\n\\[\nV(v)=\\min\\{c_v,\\ V(v0)+V(v1)\\},\n\\qquad V(v)=c_v\\text{ at a terminal node}.\n\\]\n\\head{Read the calculation from the bottom upward}\nAt $10$, the children cost $1+1=2$, better than stopping for $7$. At $11$, stopping costs $4$, better than $2+3=5$. The right subtree therefore costs $6$, better than its stop cost $8$. The left subtree costs $1+2=3$, better than $5$. At the root, $3+6=9$ beats $20$.\nThe unique optimum is\n\\[\n\\Lambda_* = \\{00,01,100,101,11\\},\\qquad J(\\Lambda_*)=9.\n\\]\nAll $26$ complete prunings are separately enumerated and give the same minimum. At equality, the rule stops, making the selector deterministic. A singleton tree needs no comparison.\nThe costs here are supplied numbers. The theorem optimizes that stated additive objective; it does not prove those numbers encode the best denoising criterion. Coupled costs that cannot be written as a sum of selected-node costs need a different optimizer.\n\\takeaway{Global optimality follows from the exact stop-or-refine decomposition of a complete pruning, not from a greedy assumption.}","]{dynamic_program}\n\\[\nV(v)=\\min\\{c_v,\\ V(v0)+V(v1)\\},\n\\qquad V(v)=c_v\\text{ at a terminal node}.\n\\]\n\\head{Read the calculation from the bottom upward}\nAt $10$, the children cost $1+1=2$, better than stopping for $7$. At $11$, stopping costs $4$, better than $2+3=5$. The right subtree therefore costs $6$, better than its stop cost $8$. The left subtree costs $1+2=3$, better than $5$. At the root, $3+6=9$ beats $20$.\nThe unique optimum is\n\\[\n\\Lambda_* = \\{00,01,100,101,11\\},\\qquad J(\\Lambda_*)=9.\n\\]\nAll $26$ complete prunings are separately enumerated and give the same minimum. At equality, the rule stops, making the selector deterministic. A singleton tree needs no comparison.\nThe costs here are supplied numbers. The theorem optimizes that stated additive objective; it does not prove those numbers encode the best denoising criterion. Coupled costs that cannot be written as a sum of selected-node costs need a different optimizer.\n\\takeaway{Global optimality follows from the exact stop-or-refine decomposition of a complete pruning, not from a greedy assumption.}","ram}\n\\[\nV(v)=\\min\\{c_v,\\ V(v0)+V(v1)\\},\n\\qquad V(v)=c_v\\text{ at a terminal node}.\n\\]\n\\head{Read the calculation from the bottom upward}\nAt $10$, the children cost $1+1=2$, better than stopping for $7$. At $11$, stopping costs $4$, better than $2+3=5$. The right subtree therefore costs $6$, better than its stop cost $8$. The left subtree costs $1+2=3$, better than $5$. At the root, $3+6=9$ beats $20$.\nThe unique optimum is\n\\[\n\\Lambda_* = \\{00,01,100,101,11\\},\\qquad J(\\Lambda_*)=9.\n\\]\nAll $26$ complete prunings are separately enumerated and give the same minimum. At equality, the rule stops, making the selector deterministic. A singleton tree needs no comparison.\nThe costs here are supplied numbers. The theorem optimizes that stated additive objective; it does not prove those numbers encode the best denoising criterion. Coupled costs that cannot be written as a sum of selected-node costs need a different optimizer.\n\\takeaway{Global optimality follows from the exact stop-or-refine decomposition of a complete pruning, not from a greedy assumption.}",")+V(v1)\\},\n\\qquad V(v)=c_v\\text{ at a terminal node}.\n\\]\n\\head{Read the calculation from the bottom upward}\nAt $10$, the children cost $1+1=2$, better than stopping for $7$. At $11$, stopping costs $4$, better than $2+3=5$. The right subtree therefore costs $6$, better than its stop cost $8$. The left subtree costs $1+2=3$, better than $5$. At the root, $3+6=9$ beats $20$.\nThe unique optimum is\n\\[\n\\Lambda_* = \\{00,01,100,101,11\\},\\qquad J(\\Lambda_*)=9.\n\\]\nAll $26$ complete prunings are separately enumerated and give the same minimum. At equality, the rule stops, making the selector deterministic. A singleton tree needs no comparison.\nThe costs here are supplied numbers. The theorem optimizes that stated additive objective; it does not prove those numbers encode the best denoising criterion. Coupled costs that cannot be written as a sum of selected-node costs need a different optimizer.\n\\takeaway{Global optimality follows from the exact stop-or-refine decomposition of a complete pruning, not from a greedy assumption.}","ad the calculation from the bottom upward}\nAt $10$, the children cost $1+1=2$, better than stopping for $7$. At $11$, stopping costs $4$, better than $2+3=5$. The right subtree therefore costs $6$, better than its stop cost $8$. The left subtree costs $1+2=3$, better than $5$. At the root, $3+6=9$ beats $20$.\nThe unique optimum is\n\\[\n\\Lambda_* = \\{00,01,100,101,11\\},\\qquad J(\\Lambda_*)=9.\n\\]\nAll $26$ complete prunings are separately enumerated and give the same minimum. At equality, the rule stops, making the selector deterministic. A singleton tree needs no comparison.\nThe costs here are supplied numbers. The theorem optimizes that stated additive objective; it does not prove those numbers encode the best denoising criterion. Coupled costs that cannot be written as a sum of selected-node costs need a different optimizer.\n\\takeaway{Global optimality follows from the exact stop-or-refine decomposition of a complete pruning, not from a greedy assumption.}","\\vpage{14}{How far can the costs move before the tree changes?}{A GLOBAL MARGIN / NOT A LOCAL HEURISTIC}{Traditional monograph: Theorem 11.3 (p. 20), Corollary 11.5 (p. 21).}\nThe best pruning on the previous page costs $9$; the cheapest competitor costs $10$. Its exact global gap is therefore $\\Gamma=1$. The plate lists every complete-pruning cost in sorted order, giving each pruning its own horizontal rank.\n\\vfig{selector_margin}\n\\head{Propagate an error budget as carefully as a cost}\nSuppose every node cost changes by at most $\\varepsilon_v=1/32$. A pruning may select as many as eight leaves, so its total cost can change by at most $8/32=1/4$. The recursive envelope proves this without guessing which pruning will be selected:\n\\[\nE(v)=\\max\\{\\varepsilon_v,\\ E(v0)+E(v1)\\},\\qquad E(\\varnothing)=\\tfrac14.\n\\]\nFor any competitor $\\Lambda$, the changed cost difference is bounded below by\n\\[\nJ'(\\Lambda)-J'(\\Lambda_","\\vpage{14}{How far can the costs move before the tree changes?}{A GLOBAL MARGIN / NOT A LOCAL HEURISTIC}{Traditional monograph: Theorem 11.3 (p. 20), Corollary 11.5 (p. 21).}\nThe best pruning on the previous page costs $9$; the cheapest competitor costs $10$. Its exact global gap is therefore $\\Gamma=1$. The plate lists every complete-pruning cost in sorted order, giving each pruning its own horizontal rank.\n\\vfig{selector_margin}\n\\head{Propagate an error budget as carefully as a cost}\nSuppose every node cost changes by at most $\\varepsilon_v=1/32$. A pruning may select as many as eight leaves, so its total cost can change by at most $8/32=1/4$. The recursive envelope proves this without guessing which pruning will be selected:\n\\[\nE(v)=\\max\\{\\varepsilon_v,\\ E(v0)+E(v1)\\},\\qquad E(\\varnothing)=\\tfrac14.\n\\]\nFor any competitor $\\Lambda$, the changed cost difference is bounded below by\n\\[\nJ'(\\Lambda)-J'(\\Lambda_*)\\ge \\Gamma-2E(\\varnothing)=\\tfrac1","\\vpage{14}{How far can the costs move before the tree changes?}{A GLOBAL MARGIN / NOT A LOCAL HEURISTIC}{Traditional monograph: Theorem 11.3 (p. 20), Corollary 11.5 (p. 21).}\nThe best pruning on the previous page costs $9$; the cheapest competitor costs $10$. Its exact global gap is therefore $\\Gamma=1$. The plate lists every complete-pruning cost in sorted order, giving each pruning its own horizontal rank.\n\\vfig{selector_margin}\n\\head{Propagate an error budget as carefully as a cost}\nSuppose every node cost changes by at most $\\varepsilon_v=1/32$. A pruning may select as many as eight leaves, so its total cost can change by at most $8/32=1/4$. The recursive envelope proves this without guessing which pruning will be selected:\n\\[\nE(v)=\\max\\{\\varepsilon_v,\\ E(v0)+E(v1)\\},\\qquad E(\\varnothing)=\\tfrac14.\n\\]\nFor any competitor $\\Lambda$, the changed cost difference is bounded below by\n\\[\nJ'(\\Lambda)-J'(\\Lambda_*)\\ge \\Gamma-2E(\\varnothing)=\\tfrac12>0.\n\\]\nThus the optimum is unchanged for \\emp","11.3 (p. 20), Corollary 11.5 (p. 21).}\nThe best pruning on the previous page costs $9$; the cheapest competitor costs $10$. Its exact global gap is therefore $\\Gamma=1$. The plate lists every complete-pruning cost in sorted order, giving each pruning its own horizontal rank.\n\\vfig{selector_margin}\n\\head{Propagate an error budget as carefully as a cost}\nSuppose every node cost changes by at most $\\varepsilon_v=1/32$. A pruning may select as many as eight leaves, so its total cost can change by at most $8/32=1/4$. The recursive envelope proves this without guessing which pruning will be selected:\n\\[\nE(v)=\\max\\{\\varepsilon_v,\\ E(v0)+E(v1)\\},\\qquad E(\\varnothing)=\\tfrac14.\n\\]\nFor any competitor $\\Lambda$, the changed cost difference is bounded below by\n\\[\nJ'(\\Lambda)-J'(\\Lambda_*)\\ge \\Gamma-2E(\\varnothing)=\\tfrac12>0.\n\\]\nThus the optimum is unchanged for \\emph{every} permitted simultaneous perturbation, not just a sampled noise realization. The envelope is over all feasible prunings, not just the five nodes in the current optimum.\n\\head{Strict inequality has content}\nIf two competing cos…","itor costs $10$. Its exact global gap is therefore $\\Gamma=1$. The plate lists every complete-pruning cost in sorted order, giving each pruning its own horizontal rank.\n\\vfig{selector_margin}\n\\head{Propagate an error budget as carefully as a cost}\nSuppose every node cost changes by at most $\\varepsilon_v=1/32$. A pruning may select as many as eight leaves, so its total cost can change by at most $8/32=1/4$. The recursive envelope proves this without guessing which pruning will be selected:\n\\[\nE(v)=\\max\\{\\varepsilon_v,\\ E(v0)+E(v1)\\},\\qquad E(\\varnothing)=\\tfrac14.\n\\]\nFor any competitor $\\Lambda$, the changed cost difference is bounded below by\n\\[\nJ'(\\Lambda)-J'(\\Lambda_*)\\ge \\Gamma-2E(\\varnothing)=\\tfrac12>0.\n\\]\nThus the optimum is unchanged for \\emph{every} permitted simultaneous perturbation, not just a sampled noise realization. The envelope is over all feasible prunings, not just the five nodes in the current optimum.\n\\head{Strict inequality has content}\nIf two competing costs are $9$ and $10$, moving them to $9.5$ and $9.5$ removes uniqueness. Equality in $2E=\\Gamma$ is therefor…","ning cost in sorted order, giving each pruning its own horizontal rank.\n\\vfig{selector_margin}\n\\head{Propagate an error budget as carefully as a cost}\nSuppose every node cost changes by at most $\\varepsilon_v=1/32$. A pruning may select as many as eight leaves, so its total cost can change by at most $8/32=1/4$. The recursive envelope proves this without guessing which pruning will be selected:\n\\[\nE(v)=\\max\\{\\varepsilon_v,\\ E(v0)+E(v1)\\},\\qquad E(\\varnothing)=\\tfrac14.\n\\]\nFor any competitor $\\Lambda$, the changed cost difference is bounded below by\n\\[\nJ'(\\Lambda)-J'(\\Lambda_*)\\ge \\Gamma-2E(\\varnothing)=\\tfrac12>0.\n\\]\nThus the optimum is unchanged for \\emph{every} permitted simultaneous perturbation, not just a sampled noise realization. The envelope is over all feasible prunings, not just the five nodes in the current optimum.\n\\head{Strict inequality has content}\nIf two competing costs are $9$ and $10$, moving them to $9.5$ and $9.5$ removes uniqueness. Equality in $2E=\\Gamma$ is therefore not a unique-selector certificate. For ordinary probability, the random node costs and the join…","ead{Propagate an error budget as carefully as a cost}\nSuppose every node cost changes by at most $\\varepsilon_v=1/32$. A pruning may select as many as eight leaves, so its total cost can change by at most $8/32=1/4$. The recursive envelope proves this without guessing which pruning will be selected:\n\\[\nE(v)=\\max\\{\\varepsilon_v,\\ E(v0)+E(v1)\\},\\qquad E(\\varnothing)=\\tfrac14.\n\\]\nFor any competitor $\\Lambda$, the changed cost difference is bounded below by\n\\[\nJ'(\\Lambda)-J'(\\Lambda_*)\\ge \\Gamma-2E(\\varnothing)=\\tfrac12>0.\n\\]\nThus the optimum is unchanged for \\emph{every} permitted simultaneous perturbation, not just a sampled noise realization. The envelope is over all feasible prunings, not just the five nodes in the current optimum.\n\\head{Strict inequality has content}\nIf two competing costs are $9$ and $10$, moving them to $9.5$ and $9.5$ removes uniqueness. Equality in $2E=\\Gamma$ is therefore not a unique-selector certificate. For ordinary probability, the random node costs and the joint good event must be measurable. Separate nodewise probability bounds do not automatically give t…","e every node cost changes by at most $\\varepsilon_v=1/32$. A pruning may select as many as eight leaves, so its total cost can change by at most $8/32=1/4$. The recursive envelope proves this without guessing which pruning will be selected:\n\\[\nE(v)=\\max\\{\\varepsilon_v,\\ E(v0)+E(v1)\\},\\qquad E(\\varnothing)=\\tfrac14.\n\\]\nFor any competitor $\\Lambda$, the changed cost difference is bounded below by\n\\[\nJ'(\\Lambda)-J'(\\Lambda_*)\\ge \\Gamma-2E(\\varnothing)=\\tfrac12>0.\n\\]\nThus the optimum is unchanged for \\emph{every} permitted simultaneous perturbation, not just a sampled noise realization. The envelope is over all feasible prunings, not just the five nodes in the current optimum.\n\\head{Strict inequality has content}\nIf two competing costs are $9$ and $10$, moving them to $9.5$ and $9.5$ removes uniqueness. Equality in $2E=\\Gamma$ is therefore not a unique-selector certificate. For ordinary probability, the random node costs and the joint good event must be measurable. Separate nodewise probability bounds do not automatically give the probability of that joint event.\n\\takeaway{The exact sele…","$, the changed cost difference is bounded below by\n\\[\nJ'(\\Lambda)-J'(\\Lambda_*)\\ge \\Gamma-2E(\\varnothing)=\\tfrac12>0.\n\\]\nThus the optimum is unchanged for \\emph{every} permitted simultaneous perturbation, not just a sampled noise realization. The envelope is over all feasible prunings, not just the five nodes in the current optimum.\n\\head{Strict inequality has content}\nIf two competing costs are $9$ and $10$, moving them to $9.5$ and $9.5$ removes uniqueness. Equality in $2E=\\Gamma$ is therefore not a unique-selector certificate. For ordinary probability, the random node costs and the joint good event must be measurable. Separate nodewise probability bounds do not automatically give the probability of that joint event.\n\\takeaway{The exact selector is stable because its global gap exceeds the total worst-case movement of both competing costs.}","hanged cost difference is bounded below by\n\\[\nJ'(\\Lambda)-J'(\\Lambda_*)\\ge \\Gamma-2E(\\varnothing)=\\tfrac12>0.\n\\]\nThus the optimum is unchanged for \\emph{every} permitted simultaneous perturbation, not just a sampled noise realization. The envelope is over all feasible prunings, not just the five nodes in the current optimum.\n\\head{Strict inequality has content}\nIf two competing costs are $9$ and $10$, moving them to $9.5$ and $9.5$ removes uniqueness. Equality in $2E=\\Gamma$ is therefore not a unique-selector certificate. For ordinary probability, the random node costs and the joint good event must be measurable. Separate nodewise probability bounds do not automatically give the probability of that joint event.\n\\takeaway{The exact selector is stable because its global gap exceeds the total worst-case movement of both competing costs.}","e is bounded below by\n\\[\nJ'(\\Lambda)-J'(\\Lambda_*)\\ge \\Gamma-2E(\\varnothing)=\\tfrac12>0.\n\\]\nThus the optimum is unchanged for \\emph{every} permitted simultaneous perturbation, not just a sampled noise realization. The envelope is over all feasible prunings, not just the five nodes in the current optimum.\n\\head{Strict inequality has content}\nIf two competing costs are $9$ and $10$, moving them to $9.5$ and $9.5$ removes uniqueness. Equality in $2E=\\Gamma$ is therefore not a unique-selector certificate. For ordinary probability, the random node costs and the joint good event must be measurable. Separate nodewise probability bounds do not automatically give the probability of that joint event.\n\\takeaway{The exact selector is stable because its global gap exceeds the total worst-case movement of both competing costs.}","ed below by\n\\[\nJ'(\\Lambda)-J'(\\Lambda_*)\\ge \\Gamma-2E(\\varnothing)=\\tfrac12>0.\n\\]\nThus the optimum is unchanged for \\emph{every} permitted simultaneous perturbation, not just a sampled noise realization. The envelope is over all feasible prunings, not just the five nodes in the current optimum.\n\\head{Strict inequality has content}\nIf two competing costs are $9$ and $10$, moving them to $9.5$ and $9.5$ removes uniqueness. Equality in $2E=\\Gamma$ is therefore not a unique-selector certificate. For ordinary probability, the random node costs and the joint good event must be measurable. Separate nodewise probability bounds do not automatically give the probability of that joint event.\n\\takeaway{The exact selector is stable because its global gap exceeds the total worst-case movement of both competing costs.}","*)\\ge \\Gamma-2E(\\varnothing)=\\tfrac12>0.\n\\]\nThus the optimum is unchanged for \\emph{every} permitted simultaneous perturbation, not just a sampled noise realization. The envelope is over all feasible prunings, not just the five nodes in the current optimum.\n\\head{Strict inequality has content}\nIf two competing costs are $9$ and $10$, moving them to $9.5$ and $9.5$ removes uniqueness. Equality in $2E=\\Gamma$ is therefore not a unique-selector certificate. For ordinary probability, the random node costs and the joint good event must be measurable. Separate nodewise probability bounds do not automatically give the probability of that joint event.\n\\takeaway{The exact selector is stable because its global gap exceeds the total worst-case movement of both competing costs.}","\\vpage{15}{Retain the jump and approximate the other packet}{AN EXACT IDENTITY / A CONDITIONAL INPUT}{Traditional monograph: Theorem 6.1 (p. 12), Corollary 6.2 (p. 12).}\nReturn to the running signal $f=s+j$. For a concrete finite calculation, use the periodic three-tap smoother\n\\[\n(Qx)[n]=\\tfrac14x[n-1]+\\tfrac12x[n]+\\tfrac14x[n+1],\\qquad n\\in\\mathbb Z/16\\mathbb Z.\n\\]\nThe plots compare smoothing the full signal with retaining its supplied jump packet. Every plotted sample is a rational number computed from the displayed formula.\n\\vfig{known_edge}\n\\head{The cancellation is the theorem}\nDefine $\\widetilde f=Qs+j$. Then, sample by sample,\n\\[\n\\boxed{f-\\widetilde f=s-Qs.}\n\\]\nBy contrast, linearity gives\n\\[\nf-Qf=(s-Qs)+(j-Qj).\n\\]\nThe second error includes a new term caused by smoothing the discontinuity. In this finite example that term is concentrated near the two jump boundaries. The identity itse","\\vpage{15}{Retain the jump and approximate the other packet}{AN EXACT IDENTITY / A CONDITIONAL INPUT}{Traditional monograph: Theorem 6.1 (p. 12), Corollary 6.2 (p. 12).}\nReturn to the running signal $f=s+j$. For a concrete finite calculation, use the periodic three-tap smoother\n\\[\n(Qx)[n]=\\tfrac14x[n-1]+\\tfrac12x[n]+\\tfrac14x[n+1],\\qquad n\\in\\mathbb Z/16\\mathbb Z.\n\\]\nThe plots compare smoothing the full signal with retaining its supplied jump packet. Every plotted sample is a rational number computed from the displayed formula.\n\\vfig{known_edge}\n\\head{The cancellation is the theorem}\nDefine $\\widetilde f=Qs+j$. Then, sample by sample,\n\\[\n\\boxed{f-\\widetilde f=s-Qs.}\n\\]\nBy contrast, linearity gives\n\\[\nf-Qf=(s-Qs)+(j-Qj).\n\\]\nThe second error includes a new term caused by smoothing the discontinuity. In this finite example that term is concentrated near the two jump boundaries. The identity itself requires no linearity of the approximation applied to $s$; linearity is used only in the comparison with $Qf$.\n\\head{The input is not discovered by subtraction","$f=s+j$. For a concrete finite calculation, use the periodic three-tap smoother\n\\[\n(Qx)[n]=\\tfrac14x[n-1]+\\tfrac12x[n]+\\tfrac14x[n+1],\\qquad n\\in\\mathbb Z/16\\mathbb Z.\n\\]\nThe plots compare smoothing the full signal with retaining its supplied jump packet. Every plotted sample is a rational number computed from the displayed formula.\n\\vfig{known_edge}\n\\head{The cancellation is the theorem}\nDefine $\\widetilde f=Qs+j$. Then, sample by sample,\n\\[\n\\boxed{f-\\widetilde f=s-Qs.}\n\\]\nBy contrast, linearity gives\n\\[\nf-Qf=(s-Qs)+(j-Qj).\n\\]\nThe second error includes a new term caused by smoothing the discontinuity. In this finite example that term is concentrated near the two jump boundaries. The identity itself requires no linearity of the approximation applied to $s$; linearity is used only in the comparison with $Qf$.\n\\head{The input is not discovered by subtraction}\nThe data here include the correct $j$. Replacing it with a wrong $\\widehat j$ retains the error $j-\\widehat j$, even when $s$ is approximated perfectly. Multiple amplitudes and overlapping shapes are allowed if their full sum is s…","e the periodic three-tap smoother\n\\[\n(Qx)[n]=\\tfrac14x[n-1]+\\tfrac12x[n]+\\tfrac14x[n+1],\\qquad n\\in\\mathbb Z/16\\mathbb Z.\n\\]\nThe plots compare smoothing the full signal with retaining its supplied jump packet. Every plotted sample is a rational number computed from the displayed formula.\n\\vfig{known_edge}\n\\head{The cancellation is the theorem}\nDefine $\\widetilde f=Qs+j$. Then, sample by sample,\n\\[\n\\boxed{f-\\widetilde f=s-Qs.}\n\\]\nBy contrast, linearity gives\n\\[\nf-Qf=(s-Qs)+(j-Qj).\n\\]\nThe second error includes a new term caused by smoothing the discontinuity. In this finite example that term is concentrated near the two jump boundaries. The identity itself requires no linearity of the approximation applied to $s$; linearity is used only in the comparison with $Qf$.\n\\head{The input is not discovered by subtraction}\nThe data here include the correct $j$. Replacing it with a wrong $\\widehat j$ retains the error $j-\\widehat j$, even when $s$ is approximated perfectly. Multiple amplitudes and overlapping shapes are allowed if their full sum is supplied, but the identity does not estimate th…","ac12x[n]+\\tfrac14x[n+1],\\qquad n\\in\\mathbb Z/16\\mathbb Z.\n\\]\nThe plots compare smoothing the full signal with retaining its supplied jump packet. Every plotted sample is a rational number computed from the displayed formula.\n\\vfig{known_edge}\n\\head{The cancellation is the theorem}\nDefine $\\widetilde f=Qs+j$. Then, sample by sample,\n\\[\n\\boxed{f-\\widetilde f=s-Qs.}\n\\]\nBy contrast, linearity gives\n\\[\nf-Qf=(s-Qs)+(j-Qj).\n\\]\nThe second error includes a new term caused by smoothing the discontinuity. In this finite example that term is concentrated near the two jump boundaries. The identity itself requires no linearity of the approximation applied to $s$; linearity is used only in the comparison with $Qf$.\n\\head{The input is not discovered by subtraction}\nThe data here include the correct $j$. Replacing it with a wrong $\\widehat j$ retains the error $j-\\widehat j$, even when $s$ is approximated perfectly. Multiple amplitudes and overlapping shapes are allowed if their full sum is supplied, but the identity does not estimate that sum from observations.\nOur finite periodic smoother is an ill…","ancellation is the theorem}\nDefine $\\widetilde f=Qs+j$. Then, sample by sample,\n\\[\n\\boxed{f-\\widetilde f=s-Qs.}\n\\]\nBy contrast, linearity gives\n\\[\nf-Qf=(s-Qs)+(j-Qj).\n\\]\nThe second error includes a new term caused by smoothing the discontinuity. In this finite example that term is concentrated near the two jump boundaries. The identity itself requires no linearity of the approximation applied to $s$; linearity is used only in the comparison with $Qf$.\n\\head{The input is not discovered by subtraction}\nThe data here include the correct $j$. Replacing it with a wrong $\\widehat j$ retains the error $j-\\widehat j$, even when $s$ is approximated perfectly. Multiple amplitudes and overlapping shapes are allowed if their full sum is supplied, but the identity does not estimate that sum from observations.\nOur finite periodic smoother is an illustration, not a new continuum approximation theorem. Its chosen boundary convention is part of the experiment.\n\\takeaway{Known-edge reconstruction removes the jump from the approximation problem. It does not remove the need to identify the jump.}","+j$. Then, sample by sample,\n\\[\n\\boxed{f-\\widetilde f=s-Qs.}\n\\]\nBy contrast, linearity gives\n\\[\nf-Qf=(s-Qs)+(j-Qj).\n\\]\nThe second error includes a new term caused by smoothing the discontinuity. In this finite example that term is concentrated near the two jump boundaries. The identity itself requires no linearity of the approximation applied to $s$; linearity is used only in the comparison with $Qf$.\n\\head{The input is not discovered by subtraction}\nThe data here include the correct $j$. Replacing it with a wrong $\\widehat j$ retains the error $j-\\widehat j$, even when $s$ is approximated perfectly. Multiple amplitudes and overlapping shapes are allowed if their full sum is supplied, but the identity does not estimate that sum from observations.\nOur finite periodic smoother is an illustration, not a new continuum approximation theorem. Its chosen boundary convention is part of the experiment.\n\\takeaway{Known-edge reconstruction removes the jump from the approximation problem. It does not remove the need to identify the jump.}","es\n\\[\nf-Qf=(s-Qs)+(j-Qj).\n\\]\nThe second error includes a new term caused by smoothing the discontinuity. In this finite example that term is concentrated near the two jump boundaries. The identity itself requires no linearity of the approximation applied to $s$; linearity is used only in the comparison with $Qf$.\n\\head{The input is not discovered by subtraction}\nThe data here include the correct $j$. Replacing it with a wrong $\\widehat j$ retains the error $j-\\widehat j$, even when $s$ is approximated perfectly. Multiple amplitudes and overlapping shapes are allowed if their full sum is supplied, but the identity does not estimate that sum from observations.\nOur finite periodic smoother is an illustration, not a new continuum approximation theorem. Its chosen boundary convention is part of the experiment.\n\\takeaway{Known-edge reconstruction removes the jump from the approximation problem. It does not remove the need to identify the jump.}","e second error includes a new term caused by smoothing the discontinuity. In this finite example that term is concentrated near the two jump boundaries. The identity itself requires no linearity of the approximation applied to $s$; linearity is used only in the comparison with $Qf$.\n\\head{The input is not discovered by subtraction}\nThe data here include the correct $j$. Replacing it with a wrong $\\widehat j$ retains the error $j-\\widehat j$, even when $s$ is approximated perfectly. Multiple amplitudes and overlapping shapes are allowed if their full sum is supplied, but the identity does not estimate that sum from observations.\nOur finite periodic smoother is an illustration, not a new continuum approximation theorem. Its chosen boundary convention is part of the experiment.\n\\takeaway{Known-edge reconstruction removes the jump from the approximation problem. It does not remove the need to identify the jump.}","erm caused by smoothing the discontinuity. In this finite example that term is concentrated near the two jump boundaries. The identity itself requires no linearity of the approximation applied to $s$; linearity is used only in the comparison with $Qf$.\n\\head{The input is not discovered by subtraction}\nThe data here include the correct $j$. Replacing it with a wrong $\\widehat j$ retains the error $j-\\widehat j$, even when $s$ is approximated perfectly. Multiple amplitudes and overlapping shapes are allowed if their full sum is supplied, but the identity does not estimate that sum from observations.\nOur finite periodic smoother is an illustration, not a new continuum approximation theorem. Its chosen boundary convention is part of the experiment.\n\\takeaway{Known-edge reconstruction removes the jump from the approximation problem. It does not remove the need to identify the jump.}","e discontinuity. In this finite example that term is concentrated near the two jump boundaries. The identity itself requires no linearity of the approximation applied to $s$; linearity is used only in the comparison with $Qf$.\n\\head{The input is not discovered by subtraction}\nThe data here include the correct $j$. Replacing it with a wrong $\\widehat j$ retains the error $j-\\widehat j$, even when $s$ is approximated perfectly. Multiple amplitudes and overlapping shapes are allowed if their full sum is supplied, but the identity does not estimate that sum from observations.\nOur finite periodic smoother is an illustration, not a new continuum approximation theorem. Its chosen boundary convention is part of the experiment.\n\\takeaway{Known-edge reconstruction removes the jump from the approximation problem. It does not remove the need to identify the jump.}","\\vpage{16}{Why a Fourier cutoff cannot reproduce this jump pointwise}{A CONTINUOUS PULSE / EXACT ENDPOINT VALUES}{Traditional monograph: Section 6 (p. 11), Section 8 (p. 13).}\nUse a separate period-one example on $[-1/2,1/2)$:\n\\[\nj(t)=\\mathbf 1_{\\{|t|<1/4\\}},\\qquad\nS_mj(t)=\\tfrac12+\\frac2\\pi\\sum_{n=1}^{m}\\frac{\\sin(\\pi n/2)}{n}\\cos(2\\pi nt).\n\\]\nThis is the exact finite Fourier formula. The drawn curves sample it numerically; the endpoint calculation below is symbolic.\n\\vfig{fourier_jump}\n\\head{One value is fixed at every cutoff}\nAt $t=\\pm1/4$, every summand vanishes because\n$\\sin(\\pi n/2)\\cos(\\pi n/2)=0$ for integer $n$. Hence\n\\[\nS_mj(\\pm1/4)=\\tfrac12 \\qquad\\text{for every }m\\ge0.\n\\]\nA larger cutoff resolves a narrower transition, but it does not make the finite trigonometric polynomial discontinuous. The two displayed cutoffs retain $11$ and $31$ signed frequency bins, respectively, not five and fifteen coeffi","\\vpage{16}{Why a Fourier cutoff cannot reproduce this jump pointwise}{A CONTINUOUS PULSE / EXACT ENDPOINT VALUES}{Traditional monograph: Section 6 (p. 11), Section 8 (p. 13).}\nUse a separate period-one example on $[-1/2,1/2)$:\n\\[\nj(t)=\\mathbf 1_{\\{|t|<1/4\\}},\\qquad\nS_mj(t)=\\tfrac12+\\frac2\\pi\\sum_{n=1}^{m}\\frac{\\sin(\\pi n/2)}{n}\\cos(2\\pi nt).\n\\]\nThis is the exact finite Fourier formula. The drawn curves sample it numerically; the endpoint calculation below is symbolic.\n\\vfig{fourier_jump}\n\\head{One value is fixed at every cutoff}\nAt $t=\\pm1/4$, every summand vanishes because\n$\\sin(\\pi n/2)\\cos(\\pi n/2)=0$ for integer $n$. Hence\n\\[\nS_mj(\\pm1/4)=\\tfrac12 \\qquad\\text{for every }m\\ge0.\n\\]\nA larger cutoff resolves a narrower transition, but it does not make the finite trigonometric polynomial discontinuous. The two displayed cutoffs retain $11$ and $31$ signed frequency bins, respectively, not five and fifteen coefficients.\n\\head{An exact known-packet comparison}\nTake $s(t)=\\frac14\\cos(2\\pi t)+\\frac1{10}\\cos(6\\pi t)$. For $m\\ge3$ this","ction 6 (p. 11), Section 8 (p. 13).}\nUse a separate period-one example on $[-1/2,1/2)$:\n\\[\nj(t)=\\mathbf 1_{\\{|t|<1/4\\}},\\qquad\nS_mj(t)=\\tfrac12+\\frac2\\pi\\sum_{n=1}^{m}\\frac{\\sin(\\pi n/2)}{n}\\cos(2\\pi nt).\n\\]\nThis is the exact finite Fourier formula. The drawn curves sample it numerically; the endpoint calculation below is symbolic.\n\\vfig{fourier_jump}\n\\head{One value is fixed at every cutoff}\nAt $t=\\pm1/4$, every summand vanishes because\n$\\sin(\\pi n/2)\\cos(\\pi n/2)=0$ for integer $n$. Hence\n\\[\nS_mj(\\pm1/4)=\\tfrac12 \\qquad\\text{for every }m\\ge0.\n\\]\nA larger cutoff resolves a narrower transition, but it does not make the finite trigonometric polynomial discontinuous. The two displayed cutoffs retain $11$ and $31$ signed frequency bins, respectively, not five and fifteen coefficients.\n\\head{An exact known-packet comparison}\nTake $s(t)=\\frac14\\cos(2\\pi t)+\\frac1{10}\\cos(6\\pi t)$. For $m\\ge3$ this smooth packet is already in the retained Fourier span. Thus $S_ms+j=s+j$ exactly in this special model, whereas $S_m(s+j)$ still replaces the pulse by $S_mj$.\nThe values assigned at the two endp…","separate period-one example on $[-1/2,1/2)$:\n\\[\nj(t)=\\mathbf 1_{\\{|t|<1/4\\}},\\qquad\nS_mj(t)=\\tfrac12+\\frac2\\pi\\sum_{n=1}^{m}\\frac{\\sin(\\pi n/2)}{n}\\cos(2\\pi nt).\n\\]\nThis is the exact finite Fourier formula. The drawn curves sample it numerically; the endpoint calculation below is symbolic.\n\\vfig{fourier_jump}\n\\head{One value is fixed at every cutoff}\nAt $t=\\pm1/4$, every summand vanishes because\n$\\sin(\\pi n/2)\\cos(\\pi n/2)=0$ for integer $n$. Hence\n\\[\nS_mj(\\pm1/4)=\\tfrac12 \\qquad\\text{for every }m\\ge0.\n\\]\nA larger cutoff resolves a narrower transition, but it does not make the finite trigonometric polynomial discontinuous. The two displayed cutoffs retain $11$ and $31$ signed frequency bins, respectively, not five and fifteen coefficients.\n\\head{An exact known-packet comparison}\nTake $s(t)=\\frac14\\cos(2\\pi t)+\\frac1{10}\\cos(6\\pi t)$. For $m\\ge3$ this smooth packet is already in the retained Fourier span. Thus $S_ms+j=s+j$ exactly in this special model, whereas $S_m(s+j)$ still replaces the pulse by $S_mj$.\nThe values assigned at the two endpoints matter for pointwise error, but do no…",":\n\\[\nj(t)=\\mathbf 1_{\\{|t|<1/4\\}},\\qquad\nS_mj(t)=\\tfrac12+\\frac2\\pi\\sum_{n=1}^{m}\\frac{\\sin(\\pi n/2)}{n}\\cos(2\\pi nt).\n\\]\nThis is the exact finite Fourier formula. The drawn curves sample it numerically; the endpoint calculation below is symbolic.\n\\vfig{fourier_jump}\n\\head{One value is fixed at every cutoff}\nAt $t=\\pm1/4$, every summand vanishes because\n$\\sin(\\pi n/2)\\cos(\\pi n/2)=0$ for integer $n$. Hence\n\\[\nS_mj(\\pm1/4)=\\tfrac12 \\qquad\\text{for every }m\\ge0.\n\\]\nA larger cutoff resolves a narrower transition, but it does not make the finite trigonometric polynomial discontinuous. The two displayed cutoffs retain $11$ and $31$ signed frequency bins, respectively, not five and fifteen coefficients.\n\\head{An exact known-packet comparison}\nTake $s(t)=\\frac14\\cos(2\\pi t)+\\frac1{10}\\cos(6\\pi t)$. For $m\\ge3$ this smooth packet is already in the retained Fourier span. Thus $S_ms+j=s+j$ exactly in this special model, whereas $S_m(s+j)$ still replaces the pulse by $S_mj$.\nThe values assigned at the two endpoints matter for pointwise error, but do not affect an $L^p$ norm for finite $p$. The …","athbf 1_{\\{|t|<1/4\\}},\\qquad\nS_mj(t)=\\tfrac12+\\frac2\\pi\\sum_{n=1}^{m}\\frac{\\sin(\\pi n/2)}{n}\\cos(2\\pi nt).\n\\]\nThis is the exact finite Fourier formula. The drawn curves sample it numerically; the endpoint calculation below is symbolic.\n\\vfig{fourier_jump}\n\\head{One value is fixed at every cutoff}\nAt $t=\\pm1/4$, every summand vanishes because\n$\\sin(\\pi n/2)\\cos(\\pi n/2)=0$ for integer $n$. Hence\n\\[\nS_mj(\\pm1/4)=\\tfrac12 \\qquad\\text{for every }m\\ge0.\n\\]\nA larger cutoff resolves a narrower transition, but it does not make the finite trigonometric polynomial discontinuous. The two displayed cutoffs retain $11$ and $31$ signed frequency bins, respectively, not five and fifteen coefficients.\n\\head{An exact known-packet comparison}\nTake $s(t)=\\frac14\\cos(2\\pi t)+\\frac1{10}\\cos(6\\pi t)$. For $m\\ge3$ this smooth packet is already in the retained Fourier span. Thus $S_ms+j=s+j$ exactly in this special model, whereas $S_m(s+j)$ still replaces the pulse by $S_mj$.\nThe values assigned at the two endpoints matter for pointwise error, but do not affect an $L^p$ norm for finite $p$. The illustrated …","lation below is symbolic.\n\\vfig{fourier_jump}\n\\head{One value is fixed at every cutoff}\nAt $t=\\pm1/4$, every summand vanishes because\n$\\sin(\\pi n/2)\\cos(\\pi n/2)=0$ for integer $n$. Hence\n\\[\nS_mj(\\pm1/4)=\\tfrac12 \\qquad\\text{for every }m\\ge0.\n\\]\nA larger cutoff resolves a narrower transition, but it does not make the finite trigonometric polynomial discontinuous. The two displayed cutoffs retain $11$ and $31$ signed frequency bins, respectively, not five and fifteen coefficients.\n\\head{An exact known-packet comparison}\nTake $s(t)=\\frac14\\cos(2\\pi t)+\\frac1{10}\\cos(6\\pi t)$. For $m\\ge3$ this smooth packet is already in the retained Fourier span. Thus $S_ms+j=s+j$ exactly in this special model, whereas $S_m(s+j)$ still replaces the pulse by $S_mj$.\nThe values assigned at the two endpoints matter for pointwise error, but do not affect an $L^p$ norm for finite $p$. The illustrated ringing is not a root enclosure or an asymptotic error certificate.\n\\takeaway{Increasing spectral resolution and retaining a known discontinuity are different operations, with different pointwise consequences.}","low is symbolic.\n\\vfig{fourier_jump}\n\\head{One value is fixed at every cutoff}\nAt $t=\\pm1/4$, every summand vanishes because\n$\\sin(\\pi n/2)\\cos(\\pi n/2)=0$ for integer $n$. Hence\n\\[\nS_mj(\\pm1/4)=\\tfrac12 \\qquad\\text{for every }m\\ge0.\n\\]\nA larger cutoff resolves a narrower transition, but it does not make the finite trigonometric polynomial discontinuous. The two displayed cutoffs retain $11$ and $31$ signed frequency bins, respectively, not five and fifteen coefficients.\n\\head{An exact known-packet comparison}\nTake $s(t)=\\frac14\\cos(2\\pi t)+\\frac1{10}\\cos(6\\pi t)$. For $m\\ge3$ this smooth packet is already in the retained Fourier span. Thus $S_ms+j=s+j$ exactly in this special model, whereas $S_m(s+j)$ still replaces the pulse by $S_mj$.\nThe values assigned at the two endpoints matter for pointwise error, but do not affect an $L^p$ norm for finite $p$. The illustrated ringing is not a root enclosure or an asymptotic error certificate.\n\\takeaway{Increasing spectral resolution and retaining a known discontinuity are different operations, with different pointwise consequences.}","se\n$\\sin(\\pi n/2)\\cos(\\pi n/2)=0$ for integer $n$. Hence\n\\[\nS_mj(\\pm1/4)=\\tfrac12 \\qquad\\text{for every }m\\ge0.\n\\]\nA larger cutoff resolves a narrower transition, but it does not make the finite trigonometric polynomial discontinuous. The two displayed cutoffs retain $11$ and $31$ signed frequency bins, respectively, not five and fifteen coefficients.\n\\head{An exact known-packet comparison}\nTake $s(t)=\\frac14\\cos(2\\pi t)+\\frac1{10}\\cos(6\\pi t)$. For $m\\ge3$ this smooth packet is already in the retained Fourier span. Thus $S_ms+j=s+j$ exactly in this special model, whereas $S_m(s+j)$ still replaces the pulse by $S_mj$.\nThe values assigned at the two endpoints matter for pointwise error, but do not affect an $L^p$ norm for finite $p$. The illustrated ringing is not a root enclosure or an asymptotic error certificate.\n\\takeaway{Increasing spectral resolution and retaining a known discontinuity are different operations, with different pointwise consequences.}","e\n\\[\nS_mj(\\pm1/4)=\\tfrac12 \\qquad\\text{for every }m\\ge0.\n\\]\nA larger cutoff resolves a narrower transition, but it does not make the finite trigonometric polynomial discontinuous. The two displayed cutoffs retain $11$ and $31$ signed frequency bins, respectively, not five and fifteen coefficients.\n\\head{An exact known-packet comparison}\nTake $s(t)=\\frac14\\cos(2\\pi t)+\\frac1{10}\\cos(6\\pi t)$. For $m\\ge3$ this smooth packet is already in the retained Fourier span. Thus $S_ms+j=s+j$ exactly in this special model, whereas $S_m(s+j)$ still replaces the pulse by $S_mj$.\nThe values assigned at the two endpoints matter for pointwise error, but do not affect an $L^p$ norm for finite $p$. The illustrated ringing is not a root enclosure or an asymptotic error certificate.\n\\takeaway{Increasing spectral resolution and retaining a known discontinuity are different operations, with different pointwise consequences.}","ff resolves a narrower transition, but it does not make the finite trigonometric polynomial discontinuous. The two displayed cutoffs retain $11$ and $31$ signed frequency bins, respectively, not five and fifteen coefficients.\n\\head{An exact known-packet comparison}\nTake $s(t)=\\frac14\\cos(2\\pi t)+\\frac1{10}\\cos(6\\pi t)$. For $m\\ge3$ this smooth packet is already in the retained Fourier span. Thus $S_ms+j=s+j$ exactly in this special model, whereas $S_m(s+j)$ still replaces the pulse by $S_mj$.\nThe values assigned at the two endpoints matter for pointwise error, but do not affect an $L^p$ norm for finite $p$. The illustrated ringing is not a root enclosure or an asymptotic error certificate.\n\\takeaway{Increasing spectral resolution and retaining a known discontinuity are different operations, with different pointwise consequences.}","ake the finite trigonometric polynomial discontinuous. The two displayed cutoffs retain $11$ and $31$ signed frequency bins, respectively, not five and fifteen coefficients.\n\\head{An exact known-packet comparison}\nTake $s(t)=\\frac14\\cos(2\\pi t)+\\frac1{10}\\cos(6\\pi t)$. For $m\\ge3$ this smooth packet is already in the retained Fourier span. Thus $S_ms+j=s+j$ exactly in this special model, whereas $S_m(s+j)$ still replaces the pulse by $S_mj$.\nThe values assigned at the two endpoints matter for pointwise error, but do not affect an $L^p$ norm for finite $p$. The illustrated ringing is not a root enclosure or an asymptotic error certificate.\n\\takeaway{Increasing spectral resolution and retaining a known discontinuity are different operations, with different pointwise consequences.}","discontinuous. The two displayed cutoffs retain $11$ and $31$ signed frequency bins, respectively, not five and fifteen coefficients.\n\\head{An exact known-packet comparison}\nTake $s(t)=\\frac14\\cos(2\\pi t)+\\frac1{10}\\cos(6\\pi t)$. For $m\\ge3$ this smooth packet is already in the retained Fourier span. Thus $S_ms+j=s+j$ exactly in this special model, whereas $S_m(s+j)$ still replaces the pulse by $S_mj$.\nThe values assigned at the two endpoints matter for pointwise error, but do not affect an $L^p$ norm for finite $p$. The illustrated ringing is not a root enclosure or an asymptotic error certificate.\n\\takeaway{Increasing spectral resolution and retaining a known discontinuity are different operations, with different pointwise consequences.}",", not five and fifteen coefficients.\n\\head{An exact known-packet comparison}\nTake $s(t)=\\frac14\\cos(2\\pi t)+\\frac1{10}\\cos(6\\pi t)$. For $m\\ge3$ this smooth packet is already in the retained Fourier span. Thus $S_ms+j=s+j$ exactly in this special model, whereas $S_m(s+j)$ still replaces the pulse by $S_mj$.\nThe values assigned at the two endpoints matter for pointwise error, but do not affect an $L^p$ norm for finite $p$. The illustrated ringing is not a root enclosure or an asymptotic error certificate.\n\\takeaway{Increasing spectral resolution and retaining a known discontinuity are different operations, with different pointwise consequences.}","coefficients.\n\\head{An exact known-packet comparison}\nTake $s(t)=\\frac14\\cos(2\\pi t)+\\frac1{10}\\cos(6\\pi t)$. For $m\\ge3$ this smooth packet is already in the retained Fourier span. Thus $S_ms+j=s+j$ exactly in this special model, whereas $S_m(s+j)$ still replaces the pulse by $S_mj$.\nThe values assigned at the two endpoints matter for pointwise error, but do not affect an $L^p$ norm for finite $p$. The illustrated ringing is not a root enclosure or an asymptotic error certificate.\n\\takeaway{Increasing spectral resolution and retaining a known discontinuity are different operations, with different pointwise consequences.}","\\vpage{17}{Small location errors, different norms}{MASK GEOMETRY / TRANSLATION BY ONE SIXTEENTH}{Traditional monograph: Theorem 7.1 (p. 13).}\nIn Lebesgue measure on the line, let\n\\[\n\\Omega=[\\tfrac14,\\tfrac34],\\qquad\n\\widehat\\Omega=[\\tfrac5{16},\\tfrac{13}{16}],\\qquad a=\\widehat a=3.\n\\]\nThe two intervals have the same length. Their disagreement consists of one missed strip and one extra strip, each of width $1/16$.