Notation by work
Sufficient Resolution
All notation · Glossary · Source coverage
390 symbol spellings in 1507 mathematical passages across 4 selected source files. Subscripts and superscripts have their own entries alongside their base symbols.
The same symbol can have several meanings here. A passage classified as a definition may define another part of its expression. Read the linked context before assigning a meaning.
All 390 entries are available below.
=Introduction passage · 492 passages
Search aliases: =
Introduction passage · Sufficient_Resolution_Traditional.tex, line 118
For a subspace L admitting orthogonal projection, P_L denotes that actual projection. On a Hilbert space, closed subspaces admit it. The notation dist(v,L)=inf_ℓ∈ Lnnv-ℓ always refers to the actual subspace, not a replacement chosen to make a theorem true. Closure is written cl L.
Exact mathematical passage
\dist(v,L)=\inf_{\ell\in L}\nn{v-\ell}
AIntroduction passage · 337 passages
Search aliases: A
Introduction passage · Sufficient_Resolution_Traditional.tex, line 2619
Multiplying δ≤γη by nn p≥0 and the strict margin γnn p<τ-1 by η>0 proves the ratio specialization. The chosen source is defined by classical choice from the existence theorem. Its raw definition takes A,p and the truth/feasibility proofs as arguments; it is a theorem-level witness rather than an extracted algorithm using only available measurements. Each chosen source is accepted, so SR-58 proves the family convergence. The selection theorem retains the feasibility premises even though the accep…
Exact mathematical passage
A,p
\clIntroduction passage · 8 passages
Search aliases: cl operatorname
Introduction passage · Sufficient_Resolution_Traditional.tex, line 119
For a subspace L admitting orthogonal projection, P_L denotes that actual projection. On a Hilbert space, closed subspaces admit it. The notation dist(v,L)=inf_ℓ∈ Lnnv-ℓ always refers to the actual subspace, not a replacement chosen to make a theorem true. Closure is written cl L.
Exact mathematical passage
\cl L
\distIntroduction passage · 12 passages
Search aliases: dist operatorname
Introduction passage · Sufficient_Resolution_Traditional.tex, line 118
For a subspace L admitting orthogonal projection, P_L denotes that actual projection. On a Hilbert space, closed subspaces admit it. The notation dist(v,L)=inf_ℓ∈ Lnnv-ℓ always refers to the actual subspace, not a replacement chosen to make a theorem true. Closure is written cl L.
Exact mathematical passage
\dist(v,L)=\inf_{\ell\in L}\nn{v-\ell}
\infIntroduction passage · 1 passage
Search aliases: inf
Introduction passage · Sufficient_Resolution_Traditional.tex, line 118
For a subspace L admitting orthogonal projection, P_L denotes that actual projection. On a Hilbert space, closed subspaces admit it. The notation dist(v,L)=inf_ℓ∈ Lnnv-ℓ always refers to the actual subspace, not a replacement chosen to make a theorem true. Closure is written cl L.
Exact mathematical passage
\dist(v,L)=\inf_{\ell\in L}\nn{v-\ell}
\inf_{\ell\in L}Introduction passage · 1 passage
Search aliases: inf_ℓ∈ L inf ell in
Introduction passage · Sufficient_Resolution_Traditional.tex, line 118
For a subspace L admitting orthogonal projection, P_L denotes that actual projection. On a Hilbert space, closed subspaces admit it. The notation dist(v,L)=inf_ℓ∈ Lnnv-ℓ always refers to the actual subspace, not a replacement chosen to make a theorem true. Closure is written cl L.
Exact mathematical passage
\dist(v,L)=\inf_{\ell\in L}\nn{v-\ell}
LIntroduction passage · 51 passages
Search aliases: L
Introduction passage · Sufficient_Resolution_Traditional.tex, line 116
For a subspace L admitting orthogonal projection, P_L denotes that actual projection. On a Hilbert space, closed subspaces admit it. The notation dist(v,L)=inf_ℓ∈ Lnnv-ℓ always refers to the actual subspace, not a replacement chosen to make a theorem true. Closure is written cl L.
Exact mathematical passage
L
\nnIntroduction passage · 267 passages
Search aliases: nn
Introduction passage · Sufficient_Resolution_Traditional.tex, line 118
For a subspace L admitting orthogonal projection, P_L denotes that actual projection. On a Hilbert space, closed subspaces admit it. The notation dist(v,L)=inf_ℓ∈ Lnnv-ℓ always refers to the actual subspace, not a replacement chosen to make a theorem true. Closure is written cl L.
Exact mathematical passage
\dist(v,L)=\inf_{\ell\in L}\nn{v-\ell}
PIntroduction passage · 66 passages
Search aliases: P
Introduction passage · Sufficient_Resolution_Traditional.tex, line 116
For a subspace L admitting orthogonal projection, P_L denotes that actual projection. On a Hilbert space, closed subspaces admit it. The notation dist(v,L)=inf_ℓ∈ Lnnv-ℓ always refers to the actual subspace, not a replacement chosen to make a theorem true. Closure is written cl L.
Exact mathematical passage
P_L
pIntroduction passage · 118 passages
Search aliases: p
Introduction passage · Sufficient_Resolution_Traditional.tex, line 2616
Multiplying δ≤γη by nn p≥0 and the strict margin γnn p<τ-1 by η>0 proves the ratio specialization. The chosen source is defined by classical choice from the existence theorem. Its raw definition takes A,p and the truth/feasibility proofs as arguments; it is a theorem-level witness rather than an extracted algorithm using only available measurements. Each chosen source is accepted, so SR-58 proves the family convergence. The selection theorem retains the feasibility premises even though the accep…
Exact mathematical passage
\nn p\ge0
P_LIntroduction passage · 1 passage
Search aliases: P_L
Introduction passage · Sufficient_Resolution_Traditional.tex, line 116
For a subspace L admitting orthogonal projection, P_L denotes that actual projection. On a Hilbert space, closed subspaces admit it. The notation dist(v,L)=inf_ℓ∈ Lnnv-ℓ always refers to the actual subspace, not a replacement chosen to make a theorem true. Closure is written cl L.
Exact mathematical passage
P_L
vIntroduction passage · 216 passages
Search aliases: v
Introduction passage · Sufficient_Resolution_Traditional.tex, line 118
For a subspace L admitting orthogonal projection, P_L denotes that actual projection. On a Hilbert space, closed subspaces admit it. The notation dist(v,L)=inf_ℓ∈ Lnnv-ℓ always refers to the actual subspace, not a replacement chosen to make a theorem true. Closure is written cl L.
Exact mathematical passage
\dist(v,L)=\inf_{\ell\in L}\nn{v-\ell}
\gammaIntroduction passage · 4 passages
Search aliases: γ gamma
Introduction passage · Sufficient_Resolution_Traditional.tex, line 2616
Multiplying δ≤γη by nn p≥0 and the strict margin γnn p<τ-1 by η>0 proves the ratio specialization. The chosen source is defined by classical choice from the existence theorem. Its raw definition takes A,p and the truth/feasibility proofs as arguments; it is a theorem-level witness rather than an extracted algorithm using only available measurements. Each chosen source is accepted, so SR-58 proves the family convergence. The selection theorem retains the feasibility premises even though the accep…
Exact mathematical passage
\delta\le\gamma\eta
\deltaIntroduction passage · 100 passages
Search aliases: δ delta
Introduction passage · Sufficient_Resolution_Traditional.tex, line 2616
Multiplying δ≤γη by nn p≥0 and the strict margin γnn p<τ-1 by η>0 proves the ratio specialization. The chosen source is defined by classical choice from the existence theorem. Its raw definition takes A,p and the truth/feasibility proofs as arguments; it is a theorem-level witness rather than an extracted algorithm using only available measurements. Each chosen source is accepted, so SR-58 proves the family convergence. The selection theorem retains the feasibility premises even though the accep…
Exact mathematical passage
\delta\le\gamma\eta
\etaIntroduction passage · 95 passages
Search aliases: η eta
Introduction passage · Sufficient_Resolution_Traditional.tex, line 2616
Multiplying δ≤γη by nn p≥0 and the strict margin γnn p<τ-1 by η>0 proves the ratio specialization. The chosen source is defined by classical choice from the existence theorem. Its raw definition takes A,p and the truth/feasibility proofs as arguments; it is a theorem-level witness rather than an extracted algorithm using only available measurements. Each chosen source is accepted, so SR-58 proves the family convergence. The selection theorem retains the feasibility premises even though the accep…
Exact mathematical passage
\delta\le\gamma\eta
\tauIntroduction passage · 47 passages
Search aliases: τ tau
Introduction passage · Sufficient_Resolution_Traditional.tex, line 2617
Multiplying δ≤γη by nn p≥0 and the strict margin γnn p<τ-1 by η>0 proves the ratio specialization. The chosen source is defined by classical choice from the existence theorem. Its raw definition takes A,p and the truth/feasibility proofs as arguments; it is a theorem-level witness rather than an extracted algorithm using only available measurements. Each chosen source is accepted, so SR-58 proves the family convergence. The selection theorem retains the feasibility premises even though the accep…
Exact mathematical passage
\gamma\nn p<\tau-1
\ellIntroduction passage · 1 passage
Search aliases: ℓ ell
Introduction passage · Sufficient_Resolution_Traditional.tex, line 118
For a subspace L admitting orthogonal projection, P_L denotes that actual projection. On a Hilbert space, closed subspaces admit it. The notation dist(v,L)=inf_ℓ∈ Lnnv-ℓ always refers to the actual subspace, not a replacement chosen to make a theorem true. Closure is written cl L.
Exact mathematical passage
\dist(v,L)=\inf_{\ell\in L}\nn{v-\ell}
\inIntroduction passage · 51 passages
Search aliases: ∈ in
Introduction passage · Sufficient_Resolution_Traditional.tex, line 118
For a subspace L admitting orthogonal projection, P_L denotes that actual projection. On a Hilbert space, closed subspaces admit it. The notation dist(v,L)=inf_ℓ∈ Lnnv-ℓ always refers to the actual subspace, not a replacement chosen to make a theorem true. Closure is written cl L.
Exact mathematical passage
\dist(v,L)=\inf_{\ell\in L}\nn{v-\ell}
\leIntroduction passage · 215 passages
Search aliases: ≤ le
Introduction passage · Sufficient_Resolution_Traditional.tex, line 2616
Multiplying δ≤γη by nn p≥0 and the strict margin γnn p<τ-1 by η>0 proves the ratio specialization. The chosen source is defined by classical choice from the existence theorem. Its raw definition takes A,p and the truth/feasibility proofs as arguments; it is a theorem-level witness rather than an extracted algorithm using only available measurements. Each chosen source is accepted, so SR-58 proves the family convergence. The selection theorem retains the feasibility premises even though the accep…
Exact mathematical passage
\delta\le\gamma\eta
\geIntroduction passage · 31 passages
Search aliases: ≥ ge
Introduction passage · Sufficient_Resolution_Traditional.tex, line 2616
Multiplying δ≤γη by nn p≥0 and the strict margin γnn p<τ-1 by η>0 proves the ratio specialization. The chosen source is defined by classical choice from the existence theorem. Its raw definition takes A,p and the truth/feasibility proofs as arguments; it is a theorem-level witness rather than an extracted algorithm using only available measurements. Each chosen source is accepted, so SR-58 proves the family convergence. The selection theorem retains the feasibility premises even though the accep…
Exact mathematical passage
\nn p\ge0
*Use in source · 52 passages
Search aliases: *
Use in source · Sufficient_Resolution_Traditional.tex, line 1547
Fix a bounded operator A:G→ V between real Hilbert spaces, without assuming closed range. For λ>0, put N_λ=A^*A+λ I, R_λ=N_λ^-1, B_λ=R_λ A^*. The inverse is constructed from coercivity. It is not supplied as a certificate premise. Here A^* is the actual Hilbert adjoint.
Exact mathematical passage
N_\lambda=A^*A+\lambda I,\qquad R_\lambda=N_\lambda^{-1},\qquad B_\lambda=R_\lambda A^*.
+Use in source · 181 passages
Search aliases: +
Use in source · Sufficient_Resolution_Traditional.tex, line 105
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
T_{u\mathbin{+\!+}v}=T_vT_u
aUse in source · 60 passages
Search aliases: a
Use in source · Sufficient_Resolution_Traditional.tex, line 143
The first contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
a_X:X\to V_X
A'Use in source · 2 passages
Search aliases: A'
Use in source · Sufficient_Resolution_Traditional.tex, line 242
Commuting presentation maps LA=A'F and δ'L=Mδ induce a map overline L: H→ H' by [v]↦[Lv]. These maps compose and preserve zero classes. Reflection of zero follows when overline L is injective.
Exact mathematical passage
LA=A'F
A^*Use in source · 52 passages
Search aliases: A^*
Use in source · Sufficient_Resolution_Traditional.tex, line 1547
Fix a bounded operator A:G→ V between real Hilbert spaces, without assuming closed range. For λ>0, put N_λ=A^*A+λ I, R_λ=N_λ^-1, B_λ=R_λ A^*. The inverse is constructed from coercivity. It is not supplied as a certificate premise. Here A^* is the actual Hilbert adjoint.
Exact mathematical passage
N_\lambda=A^*A+\lambda I,\qquad R_\lambda=N_\lambda^{-1},\qquad B_\lambda=R_\lambda A^*.
a^0Use in source · 1 passage
Search aliases: a^0
Use in source · Sufficient_Resolution_Traditional.tex, line 712
Induction proves the finite-sum formula: multiplying the old sum by a and adjoining b^n generates every term of the new sum. Multiplication by a-b telescopes, leaving a^n-b^n. Division is used only when the rates are distinct. If they coincide, all n+1 terms of H_n+1 equal a^n, including the zero-rate boundary with the usual a^0=1. For uniform bounds, the heterogeneous budget becomes D_i::w=aD_w+b^|w|ϵ, so induction gives D_w=ϵ H_|w|(a,b). Apply SR-14. The empty word has zero dynamics defect and…
Exact mathematical passage
a^0=1
a^2Use in source · 2 passages
Search aliases: a^2
Use in source · Sufficient_Resolution_Traditional.tex, line 1922
The resolvent parameter identity gives, for s,t≥ a>0, |r_v(s)-r_v(t)|≤ fracnn AnnA^*va^2|s-t|, proving continuity on positive compact intervals. The strict floor permits a comparator g with nnAg-v<ρ; objective minimization at sufficiently small positive λ gives a lower residual bracket. For large λ, nnB_λ v≤nnA^*v/λ, so the residual approaches nn v>ρ and gives an upper bracket. The intermediate value theorem provides a root; strictness gives uniqueness. The floor condition itself excludes the co…
Exact mathematical passage
|r_v(s)-r_v(t)|\le \frac{\nn A\nn{A^*v}}{a^2}|s-t|,
a^jUse in source · 1 passage
Search aliases: a^j
Use in source · Sufficient_Resolution_Traditional.tex, line 694
For H_0(a,b)=0 and H_n+1(a,b)=aH_n(a,b)+b^n, H_n(a,b)=∑_j=0^n-1a^jb^n-1-j, (a-b)H_n=a^n-b^n. If a≠ b, H_n=(a^n-b^n)/(a-b); for equal rates, H_n+1(a,a)=(n+1)a^n. Uniform step bounds give nnE_w≤ϵ H_n(a,b), nnOT_w-VU_wq≤η a^n+nn Vϵ H_n(a,b).
Exact mathematical passage
H_n(a,b)=\sum_{j=0}^{n-1}a^jb^{n-1-j},\qquad (a-b)H_n=a^n-b^n.
A^{k+2}Use in source · 2 passages
Search aliases: A^k+2
Use in source · Sufficient_Resolution_Traditional.tex, line 359
The following conditions are equivalent: E_n=A^n (∀ n), M_k=0 (∀ k), E_m+n=E_mE_n (∀ m,n). More precisely, E_n=A^n (n≤ N+1)iff M_k=0 (k<N). If M_j=0 for j<k, then E_k+2-A^k+2=M_k. Closure for every initial hidden state is equivalent to B=0.
Exact mathematical passage
E_{k+2}-A^{k+2}=M_k
A^nUse in source · 6 passages
Search aliases: A^n
Use in source · Sufficient_Resolution_Traditional.tex, line 351
The following conditions are equivalent: E_n=A^n (∀ n), M_k=0 (∀ k), E_m+n=E_mE_n (∀ m,n). More precisely, E_n=A^n (n≤ N+1)iff M_k=0 (k<N). If M_j=0 for j<k, then E_k+2-A^k+2=M_k. Closure for every initial hidden state is equivalent to B=0.
Exact mathematical passage
E_n=A^n\ (\forall n),\qquad M_k=0\ (\forall k),\qquad E_{m+n}=E_mE_n\ (\forall m,n).
a^nUse in source · 6 passages
Search aliases: a^n
Use in source · Sufficient_Resolution_Traditional.tex, line 694
For H_0(a,b)=0 and H_n+1(a,b)=aH_n(a,b)+b^n, H_n(a,b)=∑_j=0^n-1a^jb^n-1-j, (a-b)H_n=a^n-b^n. If a≠ b, H_n=(a^n-b^n)/(a-b); for equal rates, H_n+1(a,a)=(n+1)a^n. Uniform step bounds give nnE_w≤ϵ H_n(a,b), nnOT_w-VU_wq≤η a^n+nn Vϵ H_n(a,b).
Exact mathematical passage
H_n(a,b)=\sum_{j=0}^{n-1}a^jb^{n-1-j},\qquad (a-b)H_n=a^n-b^n.