\n\\vfig{mask_translation}\n\\[\n|\\Omega\\mathbin{\\triangle}\\widehat\\Omega|=\\tfrac18,\n\\qquad j-\\widehat j=3(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\n\\begin{center}\n\\begin{tabular}{c|ccc}\nerror measure & $L^1$ norm & squared $L^2$ norm & $L^\\infty$ norm","{Small location errors, different norms}{MASK GEOMETRY / TRANSLATION BY ONE SIXTEENTH}{Traditional monograph: Theorem 7.1 (p. 13).}\nIn Lebesgue measure on the line, let\n\\[\n\\Omega=[\\tfrac14,\\tfrac34],\\qquad\n\\widehat\\Omega=[\\tfrac5{16},\\tfrac{13}{16}],\\qquad a=\\widehat a=3.\n\\]\nThe two intervals have the same length. Their disagreement consists of one missed strip and one extra strip, each of width $1/16$.\n\\vfig{mask_translation}\n\\[\n|\\Omega\\mathbin{\\triangle}\\widehat\\Omega|=\\tfrac18,\n\\qquad j-\\widehat j=3(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\n\\begin{center}\n\\begin{tabular}{c|ccc}\nerror measure & $L^1$ norm & squared $L^2$ norm & $L^\\infty$ norm","norms}{MASK GEOMETRY / TRANSLATION BY ONE SIXTEENTH}{Traditional monograph: Theorem 7.1 (p. 13).}\nIn Lebesgue measure on the line, let\n\\[\n\\Omega=[\\tfrac14,\\tfrac34],\\qquad\n\\widehat\\Omega=[\\tfrac5{16},\\tfrac{13}{16}],\\qquad a=\\widehat a=3.\n\\]\nThe two intervals have the same length. Their disagreement consists of one missed strip and one extra strip, each of width $1/16$.\n\\vfig{mask_translation}\n\\[\n|\\Omega\\mathbin{\\triangle}\\widehat\\Omega|=\\tfrac18,\n\\qquad j-\\widehat j=3(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\n\\begin{center}\n\\begin{tabular}{c|ccc}\nerror measure & $L^1$ norm & squared $L^2$ norm & $L^\\infty$ norm","ehat\\Omega=[\\tfrac5{16},\\tfrac{13}{16}],\\qquad a=\\widehat a=3.\n\\]\nThe two intervals have the same length. Their disagreement consists of one missed strip and one extra strip, each of width $1/16$.\n\\vfig{mask_translation}\n\\[\n|\\Omega\\mathbin{\\triangle}\\widehat\\Omega|=\\tfrac18,\n\\qquad j-\\widehat j=3(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\n\\begin{center}\n\\begin{tabular}{c|ccc}\nerror measure & $L^1$ norm & squared $L^2$ norm & $L^\\infty$ norm","6},\\tfrac{13}{16}],\\qquad a=\\widehat a=3.\n\\]\nThe two intervals have the same length. Their disagreement consists of one missed strip and one extra strip, each of width $1/16$.\n\\vfig{mask_translation}\n\\[\n|\\Omega\\mathbin{\\triangle}\\widehat\\Omega|=\\tfrac18,\n\\qquad j-\\widehat j=3(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\n\\begin{center}\n\\begin{tabular}{c|ccc}\nerror measure & $L^1$ norm & squared $L^2$ norm & $L^\\infty$ norm","{16}],\\qquad a=\\widehat a=3.\n\\]\nThe two intervals have the same length. Their disagreement consists of one missed strip and one extra strip, each of width $1/16$.\n\\vfig{mask_translation}\n\\[\n|\\Omega\\mathbin{\\triangle}\\widehat\\Omega|=\\tfrac18,\n\\qquad j-\\widehat j=3(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\n\\begin{center}\n\\begin{tabular}{c|ccc}\nerror measure & $L^1$ norm & squared $L^2$ norm & $L^\\infty$ norm","hline\nexact value & $3/8$ & $9/8$ & $3$\n\\end{tabular}\n\\end{center}\n\\head{Area shrinks while the missed jump keeps its height}\nFor a translation by $\\delta$ with $0<|\\delta|<1/2$, the symmetric difference has measure $2|\\delta|$. At finite $p$,\n\\[\n\\|j-\\widehat j\\|_p=3(2|\\delta|)^{1/p}.\n\\]\nIt tends to zero as the displacement tends to zero. The essential supremum stays $3$ for every nonzero displacement, because the error reaches height $3$ on a set of positive measure. At exactly zero displacement all norms vanish.\nA nonempty disagreement set of measure zero would be different: it can affect the pointwise supremum while leaving the essential norm unchanged. This is why the paper distinguishes the two statements.\n\\ta","hline\nexact value & $3/8$ & $9/8$ & $3$\n\\end{tabular}\n\\end{center}\n\\head{Area shrinks while the missed jump keeps its height}\nFor a translation by $\\delta$ with $0<|\\delta|<1/2$, the symmetric difference has measure $2|\\delta|$. At finite $p$,\n\\[\n\\|j-\\widehat j\\|_p=3(2|\\delta|)^{1/p}.\n\\]\nIt tends to zero as the displacement tends to zero. The essential supremum stays $3$ for every nonzero displacement, because the error reaches height $3$ on a set of positive measure. At exactly zero displacement all norms vanish.\nA nonempty disagreement set of measure zero would be different: it can affect the pointwise supremum while leaving the essential norm unchanged. This is why the paper distinguishes the two statements.\n\\takeaway{A","hline\nexact value & $3/8$ & $9/8$ & $3$\n\\end{tabular}\n\\end{center}\n\\head{Area shrinks while the missed jump keeps its height}\nFor a translation by $\\delta$ with $0<|\\delta|<1/2$, the symmetric difference has measure $2|\\delta|$. At finite $p$,\n\\[\n\\|j-\\widehat j\\|_p=3(2|\\delta|)^{1/p}.\n\\]\nIt tends to zero as the displacement tends to zero. The essential supremum stays $3$ for every nonzero displacement, because the error reaches height $3$ on a set of positive measure. At exactly zero displacement all norms vanish.\nA nonempty disagreement set of measure zero would be different: it can affect the pointwise supremum while leaving the essential norm unchanged. This is why the paper distinguishes the two statements.\n\\takeaway{A shrin","hline\nexact value & $3/8$ & $9/8$ & $3$\n\\end{tabular}\n\\end{center}\n\\head{Area shrinks while the missed jump keeps its height}\nFor a translation by $\\delta$ with $0<|\\delta|<1/2$, the symmetric difference has measure $2|\\delta|$. At finite $p$,\n\\[\n\\|j-\\widehat j\\|_p=3(2|\\delta|)^{1/p}.\n\\]\nIt tends to zero as the displacement tends to zero. The essential supremum stays $3$ for every nonzero displacement, because the error reaches height $3$ on a set of positive measure. At exactly zero displacement all norms vanish.\nA nonempty disagreement set of measure zero would be different: it can affect the pointwise supremum while leaving the essential norm unchanged. This is why the paper distinguishes the two statements.\n\\takeaway{A shrinking boundary strip can give small integrated error and still contain a full-height jump error.}","\\end{tabular}\n\\end{center}\n\\head{Area shrinks while the missed jump keeps its height}\nFor a translation by $\\delta$ with $0<|\\delta|<1/2$, the symmetric difference has measure $2|\\delta|$. At finite $p$,\n\\[\n\\|j-\\widehat j\\|_p=3(2|\\delta|)^{1/p}.\n\\]\nIt tends to zero as the displacement tends to zero. The essential supremum stays $3$ for every nonzero displacement, because the error reaches height $3$ on a set of positive measure. At exactly zero displacement all norms vanish.\nA nonempty disagreement set of measure zero would be different: it can affect the pointwise supremum while leaving the essential norm unchanged. This is why the paper distinguishes the two statements.\n\\takeaway{A shrinking boundary strip can give small integrated error and still contain a full-height jump error.}","\\vpage{18}{Separate amplitude error from mask error}{COUNTING MEASURE / AN EXACT FINITE BUDGET}{Traditional monograph: Theorem 7.1 (p. 13).}\nOn the $16$ sample addresses, let $\\Omega=\\{6,\\ldots,11\\}$ and $\\widehat\\Omega=\\{7,\\ldots,12\\}$. Use true amplitude $a=3$ and estimate $\\widehat a=5/2$. There are six true samples, five overlapping samples, and two disagreement samples.\n\\vfig{error_budget}\nThe residual splits exactly as\n\\[\n3\\mathbf1_\\Omega-\\tfrac52\\mathbf1_{\\widehat\\Omega}\n=\\tfrac12\\mathbf1_\\Omega+\\tfrac52(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\nThe missed sample has error $3$, the five shared samples have error $1/2$, and the extra sample has error $-5/2$. With \\emph{counting measure},\n\\[\n\\|j-\\widehat j\\|_1=8,\n\\qquad \\|j-\\widehat j\\|_2^2=\\tfrac{33}{2},\n\\qquad \\|j-\\widehat j\\|_\\infty=3.\n\\]\n\\head{Read the theorem as a budget, no","\\vpage{18}{Separate amplitude error from mask error}{COUNTING MEASURE / AN EXACT FINITE BUDGET}{Traditional monograph: Theorem 7.1 (p. 13).}\nOn the $16$ sample addresses, let $\\Omega=\\{6,\\ldots,11\\}$ and $\\widehat\\Omega=\\{7,\\ldots,12\\}$. Use true amplitude $a=3$ and estimate $\\widehat a=5/2$. There are six true samples, five overlapping samples, and two disagreement samples.\n\\vfig{error_budget}\nThe residual splits exactly as\n\\[\n3\\mathbf1_\\Omega-\\tfrac52\\mathbf1_{\\widehat\\Omega}\n=\\tfrac12\\mathbf1_\\Omega+\\tfrac52(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\nThe missed sample has error $3$, the five shared samples have error $1/2$, and the extra sample has error $-5/2$. With \\emph{counting measure},\n\\[\n\\|j-\\widehat j\\|_1=8,\n\\qquad \\|j-\\widehat j\\|_2^2=\\tfrac{33}{2},\n\\qquad \\|j-\\widehat j\\|_\\infty=3.\n\\]\n\\head{Read the theorem as a budget, not an equality claim}\nHere the $L^1$ bound is at","\\vpage{18}{Separate amplitude error from mask error}{COUNTING MEASURE / AN EXACT FINITE BUDGET}{Traditional monograph: Theorem 7.1 (p. 13).}\nOn the $16$ sample addresses, let $\\Omega=\\{6,\\ldots,11\\}$ and $\\widehat\\Omega=\\{7,\\ldots,12\\}$. Use true amplitude $a=3$ and estimate $\\widehat a=5/2$. There are six true samples, five overlapping samples, and two disagreement samples.\n\\vfig{error_budget}\nThe residual splits exactly as\n\\[\n3\\mathbf1_\\Omega-\\tfrac52\\mathbf1_{\\widehat\\Omega}\n=\\tfrac12\\mathbf1_\\Omega+\\tfrac52(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\nThe missed sample has error $3$, the five shared samples have error $1/2$, and the extra sample has error $-5/2$. With \\emph{counting measure},\n\\[\n\\|j-\\widehat j\\|_1=8,\n\\qquad \\|j-\\widehat j\\|_2^2=\\tfrac{33}{2},\n\\qquad \\|j-\\widehat j\\|_\\infty=3.\n\\]\n\\head{Read the theorem as a budget, not an equality claim}\nHere the $L^1$ bound is attained:\n\\[\n|3-\\tfrac52|\\,|\\Omega|+\\tf","\\vpage{18}{Separate amplitude error from mask error}{COUNTING MEASURE / AN EXACT FINITE BUDGET}{Traditional monograph: Theorem 7.1 (p. 13).}\nOn the $16$ sample addresses, let $\\Omega=\\{6,\\ldots,11\\}$ and $\\widehat\\Omega=\\{7,\\ldots,12\\}$. Use true amplitude $a=3$ and estimate $\\widehat a=5/2$. There are six true samples, five overlapping samples, and two disagreement samples.\n\\vfig{error_budget}\nThe residual splits exactly as\n\\[\n3\\mathbf1_\\Omega-\\tfrac52\\mathbf1_{\\widehat\\Omega}\n=\\tfrac12\\mathbf1_\\Omega+\\tfrac52(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\nThe missed sample has error $3$, the five shared samples have error $1/2$, and the extra sample has error $-5/2$. With \\emph{counting measure},\n\\[\n\\|j-\\widehat j\\|_1=8,\n\\qquad \\|j-\\widehat j\\|_2^2=\\tfrac{33}{2},\n\\qquad \\|j-\\widehat j\\|_\\infty=3.\n\\]\n\\head{Read the theorem as a budget, not an equality claim}\nHere the $L^1$ bound is attained:\n\\[\n|3-\\tfrac52|\\,|\\Omega|+\\tfrac52|\\Omega\\triangle\\wide","\\vpage{18}{Separate amplitude error from mask error}{COUNTING MEASURE / AN EXACT FINITE BUDGET}{Traditional monograph: Theorem 7.1 (p. 13).}\nOn the $16$ sample addresses, let $\\Omega=\\{6,\\ldots,11\\}$ and $\\widehat\\Omega=\\{7,\\ldots,12\\}$. Use true amplitude $a=3$ and estimate $\\widehat a=5/2$. There are six true samples, five overlapping samples, and two disagreement samples.\n\\vfig{error_budget}\nThe residual splits exactly as\n\\[\n3\\mathbf1_\\Omega-\\tfrac52\\mathbf1_{\\widehat\\Omega}\n=\\tfrac12\\mathbf1_\\Omega+\\tfrac52(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\nThe missed sample has error $3$, the five shared samples have error $1/2$, and the extra sample has error $-5/2$. With \\emph{counting measure},\n\\[\n\\|j-\\widehat j\\|_1=8,\n\\qquad \\|j-\\widehat j\\|_2^2=\\tfrac{33}{2},\n\\qquad \\|j-\\widehat j\\|_\\infty=3.\n\\]\n\\head{Read the theorem as a budget, not an equality claim}\nHere the $L^1$ bound is attained:\n\\[\n|3-\\tfrac52|\\,|\\Omega|+\\tfrac52|\\Omega\\triangle\\widehat\\Omega|\n=\\tfrac12\\cdot6+\\tf","rror from mask error}{COUNTING MEASURE / AN EXACT FINITE BUDGET}{Traditional monograph: Theorem 7.1 (p. 13).}\nOn the $16$ sample addresses, let $\\Omega=\\{6,\\ldots,11\\}$ and $\\widehat\\Omega=\\{7,\\ldots,12\\}$. Use true amplitude $a=3$ and estimate $\\widehat a=5/2$. There are six true samples, five overlapping samples, and two disagreement samples.\n\\vfig{error_budget}\nThe residual splits exactly as\n\\[\n3\\mathbf1_\\Omega-\\tfrac52\\mathbf1_{\\widehat\\Omega}\n=\\tfrac12\\mathbf1_\\Omega+\\tfrac52(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\nThe missed sample has error $3$, the five shared samples have error $1/2$, and the extra sample has error $-5/2$. With \\emph{counting measure},\n\\[\n\\|j-\\widehat j\\|_1=8,\n\\qquad \\|j-\\widehat j\\|_2^2=\\tfrac{33}{2},\n\\qquad \\|j-\\widehat j\\|_\\infty=3.\n\\]\n\\head{Read the theorem as a budget, not an equality claim}\nHere the $L^1$ bound is attained:\n\\[\n|3-\\tfrac52|\\,|\\Omega|+\\tfrac52|\\Omega\\triangle\\widehat\\Omega|\n=\\tfrac12\\cdot6+\\tfrac52\\cdot2=8.\n\\]\nThe $L^2$ triangle bound is $\\frac12\\sqrt6+\\frac52\\sqrt2$, and need not be sharp. A background estimate adds another term …",",11\\}$ and $\\widehat\\Omega=\\{7,\\ldots,12\\}$. Use true amplitude $a=3$ and estimate $\\widehat a=5/2$. There are six true samples, five overlapping samples, and two disagreement samples.\n\\vfig{error_budget}\nThe residual splits exactly as\n\\[\n3\\mathbf1_\\Omega-\\tfrac52\\mathbf1_{\\widehat\\Omega}\n=\\tfrac12\\mathbf1_\\Omega+\\tfrac52(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\nThe missed sample has error $3$, the five shared samples have error $1/2$, and the extra sample has error $-5/2$. With \\emph{counting measure},\n\\[\n\\|j-\\widehat j\\|_1=8,\n\\qquad \\|j-\\widehat j\\|_2^2=\\tfrac{33}{2},\n\\qquad \\|j-\\widehat j\\|_\\infty=3.\n\\]\n\\head{Read the theorem as a budget, not an equality claim}\nHere the $L^1$ bound is attained:\n\\[\n|3-\\tfrac52|\\,|\\Omega|+\\tfrac52|\\Omega\\triangle\\widehat\\Omega|\n=\\tfrac12\\cdot6+\\tfrac52\\cdot2=8.\n\\]\nThe $L^2$ triangle bound is $\\frac12\\sqrt6+\\frac52\\sqrt2$, and need not be sharp. A background estimate adds another term $\\|s-\\widehat s\\|_p$; cancellation between terms can make the total smaller than the bound.\nNormalized RMSE divides the $L^2$ norm by $\\sqrt{16}$. A boundary-band","\\}$. Use true amplitude $a=3$ and estimate $\\widehat a=5/2$. There are six true samples, five overlapping samples, and two disagreement samples.\n\\vfig{error_budget}\nThe residual splits exactly as\n\\[\n3\\mathbf1_\\Omega-\\tfrac52\\mathbf1_{\\widehat\\Omega}\n=\\tfrac12\\mathbf1_\\Omega+\\tfrac52(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\nThe missed sample has error $3$, the five shared samples have error $1/2$, and the extra sample has error $-5/2$. With \\emph{counting measure},\n\\[\n\\|j-\\widehat j\\|_1=8,\n\\qquad \\|j-\\widehat j\\|_2^2=\\tfrac{33}{2},\n\\qquad \\|j-\\widehat j\\|_\\infty=3.\n\\]\n\\head{Read the theorem as a budget, not an equality claim}\nHere the $L^1$ bound is attained:\n\\[\n|3-\\tfrac52|\\,|\\Omega|+\\tfrac52|\\Omega\\triangle\\widehat\\Omega|\n=\\tfrac12\\cdot6+\\tfrac52\\cdot2=8.\n\\]\nThe $L^2$ triangle bound is $\\frac12\\sqrt6+\\frac52\\sqrt2$, and need not be sharp. A background estimate adds another term $\\|s-\\widehat s\\|_p$; cancellation between terms can make the total smaller than the bound.\nNormalized RMSE divides the $L^2$ norm by $\\sqrt{16}$. A boundary-band RMSE uses the band size instead. Changi…","mate $\\widehat a=5/2$. There are six true samples, five overlapping samples, and two disagreement samples.\n\\vfig{error_budget}\nThe residual splits exactly as\n\\[\n3\\mathbf1_\\Omega-\\tfrac52\\mathbf1_{\\widehat\\Omega}\n=\\tfrac12\\mathbf1_\\Omega+\\tfrac52(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\nThe missed sample has error $3$, the five shared samples have error $1/2$, and the extra sample has error $-5/2$. With \\emph{counting measure},\n\\[\n\\|j-\\widehat j\\|_1=8,\n\\qquad \\|j-\\widehat j\\|_2^2=\\tfrac{33}{2},\n\\qquad \\|j-\\widehat j\\|_\\infty=3.\n\\]\n\\head{Read the theorem as a budget, not an equality claim}\nHere the $L^1$ bound is attained:\n\\[\n|3-\\tfrac52|\\,|\\Omega|+\\tfrac52|\\Omega\\triangle\\widehat\\Omega|\n=\\tfrac12\\cdot6+\\tfrac52\\cdot2=8.\n\\]\nThe $L^2$ triangle bound is $\\frac12\\sqrt6+\\frac52\\sqrt2$, and need not be sharp. A background estimate adds another term $\\|s-\\widehat s\\|_p$; cancellation between terms can make the total smaller than the bound.\nNormalized RMSE divides the $L^2$ norm by $\\sqrt{16}$. A boundary-band RMSE uses the band size instead. Changing that denominator changes the report…","ue samples, five overlapping samples, and two disagreement samples.\n\\vfig{error_budget}\nThe residual splits exactly as\n\\[\n3\\mathbf1_\\Omega-\\tfrac52\\mathbf1_{\\widehat\\Omega}\n=\\tfrac12\\mathbf1_\\Omega+\\tfrac52(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\nThe missed sample has error $3$, the five shared samples have error $1/2$, and the extra sample has error $-5/2$. With \\emph{counting measure},\n\\[\n\\|j-\\widehat j\\|_1=8,\n\\qquad \\|j-\\widehat j\\|_2^2=\\tfrac{33}{2},\n\\qquad \\|j-\\widehat j\\|_\\infty=3.\n\\]\n\\head{Read the theorem as a budget, not an equality claim}\nHere the $L^1$ bound is attained:\n\\[\n|3-\\tfrac52|\\,|\\Omega|+\\tfrac52|\\Omega\\triangle\\widehat\\Omega|\n=\\tfrac12\\cdot6+\\tfrac52\\cdot2=8.\n\\]\nThe $L^2$ triangle bound is $\\frac12\\sqrt6+\\frac52\\sqrt2$, and need not be sharp. A background estimate adds another term $\\|s-\\widehat s\\|_p$; cancellation between terms can make the total smaller than the bound.\nNormalized RMSE divides the $L^2$ norm by $\\sqrt{16}$. A boundary-band RMSE uses the band size instead. Changing that denominator changes the reported quantity.\n\\takeaway{Mask error, ampl…","=\\tfrac12\\mathbf1_\\Omega+\\tfrac52(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\nThe missed sample has error $3$, the five shared samples have error $1/2$, and the extra sample has error $-5/2$. With \\emph{counting measure},\n\\[\n\\|j-\\widehat j\\|_1=8,\n\\qquad \\|j-\\widehat j\\|_2^2=\\tfrac{33}{2},\n\\qquad \\|j-\\widehat j\\|_\\infty=3.\n\\]\n\\head{Read the theorem as a budget, not an equality claim}\nHere the $L^1$ bound is attained:\n\\[\n|3-\\tfrac52|\\,|\\Omega|+\\tfrac52|\\Omega\\triangle\\widehat\\Omega|\n=\\tfrac12\\cdot6+\\tfrac52\\cdot2=8.\n\\]\nThe $L^2$ triangle bound is $\\frac12\\sqrt6+\\frac52\\sqrt2$, and need not be sharp. A background estimate adds another term $\\|s-\\widehat s\\|_p$; cancellation between terms can make the total smaller than the bound.\nNormalized RMSE divides the $L^2$ norm by $\\sqrt{16}$. A boundary-band RMSE uses the band size instead. Changing that denominator changes the reported quantity.\n\\takeaway{Mask error, amplitude error, and background error are separate mathematical inputs. A single reconstruction score conceals that separation.}","\\tfrac52(\\mathbf1_\\Omega-\\mathbf1_{\\widehat\\Omega}).\n\\]\nThe missed sample has error $3$, the five shared samples have error $1/2$, and the extra sample has error $-5/2$. With \\emph{counting measure},\n\\[\n\\|j-\\widehat j\\|_1=8,\n\\qquad \\|j-\\widehat j\\|_2^2=\\tfrac{33}{2},\n\\qquad \\|j-\\widehat j\\|_\\infty=3.\n\\]\n\\head{Read the theorem as a budget, not an equality claim}\nHere the $L^1$ bound is attained:\n\\[\n|3-\\tfrac52|\\,|\\Omega|+\\tfrac52|\\Omega\\triangle\\widehat\\Omega|\n=\\tfrac12\\cdot6+\\tfrac52\\cdot2=8.\n\\]\nThe $L^2$ triangle bound is $\\frac12\\sqrt6+\\frac52\\sqrt2$, and need not be sharp. A background estimate adds another term $\\|s-\\widehat s\\|_p$; cancellation between terms can make the total smaller than the bound.\nNormalized RMSE divides the $L^2$ norm by $\\sqrt{16}$. A boundary-band RMSE uses the band size instead. Changing that denominator changes the reported quantity.\n\\takeaway{Mask error, amplitude error, and background error are separate mathematical inputs. A single reconstruction score conceals that separation.}","amples have error $1/2$, and the extra sample has error $-5/2$. With \\emph{counting measure},\n\\[\n\\|j-\\widehat j\\|_1=8,\n\\qquad \\|j-\\widehat j\\|_2^2=\\tfrac{33}{2},\n\\qquad \\|j-\\widehat j\\|_\\infty=3.\n\\]\n\\head{Read the theorem as a budget, not an equality claim}\nHere the $L^1$ bound is attained:\n\\[\n|3-\\tfrac52|\\,|\\Omega|+\\tfrac52|\\Omega\\triangle\\widehat\\Omega|\n=\\tfrac12\\cdot6+\\tfrac52\\cdot2=8.\n\\]\nThe $L^2$ triangle bound is $\\frac12\\sqrt6+\\frac52\\sqrt2$, and need not be sharp. A background estimate adds another term $\\|s-\\widehat s\\|_p$; cancellation between terms can make the total smaller than the bound.\nNormalized RMSE divides the $L^2$ norm by $\\sqrt{16}$. A boundary-band RMSE uses the band size instead. Changing that denominator changes the reported quantity.\n\\takeaway{Mask error, amplitude error, and background error are separate mathematical inputs. A single reconstruction score conceals that separation.}","and the extra sample has error $-5/2$. With \\emph{counting measure},\n\\[\n\\|j-\\widehat j\\|_1=8,\n\\qquad \\|j-\\widehat j\\|_2^2=\\tfrac{33}{2},\n\\qquad \\|j-\\widehat j\\|_\\infty=3.\n\\]\n\\head{Read the theorem as a budget, not an equality claim}\nHere the $L^1$ bound is attained:\n\\[\n|3-\\tfrac52|\\,|\\Omega|+\\tfrac52|\\Omega\\triangle\\widehat\\Omega|\n=\\tfrac12\\cdot6+\\tfrac52\\cdot2=8.\n\\]\nThe $L^2$ triangle bound is $\\frac12\\sqrt6+\\frac52\\sqrt2$, and need not be sharp. A background estimate adds another term $\\|s-\\widehat s\\|_p$; cancellation between terms can make the total smaller than the bound.\nNormalized RMSE divides the $L^2$ norm by $\\sqrt{16}$. A boundary-band RMSE uses the band size instead. Changing that denominator changes the reported quantity.\n\\takeaway{Mask error, amplitude error, and background error are separate mathematical inputs. A single reconstruction score conceals that separation.}","\\qquad \\|j-\\widehat j\\|_2^2=\\tfrac{33}{2},\n\\qquad \\|j-\\widehat j\\|_\\infty=3.\n\\]\n\\head{Read the theorem as a budget, not an equality claim}\nHere the $L^1$ bound is attained:\n\\[\n|3-\\tfrac52|\\,|\\Omega|+\\tfrac52|\\Omega\\triangle\\widehat\\Omega|\n=\\tfrac12\\cdot6+\\tfrac52\\cdot2=8.\n\\]\nThe $L^2$ triangle bound is $\\frac12\\sqrt6+\\frac52\\sqrt2$, and need not be sharp. A background estimate adds another term $\\|s-\\widehat s\\|_p$; cancellation between terms can make the total smaller than the bound.\nNormalized RMSE divides the $L^2$ norm by $\\sqrt{16}$. A boundary-band RMSE uses the band size instead. Changing that denominator changes the reported quantity.\n\\takeaway{Mask error, amplitude error, and background error are separate mathematical inputs. A single reconstruction score conceals that separation.}","an equality claim}\nHere the $L^1$ bound is attained:\n\\[\n|3-\\tfrac52|\\,|\\Omega|+\\tfrac52|\\Omega\\triangle\\widehat\\Omega|\n=\\tfrac12\\cdot6+\\tfrac52\\cdot2=8.\n\\]\nThe $L^2$ triangle bound is $\\frac12\\sqrt6+\\frac52\\sqrt2$, and need not be sharp. A background estimate adds another term $\\|s-\\widehat s\\|_p$; cancellation between terms can make the total smaller than the bound.\nNormalized RMSE divides the $L^2$ norm by $\\sqrt{16}$. A boundary-band RMSE uses the band size instead. Changing that denominator changes the reported quantity.\n\\takeaway{Mask error, amplitude error, and background error are separate mathematical inputs. A single reconstruction score conceals that separation.}","laim}\nHere the $L^1$ bound is attained:\n\\[\n|3-\\tfrac52|\\,|\\Omega|+\\tfrac52|\\Omega\\triangle\\widehat\\Omega|\n=\\tfrac12\\cdot6+\\tfrac52\\cdot2=8.\n\\]\nThe $L^2$ triangle bound is $\\frac12\\sqrt6+\\frac52\\sqrt2$, and need not be sharp. A background estimate adds another term $\\|s-\\widehat s\\|_p$; cancellation between terms can make the total smaller than the bound.\nNormalized RMSE divides the $L^2$ norm by $\\sqrt{16}$. A boundary-band RMSE uses the band size instead. Changing that denominator changes the reported quantity.\n\\takeaway{Mask error, amplitude error, and background error are separate mathematical inputs. A single reconstruction score conceals that separation.}","\\vpage{19}{Geometry localizes error while data may leave it ambiguous}{A FINITE IMAGE / AN AMBIGUOUS OBSERVATION}{Traditional monograph: Section 7 (p. 12), Section 9 (p. 15).}\nOn the integer $8\\times8$ grid, choose the disk-shaped mask\n$\\Omega=\\{(i,j):(i-3)^2+(j-3)^2\\le5\\}$ and translate it one pixel to the right. Each mask contains $21$ pixels; their symmetric difference contains exactly $10$.\n\\vfig{mask_geometry}\n\\head{A geometric error calculation is not a segmentation algorithm}\nFor a common amplitude $a$, the squared counting-measure error is $10a^2$, entirely on the disagreement pixels. The full-image mean squared error is $10a^2/64$. Both are exact, but answer different normalization questions.\nNow consider the four-sample observation\n\\[\ny=(0,\\tfrac32,3,3).\n\\]\nIt is consistent with either true signal\n\\[\nu=(0,0,3,3),\\qquad v=(0,3,3,3)\n\\]\nunder deterministic sup-norm noise at most $","\\vpage{19}{Geometry localizes error while data may leave it ambiguous}{A FINITE IMAGE / AN AMBIGUOUS OBSERVATION}{Traditional monograph: Section 7 (p. 12), Section 9 (p. 15).}\nOn the integer $8\\times8$ grid, choose the disk-shaped mask\n$\\Omega=\\{(i,j):(i-3)^2+(j-3)^2\\le5\\}$ and translate it one pixel to the right. Each mask contains $21$ pixels; their symmetric difference contains exactly $10$.\n\\vfig{mask_geometry}\n\\head{A geometric error calculation is not a segmentation algorithm}\nFor a common amplitude $a$, the squared counting-measure error is $10a^2$, entirely on the disagreement pixels. The full-image mean squared error is $10a^2/64$. Both are exact, but answer different normalization questions.\nNow consider the four-sample observation\n\\[\ny=(0,\\tfrac32,3,3).\n\\]\nIt is consistent with either true signal\n\\[\nu=(0,0,3,3),\\qquad v=(0,3,3,3)\n\\]\nunder deterministic sup-norm noise at most $3/2$. Indeed $y=(u+v)/2$, and both error norms equal $3/2$. A rule receiv","\\vpage{19}{Geometry localizes error while data may leave it ambiguous}{A FINITE IMAGE / AN AMBIGUOUS OBSERVATION}{Traditional monograph: Section 7 (p. 12), Section 9 (p. 15).}\nOn the integer $8\\times8$ grid, choose the disk-shaped mask\n$\\Omega=\\{(i,j):(i-3)^2+(j-3)^2\\le5\\}$ and translate it one pixel to the right. Each mask contains $21$ pixels; their symmetric difference contains exactly $10$.\n\\vfig{mask_geometry}\n\\head{A geometric error calculation is not a segmentation algorithm}\nFor a common amplitude $a$, the squared counting-measure error is $10a^2$, entirely on the disagreement pixels. The full-image mean squared error is $10a^2/64$. Both are exact, but answer different normalization questions.\nNow consider the four-sample observation\n\\[\ny=(0,\\tfrac32,3,3).\n\\]\nIt is consistent with either true signal\n\\[\nu=(0,0,3,3),\\qquad v=(0,3,3,3)\n\\]\nunder deterministic sup-norm noise at most $3/2$. Indeed $y=(u+v)/2$, and both error norms equal $3/2$. A rule receiving only $y$ and that noise bound cannot be guaranteed to select","\\vpage{19}{Geometry localizes error while data may leave it ambiguous}{A FINITE IMAGE / AN AMBIGUOUS OBSERVATION}{Traditional monograph: Section 7 (p. 12), Section 9 (p. 15).}\nOn the integer $8\\times8$ grid, choose the disk-shaped mask\n$\\Omega=\\{(i,j):(i-3)^2+(j-3)^2\\le5\\}$ and translate it one pixel to the right. Each mask contains $21$ pixels; their symmetric difference contains exactly $10$.\n\\vfig{mask_geometry}\n\\head{A geometric error calculation is not a segmentation algorithm}\nFor a common amplitude $a$, the squared counting-measure error is $10a^2$, entirely on the disagreement pixels. The full-image mean squared error is $10a^2/64$. Both are exact, but answer different normalization questions.\nNow consider the four-sample observation\n\\[\ny=(0,\\tfrac32,3,3).\n\\]\nIt is consistent with either true signal\n\\[\nu=(0,0,3,3),\\qquad v=(0,3,3,3)\n\\]\nunder deterministic sup-norm noise at most $3/2$. Indeed $y=(u+v)/2$, and both error norms equal $3/2$. A rule receiving only $y$ and that noise bound cannot be guaranteed to select both distinct true masks correctly.\n\\head{What additional","}{Traditional monograph: Section 7 (p. 12), Section 9 (p. 15).}\nOn the integer $8\\times8$ grid, choose the disk-shaped mask\n$\\Omega=\\{(i,j):(i-3)^2+(j-3)^2\\le5\\}$ and translate it one pixel to the right. Each mask contains $21$ pixels; their symmetric difference contains exactly $10$.\n\\vfig{mask_geometry}\n\\head{A geometric error calculation is not a segmentation algorithm}\nFor a common amplitude $a$, the squared counting-measure error is $10a^2$, entirely on the disagreement pixels. The full-image mean squared error is $10a^2/64$. Both are exact, but answer different normalization questions.\nNow consider the four-sample observation\n\\[\ny=(0,\\tfrac32,3,3).\n\\]\nIt is consistent with either true signal\n\\[\nu=(0,0,3,3),\\qquad v=(0,3,3,3)\n\\]\nunder deterministic sup-norm noise at most $3/2$. Indeed $y=(u+v)/2$, and both error norms equal $3/2$. A rule receiving only $y$ and that noise bound cannot be guaranteed to select both distinct true masks correctly.\n\\head{What additional structure would have to do}\nA lower noise level, separation, contrast, geometry, or a specified prior can change the","Section 9 (p. 15).}\nOn the integer $8\\times8$ grid, choose the disk-shaped mask\n$\\Omega=\\{(i,j):(i-3)^2+(j-3)^2\\le5\\}$ and translate it one pixel to the right. Each mask contains $21$ pixels; their symmetric difference contains exactly $10$.\n\\vfig{mask_geometry}\n\\head{A geometric error calculation is not a segmentation algorithm}\nFor a common amplitude $a$, the squared counting-measure error is $10a^2$, entirely on the disagreement pixels. The full-image mean squared error is $10a^2/64$. Both are exact, but answer different normalization questions.\nNow consider the four-sample observation\n\\[\ny=(0,\\tfrac32,3,3).\n\\]\nIt is consistent with either true signal\n\\[\nu=(0,0,3,3),\\qquad v=(0,3,3,3)\n\\]\nunder deterministic sup-norm noise at most $3/2$. Indeed $y=(u+v)/2$, and both error norms equal $3/2$. A rule receiving only $y$ and that noise bound cannot be guaranteed to select both distinct true masks correctly.\n\\head{What additional structure would have to do}\nA lower noise level, separation, contrast, geometry, or a specified prior can change the identification problem. The example rules o…","Omega=\\{(i,j):(i-3)^2+(j-3)^2\\le5\\}$ and translate it one pixel to the right. Each mask contains $21$ pixels; their symmetric difference contains exactly $10$.\n\\vfig{mask_geometry}\n\\head{A geometric error calculation is not a segmentation algorithm}\nFor a common amplitude $a$, the squared counting-measure error is $10a^2$, entirely on the disagreement pixels. The full-image mean squared error is $10a^2/64$. Both are exact, but answer different normalization questions.\nNow consider the four-sample observation\n\\[\ny=(0,\\tfrac32,3,3).\n\\]\nIt is consistent with either true signal\n\\[\nu=(0,0,3,3),\\qquad v=(0,3,3,3)\n\\]\nunder deterministic sup-norm noise at most $3/2$. Indeed $y=(u+v)/2$, and both error norms equal $3/2$. A rule receiving only $y$ and that noise bound cannot be guaranteed to select both distinct true masks correctly.\n\\head{What additional structure would have to do}\nA lower noise level, separation, contrast, geometry, or a specified prior can change the identification problem. The example rules out a guarantee at the \\emph{stated} noise radius and signal pair, not all recovery…","symmetric difference contains exactly $10$.\n\\vfig{mask_geometry}\n\\head{A geometric error calculation is not a segmentation algorithm}\nFor a common amplitude $a$, the squared counting-measure error is $10a^2$, entirely on the disagreement pixels. The full-image mean squared error is $10a^2/64$. Both are exact, but answer different normalization questions.\nNow consider the four-sample observation\n\\[\ny=(0,\\tfrac32,3,3).\n\\]\nIt is consistent with either true signal\n\\[\nu=(0,0,3,3),\\qquad v=(0,3,3,3)\n\\]\nunder deterministic sup-norm noise at most $3/2$. Indeed $y=(u+v)/2$, and both error norms equal $3/2$. A rule receiving only $y$ and that noise bound cannot be guaranteed to select both distinct true masks correctly.\n\\head{What additional structure would have to do}\nA lower noise level, separation, contrast, geometry, or a specified prior can change the identification problem. The example rules out a guarantee at the \\emph{stated} noise radius and signal pair, not all recovery results. The manuscript's positive-contrast estimator is deliberately narrower than a universal segmentation method…","ead{A geometric error calculation is not a segmentation algorithm}\nFor a common amplitude $a$, the squared counting-measure error is $10a^2$, entirely on the disagreement pixels. The full-image mean squared error is $10a^2/64$. Both are exact, but answer different normalization questions.\nNow consider the four-sample observation\n\\[\ny=(0,\\tfrac32,3,3).\n\\]\nIt is consistent with either true signal\n\\[\nu=(0,0,3,3),\\qquad v=(0,3,3,3)\n\\]\nunder deterministic sup-norm noise at most $3/2$. Indeed $y=(u+v)/2$, and both error norms equal $3/2$. A rule receiving only $y$ and that noise bound cannot be guaranteed to select both distinct true masks correctly.\n\\head{What additional structure would have to do}\nA lower noise level, separation, contrast, geometry, or a specified prior can change the identification problem. The example rules out a guarantee at the \\emph{stated} noise radius and signal pair, not all recovery results. The manuscript's positive-contrast estimator is deliberately narrower than a universal segmentation method.\n\\takeaway{Exact synthesis can preserve an identified boundary. It…","amplitude $a$, the squared counting-measure error is $10a^2$, entirely on the disagreement pixels. The full-image mean squared error is $10a^2/64$. Both are exact, but answer different normalization questions.\nNow consider the four-sample observation\n\\[\ny=(0,\\tfrac32,3,3).\n\\]\nIt is consistent with either true signal\n\\[\nu=(0,0,3,3),\\qquad v=(0,3,3,3)\n\\]\nunder deterministic sup-norm noise at most $3/2$. Indeed $y=(u+v)/2$, and both error norms equal $3/2$. A rule receiving only $y$ and that noise bound cannot be guaranteed to select both distinct true masks correctly.\n\\head{What additional structure would have to do}\nA lower noise level, separation, contrast, geometry, or a specified prior can change the identification problem. The example rules out a guarantee at the \\emph{stated} noise radius and signal pair, not all recovery results. The manuscript's positive-contrast estimator is deliberately narrower than a universal segmentation method.\n\\takeaway{Exact synthesis can preserve an identified boundary. It cannot decide between two boundaries that the observation model makes indisting…",", the squared counting-measure error is $10a^2$, entirely on the disagreement pixels. The full-image mean squared error is $10a^2/64$. Both are exact, but answer different normalization questions.\nNow consider the four-sample observation\n\\[\ny=(0,\\tfrac32,3,3).\n\\]\nIt is consistent with either true signal\n\\[\nu=(0,0,3,3),\\qquad v=(0,3,3,3)\n\\]\nunder deterministic sup-norm noise at most $3/2$. Indeed $y=(u+v)/2$, and both error norms equal $3/2$. A rule receiving only $y$ and that noise bound cannot be guaranteed to select both distinct true masks correctly.\n\\head{What additional structure would have to do}\nA lower noise level, separation, contrast, geometry, or a specified prior can change the identification problem. The example rules out a guarantee at the \\emph{stated} noise radius and signal pair, not all recovery results. The manuscript's positive-contrast estimator is deliberately narrower than a universal segmentation method.\n\\takeaway{Exact synthesis can preserve an identified boundary. It cannot decide between two boundaries that the observation model makes indistinguishable.}","$10a^2$, entirely on the disagreement pixels. The full-image mean squared error is $10a^2/64$. Both are exact, but answer different normalization questions.\nNow consider the four-sample observation\n\\[\ny=(0,\\tfrac32,3,3).\n\\]\nIt is consistent with either true signal\n\\[\nu=(0,0,3,3),\\qquad v=(0,3,3,3)\n\\]\nunder deterministic sup-norm noise at most $3/2$. Indeed $y=(u+v)/2$, and both error norms equal $3/2$. A rule receiving only $y$ and that noise bound cannot be guaranteed to select both distinct true masks correctly.\n\\head{What additional structure would have to do}\nA lower noise level, separation, contrast, geometry, or a specified prior can change the identification problem. The example rules out a guarantee at the \\emph{stated} noise radius and signal pair, not all recovery results. The manuscript's positive-contrast estimator is deliberately narrower than a universal segmentation method.\n\\takeaway{Exact synthesis can preserve an identified boundary. It cannot decide between two boundaries that the observation model makes indistinguishable.}","greement pixels. The full-image mean squared error is $10a^2/64$. Both are exact, but answer different normalization questions.\nNow consider the four-sample observation\n\\[\ny=(0,\\tfrac32,3,3).\n\\]\nIt is consistent with either true signal\n\\[\nu=(0,0,3,3),\\qquad v=(0,3,3,3)\n\\]\nunder deterministic sup-norm noise at most $3/2$. Indeed $y=(u+v)/2$, and both error norms equal $3/2$. A rule receiving only $y$ and that noise bound cannot be guaranteed to select both distinct true masks correctly.\n\\head{What additional structure would have to do}\nA lower noise level, separation, contrast, geometry, or a specified prior can change the identification problem. The example rules out a guarantee at the \\emph{stated} noise radius and signal pair, not all recovery results. The manuscript's positive-contrast estimator is deliberately narrower than a universal segmentation method.\n\\takeaway{Exact synthesis can preserve an identified boundary. It cannot decide between two boundaries that the observation model makes indistinguishable.}","ISE SIGNALS VERSUS CAUCHY COEFFICIENTS}{Traditional monograph: Theorem 12.1 (p. 24).}\nThe word ``product'' must be typed. Multiplying two signal values at the same time address is not the same operation as multiplying generating series. The distinction changes which packets can interact.\n\\vfig{product_conventions}\n\\head{Pointwise multiplication preserves the same address}\nFor time-residue masks, $(P_{N,i}x)[n](P_{N,j}y)[n]=0$ when $i\\ne j$. Therefore\n\\[\nP_{N,k}(x\\cdot y)=(P_{N,k}x)\\cdot(P_{N,k}y).\n\\]\nHere the dot means multiplication at the same sample, not convolution.\n\\head{Cauchy multiplication adds degrees}\nFor formal series $A(t),B(t)$, the coefficient of $t^n$ is a sum over $a+b=n$. Consequently the output residue is $i+j\\pmod N$:\n\\[\nP_{N,k}(AB)=\\sum_{i+j\\equiv k\\ (N)}(P_{N,i}A)(P_{N,j}B).\n\\]\nFor $N=3$, take $A=1+2t+3t^2$ and $B=2+t^2$. Direct multiplication gives\n\\[\nAB=2+4t+7t^2+2t^3+3t^4,\n\\qquad P_{3,0}(AB)=2+2t^3.\n\\]\nThe $2t^3$ term comes from residues $1+2$, not $0+0$. At power zero, the empty product contributes $1$ only in residue zero. These are formal coefficient identi…","TS}{Traditional monograph: Theorem 12.1 (p. 24).}\nThe word ``product'' must be typed. Multiplying two signal values at the same time address is not the same operation as multiplying generating series. The distinction changes which packets can interact.\n\\vfig{product_conventions}\n\\head{Pointwise multiplication preserves the same address}\nFor time-residue masks, $(P_{N,i}x)[n](P_{N,j}y)[n]=0$ when $i\\ne j$. Therefore\n\\[\nP_{N,k}(x\\cdot y)=(P_{N,k}x)\\cdot(P_{N,k}y).\n\\]\nHere the dot means multiplication at the same sample, not convolution.\n\\head{Cauchy multiplication adds degrees}\nFor formal series $A(t),B(t)$, the coefficient of $t^n$ is a sum over $a+b=n$. Consequently the output residue is $i+j\\pmod N$:\n\\[\nP_{N,k}(AB)=\\sum_{i+j\\equiv k\\ (N)}(P_{N,i}A)(P_{N,j}B).\n\\]\nFor $N=3$, take $A=1+2t+3t^2$ and $B=2+t^2$. Direct multiplication gives\n\\[\nAB=2+4t+7t^2+2t^3+3t^4,\n\\qquad P_{3,0}(AB)=2+2t^3.\n\\]\nThe $2t^3$ term comes from residues $1+2$, not $0+0$. At power zero, the empty product contributes $1$ only in residue zero. These are formal coefficient identities and require no convergence assu…","raph: Theorem 12.1 (p. 24).}\nThe word ``product'' must be typed. Multiplying two signal values at the same time address is not the same operation as multiplying generating series. The distinction changes which packets can interact.\n\\vfig{product_conventions}\n\\head{Pointwise multiplication preserves the same address}\nFor time-residue masks, $(P_{N,i}x)[n](P_{N,j}y)[n]=0$ when $i\\ne j$. Therefore\n\\[\nP_{N,k}(x\\cdot y)=(P_{N,k}x)\\cdot(P_{N,k}y).\n\\]\nHere the dot means multiplication at the same sample, not convolution.\n\\head{Cauchy multiplication adds degrees}\nFor formal series $A(t),B(t)$, the coefficient of $t^n$ is a sum over $a+b=n$. Consequently the output residue is $i+j\\pmod N$:\n\\[\nP_{N,k}(AB)=\\sum_{i+j\\equiv k\\ (N)}(P_{N,i}A)(P_{N,j}B).\n\\]\nFor $N=3$, take $A=1+2t+3t^2$ and $B=2+t^2$. Direct multiplication gives\n\\[\nAB=2+4t+7t^2+2t^3+3t^4,\n\\qquad P_{3,0}(AB)=2+2t^3.\n\\]\nThe $2t^3$ term comes from residues $1+2$, not $0+0$. At power zero, the empty product contributes $1$ only in residue zero. These are formal coefficient identities and require no convergence assumption.\n\\takeaway{Bef…","he distinction changes which packets can interact.\n\\vfig{product_conventions}\n\\head{Pointwise multiplication preserves the same address}\nFor time-residue masks, $(P_{N,i}x)[n](P_{N,j}y)[n]=0$ when $i\\ne j$. Therefore\n\\[\nP_{N,k}(x\\cdot y)=(P_{N,k}x)\\cdot(P_{N,k}y).\n\\]\nHere the dot means multiplication at the same sample, not convolution.\n\\head{Cauchy multiplication adds degrees}\nFor formal series $A(t),B(t)$, the coefficient of $t^n$ is a sum over $a+b=n$. Consequently the output residue is $i+j\\pmod N$:\n\\[\nP_{N,k}(AB)=\\sum_{i+j\\equiv k\\ (N)}(P_{N,i}A)(P_{N,j}B).\n\\]\nFor $N=3$, take $A=1+2t+3t^2$ and $B=2+t^2$. Direct multiplication gives\n\\[\nAB=2+4t+7t^2+2t^3+3t^4,\n\\qquad P_{3,0}(AB)=2+2t^3.\n\\]\nThe $2t^3$ term comes from residues $1+2$, not $0+0$. At power zero, the empty product contributes $1$ only in residue zero. These are formal coefficient identities and require no convergence assumption.\n\\takeaway{Before applying a packet multiplication rule, specify whether multiplication acts on sample values or on series coefficients.}","kets can interact.\n\\vfig{product_conventions}\n\\head{Pointwise multiplication preserves the same address}\nFor time-residue masks, $(P_{N,i}x)[n](P_{N,j}y)[n]=0$ when $i\\ne j$. Therefore\n\\[\nP_{N,k}(x\\cdot y)=(P_{N,k}x)\\cdot(P_{N,k}y).\n\\]\nHere the dot means multiplication at the same sample, not convolution.\n\\head{Cauchy multiplication adds degrees}\nFor formal series $A(t),B(t)$, the coefficient of $t^n$ is a sum over $a+b=n$. Consequently the output residue is $i+j\\pmod N$:\n\\[\nP_{N,k}(AB)=\\sum_{i+j\\equiv k\\ (N)}(P_{N,i}A)(P_{N,j}B).\n\\]\nFor $N=3$, take $A=1+2t+3t^2$ and $B=2+t^2$. Direct multiplication gives\n\\[\nAB=2+4t+7t^2+2t^3+3t^4,\n\\qquad P_{3,0}(AB)=2+2t^3.\n\\]\nThe $2t^3$ term comes from residues $1+2$, not $0+0$. At power zero, the empty product contributes $1$ only in residue zero. These are formal coefficient identities and require no convergence assumption.\n\\takeaway{Before applying a packet multiplication rule, specify whether multiplication acts on sample values or on series coefficients.}","vfig{product_conventions}\n\\head{Pointwise multiplication preserves the same address}\nFor time-residue masks, $(P_{N,i}x)[n](P_{N,j}y)[n]=0$ when $i\\ne j$. Therefore\n\\[\nP_{N,k}(x\\cdot y)=(P_{N,k}x)\\cdot(P_{N,k}y).\n\\]\nHere the dot means multiplication at the same sample, not convolution.\n\\head{Cauchy multiplication adds degrees}\nFor formal series $A(t),B(t)$, the coefficient of $t^n$ is a sum over $a+b=n$. Consequently the output residue is $i+j\\pmod N$:\n\\[\nP_{N,k}(AB)=\\sum_{i+j\\equiv k\\ (N)}(P_{N,i}A)(P_{N,j}B).\n\\]\nFor $N=3$, take $A=1+2t+3t^2$ and $B=2+t^2$. Direct multiplication gives\n\\[\nAB=2+4t+7t^2+2t^3+3t^4,\n\\qquad P_{3,0}(AB)=2+2t^3.\n\\]\nThe $2t^3$ term comes from residues $1+2$, not $0+0$. At power zero, the empty product contributes $1$ only in residue zero. These are formal coefficient identities and require no convergence assumption.\n\\takeaway{Before applying a packet multiplication rule, specify whether multiplication acts on sample values or on series coefficients.}","ltiplication preserves the same address}\nFor time-residue masks, $(P_{N,i}x)[n](P_{N,j}y)[n]=0$ when $i\\ne j$. Therefore\n\\[\nP_{N,k}(x\\cdot y)=(P_{N,k}x)\\cdot(P_{N,k}y).\n\\]\nHere the dot means multiplication at the same sample, not convolution.\n\\head{Cauchy multiplication adds degrees}\nFor formal series $A(t),B(t)$, the coefficient of $t^n$ is a sum over $a+b=n$. Consequently the output residue is $i+j\\pmod N$:\n\\[\nP_{N,k}(AB)=\\sum_{i+j\\equiv k\\ (N)}(P_{N,i}A)(P_{N,j}B).\n\\]\nFor $N=3$, take $A=1+2t+3t^2$ and $B=2+t^2$. Direct multiplication gives\n\\[\nAB=2+4t+7t^2+2t^3+3t^4,\n\\qquad P_{3,0}(AB)=2+2t^3.\n\\]\nThe $2t^3$ term comes from residues $1+2$, not $0+0$. At power zero, the empty product contributes $1$ only in residue zero. These are formal coefficient identities and require no convergence assumption.\n\\takeaway{Before applying a packet multiplication rule, specify whether multiplication acts on sample values or on series coefficients.}","eserves the same address}\nFor time-residue masks, $(P_{N,i}x)[n](P_{N,j}y)[n]=0$ when $i\\ne j$. Therefore\n\\[\nP_{N,k}(x\\cdot y)=(P_{N,k}x)\\cdot(P_{N,k}y).\n\\]\nHere the dot means multiplication at the same sample, not convolution.\n\\head{Cauchy multiplication adds degrees}\nFor formal series $A(t),B(t)$, the coefficient of $t^n$ is a sum over $a+b=n$. Consequently the output residue is $i+j\\pmod N$:\n\\[\nP_{N,k}(AB)=\\sum_{i+j\\equiv k\\ (N)}(P_{N,i}A)(P_{N,j}B).\n\\]\nFor $N=3$, take $A=1+2t+3t^2$ and $B=2+t^2$. Direct multiplication gives\n\\[\nAB=2+4t+7t^2+2t^3+3t^4,\n\\qquad P_{3,0}(AB)=2+2t^3.\n\\]\nThe $2t^3$ term comes from residues $1+2$, not $0+0$. At power zero, the empty product contributes $1$ only in residue zero. These are formal coefficient identities and require no convergence assumption.\n\\takeaway{Before applying a packet multiplication rule, specify whether multiplication acts on sample values or on series coefficients.}","_{N,j}y)[n]=0$ when $i\\ne j$. Therefore\n\\[\nP_{N,k}(x\\cdot y)=(P_{N,k}x)\\cdot(P_{N,k}y).\n\\]\nHere the dot means multiplication at the same sample, not convolution.\n\\head{Cauchy multiplication adds degrees}\nFor formal series $A(t),B(t)$, the coefficient of $t^n$ is a sum over $a+b=n$. Consequently the output residue is $i+j\\pmod N$:\n\\[\nP_{N,k}(AB)=\\sum_{i+j\\equiv k\\ (N)}(P_{N,i}A)(P_{N,j}B).\n\\]\nFor $N=3$, take $A=1+2t+3t^2$ and $B=2+t^2$. Direct multiplication gives\n\\[\nAB=2+4t+7t^2+2t^3+3t^4,\n\\qquad P_{3,0}(AB)=2+2t^3.\n\\]\nThe $2t^3$ term comes from residues $1+2$, not $0+0$. At power zero, the empty product contributes $1$ only in residue zero. These are formal coefficient identities and require no convergence assumption.\n\\takeaway{Before applying a packet multiplication rule, specify whether multiplication acts on sample values or on series coefficients.}","0$ when $i\\ne j$. Therefore\n\\[\nP_{N,k}(x\\cdot y)=(P_{N,k}x)\\cdot(P_{N,k}y).\n\\]\nHere the dot means multiplication at the same sample, not convolution.\n\\head{Cauchy multiplication adds degrees}\nFor formal series $A(t),B(t)$, the coefficient of $t^n$ is a sum over $a+b=n$. Consequently the output residue is $i+j\\pmod N$:\n\\[\nP_{N,k}(AB)=\\sum_{i+j\\equiv k\\ (N)}(P_{N,i}A)(P_{N,j}B).\n\\]\nFor $N=3$, take $A=1+2t+3t^2$ and $B=2+t^2$. Direct multiplication gives\n\\[\nAB=2+4t+7t^2+2t^3+3t^4,\n\\qquad P_{3,0}(AB)=2+2t^3.\n\\]\nThe $2t^3$ term comes from residues $1+2$, not $0+0$. At power zero, the empty product contributes $1$ only in residue zero. These are formal coefficient identities and require no convergence assumption.\n\\takeaway{Before applying a packet multiplication rule, specify whether multiplication acts on sample values or on series coefficients.}","Therefore\n\\[\nP_{N,k}(x\\cdot y)=(P_{N,k}x)\\cdot(P_{N,k}y).\n\\]\nHere the dot means multiplication at the same sample, not convolution.\n\\head{Cauchy multiplication adds degrees}\nFor formal series $A(t),B(t)$, the coefficient of $t^n$ is a sum over $a+b=n$. Consequently the output residue is $i+j\\pmod N$:\n\\[\nP_{N,k}(AB)=\\sum_{i+j\\equiv k\\ (N)}(P_{N,i}A)(P_{N,j}B).\n\\]\nFor $N=3$, take $A=1+2t+3t^2$ and $B=2+t^2$. Direct multiplication gives\n\\[\nAB=2+4t+7t^2+2t^3+3t^4,\n\\qquad P_{3,0}(AB)=2+2t^3.\n\\]\nThe $2t^3$ term comes from residues $1+2$, not $0+0$. At power zero, the empty product contributes $1$ only in residue zero. These are formal coefficient identities and require no convergence assumption.\n\\takeaway{Before applying a packet multiplication rule, specify whether multiplication acts on sample values or on series coefficients.}",")\\cdot(P_{N,k}y).\n\\]\nHere the dot means multiplication at the same sample, not convolution.\n\\head{Cauchy multiplication adds degrees}\nFor formal series $A(t),B(t)$, the coefficient of $t^n$ is a sum over $a+b=n$. Consequently the output residue is $i+j\\pmod N$:\n\\[\nP_{N,k}(AB)=\\sum_{i+j\\equiv k\\ (N)}(P_{N,i}A)(P_{N,j}B).\n\\]\nFor $N=3$, take $A=1+2t+3t^2$ and $B=2+t^2$. Direct multiplication gives\n\\[\nAB=2+4t+7t^2+2t^3+3t^4,\n\\qquad P_{3,0}(AB)=2+2t^3.\n\\]\nThe $2t^3$ term comes from residues $1+2$, not $0+0$. At power zero, the empty product contributes $1$ only in residue zero. These are formal coefficient identities and require no convergence assumption.\n\\takeaway{Before applying a packet multiplication rule, specify whether multiplication acts on sample values or on series coefficients.}","e same sample, not convolution.\n\\head{Cauchy multiplication adds degrees}\nFor formal series $A(t),B(t)$, the coefficient of $t^n$ is a sum over $a+b=n$. Consequently the output residue is $i+j\\pmod N$:\n\\[\nP_{N,k}(AB)=\\sum_{i+j\\equiv k\\ (N)}(P_{N,i}A)(P_{N,j}B).\n\\]\nFor $N=3$, take $A=1+2t+3t^2$ and $B=2+t^2$. Direct multiplication gives\n\\[\nAB=2+4t+7t^2+2t^3+3t^4,\n\\qquad P_{3,0}(AB)=2+2t^3.\n\\]\nThe $2t^3$ term comes from residues $1+2$, not $0+0$. At power zero, the empty product contributes $1$ only in residue zero. These are formal coefficient identities and require no convergence assumption.\n\\takeaway{Before applying a packet multiplication rule, specify whether multiplication acts on sample values or on series coefficients.}","\\head{Cauchy multiplication adds degrees}\nFor formal series $A(t),B(t)$, the coefficient of $t^n$ is a sum over $a+b=n$. Consequently the output residue is $i+j\\pmod N$:\n\\[\nP_{N,k}(AB)=\\sum_{i+j\\equiv k\\ (N)}(P_{N,i}A)(P_{N,j}B).\n\\]\nFor $N=3$, take $A=1+2t+3t^2$ and $B=2+t^2$. Direct multiplication gives\n\\[\nAB=2+4t+7t^2+2t^3+3t^4,\n\\qquad P_{3,0}(AB)=2+2t^3.\n\\]\nThe $2t^3$ term comes from residues $1+2$, not $0+0$. At power zero, the empty product contributes $1$ only in residue zero. These are formal coefficient identities and require no convergence assumption.\n\\takeaway{Before applying a packet multiplication rule, specify whether multiplication acts on sample values or on series coefficients.}","y multiplication adds degrees}\nFor formal series $A(t),B(t)$, the coefficient of $t^n$ is a sum over $a+b=n$. Consequently the output residue is $i+j\\pmod N$:\n\\[\nP_{N,k}(AB)=\\sum_{i+j\\equiv k\\ (N)}(P_{N,i}A)(P_{N,j}B).\n\\]\nFor $N=3$, take $A=1+2t+3t^2$ and $B=2+t^2$. Direct multiplication gives\n\\[\nAB=2+4t+7t^2+2t^3+3t^4,\n\\qquad P_{3,0}(AB)=2+2t^3.\n\\]\nThe $2t^3$ term comes from residues $1+2$, not $0+0$. At power zero, the empty product contributes $1$ only in residue zero. These are formal coefficient identities and require no convergence assumption.\n\\takeaway{Before applying a packet multiplication rule, specify whether multiplication acts on sample values or on series coefficients.}","t),B(t)$, the coefficient of $t^n$ is a sum over $a+b=n$. Consequently the output residue is $i+j\\pmod N$:\n\\[\nP_{N,k}(AB)=\\sum_{i+j\\equiv k\\ (N)}(P_{N,i}A)(P_{N,j}B).\n\\]\nFor $N=3$, take $A=1+2t+3t^2$ and $B=2+t^2$. Direct multiplication gives\n\\[\nAB=2+4t+7t^2+2t^3+3t^4,\n\\qquad P_{3,0}(AB)=2+2t^3.\n\\]\nThe $2t^3$ term comes from residues $1+2$, not $0+0$. At power zero, the empty product contributes $1$ only in residue zero. These are formal coefficient identities and require no convergence assumption.\n\\takeaway{Before applying a packet multiplication rule, specify whether multiplication acts on sample values or on series coefficients.}","\\vpage{21}{Nonlinear neighbors cross channel boundaries}{FULL INPUT TUPLES / EXACT CYCLIC INDEXING}{Traditional monograph: Section 12 (p. 24).}\nFor the periodic running signal define\n\\[\nF(x)[n]=x[n]+\\tfrac12x[n]x[n-1].\n\\]\nThis is nonlinear and not a convolution operator. Nevertheless its two-channel representation is exact, provided the neighbor comes from the correct channel and index.\n\\vfig{nonlinear_shift}\nPut $e[q]=x[2q]$ and $o[q]=x[2q+1]$, with channel indices taken modulo eight. Then\n\\[\n\\begin{aligned}\nF(x)[2q]&=e[q]+\\tfrac12 e[q]o[q-1],","ar neighbors cross channel boundaries}{FULL INPUT TUPLES / EXACT CYCLIC INDEXING}{Traditional monograph: Section 12 (p. 24).}\nFor the periodic running signal define\n\\[\nF(x)[n]=x[n]+\\tfrac12x[n]x[n-1].\n\\]\nThis is nonlinear and not a convolution operator. Nevertheless its two-channel representation is exact, provided the neighbor comes from the correct channel and index.\n\\vfig{nonlinear_shift}\nPut $e[q]=x[2q]$ and $o[q]=x[2q+1]$, with channel indices taken modulo eight. Then\n\\[\n\\begin{aligned}\nF(x)[2q]&=e[q]+\\tfrac12 e[q]o[q-1],","s channel boundaries}{FULL INPUT TUPLES / EXACT CYCLIC INDEXING}{Traditional monograph: Section 12 (p. 24).}\nFor the periodic running signal define\n\\[\nF(x)[n]=x[n]+\\tfrac12x[n]x[n-1].\n\\]\nThis is nonlinear and not a convolution operator. Nevertheless its two-channel representation is exact, provided the neighbor comes from the correct channel and index.\n\\vfig{nonlinear_shift}\nPut $e[q]=x[2q]$ and $o[q]=x[2q+1]$, with channel indices taken modulo eight. Then\n\\[\n\\begin{aligned}\nF(x)[2q]&=e[q]+\\tfrac12 e[q]o[q-1],","}{Traditional monograph: Section 12 (p. 24).}\nFor the periodic running signal define\n\\[\nF(x)[n]=x[n]+\\tfrac12x[n]x[n-1].\n\\]\nThis is nonlinear and not a convolution operator. Nevertheless its two-channel representation is exact, provided the neighbor comes from the correct channel and index.\n\\vfig{nonlinear_shift}\nPut $e[q]=x[2q]$ and $o[q]=x[2q+1]$, with channel indices taken modulo eight. Then\n\\[\n\\begin{aligned}\nF(x)[2q]&=e[q]+\\tfrac12 e[q]o[q-1],\\\\\nF(x)[2q+1]&=o[q]+\\tfrac12 o[q]e[q].\n\\end{aligned}\n\\]\nThe even output needs the \\emph{previous} odd sample; the odd output needs the current even sample. This is the same carry in the address system that appeared in the one-sample delay matrix.\n\\head{Check two actual outputs}\nThe running values are $x[1]=1/4$, $x[2]=1/2$, and $x[3]=1/4$, so\n\\[\nF(x)[2]=\\tfrac12+\\tfrac12\\tfrac12\\tfrac14=\\tfrac9{16},\n\\qquad F(x)[3]=\\tfrac14+\\tfrac12\\tfrac14\\tfrac12=\\tfrac5{16}.\n\\]\nApplying the same formula independently within each decimated array would use the wrong neighbors. Keeping the full channel tuple avoids that loss of information. The supplied suit…","F(x)[2q+1]&=o[q]+\\tfrac12 o[q]e[q].\n\\end{aligned}\n\\]\nThe even output needs the \\emph{previous} odd sample; the odd output needs the current even sample. This is the same carry in the address system that appeared in the one-sample delay matrix.\n\\head{Check two actual outputs}\nThe running values are $x[1]=1/4$, $x[2]=1/2$, and $x[3]=1/4$, so\n\\[\nF(x)[2]=\\tfrac12+\\tfrac12\\tfrac12\\tfrac14=\\tfrac9{16},\n\\qquad F(x)[3]=\\tfrac14+\\tfrac12\\tfrac14\\tfrac12=\\tfrac5{16}.\n\\]\nApplying the same formula independently within each decimated array would use the wrong neighbors. Keeping the full channel tuple avoids that loss of information. The supplied suite compares every output sample in the two implementations using exact rational arithmetic.\nNo convergence, contraction, or approximation property follows merely from writing these exact coordinates. Those require hypotheses on the nonlinear map and the problem being solved.\n\\takeaway{Exact packet coordinates remain available for nonlinear maps. They do not turn","F(x)[2q+1]&=o[q]+\\tfrac12 o[q]e[q].\n\\end{aligned}\n\\]\nThe even output needs the \\emph{previous} odd sample; the odd output needs the current even sample. This is the same carry in the address system that appeared in the one-sample delay matrix.\n\\head{Check two actual outputs}\nThe running values are $x[1]=1/4$, $x[2]=1/2$, and $x[3]=1/4$, so\n\\[\nF(x)[2]=\\tfrac12+\\tfrac12\\tfrac12\\tfrac14=\\tfrac9{16},\n\\qquad F(x)[3]=\\tfrac14+\\tfrac12\\tfrac14\\tfrac12=\\tfrac5{16}.\n\\]\nApplying the same formula independently within each decimated array would use the wrong neighbors. Keeping the full channel tuple avoids that loss of information. The supplied suite compares every output sample in the two implementations using exact rational arithmetic.\nNo convergence, contraction, or approximation property follows merely from writing these exact coordinates. Those require hypotheses on the nonlinear map and the problem being solved.\n\\takeaway{Exact packet coordinates remain available for nonlinear maps. They do not turn cross-chann","F(x)[2q+1]&=o[q]+\\tfrac12 o[q]e[q].\n\\end{aligned}\n\\]\nThe even output needs the \\emph{previous} odd sample; the odd output needs the current even sample. This is the same carry in the address system that appeared in the one-sample delay matrix.\n\\head{Check two actual outputs}\nThe running values are $x[1]=1/4$, $x[2]=1/2$, and $x[3]=1/4$, so\n\\[\nF(x)[2]=\\tfrac12+\\tfrac12\\tfrac12\\tfrac14=\\tfrac9{16},\n\\qquad F(x)[3]=\\tfrac14+\\tfrac12\\tfrac14\\tfrac12=\\tfrac5{16}.\n\\]\nApplying the same formula independently within each decimated array would use the wrong neighbors. Keeping the full channel tuple avoids that loss of information. The supplied suite compares every output sample in the two implementations using exact rational arithmetic.\nNo convergence, contraction, or approximation property follows merely from writing these exact coordinates. Those require hypotheses on the nonlinear map and the problem being solved.\n\\takeaway{Exact packet coordinates remain available for nonlinear maps. They do not turn cross-channel interactions","F(x)[2q+1]&=o[q]+\\tfrac12 o[q]e[q].\n\\end{aligned}\n\\]\nThe even output needs the \\emph{previous} odd sample; the odd output needs the current even sample. This is the same carry in the address system that appeared in the one-sample delay matrix.\n\\head{Check two actual outputs}\nThe running values are $x[1]=1/4$, $x[2]=1/2$, and $x[3]=1/4$, so\n\\[\nF(x)[2]=\\tfrac12+\\tfrac12\\tfrac12\\tfrac14=\\tfrac9{16},\n\\qquad F(x)[3]=\\tfrac14+\\tfrac12\\tfrac14\\tfrac12=\\tfrac5{16}.\n\\]\nApplying the same formula independently within each decimated array would use the wrong neighbors. Keeping the full channel tuple avoids that loss of information. The supplied suite compares every output sample in the two implementations using exact rational arithmetic.\nNo convergence, contraction, or approximation property follows merely from writing these exact coordinates. Those require hypotheses on the nonlinear map and the problem being solved.\n\\takeaway{Exact packet coordinates remain available for nonlinear maps. They do not turn cross-channel interactions into independent channel problems.}","\\vpage{22}{One reflection circle, four very different impedances}{THE CAYLEY CHART / EXACT QUARTER TURNS}{Traditional monograph: Theorem 13.2 (p. 26).}\nUse normalized impedance $z$ and reflection coefficient $\\Gamma$:\n\\[\n\\Gamma=\\frac{z-1}{z+1},\\qquad z=\\frac{1+\\Gamma}{1-\\Gamma}.\n\\]\nThe circle $|\\Gamma|=1/2$ is a constant-mismatch orbit. Its four quarter-turn points map to the exact impedances shown below.\n\\vfig{cayley_orbit}\n\\begin{center}\n\\begin{tabular}{c|cccc}\n$\\Gamma$ & $1/2$ & $i/2$ & $-1/2$ & $-i/2$","CAYLEY CHART / EXACT QUARTER TURNS}{Traditional monograph: Theorem 13.2 (p. 26).}\nUse normalized impedance $z$ and reflection coefficient $\\Gamma$:\n\\[\n\\Gamma=\\frac{z-1}{z+1},\\qquad z=\\frac{1+\\Gamma}{1-\\Gamma}.\n\\]\nThe circle $|\\Gamma|=1/2$ is a constant-mismatch orbit. Its four quarter-turn points map to the exact impedances shown below.\n\\vfig{cayley_orbit}\n\\begin{center}\n\\begin{tabular}{c|cccc}\n$\\Gamma$ & $1/2$ & $i/2$ & $-1/2$ & $-i/2$","RT / EXACT QUARTER TURNS}{Traditional monograph: Theorem 13.2 (p. 26).}\nUse normalized impedance $z$ and reflection coefficient $\\Gamma$:\n\\[\n\\Gamma=\\frac{z-1}{z+1},\\qquad z=\\frac{1+\\Gamma}{1-\\Gamma}.\n\\]\nThe circle $|\\Gamma|=1/2$ is a constant-mismatch orbit. Its four quarter-turn points map to the exact impedances shown below.\n\\vfig{cayley_orbit}\n\\begin{center}\n\\begin{tabular}{c|cccc}\n$\\Gamma$ & $1/2$ & $i/2$ & $-1/2$ & $-i/2$","CT QUARTER TURNS}{Traditional monograph: Theorem 13.2 (p. 26).}\nUse normalized impedance $z$ and reflection coefficient $\\Gamma$:\n\\[\n\\Gamma=\\frac{z-1}{z+1},\\qquad z=\\frac{1+\\Gamma}{1-\\Gamma}.\n\\]\nThe circle $|\\Gamma|=1/2$ is a constant-mismatch orbit. Its four quarter-turn points map to the exact impedances shown below.\n\\vfig{cayley_orbit}\n\\begin{center}\n\\begin{tabular}{c|cccc}\n$\\Gamma$ & $1/2$ & $i/2$ & $-1/2$ & $-i/2$","ER TURNS}{Traditional monograph: Theorem 13.2 (p. 26).}\nUse normalized impedance $z$ and reflection coefficient $\\Gamma$:\n\\[\n\\Gamma=\\frac{z-1}{z+1},\\qquad z=\\frac{1+\\Gamma}{1-\\Gamma}.\n\\]\nThe circle $|\\Gamma|=1/2$ is a constant-mismatch orbit. Its four quarter-turn points map to the exact impedances shown below.\n\\vfig{cayley_orbit}\n\\begin{center}\n\\begin{tabular}{c|cccc}\n$\\Gamma$ & $1/2$ & $i/2$ & $-1/2$ & $-i/2$","{Traditional monograph: Theorem 13.2 (p. 26).}\nUse normalized impedance $z$ and reflection coefficient $\\Gamma$:\n\\[\n\\Gamma=\\frac{z-1}{z+1},\\qquad z=\\frac{1+\\Gamma}{1-\\Gamma}.\n\\]\nThe circle $|\\Gamma|=1/2$ is a constant-mismatch orbit. Its four quarter-turn points map to the exact impedances shown below.\n\\vfig{cayley_orbit}\n\\begin{center}\n\\begin{tabular}{c|cccc}\n$\\Gamma$ & $1/2$ & $i/2$ & $-1/2$ & $-i/2$","hline\n$z$ & $3$ & $3/5+4i/5$ & $1/3$ & $3/5-4i/5$\n\\end{tabular}\n\\end{center}\n\\head{The coordinates preserve structure, not Euclidean distances}\nA quarter turn is a rotation in the reflection disk. After pulling back, it is a M\\\"obius map, not a Euclidean rotation of the impedance plane. Equal reflection magnitudes need not give equal impedance magnitudes. The opposite points $1/2$ and $-1/2$ become $3$ and $1/3$. The continuous impedance locus is the circle $(\\operatorname{Re}z-5/3)^2+(\\operatorname{Im}z)^2=16/9$, also drawn in the plate.\nFinite passive impedances with $\\operatorname{Re}z\\ge0$ map to the closed disk except $\\Gamma=1$. That missing point is obtained only by adjoining impedance infini","hline\n$z$ & $3$ & $3/5+4i/5$ & $1/3$ & $3/5-4i/5$\n\\end{tabular}\n\\end{center}\n\\head{The coordinates preserve structure, not Euclidean distances}\nA quarter turn is a rotation in the reflection disk. After pulling back, it is a M\\\"obius map, not a Euclidean rotation of the impedance plane. Equal reflection magnitudes need not give equal impedance magnitudes. The opposite points $1/2$ and $-1/2$ become $3$ and $1/3$. The continuous impedance locus is the circle $(\\operatorname{Re}z-5/3)^2+(\\operatorname{Im}z)^2=16/9$, also drawn in the plate.\nFinite passive impedances with $\\operatorname{Re}z\\ge0$ map to the closed disk except $\\Gamma=1$. That missing point is obtained only by adjoining impedance infinity. Th","hline\n$z$ & $3$ & $3/5+4i/5$ & $1/3$ & $3/5-4i/5$\n\\end{tabular}\n\\end{center}\n\\head{The coordinates preserve structure, not Euclidean distances}\nA quarter turn is a rotation in the reflection disk. After pulling back, it is a M\\\"obius map, not a Euclidean rotation of the impedance plane. Equal reflection magnitudes need not give equal impedance magnitudes. The opposite points $1/2$ and $-1/2$ become $3$ and $1/3$. The continuous impedance locus is the circle $(\\operatorname{Re}z-5/3)^2+(\\operatorname{Im}z)^2=16/9$, also drawn in the plate.\nFinite passive impedances with $\\operatorname{Re}z\\ge0$ map to the closed disk except $\\Gamma=1$. That missing point is obtained only by adjoining impedance infinity. The interior ma","hline\n$z$ & $3$ & $3/5+4i/5$ & $1/3$ & $3/5-4i/5$\n\\end{tabular}\n\\end{center}\n\\head{The coordinates preserve structure, not Euclidean distances}\nA quarter turn is a rotation in the reflection disk. After pulling back, it is a M\\\"obius map, not a Euclidean rotation of the impedance plane. Equal reflection magnitudes need not give equal impedance magnitudes. The opposite points $1/2$ and $-1/2$ become $3$ and $1/3$. The continuous impedance locus is the circle $(\\operatorname{Re}z-5/3)^2+(\\operatorname{Im}z)^2=16/9$, also drawn in the plate.\nFinite passive impedances with $\\operatorname{Re}z\\ge0$ map to the closed disk except $\\Gamma=1$. That missing point is obtained only by adjoining impedance infinity. The interior maps to th","hline\n$z$ & $3$ & $3/5+4i/5$ & $1/3$ & $3/5-4i/5$\n\\end{tabular}\n\\end{center}\n\\head{The coordinates preserve structure, not Euclidean distances}\nA quarter turn is a rotation in the reflection disk. After pulling back, it is a M\\\"obius map, not a Euclidean rotation of the impedance plane. Equal reflection magnitudes need not give equal impedance magnitudes. The opposite points $1/2$ and $-1/2$ become $3$ and $1/3$. The continuous impedance locus is the circle $(\\operatorname{Re}z-5/3)^2+(\\operatorname{Im}z)^2=16/9$, also drawn in the plate.\nFinite passive impedances with $\\operatorname{Re}z\\ge0$ map to the closed disk except $\\Gamma=1$. That missing point is obtained only by adjoining impedance infinity. The interior maps to the open right","hline\n$z$ & $3$ & $3/5+4i/5$ & $1/3$ & $3/5-4i/5$\n\\end{tabular}\n\\end{center}\n\\head{The coordinates preserve structure, not Euclidean distances}\nA quarter turn is a rotation in the reflection disk. After pulling back, it is a M\\\"obius map, not a Euclidean rotation of the impedance plane. Equal reflection magnitudes need not give equal impedance magnitudes. The opposite points $1/2$ and $-1/2$ become $3$ and $1/3$. The continuous impedance locus is the circle $(\\operatorname{Re}z-5/3)^2+(\\operatorname{Im}z)^2=16/9$, also drawn in the plate.\nFinite passive impedances with $\\operatorname{Re}z\\ge0$ map to the closed disk except $\\Gamma=1$. That missing point is obtained only by adjoining impedance infinity. The interior maps to the open right half-plane.\nThe classical Smith grid supplies the geometry; the cyclic labels specify exact rotations on it. Neither a new physical law nor a new optimal sampling theorem is implied by this coordinate description.\n\\takeaway{A simple action in one chart can be nonlinear in another. Conjugacy preserves the action law, not the apparent","hline\n$z$ & $3$ & $3/5+4i/5$ & $1/3$ & $3/5-4i/5$\n\\end{tabular}\n\\end{center}\n\\head{The coordinates preserve structure, not Euclidean distances}\nA quarter turn is a rotation in the reflection disk. After pulling back, it is a M\\\"obius map, not a Euclidean rotation of the impedance plane. Equal reflection magnitudes need not give equal impedance magnitudes. The opposite points $1/2$ and $-1/2$ become $3$ and $1/3$. The continuous impedance locus is the circle $(\\operatorname{Re}z-5/3)^2+(\\operatorname{Im}z)^2=16/9$, also drawn in the plate.\nFinite passive impedances with $\\operatorname{Re}z\\ge0$ map to the closed disk except $\\Gamma=1$. That missing point is obtained only by adjoining impedance infinity. The interior maps to the open right half-plane.\nThe classical Smith grid supplies the geometry; the cyclic labels specify exact rotations on it. Neither a new physical law nor a new optimal sampling theorem is implied by this coordinate description.\n\\takeaway{A simple action in one chart can be nonlinear in another. Conjugacy preserves the action law, not the apparent shape of a","ne\n$z$ & $3$ & $3/5+4i/5$ & $1/3$ & $3/5-4i/5$\n\\end{tabular}\n\\end{center}\n\\head{The coordinates preserve structure, not Euclidean distances}\nA quarter turn is a rotation in the reflection disk. After pulling back, it is a M\\\"obius map, not a Euclidean rotation of the impedance plane. Equal reflection magnitudes need not give equal impedance magnitudes. The opposite points $1/2$ and $-1/2$ become $3$ and $1/3$. The continuous impedance locus is the circle $(\\operatorname{Re}z-5/3)^2+(\\operatorname{Im}z)^2=16/9$, also drawn in the plate.\nFinite passive impedances with $\\operatorname{Re}z\\ge0$ map to the closed disk except $\\Gamma=1$. That missing point is obtained only by adjoining impedance infinity. The interior maps to the open right half-plane.\nThe classical Smith grid supplies the geometry; the cyclic labels specify exact rotations on it. Neither a new physical law nor a new optimal sampling theorem is implied by this coordinate description.\n\\takeaway{A simple action in one chart can be nonlinear in another. Conjugacy preserves the action law, not the apparent shape of a displacem…","$3$ & $3/5+4i/5$ & $1/3$ & $3/5-4i/5$\n\\end{tabular}\n\\end{center}\n\\head{The coordinates preserve structure, not Euclidean distances}\nA quarter turn is a rotation in the reflection disk. After pulling back, it is a M\\\"obius map, not a Euclidean rotation of the impedance plane. Equal reflection magnitudes need not give equal impedance magnitudes. The opposite points $1/2$ and $-1/2$ become $3$ and $1/3$. The continuous impedance locus is the circle $(\\operatorname{Re}z-5/3)^2+(\\operatorname{Im}z)^2=16/9$, also drawn in the plate.\nFinite passive impedances with $\\operatorname{Re}z\\ge0$ map to the closed disk except $\\Gamma=1$. That missing point is obtained only by adjoining impedance infinity. The interior maps to the open right half-plane.\nThe classical Smith grid supplies the geometry; the cyclic labels specify exact rotations on it. Neither a new physical law nor a new optimal sampling theorem is implied by this coordinate description.\n\\takeaway{A simple action in one chart can be nonlinear in another. Conjugacy preserves the action law, not the apparent shape of a displacement.}","\\end{center}\n\\head{The coordinates preserve structure, not Euclidean distances}\nA quarter turn is a rotation in the reflection disk. After pulling back, it is a M\\\"obius map, not a Euclidean rotation of the impedance plane. Equal reflection magnitudes need not give equal impedance magnitudes. The opposite points $1/2$ and $-1/2$ become $3$ and $1/3$. The continuous impedance locus is the circle $(\\operatorname{Re}z-5/3)^2+(\\operatorname{Im}z)^2=16/9$, also drawn in the plate.\nFinite passive impedances with $\\operatorname{Re}z\\ge0$ map to the closed disk except $\\Gamma=1$. That missing point is obtained only by adjoining impedance infinity. The interior maps to the open right half-plane.\nThe classical Smith grid supplies the geometry; the cyclic labels specify exact rotations on it. Neither a new physical law nor a new optimal sampling theorem is implied by this coordinate description.\n\\takeaway{A simple action in one chart can be nonlinear in another. Conjugacy preserves the action law, not the apparent shape of a displacement.}","he reflection disk. After pulling back, it is a M\\\"obius map, not a Euclidean rotation of the impedance plane. Equal reflection magnitudes need not give equal impedance magnitudes. The opposite points $1/2$ and $-1/2$ become $3$ and $1/3$. The continuous impedance locus is the circle $(\\operatorname{Re}z-5/3)^2+(\\operatorname{Im}z)^2=16/9$, also drawn in the plate.\nFinite passive impedances with $\\operatorname{Re}z\\ge0$ map to the closed disk except $\\Gamma=1$. That missing point is obtained only by adjoining impedance infinity. The interior maps to the open right half-plane.\nThe classical Smith grid supplies the geometry; the cyclic labels specify exact rotations on it. Neither a new physical law nor a new optimal sampling theorem is implied by this coordinate description.\n\\takeaway{A simple action in one chart can be nonlinear in another. Conjugacy preserves the action law, not the apparent shape of a displacement.}","s map, not a Euclidean rotation of the impedance plane. Equal reflection magnitudes need not give equal impedance magnitudes. The opposite points $1/2$ and $-1/2$ become $3$ and $1/3$. The continuous impedance locus is the circle $(\\operatorname{Re}z-5/3)^2+(\\operatorname{Im}z)^2=16/9$, also drawn in the plate.\nFinite passive impedances with $\\operatorname{Re}z\\ge0$ map to the closed disk except $\\Gamma=1$. That missing point is obtained only by adjoining impedance infinity. The interior maps to the open right half-plane.\nThe classical Smith grid supplies the geometry; the cyclic labels specify exact rotations on it. Neither a new physical law nor a new optimal sampling theorem is implied by this coordinate description.\n\\takeaway{A simple action in one chart can be nonlinear in another. Conjugacy preserves the action law, not the apparent shape of a displacement.}","\\vpage{23}{A pole is a chart boundary, not a broken group action}{HOMOGENEOUS COORDINATES / NO DIVISION BY ZERO}{Traditional monograph: Theorem 13.2 (p. 26), Corollary 13.3 (p. 26).}\nRepresent impedance by a nonzero pair $[u:v]$, where $[cu:cv]=[u:v]$ for $c\\ne0$. A finite impedance is $[z:1]$; infinity is $[1:0]$. For a nonzero reflection multiplier $w$, use\n\\[\nA(w)=\\frac12\\begin{pmatrix}1+w&1-w","\\vpage{23}{A pole is a chart boundary, not a broken group action}{HOMOGENEOUS COORDINATES / NO DIVISION BY ZERO}{Traditional monograph: Theorem 13.2 (p. 26), Corollary 13.3 (p. 26).}\nRepresent impedance by a nonzero pair $[u:v]$, where $[cu:cv]=[u:v]$ for $c\\ne0$. A finite impedance is $[z:1]$; infinity is $[1:0]$. For a nonzero reflection multiplier $w$, use\n\\[\nA(w)=\\frac12\\begin{pmatrix}1+w&1-w\\\\1-w&1+w\\end{pmatrix}.\n\\]\n\\vfig{projective_poles}\n\\head{The action is defined before a ratio is taken}\nThe determinant is $w$, so the matrix cannot send a nonzero pair to zero. Direct multiplication gives\n\\[\nA(w)A(w')=A(ww'),\\qquad A(1)=I.\n\\]\nFor $w=i$, the orbit through zero is\n\\[\n[0:1]\\ \\longmapsto\\ [-i:1]\\ \\longmapsto\\ [1:0]\n\\ \\longmapsto\\ [i:1]\\ \\longmapsto\\ [0:1].\n\\]\nThe third point is a legitimate projective point, even though a finite impedance formula has a zero denominator there. The second iterate is $A(-1)$, which swaps the two coordinates. Thus a half turn exchanges $0$ and $\\infty$ and sends each finite nonzero $z$ to $1/z$.\nThe matched point $\\Gamma=0$, equivalently $z=1$, is f…","1-w&1+w\\end{pmatrix}.\n\\]\n\\vfig{projective_poles}\n\\head{The action is defined before a ratio is taken}\nThe determinant is $w$, so the matrix cannot send a nonzero pair to zero. Direct multiplication gives\n\\[\nA(w)A(w')=A(ww'),\\qquad A(1)=I.\n\\]\nFor $w=i$, the orbit through zero is\n\\[\n[0:1]\\ \\longmapsto\\ [-i:1]\\ \\longmapsto\\ [1:0]\n\\ \\longmapsto\\ [i:1]\\ \\longmapsto\\ [0:1].\n\\]\nThe third point is a legitimate projective point, even though a finite impedance formula has a zero denominator there. The second iterate is $A(-1)$, which swaps the two coordinates. Thus a half turn exchanges $0$ and $\\infty$ and sends each finite nonzero $z$ to $1/z$.\nThe matched point $\\Gamma=0$, equivalently $z=1$, is fixed by every phase multiplier. It is not part of the boundary orbit above. The exact checks test homogeneous pairs and thei","1-w&1+w\\end{pmatrix}.\n\\]\n\\vfig{projective_poles}\n\\head{The action is defined before a ratio is taken}\nThe determinant is $w$, so the matrix cannot send a nonzero pair to zero. Direct multiplication gives\n\\[\nA(w)A(w')=A(ww'),\\qquad A(1)=I.\n\\]\nFor $w=i$, the orbit through zero is\n\\[\n[0:1]\\ \\longmapsto\\ [-i:1]\\ \\longmapsto\\ [1:0]\n\\ \\longmapsto\\ [i:1]\\ \\longmapsto\\ [0:1].\n\\]\nThe third point is a legitimate projective point, even though a finite impedance formula has a zero denominator there. The second iterate is $A(-1)$, which swaps the two coordinates. Thus a half turn exchanges $0$ and $\\infty$ and sends each finite nonzero $z$ to $1/z$.\nThe matched point $\\Gamma=0$, equivalently $z=1$, is fixed by every phase multiplier. It is not part of the boundary orbit above. The exact checks test homogeneous pairs and their rescaling equivalence, rather than introducing an artificial complex value for $1/0$.\n\\takeaway{Compute with homog","1-w&1+w\\end{pmatrix}.\n\\]\n\\vfig{projective_poles}\n\\head{The action is defined before a ratio is taken}\nThe determinant is $w$, so the matrix cannot send a nonzero pair to zero. Direct multiplication gives\n\\[\nA(w)A(w')=A(ww'),\\qquad A(1)=I.\n\\]\nFor $w=i$, the orbit through zero is\n\\[\n[0:1]\\ \\longmapsto\\ [-i:1]\\ \\longmapsto\\ [1:0]\n\\ \\longmapsto\\ [i:1]\\ \\longmapsto\\ [0:1].\n\\]\nThe third point is a legitimate projective point, even though a finite impedance formula has a zero denominator there. The second iterate is $A(-1)$, which swaps the two coordinates. Thus a half turn exchanges $0$ and $\\infty$ and sends each finite nonzero $z$ to $1/z$.\nThe matched point $\\Gamma=0$, equivalently $z=1$, is fixed by every phase multiplier. It is not part of the boundary orbit above. The exact checks test homogeneous pairs and their rescaling equivalence, rather than introducing an artificial complex value for $1/0$.\n\\takeaway{Compute with homogeneous pair","1-w&1+w\\end{pmatrix}.\n\\]\n\\vfig{projective_poles}\n\\head{The action is defined before a ratio is taken}\nThe determinant is $w$, so the matrix cannot send a nonzero pair to zero. Direct multiplication gives\n\\[\nA(w)A(w')=A(ww'),\\qquad A(1)=I.\n\\]\nFor $w=i$, the orbit through zero is\n\\[\n[0:1]\\ \\longmapsto\\ [-i:1]\\ \\longmapsto\\ [1:0]\n\\ \\longmapsto\\ [i:1]\\ \\longmapsto\\ [0:1].\n\\]\nThe third point is a legitimate projective point, even though a finite impedance formula has a zero denominator there. The second iterate is $A(-1)$, which swaps the two coordinates. Thus a half turn exchanges $0$ and $\\infty$ and sends each finite nonzero $z$ to $1/z$.\nThe matched point $\\Gamma=0$, equivalently $z=1$, is fixed by every phase multiplier. It is not part of the boundary orbit above. The exact checks test homogeneous pairs and their rescaling equivalence, rather than introducing an artificial complex value for $1/0$.\n\\takeaway{Compute with homogeneous pairs first. A denominator becomes a chart-selection issue rather than an exception to the cyclic action.}","t is $w$, so the matrix cannot send a nonzero pair to zero. Direct multiplication gives\n\\[\nA(w)A(w')=A(ww'),\\qquad A(1)=I.\n\\]\nFor $w=i$, the orbit through zero is\n\\[\n[0:1]\\ \\longmapsto\\ [-i:1]\\ \\longmapsto\\ [1:0]\n\\ \\longmapsto\\ [i:1]\\ \\longmapsto\\ [0:1].\n\\]\nThe third point is a legitimate projective point, even though a finite impedance formula has a zero denominator there. The second iterate is $A(-1)$, which swaps the two coordinates. Thus a half turn exchanges $0$ and $\\infty$ and sends each finite nonzero $z$ to $1/z$.\nThe matched point $\\Gamma=0$, equivalently $z=1$, is fixed by every phase multiplier. It is not part of the boundary orbit above. The exact checks test homogeneous pairs and their rescaling equivalence, rather than introducing an artificial complex value for $1/0$.\n\\takeaway{Compute with homogeneous pairs first. A denominator becomes a chart-selection issue rather than an exception to the cyclic action.}","ltiplication gives\n\\[\nA(w)A(w')=A(ww'),\\qquad A(1)=I.\n\\]\nFor $w=i$, the orbit through zero is\n\\[\n[0:1]\\ \\longmapsto\\ [-i:1]\\ \\longmapsto\\ [1:0]\n\\ \\longmapsto\\ [i:1]\\ \\longmapsto\\ [0:1].\n\\]\nThe third point is a legitimate projective point, even though a finite impedance formula has a zero denominator there. The second iterate is $A(-1)$, which swaps the two coordinates. Thus a half turn exchanges $0$ and $\\infty$ and sends each finite nonzero $z$ to $1/z$.\nThe matched point $\\Gamma=0$, equivalently $z=1$, is fixed by every phase multiplier. It is not part of the boundary orbit above. The exact checks test homogeneous pairs and their rescaling equivalence, rather than introducing an artificial complex value for $1/0$.\n\\takeaway{Compute with homogeneous pairs first. A denominator becomes a chart-selection issue rather than an exception to the cyclic action.}","tion gives\n\\[\nA(w)A(w')=A(ww'),\\qquad A(1)=I.\n\\]\nFor $w=i$, the orbit through zero is\n\\[\n[0:1]\\ \\longmapsto\\ [-i:1]\\ \\longmapsto\\ [1:0]\n\\ \\longmapsto\\ [i:1]\\ \\longmapsto\\ [0:1].\n\\]\nThe third point is a legitimate projective point, even though a finite impedance formula has a zero denominator there. The second iterate is $A(-1)$, which swaps the two coordinates. Thus a half turn exchanges $0$ and $\\infty$ and sends each finite nonzero $z$ to $1/z$.\nThe matched point $\\Gamma=0$, equivalently $z=1$, is fixed by every phase multiplier. It is not part of the boundary orbit above. The exact checks test homogeneous pairs and their rescaling equivalence, rather than introducing an artificial complex value for $1/0$.\n\\takeaway{Compute with homogeneous pairs first. A denominator becomes a chart-selection issue rather than an exception to the cyclic action.}","(1)=I.\n\\]\nFor $w=i$, the orbit through zero is\n\\[\n[0:1]\\ \\longmapsto\\ [-i:1]\\ \\longmapsto\\ [1:0]\n\\ \\longmapsto\\ [i:1]\\ \\longmapsto\\ [0:1].\n\\]\nThe third point is a legitimate projective point, even though a finite impedance formula has a zero denominator there. The second iterate is $A(-1)$, which swaps the two coordinates. Thus a half turn exchanges $0$ and $\\infty$ and sends each finite nonzero $z$ to $1/z$.\nThe matched point $\\Gamma=0$, equivalently $z=1$, is fixed by every phase multiplier. It is not part of the boundary orbit above. The exact checks test homogeneous pairs and their rescaling equivalence, rather than introducing an artificial complex value for $1/0$.\n\\takeaway{Compute with homogeneous pairs first. A denominator becomes a chart-selection issue rather than an exception to the cyclic action.}","\\]\nFor $w=i$, the orbit through zero is\n\\[\n[0:1]\\ \\longmapsto\\ [-i:1]\\ \\longmapsto\\ [1:0]\n\\ \\longmapsto\\ [i:1]\\ \\longmapsto\\ [0:1].\n\\]\nThe third point is a legitimate projective point, even though a finite impedance formula has a zero denominator there. The second iterate is $A(-1)$, which swaps the two coordinates. Thus a half turn exchanges $0$ and $\\infty$ and sends each finite nonzero $z$ to $1/z$.\nThe matched point $\\Gamma=0$, equivalently $z=1$, is fixed by every phase multiplier. It is not part of the boundary orbit above. The exact checks test homogeneous pairs and their rescaling equivalence, rather than introducing an artificial complex value for $1/0$.\n\\takeaway{Compute with homogeneous pairs first. A denominator becomes a chart-selection issue rather than an exception to the cyclic action.}","hrough zero is\n\\[\n[0:1]\\ \\longmapsto\\ [-i:1]\\ \\longmapsto\\ [1:0]\n\\ \\longmapsto\\ [i:1]\\ \\longmapsto\\ [0:1].\n\\]\nThe third point is a legitimate projective point, even though a finite impedance formula has a zero denominator there. The second iterate is $A(-1)$, which swaps the two coordinates. Thus a half turn exchanges $0$ and $\\infty$ and sends each finite nonzero $z$ to $1/z$.\nThe matched point $\\Gamma=0$, equivalently $z=1$, is fixed by every phase multiplier. It is not part of the boundary orbit above. The exact checks test homogeneous pairs and their rescaling equivalence, rather than introducing an artificial complex value for $1/0$.\n\\takeaway{Compute with homogeneous pairs first. A denominator becomes a chart-selection issue rather than an exception to the cyclic action.}","\\longmapsto\\ [-i:1]\\ \\longmapsto\\ [1:0]\n\\ \\longmapsto\\ [i:1]\\ \\longmapsto\\ [0:1].\n\\]\nThe third point is a legitimate projective point, even though a finite impedance formula has a zero denominator there. The second iterate is $A(-1)$, which swaps the two coordinates. Thus a half turn exchanges $0$ and $\\infty$ and sends each finite nonzero $z$ to $1/z$.\nThe matched point $\\Gamma=0$, equivalently $z=1$, is fixed by every phase multiplier. It is not part of the boundary orbit above. The exact checks test homogeneous pairs and their rescaling equivalence, rather than introducing an artificial complex value for $1/0$.\n\\takeaway{Compute with homogeneous pairs first. A denominator becomes a chart-selection issue rather than an exception to the cyclic action.}","terate is $A(-1)$, which swaps the two coordinates. Thus a half turn exchanges $0$ and $\\infty$ and sends each finite nonzero $z$ to $1/z$.\nThe matched point $\\Gamma=0$, equivalently $z=1$, is fixed by every phase multiplier. It is not part of the boundary orbit above. The exact checks test homogeneous pairs and their rescaling equivalence, rather than introducing an artificial complex value for $1/0$.\n\\takeaway{Compute with homogeneous pairs first. A denominator becomes a chart-selection issue rather than an exception to the cyclic action.}","\\vpage{24}{Residue labels are not impedance averages}{DYADIC--ODD FACTORIZATION / TWELVE ADDRESSES}{Traditional monograph: Proposition 14.1 (p. 29).}\nAt phase order $12=4\\cdot3$, the map\n\\[\nk\\longmapsto(k\\bmod4,\\ k\\bmod3)\n\\]\nis a bijection from twelve residue addresses to the $4\\times3$ grid. Every cell below contains its unique original address.\n\\vfig{crt_refinement}\n\\head{The factorization is an intersection of conditions}\nFor integer sample addresses, including negative ones,\n\\[\nP_{12,k}=P_{4,k\\bmod4}\\,P_{3,k\\bmod3}.\n\\]\nThe product on the right is composition of two residue masks: both congruences must hold. Coprimality makes their intersection a single class modulo twelve. It is not a formula for multiplying or adding impedances.\n\\head{Nested samples and nested projectors are different statements}\nThe four coarse reflection rays occur at indices $0,3,6,9$ of t","\\vpage{24}{Residue labels are not impedance averages}{DYADIC--ODD FACTORIZATION / TWELVE ADDRESSES}{Traditional monograph: Proposition 14.1 (p. 29).}\nAt phase order $12=4\\cdot3$, the map\n\\[\nk\\longmapsto(k\\bmod4,\\ k\\bmod3)\n\\]\nis a bijection from twelve residue addresses to the $4\\times3$ grid. Every cell below contains its unique original address.\n\\vfig{crt_refinement}\n\\head{The factorization is an intersection of conditions}\nFor integer sample addresses, including negative ones,\n\\[\nP_{12,k}=P_{4,k\\bmod4}\\,P_{3,k\\bmod3}.\n\\]\nThe product on the right is composition of two residue masks: both congruences must hold. Coprimality makes their intersection a single class modulo twelve. It is not a formula for multiplying or adding impedances.\n\\head{Nested samples and nested projectors are different statements}\nThe four coarse reflection rays occur at indices $0,3,6,9$ of the twelve-ray grid. Thus the geometric sample","\\vpage{24}{Residue labels are not impedance averages}{DYADIC--ODD FACTORIZATION / TWELVE ADDRESSES}{Traditional monograph: Proposition 14.1 (p. 29).}\nAt phase order $12=4\\cdot3$, the map\n\\[\nk\\longmapsto(k\\bmod4,\\ k\\bmod3)\n\\]\nis a bijection from twelve residue addresses to the $4\\times3$ grid. Every cell below contains its unique original address.\n\\vfig{crt_refinement}\n\\head{The factorization is an intersection of conditions}\nFor integer sample addresses, including negative ones,\n\\[\nP_{12,k}=P_{4,k\\bmod4}\\,P_{3,k\\bmod3}.\n\\]\nThe product on the right is composition of two residue masks: both congruences must hold. Coprimality makes their intersection a single class modulo twelve. It is not a formula for multiplying or adding impedances.\n\\head{Nested samples and nested projectors are different statements}\nThe four coarse reflection rays occur at indices $0,3,6,9$ of the twelve-ray grid. Thus the geometric sample set is nested. Separately, a coarse coefficient projector is the","VE ADDRESSES}{Traditional monograph: Proposition 14.1 (p. 29).}\nAt phase order $12=4\\cdot3$, the map\n\\[\nk\\longmapsto(k\\bmod4,\\ k\\bmod3)\n\\]\nis a bijection from twelve residue addresses to the $4\\times3$ grid. Every cell below contains its unique original address.\n\\vfig{crt_refinement}\n\\head{The factorization is an intersection of conditions}\nFor integer sample addresses, including negative ones,\n\\[\nP_{12,k}=P_{4,k\\bmod4}\\,P_{3,k\\bmod3}.\n\\]\nThe product on the right is composition of two residue masks: both congruences must hold. Coprimality makes their intersection a single class modulo twelve. It is not a formula for multiplying or adding impedances.