A_cUse in source · 2 passages
Search aliases: A_c
Use in source · Sufficient_Resolution_Traditional.tex, line 1361
changing its image. Coercivity makes this source sequence Cauchy. Completeness gives a limit g∈ S, and continuity gives the image limit Ag∈Ran A. Hence the range is closed. If bounded B satisfies BA=P_S, then for g∈ S, nn g=nnBAg≤nn BnnAg. The same positive choice 1/(nn B+1) gives coercivity. Finite-dimensional image subspaces are closed in a normed space. Finally A_cg=cg, c>0, has canonical inverse B_cv=v/c. At v=c, data tend to zero with c while the recovered source remains one, ruling out a c…
Exact mathematical passage
A_cg=cg
a_iUse in source · 2 passages
Search aliases: a_i
Use in source · Sufficient_Resolution_Traditional.tex, line 655
Suppose nnT_i≤ a_i, nnU_i≤ b_i, nnδ_i≤ϵ_i. Let A_w,B_w be the products of the respective letter bounds and define D_∅=0, D_i::w=D_wa_i+B_wϵ_i. Then nnT_w≤ A_w, nnU_w≤ B_w, nnE_w≤ D_w, and D_umathbin+ +v=D_vA_u+B_vD_u. If nnO-Vq≤η, then nnOT_w-VU_wq≤η A_w+nn V D_w. For arbitrary initial x∈ X, z∈ Z, nnqT_wx-U_wz≤ D_wnn x+B_wnnqx-z.
Exact mathematical passage
\nn{T_i}\le a_i
A_SUse in source · 1 passage
Search aliases: A_S
Use in source · Sufficient_Resolution_Traditional.tex, line 1235
Restriction A_S:S→ R is a bounded linear bijection with bounded inverse. Define B=ι_S A_S^-1P_R. Then AB=P_R, BA=P_S, ABA=A, and BAB=B. Its two projections are symmetric, ker B=R^⊥, and Ran B=S. The best-image identity AB=P_R and the condition Ran B⊆ S uniquely determine B.
Exact mathematical passage
A_S:S\to R
A_S^{-1}Use in source · 3 passages
Search aliases: A_S^-1
Use in source · Sufficient_Resolution_Traditional.tex, line 1237
Restriction A_S:S→ R is a bounded linear bijection with bounded inverse. Define B=ι_S A_S^-1P_R. Then AB=P_R, BA=P_S, ABA=A, and BAB=B. Its two projections are symmetric, ker B=R^⊥, and Ran B=S. The best-image identity AB=P_R and the condition Ran B⊆ S uniquely determine B.
Exact mathematical passage
B=\iota_S A_S^{-1}P_R.
A_uUse in source · 1 passage
Search aliases: A_u
Use in source · Sufficient_Resolution_Traditional.tex, line 661
Suppose nnT_i≤ a_i, nnU_i≤ b_i, nnδ_i≤ϵ_i. Let A_w,B_w be the products of the respective letter bounds and define D_∅=0, D_i::w=D_wa_i+B_wϵ_i. Then nnT_w≤ A_w, nnU_w≤ B_w, nnE_w≤ D_w, and D_umathbin+ +v=D_vA_u+B_vD_u. If nnO-Vq≤η, then nnOT_w-VU_wq≤η A_w+nn V D_w. For arbitrary initial x∈ X, z∈ Z, nnqT_wx-U_wz≤ D_wnn x+B_wnnqx-z.
Exact mathematical passage
D_{u\mathbin{+\!+}v}=D_vA_u+B_vD_u.
A_wUse in source · 3 passages
Search aliases: A_w
Use in source · Sufficient_Resolution_Traditional.tex, line 656
Suppose nnT_i≤ a_i, nnU_i≤ b_i, nnδ_i≤ϵ_i. Let A_w,B_w be the products of the respective letter bounds and define D_∅=0, D_i::w=D_wa_i+B_wϵ_i. Then nnT_w≤ A_w, nnU_w≤ B_w, nnE_w≤ D_w, and D_umathbin+ +v=D_vA_u+B_vD_u. If nnO-Vq≤η, then nnOT_w-VU_wq≤η A_w+nn V D_w. For arbitrary initial x∈ X, z∈ Z, nnqT_wx-U_wz≤ D_wnn x+B_wnnqx-z.
Exact mathematical passage
A_w,B_w
a_XUse in source · 10 passages
Search aliases: a_X
Use in source · Sufficient_Resolution_Traditional.tex, line 143
The first contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
a_X:X\to V_X
a_YUse in source · 4 passages
Search aliases: a_Y
Use in source · Sufficient_Resolution_Traditional.tex, line 146
The first contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
\mathcal K(T)=a_YTs_X,\qquad \mathcal D(K)=s_YKa_X.
a_ZUse in source · 1 passage
Search aliases: a_Z
Use in source · Sufficient_Resolution_Traditional.tex, line 176
The splitting gives E_X^2=a_X(s_Xa_X)s_X=a_Xs_X. If E_Xv=v, then v=a_X(s_Xv); conversely E_Xa_Xx=a_Xx. Substituting the definitions yields s_Y(a_YTs_X)a_X=T, a_Y(s_YKa_X)s_X=E_YKE_X. For compatible K, take T= D(K). If another T' encodes to K, decoding proves T'= D(K). Finally, an intermediate resolution of Y has s_Ya_Y=I_Y, so K(S) K(T)=a_ZS(s_Ya_Y)Ts_X=a_ZSTs_X. Encoding I_X gives a_Xs_X=E_X, which acts as identity on compatible kernels. No inverse on all of V_X has been assumed.
Exact mathematical passage
\mathcal K(S)\mathcal K(T)=a_ZS(s_Ya_Y)Ts_X=a_ZSTs_X
BUse in source · 103 passages
Search aliases: B
Use in source · Sufficient_Resolution_Traditional.tex, line 304
Let V be visible state and H hidden state. Over a commutative ring, specify four maps and form the actual full dynamics T(y,z)=(Ay+Bz,Gy+Dz). Put C(y,z)=y, Ry=(y,0), E_n=CT^nR, and M_k=BD^kG. The return M_k contains entry into H, k hidden transitions, and return to V. Its first effect occurs after k+2 full transitions.
Exact mathematical passage
T(y,z)=(Ay+Bz,Gy+Dz).
bUse in source · 39 passages
Search aliases: b
Use in source · Sufficient_Resolution_Traditional.tex, line 655
Suppose nnT_i≤ a_i, nnU_i≤ b_i, nnδ_i≤ϵ_i. Let A_w,B_w be the products of the respective letter bounds and define D_∅=0, D_i::w=D_wa_i+B_wϵ_i. Then nnT_w≤ A_w, nnU_w≤ B_w, nnE_w≤ D_w, and D_umathbin+ +v=D_vA_u+B_vD_u. If nnO-Vq≤η, then nnOT_w-VU_wq≤η A_w+nn V D_w. For arbitrary initial x∈ X, z∈ Z, nnqT_wx-U_wz≤ D_wnn x+B_wnnqx-z.
Exact mathematical passage
\nn{U_i}\le b_i
B'Use in source · 6 passages
Search aliases: B'
Use in source · Sufficient_Resolution_Traditional.tex, line 1256
orem supplies A_S^-1 as an actual continuous linear inverse. The definition then gives AB=P_R. For g, BAg is the unique element of S with image Ag, namely P_Sg. The two inverse identities imply the generalized inverse laws. Orthogonal projections are symmetric, proving the symmetry conditions. The definition and injectivity of A_S^-1 identify the kernel and range. Finally, if B' has the same best image and values in S, then B'v-Bv∈ S∩ker A for each v, hence it is zero. This construction is the c…
Exact mathematical passage
B'
b^2Use in source · 1 passage
Search aliases: b^2
Use in source · Sufficient_Resolution_Traditional.tex, line 2576
For the source-condition rate, norm domination and p=A^*w give nny-p^2≤2ippp-y≤2nn w nnA(p-y) for an exact comparison y. Bound its image by the first conclusion. Then nnx-p≤nnx-y+nny-p and (a+b)^2≤2a^2+2b^2 prove the stated inexact rate.
Exact mathematical passage
(a+b)^2\le2a^2+2b^2
b^nUse in source · 7 passages
Search aliases: b^n
Use in source · Sufficient_Resolution_Traditional.tex, line 693
For H_0(a,b)=0 and H_n+1(a,b)=aH_n(a,b)+b^n, H_n(a,b)=∑_j=0^n-1a^jb^n-1-j, (a-b)H_n=a^n-b^n. If a≠ b, H_n=(a^n-b^n)/(a-b); for equal rates, H_n+1(a,a)=(n+1)a^n. Uniform step bounds give nnE_w≤ϵ H_n(a,b), nnOT_w-VU_wq≤η a^n+nn Vϵ H_n(a,b).
Exact mathematical passage
H_{n+1}(a,b)=aH_n(a,b)+b^n
b^{n-1}Use in source · 1 passage
Search aliases: b^n-1
Use in source · Sufficient_Resolution_Traditional.tex, line 744
For a=b=1, the response sum has n unit terms. For a=1 and b<1, it is 1+b+⋯+b^n-1≤1/(1-b). SR-15 gives the operator estimates. For initialization, add VU_w(qx-z) to the operator output error applied to x, and use nnU_w≤ b^n. The common bound concerns every allowed operation in the chosen norm. Individual spectral radii do not imply a common product bound for a noncommuting family. The nonexpansive estimate is an upper bound and does not state that every such system accumulates error linearly.
Exact mathematical passage
1+b+\cdots+b^{n-1}\le1/(1-b)
b^{n-1-j}Use in source · 1 passage
Search aliases: b^n-1-j
Use in source · Sufficient_Resolution_Traditional.tex, line 694
For H_0(a,b)=0 and H_n+1(a,b)=aH_n(a,b)+b^n, H_n(a,b)=∑_j=0^n-1a^jb^n-1-j, (a-b)H_n=a^n-b^n. If a≠ b, H_n=(a^n-b^n)/(a-b); for equal rates, H_n+1(a,a)=(n+1)a^n. Uniform step bounds give nnE_w≤ϵ H_n(a,b), nnOT_w-VU_wq≤η a^n+nn Vϵ H_n(a,b).
Exact mathematical passage
H_n(a,b)=\sum_{j=0}^{n-1}a^jb^{n-1-j},\qquad (a-b)H_n=a^n-b^n.
b^{|w|}Use in source · 1 passage
Search aliases: b^|w|
Use in source · Sufficient_Resolution_Traditional.tex, line 714
ion proves the finite-sum formula: multiplying the old sum by a and adjoining b^n generates every term of the new sum. Multiplication by a-b telescopes, leaving a^n-b^n. Division is used only when the rates are distinct. If they coincide, all n+1 terms of H_n+1 equal a^n, including the zero-rate boundary with the usual a^0=1. For uniform bounds, the heterogeneous budget becomes D_i::w=aD_w+b^|w|ϵ, so induction gives D_w=ϵ H_|w|(a,b). Apply SR-14. The empty word has zero dynamics defect and is ha…
Exact mathematical passage
D_{i::w}=aD_w+b^{|w|}\epsilon
b_1Use in source · 1 passage
Search aliases: b_1
Use in source · Sufficient_Resolution_Traditional.tex, line 1534
[A metric must survive its coordinate change] For K={(x,y,0)} and R=span{(1,0,0)}, take the compatible basis b_1=(1,0,0), b_2=(1,2,0). Its Gram matrix is (1 · 1
Exact mathematical passage
b_1=(1,0,0)
b_2Use in source · 1 passage
Search aliases: b_2
Use in source · Sufficient_Resolution_Traditional.tex, line 1534
[A metric must survive its coordinate change] For K={(x,y,0)} and R=span{(1,0,0)}, take the compatible basis b_1=(1,0,0), b_2=(1,2,0). Its Gram matrix is (1 · 1
Exact mathematical passage
b_2=(1,2,0)
B_{A,\lambda}Use in source · 1 passage
Search aliases: B_A,λ lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 1688
For two operators and parameters, R_A,λ-R_C,μ =R_A,λ(N_C,μ-N_A,λ)R_C,μ. At fixed positive λ, with ϵ=nnA-C, nnB_A,λ-B_C,λ ≤fracϵλ+ frac(nn A+nn C)ϵnn Cλ^2. If A has closed range and canonical inverse B, then B_λ v-Bv=-λ R_λ Bv. For δ A=0, nnv-AB_λ v^2 =nnc(v)^2+nni(v)^2+nnA(B_λ v-Bv)^2.
Exact mathematical passage
\nn{B_{A,\lambda}-B_{C,\lambda}} \le\frac\epsilon\lambda+ \frac{(\nn A+\nn C)\epsilon\nn C}{\lambda^2}.
B_cUse in source · 1 passage
Search aliases: B_c
Use in source · Sufficient_Resolution_Traditional.tex, line 1362
s source sequence Cauchy. Completeness gives a limit g∈ S, and continuity gives the image limit Ag∈Ran A. Hence the range is closed. If bounded B satisfies BA=P_S, then for g∈ S, nn g=nnBAg≤nn BnnAg. The same positive choice 1/(nn B+1) gives coercivity. Finite-dimensional image subspaces are closed in a normed space. Finally A_cg=cg, c>0, has canonical inverse B_cv=v/c. At v=c, data tend to zero with c while the recovered source remains one, ruling out a common constant as cdownarrow0.
Exact mathematical passage
B_cv=v/c
B_{C,\lambda}Use in source · 1 passage
Search aliases: B_C,λ lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 1688
For two operators and parameters, R_A,λ-R_C,μ =R_A,λ(N_C,μ-N_A,λ)R_C,μ. At fixed positive λ, with ϵ=nnA-C, nnB_A,λ-B_C,λ ≤fracϵλ+ frac(nn A+nn C)ϵnn Cλ^2. If A has closed range and canonical inverse B, then B_λ v-Bv=-λ R_λ Bv. For δ A=0, nnv-AB_λ v^2 =nnc(v)^2+nni(v)^2+nnA(B_λ v-Bv)^2.
Exact mathematical passage
\nn{B_{A,\lambda}-B_{C,\lambda}} \le\frac\epsilon\lambda+ \frac{(\nn A+\nn C)\epsilon\nn C}{\lambda^2}.
b_iUse in source · 1 passage
Search aliases: b_i
Use in source · Sufficient_Resolution_Traditional.tex, line 655
Suppose nnT_i≤ a_i, nnU_i≤ b_i, nnδ_i≤ϵ_i. Let A_w,B_w be the products of the respective letter bounds and define D_∅=0, D_i::w=D_wa_i+B_wϵ_i. Then nnT_w≤ A_w, nnU_w≤ B_w, nnE_w≤ D_w, and D_umathbin+ +v=D_vA_u+B_vD_u. If nnO-Vq≤η, then nnOT_w-VU_wq≤η A_w+nn V D_w. For arbitrary initial x∈ X, z∈ Z, nnqT_wx-U_wz≤ D_wnn x+B_wnnqx-z.
Exact mathematical passage
\nn{U_i}\le b_i
B_vUse in source · 1 passage
Search aliases: B_v
Use in source · Sufficient_Resolution_Traditional.tex, line 661
Suppose nnT_i≤ a_i, nnU_i≤ b_i, nnδ_i≤ϵ_i. Let A_w,B_w be the products of the respective letter bounds and define D_∅=0, D_i::w=D_wa_i+B_wϵ_i. Then nnT_w≤ A_w, nnU_w≤ B_w, nnE_w≤ D_w, and D_umathbin+ +v=D_vA_u+B_vD_u. If nnO-Vq≤η, then nnOT_w-VU_wq≤η A_w+nn V D_w. For arbitrary initial x∈ X, z∈ Z, nnqT_wx-U_wz≤ D_wnn x+B_wnnqx-z.
Exact mathematical passage
D_{u\mathbin{+\!+}v}=D_vA_u+B_vD_u.
B_wUse in source · 4 passages
Search aliases: B_w
Use in source · Sufficient_Resolution_Traditional.tex, line 656
Suppose nnT_i≤ a_i, nnU_i≤ b_i, nnδ_i≤ϵ_i. Let A_w,B_w be the products of the respective letter bounds and define D_∅=0, D_i::w=D_wa_i+B_wϵ_i. Then nnT_w≤ A_w, nnU_w≤ B_w, nnE_w≤ D_w, and D_umathbin+ +v=D_vA_u+B_vD_u. If nnO-Vq≤η, then nnOT_w-VU_wq≤η A_w+nn V D_w. For arbitrary initial x∈ X, z∈ Z, nnqT_wx-U_wz≤ D_wnn x+B_wnnqx-z.
Exact mathematical passage
A_w,B_w
B_\lambdaUse in source · 22 passages
Search aliases: B_λ lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 1547
Fix a bounded operator A:G→ V between real Hilbert spaces, without assuming closed range. For λ>0, put N_λ=A^*A+λ I, R_λ=N_λ^-1, B_λ=R_λ A^*. The inverse is constructed from coercivity. It is not supplied as a certificate premise. Here A^* is the actual Hilbert adjoint.
Exact mathematical passage
N_\lambda=A^*A+\lambda I,\qquad R_\lambda=N_\lambda^{-1},\qquad B_\lambda=R_\lambda A^*.
B_\lambda^AUse in source · 1 passage
Search aliases: B_λ^A lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 1856
For arbitrary data v, AB_λ^Av=B_λ^A^*A^*v ⟶ P_clRan Av. Adding data noise e_i→0 preserves this fitted-data limit. If v∈clRan A∖Ran A, fitted data converge to v but the corresponding reconstructed source family has no norm limit in G.
Exact mathematical passage
AB_\lambda^Av=B_\lambda^{A^*}A^*v \longrightarrow P_{\cl\Ran A}v.
B_\lambda^{A^*}Use in source · 1 passage
Search aliases: B_λ^A^* lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 1856
For arbitrary data v, AB_λ^Av=B_λ^A^*A^*v ⟶ P_clRan Av. Adding data noise e_i→0 preserves this fitted-data limit. If v∈clRan A∖Ran A, fitted data converge to v but the corresponding reconstructed source family has no norm limit in G.
Exact mathematical passage
AB_\lambda^Av=B_\lambda^{A^*}A^*v \longrightarrow P_{\cl\Ran A}v.
B_{\lambda_i}Use in source · 2 passages
Search aliases: B_λ_i lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 1788
If λ_i>0 tends to zero and nne_i/sqrtλ_i→0, then B_λ_i(Ag+e_i)⟶ P_(ker A)^⊥g. The same limit holds for computed x_i when additionally fracnn(A^*A+λ_iI)x_i-A^*(Ag+e_i)λ_i→0. For example, λ_i=s_i^2, s_i>0→0, and nne_i≤ s_i^2 give the exact-solve consistency condition.
Exact mathematical passage
B_{\lambda_i}(Ag+e_i)\longrightarrow P_{(\ker A)^\perp}g.
CUse in source · 106 passages
Search aliases: C
Use in source · Sufficient_Resolution_Traditional.tex, line 195
smallHypotheses. Commutative-ring modules; linear C:X→ Z, R:Z→ X with CR=I_Z; arbitrary endomorphisms S,T,U of X.
Exact mathematical passage
C:X\to Z
cUse in source · 50 passages
Search aliases: c
Use in source · Sufficient_Resolution_Traditional.tex, line 188
[Redundant coordinates] For X= R and V_X= R^2, take a(x)=(x,x) and s(u,v)=(u+v)/2. Then E(u,v)=((u+v)/2,(u+v)/2). The encoded scalar transformation x↦ cx has matrix frac c2(1 · 1
Exact mathematical passage
x\mapsto cx
C^*Use in source · 4 passages
Search aliases: C^*
Use in source · Sufficient_Resolution_Traditional.tex, line 1708
Multiply the difference of inverses on the left and right by their normal operators, or insert their inverse laws, to obtain the resolvent identity. SR-38 bounds its norm by nnN_C,μ-N_A,λ/(λμ). At equal parameters, C^*C-A^*A=(C^*-A^*)C+A^*(C-A), whose norm is at most (nn A+nn C)ϵ. Splitting R_AA^*-R_CC^*=R_A(A^*-C^*)+(R_A-R_C)C^* proves the stated bound. It permits rank changes and is not uniform as λ vanishes.
Exact mathematical passage
C^*C-A^*A=(C^*-A^*)C+A^*(C-A)
c^2Use in source · 2 passages
Search aliases: c^2
Use in source · Sufficient_Resolution_Traditional.tex, line 1725
[Exact scalar response and a two-coordinate example] For A_cx=cx, B_λ v=cv/(c^2+λ). At c>0, λ=c^2, the value is v/(2c), attaining the noise bound. For A(g_1,g_2)=(g_1+g_2,0,0), v=(3,4,5), and λ=1, the source is (1,1). Its squared fit is 42=16+25+1, separating obstruction, inconsistency, and regularization image bias. Its objective is 44, whereas the unregularized source (3/2,3/2) has penalized objective 91/2. Positive regularization solves its own stated objective.
Exact mathematical passage
B_\lambda v=cv/(c^2+\lambda)
C_iUse in source · 5 passages
Search aliases: C_i
Use in source · Sufficient_Resolution_Traditional.tex, line 2387
Suppose p∈ S_A, nn p≤ M, and nnC_i-A≤δ_i, nnAp-v_i≤η_i . If x_i lies within κ_i of an exact accepted source for C_i,v_i with allowance ϵ_i=η_i+δ_iM, and η_i,δ_i,κ_i→0, then x_i→ p. The model kernels and ranks may change. For arbitrary g, using p=P_S_Ag yields nnx_i-g→nnP_N_Ag and P_N_Ax_i→0.
Exact mathematical passage
\nn{C_i-A}\le\delta_i,\qquad \nn{Ap-v_i}\le\eta_i .
C_sUse in source · 7 passages
Search aliases: C_s
Use in source · Sufficient_Resolution_Traditional.tex, line 2840
Let U(t)=(t,0) and C_s(x,y)=sy. Then C_sU=0 and nnC_s=|s|. For every real M there is an operator agreeing perfectly with zero on the entire calibration subspace yet satisfying nnC_s>M. The calibration synthesis is not surjective. Thus perfect agreement on a proper source subspace does not control whole-source operator accuracy.
Exact mathematical passage
C_s(x,y)=sy
c_{\eta,\delta}Use in source · 12 passages
Search aliases: c_η,δ eta delta
Use in source · Sufficient_Resolution_Traditional.tex, line 2476
The acceptance rule can avoid an externally supplied number M by using the candidate itself in the uncertainty allowance. Put c_η,δ(x)=η+δnn x . A generalized accepted source is either zero in the small-data branch nn v≤τη, or an actual Tikhonov source whose residual lies between c_η,δ(x) and τ c_η,δ(x). This definition contains C,v,η,δ,τ,x and the positive parameter; it does not contain a true-source cap. Feasibility remains a separate question and retains a quantitative relation between model …
Exact mathematical passage
c_{\eta,\delta}(x)=\eta+\delta\nn x .
DUse in source · 51 passages
Search aliases: D
Use in source · Sufficient_Resolution_Traditional.tex, line 304
Let V be visible state and H hidden state. Over a commutative ring, specify four maps and form the actual full dynamics T(y,z)=(Ay+Bz,Gy+Dz). Put C(y,z)=y, Ry=(y,0), E_n=CT^nR, and M_k=BD^kG. The return M_k contains entry into H, k hidden transitions, and return to V. Its first effect occurs after k+2 full transitions.
Exact mathematical passage
T(y,z)=(Ay+Bz,Gy+Dz).
\mathcal DUse in source · 4 passages
Search aliases: D mathcal
Use in source · Sufficient_Resolution_Traditional.tex, line 146
The first contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
\mathcal K(T)=a_YTs_X,\qquad \mathcal D(K)=s_YKa_X.
dUse in source · 22 passages
Search aliases: d
Use in source · Sufficient_Resolution_Traditional.tex, line 388
If a monic polynomial p annihilates D, vanishing of M_k for k<deg p implies vanishing for all k. If dim H=d<∞, then E_n=A^n (∀ n) iff M_k=0 (0≤ k<d) iff E_n=A^n (0≤ n≤ d+1). The visible space need not be finite-dimensional. The case d=0 is included.
Exact mathematical passage
\dim H=d<\infty
D^2Use in source · 1 passage
Search aliases: D^2
Use in source · Sufficient_Resolution_Traditional.tex, line 906
[A linear continuous memory kernel] Take V= R, H= R^2, A=0, B(z_1,z_2)=z_1, G(y)=(0,y), and D(z_1,z_2)=(z_2,0). Since D^2=0, e^tD(z_1,z_2)=(z_1+tz_2,z_2) and M(t)=tI. The first two prepared jets agree with the reduced generator, while the third derivative differs by I. Changing D to zero leaves both boundary couplings nonzero but makes the memory kernel identically zero. Arbitrary hidden initialization still affects the visible initial derivative.
Exact mathematical passage
D^2=0
D^kUse in source · 7 passages
Search aliases: D^k
Use in source · Sufficient_Resolution_Traditional.tex, line 307
Let V be visible state and H hidden state. Over a commutative ring, specify four maps and form the actual full dynamics T(y,z)=(Ay+Bz,Gy+Dz). Put C(y,z)=y, Ry=(y,0), E_n=CT^nR, and M_k=BD^kG. The return M_k contains entry into H, k hidden transitions, and return to V. Its first effect occurs after k+2 full transitions.
Exact mathematical passage
M_k=BD^kG
D^nUse in source · 6 passages
Search aliases: D^n
Use in source · Sufficient_Resolution_Traditional.tex, line 317
For the trajectory (y_n,z_n)=T^n(y_0,z_0), z_n=D^nz_0+∑_j=0^n-1D^n-1-jGy_j, and y_n+1=Ay_n+BD^nz_0+∑_j=0^n-1M_n-1-jy_j. Prepared evolution (z_0=0) satisfies E_0=I, E_1=A, and E_n+1=AE_n+∑_j=0^n-1M_n-1-jE_j.
Exact mathematical passage
z_n=D^nz_0+\sum_{j=0}^{n-1}D^{n-1-j}Gy_j,
D^{n-1-j}Use in source · 1 passage
Search aliases: D^n-1-j
Use in source · Sufficient_Resolution_Traditional.tex, line 317
For the trajectory (y_n,z_n)=T^n(y_0,z_0), z_n=D^nz_0+∑_j=0^n-1D^n-1-jGy_j, and y_n+1=Ay_n+BD^nz_0+∑_j=0^n-1M_n-1-jy_j. Prepared evolution (z_0=0) satisfies E_0=I, E_1=A, and E_n+1=AE_n+∑_j=0^n-1M_n-1-jE_j.
Exact mathematical passage
z_n=D^nz_0+\sum_{j=0}^{n-1}D^{n-1-j}Gy_j,
D_{i::w}Use in source · 2 passages
Search aliases: D_i::w
Use in source · Sufficient_Resolution_Traditional.tex, line 657
Suppose nnT_i≤ a_i, nnU_i≤ b_i, nnδ_i≤ϵ_i. Let A_w,B_w be the products of the respective letter bounds and define D_∅=0, D_i::w=D_wa_i+B_wϵ_i. Then nnT_w≤ A_w, nnU_w≤ B_w, nnE_w≤ D_w, and D_umathbin+ +v=D_vA_u+B_vD_u. If nnO-Vq≤η, then nnOT_w-VU_wq≤η A_w+nn V D_w. For arbitrary initial x∈ X, z∈ Z, nnqT_wx-U_wz≤ D_wnn x+B_wnnqx-z.
Exact mathematical passage
D_\varnothing=0,\qquad D_{i::w}=D_wa_i+B_w\epsilon_i.
D_uUse in source · 1 passage
Search aliases: D_u
Use in source · Sufficient_Resolution_Traditional.tex, line 661
Suppose nnT_i≤ a_i, nnU_i≤ b_i, nnδ_i≤ϵ_i. Let A_w,B_w be the products of the respective letter bounds and define D_∅=0, D_i::w=D_wa_i+B_wϵ_i. Then nnT_w≤ A_w, nnU_w≤ B_w, nnE_w≤ D_w, and D_umathbin+ +v=D_vA_u+B_vD_u. If nnO-Vq≤η, then nnOT_w-VU_wq≤η A_w+nn V D_w. For arbitrary initial x∈ X, z∈ Z, nnqT_wx-U_wz≤ D_wnn x+B_wnnqx-z.
Exact mathematical passage
D_{u\mathbin{+\!+}v}=D_vA_u+B_vD_u.
D_{u\mathbin{+\!+}v}Use in source · 1 passage
Search aliases: D_umathbin+ +v mathbin
Use in source · Sufficient_Resolution_Traditional.tex, line 661
Suppose nnT_i≤ a_i, nnU_i≤ b_i, nnδ_i≤ϵ_i. Let A_w,B_w be the products of the respective letter bounds and define D_∅=0, D_i::w=D_wa_i+B_wϵ_i. Then nnT_w≤ A_w, nnU_w≤ B_w, nnE_w≤ D_w, and D_umathbin+ +v=D_vA_u+B_vD_u. If nnO-Vq≤η, then nnOT_w-VU_wq≤η A_w+nn V D_w. For arbitrary initial x∈ X, z∈ Z, nnqT_wx-U_wz≤ D_wnn x+B_wnnqx-z.
Exact mathematical passage
D_{u\mathbin{+\!+}v}=D_vA_u+B_vD_u.
D_vUse in source · 1 passage
Search aliases: D_v
Use in source · Sufficient_Resolution_Traditional.tex, line 661
Suppose nnT_i≤ a_i, nnU_i≤ b_i, nnδ_i≤ϵ_i. Let A_w,B_w be the products of the respective letter bounds and define D_∅=0, D_i::w=D_wa_i+B_wϵ_i. Then nnT_w≤ A_w, nnU_w≤ B_w, nnE_w≤ D_w, and D_umathbin+ +v=D_vA_u+B_vD_u. If nnO-Vq≤η, then nnOT_w-VU_wq≤η A_w+nn V D_w. For arbitrary initial x∈ X, z∈ Z, nnqT_wx-U_wz≤ D_wnn x+B_wnnqx-z.
Exact mathematical passage
D_{u\mathbin{+\!+}v}=D_vA_u+B_vD_u.
D_wUse in source · 7 passages
Search aliases: D_w
Use in source · Sufficient_Resolution_Traditional.tex, line 657
Suppose nnT_i≤ a_i, nnU_i≤ b_i, nnδ_i≤ϵ_i. Let A_w,B_w be the products of the respective letter bounds and define D_∅=0, D_i::w=D_wa_i+B_wϵ_i. Then nnT_w≤ A_w, nnU_w≤ B_w, nnE_w≤ D_w, and D_umathbin+ +v=D_vA_u+B_vD_u. If nnO-Vq≤η, then nnOT_w-VU_wq≤η A_w+nn V D_w. For arbitrary initial x∈ X, z∈ Z, nnqT_wx-U_wz≤ D_wnn x+B_wnnqx-z.
Exact mathematical passage
D_\varnothing=0,\qquad D_{i::w}=D_wa_i+B_w\epsilon_i.
D_\varnothingUse in source · 1 passage
Search aliases: D_∅ varnothing
Use in source · Sufficient_Resolution_Traditional.tex, line 657
Suppose nnT_i≤ a_i, nnU_i≤ b_i, nnδ_i≤ϵ_i. Let A_w,B_w be the products of the respective letter bounds and define D_∅=0, D_i::w=D_wa_i+B_wϵ_i. Then nnT_w≤ A_w, nnU_w≤ B_w, nnE_w≤ D_w, and D_umathbin+ +v=D_vA_u+B_vD_u. If nnO-Vq≤η, then nnOT_w-VU_wq≤η A_w+nn V D_w. For arbitrary initial x∈ X, z∈ Z, nnqT_wx-U_wz≤ D_wnn x+B_wnnqx-z.
Exact mathematical passage
D_\varnothing=0,\qquad D_{i::w}=D_wa_i+B_w\epsilon_i.
\degUse in source · 3 passages
Search aliases: deg
Use in source · Sufficient_Resolution_Traditional.tex, line 388
If a monic polynomial p annihilates D, vanishing of M_k for k<deg p implies vanishing for all k. If dim H=d<∞, then E_n=A^n (∀ n) iff M_k=0 (0≤ k<d) iff E_n=A^n (0≤ n≤ d+1). The visible space need not be finite-dimensional. The case d=0 is included.
Exact mathematical passage
k<\deg p
\dimUse in source · 9 passages
Search aliases: dim
Use in source · Sufficient_Resolution_Traditional.tex, line 388
If a monic polynomial p annihilates D, vanishing of M_k for k<deg p implies vanishing for all k. If dim H=d<∞, then E_n=A^n (∀ n) iff M_k=0 (0≤ k<d) iff E_n=A^n (0≤ n≤ d+1). The visible space need not be finite-dimensional. The case d=0 is included.
Exact mathematical passage
\dim H=d<\infty
\downarrowUse in source · 2 passages
Search aliases: downarrow
Use in source · Sufficient_Resolution_Traditional.tex, line 1363
the range is closed. If bounded B satisfies BA=P_S, then for g∈ S, nn g=nnBAg≤nn BnnAg. The same positive choice 1/(nn B+1) gives coercivity. Finite-dimensional image subspaces are closed in a normed space. Finally A_cg=cg, c>0, has canonical inverse B_cv=v/c. At v=c, data tend to zero with c while the recovered source remains one, ruling out a common constant as cdownarrow0.
Exact mathematical passage
c\downarrow0
EUse in source · 77 passages
Search aliases: E
Use in source · Sufficient_Resolution_Traditional.tex, line 145
The first contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
E_X=a_Xs_X
eUse in source · 65 passages
Search aliases: e
Use in source · Sufficient_Resolution_Traditional.tex, line 759
Let L be a bounded real operator on a Banach space. Its flow is the actual operator exponential e^tL. Boundedness allows differentiation of the norm-convergent exponential series and integration of continuous operator curves. Equalities in this section hold for all real t, using oriented interval integrals when t<0.
Exact mathematical passage
e^{tL}
E'Use in source · 2 passages
Search aliases: E'
Use in source · Sufficient_Resolution_Traditional.tex, line 889
SR-18 converts prepared flow equality into prepared power equality and memory flow vanishing into memory moment vanishing. SR-06 equates the corresponding algebraic conditions, and SR-07 supplies the finite moment criterion. Exact prepared flow gives its semigroup law. Conversely, if E is a semigroup, differentiating E(s+t)=E(s)E(t) in s at zero gives E'(t)=AE(t) because E'(0)=CLR=A and E(0)=I. The bounded-generator uniqueness theorem yields E(t)=e^tA.
Exact mathematical passage
E'(t)=AE(t)
e^{(a-b)s}Use in source · 1 passage
Search aliases: e^(a-b)s
Use in source · Sufficient_Resolution_Traditional.tex, line 982
Apply the integral norm inequality and submultiplicativity to SR-21. For 0≤ s≤ t, the integrand norm is at most M_Ae^b(t-s)ϵ M_Le^as. Integration gives the first estimate. Factoring e^bt and integrating e^(a-b)s gives the distinct-rate formula; for equal rates the integrand is constant e^at. The output decomposition is F(t)=(O-Vq)e^tL+VE(t), giving the stated bound. The same argument works with continuous nonnegative envelope functions u,v on the interval, replacing the scalar response by ∫_0^tv…
Exact mathematical passage
e^{(a-b)s}
E^{(k+2)}Use in source · 1 passage
Search aliases: E^(k+2)
Use in source · Sufficient_Resolution_Traditional.tex, line 877
Let E(t)=Ce^tLR. The following are equivalent: E(t)=e^tA (∀ t), M(t)=0 (∀ t), BD^kG=0 (∀ k), E(s+t)=E(s)E(t) (∀ s,t). If dim H=d, each is equivalent to BD^kG=0 for k<d, or to E^(n)(0)=A^n for n≤ d+1. If the first k return moments vanish, E^(k+2)(0)-A^k+2=BD^kG.
Exact mathematical passage
E^{(k+2)}(0)-A^{k+2}=BD^kG.
E^{(n)}Use in source · 2 passages
Search aliases: E^(n)
Use in source · Sufficient_Resolution_Traditional.tex, line 876
Let E(t)=Ce^tLR. The following are equivalent: E(t)=e^tA (∀ t), M(t)=0 (∀ t), BD^kG=0 (∀ k), E(s+t)=E(s)E(t) (∀ s,t). If dim H=d, each is equivalent to BD^kG=0 for k<d, or to E^(n)(0)=A^n for n≤ d+1. If the first k return moments vanish, E^(k+2)(0)-A^k+2=BD^kG.