\n\\head{Nested samples and nested projectors are different statements}\nThe four coarse reflection rays occur at indices $0,3,6,9$ of the twelve-ray grid. Thus the geometric sample set is nested. Separately, a coarse coefficient projector is the sum of its fine descendants in a common coefficient space.\nThere is no corresponding rule saying that a coarse impedance is the average of its descendants. The inverse Cayley map is nonlinear. Even o…","uding negative ones,\n\\[\nP_{12,k}=P_{4,k\\bmod4}\\,P_{3,k\\bmod3}.\n\\]\nThe product on the right is composition of two residue masks: both congruences must hold. Coprimality makes their intersection a single class modulo twelve. It is not a formula for multiplying or adding impedances.\n\\head{Nested samples and nested projectors are different statements}\nThe four coarse reflection rays occur at indices $0,3,6,9$ of the twelve-ray grid. Thus the geometric sample set is nested. Separately, a coarse coefficient projector is the sum of its fine descendants in a common coefficient space.\nThere is no corresponding rule saying that a coarse impedance is the average of its descendants. The inverse Cayley map is nonlinear. Even opposite reflection points have impedances $3$ and $1/3$ in the preceding example; their average $5/3$ is not the impedance $1$ of their averaged reflection coefficient zero.\n\\takeaway{CRT factors addresses. Refinement conserves coefficient packets. Neither operation authorizes linear averaging through a nonlinear chart.}","reflection rays occur at indices $0,3,6,9$ of the twelve-ray grid. Thus the geometric sample set is nested. Separately, a coarse coefficient projector is the sum of its fine descendants in a common coefficient space.\nThere is no corresponding rule saying that a coarse impedance is the average of its descendants. The inverse Cayley map is nonlinear. Even opposite reflection points have impedances $3$ and $1/3$ in the preceding example; their average $5/3$ is not the impedance $1$ of their averaged reflection coefficient zero.\n\\takeaway{CRT factors addresses. Refinement conserves coefficient packets. Neither operation authorizes linear averaging through a nonlinear chart.}","on rays occur at indices $0,3,6,9$ of the twelve-ray grid. Thus the geometric sample set is nested. Separately, a coarse coefficient projector is the sum of its fine descendants in a common coefficient space.\nThere is no corresponding rule saying that a coarse impedance is the average of its descendants. The inverse Cayley map is nonlinear. Even opposite reflection points have impedances $3$ and $1/3$ in the preceding example; their average $5/3$ is not the impedance $1$ of their averaged reflection coefficient zero.\n\\takeaway{CRT factors addresses. Refinement conserves coefficient packets. Neither operation authorizes linear averaging through a nonlinear chart.}","ve-ray grid. Thus the geometric sample set is nested. Separately, a coarse coefficient projector is the sum of its fine descendants in a common coefficient space.\nThere is no corresponding rule saying that a coarse impedance is the average of its descendants. The inverse Cayley map is nonlinear. Even opposite reflection points have impedances $3$ and $1/3$ in the preceding example; their average $5/3$ is not the impedance $1$ of their averaged reflection coefficient zero.\n\\takeaway{CRT factors addresses. Refinement conserves coefficient packets. Neither operation authorizes linear averaging through a nonlinear chart.}","tric sample set is nested. Separately, a coarse coefficient projector is the sum of its fine descendants in a common coefficient space.\nThere is no corresponding rule saying that a coarse impedance is the average of its descendants. The inverse Cayley map is nonlinear. Even opposite reflection points have impedances $3$ and $1/3$ in the preceding example; their average $5/3$ is not the impedance $1$ of their averaged reflection coefficient zero.\n\\takeaway{CRT factors addresses. Refinement conserves coefficient packets. Neither operation authorizes linear averaging through a nonlinear chart.}","\\vpage{25}{Sensitivity grows before the pole is reached}{LOCAL DERIVATIVES / FINITE INCREMENTS}{Traditional monograph: Section 14 (p. 29).}\nAway from $\\Gamma=1$, the inverse Cayley map satisfies\n\\[\n\\frac{dz}{d\\Gamma}=\\frac2{(1-\\Gamma)^2}.\n\\]\nAlong the real axis, three exact derivatives already show the scale change:\n\\[\n\\Gamma=\\tfrac12:\\ 8,\\qquad\n\\Gamma=\\tfrac9{10}:\\ 200,\\qquad\n\\Gamma=\\tfrac{99}{100}:\\ 20000.\n\\]\n\\vfig{sensitivity}\n\\head{A differential is not the full finite-error formula}\nFor a finite increment $\\varepsilon$, with both denominators nonzero,\n\\[\nz(\\Gamma+\\varepsilon)-z(\\Gamma)\n=\\frac{2\\varepsilon}{(1-\\Gamma-\\varepsilon)(1-\\Gamma)}.\n\\]\nAt $\\Gamma=9/10$ and $\\varepsilon=1/100$, the actual increment is $20/9$, while derivative times increment is $2$. Both are calculated exactly. Their difference comes from using a differential in place","\\vpage{25}{Sensitivity grows before the pole is reached}{LOCAL DERIVATIVES / FINITE INCREMENTS}{Traditional monograph: Section 14 (p. 29).}\nAway from $\\Gamma=1$, the inverse Cayley map satisfies\n\\[\n\\frac{dz}{d\\Gamma}=\\frac2{(1-\\Gamma)^2}.\n\\]\nAlong the real axis, three exact derivatives already show the scale change:\n\\[\n\\Gamma=\\tfrac12:\\ 8,\\qquad\n\\Gamma=\\tfrac9{10}:\\ 200,\\qquad\n\\Gamma=\\tfrac{99}{100}:\\ 20000.\n\\]\n\\vfig{sensitivity}\n\\head{A differential is not the full finite-error formula}\nFor a finite increment $\\varepsilon$, with both denominators nonzero,\n\\[\nz(\\Gamma+\\varepsilon)-z(\\Gamma)\n=\\frac{2\\varepsilon}{(1-\\Gamma-\\varepsilon)(1-\\Gamma)}.\n\\]\nAt $\\Gamma=9/10$ and $\\varepsilon=1/100$, the actual increment is $20/9$, while derivative times increment is $2$. Both are calculated exactly. Their difference comes from using a differential in place of a finite increment. The derivative gives a first order approximation.\n\\head{W","\\vpage{25}{Sensitivity grows before the pole is reached}{LOCAL DERIVATIVES / FINITE INCREMENTS}{Traditional monograph: Section 14 (p. 29).}\nAway from $\\Gamma=1$, the inverse Cayley map satisfies\n\\[\n\\frac{dz}{d\\Gamma}=\\frac2{(1-\\Gamma)^2}.\n\\]\nAlong the real axis, three exact derivatives already show the scale change:\n\\[\n\\Gamma=\\tfrac12:\\ 8,\\qquad\n\\Gamma=\\tfrac9{10}:\\ 200,\\qquad\n\\Gamma=\\tfrac{99}{100}:\\ 20000.\n\\]\n\\vfig{sensitivity}\n\\head{A differential is not the full finite-error formula}\nFor a finite increment $\\varepsilon$, with both denominators nonzero,\n\\[\nz(\\Gamma+\\varepsilon)-z(\\Gamma)\n=\\frac{2\\varepsilon}{(1-\\Gamma-\\varepsilon)(1-\\Gamma)}.\n\\]\nAt $\\Gamma=9/10$ and $\\varepsilon=1/100$, the actual increment is $20/9$, while derivative times increment is $2$. Both are calculated exactly. Their difference comes from using a differential in place of a finite increment. The derivative gives a first order approximation.\n\\head{What a proposed weight can and cannot certify}\nImpedance is sensitive near $\\Gamma=1$, and admittance near $\\Gamma=-1$. The symmetric weight\n\\[\nw(\\Gamma)=|1-\\Gamm…",": Section 14 (p. 29).}\nAway from $\\Gamma=1$, the inverse Cayley map satisfies\n\\[\n\\frac{dz}{d\\Gamma}=\\frac2{(1-\\Gamma)^2}.\n\\]\nAlong the real axis, three exact derivatives already show the scale change:\n\\[\n\\Gamma=\\tfrac12:\\ 8,\\qquad\n\\Gamma=\\tfrac9{10}:\\ 200,\\qquad\n\\Gamma=\\tfrac{99}{100}:\\ 20000.\n\\]\n\\vfig{sensitivity}\n\\head{A differential is not the full finite-error formula}\nFor a finite increment $\\varepsilon$, with both denominators nonzero,\n\\[\nz(\\Gamma+\\varepsilon)-z(\\Gamma)\n=\\frac{2\\varepsilon}{(1-\\Gamma-\\varepsilon)(1-\\Gamma)}.\n\\]\nAt $\\Gamma=9/10$ and $\\varepsilon=1/100$, the actual increment is $20/9$, while derivative times increment is $2$. Both are calculated exactly. Their difference comes from using a differential in place of a finite increment. The derivative gives a first order approximation.\n\\head{What a proposed weight can and cannot certify}\nImpedance is sensitive near $\\Gamma=1$, and admittance near $\\Gamma=-1$. The symmetric weight\n\\[\nw(\\Gamma)=|1-\\Gamma|^{-2}+|1+\\Gamma|^{-2}\n\\]\nreflects both chart sensitivities. Its growth motivates denser sampling near the endpoints…","inverse Cayley map satisfies\n\\[\n\\frac{dz}{d\\Gamma}=\\frac2{(1-\\Gamma)^2}.\n\\]\nAlong the real axis, three exact derivatives already show the scale change:\n\\[\n\\Gamma=\\tfrac12:\\ 8,\\qquad\n\\Gamma=\\tfrac9{10}:\\ 200,\\qquad\n\\Gamma=\\tfrac{99}{100}:\\ 20000.\n\\]\n\\vfig{sensitivity}\n\\head{A differential is not the full finite-error formula}\nFor a finite increment $\\varepsilon$, with both denominators nonzero,\n\\[\nz(\\Gamma+\\varepsilon)-z(\\Gamma)\n=\\frac{2\\varepsilon}{(1-\\Gamma-\\varepsilon)(1-\\Gamma)}.\n\\]\nAt $\\Gamma=9/10$ and $\\varepsilon=1/100$, the actual increment is $20/9$, while derivative times increment is $2$. Both are calculated exactly. Their difference comes from using a differential in place of a finite increment. The derivative gives a first order approximation.\n\\head{What a proposed weight can and cannot certify}\nImpedance is sensitive near $\\Gamma=1$, and admittance near $\\Gamma=-1$. The symmetric weight\n\\[\nw(\\Gamma)=|1-\\Gamma|^{-2}+|1+\\Gamma|^{-2}\n\\]\nreflects both chart sensitivities. Its growth motivates denser sampling near the endpoints, but does not establish an optimal grid, interpo…",", three exact derivatives already show the scale change:\n\\[\n\\Gamma=\\tfrac12:\\ 8,\\qquad\n\\Gamma=\\tfrac9{10}:\\ 200,\\qquad\n\\Gamma=\\tfrac{99}{100}:\\ 20000.\n\\]\n\\vfig{sensitivity}\n\\head{A differential is not the full finite-error formula}\nFor a finite increment $\\varepsilon$, with both denominators nonzero,\n\\[\nz(\\Gamma+\\varepsilon)-z(\\Gamma)\n=\\frac{2\\varepsilon}{(1-\\Gamma-\\varepsilon)(1-\\Gamma)}.\n\\]\nAt $\\Gamma=9/10$ and $\\varepsilon=1/100$, the actual increment is $20/9$, while derivative times increment is $2$. Both are calculated exactly. Their difference comes from using a differential in place of a finite increment. The derivative gives a first order approximation.\n\\head{What a proposed weight can and cannot certify}\nImpedance is sensitive near $\\Gamma=1$, and admittance near $\\Gamma=-1$. The symmetric weight\n\\[\nw(\\Gamma)=|1-\\Gamma|^{-2}+|1+\\Gamma|^{-2}\n\\]\nreflects both chart sensitivities. Its growth motivates denser sampling near the endpoints, but does not establish an optimal grid, interpolation rule, or measurement allocation. A finite-error estimate must also control how close the…","vatives already show the scale change:\n\\[\n\\Gamma=\\tfrac12:\\ 8,\\qquad\n\\Gamma=\\tfrac9{10}:\\ 200,\\qquad\n\\Gamma=\\tfrac{99}{100}:\\ 20000.\n\\]\n\\vfig{sensitivity}\n\\head{A differential is not the full finite-error formula}\nFor a finite increment $\\varepsilon$, with both denominators nonzero,\n\\[\nz(\\Gamma+\\varepsilon)-z(\\Gamma)\n=\\frac{2\\varepsilon}{(1-\\Gamma-\\varepsilon)(1-\\Gamma)}.\n\\]\nAt $\\Gamma=9/10$ and $\\varepsilon=1/100$, the actual increment is $20/9$, while derivative times increment is $2$. Both are calculated exactly. Their difference comes from using a differential in place of a finite increment. The derivative gives a first order approximation.\n\\head{What a proposed weight can and cannot certify}\nImpedance is sensitive near $\\Gamma=1$, and admittance near $\\Gamma=-1$. The symmetric weight\n\\[\nw(\\Gamma)=|1-\\Gamma|^{-2}+|1+\\Gamma|^{-2}\n\\]\nreflects both chart sensitivities. Its growth motivates denser sampling near the endpoints, but does not establish an optimal grid, interpolation rule, or measurement allocation. A finite-error estimate must also control how close the perturbed point c…","mma=\\tfrac12:\\ 8,\\qquad\n\\Gamma=\\tfrac9{10}:\\ 200,\\qquad\n\\Gamma=\\tfrac{99}{100}:\\ 20000.\n\\]\n\\vfig{sensitivity}\n\\head{A differential is not the full finite-error formula}\nFor a finite increment $\\varepsilon$, with both denominators nonzero,\n\\[\nz(\\Gamma+\\varepsilon)-z(\\Gamma)\n=\\frac{2\\varepsilon}{(1-\\Gamma-\\varepsilon)(1-\\Gamma)}.\n\\]\nAt $\\Gamma=9/10$ and $\\varepsilon=1/100$, the actual increment is $20/9$, while derivative times increment is $2$. Both are calculated exactly. Their difference comes from using a differential in place of a finite increment. The derivative gives a first order approximation.\n\\head{What a proposed weight can and cannot certify}\nImpedance is sensitive near $\\Gamma=1$, and admittance near $\\Gamma=-1$. The symmetric weight\n\\[\nw(\\Gamma)=|1-\\Gamma|^{-2}+|1+\\Gamma|^{-2}\n\\]\nreflects both chart sensitivities. Its growth motivates denser sampling near the endpoints, but does not establish an optimal grid, interpolation rule, or measurement allocation. A finite-error estimate must also control how close the perturbed point comes to either pole.\n\\takeaway{Use the exact …","200,\\qquad\n\\Gamma=\\tfrac{99}{100}:\\ 20000.\n\\]\n\\vfig{sensitivity}\n\\head{A differential is not the full finite-error formula}\nFor a finite increment $\\varepsilon$, with both denominators nonzero,\n\\[\nz(\\Gamma+\\varepsilon)-z(\\Gamma)\n=\\frac{2\\varepsilon}{(1-\\Gamma-\\varepsilon)(1-\\Gamma)}.\n\\]\nAt $\\Gamma=9/10$ and $\\varepsilon=1/100$, the actual increment is $20/9$, while derivative times increment is $2$. Both are calculated exactly. Their difference comes from using a differential in place of a finite increment. The derivative gives a first order approximation.\n\\head{What a proposed weight can and cannot certify}\nImpedance is sensitive near $\\Gamma=1$, and admittance near $\\Gamma=-1$. The symmetric weight\n\\[\nw(\\Gamma)=|1-\\Gamma|^{-2}+|1+\\Gamma|^{-2}\n\\]\nreflects both chart sensitivities. Its growth motivates denser sampling near the endpoints, but does not establish an optimal grid, interpolation rule, or measurement allocation. A finite-error estimate must also control how close the perturbed point comes to either pole.\n\\takeaway{Use the exact finite difference for an error budget. A larg","lon}{(1-\\Gamma-\\varepsilon)(1-\\Gamma)}.\n\\]\nAt $\\Gamma=9/10$ and $\\varepsilon=1/100$, the actual increment is $20/9$, while derivative times increment is $2$. Both are calculated exactly. Their difference comes from using a differential in place of a finite increment. The derivative gives a first order approximation.\n\\head{What a proposed weight can and cannot certify}\nImpedance is sensitive near $\\Gamma=1$, and admittance near $\\Gamma=-1$. The symmetric weight\n\\[\nw(\\Gamma)=|1-\\Gamma|^{-2}+|1+\\Gamma|^{-2}\n\\]\nreflects both chart sensitivities. Its growth motivates denser sampling near the endpoints, but does not establish an optimal grid, interpolation rule, or measurement allocation. A finite-error estimate must also control how close the perturbed point comes to either pole.\n\\takeaway{Use the exact finite difference for an error budget. A large derivative diagnoses conditioning. Choosing an optimal algorithm requires further input.}","amma)}.\n\\]\nAt $\\Gamma=9/10$ and $\\varepsilon=1/100$, the actual increment is $20/9$, while derivative times increment is $2$. Both are calculated exactly. Their difference comes from using a differential in place of a finite increment. The derivative gives a first order approximation.\n\\head{What a proposed weight can and cannot certify}\nImpedance is sensitive near $\\Gamma=1$, and admittance near $\\Gamma=-1$. The symmetric weight\n\\[\nw(\\Gamma)=|1-\\Gamma|^{-2}+|1+\\Gamma|^{-2}\n\\]\nreflects both chart sensitivities. Its growth motivates denser sampling near the endpoints, but does not establish an optimal grid, interpolation rule, or measurement allocation. A finite-error estimate must also control how close the perturbed point comes to either pole.\n\\takeaway{Use the exact finite difference for an error budget. A large derivative diagnoses conditioning. Choosing an optimal algorithm requires further input.}","arepsilon=1/100$, the actual increment is $20/9$, while derivative times increment is $2$. Both are calculated exactly. Their difference comes from using a differential in place of a finite increment. The derivative gives a first order approximation.\n\\head{What a proposed weight can and cannot certify}\nImpedance is sensitive near $\\Gamma=1$, and admittance near $\\Gamma=-1$. The symmetric weight\n\\[\nw(\\Gamma)=|1-\\Gamma|^{-2}+|1+\\Gamma|^{-2}\n\\]\nreflects both chart sensitivities. Its growth motivates denser sampling near the endpoints, but does not establish an optimal grid, interpolation rule, or measurement allocation. A finite-error estimate must also control how close the perturbed point comes to either pole.\n\\takeaway{Use the exact finite difference for an error budget. A large derivative diagnoses conditioning. Choosing an optimal algorithm requires further input.}","Finite packet tree & All $26$ complete prunings, actual projector sums, the unique cost-$9$ optimum, and a strict perturbation-margin certificate.","bottomrule\n\\end{tabularx}\n\\head{What each evidence layer means}\n\\textbf{Manuscript theorems.} The preserved v1.1.0 release maps twenty theorem-like statements to $225$ supporting Lean targets in $23$ proof modules. Its saved compiler and axiom records remain inherited evidence. No fresh Lean run is claimed for this edition. The additional Part II developments remain outside this twenty-result map.","\\textbf{Current exact examples.} The symbolic visual suite was rerun for this monograph. All $181$ rational, algebraic, symbolic, and exhaustive finite checks passed. Seven deliberately false assertions were rejected. Numerical samples draw the continuous Fourier and sensitivity curves; they are not promoted to interval enclosures or universal error bounds.","the correct packet. A finite cost optimum depends on the supplied objective. An infinite packet tree, a universal segmentation procedure, minimax recovery rates, and optimal Smith sampling require additional theorems. The core's broad edge-recovery problems are not resolved by these finite examples. V27 introduces the additional separated-jump recovery results and their explicit hypotheses.\nThe $16$-sample running signal is not the $1024$-sample manuscript experiment; the $26$-pruning example is not its $458{,}330$-pruning benchmark. Their smaller sizes allow every relevant number to be inspected exactly.\n\\takeaway{Coordinates, approximation, identification, optimization, and physical interpretation are separate layers. The examples make their interfaces visible without erasing their hypotheses.}","imum depends on the supplied objective. An infinite packet tree, a universal segmentation procedure, minimax recovery rates, and optimal Smith sampling require additional theorems. The core's broad edge-recovery problems are not resolved by these finite examples. V27 introduces the additional separated-jump recovery results and their explicit hypotheses.\nThe $16$-sample running signal is not the $1024$-sample manuscript experiment; the $26$-pruning example is not its $458{,}330$-pruning benchmark. Their smaller sizes allow every relevant number to be inspected exactly.\n\\takeaway{Coordinates, approximation, identification, optimization, and physical interpretation are separate layers. The examples make their interfaces visible without erasing their hypotheses.}","n infinite packet tree, a universal segmentation procedure, minimax recovery rates, and optimal Smith sampling require additional theorems. The core's broad edge-recovery problems are not resolved by these finite examples. V27 introduces the additional separated-jump recovery results and their explicit hypotheses.\nThe $16$-sample running signal is not the $1024$-sample manuscript experiment; the $26$-pruning example is not its $458{,}330$-pruning benchmark. Their smaller sizes allow every relevant number to be inspected exactly.\n\\takeaway{Coordinates, approximation, identification, optimization, and physical interpretation are separate layers. The examples make their interfaces visible without erasing their hypotheses.}","sal segmentation procedure, minimax recovery rates, and optimal Smith sampling require additional theorems. The core's broad edge-recovery problems are not resolved by these finite examples. V27 introduces the additional separated-jump recovery results and their explicit hypotheses.\nThe $16$-sample running signal is not the $1024$-sample manuscript experiment; the $26$-pruning example is not its $458{,}330$-pruning benchmark. Their smaller sizes allow every relevant number to be inspected exactly.\n\\takeaway{Coordinates, approximation, identification, optimization, and physical interpretation are separate layers. The examples make their interfaces visible without erasing their hypotheses.}","\\vpage{28}{Six image phases are six exact address classes}{LATTICE COORDINATES / A RECTANGULAR EXAMPLE}{Traditional monograph: Theorem E3.2 (p. 43), Theorem E3.5 (p. 44), Remark E2.2 (p. 43).}\nTake $L=2\\ZZ\\times3\\ZZ$ inside $\\ZZ^2$, with representatives\n\\[\nR=\\{0,1\\}\\times\\{0,1,2\\}.\n\\]\nEvery address has a unique expression $(i,j)=(r_1,r_2)+(2q_1,3q_2)$. The six channels are $x_r(2q_1,3q_2)=x(r_1+2q_1,r_2+3q_2)$.\n\\vfig[.80]{ext_lattice}\n\\platecaption{Each cell shows its residue label. Colors identify six disjoint classes; they do not average neighboring image pixels.}\n\\head{A negative address is still unique}\nThe point $(-1,7)$ has representative $(1,1)$ and lattice displacement $(-2,6)$, hence channel index $(-1,2)$. Floor division gives the correct answer in both coordinates.\nFor finite $p$, the channel norm uses the sum of the $p$th powers. At $p=\\infty$, it uses their maximum. Both equal the norm on","\\vpage{28}{Six image phases are six exact address classes}{LATTICE COORDINATES / A RECTANGULAR EXAMPLE}{Traditional monograph: Theorem E3.2 (p. 43), Theorem E3.5 (p. 44), Remark E2.2 (p. 43).}\nTake $L=2\\ZZ\\times3\\ZZ$ inside $\\ZZ^2$, with representatives\n\\[\nR=\\{0,1\\}\\times\\{0,1,2\\}.\n\\]\nEvery address has a unique expression $(i,j)=(r_1,r_2)+(2q_1,3q_2)$. The six channels are $x_r(2q_1,3q_2)=x(r_1+2q_1,r_2+3q_2)$.\n\\vfig[.80]{ext_lattice}\n\\platecaption{Each cell shows its residue label. Colors identify six disjoint classes; they do not average neighboring image pixels.}\n\\head{A negative address is still unique}\nThe point $(-1,7)$ has representative $(1,1)$ and lattice displacement $(-2,6)$, hence channel index $(-1,2)$. Floor division gives the correct answer in both coordinates.\nFor finite $p$, the channel norm uses the sum of the $p$th powers. At $p=\\infty$, it uses their maximum. Both equal the norm on the original la","\\vpage{28}{Six image phases are six exact address classes}{LATTICE COORDINATES / A RECTANGULAR EXAMPLE}{Traditional monograph: Theorem E3.2 (p. 43), Theorem E3.5 (p. 44), Remark E2.2 (p. 43).}\nTake $L=2\\ZZ\\times3\\ZZ$ inside $\\ZZ^2$, with representatives\n\\[\nR=\\{0,1\\}\\times\\{0,1,2\\}.\n\\]\nEvery address has a unique expression $(i,j)=(r_1,r_2)+(2q_1,3q_2)$. The six channels are $x_r(2q_1,3q_2)=x(r_1+2q_1,r_2+3q_2)$.\n\\vfig[.80]{ext_lattice}\n\\platecaption{Each cell shows its residue label. Colors identify six disjoint classes; they do not average neighboring image pixels.}\n\\head{A negative address is still unique}\nThe point $(-1,7)$ has representative $(1,1)$ and lattice displacement $(-2,6)$, hence channel index $(-1,2)$. Floor division gives the correct answer in both coordinates.\nFor finite $p$, the channel norm uses the sum of the $p$th powers. At $p=\\infty$, it uses their maximum. Both equal the norm on the original lattice because the cosets partition all addresses. In","\\vpage{28}{Six image phases are six exact address classes}{LATTICE COORDINATES / A RECTANGULAR EXAMPLE}{Traditional monograph: Theorem E3.2 (p. 43), Theorem E3.5 (p. 44), Remark E2.2 (p. 43).}\nTake $L=2\\ZZ\\times3\\ZZ$ inside $\\ZZ^2$, with representatives\n\\[\nR=\\{0,1\\}\\times\\{0,1,2\\}.\n\\]\nEvery address has a unique expression $(i,j)=(r_1,r_2)+(2q_1,3q_2)$. The six channels are $x_r(2q_1,3q_2)=x(r_1+2q_1,r_2+3q_2)$.\n\\vfig[.80]{ext_lattice}\n\\platecaption{Each cell shows its residue label. Colors identify six disjoint classes; they do not average neighboring image pixels.}\n\\head{A negative address is still unique}\nThe point $(-1,7)$ has representative $(1,1)$ and lattice displacement $(-2,6)$, hence channel index $(-1,2)$. Floor division gives the correct answer in both coordinates.\nFor finite $p$, the channel norm uses the sum of the $p$th powers. At $p=\\infty$, it uses their maximum. Both equal the norm on the original lattice because the cosets partition all addresses. In $\\ell^\\infty$, this coordinate identity does not imply that translate","\\vpage{28}{Six image phases are six exact address classes}{LATTICE COORDINATES / A RECTANGULAR EXAMPLE}{Traditional monograph: Theorem E3.2 (p. 43), Theorem E3.5 (p. 44), Remark E2.2 (p. 43).}\nTake $L=2\\ZZ\\times3\\ZZ$ inside $\\ZZ^2$, with representatives\n\\[\nR=\\{0,1\\}\\times\\{0,1,2\\}.\n\\]\nEvery address has a unique expression $(i,j)=(r_1,r_2)+(2q_1,3q_2)$. The six channels are $x_r(2q_1,3q_2)=x(r_1+2q_1,r_2+3q_2)$.\n\\vfig[.80]{ext_lattice}\n\\platecaption{Each cell shows its residue label. Colors identify six disjoint classes; they do not average neighboring image pixels.}\n\\head{A negative address is still unique}\nThe point $(-1,7)$ has representative $(1,1)$ and lattice displacement $(-2,6)$, hence channel index $(-1,2)$. Floor division gives the correct answer in both coordinates.\nFor finite $p$, the channel norm uses the sum of the $p$th powers. At $p=\\infty$, it uses their maximum. Both equal the norm on the original lattice because the cosets partition all addresses. In $\\ell^\\infty$, this coordinate identity does not imply that translates of an impulse are dense: their closure is only…","ZZ^2$, with representatives\n\\[\nR=\\{0,1\\}\\times\\{0,1,2\\}.\n\\]\nEvery address has a unique expression $(i,j)=(r_1,r_2)+(2q_1,3q_2)$. The six channels are $x_r(2q_1,3q_2)=x(r_1+2q_1,r_2+3q_2)$.\n\\vfig[.80]{ext_lattice}\n\\platecaption{Each cell shows its residue label. Colors identify six disjoint classes; they do not average neighboring image pixels.}\n\\head{A negative address is still unique}\nThe point $(-1,7)$ has representative $(1,1)$ and lattice displacement $(-2,6)$, hence channel index $(-1,2)$. Floor division gives the correct answer in both coordinates.\nFor finite $p$, the channel norm uses the sum of the $p$th powers. At $p=\\infty$, it uses their maximum. Both equal the norm on the original lattice because the cosets partition all addresses. In $\\ell^\\infty$, this coordinate identity does not imply that translates of an impulse are dense: their closure is only $c_0$.\n\\head{Refinement partitions a class again}\nFor $L'=4\\ZZ\\times6\\ZZ\\subset L$, the coarse representative $(1,2)$ has fine representatives\n\\[\n(1,2),\\ (3,2),\\ (1,5),\\ (3,5).\n\\]\nTheir original-lattice masks sum to the coars…","\\[\nR=\\{0,1\\}\\times\\{0,1,2\\}.\n\\]\nEvery address has a unique expression $(i,j)=(r_1,r_2)+(2q_1,3q_2)$. The six channels are $x_r(2q_1,3q_2)=x(r_1+2q_1,r_2+3q_2)$.\n\\vfig[.80]{ext_lattice}\n\\platecaption{Each cell shows its residue label. Colors identify six disjoint classes; they do not average neighboring image pixels.}\n\\head{A negative address is still unique}\nThe point $(-1,7)$ has representative $(1,1)$ and lattice displacement $(-2,6)$, hence channel index $(-1,2)$. Floor division gives the correct answer in both coordinates.\nFor finite $p$, the channel norm uses the sum of the $p$th powers. At $p=\\infty$, it uses their maximum. Both equal the norm on the original lattice because the cosets partition all addresses. In $\\ell^\\infty$, this coordinate identity does not imply that translates of an impulse are dense: their closure is only $c_0$.\n\\head{Refinement partitions a class again}\nFor $L'=4\\ZZ\\times6\\ZZ\\subset L$, the coarse representative $(1,2)$ has fine representatives\n\\[\n(1,2),\\ (3,2),\\ (1,5),\\ (3,5).\n\\]\nTheir original-lattice masks sum to the coarse mask. As on V4, decimated …","very address has a unique expression $(i,j)=(r_1,r_2)+(2q_1,3q_2)$. The six channels are $x_r(2q_1,3q_2)=x(r_1+2q_1,r_2+3q_2)$.\n\\vfig[.80]{ext_lattice}\n\\platecaption{Each cell shows its residue label. Colors identify six disjoint classes; they do not average neighboring image pixels.}\n\\head{A negative address is still unique}\nThe point $(-1,7)$ has representative $(1,1)$ and lattice displacement $(-2,6)$, hence channel index $(-1,2)$. Floor division gives the correct answer in both coordinates.\nFor finite $p$, the channel norm uses the sum of the $p$th powers. At $p=\\infty$, it uses their maximum. Both equal the norm on the original lattice because the cosets partition all addresses. In $\\ell^\\infty$, this coordinate identity does not imply that translates of an impulse are dense: their closure is only $c_0$.\n\\head{Refinement partitions a class again}\nFor $L'=4\\ZZ\\times6\\ZZ\\subset L$, the coarse representative $(1,2)$ has fine representatives\n\\[\n(1,2),\\ (3,2),\\ (1,5),\\ (3,5).\n\\]\nTheir original-lattice masks sum to the coarse mask. As on V4, decimated arrays are interleaved, not added…","ession $(i,j)=(r_1,r_2)+(2q_1,3q_2)$. The six channels are $x_r(2q_1,3q_2)=x(r_1+2q_1,r_2+3q_2)$.\n\\vfig[.80]{ext_lattice}\n\\platecaption{Each cell shows its residue label. Colors identify six disjoint classes; they do not average neighboring image pixels.}\n\\head{A negative address is still unique}\nThe point $(-1,7)$ has representative $(1,1)$ and lattice displacement $(-2,6)$, hence channel index $(-1,2)$. Floor division gives the correct answer in both coordinates.\nFor finite $p$, the channel norm uses the sum of the $p$th powers. At $p=\\infty$, it uses their maximum. Both equal the norm on the original lattice because the cosets partition all addresses. In $\\ell^\\infty$, this coordinate identity does not imply that translates of an impulse are dense: their closure is only $c_0$.\n\\head{Refinement partitions a class again}\nFor $L'=4\\ZZ\\times6\\ZZ\\subset L$, the coarse representative $(1,2)$ has fine representatives\n\\[\n(1,2),\\ (3,2),\\ (1,5),\\ (3,5).\n\\]\nTheir original-lattice masks sum to the coarse mask. As on V4, decimated arrays are interleaved, not added entrywise.\n\\takeaway{Dimensio…","q_1,r_2+3q_2)$.\n\\vfig[.80]{ext_lattice}\n\\platecaption{Each cell shows its residue label. Colors identify six disjoint classes; they do not average neighboring image pixels.}\n\\head{A negative address is still unique}\nThe point $(-1,7)$ has representative $(1,1)$ and lattice displacement $(-2,6)$, hence channel index $(-1,2)$. Floor division gives the correct answer in both coordinates.\nFor finite $p$, the channel norm uses the sum of the $p$th powers. At $p=\\infty$, it uses their maximum. Both equal the norm on the original lattice because the cosets partition all addresses. In $\\ell^\\infty$, this coordinate identity does not imply that translates of an impulse are dense: their closure is only $c_0$.\n\\head{Refinement partitions a class again}\nFor $L'=4\\ZZ\\times6\\ZZ\\subset L$, the coarse representative $(1,2)$ has fine representatives\n\\[\n(1,2),\\ (3,2),\\ (1,5),\\ (3,5).\n\\]\nTheir original-lattice masks sum to the coarse mask. As on V4, decimated arrays are interleaved, not added entrywise.\n\\takeaway{Dimension changes the address tuple. Unique coset decomposition still supplies exact synth…","latecaption{Each cell shows its residue label. Colors identify six disjoint classes; they do not average neighboring image pixels.}\n\\head{A negative address is still unique}\nThe point $(-1,7)$ has representative $(1,1)$ and lattice displacement $(-2,6)$, hence channel index $(-1,2)$. Floor division gives the correct answer in both coordinates.\nFor finite $p$, the channel norm uses the sum of the $p$th powers. At $p=\\infty$, it uses their maximum. Both equal the norm on the original lattice because the cosets partition all addresses. In $\\ell^\\infty$, this coordinate identity does not imply that translates of an impulse are dense: their closure is only $c_0$.\n\\head{Refinement partitions a class again}\nFor $L'=4\\ZZ\\times6\\ZZ\\subset L$, the coarse representative $(1,2)$ has fine representatives\n\\[\n(1,2),\\ (3,2),\\ (1,5),\\ (3,5).\n\\]\nTheir original-lattice masks sum to the coarse mask. As on V4, decimated arrays are interleaved, not added entrywise.\n\\takeaway{Dimension changes the address tuple. Unique coset decomposition still supplies exact synthesis and the correct norm identity.}","cell shows its residue label. Colors identify six disjoint classes; they do not average neighboring image pixels.}\n\\head{A negative address is still unique}\nThe point $(-1,7)$ has representative $(1,1)$ and lattice displacement $(-2,6)$, hence channel index $(-1,2)$. Floor division gives the correct answer in both coordinates.\nFor finite $p$, the channel norm uses the sum of the $p$th powers. At $p=\\infty$, it uses their maximum. Both equal the norm on the original lattice because the cosets partition all addresses. In $\\ell^\\infty$, this coordinate identity does not imply that translates of an impulse are dense: their closure is only $c_0$.\n\\head{Refinement partitions a class again}\nFor $L'=4\\ZZ\\times6\\ZZ\\subset L$, the coarse representative $(1,2)$ has fine representatives\n\\[\n(1,2),\\ (3,2),\\ (1,5),\\ (3,5).\n\\]\nTheir original-lattice masks sum to the coarse mask. As on V4, decimated arrays are interleaved, not added entrywise.\n\\takeaway{Dimension changes the address tuple. Unique coset decomposition still supplies exact synthesis and the correct norm identity.}","ative address is still unique}\nThe point $(-1,7)$ has representative $(1,1)$ and lattice displacement $(-2,6)$, hence channel index $(-1,2)$. Floor division gives the correct answer in both coordinates.\nFor finite $p$, the channel norm uses the sum of the $p$th powers. At $p=\\infty$, it uses their maximum. Both equal the norm on the original lattice because the cosets partition all addresses. In $\\ell^\\infty$, this coordinate identity does not imply that translates of an impulse are dense: their closure is only $c_0$.\n\\head{Refinement partitions a class again}\nFor $L'=4\\ZZ\\times6\\ZZ\\subset L$, the coarse representative $(1,2)$ has fine representatives\n\\[\n(1,2),\\ (3,2),\\ (1,5),\\ (3,5).\n\\]\nTheir original-lattice masks sum to the coarse mask. As on V4, decimated arrays are interleaved, not added entrywise.\n\\takeaway{Dimension changes the address tuple. Unique coset decomposition still supplies exact synthesis and the correct norm identity.}","channel index $(-1,2)$. Floor division gives the correct answer in both coordinates.\nFor finite $p$, the channel norm uses the sum of the $p$th powers. At $p=\\infty$, it uses their maximum. Both equal the norm on the original lattice because the cosets partition all addresses. In $\\ell^\\infty$, this coordinate identity does not imply that translates of an impulse are dense: their closure is only $c_0$.\n\\head{Refinement partitions a class again}\nFor $L'=4\\ZZ\\times6\\ZZ\\subset L$, the coarse representative $(1,2)$ has fine representatives\n\\[\n(1,2),\\ (3,2),\\ (1,5),\\ (3,5).\n\\]\nTheir original-lattice masks sum to the coarse mask. As on V4, decimated arrays are interleaved, not added entrywise.\n\\takeaway{Dimension changes the address tuple. Unique coset decomposition still supplies exact synthesis and the correct norm identity.}","ct answer in both coordinates.\nFor finite $p$, the channel norm uses the sum of the $p$th powers. At $p=\\infty$, it uses their maximum. Both equal the norm on the original lattice because the cosets partition all addresses. In $\\ell^\\infty$, this coordinate identity does not imply that translates of an impulse are dense: their closure is only $c_0$.\n\\head{Refinement partitions a class again}\nFor $L'=4\\ZZ\\times6\\ZZ\\subset L$, the coarse representative $(1,2)$ has fine representatives\n\\[\n(1,2),\\ (3,2),\\ (1,5),\\ (3,5).\n\\]\nTheir original-lattice masks sum to the coarse mask. As on V4, decimated arrays are interleaved, not added entrywise.\n\\takeaway{Dimension changes the address tuple. Unique coset decomposition still supplies exact synthesis and the correct norm identity.}","el norm uses the sum of the $p$th powers. At $p=\\infty$, it uses their maximum. Both equal the norm on the original lattice because the cosets partition all addresses. In $\\ell^\\infty$, this coordinate identity does not imply that translates of an impulse are dense: their closure is only $c_0$.\n\\head{Refinement partitions a class again}\nFor $L'=4\\ZZ\\times6\\ZZ\\subset L$, the coarse representative $(1,2)$ has fine representatives\n\\[\n(1,2),\\ (3,2),\\ (1,5),\\ (3,5).\n\\]\nTheir original-lattice masks sum to the coarse mask. As on V4, decimated arrays are interleaved, not added entrywise.\n\\takeaway{Dimension changes the address tuple. Unique coset decomposition still supplies exact synthesis and the correct norm identity.}","powers. At $p=\\infty$, it uses their maximum. Both equal the norm on the original lattice because the cosets partition all addresses. In $\\ell^\\infty$, this coordinate identity does not imply that translates of an impulse are dense: their closure is only $c_0$.\n\\head{Refinement partitions a class again}\nFor $L'=4\\ZZ\\times6\\ZZ\\subset L$, the coarse representative $(1,2)$ has fine representatives\n\\[\n(1,2),\\ (3,2),\\ (1,5),\\ (3,5).\n\\]\nTheir original-lattice masks sum to the coarse mask. As on V4, decimated arrays are interleaved, not added entrywise.\n\\takeaway{Dimension changes the address tuple. Unique coset decomposition still supplies exact synthesis and the correct norm identity.}","\\vpage{29}{A positive singular-value gap survives perturbation}{FRAME STABILITY / AN EXACT TWO-CHANNEL BANK}{Traditional monograph: Theorem E5.2 (p. 49), Theorem E5.7 (p. 50), Corollary E5.6 (p. 50).}\nConsider the constant analysis bank\n\\[\nE=\\begin{pmatrix}1&0\\\\0&2\\end{pmatrix},\\qquad\nE^*E=\\operatorname{diag}(1,4).\n\\]\nIt has frame bounds $A=1$, $B=4$. For every perturbation with operator norm at most $\\delta=1/4$, the squared-norm bounds remain at least $(1-1/4)^2=9/16$ and at most $(2+1/4)^2=81/16$.\n\\vfig[.87]{ext_frame}\n\\head{Both bounds can be attained}\nChoose $\\Delta E=\\operatorname{diag}(-1/4,1/4)$. Then $\\widetilde E=\\operatorname{diag}(3/4,9/4)$ has precisely those two frame bounds. Its inverse has operator norm $4/3$. A coefficient error $\\nu=(3/4,0)$ therefore produces signal error $(1,0)$, attaining the noise bound.\nThe strict condition $\\delta<\\sqrt A$ prevents the lower bound from vanishing. At $\\delta=1$, the perturbation $\\operatorname{diag}(-1,0)$ destroys the first channel entirely.\n\\he","0&2\\end{pmatrix},\\qquad\nE^*E=\\operatorname{diag}(1,4).\n\\]\nIt has frame bounds $A=1$, $B=4$. For every perturbation with operator norm at most $\\delta=1/4$, the squared-norm bounds remain at least $(1-1/4)^2=9/16$ and at most $(2+1/4)^2=81/16$.\n\\vfig[.87]{ext_frame}\n\\head{Both bounds can be attained}\nChoose $\\Delta E=\\operatorname{diag}(-1/4,1/4)$. Then $\\widetilde E=\\operatorname{diag}(3/4,9/4)$ has precisely those two frame bounds. Its inverse has operator norm $4/3$. A coefficient error $\\nu=(3/4,0)$ therefore produces signal error $(1,0)$, attaining the noise bound.\nThe strict condition $\\delta<\\sqrt A$ prevents the lower bound from vanishing. At $\\delta=1$, the perturbation $\\operatorname{diag}(-1,0)$ destroys the first channel entirely.\n\\head{An inverse and an approx","0&2\\end{pmatrix},\\qquad\nE^*E=\\operatorname{diag}(1,4).\n\\]\nIt has frame bounds $A=1$, $B=4$. For every perturbation with operator norm at most $\\delta=1/4$, the squared-norm bounds remain at least $(1-1/4)^2=9/16$ and at most $(2+1/4)^2=81/16$.\n\\vfig[.87]{ext_frame}\n\\head{Both bounds can be attained}\nChoose $\\Delta E=\\operatorname{diag}(-1/4,1/4)$. Then $\\widetilde E=\\operatorname{diag}(3/4,9/4)$ has precisely those two frame bounds. Its inverse has operator norm $4/3$. A coefficient error $\\nu=(3/4,0)$ therefore produces signal error $(1,0)$, attaining the noise bound.\nThe strict condition $\\delta<\\sqrt A$ prevents the lower bound from vanishing. At $\\delta=1$, the perturbation $\\operatorname{diag}(-1,0)$ destroys the first channel entirely.\n\\head{An inverse and an approximate i","0&2\\end{pmatrix},\\qquad\nE^*E=\\operatorname{diag}(1,4).\n\\]\nIt has frame bounds $A=1$, $B=4$. For every perturbation with operator norm at most $\\delta=1/4$, the squared-norm bounds remain at least $(1-1/4)^2=9/16$ and at most $(2+1/4)^2=81/16$.\n\\vfig[.87]{ext_frame}\n\\head{Both bounds can be attained}\nChoose $\\Delta E=\\operatorname{diag}(-1/4,1/4)$. Then $\\widetilde E=\\operatorname{diag}(3/4,9/4)$ has precisely those two frame bounds. Its inverse has operator norm $4/3$. A coefficient error $\\nu=(3/4,0)$ therefore produces signal error $(1,0)$, attaining the noise bound.\nThe strict condition $\\delta<\\sqrt A$ prevents the lower bound from vanishing. At $\\delta=1$, the perturbation $\\operatorname{diag}(-1,0)$ destroys the first channel entirely.\n\\head{An inverse and an approximate inverse answer different questions}\nIf the original synthesis $R=","0&2\\end{pmatrix},\\qquad\nE^*E=\\operatorname{diag}(1,4).\n\\]\nIt has frame bounds $A=1$, $B=4$. For every perturbation with operator norm at most $\\delta=1/4$, the squared-norm bounds remain at least $(1-1/4)^2=9/16$ and at most $(2+1/4)^2=81/16$.\n\\vfig[.87]{ext_frame}\n\\head{Both bounds can be attained}\nChoose $\\Delta E=\\operatorname{diag}(-1/4,1/4)$. Then $\\widetilde E=\\operatorname{diag}(3/4,9/4)$ has precisely those two frame bounds. Its inverse has operator norm $4/3$. A coefficient error $\\nu=(3/4,0)$ therefore produces signal error $(1,0)$, attaining the noise bound.\nThe strict condition $\\delta<\\sqrt A$ prevents the lower bound from vanishing. At $\\delta=1$, the perturbation $\\operatorname{diag}(-1,0)$ destroys the first channel entirely.\n\\head{An inverse and an approximate inverse answer different questions}\nIf the original synthesis $R=\\operatorname{diag}(1,1/2)$ is retained after this perturb","0&2\\end{pmatrix},\\qquad\nE^*E=\\operatorname{diag}(1,4).\n\\]\nIt has frame bounds $A=1$, $B=4$. For every perturbation with operator norm at most $\\delta=1/4$, the squared-norm bounds remain at least $(1-1/4)^2=9/16$ and at most $(2+1/4)^2=81/16$.\n\\vfig[.87]{ext_frame}\n\\head{Both bounds can be attained}\nChoose $\\Delta E=\\operatorname{diag}(-1/4,1/4)$. Then $\\widetilde E=\\operatorname{diag}(3/4,9/4)$ has precisely those two frame bounds. Its inverse has operator norm $4/3$. A coefficient error $\\nu=(3/4,0)$ therefore produces signal error $(1,0)$, attaining the noise bound.\nThe strict condition $\\delta<\\sqrt A$ prevents the lower bound from vanishing. At $\\delta=1$, the perturbation $\\operatorname{diag}(-1,0)$ destroys the first channel entirely.\n\\head{An inverse and an approximate inverse answer different questions}\nIf the original synthesis $R=\\operatorname{diag}(1,1/2)$ is retained after this perturbation, then\n\\[\nR\\widetilde E-I","0&2\\end{pmatrix},\\qquad\nE^*E=\\operatorname{diag}(1,4).\n\\]\nIt has frame bounds $A=1$, $B=4$. For every perturbation with operator norm at most $\\delta=1/4$, the squared-norm bounds remain at least $(1-1/4)^2=9/16$ and at most $(2+1/4)^2=81/16$.\n\\vfig[.87]{ext_frame}\n\\head{Both bounds can be attained}\nChoose $\\Delta E=\\operatorname{diag}(-1/4,1/4)$. Then $\\widetilde E=\\operatorname{diag}(3/4,9/4)$ has precisely those two frame bounds. Its inverse has operator norm $4/3$. A coefficient error $\\nu=(3/4,0)$ therefore produces signal error $(1,0)$, attaining the noise bound.\nThe strict condition $\\delta<\\sqrt A$ prevents the lower bound from vanishing. At $\\delta=1$, the perturbation $\\operatorname{diag}(-1,0)$ destroys the first channel entirely.\n\\head{An inverse and an approximate inverse answer different questions}\nIf the original synthesis $R=\\operatorname{diag}(1,1/2)$ is retained after this perturbation, then\n\\[\nR\\widetilde E-I=\\operatorname{diag}(-1/4,1/8).\n\\]\nIts defect norm is $1/4$. Every input has relative reconstruction error","0&2\\end{pmatrix},\\qquad\nE^*E=\\operatorname{diag}(1,4).