Exact mathematical passage
E^{(n)}(0)=A^n
e^{(s+t)D}Use in source · 1 passage
Search aliases: e^(s+t)D
Use in source · Sufficient_Resolution_Traditional.tex, line 770
frac ddte^tD=De^tD=e^tDD, e^(s+t)D=e^sDe^tD, and e^-tD is the inverse flow. Every actual curve satisfying z'=Dz+f with continuous f obeys z(t)=e^tDz(0)+∫_0^t e^(t-s)Df(s) ds. The initial value uniquely determines the homogeneous solution; two forced solutions with the same forcing and initial value also agree.
Exact mathematical passage
e^{(s+t)D}=e^{sD}e^{tD}
e^{(s+t)L}Use in source · 1 passage
Search aliases: e^(s+t)L
Use in source · Sufficient_Resolution_Traditional.tex, line 813
For bounded linear observation maps K,J on bounded generator algebras, K(e^tL)=J(e^tD) (∀ t) iff K(L^n)=J(D^n) (∀ n). In particular, Be^tDG=0 for every t exactly when BD^nG=0 for every n. For CR=I, A=CLR, one has Ce^tLR=e^tA (∀ t) iff CL^nR=A^n (∀ n), and, with Q=I-RC, Ce^(s+t)LR-(Ce^sLR)(Ce^tLR)=Ce^sLQe^tLR.
Exact mathematical passage
Ce^{(s+t)L}R-(Ce^{sL}R)(Ce^{tL}R)=Ce^{sL}Qe^{tL}R.
e^{(t-s)A}Use in source · 3 passages
Search aliases: e^(t-s)A
Use in source · Sufficient_Resolution_Traditional.tex, line 928
One has the operator identity E(t)=∫_0^t e^(t-s)Aδ e^sL ds, and δ=0 exactly when E(t)=0 for every t. Also E(s+t)=E(s)e^tL+e^sAE(t). For a second representation p:Z→ W with generator B, δ_pq=pδ_q+δ_pq, E_pq(t)=pE_q(t)+E_p(t)q.
Exact mathematical passage
E(t)=\int_0^t e^{(t-s)A}\delta e^{sL}\,ds,
e^{(t-s)D}Use in source · 5 passages
Search aliases: e^(t-s)D
Use in source · Sufficient_Resolution_Traditional.tex, line 772
frac ddte^tD=De^tD=e^tDD, e^(s+t)D=e^sDe^tD, and e^-tD is the inverse flow. Every actual curve satisfying z'=Dz+f with continuous f obeys z(t)=e^tDz(0)+∫_0^t e^(t-s)Df(s) ds. The initial value uniquely determines the homogeneous solution; two forced solutions with the same forcing and initial value also agree.
Exact mathematical passage
z(t)=e^{tD}z(0)+\int_0^t e^{(t-s)D}f(s)\,ds.
e^{-rt}Use in source · 3 passages
Search aliases: e^-rt
Use in source · Sufficient_Resolution_Traditional.tex, line 999
If nne^tL≤ M_L and nne^tA≤ M_Ae^-rt with r>0, then nnE(t)≤fracM_LM_Aϵr, nnF(t)≤η M_L+nn VfracM_LM_Aϵr. For arbitrary x,z, nnOe^tLx-Ve^tAz · ≤(η M_L+nn VfracM_LM_Aϵr)nn x
Exact mathematical passage
\nn{e^{tA}}\le M_Ae^{-rt}
e^{-tD}Use in source · 1 passage
Search aliases: e^-tD
Use in source · Sufficient_Resolution_Traditional.tex, line 770
frac ddte^tD=De^tD=e^tDD, e^(s+t)D=e^sDe^tD, and e^-tD is the inverse flow. Every actual curve satisfying z'=Dz+f with continuous f obeys z(t)=e^tDz(0)+∫_0^t e^(t-s)Df(s) ds. The initial value uniquely determines the homogeneous solution; two forced solutions with the same forcing and initial value also agree.
Exact mathematical passage
e^{-tD}
e^{as}Use in source · 3 passages
Search aliases: e^as
Use in source · Sufficient_Resolution_Traditional.tex, line 958
smallHypotheses. Real Banach spaces X,Z; bounded autonomous real generators L,A and bounded representation/readouts q,O,V; real rates a,b of either sign; t≥0 and M_L,M_A,ϵ,η≥0. For every s≥0, nne^sL≤ M_Le^as and nne^sA≤ M_Ae^bs; also nnqL-Aq≤ϵ and nnO-Vq≤η.
Exact mathematical passage
\nn{e^{sL}}\le M_Le^{as}
e^{at}Use in source · 4 passages
Search aliases: e^at
Use in source · Sufficient_Resolution_Traditional.tex, line 963
For nnδ≤ϵ, put R(a,b,t)=∫_0^t e^b(t-s)e^as ds =(e^at-e^bt)/(a-b), · a≠ b, te^at, · a=b. Then nnE(t)≤ M_LM_Aϵ R(a,b,t). If nnO-Vq≤η, the output defect F(t)=Oe^tL-Ve^tAq obeys nnF(t)≤η M_Le^at+nn V M_LM_Aϵ R(a,b,t).
Exact mathematical passage
R(a,b,t)=\int_0^t e^{b(t-s)}e^{as}\,ds =\begin{cases}(e^{at}-e^{bt})/(a-b),&a\ne b,\\ te^{at},&a=b.\end{cases}
e^{b(t-s)}Use in source · 2 passages
Search aliases: e^b(t-s)
Use in source · Sufficient_Resolution_Traditional.tex, line 963
For nnδ≤ϵ, put R(a,b,t)=∫_0^t e^b(t-s)e^as ds =(e^at-e^bt)/(a-b), · a≠ b, te^at, · a=b. Then nnE(t)≤ M_LM_Aϵ R(a,b,t). If nnO-Vq≤η, the output defect F(t)=Oe^tL-Ve^tAq obeys nnF(t)≤η M_Le^at+nn V M_LM_Aϵ R(a,b,t).
Exact mathematical passage
R(a,b,t)=\int_0^t e^{b(t-s)}e^{as}\,ds =\begin{cases}(e^{at}-e^{bt})/(a-b),&a\ne b,\\ te^{at},&a=b.\end{cases}
e^{bs}Use in source · 1 passage
Search aliases: e^bs
Use in source · Sufficient_Resolution_Traditional.tex, line 958
smallHypotheses. Real Banach spaces X,Z; bounded autonomous real generators L,A and bounded representation/readouts q,O,V; real rates a,b of either sign; t≥0 and M_L,M_A,ϵ,η≥0. For every s≥0, nne^sL≤ M_Le^as and nne^sA≤ M_Ae^bs; also nnqL-Aq≤ϵ and nnO-Vq≤η.
Exact mathematical passage
\nn{e^{sA}}\le M_Ae^{bs}
e^{bt}Use in source · 3 passages
Search aliases: e^bt
Use in source · Sufficient_Resolution_Traditional.tex, line 963
For nnδ≤ϵ, put R(a,b,t)=∫_0^t e^b(t-s)e^as ds =(e^at-e^bt)/(a-b), · a≠ b, te^at, · a=b. Then nnE(t)≤ M_LM_Aϵ R(a,b,t). If nnO-Vq≤η, the output defect F(t)=Oe^tL-Ve^tAq obeys nnF(t)≤η M_Le^at+nn V M_LM_Aϵ R(a,b,t).
Exact mathematical passage
R(a,b,t)=\int_0^t e^{b(t-s)}e^{as}\,ds =\begin{cases}(e^{at}-e^{bt})/(a-b),&a\ne b,\\ te^{at},&a=b.\end{cases}
e^{sA}Use in source · 3 passages
Search aliases: e^sA
Use in source · Sufficient_Resolution_Traditional.tex, line 932
One has the operator identity E(t)=∫_0^t e^(t-s)Aδ e^sL ds, and δ=0 exactly when E(t)=0 for every t. Also E(s+t)=E(s)e^tL+e^sAE(t). For a second representation p:Z→ W with generator B, δ_pq=pδ_q+δ_pq, E_pq(t)=pE_q(t)+E_p(t)q.
Exact mathematical passage
E(s+t)=E(s)e^{tL}+e^{sA}E(t).
e^{sD}Use in source · 1 passage
Search aliases: e^sD
Use in source · Sufficient_Resolution_Traditional.tex, line 770
frac ddte^tD=De^tD=e^tDD, e^(s+t)D=e^sDe^tD, and e^-tD is the inverse flow. Every actual curve satisfying z'=Dz+f with continuous f obeys z(t)=e^tDz(0)+∫_0^t e^(t-s)Df(s) ds. The initial value uniquely determines the homogeneous solution; two forced solutions with the same forcing and initial value also agree.
Exact mathematical passage
e^{(s+t)D}=e^{sD}e^{tD}
e^{sL}Use in source · 5 passages
Search aliases: e^sL
Use in source · Sufficient_Resolution_Traditional.tex, line 813
For bounded linear observation maps K,J on bounded generator algebras, K(e^tL)=J(e^tD) (∀ t) iff K(L^n)=J(D^n) (∀ n). In particular, Be^tDG=0 for every t exactly when BD^nG=0 for every n. For CR=I, A=CLR, one has Ce^tLR=e^tA (∀ t) iff CL^nR=A^n (∀ n), and, with Q=I-RC, Ce^(s+t)LR-(Ce^sLR)(Ce^tLR)=Ce^sLQe^tLR.
Exact mathematical passage
Ce^{(s+t)L}R-(Ce^{sL}R)(Ce^{tL}R)=Ce^{sL}Qe^{tL}R.
e^{tA}Use in source · 9 passages
Search aliases: e^tA
Use in source · Sufficient_Resolution_Traditional.tex, line 808
For bounded linear observation maps K,J on bounded generator algebras, K(e^tL)=J(e^tD) (∀ t) iff K(L^n)=J(D^n) (∀ n). In particular, Be^tDG=0 for every t exactly when BD^nG=0 for every n. For CR=I, A=CLR, one has Ce^tLR=e^tA (∀ t) iff CL^nR=A^n (∀ n), and, with Q=I-RC, Ce^(s+t)LR-(Ce^sLR)(Ce^tLR)=Ce^sLQe^tLR.
Exact mathematical passage
Ce^{tL}R=e^{tA}\ (\forall t) \iff CL^nR=A^n\ (\forall n),
e^{tD}Use in source · 9 passages
Search aliases: e^tD
Use in source · Sufficient_Resolution_Traditional.tex, line 769
frac ddte^tD=De^tD=e^tDD, e^(s+t)D=e^sDe^tD, and e^-tD is the inverse flow. Every actual curve satisfying z'=Dz+f with continuous f obeys z(t)=e^tDz(0)+∫_0^t e^(t-s)Df(s) ds. The initial value uniquely determines the homogeneous solution; two forced solutions with the same forcing and initial value also agree.
Exact mathematical passage
\frac d{dt}e^{tD}=De^{tD}=e^{tD}D
e^{tL}Use in source · 16 passages
Search aliases: e^tL
Use in source · Sufficient_Resolution_Traditional.tex, line 759
Let L be a bounded real operator on a Banach space. Its flow is the actual operator exponential e^tL. Boundedness allows differentiation of the norm-convergent exponential series and integration of continuous operator curves. Equalities in this section hold for all real t, using oriented interval integrals when t<0.
Exact mathematical passage
e^{tL}
e^uUse in source · 1 passage
Search aliases: e^u
Use in source · Sufficient_Resolution_Traditional.tex, line 1053
The scalar exponential is the actual flow by the homogeneous ODE uniqueness theorem. For a>b, the response formula immediately gives the equality; for a=b, both sides vanish. In the second example, the generator defect is multiplication by ϵ, while the evolution defect is multiplication by e^ϵ t-1. The elementary exponential inequality e^u≥1+u for u≥0 gives the lower bound. Taking t sufficiently large makes it exceed any prescribed real bound. Thus a small generator defect alone does not supply …
Exact mathematical passage
e^u\ge1+u
e^{\epsilon t}Use in source · 2 passages
Search aliases: e^ϵ t epsilon
Use in source · Sufficient_Resolution_Traditional.tex, line 1040
For scalar generators a≥ b, |e^at-e^bt|=(a-b)R(a,b,t), so the response estimate is attained. For every ϵ>0, choosing L=ϵ I, A=0, q=I gives nnδ=ϵ, nnE(t)=e^ϵ t-1≥ϵ t. Its evolution defect exceeds every fixed bound at some nonnegative time.
Exact mathematical passage
\nn\delta=\epsilon,\qquad \nn{E(t)}=e^{\epsilon t}-1\ge\epsilon t.
E_0Use in source · 2 passages
Search aliases: E_0
Use in source · Sufficient_Resolution_Traditional.tex, line 325
For the trajectory (y_n,z_n)=T^n(y_0,z_0), z_n=D^nz_0+∑_j=0^n-1D^n-1-jGy_j, and y_n+1=Ay_n+BD^nz_0+∑_j=0^n-1M_n-1-jy_j. Prepared evolution (z_0=0) satisfies E_0=I, E_1=A, and E_n+1=AE_n+∑_j=0^n-1M_n-1-jE_j.
Exact mathematical passage
E_0=I
E_1Use in source · 3 passages
Search aliases: E_1
Use in source · Sufficient_Resolution_Traditional.tex, line 325
For the trajectory (y_n,z_n)=T^n(y_0,z_0), z_n=D^nz_0+∑_j=0^n-1D^n-1-jGy_j, and y_n+1=Ay_n+BD^nz_0+∑_j=0^n-1M_n-1-jy_j. Prepared evolution (z_0=0) satisfies E_0=I, E_1=A, and E_n+1=AE_n+∑_j=0^n-1M_n-1-jE_j.
Exact mathematical passage
E_1=A
E_2Use in source · 1 passage
Search aliases: E_2
Use in source · Sufficient_Resolution_Traditional.tex, line 451
[Delayed return and disconnected return] On Q^3, set T(y,z_1,z_2)=(z_1,z_2,y) and retain y. Then A=0, M_0=0, M_1=I, and E_1=E_2=0 while E_3=I. Two shallow exact checks cannot certify all depths without the dimension budget. For another system, let A=D=I, B(z_1,z_2)=z_1, and G(y)=(0,y). Both couplings are nonzero but every BD^kG vanishes. Prepared closure is exact, while an arbitrary hidden initial first coordinate affects the next visible state.
Exact mathematical passage
E_1=E_2=0
E_3Use in source · 1 passage
Search aliases: E_3
Use in source · Sufficient_Resolution_Traditional.tex, line 451
[Delayed return and disconnected return] On Q^3, set T(y,z_1,z_2)=(z_1,z_2,y) and retain y. Then A=0, M_0=0, M_1=I, and E_1=E_2=0 while E_3=I. Two shallow exact checks cannot certify all depths without the dimension budget. For another system, let A=D=I, B(z_1,z_2)=z_1, and G(y)=(0,y). Both couplings are nonzero but every BD^kG vanishes. Prepared closure is exact, while an arbitrary hidden initial first coordinate affects the next visible state.
Exact mathematical passage
E_3=I
e_iUse in source · 9 passages
Search aliases: e_i
Use in source · Sufficient_Resolution_Traditional.tex, line 1787
If λ_i>0 tends to zero and nne_i/sqrtλ_i→0, then B_λ_i(Ag+e_i)⟶ P_(ker A)^⊥g. The same limit holds for computed x_i when additionally fracnn(A^*A+λ_iI)x_i-A^*(Ag+e_i)λ_i→0. For example, λ_i=s_i^2, s_i>0→0, and nne_i≤ s_i^2 give the exact-solve consistency condition.
Exact mathematical passage
\nn{e_i}/\sqrt{\lambda_i}\to0
E_{i::w}Use in source · 1 passage
Search aliases: E_i::w
Use in source · Sufficient_Resolution_Traditional.tex, line 624
One has E_∅=0, E_i::w=E_wT_i+U_wδ_i, E_umathbin+ +v=E_vT_u+U_vE_u. For w=[i_1,…,i_n] this yields E_w=∑_j=1^n U_[i_j+1,…,i_n]δ_i_jT_[i_1,…,i_j-1]. The output defect is (O-Vq)T_w+VE_w.
Exact mathematical passage
E_{i::w}=E_wT_i+U_w\delta_i,\qquad E_{u\mathbin{+\!+}v}=E_vT_u+U_vE_u.
E_jUse in source · 2 passages
Search aliases: E_j
Use in source · Sufficient_Resolution_Traditional.tex, line 326
For the trajectory (y_n,z_n)=T^n(y_0,z_0), z_n=D^nz_0+∑_j=0^n-1D^n-1-jGy_j, and y_n+1=Ay_n+BD^nz_0+∑_j=0^n-1M_n-1-jy_j. Prepared evolution (z_0=0) satisfies E_0=I, E_1=A, and E_n+1=AE_n+∑_j=0^n-1M_n-1-jE_j.
Exact mathematical passage
E_{n+1}=AE_n+\sum_{j=0}^{n-1}M_{n-1-j}E_j.
E_{k+1}Use in source · 1 passage
Search aliases: E_k+1
Use in source · Sufficient_Resolution_Traditional.tex, line 369
If all memory maps vanish, the recurrence in SR-05 reduces to E_n+1=AE_n, proving E_n=A^n by induction. Conversely, suppose exact powers hold. Induct on k. In the recurrence for E_k+2, every term except M_kE_0 involves a previously vanishing memory map. Thus E_k+2=AE_k+1+M_k, forcing M_k=0. The same induction restricted to k<N proves the finite-prefix equivalence and, before setting the last term to zero, the first-return identity.
Exact mathematical passage
E_{k+2}=AE_{k+1}+M_k
E_{k+2}Use in source · 3 passages
Search aliases: E_k+2
Use in source · Sufficient_Resolution_Traditional.tex, line 359
The following conditions are equivalent: E_n=A^n (∀ n), M_k=0 (∀ k), E_m+n=E_mE_n (∀ m,n). More precisely, E_n=A^n (n≤ N+1)iff M_k=0 (k<N). If M_j=0 for j<k, then E_k+2-A^k+2=M_k. Closure for every initial hidden state is equivalent to B=0.
Exact mathematical passage
E_{k+2}-A^{k+2}=M_k
E_mUse in source · 1 passage
Search aliases: E_m
Use in source · Sufficient_Resolution_Traditional.tex, line 351
The following conditions are equivalent: E_n=A^n (∀ n), M_k=0 (∀ k), E_m+n=E_mE_n (∀ m,n). More precisely, E_n=A^n (n≤ N+1)iff M_k=0 (k<N). If M_j=0 for j<k, then E_k+2-A^k+2=M_k. Closure for every initial hidden state is equivalent to B=0.
Exact mathematical passage
E_n=A^n\ (\forall n),\qquad M_k=0\ (\forall k),\qquad E_{m+n}=E_mE_n\ (\forall m,n).
E_{m+n}Use in source · 1 passage
Search aliases: E_m+n
Use in source · Sufficient_Resolution_Traditional.tex, line 351
The following conditions are equivalent: E_n=A^n (∀ n), M_k=0 (∀ k), E_m+n=E_mE_n (∀ m,n). More precisely, E_n=A^n (n≤ N+1)iff M_k=0 (k<N). If M_j=0 for j<k, then E_k+2-A^k+2=M_k. Closure for every initial hidden state is equivalent to B=0.
Exact mathematical passage
E_n=A^n\ (\forall n),\qquad M_k=0\ (\forall k),\qquad E_{m+n}=E_mE_n\ (\forall m,n).
E_nUse in source · 8 passages
Search aliases: E_n
Use in source · Sufficient_Resolution_Traditional.tex, line 307
Let V be visible state and H hidden state. Over a commutative ring, specify four maps and form the actual full dynamics T(y,z)=(Ay+Bz,Gy+Dz). Put C(y,z)=y, Ry=(y,0), E_n=CT^nR, and M_k=BD^kG. The return M_k contains entry into H, k hidden transitions, and return to V. Its first effect occurs after k+2 full transitions.
Exact mathematical passage
E_n=CT^nR
E_{n+1}Use in source · 3 passages
Search aliases: E_n+1
Use in source · Sufficient_Resolution_Traditional.tex, line 326
For the trajectory (y_n,z_n)=T^n(y_0,z_0), z_n=D^nz_0+∑_j=0^n-1D^n-1-jGy_j, and y_n+1=Ay_n+BD^nz_0+∑_j=0^n-1M_n-1-jy_j. Prepared evolution (z_0=0) satisfies E_0=I, E_1=A, and E_n+1=AE_n+∑_j=0^n-1M_n-1-jE_j.
Exact mathematical passage
E_{n+1}=AE_n+\sum_{j=0}^{n-1}M_{n-1-j}E_j.
E_pUse in source · 2 passages
Search aliases: E_p
Use in source · Sufficient_Resolution_Traditional.tex, line 936
One has the operator identity E(t)=∫_0^t e^(t-s)Aδ e^sL ds, and δ=0 exactly when E(t)=0 for every t. Also E(s+t)=E(s)e^tL+e^sAE(t). For a second representation p:Z→ W with generator B, δ_pq=pδ_q+δ_pq, E_pq(t)=pE_q(t)+E_p(t)q.
Exact mathematical passage
\delta_{pq}=p\delta_q+\delta_pq,\qquad E_{pq}(t)=pE_q(t)+E_p(t)q.
E_{pq}Use in source · 2 passages
Search aliases: E_pq
Use in source · Sufficient_Resolution_Traditional.tex, line 936
One has the operator identity E(t)=∫_0^t e^(t-s)Aδ e^sL ds, and δ=0 exactly when E(t)=0 for every t. Also E(s+t)=E(s)e^tL+e^sAE(t). For a second representation p:Z→ W with generator B, δ_pq=pδ_q+δ_pq, E_pq(t)=pE_q(t)+E_p(t)q.
Exact mathematical passage
\delta_{pq}=p\delta_q+\delta_pq,\qquad E_{pq}(t)=pE_q(t)+E_p(t)q.
E_qUse in source · 2 passages
Search aliases: E_q
Use in source · Sufficient_Resolution_Traditional.tex, line 936
One has the operator identity E(t)=∫_0^t e^(t-s)Aδ e^sL ds, and δ=0 exactly when E(t)=0 for every t. Also E(s+t)=E(s)e^tL+e^sAE(t). For a second representation p:Z→ W with generator B, δ_pq=pδ_q+δ_pq, E_pq(t)=pE_q(t)+E_p(t)q.
Exact mathematical passage
\delta_{pq}=p\delta_q+\delta_pq,\qquad E_{pq}(t)=pE_q(t)+E_p(t)q.
E_uUse in source · 1 passage
Search aliases: E_u
Use in source · Sufficient_Resolution_Traditional.tex, line 624
One has E_∅=0, E_i::w=E_wT_i+U_wδ_i, E_umathbin+ +v=E_vT_u+U_vE_u. For w=[i_1,…,i_n] this yields E_w=∑_j=1^n U_[i_j+1,…,i_n]δ_i_jT_[i_1,…,i_j-1]. The output defect is (O-Vq)T_w+VE_w.
Exact mathematical passage
E_{i::w}=E_wT_i+U_w\delta_i,\qquad E_{u\mathbin{+\!+}v}=E_vT_u+U_vE_u.
E_{u\mathbin{+\!+}v}Use in source · 1 passage
Search aliases: E_umathbin+ +v mathbin
Use in source · Sufficient_Resolution_Traditional.tex, line 624
One has E_∅=0, E_i::w=E_wT_i+U_wδ_i, E_umathbin+ +v=E_vT_u+U_vE_u. For w=[i_1,…,i_n] this yields E_w=∑_j=1^n U_[i_j+1,…,i_n]δ_i_jT_[i_1,…,i_j-1]. The output defect is (O-Vq)T_w+VE_w.
Exact mathematical passage
E_{i::w}=E_wT_i+U_w\delta_i,\qquad E_{u\mathbin{+\!+}v}=E_vT_u+U_vE_u.
E_vUse in source · 1 passage
Search aliases: E_v
Use in source · Sufficient_Resolution_Traditional.tex, line 624
One has E_∅=0, E_i::w=E_wT_i+U_wδ_i, E_umathbin+ +v=E_vT_u+U_vE_u. For w=[i_1,…,i_n] this yields E_w=∑_j=1^n U_[i_j+1,…,i_n]δ_i_jT_[i_1,…,i_j-1]. The output defect is (O-Vq)T_w+VE_w.
Exact mathematical passage
E_{i::w}=E_wT_i+U_w\delta_i,\qquad E_{u\mathbin{+\!+}v}=E_vT_u+U_vE_u.
E_wUse in source · 9 passages
Search aliases: E_w
Use in source · Sufficient_Resolution_Traditional.tex, line 611
Let q:X→ Z be a proposed representation, with true operations T_i, reduced operations U_i, and readouts O:X→ Y, V:Z→ Y. Define the step and word defects δ_i=qT_i-U_iq, E_w=qT_w-U_wq. Algebraic identities require only the stated commutative-ring structure. Quantitative estimates below use continuous linear maps over a nontrivially normed field and their whole-map operator norms.
Exact mathematical passage
\delta_i=qT_i-U_iq,\qquad E_w=qT_w-U_wq.
E_XUse in source · 10 passages
Search aliases: E_X
Use in source · Sufficient_Resolution_Traditional.tex, line 145
The first contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
E_X=a_Xs_X
E_X^2Use in source · 2 passages
Search aliases: E_X^2
Use in source · Sufficient_Resolution_Traditional.tex, line 156
E_X^2=E_X, and E_Xv=v exactly when v∈Ran a_X. Moreover D K(T)=T, K D(K)=E_YKE_X. Consequently a compatible K is the encoding of a unique map T:X→ Y. Encoding preserves composition. Its identity on resolved coordinates is E_X.
Exact mathematical passage
E_X^2=E_X
E_YUse in source · 3 passages
Search aliases: E_Y
Use in source · Sufficient_Resolution_Traditional.tex, line 149
contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
E_YKE_X=K
E_\lambdaUse in source · 6 passages
Search aliases: E_λ lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 1819
Let E_λ=B_λ A-P_(ker A)^⊥. Always nnE_λ≤1. If Ran A is nonclosed, then nnE_λ=1 for every positive parameter. For a positive family tending to zero along a nontrivial filter, E_λ→0 in operator normiffRan A is closed. Under source coercivity α>0, nnE_λ≤λ/(α^2+λ).
Exact mathematical passage
E_\lambda=B_\lambda A-P_{(\ker A)^\perp}
E_\varnothingUse in source · 1 passage
Search aliases: E_∅ varnothing
Use in source · Sufficient_Resolution_Traditional.tex, line 623
One has E_∅=0, E_i::w=E_wT_i+U_wδ_i, E_umathbin+ +v=E_vT_u+U_vE_u. For w=[i_1,…,i_n] this yields E_w=∑_j=1^n U_[i_j+1,…,i_n]δ_i_jT_[i_1,…,i_j-1]. The output defect is (O-Vq)T_w+VE_w.
Exact mathematical passage
E_\varnothing=0
FUse in source · 26 passages
Search aliases: F
Use in source · Sufficient_Resolution_Traditional.tex, line 242
Commuting presentation maps LA=A'F and δ'L=Mδ induce a map overline L: H→ H' by [v]↦[Lv]. These maps compose and preserve zero classes. Reflection of zero follows when overline L is injective.
Exact mathematical passage
LA=A'F
fUse in source · 16 passages
Search aliases: f
Use in source · Sufficient_Resolution_Traditional.tex, line 771
frac ddte^tD=De^tD=e^tDD, e^(s+t)D=e^sDe^tD, and e^-tD is the inverse flow. Every actual curve satisfying z'=Dz+f with continuous f obeys z(t)=e^tDz(0)+∫_0^t e^(t-s)Df(s) ds. The initial value uniquely determines the homogeneous solution; two forced solutions with the same forcing and initial value also agree.
Exact mathematical passage
z'=Dz+f
f_iUse in source · 1 passage
Search aliases: f_i
Use in source · Sufficient_Resolution_Traditional.tex, line 2739
For recovery assume G,V are real Hilbert spaces, A is a fixed bounded true operator, τ≥0 is fixed, and p∈(ker A)^⊥ is fixed. Retain the whole calibration bounds and the true data bounds nnAp-v_i≤η_i. Put δ_i=(η_rm cal,i+f_i)nn R. If the true data, calibration-truth and model-fit allowances tend to zero, then δ_i→0. Generalized accepted sources using this generated allowance converge to a fixed identifiable true source p. For an arbitrary true source g, apply this with p=P_S_Ag and genuine data b…
Exact mathematical passage
\delta_i=(\eta_{{\rm cal},i}+f_i)\nn R
F_\lambdaUse in source · 4 passages
Search aliases: F_λ lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 1593
For F_λ(v,x)=nnAx-v^2+λnn x^2, F_λ(v,x)=F_λ(v,x_λ) +nnA(x-x_λ)^2+λnnx-x_λ^2. Thus x_λ is the unique minimizer and lies in (ker A)^⊥. The whole-map bounds are nnR_λ≤1/λ, nnB_λ≤frac12sqrtλ, nnAB_λ v≤nn v, nnAB_λ v-v≤nn v. For r=N_λ x-A^*v, one has x-x_λ=R_λ r and nnx-x_λ≤nn r/λ.
Exact mathematical passage
F_\lambda(v,x)=\nn{Ax-v}^2+\lambda\nn x^2
GUse in source · 34 passages
Search aliases: G
Use in source · Sufficient_Resolution_Traditional.tex, line 233
smallHypotheses. Commutative-ring modules and a linear presentation GxrightarrowAVxrightarrowδW with δ A=0; a compatible vector v∈kerδ.
Exact mathematical passage
G\xrightarrow{A}V\xrightarrow{\delta}W
gUse in source · 103 passages
Search aliases: g
Use in source · Sufficient_Resolution_Traditional.tex, line 238
For δ A=0, define K=kerδ, R=Ran A⊆ K, and H=K/R. A compatible v has class zero exactly when v=Ag for some g. All classes vanish exactly when R=K. If A is injective, a zero class has a unique source.
Exact mathematical passage
v=Ag
g^2Use in source · 1 passage
Search aliases: g^2
Use in source · Sufficient_Resolution_Traditional.tex, line 1955
Objective minimality gives r_v(λ)^2+λnnB_λ v^2 ≤nnAg-v^2+λnn g^2. The residual ordering and positive λ imply norm domination. For the source estimate, expand nnx-p^2 and use nn x≤nn p to obtain nnx-p^2≤2ip pp-x. Write p=(p-A^*w)+A^*w. The first pairing is at most nnp-A^*wnnp-x≤2nn pnnp-A^*w; the adjoint identity bounds the second by nn wnnAp-Ax. This yields the displayed estimate. For identifiable p, the density in SR-41 permits w with p-A^*w arbitrarily small (take w=A u from the normal range).…
Exact mathematical passage
r_v(\lambda)^2+\lambda\nn{B_\lambda v}^2 \le\nn{Ag-v}^2+\lambda\nn g^2
g_1Use in source · 2 passages
Search aliases: g_1
Use in source · Sufficient_Resolution_Traditional.tex, line 1371
[Source and image are different optima] Let A(g_1,g_2)=(g_1+g_2,0,0) and v=(3,4,5). The best image is (3,0,0), with the same squared local error 41. The minimum-norm source is (3/2,3/2) of squared norm 9/2. The source (3,0) gives the same best image but has squared norm 9. Their difference (3/2,-3/2) is kernel ambiguity, not local obstruction.
Exact mathematical passage
A(g_1,g_2)=(g_1+g_2,0,0)
g_2Use in source · 2 passages
Search aliases: g_2
Use in source · Sufficient_Resolution_Traditional.tex, line 1371
[Source and image are different optima] Let A(g_1,g_2)=(g_1+g_2,0,0) and v=(3,4,5). The best image is (3,0,0), with the same squared local error 41. The minimum-norm source is (3/2,3/2) of squared norm 9/2. The source (3,0) gives the same best image but has squared norm 9. Their difference (3/2,-3/2) is kernel ambiguity, not local obstruction.
Exact mathematical passage
A(g_1,g_2)=(g_1+g_2,0,0)
g_nUse in source · 3 passages
Search aliases: g_n
Use in source · Sufficient_Resolution_Traditional.tex, line 1354
Closed range gives the bounded constructor and coercivity by SR-29--SR-31. Conversely suppose coercivity holds. If Ag_n converges in V, replace g_n by P_Sg_n without changing its image. Coercivity makes this source sequence Cauchy. Completeness gives a limit g∈ S, and continuity gives the image limit Ag∈Ran A. Hence the range is closed. If bounded B satisfies BA=P_S, then for g∈ S, nn g=nnBAg≤nn BnnAg. The same positive choice 1/(nn B+1) gives coercivity. Finite-dimensional image subspaces are c…
Exact mathematical passage
Ag_n
HUse in source · 32 passages
Search aliases: H
Use in source · Sufficient_Resolution_Traditional.tex, line 302
Let V be visible state and H hidden state. Over a commutative ring, specify four maps and form the actual full dynamics T(y,z)=(Ay+Bz,Gy+Dz). Put C(y,z)=y, Ry=(y,0), E_n=CT^nR, and M_k=BD^kG. The return M_k contains entry into H, k hidden transitions, and return to V. Its first effect occurs after k+2 full transitions.
Exact mathematical passage
H
\mathcal HUse in source · 4 passages
Search aliases: H mathcal
Use in source · Sufficient_Resolution_Traditional.tex, line 238
For δ A=0, define K=kerδ, R=Ran A⊆ K, and H=K/R. A compatible v has class zero exactly when v=Ag for some g. All classes vanish exactly when R=K. If A is injective, a zero class has a unique source.
Exact mathematical passage
\mathcal H=K/R
hUse in source · 20 passages
Search aliases: h
Use in source · Sufficient_Resolution_Traditional.tex, line 594
[A noncommuting minimal model] On Q^3, let O(x,y,h)=x, T_A(x,y,h)=(y,0,h), and T_B(x,y,h)=(0,x,h). The two operations do not commute: on (1,0,0) their two compositions give (1,0,0) and (0,0,0). The map q(x,y,h)=(x,y) is generated by O and OT_A, intertwines both operations, and factors O. It therefore has a depth-one certificate. The hidden coordinate h is invisible to every future word, and the minimum exact state dimension is two. The current observation alone has dimension one and misses the d…
Exact mathematical passage
O(x,y,h)=x
\mathcal H'Use in source · 1 passage
Search aliases: H' mathcal
Use in source · Sufficient_Resolution_Traditional.tex, line 243
Commuting presentation maps LA=A'F and δ'L=Mδ induce a map overline L: H→ H' by [v]↦[Lv]. These maps compose and preserve zero classes. Reflection of zero follows when overline L is injective.
Exact mathematical passage
\overline L:\mathcal H\to\mathcal H'
H_0Use in source · 2 passages
Search aliases: H_0
Use in source · Sufficient_Resolution_Traditional.tex, line 692
For H_0(a,b)=0 and H_n+1(a,b)=aH_n(a,b)+b^n, H_n(a,b)=∑_j=0^n-1a^jb^n-1-j, (a-b)H_n=a^n-b^n. If a≠ b, H_n=(a^n-b^n)/(a-b); for equal rates, H_n+1(a,a)=(n+1)a^n. Uniform step bounds give nnE_w≤ϵ H_n(a,b), nnOT_w-VU_wq≤η a^n+nn Vϵ H_n(a,b).
Exact mathematical passage
H_0(a,b)=0
H_kUse in source · 9 passages
Search aliases: H_k
Use in source · Sufficient_Resolution_Traditional.tex, line 536
Let H_kx=(OT_wx)_|w|≤ k and Z_k=Ran H_k. Then ker H_k=N and X/N≅ Z_k. The formulas U_i(H_kx)=H_kT_ix, V(H_kx)=Ox define actual linear maps on Z_k and preserve every finite future word. This model has minimum exact dimension.