\n\\]\nIt has frame bounds $A=1$, $B=4$. For every perturbation with operator norm at most $\\delta=1/4$, the squared-norm bounds remain at least $(1-1/4)^2=9/16$ and at most $(2+1/4)^2=81/16$.\n\\vfig[.87]{ext_frame}\n\\head{Both bounds can be attained}\nChoose $\\Delta E=\\operatorname{diag}(-1/4,1/4)$. Then $\\widetilde E=\\operatorname{diag}(3/4,9/4)$ has precisely those two frame bounds. Its inverse has operator norm $4/3$. A coefficient error $\\nu=(3/4,0)$ therefore produces signal error $(1,0)$, attaining the noise bound.\nThe strict condition $\\delta<\\sqrt A$ prevents the lower bound from vanishing. At $\\delta=1$, the perturbation $\\operatorname{diag}(-1,0)$ destroys the first channel entirely.\n\\head{An inverse and an approximate inverse answer different questions}\nIf the original synthesis $R=\\operatorname{diag}(1,1/2)$ is retained after this perturbation, then\n\\[\nR\\widetilde E-I=\\operatorname{diag}(-1/4,1/8).\n\\]\nIts defect norm is $1/4$. Every input has relative reconstruction error at most that value, with equality on the first co","me bounds $A=1$, $B=4$. For every perturbation with operator norm at most $\\delta=1/4$, the squared-norm bounds remain at least $(1-1/4)^2=9/16$ and at most $(2+1/4)^2=81/16$.\n\\vfig[.87]{ext_frame}\n\\head{Both bounds can be attained}\nChoose $\\Delta E=\\operatorname{diag}(-1/4,1/4)$. Then $\\widetilde E=\\operatorname{diag}(3/4,9/4)$ has precisely those two frame bounds. Its inverse has operator norm $4/3$. A coefficient error $\\nu=(3/4,0)$ therefore produces signal error $(1,0)$, attaining the noise bound.\nThe strict condition $\\delta<\\sqrt A$ prevents the lower bound from vanishing. At $\\delta=1$, the perturbation $\\operatorname{diag}(-1,0)$ destroys the first channel entirely.\n\\head{An inverse and an approximate inverse answer different questions}\nIf the original synthesis $R=\\operatorname{diag}(1,1/2)$ is retained after this perturbation, then\n\\[\nR\\widetilde E-I=\\operatorname{diag}(-1/4,1/8).\n\\]\nIts defect norm is $1/4$. Every input has relative reconstruction error at most that value, with equality on the first coordinate. For frequency-dependent banks the theorem uses an essential s…","every perturbation with operator norm at most $\\delta=1/4$, the squared-norm bounds remain at least $(1-1/4)^2=9/16$ and at most $(2+1/4)^2=81/16$.\n\\vfig[.87]{ext_frame}\n\\head{Both bounds can be attained}\nChoose $\\Delta E=\\operatorname{diag}(-1/4,1/4)$. Then $\\widetilde E=\\operatorname{diag}(3/4,9/4)$ has precisely those two frame bounds. Its inverse has operator norm $4/3$. A coefficient error $\\nu=(3/4,0)$ therefore produces signal error $(1,0)$, attaining the noise bound.\nThe strict condition $\\delta<\\sqrt A$ prevents the lower bound from vanishing. At $\\delta=1$, the perturbation $\\operatorname{diag}(-1,0)$ destroys the first channel entirely.\n\\head{An inverse and an approximate inverse answer different questions}\nIf the original synthesis $R=\\operatorname{diag}(1,1/2)$ is retained after this perturbation, then\n\\[\nR\\widetilde E-I=\\operatorname{diag}(-1/4,1/8).\n\\]\nIts defect norm is $1/4$. Every input has relative reconstruction error at most that value, with equality on the first coordinate. For frequency-dependent banks the theorem uses an essential supremum on the unit circle, …","$\\delta=1/4$, the squared-norm bounds remain at least $(1-1/4)^2=9/16$ and at most $(2+1/4)^2=81/16$.\n\\vfig[.87]{ext_frame}\n\\head{Both bounds can be attained}\nChoose $\\Delta E=\\operatorname{diag}(-1/4,1/4)$. Then $\\widetilde E=\\operatorname{diag}(3/4,9/4)$ has precisely those two frame bounds. Its inverse has operator norm $4/3$. A coefficient error $\\nu=(3/4,0)$ therefore produces signal error $(1,0)$, attaining the noise bound.\nThe strict condition $\\delta<\\sqrt A$ prevents the lower bound from vanishing. At $\\delta=1$, the perturbation $\\operatorname{diag}(-1,0)$ destroys the first channel entirely.\n\\head{An inverse and an approximate inverse answer different questions}\nIf the original synthesis $R=\\operatorname{diag}(1,1/2)$ is retained after this perturbation, then\n\\[\nR\\widetilde E-I=\\operatorname{diag}(-1/4,1/8).\n\\]\nIts defect norm is $1/4$. Every input has relative reconstruction error at most that value, with equality on the first coordinate. For frequency-dependent banks the theorem uses an essential supremum on the unit circle, not a few sampled frequencies.\n\\takeaway{A low…","1-1/4)^2=9/16$ and at most $(2+1/4)^2=81/16$.\n\\vfig[.87]{ext_frame}\n\\head{Both bounds can be attained}\nChoose $\\Delta E=\\operatorname{diag}(-1/4,1/4)$. Then $\\widetilde E=\\operatorname{diag}(3/4,9/4)$ has precisely those two frame bounds. Its inverse has operator norm $4/3$. A coefficient error $\\nu=(3/4,0)$ therefore produces signal error $(1,0)$, attaining the noise bound.\nThe strict condition $\\delta<\\sqrt A$ prevents the lower bound from vanishing. At $\\delta=1$, the perturbation $\\operatorname{diag}(-1,0)$ destroys the first channel entirely.\n\\head{An inverse and an approximate inverse answer different questions}\nIf the original synthesis $R=\\operatorname{diag}(1,1/2)$ is retained after this perturbation, then\n\\[\nR\\widetilde E-I=\\operatorname{diag}(-1/4,1/8).\n\\]\nIts defect norm is $1/4$. Every input has relative reconstruction error at most that value, with equality on the first coordinate. For frequency-dependent banks the theorem uses an essential supremum on the unit circle, not a few sampled frequencies.\n\\takeaway{A lower singular-value gap quantifies stable recovery. A reco…","frame}\n\\head{Both bounds can be attained}\nChoose $\\Delta E=\\operatorname{diag}(-1/4,1/4)$. Then $\\widetilde E=\\operatorname{diag}(3/4,9/4)$ has precisely those two frame bounds. Its inverse has operator norm $4/3$. A coefficient error $\\nu=(3/4,0)$ therefore produces signal error $(1,0)$, attaining the noise bound.\nThe strict condition $\\delta<\\sqrt A$ prevents the lower bound from vanishing. At $\\delta=1$, the perturbation $\\operatorname{diag}(-1,0)$ destroys the first channel entirely.\n\\head{An inverse and an approximate inverse answer different questions}\nIf the original synthesis $R=\\operatorname{diag}(1,1/2)$ is retained after this perturbation, then\n\\[\nR\\widetilde E-I=\\operatorname{diag}(-1/4,1/8).\n\\]\nIts defect norm is $1/4$. Every input has relative reconstruction error at most that value, with equality on the first coordinate. For frequency-dependent banks the theorem uses an essential supremum on the unit circle, not a few sampled frequencies.\n\\takeaway{A lower singular-value gap quantifies stable recovery. A reconstruction defect quantifies the error of the actual synthesi…","be attained}\nChoose $\\Delta E=\\operatorname{diag}(-1/4,1/4)$. Then $\\widetilde E=\\operatorname{diag}(3/4,9/4)$ has precisely those two frame bounds. Its inverse has operator norm $4/3$. A coefficient error $\\nu=(3/4,0)$ therefore produces signal error $(1,0)$, attaining the noise bound.\nThe strict condition $\\delta<\\sqrt A$ prevents the lower bound from vanishing. At $\\delta=1$, the perturbation $\\operatorname{diag}(-1,0)$ destroys the first channel entirely.\n\\head{An inverse and an approximate inverse answer different questions}\nIf the original synthesis $R=\\operatorname{diag}(1,1/2)$ is retained after this perturbation, then\n\\[\nR\\widetilde E-I=\\operatorname{diag}(-1/4,1/8).\n\\]\nIts defect norm is $1/4$. Every input has relative reconstruction error at most that value, with equality on the first coordinate. For frequency-dependent banks the theorem uses an essential supremum on the unit circle, not a few sampled frequencies.\n\\takeaway{A lower singular-value gap quantifies stable recovery. A reconstruction defect quantifies the error of the actual synthesis bank being used.}","s operator norm $4/3$. A coefficient error $\\nu=(3/4,0)$ therefore produces signal error $(1,0)$, attaining the noise bound.\nThe strict condition $\\delta<\\sqrt A$ prevents the lower bound from vanishing. At $\\delta=1$, the perturbation $\\operatorname{diag}(-1,0)$ destroys the first channel entirely.\n\\head{An inverse and an approximate inverse answer different questions}\nIf the original synthesis $R=\\operatorname{diag}(1,1/2)$ is retained after this perturbation, then\n\\[\nR\\widetilde E-I=\\operatorname{diag}(-1/4,1/8).\n\\]\nIts defect norm is $1/4$. Every input has relative reconstruction error at most that value, with equality on the first coordinate. For frequency-dependent banks the theorem uses an essential supremum on the unit circle, not a few sampled frequencies.\n\\takeaway{A lower singular-value gap quantifies stable recovery. A reconstruction defect quantifies the error of the actual synthesis bank being used.}","s signal error $(1,0)$, attaining the noise bound.\nThe strict condition $\\delta<\\sqrt A$ prevents the lower bound from vanishing. At $\\delta=1$, the perturbation $\\operatorname{diag}(-1,0)$ destroys the first channel entirely.\n\\head{An inverse and an approximate inverse answer different questions}\nIf the original synthesis $R=\\operatorname{diag}(1,1/2)$ is retained after this perturbation, then\n\\[\nR\\widetilde E-I=\\operatorname{diag}(-1/4,1/8).\n\\]\nIts defect norm is $1/4$. Every input has relative reconstruction error at most that value, with equality on the first coordinate. For frequency-dependent banks the theorem uses an essential supremum on the unit circle, not a few sampled frequencies.\n\\takeaway{A lower singular-value gap quantifies stable recovery. A reconstruction defect quantifies the error of the actual synthesis bank being used.}","$\\delta<\\sqrt A$ prevents the lower bound from vanishing. At $\\delta=1$, the perturbation $\\operatorname{diag}(-1,0)$ destroys the first channel entirely.\n\\head{An inverse and an approximate inverse answer different questions}\nIf the original synthesis $R=\\operatorname{diag}(1,1/2)$ is retained after this perturbation, then\n\\[\nR\\widetilde E-I=\\operatorname{diag}(-1/4,1/8).\n\\]\nIts defect norm is $1/4$. Every input has relative reconstruction error at most that value, with equality on the first coordinate. For frequency-dependent banks the theorem uses an essential supremum on the unit circle, not a few sampled frequencies.\n\\takeaway{A lower singular-value gap quantifies stable recovery. A reconstruction defect quantifies the error of the actual synthesis bank being used.}","\\vpage{30}{Nonunitary tree bounds propagate along branches}{LOCAL FRAME CONSTANTS / A MIXED-DEPTH CERTIFICATE}{Traditional monograph: Proposition E6.4 (p. 53).}\nSuppose the root split has nonnegative squared-norm bounds $A=1/2$ and $B=2$. Retain the left child. Split the right child with bounds $3/4$ and $5/4$, then retain both grandchildren.\n\\vfig[.85]{ext_tree}\n\\head{Start with one at every retained node}\nThe left subtree has bounds $a_0=b_0=1$. The right subtree has $a_1=3/4$ and $b_1=5/4$. The recursive certificate gives\n\\[\na_{\\rm root}=\\tfrac12\\min\\{1,\\tfrac34\\}=\\tfrac38,\n\\qquad\nb_{\\rm root}=2\\max\\{1,\\tfrac54\\}=\\tfrac52.\n\\]\nConsequently the total leaf coefficient energy lies between $3/8$ and $5/2$ times the root energy. These are guaranteed bounds; no assertion that every admissible collection of local maps attains both constants is needed.\n\\head{Why a minimum and maximum appear}\nAfter the root split, its e","\\vpage{30}{Nonunitary tree bounds propagate along branches}{LOCAL FRAME CONSTANTS / A MIXED-DEPTH CERTIFICATE}{Traditional monograph: Proposition E6.4 (p. 53).}\nSuppose the root split has nonnegative squared-norm bounds $A=1/2$ and $B=2$. Retain the left child. Split the right child with bounds $3/4$ and $5/4$, then retain both grandchildren.\n\\vfig[.85]{ext_tree}\n\\head{Start with one at every retained node}\nThe left subtree has bounds $a_0=b_0=1$. The right subtree has $a_1=3/4$ and $b_1=5/4$. The recursive certificate gives\n\\[\na_{\\rm root}=\\tfrac12\\min\\{1,\\tfrac34\\}=\\tfrac38,\n\\qquad\nb_{\\rm root}=2\\max\\{1,\\tfrac54\\}=\\tfrac52.\n\\]\nConsequently the total leaf coefficient energy lies between $3/8$ and $5/2$ times the root energy. These are guaranteed bounds; no assertion that every admissible collection of local maps attains both constants is needed.\n\\head{Why a minimum and maximum appear}\nAfter the root split, its energy may","\\vpage{30}{Nonunitary tree bounds propagate along branches}{LOCAL FRAME CONSTANTS / A MIXED-DEPTH CERTIFICATE}{Traditional monograph: Proposition E6.4 (p. 53).}\nSuppose the root split has nonnegative squared-norm bounds $A=1/2$ and $B=2$. Retain the left child. Split the right child with bounds $3/4$ and $5/4$, then retain both grandchildren.\n\\vfig[.85]{ext_tree}\n\\head{Start with one at every retained node}\nThe left subtree has bounds $a_0=b_0=1$. The right subtree has $a_1=3/4$ and $b_1=5/4$. The recursive certificate gives\n\\[\na_{\\rm root}=\\tfrac12\\min\\{1,\\tfrac34\\}=\\tfrac38,\n\\qquad\nb_{\\rm root}=2\\max\\{1,\\tfrac54\\}=\\tfrac52.\n\\]\nConsequently the total leaf coefficient energy lies between $3/8$ and $5/2$ times the root energy. These are guaranteed bounds; no assertion that every admissible collection of local maps attains both constants is needed.\n\\head{Why a minimum and maximum appear}\nAfter the root split, its energy may distribute unevenly between the two children. The lower estimate","\\vpage{30}{Nonunitary tree bounds propagate along branches}{LOCAL FRAME CONSTANTS / A MIXED-DEPTH CERTIFICATE}{Traditional monograph: Proposition E6.4 (p. 53).}\nSuppose the root split has nonnegative squared-norm bounds $A=1/2$ and $B=2$. Retain the left child. Split the right child with bounds $3/4$ and $5/4$, then retain both grandchildren.\n\\vfig[.85]{ext_tree}\n\\head{Start with one at every retained node}\nThe left subtree has bounds $a_0=b_0=1$. The right subtree has $a_1=3/4$ and $b_1=5/4$. The recursive certificate gives\n\\[\na_{\\rm root}=\\tfrac12\\min\\{1,\\tfrac34\\}=\\tfrac38,\n\\qquad\nb_{\\rm root}=2\\max\\{1,\\tfrac54\\}=\\tfrac52.\n\\]\nConsequently the total leaf coefficient energy lies between $3/8$ and $5/2$ times the root energy. These are guaranteed bounds; no assertion that every admissible collection of local maps attains both constants is needed.\n\\head{Why a minimum and maximum appear}\nAfter the root split, its energy may distribute unevenly between the two children. The lower estimate must surv","ate along branches}{LOCAL FRAME CONSTANTS / A MIXED-DEPTH CERTIFICATE}{Traditional monograph: Proposition E6.4 (p. 53).}\nSuppose the root split has nonnegative squared-norm bounds $A=1/2$ and $B=2$. Retain the left child. Split the right child with bounds $3/4$ and $5/4$, then retain both grandchildren.\n\\vfig[.85]{ext_tree}\n\\head{Start with one at every retained node}\nThe left subtree has bounds $a_0=b_0=1$. The right subtree has $a_1=3/4$ and $b_1=5/4$. The recursive certificate gives\n\\[\na_{\\rm root}=\\tfrac12\\min\\{1,\\tfrac34\\}=\\tfrac38,\n\\qquad\nb_{\\rm root}=2\\max\\{1,\\tfrac54\\}=\\tfrac52.\n\\]\nConsequently the total leaf coefficient energy lies between $3/8$ and $5/2$ times the root energy. These are guaranteed bounds; no assertion that every admissible collection of local maps attains both constants is needed.\n\\head{Why a minimum and maximum appear}\nAfter the root split, its energy may distribute unevenly between the two children. The lower estimate must survive the least favorable child lower bound. The upper estimate must survive the largest child upper bound. Replacing these by an un…","STANTS / A MIXED-DEPTH CERTIFICATE}{Traditional monograph: Proposition E6.4 (p. 53).}\nSuppose the root split has nonnegative squared-norm bounds $A=1/2$ and $B=2$. Retain the left child. Split the right child with bounds $3/4$ and $5/4$, then retain both grandchildren.\n\\vfig[.85]{ext_tree}\n\\head{Start with one at every retained node}\nThe left subtree has bounds $a_0=b_0=1$. The right subtree has $a_1=3/4$ and $b_1=5/4$. The recursive certificate gives\n\\[\na_{\\rm root}=\\tfrac12\\min\\{1,\\tfrac34\\}=\\tfrac38,\n\\qquad\nb_{\\rm root}=2\\max\\{1,\\tfrac54\\}=\\tfrac52.\n\\]\nConsequently the total leaf coefficient energy lies between $3/8$ and $5/2$ times the root energy. These are guaranteed bounds; no assertion that every admissible collection of local maps attains both constants is needed.\n\\head{Why a minimum and maximum appear}\nAfter the root split, its energy may distribute unevenly between the two children. The lower estimate must survive the least favorable child lower bound. The upper estimate must survive the largest child upper bound. Replacing these by an unweighted average would presume an e…","ED-DEPTH CERTIFICATE}{Traditional monograph: Proposition E6.4 (p. 53).}\nSuppose the root split has nonnegative squared-norm bounds $A=1/2$ and $B=2$. Retain the left child. Split the right child with bounds $3/4$ and $5/4$, then retain both grandchildren.\n\\vfig[.85]{ext_tree}\n\\head{Start with one at every retained node}\nThe left subtree has bounds $a_0=b_0=1$. The right subtree has $a_1=3/4$ and $b_1=5/4$. The recursive certificate gives\n\\[\na_{\\rm root}=\\tfrac12\\min\\{1,\\tfrac34\\}=\\tfrac38,\n\\qquad\nb_{\\rm root}=2\\max\\{1,\\tfrac54\\}=\\tfrac52.\n\\]\nConsequently the total leaf coefficient energy lies between $3/8$ and $5/2$ times the root energy. These are guaranteed bounds; no assertion that every admissible collection of local maps attains both constants is needed.\n\\head{Why a minimum and maximum appear}\nAfter the root split, its energy may distribute unevenly between the two children. The lower estimate must survive the least favorable child lower bound. The upper estimate must survive the largest child upper bound. Replacing these by an unweighted average would presume an energy distribu…","Proposition E6.4 (p. 53).}\nSuppose the root split has nonnegative squared-norm bounds $A=1/2$ and $B=2$. Retain the left child. Split the right child with bounds $3/4$ and $5/4$, then retain both grandchildren.\n\\vfig[.85]{ext_tree}\n\\head{Start with one at every retained node}\nThe left subtree has bounds $a_0=b_0=1$. The right subtree has $a_1=3/4$ and $b_1=5/4$. The recursive certificate gives\n\\[\na_{\\rm root}=\\tfrac12\\min\\{1,\\tfrac34\\}=\\tfrac38,\n\\qquad\nb_{\\rm root}=2\\max\\{1,\\tfrac54\\}=\\tfrac52.\n\\]\nConsequently the total leaf coefficient energy lies between $3/8$ and $5/2$ times the root energy. These are guaranteed bounds; no assertion that every admissible collection of local maps attains both constants is needed.\n\\head{Why a minimum and maximum appear}\nAfter the root split, its energy may distribute unevenly between the two children. The lower estimate must survive the least favorable child lower bound. The upper estimate must survive the largest child upper bound. Replacing these by an unweighted average would presume an energy distribution that the hypotheses do not supply.\nWhen …","/4$ and $5/4$, then retain both grandchildren.\n\\vfig[.85]{ext_tree}\n\\head{Start with one at every retained node}\nThe left subtree has bounds $a_0=b_0=1$. The right subtree has $a_1=3/4$ and $b_1=5/4$. The recursive certificate gives\n\\[\na_{\\rm root}=\\tfrac12\\min\\{1,\\tfrac34\\}=\\tfrac38,\n\\qquad\nb_{\\rm root}=2\\max\\{1,\\tfrac54\\}=\\tfrac52.\n\\]\nConsequently the total leaf coefficient energy lies between $3/8$ and $5/2$ times the root energy. These are guaranteed bounds; no assertion that every admissible collection of local maps attains both constants is needed.\n\\head{Why a minimum and maximum appear}\nAfter the root split, its energy may distribute unevenly between the two children. The lower estimate must survive the least favorable child lower bound. The upper estimate must survive the largest child upper bound. Replacing these by an unweighted average would presume an energy distribution that the hypotheses do not supply.\nWhen all local maps are unitary, every factor is one and the recursion recovers exact energy conservation. For general local maps, a positive lower bound alone establish…","/4$, then retain both grandchildren.\n\\vfig[.85]{ext_tree}\n\\head{Start with one at every retained node}\nThe left subtree has bounds $a_0=b_0=1$. The right subtree has $a_1=3/4$ and $b_1=5/4$. The recursive certificate gives\n\\[\na_{\\rm root}=\\tfrac12\\min\\{1,\\tfrac34\\}=\\tfrac38,\n\\qquad\nb_{\\rm root}=2\\max\\{1,\\tfrac54\\}=\\tfrac52.\n\\]\nConsequently the total leaf coefficient energy lies between $3/8$ and $5/2$ times the root energy. These are guaranteed bounds; no assertion that every admissible collection of local maps attains both constants is needed.\n\\head{Why a minimum and maximum appear}\nAfter the root split, its energy may distribute unevenly between the two children. The lower estimate must survive the least favorable child lower bound. The upper estimate must survive the largest child upper bound. Replacing these by an unweighted average would presume an energy distribution that the hypotheses do not supply.\nWhen all local maps are unitary, every factor is one and the recursion recovers exact energy conservation. For general local maps, a positive lower bound alone establishes injecti…","\\vpage{31}{Derivative jumps have their own packets}{TRACE DATA / A STEP AND A RAMP}{Traditional monograph: Theorem E7.2 (p. 56), Corollary E7.3 (p. 57).}\nOn $[0,1]$, let $H_\\tau=\\mathbf1_{[\\tau,1]}$ and take\n\\[\nf(t)=t^2+2H_{1/3}(t)-3(t-2/3)_+.\n\\]\nThe function has a value jump $2$ at $1/3$ and a first-derivative jump $-3$ at $2/3$. Its packet and remainder are\n\\[\nj_f(t)=2\\phi_{0,1/3}(t)-3\\phi_{1,2/3}(t),\\qquad s_f(t)=t^2.\n\\]\n\\vfig[.92]{ext_trace}\n\\platecaption{The displayed curves are exact polynomial pieces. The solid jump is not connected across its discontinuity. The dashed curve is the global smooth remainder.}\n\\head{The normalization makes the trace coefficient literal}\nThe general packet is $\\phi_{m,\\tau}(t)=(t-\\tau)_+^m/m!$. Its derivative of order $m$ has a unit trace jump, while its other classical trace jumps vanish. Thus its multiplier is ex","\\vpage{31}{Derivative jumps have their own packets}{TRACE DATA / A STEP AND A RAMP}{Traditional monograph: Theorem E7.2 (p. 56), Corollary E7.3 (p. 57).}\nOn $[0,1]$, let $H_\\tau=\\mathbf1_{[\\tau,1]}$ and take\n\\[\nf(t)=t^2+2H_{1/3}(t)-3(t-2/3)_+.\n\\]\nThe function has a value jump $2$ at $1/3$ and a first-derivative jump $-3$ at $2/3$. Its packet and remainder are\n\\[\nj_f(t)=2\\phi_{0,1/3}(t)-3\\phi_{1,2/3}(t),\\qquad s_f(t)=t^2.\n\\]\n\\vfig[.92]{ext_trace}\n\\platecaption{The displayed curves are exact polynomial pieces. The solid jump is not connected across its discontinuity. The dashed curve is the global smooth remainder.}\n\\head{The normalization makes the trace coefficient literal}\nThe general packet is $\\phi_{m,\\tau}(t)=(t-\\tau)_+^m/m!$. Its derivative of order $m$ has a unit trace jump, while its other classical trace jumps vanish. Thus its multiplier is exactly the jump being removed, with","\\vpage{31}{Derivative jumps have their own packets}{TRACE DATA / A STEP AND A RAMP}{Traditional monograph: Theorem E7.2 (p. 56), Corollary E7.3 (p. 57).}\nOn $[0,1]$, let $H_\\tau=\\mathbf1_{[\\tau,1]}$ and take\n\\[\nf(t)=t^2+2H_{1/3}(t)-3(t-2/3)_+.\n\\]\nThe function has a value jump $2$ at $1/3$ and a first-derivative jump $-3$ at $2/3$. Its packet and remainder are\n\\[\nj_f(t)=2\\phi_{0,1/3}(t)-3\\phi_{1,2/3}(t),\\qquad s_f(t)=t^2.\n\\]\n\\vfig[.92]{ext_trace}\n\\platecaption{The displayed curves are exact polynomial pieces. The solid jump is not connected across its discontinuity. The dashed curve is the global smooth remainder.}\n\\head{The normalization makes the trace coefficient literal}\nThe general packet is $\\phi_{m,\\tau}(t)=(t-\\tau)_+^m/m!$. Its derivative of order $m$ has a unit trace jump, while its other classical trace jumps vanish. Thus its multiplier is exactly the jump being removed, with no hidden factorial.\nHere the remainder has ma","\\vpage{31}{Derivative jumps have their own packets}{TRACE DATA / A STEP AND A RAMP}{Traditional monograph: Theorem E7.2 (p. 56), Corollary E7.3 (p. 57).}\nOn $[0,1]$, let $H_\\tau=\\mathbf1_{[\\tau,1]}$ and take\n\\[\nf(t)=t^2+2H_{1/3}(t)-3(t-2/3)_+.\n\\]\nThe function has a value jump $2$ at $1/3$ and a first-derivative jump $-3$ at $2/3$. Its packet and remainder are\n\\[\nj_f(t)=2\\phi_{0,1/3}(t)-3\\phi_{1,2/3}(t),\\qquad s_f(t)=t^2.\n\\]\n\\vfig[.92]{ext_trace}\n\\platecaption{The displayed curves are exact polynomial pieces. The solid jump is not connected across its discontinuity. The dashed curve is the global smooth remainder.}\n\\head{The normalization makes the trace coefficient literal}\nThe general packet is $\\phi_{m,\\tau}(t)=(t-\\tau)_+^m/m!$. Its derivative of order $m$ has a unit trace jump, while its other classical trace jumps vanish. Thus its multiplier is exactly the jump being removed, with no hidden factorial.\nHere the remainder has matching value and first-derivative t","\\vpage{31}{Derivative jumps have their own packets}{TRACE DATA / A STEP AND A RAMP}{Traditional monograph: Theorem E7.2 (p. 56), Corollary E7.3 (p. 57).}\nOn $[0,1]$, let $H_\\tau=\\mathbf1_{[\\tau,1]}$ and take\n\\[\nf(t)=t^2+2H_{1/3}(t)-3(t-2/3)_+.\n\\]\nThe function has a value jump $2$ at $1/3$ and a first-derivative jump $-3$ at $2/3$. Its packet and remainder are\n\\[\nj_f(t)=2\\phi_{0,1/3}(t)-3\\phi_{1,2/3}(t),\\qquad s_f(t)=t^2.\n\\]\n\\vfig[.92]{ext_trace}\n\\platecaption{The displayed curves are exact polynomial pieces. The solid jump is not connected across its discontinuity. The dashed curve is the global smooth remainder.}\n\\head{The normalization makes the trace coefficient literal}\nThe general packet is $\\phi_{m,\\tau}(t)=(t-\\tau)_+^m/m!$. Its derivative of order $m$ has a unit trace jump, while its other classical trace jumps vanish. Thus its multiplier is exactly the jump being removed, with no hidden factorial.\nHere the remainder has matching value and first-derivative traces and","\\vpage{31}{Derivative jumps have their own packets}{TRACE DATA / A STEP AND A RAMP}{Traditional monograph: Theorem E7.2 (p. 56), Corollary E7.3 (p. 57).}\nOn $[0,1]$, let $H_\\tau=\\mathbf1_{[\\tau,1]}$ and take\n\\[\nf(t)=t^2+2H_{1/3}(t)-3(t-2/3)_+.\n\\]\nThe function has a value jump $2$ at $1/3$ and a first-derivative jump $-3$ at $2/3$. Its packet and remainder are\n\\[\nj_f(t)=2\\phi_{0,1/3}(t)-3\\phi_{1,2/3}(t),\\qquad s_f(t)=t^2.\n\\]\n\\vfig[.92]{ext_trace}\n\\platecaption{The displayed curves are exact polynomial pieces. The solid jump is not connected across its discontinuity. The dashed curve is the global smooth remainder.}\n\\head{The normalization makes the trace coefficient literal}\nThe general packet is $\\phi_{m,\\tau}(t)=(t-\\tau)_+^m/m!$. Its derivative of order $m$ has a unit trace jump, while its other classical trace jumps vanish. Thus its multiplier is exactly the jump being removed, with no hidden factorial.\nHere the remainder has matching value and first-derivative traces and belongs to $W^{2,p}(0,1)$ for ev","\\vpage{31}{Derivative jumps have their own packets}{TRACE DATA / A STEP AND A RAMP}{Traditional monograph: Theorem E7.2 (p. 56), Corollary E7.3 (p. 57).}\nOn $[0,1]$, let $H_\\tau=\\mathbf1_{[\\tau,1]}$ and take\n\\[\nf(t)=t^2+2H_{1/3}(t)-3(t-2/3)_+.\n\\]\nThe function has a value jump $2$ at $1/3$ and a first-derivative jump $-3$ at $2/3$. Its packet and remainder are\n\\[\nj_f(t)=2\\phi_{0,1/3}(t)-3\\phi_{1,2/3}(t),\\qquad s_f(t)=t^2.\n\\]\n\\vfig[.92]{ext_trace}\n\\platecaption{The displayed curves are exact polynomial pieces. The solid jump is not connected across its discontinuity. The dashed curve is the global smooth remainder.}\n\\head{The normalization makes the trace coefficient literal}\nThe general packet is $\\phi_{m,\\tau}(t)=(t-\\tau)_+^m/m!$. Its derivative of order $m$ has a unit trace jump, while its other classical trace jumps vanish. Thus its multiplier is exactly the jump being removed, with no hidden factorial.\nHere the remainder has matching value and first-derivative traces and belongs to $W^{2,p}(0,1)$ for every $1\\le","\\vpage{31}{Derivative jumps have their own packets}{TRACE DATA / A STEP AND A RAMP}{Traditional monograph: Theorem E7.2 (p. 56), Corollary E7.3 (p. 57).}\nOn $[0,1]$, let $H_\\tau=\\mathbf1_{[\\tau,1]}$ and take\n\\[\nf(t)=t^2+2H_{1/3}(t)-3(t-2/3)_+.\n\\]\nThe function has a value jump $2$ at $1/3$ and a first-derivative jump $-3$ at $2/3$. Its packet and remainder are\n\\[\nj_f(t)=2\\phi_{0,1/3}(t)-3\\phi_{1,2/3}(t),\\qquad s_f(t)=t^2.\n\\]\n\\vfig[.92]{ext_trace}\n\\platecaption{The displayed curves are exact polynomial pieces. The solid jump is not connected across its discontinuity. The dashed curve is the global smooth remainder.}\n\\head{The normalization makes the trace coefficient literal}\nThe general packet is $\\phi_{m,\\tau}(t)=(t-\\tau)_+^m/m!$. Its derivative of order $m$ has a unit trace jump, while its other classical trace jumps vanish. Thus its multiplier is exactly the jump being removed, with no hidden factorial.\nHere the remainder has matching value and first-derivative traces and belongs to $W^{2,p}(0,1)$ for every $1\\le p\\le\\infty$. Its second weak derivative is $2$. The singular packet a…","vative jump $-3$ at $2/3$. Its packet and remainder are\n\\[\nj_f(t)=2\\phi_{0,1/3}(t)-3\\phi_{1,2/3}(t),\\qquad s_f(t)=t^2.\n\\]\n\\vfig[.92]{ext_trace}\n\\platecaption{The displayed curves are exact polynomial pieces. The solid jump is not connected across its discontinuity. The dashed curve is the global smooth remainder.}\n\\head{The normalization makes the trace coefficient literal}\nThe general packet is $\\phi_{m,\\tau}(t)=(t-\\tau)_+^m/m!$. Its derivative of order $m$ has a unit trace jump, while its other classical trace jumps vanish. Thus its multiplier is exactly the jump being removed, with no hidden factorial.\nHere the remainder has matching value and first-derivative traces and belongs to $W^{2,p}(0,1)$ for every $1\\le p\\le\\infty$. Its second weak derivative is $2$. The singular packet accounts for the distributional boundary terms that would otherwise arise at the breakpoints.\n\\head{What the general theorem requires}\nEach piece must lie in the stated Sobolev class and possess the specified one-sided traces. Subtracting the finite trace packet then joins those pieces into a global Sobole…","_f(t)=2\\phi_{0,1/3}(t)-3\\phi_{1,2/3}(t),\\qquad s_f(t)=t^2.\n\\]\n\\vfig[.92]{ext_trace}\n\\platecaption{The displayed curves are exact polynomial pieces. The solid jump is not connected across its discontinuity. The dashed curve is the global smooth remainder.}\n\\head{The normalization makes the trace coefficient literal}\nThe general packet is $\\phi_{m,\\tau}(t)=(t-\\tau)_+^m/m!$. Its derivative of order $m$ has a unit trace jump, while its other classical trace jumps vanish. Thus its multiplier is exactly the jump being removed, with no hidden factorial.\nHere the remainder has matching value and first-derivative traces and belongs to $W^{2,p}(0,1)$ for every $1\\le p\\le\\infty$. Its second weak derivative is $2$. The singular packet accounts for the distributional boundary terms that would otherwise arise at the breakpoints.\n\\head{What the general theorem requires}\nEach piece must lie in the stated Sobolev class and possess the specified one-sided traces. Subtracting the finite trace packet then joins those pieces into a global Sobolev remainder. This is a theorem about finitely many declared …","l smooth remainder.}\n\\head{The normalization makes the trace coefficient literal}\nThe general packet is $\\phi_{m,\\tau}(t)=(t-\\tau)_+^m/m!$. Its derivative of order $m$ has a unit trace jump, while its other classical trace jumps vanish. Thus its multiplier is exactly the jump being removed, with no hidden factorial.\nHere the remainder has matching value and first-derivative traces and belongs to $W^{2,p}(0,1)$ for every $1\\le p\\le\\infty$. Its second weak derivative is $2$. The singular packet accounts for the distributional boundary terms that would otherwise arise at the breakpoints.\n\\head{What the general theorem requires}\nEach piece must lie in the stated Sobolev class and possess the specified one-sided traces. Subtracting the finite trace packet then joins those pieces into a global Sobolev remainder. This is a theorem about finitely many declared breakpoints, not a representation of every possible singular measure or arbitrary image boundary.\n\\takeaway{A value jump uses a step. A slope jump uses a ramp. The derivative order determines the packet and its normalization.}","d{The normalization makes the trace coefficient literal}\nThe general packet is $\\phi_{m,\\tau}(t)=(t-\\tau)_+^m/m!$. Its derivative of order $m$ has a unit trace jump, while its other classical trace jumps vanish. Thus its multiplier is exactly the jump being removed, with no hidden factorial.\nHere the remainder has matching value and first-derivative traces and belongs to $W^{2,p}(0,1)$ for every $1\\le p\\le\\infty$. Its second weak derivative is $2$. The singular packet accounts for the distributional boundary terms that would otherwise arise at the breakpoints.\n\\head{What the general theorem requires}\nEach piece must lie in the stated Sobolev class and possess the specified one-sided traces. Subtracting the finite trace packet then joins those pieces into a global Sobolev remainder. This is a theorem about finitely many declared breakpoints, not a representation of every possible singular measure or arbitrary image boundary.\n\\takeaway{A value jump uses a step. A slope jump uses a ramp. The derivative order determines the packet and its normalization.}","iteral}\nThe general packet is $\\phi_{m,\\tau}(t)=(t-\\tau)_+^m/m!$. Its derivative of order $m$ has a unit trace jump, while its other classical trace jumps vanish. Thus its multiplier is exactly the jump being removed, with no hidden factorial.\nHere the remainder has matching value and first-derivative traces and belongs to $W^{2,p}(0,1)$ for every $1\\le p\\le\\infty$. Its second weak derivative is $2$. The singular packet accounts for the distributional boundary terms that would otherwise arise at the breakpoints.\n\\head{What the general theorem requires}\nEach piece must lie in the stated Sobolev class and possess the specified one-sided traces. Subtracting the finite trace packet then joins those pieces into a global Sobolev remainder. This is a theorem about finitely many declared breakpoints, not a representation of every possible singular measure or arbitrary image boundary.\n\\takeaway{A value jump uses a step. A slope jump uses a ramp. The derivative order determines the packet and its normalization.}","\\vpage{32}{The approximation operator also acts on packet error}{ESTIMATED PACKETS / AN EXACT AMPLIFICATION EXAMPLE}{Traditional monograph: Theorem E7.8 (p. 58).}\nLet $Q$ be a bounded \\emph{linear} operator and set $e=j-\\widehat j$. Substituting $f=s+j$ into $\\widehat f=Q(f-\\widehat j)+\\widehat j$ gives\n\\[\nf-\\widehat f=(I-Q)s+(I-Q)e.\n\\]\nThe second term includes the action of $Q$ on the packet error. Linearity is required for this split.\n\\vfig[.80]{ext_amplification}\n\\head{An oblique projection can increase the error}\nIn $\\RR^2$, choose\n\\[\nQ=\\begin{pmatrix}1&2","l monograph: Theorem E7.8 (p. 58).}\nLet $Q$ be a bounded \\emph{linear} operator and set $e=j-\\widehat j$. Substituting $f=s+j$ into $\\widehat f=Q(f-\\widehat j)+\\widehat j$ gives\n\\[\nf-\\widehat f=(I-Q)s+(I-Q)e.\n\\]\nThe second term includes the action of $Q$ on the packet error. Linearity is required for this split.\n\\vfig[.80]{ext_amplification}\n\\head{An oblique projection can increase the error}\nIn $\\RR^2$, choose\n\\[\nQ=\\begin{pmatrix}1&2","rem E7.8 (p. 58).}\nLet $Q$ be a bounded \\emph{linear} operator and set $e=j-\\widehat j$. Substituting $f=s+j$ into $\\widehat f=Q(f-\\widehat j)+\\widehat j$ gives\n\\[\nf-\\widehat f=(I-Q)s+(I-Q)e.\n\\]\nThe second term includes the action of $Q$ on the packet error. Linearity is required for this split.\n\\vfig[.80]{ext_amplification}\n\\head{An oblique projection can increase the error}\nIn $\\RR^2$, choose\n\\[\nQ=\\begin{pmatrix}1&2\\\\0&0\\end{pmatrix},\\qquad\ns=(1,0),\\qquad e=(0,1).\n\\]\nThen $Q^2=Q$ and $Qs=s$, but $(I-Q)e=(-2,1)$. The smooth error is zero while the unit packet error becomes a reconstruction error of norm $\\sqrt5$.\nFor a fully specified realization, take $j=e$ and $\\widehat j=0$. Then $f=(1,1)$, $\\widehat f=Qf=(3,0)$, and their difference is exactly $(-2,1)$.\n\\head{Use the actual operator bound}\nThe always-valid estimate in this setting is\n\\[\n\\|f-\\widehat f\\|\\leq\\|s-Qs\\|+(1+\\|Q\\|)\\|e\\|.\n\\]\nHere $\\|Q\\|=\\sqrt5$, so the bound is conservative. For an orthogonal projection, $\\|I-Q\\|\\le1$ gives a sharper estimate. The operator's geometry determines which bound can be used.\n\\takeaway{Recover…","0&0\\end{pmatrix},\\qquad\ns=(1,0),\\qquad e=(0,1).\n\\]\nThen $Q^2=Q$ and $Qs=s$, but $(I-Q)e=(-2,1)$. The smooth error is zero while the unit packet error becomes a reconstruction error of norm $\\sqrt5$.\nFor a fully specified realization, take $j=e$ and $\\widehat j=0$. Then $f=(1,1)$, $\\widehat f=Qf=(3,0)$, and their difference is exactly $(-2,1)$.\n\\head{Use the actual operator bound}\nThe always-valid estimate in this setting is\n\\[\n\\|f-\\widehat f\\|\\leq\\|s-Qs\\|+(1+\\|Q\\|)\\|e\\|.\n\\]\nHere $\\|Q\\|=\\sqrt5$, so the bound is conservative. For an orthogonal projection, $\\|I-Q\\|\\le1$ gives a sharper estimate. The operator's geometry determines which bound can be used.\n\\takeaway{Recovering the packet and approximating the background are separate tasks, but the chosen ap","0&0\\end{pmatrix},\\qquad\ns=(1,0),\\qquad e=(0,1).\n\\]\nThen $Q^2=Q$ and $Qs=s$, but $(I-Q)e=(-2,1)$. The smooth error is zero while the unit packet error becomes a reconstruction error of norm $\\sqrt5$.\nFor a fully specified realization, take $j=e$ and $\\widehat j=0$. Then $f=(1,1)$, $\\widehat f=Qf=(3,0)$, and their difference is exactly $(-2,1)$.\n\\head{Use the actual operator bound}\nThe always-valid estimate in this setting is\n\\[\n\\|f-\\widehat f\\|\\leq\\|s-Qs\\|+(1+\\|Q\\|)\\|e\\|.\n\\]\nHere $\\|Q\\|=\\sqrt5$, so the bound is conservative. For an orthogonal projection, $\\|I-Q\\|\\le1$ gives a sharper estimate. The operator's geometry determines which bound can be used.\n\\takeaway{Recovering the packet and approximating the background are separate tasks, but the chosen approximation","0&0\\end{pmatrix},\\qquad\ns=(1,0),\\qquad e=(0,1).\n\\]\nThen $Q^2=Q$ and $Qs=s$, but $(I-Q)e=(-2,1)$. The smooth error is zero while the unit packet error becomes a reconstruction error of norm $\\sqrt5$.\nFor a fully specified realization, take $j=e$ and $\\widehat j=0$. Then $f=(1,1)$, $\\widehat f=Qf=(3,0)$, and their difference is exactly $(-2,1)$.\n\\head{Use the actual operator bound}\nThe always-valid estimate in this setting is\n\\[\n\\|f-\\widehat f\\|\\leq\\|s-Qs\\|+(1+\\|Q\\|)\\|e\\|.\n\\]\nHere $\\|Q\\|=\\sqrt5$, so the bound is conservative. For an orthogonal projection, $\\|I-Q\\|\\le1$ gives a sharper estimate. The operator's geometry determines which bound can be used.\n\\takeaway{Recovering the packet and approximating the background are separate tasks, but the chosen approximation operator couples the","0&0\\end{pmatrix},\\qquad\ns=(1,0),\\qquad e=(0,1).\n\\]\nThen $Q^2=Q$ and $Qs=s$, but $(I-Q)e=(-2,1)$. The smooth error is zero while the unit packet error becomes a reconstruction error of norm $\\sqrt5$.\nFor a fully specified realization, take $j=e$ and $\\widehat j=0$. Then $f=(1,1)$, $\\widehat f=Qf=(3,0)$, and their difference is exactly $(-2,1)$.\n\\head{Use the actual operator bound}\nThe always-valid estimate in this setting is\n\\[\n\\|f-\\widehat f\\|\\leq\\|s-Qs\\|+(1+\\|Q\\|)\\|e\\|.\n\\]\nHere $\\|Q\\|=\\sqrt5$, so the bound is conservative. For an orthogonal projection, $\\|I-Q\\|\\le1$ gives a sharper estimate. The operator's geometry determines which bound can be used.\n\\takeaway{Recovering the packet and approximating the background are separate tasks, but the chosen approximation operator couples their final errors.}","),\\qquad e=(0,1).\n\\]\nThen $Q^2=Q$ and $Qs=s$, but $(I-Q)e=(-2,1)$. The smooth error is zero while the unit packet error becomes a reconstruction error of norm $\\sqrt5$.\nFor a fully specified realization, take $j=e$ and $\\widehat j=0$. Then $f=(1,1)$, $\\widehat f=Qf=(3,0)$, and their difference is exactly $(-2,1)$.\n\\head{Use the actual operator bound}\nThe always-valid estimate in this setting is\n\\[\n\\|f-\\widehat f\\|\\leq\\|s-Qs\\|+(1+\\|Q\\|)\\|e\\|.\n\\]\nHere $\\|Q\\|=\\sqrt5$, so the bound is conservative. For an orthogonal projection, $\\|I-Q\\|\\le1$ gives a sharper estimate. The operator's geometry determines which bound can be used.\n\\takeaway{Recovering the packet and approximating the background are separate tasks, but the chosen approximation operator couples their final errors.}",")e=(-2,1)$. The smooth error is zero while the unit packet error becomes a reconstruction error of norm $\\sqrt5$.\nFor a fully specified realization, take $j=e$ and $\\widehat j=0$. Then $f=(1,1)$, $\\widehat f=Qf=(3,0)$, and their difference is exactly $(-2,1)$.\n\\head{Use the actual operator bound}\nThe always-valid estimate in this setting is\n\\[\n\\|f-\\widehat f\\|\\leq\\|s-Qs\\|+(1+\\|Q\\|)\\|e\\|.\n\\]\nHere $\\|Q\\|=\\sqrt5$, so the bound is conservative. For an orthogonal projection, $\\|I-Q\\|\\le1$ gives a sharper estimate. The operator's geometry determines which bound can be used.\n\\takeaway{Recovering the packet and approximating the background are separate tasks, but the chosen approximation operator couples their final errors.}","econstruction error of norm $\\sqrt5$.\nFor a fully specified realization, take $j=e$ and $\\widehat j=0$. Then $f=(1,1)$, $\\widehat f=Qf=(3,0)$, and their difference is exactly $(-2,1)$.\n\\head{Use the actual operator bound}\nThe always-valid estimate in this setting is\n\\[\n\\|f-\\widehat f\\|\\leq\\|s-Qs\\|+(1+\\|Q\\|)\\|e\\|.\n\\]\nHere $\\|Q\\|=\\sqrt5$, so the bound is conservative. For an orthogonal projection, $\\|I-Q\\|\\le1$ gives a sharper estimate. The operator's geometry determines which bound can be used.\n\\takeaway{Recovering the packet and approximating the background are separate tasks, but the chosen approximation operator couples their final errors.}","\\vpage{33}{An overlap matrix is an exact error ledger}{REGION VALUES / EVERY MISCLASSIFICATION HAS A WEIGHT}{Traditional monograph: Theorem E8.1 (p. 59), Theorem E8.4 (p. 61).}\nLet two piecewise-constant signals have region values $a=(0,2,5)$ and $b=(0,2,5)$ on a counting-measure domain of twelve samples. Suppose their overlap counts are\n\\[\nM=\\begin{pmatrix}3&1&0","\\vpage{33}{An overlap matrix is an exact error ledger}{REGION VALUES / EVERY MISCLASSIFICATION HAS A WEIGHT}{Traditional monograph: Theorem E8.1 (p. 59), Theorem E8.4 (p. 61).}\nLet two piecewise-constant signals have region values $a=(0,2,5)$ and $b=(0,2,5)$ on a counting-measure domain of twelve samples. Suppose their overlap counts are\n\\[\nM=\\begin{pmatrix}3&1&0\\\\0&2&1\\\\1&0&4\\end{pmatrix}.\n\\]\nThe row totals $(4,3,5)$ are the true region sizes. The column totals are also $(4,3,5)$, although three samples are assigned to the wrong values.\n\\vfig[.80]{ext_overlap}\n\\head{Equal error counts need not mean equal reconstruction error}\nThe three off-diagonal entries contribute one sample each. Their value differences have magnitudes $2$, $3$, and $5$, so\n\\[\n\\|j-\\widehat j\\|_1=2+3+5=10,\\qquad\n\\|j-\\widehat j\\|_2^2=4+9+25=38.\n\\]\nThe counting-measure supremum error is $5$. The full-domain mean squared error is $38/12=19/6$. A mismatch fraction of $3/12$ reports a different statistic and omits all three amplitude weights.\n\\head{The formula requires partitions}\nFor measurable partitions $(E_i","1&0&4\\end{pmatrix}.\n\\]\nThe row totals $(4,3,5)$ are the true region sizes. The column totals are also $(4,3,5)$, although three samples are assigned to the wrong values.\n\\vfig[.80]{ext_overlap}\n\\head{Equal error counts need not mean equal reconstruction error}\nThe three off-diagonal entries contribute one sample each. Their value differences have magnitudes $2$, $3$, and $5$, so\n\\[\n\\|j-\\widehat j\\|_1=2+3+5=10,\\qquad\n\\|j-\\widehat j\\|_2^2=4+9+25=38.\n\\]\nThe counting-measure supremum error is $5$. The full-domain mean squared error is $38/12=19/6$. A mismatch fraction of $3/12$ reports a different statistic and omits all three amplitude weights.\n\\head{The formula requires partitions}\nFor measurable partitions $(E_i)$ and $(F_j)$, with $M_{i","1&0&4\\end{pmatrix}.