Exact mathematical passage
H_kx=(OT_wx)_{|w|\le k}
H_nUse in source · 4 passages
Search aliases: H_n
Use in source · Sufficient_Resolution_Traditional.tex, line 693
For H_0(a,b)=0 and H_n+1(a,b)=aH_n(a,b)+b^n, H_n(a,b)=∑_j=0^n-1a^jb^n-1-j, (a-b)H_n=a^n-b^n. If a≠ b, H_n=(a^n-b^n)/(a-b); for equal rates, H_n+1(a,a)=(n+1)a^n. Uniform step bounds give nnE_w≤ϵ H_n(a,b), nnOT_w-VU_wq≤η a^n+nn Vϵ H_n(a,b).
Exact mathematical passage
H_{n+1}(a,b)=aH_n(a,b)+b^n
H_{n+1}Use in source · 3 passages
Search aliases: H_n+1
Use in source · Sufficient_Resolution_Traditional.tex, line 693
For H_0(a,b)=0 and H_n+1(a,b)=aH_n(a,b)+b^n, H_n(a,b)=∑_j=0^n-1a^jb^n-1-j, (a-b)H_n=a^n-b^n. If a≠ b, H_n=(a^n-b^n)/(a-b); for equal rates, H_n+1(a,a)=(n+1)a^n. Uniform step bounds give nnE_w≤ϵ H_n(a,b), nnOT_w-VU_wq≤η a^n+nn Vϵ H_n(a,b).
Exact mathematical passage
H_{n+1}(a,b)=aH_n(a,b)+b^n
H_rUse in source · 3 passages
Search aliases: H_r
Use in source · Sufficient_Resolution_Traditional.tex, line 566
For a finite operation alphabet, suppose a proposed representation satisfies q=F H_r, qT_i=U_iq, Vq=O. Then ker q=N, regardless of whether r reaches the a priori horizon. If q is surjective, it gives an equivalence Z≅ X/N preserving the observation and intertwining every operation. Without surjectivity, the equivalence applies to Ran q.
Exact mathematical passage
q=F H_r,\qquad qT_i=U_iq,\qquad Vq=O.
H_{|w|}Use in source · 1 passage
Search aliases: H_|w|
Use in source · Sufficient_Resolution_Traditional.tex, line 715
ld sum by a and adjoining b^n generates every term of the new sum. Multiplication by a-b telescopes, leaving a^n-b^n. Division is used only when the rates are distinct. If they coincide, all n+1 terms of H_n+1 equal a^n, including the zero-rate boundary with the usual a^0=1. For uniform bounds, the heterogeneous budget becomes D_i::w=aD_w+b^|w|ϵ, so induction gives D_w=ϵ H_|w|(a,b). Apply SR-14. The empty word has zero dynamics defect and is handled by H_0=0, with no negative exponent required.
Exact mathematical passage
D_w=\epsilon H_{|w|}(a,b)
IUse in source · 40 passages
Search aliases: I
Use in source · Sufficient_Resolution_Traditional.tex, line 104
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
T_\varnothing=I
iUse in source · 110 passages
Search aliases: i
Use in source · Sufficient_Resolution_Traditional.tex, line 103
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
w=[i_1,\ldots,i_n]
i^2Use in source · 1 passage
Search aliases: i^2
Use in source · Sufficient_Resolution_Traditional.tex, line 1080
The components lie in R, K∩ R^⊥, and K^⊥, respectively, are pairwise orthogonal, and give the unique decomposition v=r(v)+o(v)+i(v). Furthermore, nn v^2=nn r^2+nn o^2+nn i^2, dist(v,R)^2=nn o^2+nn i^2. For any u∈ R, nnv-u^2=dist(v,R)^2+nnr(v)-u^2. Thus r(v) is the unique nearest represented vector.
Exact mathematical passage
\nn v^2=\nn r^2+\nn o^2+\nn i^2,\qquad \dist(v,R)^2=\nn o^2+\nn i^2.
i_1Use in source · 4 passages
Search aliases: i_1
Use in source · Sufficient_Resolution_Traditional.tex, line 103
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
w=[i_1,\ldots,i_n]
I_GUse in source · 2 passages
Search aliases: I_G
Use in source · Sufficient_Resolution_Traditional.tex, line 2679
The final link connects justified calibration measurements with the whole-operator premise used above. It does not infer the calibration truth allowance from the model's own fit. Let K be a real normed calibration coordinate space, G a real normed source space and V a real normed response space. A split frame has bounded synthesis U:K→ G and analysis R:G→ K with UR=I_G. Its coordinates may be redundant: U need not be injective or square. The reconstruction identity requires coverage of every sou…
Exact mathematical passage
UR=I_G
i_jUse in source · 1 passage
Search aliases: i_j
Use in source · Sufficient_Resolution_Traditional.tex, line 629
One has E_∅=0, E_i::w=E_wT_i+U_wδ_i, E_umathbin+ +v=E_vT_u+U_vE_u. For w=[i_1,…,i_n] this yields E_w=∑_j=1^n U_[i_j+1,…,i_n]δ_i_jT_[i_1,…,i_j-1]. The output defect is (O-Vq)T_w+VE_w.
Exact mathematical passage
E_w=\sum_{j=1}^n U_{[i_{j+1},\ldots,i_n]}\delta_{i_j}T_{[i_1,\ldots,i_{j-1}]}.
i_{j+1}Use in source · 1 passage
Search aliases: i_j+1
Use in source · Sufficient_Resolution_Traditional.tex, line 629
One has E_∅=0, E_i::w=E_wT_i+U_wδ_i, E_umathbin+ +v=E_vT_u+U_vE_u. For w=[i_1,…,i_n] this yields E_w=∑_j=1^n U_[i_j+1,…,i_n]δ_i_jT_[i_1,…,i_j-1]. The output defect is (O-Vq)T_w+VE_w.
Exact mathematical passage
E_w=\sum_{j=1}^n U_{[i_{j+1},\ldots,i_n]}\delta_{i_j}T_{[i_1,\ldots,i_{j-1}]}.
i_{j-1}Use in source · 1 passage
Search aliases: i_j-1
Use in source · Sufficient_Resolution_Traditional.tex, line 629
One has E_∅=0, E_i::w=E_wT_i+U_wδ_i, E_umathbin+ +v=E_vT_u+U_vE_u. For w=[i_1,…,i_n] this yields E_w=∑_j=1^n U_[i_j+1,…,i_n]δ_i_jT_[i_1,…,i_j-1]. The output defect is (O-Vq)T_w+VE_w.
Exact mathematical passage
E_w=\sum_{j=1}^n U_{[i_{j+1},\ldots,i_n]}\delta_{i_j}T_{[i_1,\ldots,i_{j-1}]}.
i_nUse in source · 4 passages
Search aliases: i_n
Use in source · Sufficient_Resolution_Traditional.tex, line 103
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
w=[i_1,\ldots,i_n]
I_{\RR^2}Use in source · 1 passage
Search aliases: I_RR^2 I_ R^2 RR mathbb
Use in source · Sufficient_Resolution_Traditional.tex, line 2859
The vector (0,1) is outside the range of U, proving nonsurjectivity. A split reconstruction UR=I_ R^2 would make every source a range vector, so no such reconstruction exists here. Nor can any approximate reconstruction for this synthesis have whole defect below one: the range of UR has second coordinate zero, and therefore (I-UR)(0,1) has second coordinate one and norm at least one. This last observation explains the direction-coverage role of the strict absorption threshold in SR-63.
Exact mathematical passage
UR=I_{\RR^2}
I_XUse in source · 2 passages
Search aliases: I_X
Use in source · Sufficient_Resolution_Traditional.tex, line 144
The first contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
s_Xa_X=I_X
I_YUse in source · 1 passage
Search aliases: I_Y
Use in source · Sufficient_Resolution_Traditional.tex, line 175
The splitting gives E_X^2=a_X(s_Xa_X)s_X=a_Xs_X. If E_Xv=v, then v=a_X(s_Xv); conversely E_Xa_Xx=a_Xx. Substituting the definitions yields s_Y(a_YTs_X)a_X=T, a_Y(s_YKa_X)s_X=E_YKE_X. For compatible K, take T= D(K). If another T' encodes to K, decoding proves T'= D(K). Finally, an intermediate resolution of Y has s_Ya_Y=I_Y, so K(S) K(T)=a_ZS(s_Ya_Y)Ts_X=a_ZSTs_X. Encoding I_X gives a_Xs_X=E_X, which acts as identity on compatible kernels. No inverse on all of V_X has been assumed.
Exact mathematical passage
s_Ya_Y=I_Y
I_ZUse in source · 1 passage
Search aliases: I_Z
Use in source · Sufficient_Resolution_Traditional.tex, line 195
smallHypotheses. Commutative-ring modules; linear C:X→ Z, R:Z→ X with CR=I_Z; arbitrary endomorphisms S,T,U of X.
Exact mathematical passage
CR=I_Z
\iffUse in source · 9 passages
Search aliases: iff
Use in source · Sufficient_Resolution_Traditional.tex, line 252
The chain condition puts the actual range of A inside K. The zero class criterion for a quotient is membership in the submodule divided out, giving [v]=0iff v∈ Riff∃ g,Ag=v. Every compatible class is zero exactly when K⊆ R; the reverse inclusion is already known. Injectivity of A gives uniqueness of any preimage.
Exact mathematical passage
[v]=0\iff v\in R\iff\exists g,Ag=v
\ipUse in source · 10 passages
Search aliases: ip
Use in source · Sufficient_Resolution_Traditional.tex, line 1566
N_λ is a bounded bijection with bounded inverse R_λ. For x_λ=B_λ v, N_λ x_λ=A^*v, A^*(Ax_λ-v)+λ x_λ=0, and nnAx_λ^2+λnnx_λ^2 =ip vAx_λ.
Exact mathematical passage
\nn{Ax_\lambda}^2+\lambda\nn{x_\lambda}^2 =\ip v{Ax_\lambda}.
JUse in source · 3 passages
Search aliases: J
Use in source · Sufficient_Resolution_Traditional.tex, line 801
For bounded linear observation maps K,J on bounded generator algebras, K(e^tL)=J(e^tD) (∀ t) iff K(L^n)=J(D^n) (∀ n). In particular, Be^tDG=0 for every t exactly when BD^nG=0 for every n. For CR=I, A=CLR, one has Ce^tLR=e^tA (∀ t) iff CL^nR=A^n (∀ n), and, with Q=I-RC, Ce^(s+t)LR-(Ce^sLR)(Ce^tLR)=Ce^sLQe^tLR.
Exact mathematical passage
K,J
jUse in source · 9 passages
Search aliases: j
Use in source · Sufficient_Resolution_Traditional.tex, line 317
For the trajectory (y_n,z_n)=T^n(y_0,z_0), z_n=D^nz_0+∑_j=0^n-1D^n-1-jGy_j, and y_n+1=Ay_n+BD^nz_0+∑_j=0^n-1M_n-1-jy_j. Prepared evolution (z_0=0) satisfies E_0=I, E_1=A, and E_n+1=AE_n+∑_j=0^n-1M_n-1-jE_j.
Exact mathematical passage
z_n=D^nz_0+\sum_{j=0}^{n-1}D^{n-1-j}Gy_j,
KUse in source · 64 passages
Search aliases: K
Use in source · Sufficient_Resolution_Traditional.tex, line 146
The first contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
\mathcal K(T)=a_YTs_X,\qquad \mathcal D(K)=s_YKa_X.
\mathcal KUse in source · 3 passages
Search aliases: K mathcal
Use in source · Sufficient_Resolution_Traditional.tex, line 146
The first contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
\mathcal K(T)=a_YTs_X,\qquad \mathcal D(K)=s_YKa_X.
kUse in source · 61 passages
Search aliases: k
Use in source · Sufficient_Resolution_Traditional.tex, line 307
Let V be visible state and H hidden state. Over a commutative ring, specify four maps and form the actual full dynamics T(y,z)=(Ay+Bz,Gy+Dz). Put C(y,z)=y, Ry=(y,0), E_n=CT^nR, and M_k=BD^kG. The return M_k contains entry into H, k hidden transitions, and return to V. Its first effect occurs after k+2 full transitions.
Exact mathematical passage
M_k=BD^kG
k^2Use in source · 3 passages
Search aliases: k^2
Use in source · Sufficient_Resolution_Traditional.tex, line 1287
ABv=P_Rv, so SR-25's candidate identity with u=Ag proves the first formula. Equality to the minimum image error is equivalent to A(g-Bv)=0, yielding exactly the affine family of minimizers. Since Bv∈ S and k∈ N_A, Pythagoras gives nnBv+k^2=nnBv^2+nn k^2. It proves minimum norm and uniqueness. The identity BA=P_S proves the exact recovery criterion. For noisy data, linearity gives B(Ag+e)-g=Be-P_N_Ag. The first term belongs to S and the second to N_A, so they are orthogonal. This last identity do…
Exact mathematical passage
\nn{Bv+k}^2=\nn{Bv}^2+\nn k^2
K^\perpUse in source · 6 passages
Search aliases: K^⊥ perp
Use in source · Sufficient_Resolution_Traditional.tex, line 1077
The components lie in R, K∩ R^⊥, and K^⊥, respectively, are pairwise orthogonal, and give the unique decomposition v=r(v)+o(v)+i(v). Furthermore, nn v^2=nn r^2+nn o^2+nn i^2, dist(v,R)^2=nn o^2+nn i^2. For any u∈ R, nnv-u^2=dist(v,R)^2+nnr(v)-u^2. Thus r(v) is the unique nearest represented vector.
Exact mathematical passage
K^\perp
k_1Use in source · 2 passages
Search aliases: k_1
Use in source · Sufficient_Resolution_Traditional.tex, line 2718
Redundancy causes no logical problem. For instance, let K= R^2, G= R, U(k_1,k_2)=k_1 and Rg=(g,0), with the standard product norm. Then UR=I, nn U=nn R=1, but U has a nontrivial kernel. The frame spans the source even though it has an unused coordinate. A true scalar map A=a and model C=c with measured row Y(k_1,k_2)=yk_1 give |y-a|≤η_rm cal and |y-c|≤ f, hence |c-a|≤η_rm cal+f. The same estimate comes from the general factorization rather than assuming the coordinates are square and invertible.
Exact mathematical passage
U(k_1,k_2)=k_1
k_2Use in source · 2 passages
Search aliases: k_2
Use in source · Sufficient_Resolution_Traditional.tex, line 2718
Redundancy causes no logical problem. For instance, let K= R^2, G= R, U(k_1,k_2)=k_1 and Rg=(g,0), with the standard product norm. Then UR=I, nn U=nn R=1, but U has a nontrivial kernel. The frame spans the source even though it has an unused coordinate. A true scalar map A=a and model C=c with measured row Y(k_1,k_2)=yk_1 give |y-a|≤η_rm cal and |y-c|≤ f, hence |c-a|≤η_rm cal+f. The same estimate comes from the general factorization rather than assuming the coordinates are square and invertible.
Exact mathematical passage
U(k_1,k_2)=k_1
\kerUse in source · 49 passages
Search aliases: ker
Use in source · Sufficient_Resolution_Traditional.tex, line 233
smallHypotheses. Commutative-ring modules and a linear presentation GxrightarrowAVxrightarrowδW with δ A=0; a compatible vector v∈kerδ.
Exact mathematical passage
v\in\ker\delta
lUse in source · 12 passages
Search aliases: l
Use in source · Sufficient_Resolution_Traditional.tex, line 2684
The observed response Y:K→ V is an operator, and the assumptions nnY-AU≤η_rm cal and nnY-CU≤ f are operator-norm enclosures. Individual calibration samples do not automatically give these enclosures. A mathematical or experimental justification of η_rm cal remains a separate input.
Exact mathematical passage
\nn{Y-AU}\le\eta_{\rm cal}
L^nUse in source · 5 passages
Search aliases: L^n
Use in source · Sufficient_Resolution_Traditional.tex, line 802
For bounded linear observation maps K,J on bounded generator algebras, K(e^tL)=J(e^tD) (∀ t) iff K(L^n)=J(D^n) (∀ n). In particular, Be^tDG=0 for every t exactly when BD^nG=0 for every n. For CR=I, A=CLR, one has Ce^tLR=e^tA (∀ t) iff CL^nR=A^n (∀ n), and, with Q=I-RC, Ce^(s+t)LR-(Ce^sLR)(Ce^tLR)=Ce^sLQe^tLR.
Exact mathematical passage
K(e^{tL})=J(e^{tD})\ (\forall t) \iff K(L^n)=J(D^n)\ (\forall n).
MUse in source · 55 passages
Search aliases: M
Use in source · Sufficient_Resolution_Traditional.tex, line 242
Commuting presentation maps LA=A'F and δ'L=Mδ induce a map overline L: H→ H' by [v]↦[Lv]. These maps compose and preserve zero classes. Reflection of zero follows when overline L is injective.
Exact mathematical passage
\delta'L=M\delta
mUse in source · 12 passages
Search aliases: m
Use in source · Sufficient_Resolution_Traditional.tex, line 351
The following conditions are equivalent: E_n=A^n (∀ n), M_k=0 (∀ k), E_m+n=E_mE_n (∀ m,n). More precisely, E_n=A^n (n≤ N+1)iff M_k=0 (k<N). If M_j=0 for j<k, then E_k+2-A^k+2=M_k. Closure for every initial hidden state is equivalent to B=0.
Exact mathematical passage
E_n=A^n\ (\forall n),\qquad M_k=0\ (\forall k),\qquad E_{m+n}=E_mE_n\ (\forall m,n).
M_0Use in source · 1 passage
Search aliases: M_0
Use in source · Sufficient_Resolution_Traditional.tex, line 451
[Delayed return and disconnected return] On Q^3, set T(y,z_1,z_2)=(z_1,z_2,y) and retain y. Then A=0, M_0=0, M_1=I, and E_1=E_2=0 while E_3=I. Two shallow exact checks cannot certify all depths without the dimension budget. For another system, let A=D=I, B(z_1,z_2)=z_1, and G(y)=(0,y). Both couplings are nonzero but every BD^kG vanishes. Prepared closure is exact, while an arbitrary hidden initial first coordinate affects the next visible state.