\n\\]\nThe row totals $(4,3,5)$ are the true region sizes. The column totals are also $(4,3,5)$, although three samples are assigned to the wrong values.\n\\vfig[.80]{ext_overlap}\n\\head{Equal error counts need not mean equal reconstruction error}\nThe three off-diagonal entries contribute one sample each. Their value differences have magnitudes $2$, $3$, and $5$, so\n\\[\n\\|j-\\widehat j\\|_1=2+3+5=10,\\qquad\n\\|j-\\widehat j\\|_2^2=4+9+25=38.\n\\]\nThe counting-measure supremum error is $5$. The full-domain mean squared error is $38/12=19/6$. A mismatch fraction of $3/12$ reports a different statistic and omits all three amplitude weights.\n\\head{The formula requires partitions}\nFor measurable partitions $(E_i)$ and $(F_j)$, with $M_{ij}=\\mu(E_i\\cap F_j)$,\n\\[\n\\|j-\\widehat j\\|_p^p=\\sum_{i,j}|a_i-b_j","1&0&4\\end{pmatrix}.\n\\]\nThe row totals $(4,3,5)$ are the true region sizes. The column totals are also $(4,3,5)$, although three samples are assigned to the wrong values.\n\\vfig[.80]{ext_overlap}\n\\head{Equal error counts need not mean equal reconstruction error}\nThe three off-diagonal entries contribute one sample each. Their value differences have magnitudes $2$, $3$, and $5$, so\n\\[\n\\|j-\\widehat j\\|_1=2+3+5=10,\\qquad\n\\|j-\\widehat j\\|_2^2=4+9+25=38.\n\\]\nThe counting-measure supremum error is $5$. The full-domain mean squared error is $38/12=19/6$. A mismatch fraction of $3/12$ reports a different statistic and omits all three amplitude weights.\n\\head{The formula requires partitions}\nFor measurable partitions $(E_i)$ and $(F_j)$, with $M_{ij}=\\mu(E_i\\cap F_j)$,\n\\[\n\\|j-\\widehat j\\|_p^p=\\sum_{i,j}|a_i-b_j|^pM_{ij},\\qquad 1\\le p<\\infty.\n\\]\nEach location belongs to exactly one intersection. Overlapping indicator packets that are summed at the same location must first be converted to their actual combined value regions before this partition formula is app","1&0&4\\end{pmatrix}.\n\\]\nThe row totals $(4,3,5)$ are the true region sizes. The column totals are also $(4,3,5)$, although three samples are assigned to the wrong values.\n\\vfig[.80]{ext_overlap}\n\\head{Equal error counts need not mean equal reconstruction error}\nThe three off-diagonal entries contribute one sample each. Their value differences have magnitudes $2$, $3$, and $5$, so\n\\[\n\\|j-\\widehat j\\|_1=2+3+5=10,\\qquad\n\\|j-\\widehat j\\|_2^2=4+9+25=38.\n\\]\nThe counting-measure supremum error is $5$. The full-domain mean squared error is $38/12=19/6$. A mismatch fraction of $3/12$ reports a different statistic and omits all three amplitude weights.\n\\head{The formula requires partitions}\nFor measurable partitions $(E_i)$ and $(F_j)$, with $M_{ij}=\\mu(E_i\\cap F_j)$,\n\\[\n\\|j-\\widehat j\\|_p^p=\\sum_{i,j}|a_i-b_j|^pM_{ij},\\qquad 1\\le p<\\infty.\n\\]\nEach location belongs to exactly one intersection. Overlapping indicator packets that are summed at the same location must first be converted to their actual combined value regions before this partition formula is applied.","1&0&4\\end{pmatrix}.\n\\]\nThe row totals $(4,3,5)$ are the true region sizes. The column totals are also $(4,3,5)$, although three samples are assigned to the wrong values.\n\\vfig[.80]{ext_overlap}\n\\head{Equal error counts need not mean equal reconstruction error}\nThe three off-diagonal entries contribute one sample each. Their value differences have magnitudes $2$, $3$, and $5$, so\n\\[\n\\|j-\\widehat j\\|_1=2+3+5=10,\\qquad\n\\|j-\\widehat j\\|_2^2=4+9+25=38.\n\\]\nThe counting-measure supremum error is $5$. The full-domain mean squared error is $38/12=19/6$. A mismatch fraction of $3/12$ reports a different statistic and omits all three amplitude weights.\n\\head{The formula requires partitions}\nFor measurable partitions $(E_i)$ and $(F_j)$, with $M_{ij}=\\mu(E_i\\cap F_j)$,\n\\[\n\\|j-\\widehat j\\|_p^p=\\sum_{i,j}|a_i-b_j|^pM_{ij},\\qquad 1\\le p<\\infty.\n\\]\nEach location belongs to exactly one intersection. Overlapping indicator packets that are summed at the same location must first be converted to their actual combined value regions before this partition formula is applied.\n\\takeawa","1&0&4\\end{pmatrix}.\n\\]\nThe row totals $(4,3,5)$ are the true region sizes. The column totals are also $(4,3,5)$, although three samples are assigned to the wrong values.\n\\vfig[.80]{ext_overlap}\n\\head{Equal error counts need not mean equal reconstruction error}\nThe three off-diagonal entries contribute one sample each. Their value differences have magnitudes $2$, $3$, and $5$, so\n\\[\n\\|j-\\widehat j\\|_1=2+3+5=10,\\qquad\n\\|j-\\widehat j\\|_2^2=4+9+25=38.\n\\]\nThe counting-measure supremum error is $5$. The full-domain mean squared error is $38/12=19/6$. A mismatch fraction of $3/12$ reports a different statistic and omits all three amplitude weights.\n\\head{The formula requires partitions}\nFor measurable partitions $(E_i)$ and $(F_j)$, with $M_{ij}=\\mu(E_i\\cap F_j)$,\n\\[\n\\|j-\\widehat j\\|_p^p=\\sum_{i,j}|a_i-b_j|^pM_{ij},\\qquad 1\\le p<\\infty.\n\\]\nEach location belongs to exactly one intersection. Overlapping indicator packets that are summed at the same location must first be converted to their actual combined value regions before this partition formula is applied.\n\\takeaway{Geometry supplies the …","e also $(4,3,5)$, although three samples are assigned to the wrong values.\n\\vfig[.80]{ext_overlap}\n\\head{Equal error counts need not mean equal reconstruction error}\nThe three off-diagonal entries contribute one sample each. Their value differences have magnitudes $2$, $3$, and $5$, so\n\\[\n\\|j-\\widehat j\\|_1=2+3+5=10,\\qquad\n\\|j-\\widehat j\\|_2^2=4+9+25=38.\n\\]\nThe counting-measure supremum error is $5$. The full-domain mean squared error is $38/12=19/6$. A mismatch fraction of $3/12$ reports a different statistic and omits all three amplitude weights.\n\\head{The formula requires partitions}\nFor measurable partitions $(E_i)$ and $(F_j)$, with $M_{ij}=\\mu(E_i\\cap F_j)$,\n\\[\n\\|j-\\widehat j\\|_p^p=\\sum_{i,j}|a_i-b_j|^pM_{ij},\\qquad 1\\le p<\\infty.\n\\]\nEach location belongs to exactly one intersection. Overlapping indicator packets that are summed at the same location must first be converted to their actual combined value regions before this partition formula is applied.\n\\takeaway{Geometry supplies the overlap counts. Signal values supply the weights. Their combination gives the norm relevant to …","e assigned to the wrong values.\n\\vfig[.80]{ext_overlap}\n\\head{Equal error counts need not mean equal reconstruction error}\nThe three off-diagonal entries contribute one sample each. Their value differences have magnitudes $2$, $3$, and $5$, so\n\\[\n\\|j-\\widehat j\\|_1=2+3+5=10,\\qquad\n\\|j-\\widehat j\\|_2^2=4+9+25=38.\n\\]\nThe counting-measure supremum error is $5$. The full-domain mean squared error is $38/12=19/6$. A mismatch fraction of $3/12$ reports a different statistic and omits all three amplitude weights.\n\\head{The formula requires partitions}\nFor measurable partitions $(E_i)$ and $(F_j)$, with $M_{ij}=\\mu(E_i\\cap F_j)$,\n\\[\n\\|j-\\widehat j\\|_p^p=\\sum_{i,j}|a_i-b_j|^pM_{ij},\\qquad 1\\le p<\\infty.\n\\]\nEach location belongs to exactly one intersection. Overlapping indicator packets that are summed at the same location must first be converted to their actual combined value regions before this partition formula is applied.\n\\takeaway{Geometry supplies the overlap counts. Signal values supply the weights. Their combination gives the norm relevant to reconstruction.}","[.80]{ext_overlap}\n\\head{Equal error counts need not mean equal reconstruction error}\nThe three off-diagonal entries contribute one sample each. Their value differences have magnitudes $2$, $3$, and $5$, so\n\\[\n\\|j-\\widehat j\\|_1=2+3+5=10,\\qquad\n\\|j-\\widehat j\\|_2^2=4+9+25=38.\n\\]\nThe counting-measure supremum error is $5$. The full-domain mean squared error is $38/12=19/6$. A mismatch fraction of $3/12$ reports a different statistic and omits all three amplitude weights.\n\\head{The formula requires partitions}\nFor measurable partitions $(E_i)$ and $(F_j)$, with $M_{ij}=\\mu(E_i\\cap F_j)$,\n\\[\n\\|j-\\widehat j\\|_p^p=\\sum_{i,j}|a_i-b_j|^pM_{ij},\\qquad 1\\le p<\\infty.\n\\]\nEach location belongs to exactly one intersection. Overlapping indicator packets that are summed at the same location must first be converted to their actual combined value regions before this partition formula is applied.\n\\takeaway{Geometry supplies the overlap counts. Signal values supply the weights. Their combination gives the norm relevant to reconstruction.}","ch. Their value differences have magnitudes $2$, $3$, and $5$, so\n\\[\n\\|j-\\widehat j\\|_1=2+3+5=10,\\qquad\n\\|j-\\widehat j\\|_2^2=4+9+25=38.\n\\]\nThe counting-measure supremum error is $5$. The full-domain mean squared error is $38/12=19/6$. A mismatch fraction of $3/12$ reports a different statistic and omits all three amplitude weights.\n\\head{The formula requires partitions}\nFor measurable partitions $(E_i)$ and $(F_j)$, with $M_{ij}=\\mu(E_i\\cap F_j)$,\n\\[\n\\|j-\\widehat j\\|_p^p=\\sum_{i,j}|a_i-b_j|^pM_{ij},\\qquad 1\\le p<\\infty.\n\\]\nEach location belongs to exactly one intersection. Overlapping indicator packets that are summed at the same location must first be converted to their actual combined value regions before this partition formula is applied.\n\\takeaway{Geometry supplies the overlap counts. Signal values supply the weights. Their combination gives the norm relevant to reconstruction.}","lue differences have magnitudes $2$, $3$, and $5$, so\n\\[\n\\|j-\\widehat j\\|_1=2+3+5=10,\\qquad\n\\|j-\\widehat j\\|_2^2=4+9+25=38.\n\\]\nThe counting-measure supremum error is $5$. The full-domain mean squared error is $38/12=19/6$. A mismatch fraction of $3/12$ reports a different statistic and omits all three amplitude weights.\n\\head{The formula requires partitions}\nFor measurable partitions $(E_i)$ and $(F_j)$, with $M_{ij}=\\mu(E_i\\cap F_j)$,\n\\[\n\\|j-\\widehat j\\|_p^p=\\sum_{i,j}|a_i-b_j|^pM_{ij},\\qquad 1\\le p<\\infty.\n\\]\nEach location belongs to exactly one intersection. Overlapping indicator packets that are summed at the same location must first be converted to their actual combined value regions before this partition formula is applied.\n\\takeaway{Geometry supplies the overlap counts. Signal values supply the weights. Their combination gives the norm relevant to reconstruction.}","s have magnitudes $2$, $3$, and $5$, so\n\\[\n\\|j-\\widehat j\\|_1=2+3+5=10,\\qquad\n\\|j-\\widehat j\\|_2^2=4+9+25=38.\n\\]\nThe counting-measure supremum error is $5$. The full-domain mean squared error is $38/12=19/6$. A mismatch fraction of $3/12$ reports a different statistic and omits all three amplitude weights.\n\\head{The formula requires partitions}\nFor measurable partitions $(E_i)$ and $(F_j)$, with $M_{ij}=\\mu(E_i\\cap F_j)$,\n\\[\n\\|j-\\widehat j\\|_p^p=\\sum_{i,j}|a_i-b_j|^pM_{ij},\\qquad 1\\le p<\\infty.\n\\]\nEach location belongs to exactly one intersection. Overlapping indicator packets that are summed at the same location must first be converted to their actual combined value regions before this partition formula is applied.\n\\takeaway{Geometry supplies the overlap counts. Signal values supply the weights. Their combination gives the norm relevant to reconstruction.}","and $5$, so\n\\[\n\\|j-\\widehat j\\|_1=2+3+5=10,\\qquad\n\\|j-\\widehat j\\|_2^2=4+9+25=38.\n\\]\nThe counting-measure supremum error is $5$. The full-domain mean squared error is $38/12=19/6$. A mismatch fraction of $3/12$ reports a different statistic and omits all three amplitude weights.\n\\head{The formula requires partitions}\nFor measurable partitions $(E_i)$ and $(F_j)$, with $M_{ij}=\\mu(E_i\\cap F_j)$,\n\\[\n\\|j-\\widehat j\\|_p^p=\\sum_{i,j}|a_i-b_j|^pM_{ij},\\qquad 1\\le p<\\infty.\n\\]\nEach location belongs to exactly one intersection. Overlapping indicator packets that are summed at the same location must first be converted to their actual combined value regions before this partition formula is applied.\n\\takeaway{Geometry supplies the overlap counts. Signal values supply the weights. Their combination gives the norm relevant to reconstruction.}","\\vpage{34}{A shifted step and a shifted ramp have different moduli}{LOCATION STABILITY / A ONE-SIXTEENTH DISPLACEMENT}{Traditional monograph: Corollary E8.2 (p. 60), Theorem E8.3 (p. 60), Theorem E8.7 (p. 62).}\nSet $\\tau=1/4$, $\\widehat\\tau=5/16$, and $\\delta=1/16$. A unit step differs only on the interval between its two locations:\n\\[\n\\|H_\\tau-H_{\\widehat\\tau}\\|_2=\\sqrt\\delta=\\tfrac14,\n\\qquad \\|H_\\tau-H_{\\widehat\\tau}\\|_\\infty=1.\n\\]\n\\vfig[.89]{ext_location}\n\\head{For a ramp, the error height shrinks too}\nWrite $r_\\tau(t)=(t-\\tau)_+$. The difference grows linearly from zero to $\\delta$ on $[\\tau,\\widehat\\tau]$ and stays at $\\delta$ on $[\\widehat\\tau,1]$. Direct integration gives\n\\[\n\\|r_\\tau-r_{\\widehat\\tau}\\|_2^2\n=\\frac{\\delta^3}{3}+(1-\\widehat\\tau)\\delta^2\n=\\frac{17}{6144}<\\frac1{256}=\\delta^2.\n\\]\nThus the $L^2$ ramp error is at most $\\delta$, and its supremum error is exactly $\\delta$. Positive-order truncate","\\vpage{34}{A shifted step and a shifted ramp have different moduli}{LOCATION STABILITY / A ONE-SIXTEENTH DISPLACEMENT}{Traditional monograph: Corollary E8.2 (p. 60), Theorem E8.3 (p. 60), Theorem E8.7 (p. 62).}\nSet $\\tau=1/4$, $\\widehat\\tau=5/16$, and $\\delta=1/16$. A unit step differs only on the interval between its two locations:\n\\[\n\\|H_\\tau-H_{\\widehat\\tau}\\|_2=\\sqrt\\delta=\\tfrac14,\n\\qquad \\|H_\\tau-H_{\\widehat\\tau}\\|_\\infty=1.\n\\]\n\\vfig[.89]{ext_location}\n\\head{For a ramp, the error height shrinks too}\nWrite $r_\\tau(t)=(t-\\tau)_+$. The difference grows linearly from zero to $\\delta$ on $[\\tau,\\widehat\\tau]$ and stays at $\\delta$ on $[\\widehat\\tau,1]$. Direct integration gives\n\\[\n\\|r_\\tau-r_{\\widehat\\tau}\\|_2^2\n=\\frac{\\delta^3}{3}+(1-\\widehat\\tau)\\delta^2\n=\\frac{17}{6144}<\\frac1{256}=\\delta^2.\n\\]\nThus the $L^2$ ramp error is at most $\\delta$, and its supremum error is exactly $\\delta$. Positive-order truncated powers are Lipschit","\\vpage{34}{A shifted step and a shifted ramp have different moduli}{LOCATION STABILITY / A ONE-SIXTEENTH DISPLACEMENT}{Traditional monograph: Corollary E8.2 (p. 60), Theorem E8.3 (p. 60), Theorem E8.7 (p. 62).}\nSet $\\tau=1/4$, $\\widehat\\tau=5/16$, and $\\delta=1/16$. A unit step differs only on the interval between its two locations:\n\\[\n\\|H_\\tau-H_{\\widehat\\tau}\\|_2=\\sqrt\\delta=\\tfrac14,\n\\qquad \\|H_\\tau-H_{\\widehat\\tau}\\|_\\infty=1.\n\\]\n\\vfig[.89]{ext_location}\n\\head{For a ramp, the error height shrinks too}\nWrite $r_\\tau(t)=(t-\\tau)_+$. The difference grows linearly from zero to $\\delta$ on $[\\tau,\\widehat\\tau]$ and stays at $\\delta$ on $[\\widehat\\tau,1]$. Direct integration gives\n\\[\n\\|r_\\tau-r_{\\widehat\\tau}\\|_2^2\n=\\frac{\\delta^3}{3}+(1-\\widehat\\tau)\\delta^2\n=\\frac{17}{6144}<\\frac1{256}=\\delta^2.\n\\]\nThus the $L^2$ ramp error is at most $\\delta$, and its supremum error is exactly $\\delta$. Positive-order truncated powers are Lipschitz in location, with","\\vpage{34}{A shifted step and a shifted ramp have different moduli}{LOCATION STABILITY / A ONE-SIXTEENTH DISPLACEMENT}{Traditional monograph: Corollary E8.2 (p. 60), Theorem E8.3 (p. 60), Theorem E8.7 (p. 62).}\nSet $\\tau=1/4$, $\\widehat\\tau=5/16$, and $\\delta=1/16$. A unit step differs only on the interval between its two locations:\n\\[\n\\|H_\\tau-H_{\\widehat\\tau}\\|_2=\\sqrt\\delta=\\tfrac14,\n\\qquad \\|H_\\tau-H_{\\widehat\\tau}\\|_\\infty=1.\n\\]\n\\vfig[.89]{ext_location}\n\\head{For a ramp, the error height shrinks too}\nWrite $r_\\tau(t)=(t-\\tau)_+$. The difference grows linearly from zero to $\\delta$ on $[\\tau,\\widehat\\tau]$ and stays at $\\delta$ on $[\\widehat\\tau,1]$. Direct integration gives\n\\[\n\\|r_\\tau-r_{\\widehat\\tau}\\|_2^2\n=\\frac{\\delta^3}{3}+(1-\\widehat\\tau)\\delta^2\n=\\frac{17}{6144}<\\frac1{256}=\\delta^2.\n\\]\nThus the $L^2$ ramp error is at most $\\delta$, and its supremum error is exactly $\\delta$. Positive-order truncated powers are Lipschitz in location, with the order-dependent constants supplied by the theorem.\n\\head{The geometric analogue needs a tube hypothesis}\nIf $E\\triangle F\\subseteq …","{Traditional monograph: Corollary E8.2 (p. 60), Theorem E8.3 (p. 60), Theorem E8.7 (p. 62).}\nSet $\\tau=1/4$, $\\widehat\\tau=5/16$, and $\\delta=1/16$. A unit step differs only on the interval between its two locations:\n\\[\n\\|H_\\tau-H_{\\widehat\\tau}\\|_2=\\sqrt\\delta=\\tfrac14,\n\\qquad \\|H_\\tau-H_{\\widehat\\tau}\\|_\\infty=1.\n\\]\n\\vfig[.89]{ext_location}\n\\head{For a ramp, the error height shrinks too}\nWrite $r_\\tau(t)=(t-\\tau)_+$. The difference grows linearly from zero to $\\delta$ on $[\\tau,\\widehat\\tau]$ and stays at $\\delta$ on $[\\widehat\\tau,1]$. Direct integration gives\n\\[\n\\|r_\\tau-r_{\\widehat\\tau}\\|_2^2\n=\\frac{\\delta^3}{3}+(1-\\widehat\\tau)\\delta^2\n=\\frac{17}{6144}<\\frac1{256}=\\delta^2.\n\\]\nThus the $L^2$ ramp error is at most $\\delta$, and its supremum error is exactly $\\delta$. Positive-order truncated powers are Lipschitz in location, with the order-dependent constants supplied by the theorem.\n\\head{The geometric analogue needs a tube hypothesis}\nIf $E\\triangle F\\subseteq N_\\delta(\\partial E)$ and the tube measure is bounded by $C\\delta$ for the stated range $0<\\delta<\\delta_0$, then an e…","), Theorem E8.7 (p. 62).}\nSet $\\tau=1/4$, $\\widehat\\tau=5/16$, and $\\delta=1/16$. A unit step differs only on the interval between its two locations:\n\\[\n\\|H_\\tau-H_{\\widehat\\tau}\\|_2=\\sqrt\\delta=\\tfrac14,\n\\qquad \\|H_\\tau-H_{\\widehat\\tau}\\|_\\infty=1.\n\\]\n\\vfig[.89]{ext_location}\n\\head{For a ramp, the error height shrinks too}\nWrite $r_\\tau(t)=(t-\\tau)_+$. The difference grows linearly from zero to $\\delta$ on $[\\tau,\\widehat\\tau]$ and stays at $\\delta$ on $[\\widehat\\tau,1]$. Direct integration gives\n\\[\n\\|r_\\tau-r_{\\widehat\\tau}\\|_2^2\n=\\frac{\\delta^3}{3}+(1-\\widehat\\tau)\\delta^2\n=\\frac{17}{6144}<\\frac1{256}=\\delta^2.\n\\]\nThus the $L^2$ ramp error is at most $\\delta$, and its supremum error is exactly $\\delta$. Positive-order truncated powers are Lipschitz in location, with the order-dependent constants supplied by the theorem.\n\\head{The geometric analogue needs a tube hypothesis}\nIf $E\\triangle F\\subseteq N_\\delta(\\partial E)$ and the tube measure is bounded by $C\\delta$ for the stated range $0<\\delta<\\delta_0$, then an equal-amplitude packet has finite-$p$ error at most $|a|(C\\delta)^{1…","8.7 (p. 62).}\nSet $\\tau=1/4$, $\\widehat\\tau=5/16$, and $\\delta=1/16$. A unit step differs only on the interval between its two locations:\n\\[\n\\|H_\\tau-H_{\\widehat\\tau}\\|_2=\\sqrt\\delta=\\tfrac14,\n\\qquad \\|H_\\tau-H_{\\widehat\\tau}\\|_\\infty=1.\n\\]\n\\vfig[.89]{ext_location}\n\\head{For a ramp, the error height shrinks too}\nWrite $r_\\tau(t)=(t-\\tau)_+$. The difference grows linearly from zero to $\\delta$ on $[\\tau,\\widehat\\tau]$ and stays at $\\delta$ on $[\\widehat\\tau,1]$. Direct integration gives\n\\[\n\\|r_\\tau-r_{\\widehat\\tau}\\|_2^2\n=\\frac{\\delta^3}{3}+(1-\\widehat\\tau)\\delta^2\n=\\frac{17}{6144}<\\frac1{256}=\\delta^2.\n\\]\nThus the $L^2$ ramp error is at most $\\delta$, and its supremum error is exactly $\\delta$. Positive-order truncated powers are Lipschitz in location, with the order-dependent constants supplied by the theorem.\n\\head{The geometric analogue needs a tube hypothesis}\nIf $E\\triangle F\\subseteq N_\\delta(\\partial E)$ and the tube measure is bounded by $C\\delta$ for the stated range $0<\\delta<\\delta_0$, then an equal-amplitude packet has finite-$p$ error at most $|a|(C\\delta)^{1/p}$. A mere…","ehat\\tau=5/16$, and $\\delta=1/16$. A unit step differs only on the interval between its two locations:\n\\[\n\\|H_\\tau-H_{\\widehat\\tau}\\|_2=\\sqrt\\delta=\\tfrac14,\n\\qquad \\|H_\\tau-H_{\\widehat\\tau}\\|_\\infty=1.\n\\]\n\\vfig[.89]{ext_location}\n\\head{For a ramp, the error height shrinks too}\nWrite $r_\\tau(t)=(t-\\tau)_+$. The difference grows linearly from zero to $\\delta$ on $[\\tau,\\widehat\\tau]$ and stays at $\\delta$ on $[\\widehat\\tau,1]$. Direct integration gives\n\\[\n\\|r_\\tau-r_{\\widehat\\tau}\\|_2^2\n=\\frac{\\delta^3}{3}+(1-\\widehat\\tau)\\delta^2\n=\\frac{17}{6144}<\\frac1{256}=\\delta^2.\n\\]\nThus the $L^2$ ramp error is at most $\\delta$, and its supremum error is exactly $\\delta$. Positive-order truncated powers are Lipschitz in location, with the order-dependent constants supplied by the theorem.\n\\head{The geometric analogue needs a tube hypothesis}\nIf $E\\triangle F\\subseteq N_\\delta(\\partial E)$ and the tube measure is bounded by $C\\delta$ for the stated range $0<\\delta<\\delta_0$, then an equal-amplitude packet has finite-$p$ error at most $|a|(C\\delta)^{1/p}$. A mere visual impression that two boundar…","6$, and $\\delta=1/16$. A unit step differs only on the interval between its two locations:\n\\[\n\\|H_\\tau-H_{\\widehat\\tau}\\|_2=\\sqrt\\delta=\\tfrac14,\n\\qquad \\|H_\\tau-H_{\\widehat\\tau}\\|_\\infty=1.\n\\]\n\\vfig[.89]{ext_location}\n\\head{For a ramp, the error height shrinks too}\nWrite $r_\\tau(t)=(t-\\tau)_+$. The difference grows linearly from zero to $\\delta$ on $[\\tau,\\widehat\\tau]$ and stays at $\\delta$ on $[\\widehat\\tau,1]$. Direct integration gives\n\\[\n\\|r_\\tau-r_{\\widehat\\tau}\\|_2^2\n=\\frac{\\delta^3}{3}+(1-\\widehat\\tau)\\delta^2\n=\\frac{17}{6144}<\\frac1{256}=\\delta^2.\n\\]\nThus the $L^2$ ramp error is at most $\\delta$, and its supremum error is exactly $\\delta$. Positive-order truncated powers are Lipschitz in location, with the order-dependent constants supplied by the theorem.\n\\head{The geometric analogue needs a tube hypothesis}\nIf $E\\triangle F\\subseteq N_\\delta(\\partial E)$ and the tube measure is bounded by $C\\delta$ for the stated range $0<\\delta<\\delta_0$, then an equal-amplitude packet has finite-$p$ error at most $|a|(C\\delta)^{1/p}$. A mere visual impression that two boundaries are clos…","y on the interval between its two locations:\n\\[\n\\|H_\\tau-H_{\\widehat\\tau}\\|_2=\\sqrt\\delta=\\tfrac14,\n\\qquad \\|H_\\tau-H_{\\widehat\\tau}\\|_\\infty=1.\n\\]\n\\vfig[.89]{ext_location}\n\\head{For a ramp, the error height shrinks too}\nWrite $r_\\tau(t)=(t-\\tau)_+$. The difference grows linearly from zero to $\\delta$ on $[\\tau,\\widehat\\tau]$ and stays at $\\delta$ on $[\\widehat\\tau,1]$. Direct integration gives\n\\[\n\\|r_\\tau-r_{\\widehat\\tau}\\|_2^2\n=\\frac{\\delta^3}{3}+(1-\\widehat\\tau)\\delta^2\n=\\frac{17}{6144}<\\frac1{256}=\\delta^2.\n\\]\nThus the $L^2$ ramp error is at most $\\delta$, and its supremum error is exactly $\\delta$. Positive-order truncated powers are Lipschitz in location, with the order-dependent constants supplied by the theorem.\n\\head{The geometric analogue needs a tube hypothesis}\nIf $E\\triangle F\\subseteq N_\\delta(\\partial E)$ and the tube measure is bounded by $C\\delta$ for the stated range $0<\\delta<\\delta_0$, then an equal-amplitude packet has finite-$p$ error at most $|a|(C\\delta)^{1/p}$. A mere visual impression that two boundaries are close does not establish the inclusion or the tube…","u}\\|_\\infty=1.\n\\]\n\\vfig[.89]{ext_location}\n\\head{For a ramp, the error height shrinks too}\nWrite $r_\\tau(t)=(t-\\tau)_+$. The difference grows linearly from zero to $\\delta$ on $[\\tau,\\widehat\\tau]$ and stays at $\\delta$ on $[\\widehat\\tau,1]$. Direct integration gives\n\\[\n\\|r_\\tau-r_{\\widehat\\tau}\\|_2^2\n=\\frac{\\delta^3}{3}+(1-\\widehat\\tau)\\delta^2\n=\\frac{17}{6144}<\\frac1{256}=\\delta^2.\n\\]\nThus the $L^2$ ramp error is at most $\\delta$, and its supremum error is exactly $\\delta$. Positive-order truncated powers are Lipschitz in location, with the order-dependent constants supplied by the theorem.\n\\head{The geometric analogue needs a tube hypothesis}\nIf $E\\triangle F\\subseteq N_\\delta(\\partial E)$ and the tube measure is bounded by $C\\delta$ for the stated range $0<\\delta<\\delta_0$, then an equal-amplitude packet has finite-$p$ error at most $|a|(C\\delta)^{1/p}$. A mere visual impression that two boundaries are close does not establish the inclusion or the tube estimate.\n\\takeaway{A displaced value jump keeps its full height on a smaller set. A displaced derivative-jump packet also decrea…","{ext_location}\n\\head{For a ramp, the error height shrinks too}\nWrite $r_\\tau(t)=(t-\\tau)_+$. The difference grows linearly from zero to $\\delta$ on $[\\tau,\\widehat\\tau]$ and stays at $\\delta$ on $[\\widehat\\tau,1]$. Direct integration gives\n\\[\n\\|r_\\tau-r_{\\widehat\\tau}\\|_2^2\n=\\frac{\\delta^3}{3}+(1-\\widehat\\tau)\\delta^2\n=\\frac{17}{6144}<\\frac1{256}=\\delta^2.\n\\]\nThus the $L^2$ ramp error is at most $\\delta$, and its supremum error is exactly $\\delta$. Positive-order truncated powers are Lipschitz in location, with the order-dependent constants supplied by the theorem.\n\\head{The geometric analogue needs a tube hypothesis}\nIf $E\\triangle F\\subseteq N_\\delta(\\partial E)$ and the tube measure is bounded by $C\\delta$ for the stated range $0<\\delta<\\delta_0$, then an equal-amplitude packet has finite-$p$ error at most $|a|(C\\delta)^{1/p}$. A mere visual impression that two boundaries are close does not establish the inclusion or the tube estimate.\n\\takeaway{A displaced value jump keeps its full height on a smaller set. A displaced derivative-jump packet also decreases in height.}","eight shrinks too}\nWrite $r_\\tau(t)=(t-\\tau)_+$. The difference grows linearly from zero to $\\delta$ on $[\\tau,\\widehat\\tau]$ and stays at $\\delta$ on $[\\widehat\\tau,1]$. Direct integration gives\n\\[\n\\|r_\\tau-r_{\\widehat\\tau}\\|_2^2\n=\\frac{\\delta^3}{3}+(1-\\widehat\\tau)\\delta^2\n=\\frac{17}{6144}<\\frac1{256}=\\delta^2.\n\\]\nThus the $L^2$ ramp error is at most $\\delta$, and its supremum error is exactly $\\delta$. Positive-order truncated powers are Lipschitz in location, with the order-dependent constants supplied by the theorem.\n\\head{The geometric analogue needs a tube hypothesis}\nIf $E\\triangle F\\subseteq N_\\delta(\\partial E)$ and the tube measure is bounded by $C\\delta$ for the stated range $0<\\delta<\\delta_0$, then an equal-amplitude packet has finite-$p$ error at most $|a|(C\\delta)^{1/p}$. A mere visual impression that two boundaries are close does not establish the inclusion or the tube estimate.\n\\takeaway{A displaced value jump keeps its full height on a smaller set. A displaced derivative-jump packet also decreases in height.}","ion gives\n\\[\n\\|r_\\tau-r_{\\widehat\\tau}\\|_2^2\n=\\frac{\\delta^3}{3}+(1-\\widehat\\tau)\\delta^2\n=\\frac{17}{6144}<\\frac1{256}=\\delta^2.\n\\]\nThus the $L^2$ ramp error is at most $\\delta$, and its supremum error is exactly $\\delta$. Positive-order truncated powers are Lipschitz in location, with the order-dependent constants supplied by the theorem.\n\\head{The geometric analogue needs a tube hypothesis}\nIf $E\\triangle F\\subseteq N_\\delta(\\partial E)$ and the tube measure is bounded by $C\\delta$ for the stated range $0<\\delta<\\delta_0$, then an equal-amplitude packet has finite-$p$ error at most $|a|(C\\delta)^{1/p}$. A mere visual impression that two boundaries are close does not establish the inclusion or the tube estimate.\n\\takeaway{A displaced value jump keeps its full height on a smaller set. A displaced derivative-jump packet also decreases in height.}",")\\delta^2\n=\\frac{17}{6144}<\\frac1{256}=\\delta^2.\n\\]\nThus the $L^2$ ramp error is at most $\\delta$, and its supremum error is exactly $\\delta$. Positive-order truncated powers are Lipschitz in location, with the order-dependent constants supplied by the theorem.\n\\head{The geometric analogue needs a tube hypothesis}\nIf $E\\triangle F\\subseteq N_\\delta(\\partial E)$ and the tube measure is bounded by $C\\delta$ for the stated range $0<\\delta<\\delta_0$, then an equal-amplitude packet has finite-$p$ error at most $|a|(C\\delta)^{1/p}$. A mere visual impression that two boundaries are close does not establish the inclusion or the tube estimate.\n\\takeaway{A displaced value jump keeps its full height on a smaller set. A displaced derivative-jump packet also decreases in height.}","c1{256}=\\delta^2.\n\\]\nThus the $L^2$ ramp error is at most $\\delta$, and its supremum error is exactly $\\delta$. Positive-order truncated powers are Lipschitz in location, with the order-dependent constants supplied by the theorem.\n\\head{The geometric analogue needs a tube hypothesis}\nIf $E\\triangle F\\subseteq N_\\delta(\\partial E)$ and the tube measure is bounded by $C\\delta$ for the stated range $0<\\delta<\\delta_0$, then an equal-amplitude packet has finite-$p$ error at most $|a|(C\\delta)^{1/p}$. A mere visual impression that two boundaries are close does not establish the inclusion or the tube estimate.\n\\takeaway{A displaced value jump keeps its full height on a smaller set. A displaced derivative-jump packet also decreases in height.}","ta$, and its supremum error is exactly $\\delta$. Positive-order truncated powers are Lipschitz in location, with the order-dependent constants supplied by the theorem.\n\\head{The geometric analogue needs a tube hypothesis}\nIf $E\\triangle F\\subseteq N_\\delta(\\partial E)$ and the tube measure is bounded by $C\\delta$ for the stated range $0<\\delta<\\delta_0$, then an equal-amplitude packet has finite-$p$ error at most $|a|(C\\delta)^{1/p}$. A mere visual impression that two boundaries are close does not establish the inclusion or the tube estimate.\n\\takeaway{A displaced value jump keeps its full height on a smaller set. A displaced derivative-jump packet also decreases in height.}","mum error is exactly $\\delta$. Positive-order truncated powers are Lipschitz in location, with the order-dependent constants supplied by the theorem.\n\\head{The geometric analogue needs a tube hypothesis}\nIf $E\\triangle F\\subseteq N_\\delta(\\partial E)$ and the tube measure is bounded by $C\\delta$ for the stated range $0<\\delta<\\delta_0$, then an equal-amplitude packet has finite-$p$ error at most $|a|(C\\delta)^{1/p}$. A mere visual impression that two boundaries are close does not establish the inclusion or the tube estimate.\n\\takeaway{A displaced value jump keeps its full height on a smaller set. A displaced derivative-jump packet also decreases in height.}","\\vpage{35}{A strict contrast gap locates every separated jump cell}{DETERMINISTIC RECOVERY / A NUMERICAL CERTIFICATE}{Traditional monograph: Lemma E9.1 (p. 63), Theorem E9.2 (p. 64), Example E9.7 (p. 65).}\nAssume a real signal is a Lipschitz background plus jumps. No jump occurs at a sample point, and each grid cell contains at most one jump. With spacing $h$, background constant $M$, and sample noise bounded by $\\varepsilon$, write\n\\[\nb_h=Mh+2\\varepsilon.\n\\]\nA no-jump contrast has magnitude at most $b_h$. A jump contrast differs from its true amplitude by at most $b_h$.\n\\vfig[.89]{ext_threshold}\n\\head{Use the manuscript's exact numerical inputs}\nFor $h=1/256$, $M=1/2$, $\\varepsilon=1/500$, and $a_{\\min}=2/25$,\n\\[\nb_h=\\frac{381}{64000}<\\frac3{500},\\qquad\n2b_h<\\frac2{25}.\n\\]\nThe threshold $\\theta=1/25$ lies strictly between $b_h$ and $a_{\\min}-b_h$. Thresholding $|y_{k+1}-y_k|$ therefore identifies exactly the jump-containing cells. Their midpoints locate each jump within $1/512$, and each estimated amplitude differs by at most $381/64000$.\n\\hea","\\vpage{35}{A strict contrast gap locates every separated jump cell}{DETERMINISTIC RECOVERY / A NUMERICAL CERTIFICATE}{Traditional monograph: Lemma E9.1 (p. 63), Theorem E9.2 (p. 64), Example E9.7 (p. 65).}\nAssume a real signal is a Lipschitz background plus jumps. No jump occurs at a sample point, and each grid cell contains at most one jump. With spacing $h$, background constant $M$, and sample noise bounded by $\\varepsilon$, write\n\\[\nb_h=Mh+2\\varepsilon.\n\\]\nA no-jump contrast has magnitude at most $b_h$. A jump contrast differs from its true amplitude by at most $b_h$.\n\\vfig[.89]{ext_threshold}\n\\head{Use the manuscript's exact numerical inputs}\nFor $h=1/256$, $M=1/2$, $\\varepsilon=1/500$, and $a_{\\min}=2/25$,\n\\[\nb_h=\\frac{381}{64000}<\\frac3{500},\\qquad\n2b_h<\\frac2{25}.\n\\]\nThe threshold $\\theta=1/25$ lies strictly between $b_h$ and $a_{\\min}-b_h$. Thresholding $|y_{k+1}-y_k|$ therefore identifies exactly the jump-containing cells. Their midpoints locate each jump within $1/512$, and each estimated amplitude differs by at most $381/64000$.\n\\head{The recovered object is","ct contrast gap locates every separated jump cell}{DETERMINISTIC RECOVERY / A NUMERICAL CERTIFICATE}{Traditional monograph: Lemma E9.1 (p. 63), Theorem E9.2 (p. 64), Example E9.7 (p. 65).}\nAssume a real signal is a Lipschitz background plus jumps. No jump occurs at a sample point, and each grid cell contains at most one jump. With spacing $h$, background constant $M$, and sample noise bounded by $\\varepsilon$, write\n\\[\nb_h=Mh+2\\varepsilon.\n\\]\nA no-jump contrast has magnitude at most $b_h$. A jump contrast differs from its true amplitude by at most $b_h$.\n\\vfig[.89]{ext_threshold}\n\\head{Use the manuscript's exact numerical inputs}\nFor $h=1/256$, $M=1/2$, $\\varepsilon=1/500$, and $a_{\\min}=2/25$,\n\\[\nb_h=\\frac{381}{64000}<\\frac3{500},\\qquad\n2b_h<\\frac2{25}.\n\\]\nThe threshold $\\theta=1/25$ lies strictly between $b_h$ and $a_{\\min}-b_h$. Thresholding $|y_{k+1}-y_k|$ therefore identifies exactly the jump-containing cells. Their midpoints locate each jump within $1/512$, and each estimated amplitude differs by at most $381/64000$.\n\\head{The recovered object is a cell, then an estimate}\nThe t…","s every separated jump cell}{DETERMINISTIC RECOVERY / A NUMERICAL CERTIFICATE}{Traditional monograph: Lemma E9.1 (p. 63), Theorem E9.2 (p. 64), Example E9.7 (p. 65).}\nAssume a real signal is a Lipschitz background plus jumps. No jump occurs at a sample point, and each grid cell contains at most one jump. With spacing $h$, background constant $M$, and sample noise bounded by $\\varepsilon$, write\n\\[\nb_h=Mh+2\\varepsilon.\n\\]\nA no-jump contrast has magnitude at most $b_h$. A jump contrast differs from its true amplitude by at most $b_h$.\n\\vfig[.89]{ext_threshold}\n\\head{Use the manuscript's exact numerical inputs}\nFor $h=1/256$, $M=1/2$, $\\varepsilon=1/500$, and $a_{\\min}=2/25$,\n\\[\nb_h=\\frac{381}{64000}<\\frac3{500},\\qquad\n2b_h<\\frac2{25}.\n\\]\nThe threshold $\\theta=1/25$ lies strictly between $b_h$ and $a_{\\min}-b_h$. Thresholding $|y_{k+1}-y_k|$ therefore identifies exactly the jump-containing cells. Their midpoints locate each jump within $1/512$, and each estimated amplitude differs by at most $381/64000$.\n\\head{The recovered object is a cell, then an estimate}\nThe theorem does not claim …","ERTIFICATE}{Traditional monograph: Lemma E9.1 (p. 63), Theorem E9.2 (p. 64), Example E9.7 (p. 65).}\nAssume a real signal is a Lipschitz background plus jumps. No jump occurs at a sample point, and each grid cell contains at most one jump. With spacing $h$, background constant $M$, and sample noise bounded by $\\varepsilon$, write\n\\[\nb_h=Mh+2\\varepsilon.\n\\]\nA no-jump contrast has magnitude at most $b_h$. A jump contrast differs from its true amplitude by at most $b_h$.\n\\vfig[.89]{ext_threshold}\n\\head{Use the manuscript's exact numerical inputs}\nFor $h=1/256$, $M=1/2$, $\\varepsilon=1/500$, and $a_{\\min}=2/25$,\n\\[\nb_h=\\frac{381}{64000}<\\frac3{500},\\qquad\n2b_h<\\frac2{25}.\n\\]\nThe threshold $\\theta=1/25$ lies strictly between $b_h$ and $a_{\\min}-b_h$. Thresholding $|y_{k+1}-y_k|$ therefore identifies exactly the jump-containing cells. Their midpoints locate each jump within $1/512$, and each estimated amplitude differs by at most $381/64000$.\n\\head{The recovered object is a cell, then an estimate}\nThe theorem does not claim the exact real-valued breakpoint from a finite grid. It supplies th…","2 (p. 64), Example E9.7 (p. 65).}\nAssume a real signal is a Lipschitz background plus jumps. No jump occurs at a sample point, and each grid cell contains at most one jump. With spacing $h$, background constant $M$, and sample noise bounded by $\\varepsilon$, write\n\\[\nb_h=Mh+2\\varepsilon.\n\\]\nA no-jump contrast has magnitude at most $b_h$. A jump contrast differs from its true amplitude by at most $b_h$.\n\\vfig[.89]{ext_threshold}\n\\head{Use the manuscript's exact numerical inputs}\nFor $h=1/256$, $M=1/2$, $\\varepsilon=1/500$, and $a_{\\min}=2/25$,\n\\[\nb_h=\\frac{381}{64000}<\\frac3{500},\\qquad\n2b_h<\\frac2{25}.\n\\]\nThe threshold $\\theta=1/25$ lies strictly between $b_h$ and $a_{\\min}-b_h$. Thresholding $|y_{k+1}-y_k|$ therefore identifies exactly the jump-containing cells. Their midpoints locate each jump within $1/512$, and each estimated amplitude differs by at most $381/64000$.\n\\head{The recovered object is a cell, then an estimate}\nThe theorem does not claim the exact real-valued breakpoint from a finite grid. It supplies the correct cell and a controlled representative of that cell. For r…","mps. No jump occurs at a sample point, and each grid cell contains at most one jump. With spacing $h$, background constant $M$, and sample noise bounded by $\\varepsilon$, write\n\\[\nb_h=Mh+2\\varepsilon.\n\\]\nA no-jump contrast has magnitude at most $b_h$. A jump contrast differs from its true amplitude by at most $b_h$.\n\\vfig[.89]{ext_threshold}\n\\head{Use the manuscript's exact numerical inputs}\nFor $h=1/256$, $M=1/2$, $\\varepsilon=1/500$, and $a_{\\min}=2/25$,\n\\[\nb_h=\\frac{381}{64000}<\\frac3{500},\\qquad\n2b_h<\\frac2{25}.\n\\]\nThe threshold $\\theta=1/25$ lies strictly between $b_h$ and $a_{\\min}-b_h$. Thresholding $|y_{k+1}-y_k|$ therefore identifies exactly the jump-containing cells. Their midpoints locate each jump within $1/512$, and each estimated amplitude differs by at most $381/64000$.\n\\head{The recovered object is a cell, then an estimate}\nThe theorem does not claim the exact real-valued breakpoint from a finite grid. It supplies the correct cell and a controlled representative of that cell. For real jumps, the amplitude error is too small to reverse the sign. Two opposite jumps in o…","p occurs at a sample point, and each grid cell contains at most one jump. With spacing $h$, background constant $M$, and sample noise bounded by $\\varepsilon$, write\n\\[\nb_h=Mh+2\\varepsilon.\n\\]\nA no-jump contrast has magnitude at most $b_h$. A jump contrast differs from its true amplitude by at most $b_h$.\n\\vfig[.89]{ext_threshold}\n\\head{Use the manuscript's exact numerical inputs}\nFor $h=1/256$, $M=1/2$, $\\varepsilon=1/500$, and $a_{\\min}=2/25$,\n\\[\nb_h=\\frac{381}{64000}<\\frac3{500},\\qquad\n2b_h<\\frac2{25}.\n\\]\nThe threshold $\\theta=1/25$ lies strictly between $b_h$ and $a_{\\min}-b_h$. Thresholding $|y_{k+1}-y_k|$ therefore identifies exactly the jump-containing cells. Their midpoints locate each jump within $1/512$, and each estimated amplitude differs by at most $381/64000$.\n\\head{The recovered object is a cell, then an estimate}\nThe theorem does not claim the exact real-valued breakpoint from a finite grid. It supplies the correct cell and a controlled representative of that cell. For real jumps, the amplitude error is too small to reverse the sign. Two opposite jumps in one cell cou…","at a sample point, and each grid cell contains at most one jump. With spacing $h$, background constant $M$, and sample noise bounded by $\\varepsilon$, write\n\\[\nb_h=Mh+2\\varepsilon.\n\\]\nA no-jump contrast has magnitude at most $b_h$. A jump contrast differs from its true amplitude by at most $b_h$.\n\\vfig[.89]{ext_threshold}\n\\head{Use the manuscript's exact numerical inputs}\nFor $h=1/256$, $M=1/2$, $\\varepsilon=1/500$, and $a_{\\min}=2/25$,\n\\[\nb_h=\\frac{381}{64000}<\\frac3{500},\\qquad\n2b_h<\\frac2{25}.\n\\]\nThe threshold $\\theta=1/25$ lies strictly between $b_h$ and $a_{\\min}-b_h$. Thresholding $|y_{k+1}-y_k|$ therefore identifies exactly the jump-containing cells. Their midpoints locate each jump within $1/512$, and each estimated amplitude differs by at most $381/64000$.\n\\head{The recovered object is a cell, then an estimate}\nThe theorem does not claim the exact real-valued breakpoint from a finite grid. It supplies the correct cell and a controlled representative of that cell. For real jumps, the amplitude error is too small to reverse the sign. Two opposite jumps in one cell could cancel…","ch grid cell contains at most one jump. With spacing $h$, background constant $M$, and sample noise bounded by $\\varepsilon$, write\n\\[\nb_h=Mh+2\\varepsilon.\n\\]\nA no-jump contrast has magnitude at most $b_h$. A jump contrast differs from its true amplitude by at most $b_h$.\n\\vfig[.89]{ext_threshold}\n\\head{Use the manuscript's exact numerical inputs}\nFor $h=1/256$, $M=1/2$, $\\varepsilon=1/500$, and $a_{\\min}=2/25$,\n\\[\nb_h=\\frac{381}{64000}<\\frac3{500},\\qquad\n2b_h<\\frac2{25}.\n\\]\nThe threshold $\\theta=1/25$ lies strictly between $b_h$ and $a_{\\min}-b_h$. Thresholding $|y_{k+1}-y_k|$ therefore identifies exactly the jump-containing cells. Their midpoints locate each jump within $1/512$, and each estimated amplitude differs by at most $381/64000$.\n\\head{The recovered object is a cell, then an estimate}\nThe theorem does not claim the exact real-valued breakpoint from a finite grid. It supplies the correct cell and a controlled representative of that cell. For real jumps, the amplitude error is too small to reverse the sign. Two opposite jumps in one cell could cancel in the contrast, which i…","ins at most one jump. With spacing $h$, background constant $M$, and sample noise bounded by $\\varepsilon$, write\n\\[\nb_h=Mh+2\\varepsilon.\n\\]\nA no-jump contrast has magnitude at most $b_h$. A jump contrast differs from its true amplitude by at most $b_h$.\n\\vfig[.89]{ext_threshold}\n\\head{Use the manuscript's exact numerical inputs}\nFor $h=1/256$, $M=1/2$, $\\varepsilon=1/500$, and $a_{\\min}=2/25$,\n\\[\nb_h=\\frac{381}{64000}<\\frac3{500},\\qquad\n2b_h<\\frac2{25}.\n\\]\nThe threshold $\\theta=1/25$ lies strictly between $b_h$ and $a_{\\min}-b_h$. Thresholding $|y_{k+1}-y_k|$ therefore identifies exactly the jump-containing cells. Their midpoints locate each jump within $1/512$, and each estimated amplitude differs by at most $381/64000$.\n\\head{The recovered object is a cell, then an estimate}\nThe theorem does not claim the exact real-valued breakpoint from a finite grid. It supplies the correct cell and a controlled representative of that cell. For real jumps, the amplitude error is too small to reverse the sign. Two opposite jumps in one cell could cancel in the contrast, which is why the separati…","oise bounded by $\\varepsilon$, write\n\\[\nb_h=Mh+2\\varepsilon.\n\\]\nA no-jump contrast has magnitude at most $b_h$. A jump contrast differs from its true amplitude by at most $b_h$.\n\\vfig[.89]{ext_threshold}\n\\head{Use the manuscript's exact numerical inputs}\nFor $h=1/256$, $M=1/2$, $\\varepsilon=1/500$, and $a_{\\min}=2/25$,\n\\[\nb_h=\\frac{381}{64000}<\\frac3{500},\\qquad\n2b_h<\\frac2{25}.\n\\]\nThe threshold $\\theta=1/25$ lies strictly between $b_h$ and $a_{\\min}-b_h$. Thresholding $|y_{k+1}-y_k|$ therefore identifies exactly the jump-containing cells. Their midpoints locate each jump within $1/512$, and each estimated amplitude differs by at most $381/64000$.\n\\head{The recovered object is a cell, then an estimate}\nThe theorem does not claim the exact real-valued breakpoint from a finite grid. It supplies the correct cell and a controlled representative of that cell. For real jumps, the amplitude error is too small to reverse the sign. Two opposite jumps in one cell could cancel in the contrast, which is why the separation assumption cannot be dropped.\nFor $Q$ jumps the finite-$p$ packet bound co…","\\[\nb_h=Mh+2\\varepsilon.\n\\]\nA no-jump contrast has magnitude at most $b_h$. A jump contrast differs from its true amplitude by at most $b_h$.\n\\vfig[.89]{ext_threshold}\n\\head{Use the manuscript's exact numerical inputs}\nFor $h=1/256$, $M=1/2$, $\\varepsilon=1/500$, and $a_{\\min}=2/25$,\n\\[\nb_h=\\frac{381}{64000}<\\frac3{500},\\qquad\n2b_h<\\frac2{25}.