Exact mathematical passage
M_0=0
M_1Use in source · 1 passage
Search aliases: M_1
Use in source · Sufficient_Resolution_Traditional.tex, line 451
[Delayed return and disconnected return] On Q^3, set T(y,z_1,z_2)=(z_1,z_2,y) and retain y. Then A=0, M_0=0, M_1=I, and E_1=E_2=0 while E_3=I. Two shallow exact checks cannot certify all depths without the dimension budget. For another system, let A=D=I, B(z_1,z_2)=z_1, and G(y)=(0,y). Both couplings are nonzero but every BD^kG vanishes. Prepared closure is exact, while an arbitrary hidden initial first coordinate affects the next visible state.
Exact mathematical passage
M_1=I
M_AUse in source · 8 passages
Search aliases: M_A
Use in source · Sufficient_Resolution_Traditional.tex, line 958
smallHypotheses. Real Banach spaces X,Z; bounded autonomous real generators L,A and bounded representation/readouts q,O,V; real rates a,b of either sign; t≥0 and M_L,M_A,ϵ,η≥0. For every s≥0, nne^sL≤ M_Le^as and nne^sA≤ M_Ae^bs; also nnqL-Aq≤ϵ and nnO-Vq≤η.
Exact mathematical passage
M_L,M_A,\epsilon,\eta\ge0
M_jUse in source · 1 passage
Search aliases: M_j
Use in source · Sufficient_Resolution_Traditional.tex, line 359
The following conditions are equivalent: E_n=A^n (∀ n), M_k=0 (∀ k), E_m+n=E_mE_n (∀ m,n). More precisely, E_n=A^n (n≤ N+1)iff M_k=0 (k<N). If M_j=0 for j<k, then E_k+2-A^k+2=M_k. Closure for every initial hidden state is equivalent to B=0.
Exact mathematical passage
M_j=0
M_kUse in source · 10 passages
Search aliases: M_k
Use in source · Sufficient_Resolution_Traditional.tex, line 307
Let V be visible state and H hidden state. Over a commutative ring, specify four maps and form the actual full dynamics T(y,z)=(Ay+Bz,Gy+Dz). Put C(y,z)=y, Ry=(y,0), E_n=CT^nR, and M_k=BD^kG. The return M_k contains entry into H, k hidden transitions, and return to V. Its first effect occurs after k+2 full transitions.
Exact mathematical passage
M_k=BD^kG
M_LUse in source · 8 passages
Search aliases: M_L
Use in source · Sufficient_Resolution_Traditional.tex, line 958
smallHypotheses. Real Banach spaces X,Z; bounded autonomous real generators L,A and bounded representation/readouts q,O,V; real rates a,b of either sign; t≥0 and M_L,M_A,ϵ,η≥0. For every s≥0, nne^sL≤ M_Le^as and nne^sA≤ M_Ae^bs; also nnqL-Aq≤ϵ and nnO-Vq≤η.
Exact mathematical passage
M_L,M_A,\epsilon,\eta\ge0
M_{n-1-j}Use in source · 2 passages
Search aliases: M_n-1-j
Use in source · Sufficient_Resolution_Traditional.tex, line 321
For the trajectory (y_n,z_n)=T^n(y_0,z_0), z_n=D^nz_0+∑_j=0^n-1D^n-1-jGy_j, and y_n+1=Ay_n+BD^nz_0+∑_j=0^n-1M_n-1-jy_j. Prepared evolution (z_0=0) satisfies E_0=I, E_1=A, and E_n+1=AE_n+∑_j=0^n-1M_n-1-jE_j.
Exact mathematical passage
y_{n+1}=Ay_n+BD^nz_0+\sum_{j=0}^{n-1}M_{n-1-j}y_j.
\mathbinUse in source · 4 passages
Search aliases: mathbin
Use in source · Sufficient_Resolution_Traditional.tex, line 105
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
T_{u\mathbin{+\!+}v}=T_vT_u
\maxUse in source · 2 passages
Search aliases: max
Use in source · Sufficient_Resolution_Traditional.tex, line 2243
Suppose a record supplies λ>0, nonnegative ρ,ξ,κ,m,ω, and nnr_λ(x)^2≤ρ^2, ρ^2≤(λκ)^2, ρ^2≤4λξ^2, max(0,m-ω)^2≤nnAx-v^2≤(m+ω)^2, η+ω+ξ≤ m≤τη-ω-ξ . Then x is near accepted with source budget κ. Families of such records give SR-51 when the genuine data allowances and source budgets tend to zero.
Exact mathematical passage
\max(0,m-\omega)^2\le\nn{Ax-v}^2\le(m+\omega)^2, \qquad \eta+\omega+\xi\le m\le\tau\eta-\omega-\xi .
NUse in source · 71 passages
Search aliases: N
Use in source · Sufficient_Resolution_Traditional.tex, line 272
Let N=⋂_wker(OT_w). It is the largest submodule contained in ker O and invariant under every T_i. The operations and observation descend to X/N, preserving every future observation. In particular, [x]=[y]iff OT_wx=OT_wy for every finite word w. If q:X→ Z satisfies qT_i=U_iq and Vq=O, then ker q⊆ N.
Exact mathematical passage
N=\bigcap_w\ker(OT_w)
nUse in source · 57 passages
Search aliases: n
Use in source · Sufficient_Resolution_Traditional.tex, line 103
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
w=[i_1,\ldots,i_n]
N_0Use in source · 1 passage
Search aliases: N_0
Use in source · Sufficient_Resolution_Traditional.tex, line 474
N_0=ker O and N_r+1=ker O∩⋂_i T_i^-1N_r⊆ N_r. One plateau N_r+1=N_r implies N_r=N. In all cases N_k=N. Thus equality of future observations through word length k is equivalent to equality for every finite word.
Exact mathematical passage
N_0=\ker O
N_AUse in source · 17 passages
Search aliases: N_A
Use in source · Sufficient_Resolution_Traditional.tex, line 1226
Let A:G→ V be bounded between real Hilbert spaces, with actual closed range R=Ran A. Write N_A=ker A and S=N_A^⊥. Closedness is a substantive hypothesis for a bounded exact inverse. It will be removed for regularized recovery in the following sections.
Exact mathematical passage
N_A=\ker A
N_{A,\lambda}Use in source · 2 passages
Search aliases: N_A,λ lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 1683
For two operators and parameters, R_A,λ-R_C,μ =R_A,λ(N_C,μ-N_A,λ)R_C,μ. At fixed positive λ, with ϵ=nnA-C, nnB_A,λ-B_C,λ ≤fracϵλ+ frac(nn A+nn C)ϵnn Cλ^2. If A has closed range and canonical inverse B, then B_λ v-Bv=-λ R_λ Bv. For δ A=0, nnv-AB_λ v^2 =nnc(v)^2+nni(v)^2+nnA(B_λ v-Bv)^2.
Exact mathematical passage
R_{A,\lambda}-R_{C,\mu} =R_{A,\lambda}(N_{C,\mu}-N_{A,\lambda})R_{C,\mu}.
N_A^\perpUse in source · 2 passages
Search aliases: N_A^⊥ perp
Use in source · Sufficient_Resolution_Traditional.tex, line 1226
Let A:G→ V be bounded between real Hilbert spaces, with actual closed range R=Ran A. Write N_A=ker A and S=N_A^⊥. Closedness is a substantive hypothesis for a bounded exact inverse. It will be removed for regularized recovery in the following sections.
Exact mathematical passage
S=N_A^\perp
N_{C,\mu}Use in source · 2 passages
Search aliases: N_C,μ mu
Use in source · Sufficient_Resolution_Traditional.tex, line 1683
For two operators and parameters, R_A,λ-R_C,μ =R_A,λ(N_C,μ-N_A,λ)R_C,μ. At fixed positive λ, with ϵ=nnA-C, nnB_A,λ-B_C,λ ≤fracϵλ+ frac(nn A+nn C)ϵnn Cλ^2. If A has closed range and canonical inverse B, then B_λ v-Bv=-λ R_λ Bv. For δ A=0, nnv-AB_λ v^2 =nnc(v)^2+nni(v)^2+nnA(B_λ v-Bv)^2.
Exact mathematical passage
R_{A,\lambda}-R_{C,\mu} =R_{A,\lambda}(N_{C,\mu}-N_{A,\lambda})R_{C,\mu}.
N_kUse in source · 2 passages
Search aliases: N_k
Use in source · Sufficient_Resolution_Traditional.tex, line 478
N_0=ker O and N_r+1=ker O∩⋂_i T_i^-1N_r⊆ N_r. One plateau N_r+1=N_r implies N_r=N. In all cases N_k=N. Thus equality of future observations through word length k is equivalent to equality for every finite word.
Exact mathematical passage
N_k=N
N_rUse in source · 7 passages
Search aliases: N_r
Use in source · Sufficient_Resolution_Traditional.tex, line 463
Now work over a field and, for dimension statements, assume dim X<∞. The allowed operations need not commute. Define N_r=⋂_|w|≤ rker(OT_w), k=dimker O. A finite horizon means bounded word length. It gives a finite enumeration of constraints only when the operation alphabet is finite.
Exact mathematical passage
N_r=\bigcap_{|w|\le r}\ker(OT_w),\qquad k=\dim\ker O.
N_{r+1}Use in source · 2 passages
Search aliases: N_r+1
Use in source · Sufficient_Resolution_Traditional.tex, line 475
N_0=ker O and N_r+1=ker O∩⋂_i T_i^-1N_r⊆ N_r. One plateau N_r+1=N_r implies N_r=N. In all cases N_k=N. Thus equality of future observations through word length k is equivalent to equality for every finite word.
Exact mathematical passage
N_{r+1}=\ker O\cap\bigcap_i T_i^{-1}N_r\subseteq N_r.
N_\lambdaUse in source · 7 passages
Search aliases: N_λ lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 1547
Fix a bounded operator A:G→ V between real Hilbert spaces, without assuming closed range. For λ>0, put N_λ=A^*A+λ I, R_λ=N_λ^-1, B_λ=R_λ A^*. The inverse is constructed from coercivity. It is not supplied as a certificate premise. Here A^* is the actual Hilbert adjoint.
Exact mathematical passage
N_\lambda=A^*A+\lambda I,\qquad R_\lambda=N_\lambda^{-1},\qquad B_\lambda=R_\lambda A^*.
N_\lambda^{-1}Use in source · 4 passages
Search aliases: N_λ^-1 lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 1547
Fix a bounded operator A:G→ V between real Hilbert spaces, without assuming closed range. For λ>0, put N_λ=A^*A+λ I, R_λ=N_λ^-1, B_λ=R_λ A^*. The inverse is constructed from coercivity. It is not supplied as a certificate premise. Here A^* is the actual Hilbert adjoint.
Exact mathematical passage
N_\lambda=A^*A+\lambda I,\qquad R_\lambda=N_\lambda^{-1},\qquad B_\lambda=R_\lambda A^*.
OUse in source · 42 passages
Search aliases: O
Use in source · Sufficient_Resolution_Traditional.tex, line 268
smallHypotheses. Commutative-ring modules; a linear observation O:X→ Y and an arbitrary family of allowed endomorphisms T_i:X→ X; every finite chronological word is included.
Exact mathematical passage
O:X\to Y
oUse in source · 11 passages
Search aliases: o
Use in source · Sufficient_Resolution_Traditional.tex, line 1066
Let R⊆ K be subspaces of a real inner-product space V, both admitting orthogonal projections. Define the actual maps r(v)=P_Rv, o(v)=(P_K-P_R)v, i(v)=(I-P_K)v. They distinguish realized data, compatible obstruction, and incompatibility. In a globalization presentation use R=Ran A, K=kerδ with δ A=0.
Exact mathematical passage
r(v)=P_Rv,\quad o(v)=(P_K-P_R)v,\quad i(v)=(I-P_K)v.
o^2Use in source · 1 passage
Search aliases: o^2
Use in source · Sufficient_Resolution_Traditional.tex, line 1080
The components lie in R, K∩ R^⊥, and K^⊥, respectively, are pairwise orthogonal, and give the unique decomposition v=r(v)+o(v)+i(v). Furthermore, nn v^2=nn r^2+nn o^2+nn i^2, dist(v,R)^2=nn o^2+nn i^2. For any u∈ R, nnv-u^2=dist(v,R)^2+nnr(v)-u^2. Thus r(v) is the unique nearest represented vector.
Exact mathematical passage
\nn v^2=\nn r^2+\nn o^2+\nn i^2,\qquad \dist(v,R)^2=\nn o^2+\nn i^2.
o_0Use in source · 2 passages
Search aliases: o_0
Use in source · Sufficient_Resolution_Traditional.tex, line 1184
In finite dimension some σ>0 satisfies SR-27. Under δ A=0, ∃ c>0 ∀ v:dist(v,Ran A)≤ cnnδ v iffRan A=kerδ. For projected R_0⊆ R_1⊆ K in a fixed metric, put g(v)=(P_R_1-P_R_0)v. Then nno_0(v)^2=nno_1(v)^2+nng(v)^2, dist(v,R_0)^2=dist(v,R_1)^2+nng(v)^2. No uniform conditioning constant exists across all nonzero scalar compatibility maps.
Exact mathematical passage
\nn{o_0(v)}^2=\nn{o_1(v)}^2+\nn{g(v)}^2, \quad \dist(v,R_0)^2=\dist(v,R_1)^2+\nn{g(v)}^2.
o_1Use in source · 2 passages
Search aliases: o_1
Use in source · Sufficient_Resolution_Traditional.tex, line 1184
In finite dimension some σ>0 satisfies SR-27. Under δ A=0, ∃ c>0 ∀ v:dist(v,Ran A)≤ cnnδ v iffRan A=kerδ. For projected R_0⊆ R_1⊆ K in a fixed metric, put g(v)=(P_R_1-P_R_0)v. Then nno_0(v)^2=nno_1(v)^2+nng(v)^2, dist(v,R_0)^2=dist(v,R_1)^2+nng(v)^2. No uniform conditioning constant exists across all nonzero scalar compatibility maps.
Exact mathematical passage
\nn{o_0(v)}^2=\nn{o_1(v)}^2+\nn{g(v)}^2, \quad \dist(v,R_0)^2=\dist(v,R_1)^2+\nn{g(v)}^2.
\overline LUse in source · 8 passages
Search aliases: overline L overline
Use in source · Sufficient_Resolution_Traditional.tex, line 243
Commuting presentation maps LA=A'F and δ'L=Mδ induce a map overline L: H→ H' by [v]↦[Lv]. These maps compose and preserve zero classes. Reflection of zero follows when overline L is injective.
Exact mathematical passage
\overline L:\mathcal H\to\mathcal H'
P_{(\ker A)^\perp}Use in source · 8 passages
Search aliases: P_(ker A)^⊥ ker perp
Use in source · Sufficient_Resolution_Traditional.tex, line 1637
Write p=P_(ker A)^⊥g, k=P_ker Ag. Then B_λ Ag+λ R_λ g=g, B_λ(Ag+e)-g=-λ R_λ g+B_λ e, and nnB_λ(Ag+e)-g^2 =nn-λ R_λ p+B_λ e^2+nn k^2. If A has source coercivity α>0, this is at most (fracλα^2+λnn p +fracnn e2sqrtλ)^2+nn k^2. For a supplied source condition g=A^*Aw, nnB_λ Ag-g≤λnn w without requiring closed range.
Exact mathematical passage
p=P_{(\ker A)^\perp}g
P_{\cl\Ran A}Use in source · 1 passage
Search aliases: P_clRan A cl Ran operatorname operatorname
Use in source · Sufficient_Resolution_Traditional.tex, line 1856
For arbitrary data v, AB_λ^Av=B_λ^A^*A^*v ⟶ P_clRan Av. Adding data noise e_i→0 preserves this fitted-data limit. If v∈clRan A∖Ran A, fitted data converge to v but the corresponding reconstructed source family has no norm limit in G.
Exact mathematical passage
AB_\lambda^Av=B_\lambda^{A^*}A^*v \longrightarrow P_{\cl\Ran A}v.
P_KUse in source · 7 passages
Search aliases: P_K
Use in source · Sufficient_Resolution_Traditional.tex, line 1066
Let R⊆ K be subspaces of a real inner-product space V, both admitting orthogonal projections. Define the actual maps r(v)=P_Rv, o(v)=(P_K-P_R)v, i(v)=(I-P_K)v. They distinguish realized data, compatible obstruction, and incompatibility. In a globalization presentation use R=Ran A, K=kerδ with δ A=0.
Exact mathematical passage
r(v)=P_Rv,\quad o(v)=(P_K-P_R)v,\quad i(v)=(I-P_K)v.
P_{\ker A}Use in source · 3 passages
Search aliases: P_ker A ker
Use in source · Sufficient_Resolution_Traditional.tex, line 1637
Write p=P_(ker A)^⊥g, k=P_ker Ag. Then B_λ Ag+λ R_λ g=g, B_λ(Ag+e)-g=-λ R_λ g+B_λ e, and nnB_λ(Ag+e)-g^2 =nn-λ R_λ p+B_λ e^2+nn k^2. If A has source coercivity α>0, this is at most (fracλα^2+λnn p +fracnn e2sqrtλ)^2+nn k^2. For a supplied source condition g=A^*Aw, nnB_λ Ag-g≤λnn w without requiring closed range.
Exact mathematical passage
k=P_{\ker A}g
P_{N_A}Use in source · 13 passages
Search aliases: P_N_A
Use in source · Sufficient_Resolution_Traditional.tex, line 1276
For every source g, nnv-Ag^2=dist(v,R)^2+nnA(g-Bv)^2. All least-squares minimizers are Bv+k with k∈ker A. Among them Bv is the unique minimum-norm source. Exact source recovery BAg=g holds exactly for g∈ S. For data Ag+e, nnB(Ag+e)-g^2=nnBe^2+nnP_N_Ag^2.
Exact mathematical passage
\nn{B(Ag+e)-g}^2=\nn{Be}^2+\nn{P_{N_A}g}^2.
P_RUse in source · 18 passages
Search aliases: P_R
Use in source · Sufficient_Resolution_Traditional.tex, line 1066
Let R⊆ K be subspaces of a real inner-product space V, both admitting orthogonal projections. Define the actual maps r(v)=P_Rv, o(v)=(P_K-P_R)v, i(v)=(I-P_K)v. They distinguish realized data, compatible obstruction, and incompatibility. In a globalization presentation use R=Ran A, K=kerδ with δ A=0.