\n\\]\nThe threshold $\\theta=1/25$ lies strictly between $b_h$ and $a_{\\min}-b_h$. Thresholding $|y_{k+1}-y_k|$ therefore identifies exactly the jump-containing cells. Their midpoints locate each jump within $1/512$, and each estimated amplitude differs by at most $381/64000$.\n\\head{The recovered object is a cell, then an estimate}\nThe theorem does not claim the exact real-valued breakpoint from a finite grid. It supplies the correct cell and a controlled representative of that cell. For real jumps, the amplitude error is too small to reverse the sign. Two opposite jumps in one cell could cancel in the contrast, which is why the separation assumption cannot be dropped.\nFor $Q$ jumps the finite-$p$ packet bound contains both $Qb_h$ and the location t…","+2\\varepsilon.\n\\]\nA no-jump contrast has magnitude at most $b_h$. A jump contrast differs from its true amplitude by at most $b_h$.\n\\vfig[.89]{ext_threshold}\n\\head{Use the manuscript's exact numerical inputs}\nFor $h=1/256$, $M=1/2$, $\\varepsilon=1/500$, and $a_{\\min}=2/25$,\n\\[\nb_h=\\frac{381}{64000}<\\frac3{500},\\qquad\n2b_h<\\frac2{25}.\n\\]\nThe threshold $\\theta=1/25$ lies strictly between $b_h$ and $a_{\\min}-b_h$. Thresholding $|y_{k+1}-y_k|$ therefore identifies exactly the jump-containing cells. Their midpoints locate each jump within $1/512$, and each estimated amplitude differs by at most $381/64000$.\n\\head{The recovered object is a cell, then an estimate}\nThe theorem does not claim the exact real-valued breakpoint from a finite grid. It supplies the correct cell and a controlled representative of that cell. For real jumps, the amplitude error is too small to reverse the sign. Two opposite jumps in one cell could cancel in the contrast, which is why the separation assumption cannot be dropped.\nFor $Q$ jumps the finite-$p$ packet bound contains both $Qb_h$ and the location term $(\\tf…","ontrast has magnitude at most $b_h$. A jump contrast differs from its true amplitude by at most $b_h$.\n\\vfig[.89]{ext_threshold}\n\\head{Use the manuscript's exact numerical inputs}\nFor $h=1/256$, $M=1/2$, $\\varepsilon=1/500$, and $a_{\\min}=2/25$,\n\\[\nb_h=\\frac{381}{64000}<\\frac3{500},\\qquad\n2b_h<\\frac2{25}.\n\\]\nThe threshold $\\theta=1/25$ lies strictly between $b_h$ and $a_{\\min}-b_h$. Thresholding $|y_{k+1}-y_k|$ therefore identifies exactly the jump-containing cells. Their midpoints locate each jump within $1/512$, and each estimated amplitude differs by at most $381/64000$.\n\\head{The recovered object is a cell, then an estimate}\nThe theorem does not claim the exact real-valued breakpoint from a finite grid. It supplies the correct cell and a controlled representative of that cell. For real jumps, the amplitude error is too small to reverse the sign. Two opposite jumps in one cell could cancel in the contrast, which is why the separation assumption cannot be dropped.\nFor $Q$ jumps the finite-$p$ packet bound contains both $Qb_h$ and the location term $(\\tfrac h2\\sum_q|a_q|^p)^{1/p}$. …","]{ext_threshold}\n\\head{Use the manuscript's exact numerical inputs}\nFor $h=1/256$, $M=1/2$, $\\varepsilon=1/500$, and $a_{\\min}=2/25$,\n\\[\nb_h=\\frac{381}{64000}<\\frac3{500},\\qquad\n2b_h<\\frac2{25}.\n\\]\nThe threshold $\\theta=1/25$ lies strictly between $b_h$ and $a_{\\min}-b_h$. Thresholding $|y_{k+1}-y_k|$ therefore identifies exactly the jump-containing cells. Their midpoints locate each jump within $1/512$, and each estimated amplitude differs by at most $381/64000$.\n\\head{The recovered object is a cell, then an estimate}\nThe theorem does not claim the exact real-valued breakpoint from a finite grid. It supplies the correct cell and a controlled representative of that cell. For real jumps, the amplitude error is too small to reverse the sign. Two opposite jumps in one cell could cancel in the contrast, which is why the separation assumption cannot be dropped.\nFor $Q$ jumps the finite-$p$ packet bound contains both $Qb_h$ and the location term $(\\tfrac h2\\sum_q|a_q|^p)^{1/p}$. A growing number of jumps or shrinking separation requires its own analysis.\n\\takeaway{Exact detection means exa…","al inputs}\nFor $h=1/256$, $M=1/2$, $\\varepsilon=1/500$, and $a_{\\min}=2/25$,\n\\[\nb_h=\\frac{381}{64000}<\\frac3{500},\\qquad\n2b_h<\\frac2{25}.\n\\]\nThe threshold $\\theta=1/25$ lies strictly between $b_h$ and $a_{\\min}-b_h$. Thresholding $|y_{k+1}-y_k|$ therefore identifies exactly the jump-containing cells. Their midpoints locate each jump within $1/512$, and each estimated amplitude differs by at most $381/64000$.\n\\head{The recovered object is a cell, then an estimate}\nThe theorem does not claim the exact real-valued breakpoint from a finite grid. It supplies the correct cell and a controlled representative of that cell. For real jumps, the amplitude error is too small to reverse the sign. Two opposite jumps in one cell could cancel in the contrast, which is why the separation assumption cannot be dropped.\nFor $Q$ jumps the finite-$p$ packet bound contains both $Qb_h$ and the location term $(\\tfrac h2\\sum_q|a_q|^p)^{1/p}$. A growing number of jumps or shrinking separation requires its own analysis.\n\\takeaway{Exact detection means exact cell identification under a strict margin. Location an…","{The recovered object is a cell, then an estimate}\nThe theorem does not claim the exact real-valued breakpoint from a finite grid. It supplies the correct cell and a controlled representative of that cell. For real jumps, the amplitude error is too small to reverse the sign. Two opposite jumps in one cell could cancel in the contrast, which is why the separation assumption cannot be dropped.\nFor $Q$ jumps the finite-$p$ packet bound contains both $Qb_h$ and the location term $(\\tfrac h2\\sum_q|a_q|^p)^{1/p}$. A growing number of jumps or shrinking separation requires its own analysis.\n\\takeaway{Exact detection means exact cell identification under a strict margin. Location and amplitude remain quantitative estimates with explicit errors.}","is a cell, then an estimate}\nThe theorem does not claim the exact real-valued breakpoint from a finite grid. It supplies the correct cell and a controlled representative of that cell. For real jumps, the amplitude error is too small to reverse the sign. Two opposite jumps in one cell could cancel in the contrast, which is why the separation assumption cannot be dropped.\nFor $Q$ jumps the finite-$p$ packet bound contains both $Qb_h$ and the location term $(\\tfrac h2\\sum_q|a_q|^p)^{1/p}$. A growing number of jumps or shrinking separation requires its own analysis.\n\\takeaway{Exact detection means exact cell identification under a strict margin. Location and amplitude remain quantitative estimates with explicit errors.}","he theorem does not claim the exact real-valued breakpoint from a finite grid. It supplies the correct cell and a controlled representative of that cell. For real jumps, the amplitude error is too small to reverse the sign. Two opposite jumps in one cell could cancel in the contrast, which is why the separation assumption cannot be dropped.\nFor $Q$ jumps the finite-$p$ packet bound contains both $Qb_h$ and the location term $(\\tfrac h2\\sum_q|a_q|^p)^{1/p}$. A growing number of jumps or shrinking separation requires its own analysis.\n\\takeaway{Exact detection means exact cell identification under a strict margin. Location and amplitude remain quantitative estimates with explicit errors.}","exact real-valued breakpoint from a finite grid. It supplies the correct cell and a controlled representative of that cell. For real jumps, the amplitude error is too small to reverse the sign. Two opposite jumps in one cell could cancel in the contrast, which is why the separation assumption cannot be dropped.\nFor $Q$ jumps the finite-$p$ packet bound contains both $Qb_h$ and the location term $(\\tfrac h2\\sum_q|a_q|^p)^{1/p}$. A growing number of jumps or shrinking separation requires its own analysis.\n\\takeaway{Exact detection means exact cell identification under a strict margin. Location and amplitude remain quantitative estimates with explicit errors.}","\\vpage{36}{Repeated observations shrink the noise side of the gap}{SUBGAUSSIAN RECOVERY / A CONSERVATIVE SAMPLE BUDGET}{Traditional monograph: Theorem E9.8 (p. 67), Corollary E9.9 (p. 67), Corollary E9.11 (p. 67).}\nSuppose all noise observations are independent and centered, with moment-generating bound $\\mathbb E e^{\\lambda\\eta}\\le e^{\\sigma^2\\lambda^2/2}$. Averaging $R$ observations at each of $n+1$ sample sites replaces $\\sigma$ by $\\sigma/\\sqrt R$. A sufficient recovery condition is\n\\[\na_{\\min}>2Mh+\\frac{4\\sigma}{\\sqrt R}\\sqrt{\\log(2n/\\delta)}.\n\\]\nThe probability of simultaneous recovery is then at least $1-\\delta$, subject to the same separated-jump sampling model.\n\\vfig[.86]{ext_repetition}\n\\head{One explicit conservative choice}\nTake $n=256$, $M=1/2$, $\\sigma=1/50$, $\\delta=1/20$, and $a_{\\min}=2/25$. Since $\\log(10240)<10<(16/5)^2$, choosing $R=16$ makes the right-hand side strictly less than\n\\[\n\\frac1{256}+\\frac1{50}\\frac{16}{5}\n=\\frac{2173}{32000}<\\frac2{25}.\n\\]\nThus sixteen independent observations per sample site suffice for this","\\vpage{36}{Repeated observations shrink the noise side of the gap}{SUBGAUSSIAN RECOVERY / A CONSERVATIVE SAMPLE BUDGET}{Traditional monograph: Theorem E9.8 (p. 67), Corollary E9.9 (p. 67), Corollary E9.11 (p. 67).}\nSuppose all noise observations are independent and centered, with moment-generating bound $\\mathbb E e^{\\lambda\\eta}\\le e^{\\sigma^2\\lambda^2/2}$. Averaging $R$ observations at each of $n+1$ sample sites replaces $\\sigma$ by $\\sigma/\\sqrt R$. A sufficient recovery condition is\n\\[\na_{\\min}>2Mh+\\frac{4\\sigma}{\\sqrt R}\\sqrt{\\log(2n/\\delta)}.\n\\]\nThe probability of simultaneous recovery is then at least $1-\\delta$, subject to the same separated-jump sampling model.\n\\vfig[.86]{ext_repetition}\n\\head{One explicit conservative choice}\nTake $n=256$, $M=1/2$, $\\sigma=1/50$, $\\delta=1/20$, and $a_{\\min}=2/25$. Since $\\log(10240)<10<(16/5)^2$, choosing $R=16$ makes the right-hand side strictly less than\n\\[\n\\frac1{256}+\\frac1{50}\\frac{16}{5}\n=\\frac{2173}{32000}<\\frac2{25}.\n\\]\nThus sixteen independent observations per sample site suffice for this bound. It is a","\\vpage{36}{Repeated observations shrink the noise side of the gap}{SUBGAUSSIAN RECOVERY / A CONSERVATIVE SAMPLE BUDGET}{Traditional monograph: Theorem E9.8 (p. 67), Corollary E9.9 (p. 67), Corollary E9.11 (p. 67).}\nSuppose all noise observations are independent and centered, with moment-generating bound $\\mathbb E e^{\\lambda\\eta}\\le e^{\\sigma^2\\lambda^2/2}$. Averaging $R$ observations at each of $n+1$ sample sites replaces $\\sigma$ by $\\sigma/\\sqrt R$. A sufficient recovery condition is\n\\[\na_{\\min}>2Mh+\\frac{4\\sigma}{\\sqrt R}\\sqrt{\\log(2n/\\delta)}.\n\\]\nThe probability of simultaneous recovery is then at least $1-\\delta$, subject to the same separated-jump sampling model.\n\\vfig[.86]{ext_repetition}\n\\head{One explicit conservative choice}\nTake $n=256$, $M=1/2$, $\\sigma=1/50$, $\\delta=1/20$, and $a_{\\min}=2/25$. Since $\\log(10240)<10<(16/5)^2$, choosing $R=16$ makes the right-hand side strictly less than\n\\[\n\\frac1{256}+\\frac1{50}\\frac{16}{5}\n=\\frac{2173}{32000}<\\frac2{25}.\n\\]\nThus sixteen independent observations per sample site suffice for this bound. It is a sufficient budget, not a cl…","ions shrink the noise side of the gap}{SUBGAUSSIAN RECOVERY / A CONSERVATIVE SAMPLE BUDGET}{Traditional monograph: Theorem E9.8 (p. 67), Corollary E9.9 (p. 67), Corollary E9.11 (p. 67).}\nSuppose all noise observations are independent and centered, with moment-generating bound $\\mathbb E e^{\\lambda\\eta}\\le e^{\\sigma^2\\lambda^2/2}$. Averaging $R$ observations at each of $n+1$ sample sites replaces $\\sigma$ by $\\sigma/\\sqrt R$. A sufficient recovery condition is\n\\[\na_{\\min}>2Mh+\\frac{4\\sigma}{\\sqrt R}\\sqrt{\\log(2n/\\delta)}.\n\\]\nThe probability of simultaneous recovery is then at least $1-\\delta$, subject to the same separated-jump sampling model.\n\\vfig[.86]{ext_repetition}\n\\head{One explicit conservative choice}\nTake $n=256$, $M=1/2$, $\\sigma=1/50$, $\\delta=1/20$, and $a_{\\min}=2/25$. Since $\\log(10240)<10<(16/5)^2$, choosing $R=16$ makes the right-hand side strictly less than\n\\[\n\\frac1{256}+\\frac1{50}\\frac{16}{5}\n=\\frac{2173}{32000}<\\frac2{25}.\n\\]\nThus sixteen independent observations per sample site suffice for this bound. It is a sufficient budget, not a claim of optimal sample comple…","the noise side of the gap}{SUBGAUSSIAN RECOVERY / A CONSERVATIVE SAMPLE BUDGET}{Traditional monograph: Theorem E9.8 (p. 67), Corollary E9.9 (p. 67), Corollary E9.11 (p. 67).}\nSuppose all noise observations are independent and centered, with moment-generating bound $\\mathbb E e^{\\lambda\\eta}\\le e^{\\sigma^2\\lambda^2/2}$. Averaging $R$ observations at each of $n+1$ sample sites replaces $\\sigma$ by $\\sigma/\\sqrt R$. A sufficient recovery condition is\n\\[\na_{\\min}>2Mh+\\frac{4\\sigma}{\\sqrt R}\\sqrt{\\log(2n/\\delta)}.\n\\]\nThe probability of simultaneous recovery is then at least $1-\\delta$, subject to the same separated-jump sampling model.\n\\vfig[.86]{ext_repetition}\n\\head{One explicit conservative choice}\nTake $n=256$, $M=1/2$, $\\sigma=1/50$, $\\delta=1/20$, and $a_{\\min}=2/25$. Since $\\log(10240)<10<(16/5)^2$, choosing $R=16$ makes the right-hand side strictly less than\n\\[\n\\frac1{256}+\\frac1{50}\\frac{16}{5}\n=\\frac{2173}{32000}<\\frac2{25}.\n\\]\nThus sixteen independent observations per sample site suffice for this bound. It is a sufficient budget, not a claim of optimal sample complexity. The ex…","NSERVATIVE SAMPLE BUDGET}{Traditional monograph: Theorem E9.8 (p. 67), Corollary E9.9 (p. 67), Corollary E9.11 (p. 67).}\nSuppose all noise observations are independent and centered, with moment-generating bound $\\mathbb E e^{\\lambda\\eta}\\le e^{\\sigma^2\\lambda^2/2}$. Averaging $R$ observations at each of $n+1$ sample sites replaces $\\sigma$ by $\\sigma/\\sqrt R$. A sufficient recovery condition is\n\\[\na_{\\min}>2Mh+\\frac{4\\sigma}{\\sqrt R}\\sqrt{\\log(2n/\\delta)}.\n\\]\nThe probability of simultaneous recovery is then at least $1-\\delta$, subject to the same separated-jump sampling model.\n\\vfig[.86]{ext_repetition}\n\\head{One explicit conservative choice}\nTake $n=256$, $M=1/2$, $\\sigma=1/50$, $\\delta=1/20$, and $a_{\\min}=2/25$. Since $\\log(10240)<10<(16/5)^2$, choosing $R=16$ makes the right-hand side strictly less than\n\\[\n\\frac1{256}+\\frac1{50}\\frac{16}{5}\n=\\frac{2173}{32000}<\\frac2{25}.\n\\]\nThus sixteen independent observations per sample site suffice for this bound. It is a sufficient budget, not a claim of optimal sample complexity. The exact checker certifies $e^{10}>10240$ using a rational …","ppose all noise observations are independent and centered, with moment-generating bound $\\mathbb E e^{\\lambda\\eta}\\le e^{\\sigma^2\\lambda^2/2}$. Averaging $R$ observations at each of $n+1$ sample sites replaces $\\sigma$ by $\\sigma/\\sqrt R$. A sufficient recovery condition is\n\\[\na_{\\min}>2Mh+\\frac{4\\sigma}{\\sqrt R}\\sqrt{\\log(2n/\\delta)}.\n\\]\nThe probability of simultaneous recovery is then at least $1-\\delta$, subject to the same separated-jump sampling model.\n\\vfig[.86]{ext_repetition}\n\\head{One explicit conservative choice}\nTake $n=256$, $M=1/2$, $\\sigma=1/50$, $\\delta=1/20$, and $a_{\\min}=2/25$. Since $\\log(10240)<10<(16/5)^2$, choosing $R=16$ makes the right-hand side strictly less than\n\\[\n\\frac1{256}+\\frac1{50}\\frac{16}{5}\n=\\frac{2173}{32000}<\\frac2{25}.\n\\]\nThus sixteen independent observations per sample site suffice for this bound. It is a sufficient budget, not a claim of optimal sample complexity. The exact checker certifies $e^{10}>10240$ using a rational Taylor partial sum, so the displayed logarithm bound is not based on rounded numerical logarithms.\n\\head{Dependence between…","a^2/2}$. Averaging $R$ observations at each of $n+1$ sample sites replaces $\\sigma$ by $\\sigma/\\sqrt R$. A sufficient recovery condition is\n\\[\na_{\\min}>2Mh+\\frac{4\\sigma}{\\sqrt R}\\sqrt{\\log(2n/\\delta)}.\n\\]\nThe probability of simultaneous recovery is then at least $1-\\delta$, subject to the same separated-jump sampling model.\n\\vfig[.86]{ext_repetition}\n\\head{One explicit conservative choice}\nTake $n=256$, $M=1/2$, $\\sigma=1/50$, $\\delta=1/20$, and $a_{\\min}=2/25$. Since $\\log(10240)<10<(16/5)^2$, choosing $R=16$ makes the right-hand side strictly less than\n\\[\n\\frac1{256}+\\frac1{50}\\frac{16}{5}\n=\\frac{2173}{32000}<\\frac2{25}.\n\\]\nThus sixteen independent observations per sample site suffice for this bound. It is a sufficient budget, not a claim of optimal sample complexity. The exact checker certifies $e^{10}>10240$ using a rational Taylor partial sum, so the displayed logarithm bound is not based on rounded numerical logarithms.\n\\head{Dependence between adjacent contrasts is allowed}\nNeighboring differences share one sample. The proof does not require those differences to be independen…","Averaging $R$ observations at each of $n+1$ sample sites replaces $\\sigma$ by $\\sigma/\\sqrt R$. A sufficient recovery condition is\n\\[\na_{\\min}>2Mh+\\frac{4\\sigma}{\\sqrt R}\\sqrt{\\log(2n/\\delta)}.\n\\]\nThe probability of simultaneous recovery is then at least $1-\\delta$, subject to the same separated-jump sampling model.\n\\vfig[.86]{ext_repetition}\n\\head{One explicit conservative choice}\nTake $n=256$, $M=1/2$, $\\sigma=1/50$, $\\delta=1/20$, and $a_{\\min}=2/25$. Since $\\log(10240)<10<(16/5)^2$, choosing $R=16$ makes the right-hand side strictly less than\n\\[\n\\frac1{256}+\\frac1{50}\\frac{16}{5}\n=\\frac{2173}{32000}<\\frac2{25}.\n\\]\nThus sixteen independent observations per sample site suffice for this bound. It is a sufficient budget, not a claim of optimal sample complexity. The exact checker certifies $e^{10}>10240$ using a rational Taylor partial sum, so the displayed logarithm bound is not based on rounded numerical logarithms.\n\\head{Dependence between adjacent contrasts is allowed}\nNeighboring differences share one sample. The proof does not require those differences to be independent: it bou…","$R$ observations at each of $n+1$ sample sites replaces $\\sigma$ by $\\sigma/\\sqrt R$. A sufficient recovery condition is\n\\[\na_{\\min}>2Mh+\\frac{4\\sigma}{\\sqrt R}\\sqrt{\\log(2n/\\delta)}.\n\\]\nThe probability of simultaneous recovery is then at least $1-\\delta$, subject to the same separated-jump sampling model.\n\\vfig[.86]{ext_repetition}\n\\head{One explicit conservative choice}\nTake $n=256$, $M=1/2$, $\\sigma=1/50$, $\\delta=1/20$, and $a_{\\min}=2/25$. Since $\\log(10240)<10<(16/5)^2$, choosing $R=16$ makes the right-hand side strictly less than\n\\[\n\\frac1{256}+\\frac1{50}\\frac{16}{5}\n=\\frac{2173}{32000}<\\frac2{25}.\n\\]\nThus sixteen independent observations per sample site suffice for this bound. It is a sufficient budget, not a claim of optimal sample complexity. The exact checker certifies $e^{10}>10240$ using a rational Taylor partial sum, so the displayed logarithm bound is not based on rounded numerical logarithms.\n\\head{Dependence between adjacent contrasts is allowed}\nNeighboring differences share one sample. The proof does not require those differences to be independent: it bounds each t…","ns at each of $n+1$ sample sites replaces $\\sigma$ by $\\sigma/\\sqrt R$. A sufficient recovery condition is\n\\[\na_{\\min}>2Mh+\\frac{4\\sigma}{\\sqrt R}\\sqrt{\\log(2n/\\delta)}.\n\\]\nThe probability of simultaneous recovery is then at least $1-\\delta$, subject to the same separated-jump sampling model.\n\\vfig[.86]{ext_repetition}\n\\head{One explicit conservative choice}\nTake $n=256$, $M=1/2$, $\\sigma=1/50$, $\\delta=1/20$, and $a_{\\min}=2/25$. Since $\\log(10240)<10<(16/5)^2$, choosing $R=16$ makes the right-hand side strictly less than\n\\[\n\\frac1{256}+\\frac1{50}\\frac{16}{5}\n=\\frac{2173}{32000}<\\frac2{25}.\n\\]\nThus sixteen independent observations per sample site suffice for this bound. It is a sufficient budget, not a claim of optimal sample complexity. The exact checker certifies $e^{10}>10240$ using a rational Taylor partial sum, so the displayed logarithm bound is not based on rounded numerical logarithms.\n\\head{Dependence between adjacent contrasts is allowed}\nNeighboring differences share one sample. The proof does not require those differences to be independent: it bounds each tail and applie…","sample sites replaces $\\sigma$ by $\\sigma/\\sqrt R$. A sufficient recovery condition is\n\\[\na_{\\min}>2Mh+\\frac{4\\sigma}{\\sqrt R}\\sqrt{\\log(2n/\\delta)}.\n\\]\nThe probability of simultaneous recovery is then at least $1-\\delta$, subject to the same separated-jump sampling model.\n\\vfig[.86]{ext_repetition}\n\\head{One explicit conservative choice}\nTake $n=256$, $M=1/2$, $\\sigma=1/50$, $\\delta=1/20$, and $a_{\\min}=2/25$. Since $\\log(10240)<10<(16/5)^2$, choosing $R=16$ makes the right-hand side strictly less than\n\\[\n\\frac1{256}+\\frac1{50}\\frac{16}{5}\n=\\frac{2173}{32000}<\\frac2{25}.\n\\]\nThus sixteen independent observations per sample site suffice for this bound. It is a sufficient budget, not a claim of optimal sample complexity. The exact checker certifies $e^{10}>10240$ using a rational Taylor partial sum, so the displayed logarithm bound is not based on rounded numerical logarithms.\n\\head{Dependence between adjacent contrasts is allowed}\nNeighboring differences share one sample. The proof does not require those differences to be independent: it bounds each tail and applies a union bound acro…","$\\sigma$ by $\\sigma/\\sqrt R$. A sufficient recovery condition is\n\\[\na_{\\min}>2Mh+\\frac{4\\sigma}{\\sqrt R}\\sqrt{\\log(2n/\\delta)}.\n\\]\nThe probability of simultaneous recovery is then at least $1-\\delta$, subject to the same separated-jump sampling model.\n\\vfig[.86]{ext_repetition}\n\\head{One explicit conservative choice}\nTake $n=256$, $M=1/2$, $\\sigma=1/50$, $\\delta=1/20$, and $a_{\\min}=2/25$. Since $\\log(10240)<10<(16/5)^2$, choosing $R=16$ makes the right-hand side strictly less than\n\\[\n\\frac1{256}+\\frac1{50}\\frac{16}{5}\n=\\frac{2173}{32000}<\\frac2{25}.\n\\]\nThus sixteen independent observations per sample site suffice for this bound. It is a sufficient budget, not a claim of optimal sample complexity. The exact checker certifies $e^{10}>10240$ using a rational Taylor partial sum, so the displayed logarithm bound is not based on rounded numerical logarithms.\n\\head{Dependence between adjacent contrasts is allowed}\nNeighboring differences share one sample. The proof does not require those differences to be independent: it bounds each tail and applies a union bound across all cells. Independ…","icient recovery condition is\n\\[\na_{\\min}>2Mh+\\frac{4\\sigma}{\\sqrt R}\\sqrt{\\log(2n/\\delta)}.\n\\]\nThe probability of simultaneous recovery is then at least $1-\\delta$, subject to the same separated-jump sampling model.\n\\vfig[.86]{ext_repetition}\n\\head{One explicit conservative choice}\nTake $n=256$, $M=1/2$, $\\sigma=1/50$, $\\delta=1/20$, and $a_{\\min}=2/25$. Since $\\log(10240)<10<(16/5)^2$, choosing $R=16$ makes the right-hand side strictly less than\n\\[\n\\frac1{256}+\\frac1{50}\\frac{16}{5}\n=\\frac{2173}{32000}<\\frac2{25}.\n\\]\nThus sixteen independent observations per sample site suffice for this bound. It is a sufficient budget, not a claim of optimal sample complexity. The exact checker certifies $e^{10}>10240$ using a rational Taylor partial sum, so the displayed logarithm bound is not based on rounded numerical logarithms.\n\\head{Dependence between adjacent contrasts is allowed}\nNeighboring differences share one sample. The proof does not require those differences to be independent: it bounds each tail and applies a union bound across all cells. Independence of the underlying sample noises…","sigma}{\\sqrt R}\\sqrt{\\log(2n/\\delta)}.\n\\]\nThe probability of simultaneous recovery is then at least $1-\\delta$, subject to the same separated-jump sampling model.\n\\vfig[.86]{ext_repetition}\n\\head{One explicit conservative choice}\nTake $n=256$, $M=1/2$, $\\sigma=1/50$, $\\delta=1/20$, and $a_{\\min}=2/25$. Since $\\log(10240)<10<(16/5)^2$, choosing $R=16$ makes the right-hand side strictly less than\n\\[\n\\frac1{256}+\\frac1{50}\\frac{16}{5}\n=\\frac{2173}{32000}<\\frac2{25}.\n\\]\nThus sixteen independent observations per sample site suffice for this bound. It is a sufficient budget, not a claim of optimal sample complexity. The exact checker certifies $e^{10}>10240$ using a rational Taylor partial sum, so the displayed logarithm bound is not based on rounded numerical logarithms.\n\\head{Dependence between adjacent contrasts is allowed}\nNeighboring differences share one sample. The proof does not require those differences to be independent: it bounds each tail and applies a union bound across all cells. Independence of the underlying sample noises supplies the stated subGaussian difference proxy.\nFo…","1/2$, $\\sigma=1/50$, $\\delta=1/20$, and $a_{\\min}=2/25$. Since $\\log(10240)<10<(16/5)^2$, choosing $R=16$ makes the right-hand side strictly less than\n\\[\n\\frac1{256}+\\frac1{50}\\frac{16}{5}\n=\\frac{2173}{32000}<\\frac2{25}.\n\\]\nThus sixteen independent observations per sample site suffice for this bound. It is a sufficient budget, not a claim of optimal sample complexity. The exact checker certifies $e^{10}>10240$ using a rational Taylor partial sum, so the displayed logarithm bound is not based on rounded numerical logarithms.\n\\head{Dependence between adjacent contrasts is allowed}\nNeighboring differences share one sample. The proof does not require those differences to be independent: it bounds each tail and applies a union bound across all cells. Independence of the underlying sample noises supplies the stated subGaussian difference proxy.\nFor a fixed separated signal model, a noise scale satisfying $\\sigma_n\\sqrt{\\log n}\\to0$ gives the stated recovery consistency in probability. A fixed positive noise scale with one observation per site does not satisfy that condition.\n\\takeaway{More…","ical logarithms.\n\\head{Dependence between adjacent contrasts is allowed}\nNeighboring differences share one sample. The proof does not require those differences to be independent: it bounds each tail and applies a union bound across all cells. Independence of the underlying sample noises supplies the stated subGaussian difference proxy.\nFor a fixed separated signal model, a noise scale satisfying $\\sigma_n\\sqrt{\\log n}\\to0$ gives the stated recovery consistency in probability. A fixed positive noise scale with one observation per site does not satisfy that condition.\n\\takeaway{More grid points and more repeated observations affect different terms. Finer spacing reduces background variation; averaging reduces noise.}","\\vpage{37}{Haar sees a jump only through its position in the support}{SCALE SEPARATION / A QUANTIFIED THRESHOLD}{Traditional monograph: Theorem E10.1 (p. 68), Theorem E10.2 (p. 69), Corollary E10.3 (p. 69).}\nUse a normalized Haar wavelet on an available interval of length $L$ on the real line, positive on its left half and negative on its right half. Here $H_\\tau$ is the half-line step $\\mathbf1_{[\\tau,\\infty)}$, so there is no artificial second endpoint jump.\nIf $u$ is the relative jump position in the support, then\n\\[\n|\\langle aH_\\tau,\\psi\\rangle|\n=|a|\\sqrt L\\min\\{u,1-u\\}.\n\\]\nThe coefficient vanishes when the jump lies exactly at a support boundary.\n\\vfig[.86]{ext_haar_margin}\n\\head{A worked gap with all three contributions}\nTake $L=1/4$, $M=1$, noise bound $\\varepsilon=1/100$, jump size $|a|=1$, and relative boundary distance at least $\\rho=1/4$. The smooth and noise contributions together are bounded by\n\\[\n\\frac M4L^{3/2}+\\varepsilon\\sqrt L\n=\\frac1{32}+\\fra","\\vpage{37}{Haar sees a jump only through its position in the support}{SCALE SEPARATION / A QUANTIFIED THRESHOLD}{Traditional monograph: Theorem E10.1 (p. 68), Theorem E10.2 (p. 69), Corollary E10.3 (p. 69).}\nUse a normalized Haar wavelet on an available interval of length $L$ on the real line, positive on its left half and negative on its right half. Here $H_\\tau$ is the half-line step $\\mathbf1_{[\\tau,\\infty)}$, so there is no artificial second endpoint jump.\nIf $u$ is the relative jump position in the support, then\n\\[\n|\\langle aH_\\tau,\\psi\\rangle|\n=|a|\\sqrt L\\min\\{u,1-u\\}.\n\\]\nThe coefficient vanishes when the jump lies exactly at a support boundary.\n\\vfig[.86]{ext_haar_margin}\n\\head{A worked gap with all three contributions}\nTake $L=1/4$, $M=1$, noise bound $\\varepsilon=1/100$, jump size $|a|=1$, and relative boundary distance at least $\\rho=1/4$. The smooth and noise contributions together are bounded by\n\\[\n\\frac M4L^{3/2}+\\varepsilon\\sqrt L\n=\\frac1{32}+\\frac1{200}=\\frac{29}{800}.\n\\]\nThe jump contributes at least $|a|\\rho\\sqrt L=1/8$. The observe","\\vpage{37}{Haar sees a jump only through its position in the support}{SCALE SEPARATION / A QUANTIFIED THRESHOLD}{Traditional monograph: Theorem E10.1 (p. 68), Theorem E10.2 (p. 69), Corollary E10.3 (p. 69).}\nUse a normalized Haar wavelet on an available interval of length $L$ on the real line, positive on its left half and negative on its right half. Here $H_\\tau$ is the half-line step $\\mathbf1_{[\\tau,\\infty)}$, so there is no artificial second endpoint jump.\nIf $u$ is the relative jump position in the support, then\n\\[\n|\\langle aH_\\tau,\\psi\\rangle|\n=|a|\\sqrt L\\min\\{u,1-u\\}.\n\\]\nThe coefficient vanishes when the jump lies exactly at a support boundary.\n\\vfig[.86]{ext_haar_margin}\n\\head{A worked gap with all three contributions}\nTake $L=1/4$, $M=1$, noise bound $\\varepsilon=1/100$, jump size $|a|=1$, and relative boundary distance at least $\\rho=1/4$. The smooth and noise contributions together are bounded by\n\\[\n\\frac M4L^{3/2}+\\varepsilon\\sqrt L\n=\\frac1{32}+\\frac1{200}=\\frac{29}{800}.\n\\]\nThe jump contributes at least $|a|\\rho\\sqrt L=1/8$. The observed coefficient therefore has magnitu…","{SCALE SEPARATION / A QUANTIFIED THRESHOLD}{Traditional monograph: Theorem E10.1 (p. 68), Theorem E10.2 (p. 69), Corollary E10.3 (p. 69).}\nUse a normalized Haar wavelet on an available interval of length $L$ on the real line, positive on its left half and negative on its right half. Here $H_\\tau$ is the half-line step $\\mathbf1_{[\\tau,\\infty)}$, so there is no artificial second endpoint jump.\nIf $u$ is the relative jump position in the support, then\n\\[\n|\\langle aH_\\tau,\\psi\\rangle|\n=|a|\\sqrt L\\min\\{u,1-u\\}.\n\\]\nThe coefficient vanishes when the jump lies exactly at a support boundary.\n\\vfig[.86]{ext_haar_margin}\n\\head{A worked gap with all three contributions}\nTake $L=1/4$, $M=1$, noise bound $\\varepsilon=1/100$, jump size $|a|=1$, and relative boundary distance at least $\\rho=1/4$. The smooth and noise contributions together are bounded by\n\\[\n\\frac M4L^{3/2}+\\varepsilon\\sqrt L\n=\\frac1{32}+\\frac1{200}=\\frac{29}{800}.\n\\]\nThe jump contributes at least $|a|\\rho\\sqrt L=1/8$. The observed coefficient therefore has magnitude at least $71/800$. The threshold $1/16=50/800$ separates every suc…","monograph: Theorem E10.1 (p. 68), Theorem E10.2 (p. 69), Corollary E10.3 (p. 69).}\nUse a normalized Haar wavelet on an available interval of length $L$ on the real line, positive on its left half and negative on its right half. Here $H_\\tau$ is the half-line step $\\mathbf1_{[\\tau,\\infty)}$, so there is no artificial second endpoint jump.\nIf $u$ is the relative jump position in the support, then\n\\[\n|\\langle aH_\\tau,\\psi\\rangle|\n=|a|\\sqrt L\\min\\{u,1-u\\}.\n\\]\nThe coefficient vanishes when the jump lies exactly at a support boundary.\n\\vfig[.86]{ext_haar_margin}\n\\head{A worked gap with all three contributions}\nTake $L=1/4$, $M=1$, noise bound $\\varepsilon=1/100$, jump size $|a|=1$, and relative boundary distance at least $\\rho=1/4$. The smooth and noise contributions together are bounded by\n\\[\n\\frac M4L^{3/2}+\\varepsilon\\sqrt L\n=\\frac1{32}+\\frac1{200}=\\frac{29}{800}.\n\\]\nThe jump contributes at least $|a|\\rho\\sqrt L=1/8$. The observed coefficient therefore has magnitude at least $71/800$. The threshold $1/16=50/800$ separates every such jump support from every support containing no jump.\n\\h…","ght half. Here $H_\\tau$ is the half-line step $\\mathbf1_{[\\tau,\\infty)}$, so there is no artificial second endpoint jump.\nIf $u$ is the relative jump position in the support, then\n\\[\n|\\langle aH_\\tau,\\psi\\rangle|\n=|a|\\sqrt L\\min\\{u,1-u\\}.\n\\]\nThe coefficient vanishes when the jump lies exactly at a support boundary.\n\\vfig[.86]{ext_haar_margin}\n\\head{A worked gap with all three contributions}\nTake $L=1/4$, $M=1$, noise bound $\\varepsilon=1/100$, jump size $|a|=1$, and relative boundary distance at least $\\rho=1/4$. The smooth and noise contributions together are bounded by\n\\[\n\\frac M4L^{3/2}+\\varepsilon\\sqrt L\n=\\frac1{32}+\\frac1{200}=\\frac{29}{800}.\n\\]\nThe jump contributes at least $|a|\\rho\\sqrt L=1/8$. The observed coefficient therefore has magnitude at least $71/800$. The threshold $1/16=50/800$ separates every such jump support from every support containing no jump.\n\\head{The available support is part of the hypothesis}\nA fixed extension is needed if the observation domain ends inside a desired wavelet support. Also, the selected support must contain only the specified jump. An unmo…","Here $H_\\tau$ is the half-line step $\\mathbf1_{[\\tau,\\infty)}$, so there is no artificial second endpoint jump.\nIf $u$ is the relative jump position in the support, then\n\\[\n|\\langle aH_\\tau,\\psi\\rangle|\n=|a|\\sqrt L\\min\\{u,1-u\\}.\n\\]\nThe coefficient vanishes when the jump lies exactly at a support boundary.\n\\vfig[.86]{ext_haar_margin}\n\\head{A worked gap with all three contributions}\nTake $L=1/4$, $M=1$, noise bound $\\varepsilon=1/100$, jump size $|a|=1$, and relative boundary distance at least $\\rho=1/4$. The smooth and noise contributions together are bounded by\n\\[\n\\frac M4L^{3/2}+\\varepsilon\\sqrt L\n=\\frac1{32}+\\frac1{200}=\\frac{29}{800}.\n\\]\nThe jump contributes at least $|a|\\rho\\sqrt L=1/8$. The observed coefficient therefore has magnitude at least $71/800$. The threshold $1/16=50/800$ separates every such jump support from every support containing no jump.\n\\head{The available support is part of the hypothesis}\nA fixed extension is needed if the observation domain ends inside a desired wavelet support. Also, the selected support must contain only the specified jump. An unmodeled seco…","he half-line step $\\mathbf1_{[\\tau,\\infty)}$, so there is no artificial second endpoint jump.\nIf $u$ is the relative jump position in the support, then\n\\[\n|\\langle aH_\\tau,\\psi\\rangle|\n=|a|\\sqrt L\\min\\{u,1-u\\}.\n\\]\nThe coefficient vanishes when the jump lies exactly at a support boundary.\n\\vfig[.86]{ext_haar_margin}\n\\head{A worked gap with all three contributions}\nTake $L=1/4$, $M=1$, noise bound $\\varepsilon=1/100$, jump size $|a|=1$, and relative boundary distance at least $\\rho=1/4$. The smooth and noise contributions together are bounded by\n\\[\n\\frac M4L^{3/2}+\\varepsilon\\sqrt L\n=\\frac1{32}+\\frac1{200}=\\frac{29}{800}.\n\\]\nThe jump contributes at least $|a|\\rho\\sqrt L=1/8$. The observed coefficient therefore has magnitude at least $71/800$. The threshold $1/16=50/800$ separates every such jump support from every support containing no jump.\n\\head{The available support is part of the hypothesis}\nA fixed extension is needed if the observation domain ends inside a desired wavelet support. Also, the selected support must contain only the specified jump. An unmodeled second jump can reinfo…","tau,\\infty)}$, so there is no artificial second endpoint jump.\nIf $u$ is the relative jump position in the support, then\n\\[\n|\\langle aH_\\tau,\\psi\\rangle|\n=|a|\\sqrt L\\min\\{u,1-u\\}.\n\\]\nThe coefficient vanishes when the jump lies exactly at a support boundary.\n\\vfig[.86]{ext_haar_margin}\n\\head{A worked gap with all three contributions}\nTake $L=1/4$, $M=1$, noise bound $\\varepsilon=1/100$, jump size $|a|=1$, and relative boundary distance at least $\\rho=1/4$. The smooth and noise contributions together are bounded by\n\\[\n\\frac M4L^{3/2}+\\varepsilon\\sqrt L\n=\\frac1{32}+\\frac1{200}=\\frac{29}{800}.\n\\]\nThe jump contributes at least $|a|\\rho\\sqrt L=1/8$. The observed coefficient therefore has magnitude at least $71/800$. The threshold $1/16=50/800$ separates every such jump support from every support containing no jump.\n\\head{The available support is part of the hypothesis}\nA fixed extension is needed if the observation domain ends inside a desired wavelet support. Also, the selected support must contain only the specified jump. An unmodeled second jump can reinforce or cancel the coefficient.\n…","ndpoint jump.\nIf $u$ is the relative jump position in the support, then\n\\[\n|\\langle aH_\\tau,\\psi\\rangle|\n=|a|\\sqrt L\\min\\{u,1-u\\}.\n\\]\nThe coefficient vanishes when the jump lies exactly at a support boundary.\n\\vfig[.86]{ext_haar_margin}\n\\head{A worked gap with all three contributions}\nTake $L=1/4$, $M=1$, noise bound $\\varepsilon=1/100$, jump size $|a|=1$, and relative boundary distance at least $\\rho=1/4$. The smooth and noise contributions together are bounded by\n\\[\n\\frac M4L^{3/2}+\\varepsilon\\sqrt L\n=\\frac1{32}+\\frac1{200}=\\frac{29}{800}.\n\\]\nThe jump contributes at least $|a|\\rho\\sqrt L=1/8$. The observed coefficient therefore has magnitude at least $71/800$. The threshold $1/16=50/800$ separates every such jump support from every support containing no jump.\n\\head{The available support is part of the hypothesis}\nA fixed extension is needed if the observation domain ends inside a desired wavelet support. Also, the selected support must contain only the specified jump. An unmodeled second jump can reinforce or cancel the coefficient.\n\\takeaway{Scale determines the smooth contributio…","\\[\n|\\langle aH_\\tau,\\psi\\rangle|\n=|a|\\sqrt L\\min\\{u,1-u\\}.\n\\]\nThe coefficient vanishes when the jump lies exactly at a support boundary.\n\\vfig[.86]{ext_haar_margin}\n\\head{A worked gap with all three contributions}\nTake $L=1/4$, $M=1$, noise bound $\\varepsilon=1/100$, jump size $|a|=1$, and relative boundary distance at least $\\rho=1/4$. The smooth and noise contributions together are bounded by\n\\[\n\\frac M4L^{3/2}+\\varepsilon\\sqrt L\n=\\frac1{32}+\\frac1{200}=\\frac{29}{800}.\n\\]\nThe jump contributes at least $|a|\\rho\\sqrt L=1/8$. The observed coefficient therefore has magnitude at least $71/800$. The threshold $1/16=50/800$ separates every such jump support from every support containing no jump.\n\\head{The available support is part of the hypothesis}\nA fixed extension is needed if the observation domain ends inside a desired wavelet support. Also, the selected support must contain only the specified jump. An unmodeled second jump can reinforce or cancel the coefficient.\n\\takeaway{Scale determines the smooth contribution, but alignment determines the jump contribution. A detection guarantee…","tly at a support boundary.\n\\vfig[.86]{ext_haar_margin}\n\\head{A worked gap with all three contributions}\nTake $L=1/4$, $M=1$, noise bound $\\varepsilon=1/100$, jump size $|a|=1$, and relative boundary distance at least $\\rho=1/4$. The smooth and noise contributions together are bounded by\n\\[\n\\frac M4L^{3/2}+\\varepsilon\\sqrt L\n=\\frac1{32}+\\frac1{200}=\\frac{29}{800}.\n\\]\nThe jump contributes at least $|a|\\rho\\sqrt L=1/8$. The observed coefficient therefore has magnitude at least $71/800$. The threshold $1/16=50/800$ separates every such jump support from every support containing no jump.\n\\head{The available support is part of the hypothesis}\nA fixed extension is needed if the observation domain ends inside a desired wavelet support. Also, the selected support must contain only the specified jump. An unmodeled second jump can reinforce or cancel the coefficient.\n\\takeaway{Scale determines the smooth contribution, but alignment determines the jump contribution. A detection guarantee needs both.}","ll three contributions}\nTake $L=1/4$, $M=1$, noise bound $\\varepsilon=1/100$, jump size $|a|=1$, and relative boundary distance at least $\\rho=1/4$. The smooth and noise contributions together are bounded by\n\\[\n\\frac M4L^{3/2}+\\varepsilon\\sqrt L\n=\\frac1{32}+\\frac1{200}=\\frac{29}{800}.\n\\]\nThe jump contributes at least $|a|\\rho\\sqrt L=1/8$. The observed coefficient therefore has magnitude at least $71/800$. The threshold $1/16=50/800$ separates every such jump support from every support containing no jump.\n\\head{The available support is part of the hypothesis}\nA fixed extension is needed if the observation domain ends inside a desired wavelet support. Also, the selected support must contain only the specified jump. An unmodeled second jump can reinforce or cancel the coefficient.\n\\takeaway{Scale determines the smooth contribution, but alignment determines the jump contribution. A detection guarantee needs both.}","Take $L=1/4$, $M=1$, noise bound $\\varepsilon=1/100$, jump size $|a|=1$, and relative boundary distance at least $\\rho=1/4$. The smooth and noise contributions together are bounded by\n\\[\n\\frac M4L^{3/2}+\\varepsilon\\sqrt L\n=\\frac1{32}+\\frac1{200}=\\frac{29}{800}.\n\\]\nThe jump contributes at least $|a|\\rho\\sqrt L=1/8$. The observed coefficient therefore has magnitude at least $71/800$. The threshold $1/16=50/800$ separates every such jump support from every support containing no jump.\n\\head{The available support is part of the hypothesis}\nA fixed extension is needed if the observation domain ends inside a desired wavelet support. Also, the selected support must contain only the specified jump. An unmodeled second jump can reinforce or cancel the coefficient.\n\\takeaway{Scale determines the smooth contribution, but alignment determines the jump contribution. A detection guarantee needs both.}","\\vpage{38}{Two half-shifted grids remove exact alignment failure}{HAAR SUPPORT GEOMETRY / A QUARTER-CELL GUARANTEE}{Traditional monograph: Theorem E10.4 (p. 69), Corollary E10.5 (p. 70).}\nNormalize the support length to $L=1$. The first grid has boundaries at integers; the second has boundaries at half-integers. For a point at position $u$ modulo one, consider its distance to the nearest boundary of each grid.\n\\vfig[.89]{ext_two_grids}\n\\head{At least one distance is one quarter or larger}\nIf $1/4\\le u\\le3/4$, the original cell works. Outside that range, a shifted cell works. At $u=1/8$, the original margin is $1/8$ but the shifted cell $[-1/2,1/2)$ gives margin $3/8$. At $u=1/4$, both margins equal $1/4$, showing that the uniform constant cannot be increased for these two grids.\nRestoring $L$, every jump has a support with boundary distance at least $L/4$ in one grid. If $L$ is smaller than the minimum jump sep","\\vpage{38}{Two half-shifted grids remove exact alignment failure}{HAAR SUPPORT GEOMETRY / A QUARTER-CELL GUARANTEE}{Traditional monograph: Theorem E10.4 (p. 69), Corollary E10.5 (p. 70).}\nNormalize the support length to $L=1$. The first grid has boundaries at integers; the second has boundaries at half-integers. For a point at position $u$ modulo one, consider its distance to the nearest boundary of each grid.\n\\vfig[.89]{ext_two_grids}\n\\head{At least one distance is one quarter or larger}\nIf $1/4\\le u\\le3/4$, the original cell works. Outside that range, a shifted cell works. At $u=1/8$, the original margin is $1/8$ but the shifted cell $[-1/2,1/2)$ gives margin $3/8$. At $u=1/4$, both margins equal $1/4$, showing that the uniform constant cannot be increased for these two grids.\nRestoring $L$, every jump has a support with boundary distance at least $L/4$ in one grid. If $L$ is smaller than the minimum jump separation, that support contains no other jump.\n\\head{Insert the geometric guarantee into the analytic estimate}\nThe a","R-CELL GUARANTEE}{Traditional monograph: Theorem E10.4 (p. 69), Corollary E10.5 (p. 70).}\nNormalize the support length to $L=1$. The first grid has boundaries at integers; the second has boundaries at half-integers. For a point at position $u$ modulo one, consider its distance to the nearest boundary of each grid.\n\\vfig[.89]{ext_two_grids}\n\\head{At least one distance is one quarter or larger}\nIf $1/4\\le u\\le3/4$, the original cell works. Outside that range, a shifted cell works. At $u=1/8$, the original margin is $1/8$ but the shifted cell $[-1/2,1/2)$ gives margin $3/8$. At $u=1/4$, both margins equal $1/4$, showing that the uniform constant cannot be increased for these two grids.\nRestoring $L$, every jump has a support with boundary distance at least $L/4$ in one grid. If $L$ is smaller than the minimum jump separation, that support contains no other jump.\n\\head{Insert the geometric guarantee into the analytic estimate}\nThe alignment-independent gap condition is\n\\[\n\\frac{|a|}{4}>2\\varepsilon+\\frac M2L.\n\\]\nFor the numbers on V37, the right side is $1/50+1/8=29/200$, strictly below …","}\nNormalize the support length to $L=1$. The first grid has boundaries at integers; the second has boundaries at half-integers. For a point at position $u$ modulo one, consider its distance to the nearest boundary of each grid.\n\\vfig[.89]{ext_two_grids}\n\\head{At least one distance is one quarter or larger}\nIf $1/4\\le u\\le3/4$, the original cell works. Outside that range, a shifted cell works. At $u=1/8$, the original margin is $1/8$ but the shifted cell $[-1/2,1/2)$ gives margin $3/8$. At $u=1/4$, both margins equal $1/4$, showing that the uniform constant cannot be increased for these two grids.\nRestoring $L$, every jump has a support with boundary distance at least $L/4$ in one grid. If $L$ is smaller than the minimum jump separation, that support contains no other jump.\n\\head{Insert the geometric guarantee into the analytic estimate}\nThe alignment-independent gap condition is\n\\[\n\\frac{|a|}{4}>2\\varepsilon+\\frac M2L.\n\\]\nFor the numbers on V37, the right side is $1/50+1/8=29/200$, strictly below $1/4$. Hence one of the two grids supplies a detectable support for every jump, regardle…","o $L=1$. The first grid has boundaries at integers; the second has boundaries at half-integers. For a point at position $u$ modulo one, consider its distance to the nearest boundary of each grid.\n\\vfig[.89]{ext_two_grids}\n\\head{At least one distance is one quarter or larger}\nIf $1/4\\le u\\le3/4$, the original cell works. Outside that range, a shifted cell works. At $u=1/8$, the original margin is $1/8$ but the shifted cell $[-1/2,1/2)$ gives margin $3/8$. At $u=1/4$, both margins equal $1/4$, showing that the uniform constant cannot be increased for these two grids.\nRestoring $L$, every jump has a support with boundary distance at least $L/4$ in one grid. If $L$ is smaller than the minimum jump separation, that support contains no other jump.\n\\head{Insert the geometric guarantee into the analytic estimate}\nThe alignment-independent gap condition is\n\\[\n\\frac{|a|}{4}>2\\varepsilon+\\frac M2L.\n\\]\nFor the numbers on V37, the right side is $1/50+1/8=29/200$, strictly below $1/4$. Hence one of the two grids supplies a detectable support for every jump, regardless of its position relative to t…","boundaries at integers; the second has boundaries at half-integers. For a point at position $u$ modulo one, consider its distance to the nearest boundary of each grid.\n\\vfig[.89]{ext_two_grids}\n\\head{At least one distance is one quarter or larger}\nIf $1/4\\le u\\le3/4$, the original cell works. Outside that range, a shifted cell works. At $u=1/8$, the original margin is $1/8$ but the shifted cell $[-1/2,1/2)$ gives margin $3/8$. At $u=1/4$, both margins equal $1/4$, showing that the uniform constant cannot be increased for these two grids.\nRestoring $L$, every jump has a support with boundary distance at least $L/4$ in one grid. If $L$ is smaller than the minimum jump separation, that support contains no other jump.\n\\head{Insert the geometric guarantee into the analytic estimate}\nThe alignment-independent gap condition is\n\\[\n\\frac{|a|}{4}>2\\varepsilon+\\frac M2L.\n\\]\nFor the numbers on V37, the right side is $1/50+1/8=29/200$, strictly below $1/4$. Hence one of the two grids supplies a detectable support for every jump, regardless of its position relative to the original grid.\nThis certi…","he second has boundaries at half-integers. For a point at position $u$ modulo one, consider its distance to the nearest boundary of each grid.\n\\vfig[.89]{ext_two_grids}\n\\head{At least one distance is one quarter or larger}\nIf $1/4\\le u\\le3/4$, the original cell works. Outside that range, a shifted cell works. At $u=1/8$, the original margin is $1/8$ but the shifted cell $[-1/2,1/2)$ gives margin $3/8$. At $u=1/4$, both margins equal $1/4$, showing that the uniform constant cannot be increased for these two grids.\nRestoring $L$, every jump has a support with boundary distance at least $L/4$ in one grid. If $L$ is smaller than the minimum jump separation, that support contains no other jump.\n\\head{Insert the geometric guarantee into the analytic estimate}\nThe alignment-independent gap condition is\n\\[\n\\frac{|a|}{4}>2\\varepsilon+\\frac M2L.\n\\]\nFor the numbers on V37, the right side is $1/50+1/8=29/200$, strictly below $1/4$. Hence one of the two grids supplies a detectable support for every jump, regardless of its position relative to the original grid.\nThis certifies a useful support per…","has boundaries at half-integers. For a point at position $u$ modulo one, consider its distance to the nearest boundary of each grid.\n\\vfig[.89]{ext_two_grids}\n\\head{At least one distance is one quarter or larger}\nIf $1/4\\le u\\le3/4$, the original cell works. Outside that range, a shifted cell works. At $u=1/8$, the original margin is $1/8$ but the shifted cell $[-1/2,1/2)$ gives margin $3/8$. At $u=1/4$, both margins equal $1/4$, showing that the uniform constant cannot be increased for these two grids.\nRestoring $L$, every jump has a support with boundary distance at least $L/4$ in one grid. If $L$ is smaller than the minimum jump separation, that support contains no other jump.\n\\head{Insert the geometric guarantee into the analytic estimate}\nThe alignment-independent gap condition is\n\\[\n\\frac{|a|}{4}>2\\varepsilon+\\frac M2L.\n\\]\nFor the numbers on V37, the right side is $1/50+1/8=29/200$, strictly below $1/4$. Hence one of the two grids supplies a detectable support for every jump, regardless of its position relative to the original grid.\nThis certifies a useful support per jump. It …","ers. For a point at position $u$ modulo one, consider its distance to the nearest boundary of each grid.\n\\vfig[.89]{ext_two_grids}\n\\head{At least one distance is one quarter or larger}\nIf $1/4\\le u\\le3/4$, the original cell works. Outside that range, a shifted cell works. At $u=1/8$, the original margin is $1/8$ but the shifted cell $[-1/2,1/2)$ gives margin $3/8$. At $u=1/4$, both margins equal $1/4$, showing that the uniform constant cannot be increased for these two grids.\nRestoring $L$, every jump has a support with boundary distance at least $L/4$ in one grid. If $L$ is smaller than the minimum jump separation, that support contains no other jump.\n\\head{Insert the geometric guarantee into the analytic estimate}\nThe alignment-independent gap condition is\n\\[\n\\frac{|a|}{4}>2\\varepsilon+\\frac M2L.\n\\]\nFor the numbers on V37, the right side is $1/50+1/8=29/200$, strictly below $1/4$. Hence one of the two grids supplies a detectable support for every jump, regardless of its position relative to the original grid.\nThis certifies a useful support per jump. It does not by itself specify a…","f each grid.\n\\vfig[.89]{ext_two_grids}\n\\head{At least one distance is one quarter or larger}\nIf $1/4\\le u\\le3/4$, the original cell works. Outside that range, a shifted cell works. At $u=1/8$, the original margin is $1/8$ but the shifted cell $[-1/2,1/2)$ gives margin $3/8$. At $u=1/4$, both margins equal $1/4$, showing that the uniform constant cannot be increased for these two grids.\nRestoring $L$, every jump has a support with boundary distance at least $L/4$ in one grid. If $L$ is smaller than the minimum jump separation, that support contains no other jump.\n\\head{Insert the geometric guarantee into the analytic estimate}\nThe alignment-independent gap condition is\n\\[\n\\frac{|a|}{4}>2\\varepsilon+\\frac M2L.\n\\]\nFor the numbers on V37, the right side is $1/50+1/8=29/200$, strictly below $1/4$. Hence one of the two grids supplies a detectable support for every jump, regardless of its position relative to the original grid.\nThis certifies a useful support per jump. It does not by itself specify a unique rule for merging duplicate detections from two grids into exact breakpoint estimates…","ance is one quarter or larger}\nIf $1/4\\le u\\le3/4$, the original cell works. Outside that range, a shifted cell works. At $u=1/8$, the original margin is $1/8$ but the shifted cell $[-1/2,1/2)$ gives margin $3/8$. At $u=1/4$, both margins equal $1/4$, showing that the uniform constant cannot be increased for these two grids.\nRestoring $L$, every jump has a support with boundary distance at least $L/4$ in one grid. If $L$ is smaller than the minimum jump separation, that support contains no other jump.\n\\head{Insert the geometric guarantee into the analytic estimate}\nThe alignment-independent gap condition is\n\\[\n\\frac{|a|}{4}>2\\varepsilon+\\frac M2L.\n\\]\nFor the numbers on V37, the right side is $1/50+1/8=29/200$, strictly below $1/4$. Hence one of the two grids supplies a detectable support for every jump, regardless of its position relative to the original grid.\nThis certifies a useful support per jump. It does not by itself specify a unique rule for merging duplicate detections from two grids into exact breakpoint estimates.\n\\takeaway{A second grid repairs a geometric blind spot. The …","larger}\nIf $1/4\\le u\\le3/4$, the original cell works. Outside that range, a shifted cell works. At $u=1/8$, the original margin is $1/8$ but the shifted cell $[-1/2,1/2)$ gives margin $3/8$. At $u=1/4$, both margins equal $1/4$, showing that the uniform constant cannot be increased for these two grids.\nRestoring $L$, every jump has a support with boundary distance at least $L/4$ in one grid. If $L$ is smaller than the minimum jump separation, that support contains no other jump.\n\\head{Insert the geometric guarantee into the analytic estimate}\nThe alignment-independent gap condition is\n\\[\n\\frac{|a|}{4}>2\\varepsilon+\\frac M2L.\n\\]\nFor the numbers on V37, the right side is $1/50+1/8=29/200$, strictly below $1/4$. Hence one of the two grids supplies a detectable support for every jump, regardless of its position relative to the original grid.\nThis certifies a useful support per jump. It does not by itself specify a unique rule for merging duplicate detections from two grids into exact breakpoint estimates.\n\\takeaway{A second grid repairs a geometric blind spot. The contrast inequality and","$u=1/4$, both margins equal $1/4$, showing that the uniform constant cannot be increased for these two grids.\nRestoring $L$, every jump has a support with boundary distance at least $L/4$ in one grid. If $L$ is smaller than the minimum jump separation, that support contains no other jump.\n\\head{Insert the geometric guarantee into the analytic estimate}\nThe alignment-independent gap condition is\n\\[\n\\frac{|a|}{4}>2\\varepsilon+\\frac M2L.\n\\]\nFor the numbers on V37, the right side is $1/50+1/8=29/200$, strictly below $1/4$. Hence one of the two grids supplies a detectable support for every jump, regardless of its position relative to the original grid.\nThis certifies a useful support per jump. It does not by itself specify a unique rule for merging duplicate detections from two grids into exact breakpoint estimates.\n\\takeaway{A second grid repairs a geometric blind spot. The contrast inequality and separation hypotheses still supply the recovery guarantee.}","sed for these two grids.\nRestoring $L$, every jump has a support with boundary distance at least $L/4$ in one grid. If $L$ is smaller than the minimum jump separation, that support contains no other jump.\n\\head{Insert the geometric guarantee into the analytic estimate}\nThe alignment-independent gap condition is\n\\[\n\\frac{|a|}{4}>2\\varepsilon+\\frac M2L.\n\\]\nFor the numbers on V37, the right side is $1/50+1/8=29/200$, strictly below $1/4$. Hence one of the two grids supplies a detectable support for every jump, regardless of its position relative to the original grid.\nThis certifies a useful support per jump. It does not by itself specify a unique rule for merging duplicate detections from two grids into exact breakpoint estimates.\n\\takeaway{A second grid repairs a geometric blind spot. The contrast inequality and separation hypotheses still supply the recovery guarantee.}","$L$, every jump has a support with boundary distance at least $L/4$ in one grid. If $L$ is smaller than the minimum jump separation, that support contains no other jump.\n\\head{Insert the geometric guarantee into the analytic estimate}\nThe alignment-independent gap condition is\n\\[\n\\frac{|a|}{4}>2\\varepsilon+\\frac M2L.\n\\]\nFor the numbers on V37, the right side is $1/50+1/8=29/200$, strictly below $1/4$. Hence one of the two grids supplies a detectable support for every jump, regardless of its position relative to the original grid.\nThis certifies a useful support per jump. It does not by itself specify a unique rule for merging duplicate detections from two grids into exact breakpoint estimates.\n\\takeaway{A second grid repairs a geometric blind spot. The contrast inequality and separation hypotheses still supply the recovery guarantee.}","\\vpage{39}{Higher differences annihilate polynomial interiors}{BOUNDARY STENCILS / SIGNAL RESPONSE AND NOISE GAIN}{Traditional monograph: Theorem E10.6 (p. 71), Proposition E10.7 (p. 71), Corollary E10.8 (p. 71).}\nLet $x(n)=n+2(n-3)_+$ on the integer line. Its slope changes from one to three at the breakpoint $3$. The second forward difference is\n\\[\n\\Delta^2x(n)=x(n+2)-2x(n+1)+x(n).\n\\]\n\\vfig[.87]{ext_differences}\n\\head{An exact small stencil calculation}\nThe values at $n=0,\\ldots,6$ are $(0,1,2,3,6,9,12)$. The second differences at $n=0,\\ldots,4$ are $(0,0,2,0,0)$. The only nonzero stencil straddles the slope change.\nFor a piecewise polynomial of degree at most $d$, $\\Delta^{d+1}$ vanishes whenever its complete stencil lies inside one polynomial piece. Candidate responses are confined to at most $d+1$ starting addresses before each breakpoint. They are boundary bands, not necessarily exact breakpoint locations.\n\\head{Dif","\\vpage{39}{Higher differences annihilate polynomial interiors}{BOUNDARY STENCILS / SIGNAL RESPONSE AND NOISE GAIN}{Traditional monograph: Theorem E10.6 (p. 71), Proposition E10.7 (p. 71), Corollary E10.8 (p. 71).}\nLet $x(n)=n+2(n-3)_+$ on the integer line. Its slope changes from one to three at the breakpoint $3$. The second forward difference is\n\\[\n\\Delta^2x(n)=x(n+2)-2x(n+1)+x(n).\n\\]\n\\vfig[.87]{ext_differences}\n\\head{An exact small stencil calculation}\nThe values at $n=0,\\ldots,6$ are $(0,1,2,3,6,9,12)$. The second differences at $n=0,\\ldots,4$ are $(0,0,2,0,0)$. The only nonzero stencil straddles the slope change.\nFor a piecewise polynomial of degree at most $d$, $\\Delta^{d+1}$ vanishes whenever its complete stencil lies inside one polynomial piece. Candidate responses are confined to at most $d+1$ starting addresses before each breakpoint. They are boundary bands, not necessarily exact breakpoint locations.\n\\head{Differencing also acts on noise}\nThe absolute binomial coefficients of $\\Delta^m$","\\vpage{39}{Higher differences annihilate polynomial interiors}{BOUNDARY STENCILS / SIGNAL RESPONSE AND NOISE GAIN}{Traditional monograph: Theorem E10.6 (p. 71), Proposition E10.7 (p. 71), Corollary E10.8 (p. 71).}\nLet $x(n)=n+2(n-3)_+$ on the integer line. Its slope changes from one to three at the breakpoint $3$. The second forward difference is\n\\[\n\\Delta^2x(n)=x(n+2)-2x(n+1)+x(n).\n\\]\n\\vfig[.87]{ext_differences}\n\\head{An exact small stencil calculation}\nThe values at $n=0,\\ldots,6$ are $(0,1,2,3,6,9,12)$. The second differences at $n=0,\\ldots,4$ are $(0,0,2,0,0)$. The only nonzero stencil straddles the slope change.\nFor a piecewise polynomial of degree at most $d$, $\\Delta^{d+1}$ vanishes whenever its complete stencil lies inside one polynomial piece. Candidate responses are confined to at most $d+1$ starting addresses before each breakpoint. They are boundary bands, not necessarily exact breakpoint locations.\n\\head{Differencing also acts on noise}\nThe absolute binomial coefficients of $\\Delta^m$ sum to $2^m$. With $\\|\\eta\\|_\\infty\\le1/16$, second-difference noise is b","ENCILS / SIGNAL RESPONSE AND NOISE GAIN}{Traditional monograph: Theorem E10.6 (p. 71), Proposition E10.7 (p. 71), Corollary E10.8 (p. 71).}\nLet $x(n)=n+2(n-3)_+$ on the integer line. Its slope changes from one to three at the breakpoint $3$. The second forward difference is\n\\[\n\\Delta^2x(n)=x(n+2)-2x(n+1)+x(n).\n\\]\n\\vfig[.87]{ext_differences}\n\\head{An exact small stencil calculation}\nThe values at $n=0,\\ldots,6$ are $(0,1,2,3,6,9,12)$. The second differences at $n=0,\\ldots,4$ are $(0,0,2,0,0)$. The only nonzero stencil straddles the slope change.\nFor a piecewise polynomial of degree at most $d$, $\\Delta^{d+1}$ vanishes whenever its complete stencil lies inside one polynomial piece. Candidate responses are confined to at most $d+1$ starting addresses before each breakpoint. They are boundary bands, not necessarily exact breakpoint locations.\n\\head{Differencing also acts on noise}\nThe absolute binomial coefficients of $\\Delta^m$ sum to $2^m$. With $\\|\\eta\\|_\\infty\\le1/16$, second-difference noise is bounded by $4/16=1/4$. The true response $2$ remains at least $7/4$, while all responses …","PONSE AND NOISE GAIN}{Traditional monograph: Theorem E10.6 (p. 71), Proposition E10.7 (p. 71), Corollary E10.8 (p. 71).}\nLet $x(n)=n+2(n-3)_+$ on the integer line. Its slope changes from one to three at the breakpoint $3$. The second forward difference is\n\\[\n\\Delta^2x(n)=x(n+2)-2x(n+1)+x(n).\n\\]\n\\vfig[.87]{ext_differences}\n\\head{An exact small stencil calculation}\nThe values at $n=0,\\ldots,6$ are $(0,1,2,3,6,9,12)$. The second differences at $n=0,\\ldots,4$ are $(0,0,2,0,0)$. The only nonzero stencil straddles the slope change.\nFor a piecewise polynomial of degree at most $d$, $\\Delta^{d+1}$ vanishes whenever its complete stencil lies inside one polynomial piece. Candidate responses are confined to at most $d+1$ starting addresses before each breakpoint. They are boundary bands, not necessarily exact breakpoint locations.\n\\head{Differencing also acts on noise}\nThe absolute binomial coefficients of $\\Delta^m$ sum to $2^m$. With $\\|\\eta\\|_\\infty\\le1/16$, second-difference noise is bounded by $4/16=1/4$. The true response $2$ remains at least $7/4$, while all responses away from the break…","heorem E10.6 (p. 71), Proposition E10.7 (p. 71), Corollary E10.8 (p. 71).}\nLet $x(n)=n+2(n-3)_+$ on the integer line. Its slope changes from one to three at the breakpoint $3$. The second forward difference is\n\\[\n\\Delta^2x(n)=x(n+2)-2x(n+1)+x(n).\n\\]\n\\vfig[.87]{ext_differences}\n\\head{An exact small stencil calculation}\nThe values at $n=0,\\ldots,6$ are $(0,1,2,3,6,9,12)$. The second differences at $n=0,\\ldots,4$ are $(0,0,2,0,0)$. The only nonzero stencil straddles the slope change.\nFor a piecewise polynomial of degree at most $d$, $\\Delta^{d+1}$ vanishes whenever its complete stencil lies inside one polynomial piece. Candidate responses are confined to at most $d+1$ starting addresses before each breakpoint. They are boundary bands, not necessarily exact breakpoint locations.\n\\head{Differencing also acts on noise}\nThe absolute binomial coefficients of $\\Delta^m$ sum to $2^m$. With $\\|\\eta\\|_\\infty\\le1/16$, second-difference noise is bounded by $4/16=1/4$. The true response $2$ remains at least $7/4$, while all responses away from the breakpoint band remain at most $1/4$. Threshold one…","), Proposition E10.7 (p. 71), Corollary E10.8 (p. 71).}\nLet $x(n)=n+2(n-3)_+$ on the integer line. Its slope changes from one to three at the breakpoint $3$. The second forward difference is\n\\[\n\\Delta^2x(n)=x(n+2)-2x(n+1)+x(n).\n\\]\n\\vfig[.87]{ext_differences}\n\\head{An exact small stencil calculation}\nThe values at $n=0,\\ldots,6$ are $(0,1,2,3,6,9,12)$. The second differences at $n=0,\\ldots,4$ are $(0,0,2,0,0)$. The only nonzero stencil straddles the slope change.\nFor a piecewise polynomial of degree at most $d$, $\\Delta^{d+1}$ vanishes whenever its complete stencil lies inside one polynomial piece. Candidate responses are confined to at most $d+1$ starting addresses before each breakpoint. They are boundary bands, not necessarily exact breakpoint locations.\n\\head{Differencing also acts on noise}\nThe absolute binomial coefficients of $\\Delta^m$ sum to $2^m$. With $\\|\\eta\\|_\\infty\\le1/16$, second-difference noise is bounded by $4/16=1/4$. The true response $2$ remains at least $7/4$, while all responses away from the breakpoint band remain at most $1/4$. Threshold one separates them.\nTh…","ges from one to three at the breakpoint $3$. The second forward difference is\n\\[\n\\Delta^2x(n)=x(n+2)-2x(n+1)+x(n).\n\\]\n\\vfig[.87]{ext_differences}\n\\head{An exact small stencil calculation}\nThe values at $n=0,\\ldots,6$ are $(0,1,2,3,6,9,12)$. The second differences at $n=0,\\ldots,4$ are $(0,0,2,0,0)$. The only nonzero stencil straddles the slope change.\nFor a piecewise polynomial of degree at most $d$, $\\Delta^{d+1}$ vanishes whenever its complete stencil lies inside one polynomial piece. Candidate responses are confined to at most $d+1$ starting addresses before each breakpoint. They are boundary bands, not necessarily exact breakpoint locations.\n\\head{Differencing also acts on noise}\nThe absolute binomial coefficients of $\\Delta^m$ sum to $2^m$. With $\\|\\eta\\|_\\infty\\le1/16$, second-difference noise is bounded by $4/16=1/4$. The true response $2$ remains at least $7/4$, while all responses away from the breakpoint band remain at most $1/4$. Threshold one separates them.\nThe general robust-band statement assumes that each breakpoint has at least one sufficiently large true response. A","rom one to three at the breakpoint $3$. The second forward difference is\n\\[\n\\Delta^2x(n)=x(n+2)-2x(n+1)+x(n).\n\\]\n\\vfig[.87]{ext_differences}\n\\head{An exact small stencil calculation}\nThe values at $n=0,\\ldots,6$ are $(0,1,2,3,6,9,12)$. The second differences at $n=0,\\ldots,4$ are $(0,0,2,0,0)$. The only nonzero stencil straddles the slope change.\nFor a piecewise polynomial of degree at most $d$, $\\Delta^{d+1}$ vanishes whenever its complete stencil lies inside one polynomial piece. Candidate responses are confined to at most $d+1$ starting addresses before each breakpoint. They are boundary bands, not necessarily exact breakpoint locations.\n\\head{Differencing also acts on noise}\nThe absolute binomial coefficients of $\\Delta^m$ sum to $2^m$. With $\\|\\eta\\|_\\infty\\le1/16$, second-difference noise is bounded by $4/16=1/4$. The true response $2$ remains at least $7/4$, while all responses away from the breakpoint band remain at most $1/4$. Threshold one separates them.\nThe general robust-band statement assumes that each breakpoint has at least one sufficiently large true response. A brea…","erences}\n\\head{An exact small stencil calculation}\nThe values at $n=0,\\ldots,6$ are $(0,1,2,3,6,9,12)$. The second differences at $n=0,\\ldots,4$ are $(0,0,2,0,0)$. The only nonzero stencil straddles the slope change.\nFor a piecewise polynomial of degree at most $d$, $\\Delta^{d+1}$ vanishes whenever its complete stencil lies inside one polynomial piece. Candidate responses are confined to at most $d+1$ starting addresses before each breakpoint. They are boundary bands, not necessarily exact breakpoint locations.\n\\head{Differencing also acts on noise}\nThe absolute binomial coefficients of $\\Delta^m$ sum to $2^m$. With $\\|\\eta\\|_\\infty\\le1/16$, second-difference noise is bounded by $4/16=1/4$. The true response $2$ remains at least $7/4$, while all responses away from the breakpoint band remain at most $1/4$. Threshold one separates them.\nThe general robust-band statement assumes that each breakpoint has at least one sufficiently large true response. A breakpoint can fail that assumption through its amplitude or local configuration. Annihilation away from boundaries does not prove a low…","les the slope change.\nFor a piecewise polynomial of degree at most $d$, $\\Delta^{d+1}$ vanishes whenever its complete stencil lies inside one polynomial piece. Candidate responses are confined to at most $d+1$ starting addresses before each breakpoint. They are boundary bands, not necessarily exact breakpoint locations.\n\\head{Differencing also acts on noise}\nThe absolute binomial coefficients of $\\Delta^m$ sum to $2^m$. With $\\|\\eta\\|_\\infty\\le1/16$, second-difference noise is bounded by $4/16=1/4$. The true response $2$ remains at least $7/4$, while all responses away from the breakpoint band remain at most $1/4$. Threshold one separates them.\nThe general robust-band statement assumes that each breakpoint has at least one sufficiently large true response. A breakpoint can fail that assumption through its amplitude or local configuration. Annihilation away from boundaries does not prove a lower bound at every boundary.\n\\takeaway{A vanishing interior response is structural. Reliable detection also requires the boundary response to exceed the amplified noise.}","ge.\nFor a piecewise polynomial of degree at most $d$, $\\Delta^{d+1}$ vanishes whenever its complete stencil lies inside one polynomial piece. Candidate responses are confined to at most $d+1$ starting addresses before each breakpoint. They are boundary bands, not necessarily exact breakpoint locations.\n\\head{Differencing also acts on noise}\nThe absolute binomial coefficients of $\\Delta^m$ sum to $2^m$. With $\\|\\eta\\|_\\infty\\le1/16$, second-difference noise is bounded by $4/16=1/4$. The true response $2$ remains at least $7/4$, while all responses away from the breakpoint band remain at most $1/4$. Threshold one separates them.\nThe general robust-band statement assumes that each breakpoint has at least one sufficiently large true response. A breakpoint can fail that assumption through its amplitude or local configuration. Annihilation away from boundaries does not prove a lower bound at every boundary.\n\\takeaway{A vanishing interior response is structural. Reliable detection also requires the boundary response to exceed the amplified noise.}","ecewise polynomial of degree at most $d$, $\\Delta^{d+1}$ vanishes whenever its complete stencil lies inside one polynomial piece. Candidate responses are confined to at most $d+1$ starting addresses before each breakpoint. They are boundary bands, not necessarily exact breakpoint locations.\n\\head{Differencing also acts on noise}\nThe absolute binomial coefficients of $\\Delta^m$ sum to $2^m$. With $\\|\\eta\\|_\\infty\\le1/16$, second-difference noise is bounded by $4/16=1/4$. The true response $2$ remains at least $7/4$, while all responses away from the breakpoint band remain at most $1/4$. Threshold one separates them.\nThe general robust-band statement assumes that each breakpoint has at least one sufficiently large true response. A breakpoint can fail that assumption through its amplitude or local configuration. Annihilation away from boundaries does not prove a lower bound at every boundary.\n\\takeaway{A vanishing interior response is structural. Reliable detection also requires the boundary response to exceed the amplified noise.}","s whenever its complete stencil lies inside one polynomial piece. Candidate responses are confined to at most $d+1$ starting addresses before each breakpoint. They are boundary bands, not necessarily exact breakpoint locations.\n\\head{Differencing also acts on noise}\nThe absolute binomial coefficients of $\\Delta^m$ sum to $2^m$. With $\\|\\eta\\|_\\infty\\le1/16$, second-difference noise is bounded by $4/16=1/4$. The true response $2$ remains at least $7/4$, while all responses away from the breakpoint band remain at most $1/4$. Threshold one separates them.\nThe general robust-band statement assumes that each breakpoint has at least one sufficiently large true response. A breakpoint can fail that assumption through its amplitude or local configuration. Annihilation away from boundaries does not prove a lower bound at every boundary.\n\\takeaway{A vanishing interior response is structural. Reliable detection also requires the boundary response to exceed the amplified noise.}","l lies inside one polynomial piece. Candidate responses are confined to at most $d+1$ starting addresses before each breakpoint. They are boundary bands, not necessarily exact breakpoint locations.\n\\head{Differencing also acts on noise}\nThe absolute binomial coefficients of $\\Delta^m$ sum to $2^m$. With $\\|\\eta\\|_\\infty\\le1/16$, second-difference noise is bounded by $4/16=1/4$. The true response $2$ remains at least $7/4$, while all responses away from the breakpoint band remain at most $1/4$. Threshold one separates them.\nThe general robust-band statement assumes that each breakpoint has at least one sufficiently large true response. A breakpoint can fail that assumption through its amplitude or local configuration. Annihilation away from boundaries does not prove a lower bound at every boundary.\n\\takeaway{A vanishing interior response is structural. Reliable detection also requires the boundary response to exceed the amplified noise.}","ynomial piece. Candidate responses are confined to at most $d+1$ starting addresses before each breakpoint. They are boundary bands, not necessarily exact breakpoint locations.\n\\head{Differencing also acts on noise}\nThe absolute binomial coefficients of $\\Delta^m$ sum to $2^m$. With $\\|\\eta\\|_\\infty\\le1/16$, second-difference noise is bounded by $4/16=1/4$. The true response $2$ remains at least $7/4$, while all responses away from the breakpoint band remain at most $1/4$. Threshold one separates them.\nThe general robust-band statement assumes that each breakpoint has at least one sufficiently large true response. A breakpoint can fail that assumption through its amplitude or local configuration. Annihilation away from boundaries does not prove a lower bound at every boundary.\n\\takeaway{A vanishing interior response is structural. Reliable detection also requires the boundary response to exceed the amplified noise.}","g addresses before each breakpoint. They are boundary bands, not necessarily exact breakpoint locations.\n\\head{Differencing also acts on noise}\nThe absolute binomial coefficients of $\\Delta^m$ sum to $2^m$. With $\\|\\eta\\|_\\infty\\le1/16$, second-difference noise is bounded by $4/16=1/4$. The true response $2$ remains at least $7/4$, while all responses away from the breakpoint band remain at most $1/4$. Threshold one separates them.\nThe general robust-band statement assumes that each breakpoint has at least one sufficiently large true response. A breakpoint can fail that assumption through its amplitude or local configuration. Annihilation away from boundaries does not prove a lower bound at every boundary.\n\\takeaway{A vanishing interior response is structural. Reliable detection also requires the boundary response to exceed the amplified noise.}","\\vpage{40}{Conditional Gibbs control closes the reconstruction estimate}{THE TWO ERRORS / A COMPLETE READING CONTRACT}{Traditional monograph: Theorem E11.2 (p. 72), Theorem E11.3 (p. 73).}\nThe exact retained-packet approximation shares the same discontinuity as its target. If $f=s+j$ and $Q_ms\\to s$ uniformly, then\n\\[\n\\|f-(Q_ms+j)\\|_\\infty=\\|s-Q_ms\\|_\\infty\\longrightarrow0.\n\\]\nThe approximants themselves may be discontinuous. There is no conflict with the failure of continuous Fourier polynomials to converge uniformly to a jump.\n\\begin{center}\\includegraphics[width=.78\\linewidth]{visual_graphics/annals_gibbs_comparison.pdf}\\end{center}\n\\platecaption{Preserved vector figure from the expanded source. It illustrates the choice of approximation target; it is not an independent","\\vpage{40}{Conditional Gibbs control closes the reconstruction estimate}{THE TWO ERRORS / A COMPLETE READING CONTRACT}{Traditional monograph: Theorem E11.2 (p. 72), Theorem E11.3 (p. 73).}\nThe exact retained-packet approximation shares the same discontinuity as its target. If $f=s+j$ and $Q_ms\\to s$ uniformly, then\n\\[\n\\|f-(Q_ms+j)\\|_\\infty=\\|s-Q_ms\\|_\\infty\\longrightarrow0.\n\\]\nThe approximants themselves may be discontinuous. There is no conflict with the failure of continuous Fourier polynomials to converge uniformly to a jump.\n\\begin{center}\\includegraphics[width=.78\\linewidth]{visual_graphics/annals_gibbs_comparison.pdf}\\end{center}\n\\platecaption{Preserved vector figure from the expanded source. It illustrates the choice of approximation target; it is not an independent universal recov","\\vpage{40}{Conditional Gibbs control closes the reconstruction estimate}{THE TWO ERRORS / A COMPLETE READING CONTRACT}{Traditional monograph: Theorem E11.2 (p. 72), Theorem E11.3 (p. 73).}\nThe exact retained-packet approximation shares the same discontinuity as its target. If $f=s+j$ and $Q_ms\\to s$ uniformly, then\n\\[\n\\|f-(Q_ms+j)\\|_\\infty=\\|s-Q_ms\\|_\\infty\\longrightarrow0.\n\\]\nThe approximants themselves may be discontinuous. There is no conflict with the failure of continuous Fourier polynomials to converge uniformly to a jump.\n\\begin{center}\\includegraphics[width=.78\\linewidth]{visual_graphics/annals_gibbs_comparison.pdf}\\end{center}\n\\platecaption{Preserved vector figure from the expanded source. It illustrates the choice of approximation target; it is not an independent universal recovery certificate.}\n\\head{Estimated packets need the operator norm too}\nIf the b","rge uniformly to a jump.\n\\begin{center}\\includegraphics[width=.78\\linewidth]{visual_graphics/annals_gibbs_comparison.pdf}\\end{center}\n\\platecaption{Preserved vector figure from the expanded source. It illustrates the choice of approximation target; it is not an independent universal recovery certificate.}\n\\head{Estimated packets need the operator norm too}\nIf the bounded linear operators satisfy $\\sup_m\\|Q_m\\|_{L^p\\to L^p}\\le C_Q$, then\n\\[\n\\|f-\\widehat f_m\\|_p\n\\le\\|s-Q_ms\\|_p+(1+C_Q)\\|j-\\widehat j_m\\|_p.\n\\]\nFor an orthogonal $L^2$ projection one may use $\\|I-Q_m\\|\\le1$, sharpening the packet multiplier to one. Such a bound must be justified in the norm actually being used. In particular, uniform boundedness of Fourier partial sums in the supremum norm is not supplied by this theorem.\n\\head{What the complete guide establishes}\nV1 to V26 preserve the original worked layer. V27 to V40 explain the additional traditional dev","ludegraphics[width=.78\\linewidth]{visual_graphics/annals_gibbs_comparison.pdf}\\end{center}\n\\platecaption{Preserved vector figure from the expanded source. It illustrates the choice of approximation target; it is not an independent universal recovery certificate.}\n\\head{Estimated packets need the operator norm too}\nIf the bounded linear operators satisfy $\\sup_m\\|Q_m\\|_{L^p\\to L^p}\\le C_Q$, then\n\\[\n\\|f-\\widehat f_m\\|_p\n\\le\\|s-Q_ms\\|_p+(1+C_Q)\\|j-\\widehat j_m\\|_p.\n\\]\nFor an orthogonal $L^2$ projection one may use $\\|I-Q_m\\|\\le1$, sharpening the packet multiplier to one. Such a bound must be justified in the norm actually being used. In particular, uniform boundedness of Fourier partial sums in the supremum norm is not supplied by this theorem.\n\\head{What the complete guide establishes}\nV1 to V26 preserve the original worked layer. V27 to V40 explain the additional traditional developments, outside the inherited twenty-result Lean map. The new exact exampl","}\n\\platecaption{Preserved vector figure from the expanded source. It illustrates the choice of approximation target; it is not an independent universal recovery certificate.}\n\\head{Estimated packets need the operator norm too}\nIf the bounded linear operators satisfy $\\sup_m\\|Q_m\\|_{L^p\\to L^p}\\le C_Q$, then\n\\[\n\\|f-\\widehat f_m\\|_p\n\\le\\|s-Q_ms\\|_p+(1+C_Q)\\|j-\\widehat j_m\\|_p.\n\\]\nFor an orthogonal $L^2$ projection one may use $\\|I-Q_m\\|\\le1$, sharpening the packet multiplier to one. Such a bound must be justified in the norm actually being used. In particular, uniform boundedness of Fourier partial sums in the supremum norm is not supplied by this theorem.\n\\head{What the complete guide establishes}\nV1 to V26 preserve the original worked layer. V27 to V40 explain the additional traditional developments, outside the inherited twenty-result Lean map. The new exact examples have their own finite","tor figure from the expanded source. It illustrates the choice of approximation target; it is not an independent universal recovery certificate.}\n\\head{Estimated packets need the operator norm too}\nIf the bounded linear operators satisfy $\\sup_m\\|Q_m\\|_{L^p\\to L^p}\\le C_Q$, then\n\\[\n\\|f-\\widehat f_m\\|_p\n\\le\\|s-Q_ms\\|_p+(1+C_Q)\\|j-\\widehat j_m\\|_p.\n\\]\nFor an orthogonal $L^2$ projection one may use $\\|I-Q_m\\|\\le1$, sharpening the packet multiplier to one. Such a bound must be justified in the norm actually being used. In particular, uniform boundedness of Fourier partial sums in the supremum norm is not supplied by this theorem.\n\\head{What the complete guide establishes}\nV1 to V26 preserve the original worked layer. V27 to V40 explain the additional traditional developments, outside the inherited twenty-result Lean map. The new exact examples have their own finite arithmetic record. They do not extend t","\\begin{tikzpicture}[x=1cm,y=1cm,>=Latex,font=\\sffamily\\small,line cap=round,line join=round]\n\\node[] at (6.40000,4.85000) {FOUR PHASES: EVERY ADDRESS APPEARS ONCE};\n\\node[anchor=east] at (0.75000,3.75000) {$r=0$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (1.10000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (2.20000,3.74500) {$n=0:\\ 0$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (3.30000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (4.40000,3.74500) {$n=4:\\ 0$};\n\\draw[fill=black!12,draw=black!40,line width=.35pt] (5.50000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (6.60000,3.74500) {$n=8:\\ 3$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (7.70000,3.42000)","\\begin{tikzpicture}[x=1cm,y=1cm,>=Latex,font=\\sffamily\\small,line cap=round,line join=round]\n\\node[] at (6.40000,4.85000) {FOUR PHASES: EVERY ADDRESS APPEARS ONCE};\n\\node[anchor=east] at (0.75000,3.75000) {$r=0$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (1.10000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (2.20000,3.74500) {$n=0:\\ 0$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (3.30000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (4.40000,3.74500) {$n=4:\\ 0$};\n\\draw[fill=black!12,draw=black!40,line width=.35pt] (5.50000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (6.60000,3.74500) {$n=8:\\ 3$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (7.70000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (8.80000,3.74500) {$n=12:\\ 0$};\n\\node[anchor=east] at (0.75000,2.95000) {$r=1$};\n\\draw[fill=black!0,","ound]\n\\node[] at (6.40000,4.85000) {FOUR PHASES: EVERY ADDRESS APPEARS ONCE};\n\\node[anchor=east] at (0.75000,3.75000) {$r=0$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (1.10000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (2.20000,3.74500) {$n=0:\\ 0$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (3.30000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (4.40000,3.74500) {$n=4:\\ 0$};\n\\draw[fill=black!12,draw=black!40,line width=.35pt] (5.50000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (6.60000,3.74500) {$n=8:\\ 3$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (7.70000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (8.80000,3.74500) {$n=12:\\ 0$};\n\\node[anchor=east] at (0.75000,2.95000) {$r=1$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (1.10000,2.62000) rectangle ++(2.20000,0.65000);\n\\node[] at (2.20000,2.94500) {$n=1:\\ \\frac{1}{4}$};\n\\draw[fil","!0,draw=black!40,line width=.35pt] (1.10000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (2.20000,3.74500) {$n=0:\\ 0$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (3.30000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (4.40000,3.74500) {$n=4:\\ 0$};\n\\draw[fill=black!12,draw=black!40,line width=.35pt] (5.50000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (6.60000,3.74500) {$n=8:\\ 3$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (7.70000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (8.80000,3.74500) {$n=12:\\ 0$};\n\\node[anchor=east] at (0.75000,2.95000) {$r=1$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (1.10000,2.62000) rectangle ++(2.20000,0.65000);\n\\node[] at (2.20000,2.94500) {$n=1:\\ \\frac{1}{4}$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (3.30000,2.62000) rectangle ++(2.20000,0.65000);\n\\node[] at (4.40000,2.94500) {$n=5:\\ - \\frac{1}{4}$}","!0,draw=black!40,line width=.35pt] (3.30000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (4.40000,3.74500) {$n=4:\\ 0$};\n\\draw[fill=black!12,draw=black!40,line width=.35pt] (5.50000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (6.60000,3.74500) {$n=8:\\ 3$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (7.70000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (8.80000,3.74500) {$n=12:\\ 0$};\n\\node[anchor=east] at (0.75000,2.95000) {$r=1$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (1.10000,2.62000) rectangle ++(2.20000,0.65000);\n\\node[] at (2.20000,2.94500) {$n=1:\\ \\frac{1}{4}$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (3.30000,2.62000) rectangle ++(2.20000,0.65000);\n\\node[] at (4.40000,2.94500) {$n=5:\\ - \\frac{1}{4}$};\n\\draw[fill=black!12,draw=black!40,line width=.35pt] (5.50000,2.62000) rectangle ++(2.20000,0.65000);\n\\node[] at (6.60000,2.94500) {$n=9:\\ \\fr","ectangle ++(2.20000,0.65000);\n\\node[] at (4.40000,3.74500) {$n=4:\\ 0$};\n\\draw[fill=black!12,draw=black!40,line width=.35pt] (5.50000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (6.60000,3.74500) {$n=8:\\ 3$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (7.70000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (8.80000,3.74500) {$n=12:\\ 0$};\n\\node[anchor=east] at (0.75000,2.95000) {$r=1$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (1.10000,2.62000) rectangle ++(2.20000,0.65000);\n\\node[] at (2.20000,2.94500) {$n=1:\\ \\frac{1}{4}$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (3.30000,2.62000) rectangle ++(2.20000,0.65000);\n\\node[] at (4.40000,2.94500) {$n=5:\\ - \\frac{1}{4}$};\n\\draw[fill=black!12,draw=black!40,line width=.35pt] (5.50000,2.62000) rectangle ++(2.20000,0.65000);\n\\node[] at (6.60000,2.94500) {$n=9:\\ \\frac{13}{4}$};\n\\draw[fill=black!0,draw=black!40,lin","00) rectangle ++(2.20000,0.65000);\n\\node[] at (6.60000,3.74500) {$n=8:\\ 3$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (7.70000,3.42000) rectangle ++(2.20000,0.65000);\n\\node[] at (8.80000,3.74500) {$n=12:\\ 0$};\n\\node[anchor=east] at (0.75000,2.95000) {$r=1$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (1.10000,2.62000) rectangle ++(2.20000,0.65000);\n\\node[] at (2.20000,2.94500) {$n=1:\\ \\frac{1}{4}$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (3.30000,2.62000) rectangle 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join=round]\n\\node[anchor=west] at (0.20000,3.78000) {coarse $P_{2,0}$};\n\\draw[fill=black!22,draw=black!40,line width=.35pt] (3.00000,3.50000) rectangle ++(0.76000,0.56000);\n\\node[] at (3.38000,3.78000) {$0$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (3.76000,3.50000) rectangle ++(0.76000,0.56000);\n\\draw[fill=black!22,draw=black!40,line width=.35pt] (4.52000,3.50000) rectangle ++(0.76000,0.56000);\n\\node[] at (4.90000,3.78000) {$2$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (5.28000,3.50000) rectangle ++(0.76000,0.56000);\n\\draw[fill=black!22,dr","\\begin{tikzpicture}[x=1cm,y=1cm,>=Latex,font=\\sffamily\\small,line cap=round,line join=round]\n\\node[anchor=west] at (0.20000,3.78000) {coarse $P_{2,0}$};\n\\draw[fill=black!22,draw=black!40,line width=.35pt] (3.00000,3.50000) rectangle ++(0.76000,0.56000);\n\\node[] at (3.38000,3.78000) {$0$};\n\\draw[fill=black!0,draw=black!40,line width=.35pt] (3.76000,3.50000) rectangle 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