Exact mathematical passage
r(v)=P_Rv,\quad o(v)=(P_K-P_R)v,\quad i(v)=(I-P_K)v.
P_{R_0}Use in source · 2 passages
Search aliases: P_R_0
Use in source · Sufficient_Resolution_Traditional.tex, line 1183
In finite dimension some σ>0 satisfies SR-27. Under δ A=0, ∃ c>0 ∀ v:dist(v,Ran A)≤ cnnδ v iffRan A=kerδ. For projected R_0⊆ R_1⊆ K in a fixed metric, put g(v)=(P_R_1-P_R_0)v. Then nno_0(v)^2=nno_1(v)^2+nng(v)^2, dist(v,R_0)^2=dist(v,R_1)^2+nng(v)^2. No uniform conditioning constant exists across all nonzero scalar compatibility maps.
Exact mathematical passage
g(v)=(P_{R_1}-P_{R_0})v
P_{R_1}Use in source · 2 passages
Search aliases: P_R_1
Use in source · Sufficient_Resolution_Traditional.tex, line 1183
In finite dimension some σ>0 satisfies SR-27. Under δ A=0, ∃ c>0 ∀ v:dist(v,Ran A)≤ cnnδ v iffRan A=kerδ. For projected R_0⊆ R_1⊆ K in a fixed metric, put g(v)=(P_R_1-P_R_0)v. Then nno_0(v)^2=nno_1(v)^2+nng(v)^2, dist(v,R_0)^2=dist(v,R_1)^2+nng(v)^2. No uniform conditioning constant exists across all nonzero scalar compatibility maps.
Exact mathematical passage
g(v)=(P_{R_1}-P_{R_0})v
P_SUse in source · 10 passages
Search aliases: P_S
Use in source · Sufficient_Resolution_Traditional.tex, line 1240
Restriction A_S:S→ R is a bounded linear bijection with bounded inverse. Define B=ι_S A_S^-1P_R. Then AB=P_R, BA=P_S, ABA=A, and BAB=B. Its two projections are symmetric, ker B=R^⊥, and Ran B=S. The best-image identity AB=P_R and the condition Ran B⊆ S uniquely determine B.
Exact mathematical passage
BA=P_S
P_{S_A}Use in source · 7 passages
Search aliases: P_S_A
Use in source · Sufficient_Resolution_Traditional.tex, line 2187
Call x near accepted with budget κ when some exact accepted source y satisfies nnx-y≤κ. If nnAg-v_i≤η_i, η_i→0, and near-accepted x_i have κ_i→0, then x_i⟶ P_S_Ag, nnx_i-g⟶nnP_N_Ag. At every index nnP_N_Ax_i≤κ_i; therefore the computed true-kernel component tends to zero. No separate parameter schedule is assumed.
Exact mathematical passage
x_i\longrightarrow P_{S_A}g,\qquad \nn{x_i-g}\longrightarrow\nn{P_{N_A}g}.
\primeUse in source · 1 passage
Search aliases: prime
Use in source · Sufficient_Resolution_Traditional.tex, line 765
smallHypotheses. Real Banach space; bounded autonomous D; continuous forcing f; differentiable actual z with z^prime=Dz+f at every real time.
Exact mathematical passage
{}^{\prime}
QUse in source · 18 passages
Search aliases: Q
Use in source · Sufficient_Resolution_Traditional.tex, line 199
Set P=RC, Q=I-P, Φ(T)=CTR, and Δ(S,T)=Φ(ST)-Φ(S)Φ(T). Then Δ(S,T)=CSQTR, and associativity gives Δ(S,T)Φ(U)+Δ(ST,U) =Φ(S)Δ(T,U)+Δ(S,TU). For two successive split compressions, Δ_21(S,T)=Φ_2(Δ_1(S,T)) +Δ_2(Φ_1(S),Φ_1(T)).
Exact mathematical passage
Q=I-P
qUse in source · 67 passages
Search aliases: q
Use in source · Sufficient_Resolution_Traditional.tex, line 278
Let N=⋂_wker(OT_w). It is the largest submodule contained in ker O and invariant under every T_i. The operations and observation descend to X/N, preserving every future observation. In particular, [x]=[y]iff OT_wx=OT_wy for every finite word w. If q:X→ Z satisfies qT_i=U_iq and Vq=O, then ker q⊆ N.
Exact mathematical passage
q:X\to Z
q_nUse in source · 3 passages
Search aliases: q_n
Use in source · Sufficient_Resolution_Traditional.tex, line 399
Divide X^n by the monic p to obtain a remainder q_n of degree below deg p. Evaluation at D gives D^n=q_n(D) because p(D)=0. Writing q_n(D) as a finite linear combination of the powers D^k below deg p shows BD^nG=0. For finite-dimensional H, Cayley--Hamilton provides the monic characteristic polynomial of D with degree d. SR-06 then yields both displayed equivalences. In dimension zero, all maps through H vanish and the empty list of memory tests is sufficient. If exact closure fails, the contrap…
Exact mathematical passage
q_n
\QQUse in source · 2 passages
Search aliases: QQ Q mathbb
Use in source · Sufficient_Resolution_Traditional.tex, line 450
[Delayed return and disconnected return] On Q^3, set T(y,z_1,z_2)=(z_1,z_2,y) and retain y. Then A=0, M_0=0, M_1=I, and E_1=E_2=0 while E_3=I. Two shallow exact checks cannot certify all depths without the dimension budget. For another system, let A=D=I, B(z_1,z_2)=z_1, and G(y)=(0,y). Both couplings are nonzero but every BD^kG vanishes. Prepared closure is exact, while an arbitrary hidden initial first coordinate affects the next visible state.
Exact mathematical passage
\QQ^3
\QQ^3Use in source · 2 passages
Search aliases: QQ^3 Q^3 QQ mathbb
Use in source · Sufficient_Resolution_Traditional.tex, line 450
[Delayed return and disconnected return] On Q^3, set T(y,z_1,z_2)=(z_1,z_2,y) and retain y. Then A=0, M_0=0, M_1=I, and E_1=E_2=0 while E_3=I. Two shallow exact checks cannot certify all depths without the dimension budget. For another system, let A=D=I, B(z_1,z_2)=z_1, and G(y)=(0,y). Both couplings are nonzero but every BD^kG vanishes. Prepared closure is exact, while an arbitrary hidden initial first coordinate affects the next visible state.
Exact mathematical passage
\QQ^3
RUse in source · 144 passages
Search aliases: R
Use in source · Sufficient_Resolution_Traditional.tex, line 195
smallHypotheses. Commutative-ring modules; linear C:X→ Z, R:Z→ X with CR=I_Z; arbitrary endomorphisms S,T,U of X.
Exact mathematical passage
R:Z\to X
rUse in source · 58 passages
Search aliases: r
Use in source · Sufficient_Resolution_Traditional.tex, line 463
Now work over a field and, for dimension statements, assume dim X<∞. The allowed operations need not commute. Define N_r=⋂_|w|≤ rker(OT_w), k=dimker O. A finite horizon means bounded word length. It gives a finite enumeration of constraints only when the operation alphabet is finite.
Exact mathematical passage
N_r=\bigcap_{|w|\le r}\ker(OT_w),\qquad k=\dim\ker O.
R'Use in source · 1 passage
Search aliases: R'
Use in source · Sufficient_Resolution_Traditional.tex, line 257
For transport, if v∈ K, then δ'Lv=Mδ v=0. If v=Ag, then Lv=A'Fg lies in R'. Hence L descends to the actual quotients. Evaluating on representatives proves the composition law. A linear map sends zero to zero. If its induced quotient map is injective, overline L[v]=0=overline L[0] implies [v]=0. The commuting squares alone provide no such injectivity and do not assert that either obstruction quotient vanishes.
Exact mathematical passage
R'
r^2Use in source · 2 passages
Search aliases: r^2
Use in source · Sufficient_Resolution_Traditional.tex, line 1080
The components lie in R, K∩ R^⊥, and K^⊥, respectively, are pairwise orthogonal, and give the unique decomposition v=r(v)+o(v)+i(v). Furthermore, nn v^2=nn r^2+nn o^2+nn i^2, dist(v,R)^2=nn o^2+nn i^2. For any u∈ R, nnv-u^2=dist(v,R)^2+nnr(v)-u^2. Thus r(v) is the unique nearest represented vector.
Exact mathematical passage
\nn v^2=\nn r^2+\nn o^2+\nn i^2,\qquad \dist(v,R)^2=\nn o^2+\nn i^2.
r^AUse in source · 1 passage
Search aliases: r^A
Use in source · Sufficient_Resolution_Traditional.tex, line 2369
s cancel. Insert and subtract A^*Cx to obtain the displayed identity. Adjoint isometry gives nnA^*-C^*=nnA-C≤δ and nnA^*=nn A. Submultiplicativity therefore bounds its two terms by δnnCx-v and nn A δnn x. The triangle inequality gives nn A≤nn C+δ, yielding a bound written in terms of available-model quantities and the allowance. Finally apply nnr^A≤nnr^C+nnr^A-r^C.
Exact mathematical passage
\nn{r^A}\le\nn{r^C}+\nn{r^A-r^C}
r^A_\lambdaUse in source · 2 passages
Search aliases: r^A_λ lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 2347
For the normal residuals formed with the same parameter, r^A_λ(x)-r^C_λ(x) =(A^*-C^*)(Cx-v)+A^*(A-C)x . If nnC-A≤δ, then nnr^A_λ(x)-r^C_λ(x) ≤δ(nnCx-v+(nn C+δ)nn x). Hence a computed-model normal allowance nnr^C_λ(x)≤ρ gives the corresponding true normal allowance by adding this expression.
Exact mathematical passage
r^A_\lambda(x)-r^C_\lambda(x) =(A^*-C^*)(Cx-v)+A^*(A-C)x .
r^CUse in source · 1 passage
Search aliases: r^C
Use in source · Sufficient_Resolution_Traditional.tex, line 2369
s cancel. Insert and subtract A^*Cx to obtain the displayed identity. Adjoint isometry gives nnA^*-C^*=nnA-C≤δ and nnA^*=nn A. Submultiplicativity therefore bounds its two terms by δnnCx-v and nn A δnn x. The triangle inequality gives nn A≤nn C+δ, yielding a bound written in terms of available-model quantities and the allowance. Finally apply nnr^A≤nnr^C+nnr^A-r^C.
Exact mathematical passage
\nn{r^A}\le\nn{r^C}+\nn{r^A-r^C}
r^C_\lambdaUse in source · 3 passages
Search aliases: r^C_λ lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 2347
For the normal residuals formed with the same parameter, r^A_λ(x)-r^C_λ(x) =(A^*-C^*)(Cx-v)+A^*(A-C)x . If nnC-A≤δ, then nnr^A_λ(x)-r^C_λ(x) ≤δ(nnCx-v+(nn C+δ)nn x). Hence a computed-model normal allowance nnr^C_λ(x)≤ρ gives the corresponding true normal allowance by adding this expression.
Exact mathematical passage
r^A_\lambda(x)-r^C_\lambda(x) =(A^*-C^*)(Cx-v)+A^*(A-C)x .
R^\perpUse in source · 5 passages
Search aliases: R^⊥ perp
Use in source · Sufficient_Resolution_Traditional.tex, line 1077
The components lie in R, K∩ R^⊥, and K^⊥, respectively, are pairwise orthogonal, and give the unique decomposition v=r(v)+o(v)+i(v). Furthermore, nn v^2=nn r^2+nn o^2+nn i^2, dist(v,R)^2=nn o^2+nn i^2. For any u∈ R, nnv-u^2=dist(v,R)^2+nnr(v)-u^2. Thus r(v) is the unique nearest represented vector.
Exact mathematical passage
K\cap R^\perp
R_0Use in source · 5 passages
Search aliases: R_0
Use in source · Sufficient_Resolution_Traditional.tex, line 1182
In finite dimension some σ>0 satisfies SR-27. Under δ A=0, ∃ c>0 ∀ v:dist(v,Ran A)≤ cnnδ v iffRan A=kerδ. For projected R_0⊆ R_1⊆ K in a fixed metric, put g(v)=(P_R_1-P_R_0)v. Then nno_0(v)^2=nno_1(v)^2+nng(v)^2, dist(v,R_0)^2=dist(v,R_1)^2+nng(v)^2. No uniform conditioning constant exists across all nonzero scalar compatibility maps.
Exact mathematical passage
R_0\subseteq R_1\subseteq K
R_1Use in source · 5 passages
Search aliases: R_1
Use in source · Sufficient_Resolution_Traditional.tex, line 1182
In finite dimension some σ>0 satisfies SR-27. Under δ A=0, ∃ c>0 ∀ v:dist(v,Ran A)≤ cnnδ v iffRan A=kerδ. For projected R_0⊆ R_1⊆ K in a fixed metric, put g(v)=(P_R_1-P_R_0)v. Then nno_0(v)^2=nno_1(v)^2+nng(v)^2, dist(v,R_0)^2=dist(v,R_1)^2+nng(v)^2. No uniform conditioning constant exists across all nonzero scalar compatibility maps.
Exact mathematical passage
R_0\subseteq R_1\subseteq K
R_AUse in source · 1 passage
Search aliases: R_A
Use in source · Sufficient_Resolution_Traditional.tex, line 1710
Multiply the difference of inverses on the left and right by their normal operators, or insert their inverse laws, to obtain the resolvent identity. SR-38 bounds its norm by nnN_C,μ-N_A,λ/(λμ). At equal parameters, C^*C-A^*A=(C^*-A^*)C+A^*(C-A), whose norm is at most (nn A+nn C)ϵ. Splitting R_AA^*-R_CC^*=R_A(A^*-C^*)+(R_A-R_C)C^* proves the stated bound. It permits rank changes and is not uniform as λ vanishes.
Exact mathematical passage
R_AA^*-R_CC^*=R_A(A^*-C^*)+(R_A-R_C)C^*
R_{A,\lambda}Use in source · 1 passage
Search aliases: R_A,λ lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 1683
For two operators and parameters, R_A,λ-R_C,μ =R_A,λ(N_C,μ-N_A,λ)R_C,μ. At fixed positive λ, with ϵ=nnA-C, nnB_A,λ-B_C,λ ≤fracϵλ+ frac(nn A+nn C)ϵnn Cλ^2. If A has closed range and canonical inverse B, then B_λ v-Bv=-λ R_λ Bv. For δ A=0, nnv-AB_λ v^2 =nnc(v)^2+nni(v)^2+nnA(B_λ v-Bv)^2.
Exact mathematical passage
R_{A,\lambda}-R_{C,\mu} =R_{A,\lambda}(N_{C,\mu}-N_{A,\lambda})R_{C,\mu}.
R_CUse in source · 1 passage
Search aliases: R_C
Use in source · Sufficient_Resolution_Traditional.tex, line 1710
Multiply the difference of inverses on the left and right by their normal operators, or insert their inverse laws, to obtain the resolvent identity. SR-38 bounds its norm by nnN_C,μ-N_A,λ/(λμ). At equal parameters, C^*C-A^*A=(C^*-A^*)C+A^*(C-A), whose norm is at most (nn A+nn C)ϵ. Splitting R_AA^*-R_CC^*=R_A(A^*-C^*)+(R_A-R_C)C^* proves the stated bound. It permits rank changes and is not uniform as λ vanishes.
Exact mathematical passage
R_AA^*-R_CC^*=R_A(A^*-C^*)+(R_A-R_C)C^*
R_{C,\mu}Use in source · 1 passage
Search aliases: R_C,μ mu
Use in source · Sufficient_Resolution_Traditional.tex, line 1683
For two operators and parameters, R_A,λ-R_C,μ =R_A,λ(N_C,μ-N_A,λ)R_C,μ. At fixed positive λ, with ϵ=nnA-C, nnB_A,λ-B_C,λ ≤fracϵλ+ frac(nn A+nn C)ϵnn Cλ^2. If A has closed range and canonical inverse B, then B_λ v-Bv=-λ R_λ Bv. For δ A=0, nnv-AB_λ v^2 =nnc(v)^2+nni(v)^2+nnA(B_λ v-Bv)^2.
Exact mathematical passage
R_{A,\lambda}-R_{C,\mu} =R_{A,\lambda}(N_{C,\mu}-N_{A,\lambda})R_{C,\mu}.
R_EUse in source · 6 passages
Search aliases: R_E
Use in source · Sufficient_Resolution_Traditional.tex, line 2789
For split frames (U_E,R_E):Jrightleftarrows K and (U_F,R_F):Krightleftarrows G, the composite is split, with analysis R_ER_F and nnR_ER_F≤nnR_EnnR_F. The corresponding calibration allowance is at most (η_rm cal+f)nnR_EnnR_F.
Exact mathematical passage
(U_E,R_E):J\rightleftarrows K
R_FUse in source · 6 passages
Search aliases: R_F
Use in source · Sufficient_Resolution_Traditional.tex, line 2790
For split frames (U_E,R_E):Jrightleftarrows K and (U_F,R_F):Krightleftarrows G, the composite is split, with analysis R_ER_F and nnR_ER_F≤nnR_EnnR_F. The corresponding calibration allowance is at most (η_rm cal+f)nnR_EnnR_F.
Exact mathematical passage
(U_F,R_F):K\rightleftarrows G
r_vUse in source · 9 passages
Search aliases: r_v
Use in source · Sufficient_Resolution_Traditional.tex, line 1888
Still fix a known bounded A between real Hilbert spaces. Define the actual residual curve on positive parameters r_v(λ)=nnAB_λ v-v. Its parameter is mathematical data, not an assumed noise estimate. Let η be a justified true-data allowance and τ>1 a band multiplier. A total rule must include a small-data branch.
Exact mathematical passage
r_v(\lambda)=\nn{AB_\lambda v-v}
R_\lambdaUse in source · 13 passages
Search aliases: R_λ lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 1547
Fix a bounded operator A:G→ V between real Hilbert spaces, without assuming closed range. For λ>0, put N_λ=A^*A+λ I, R_λ=N_λ^-1, B_λ=R_λ A^*. The inverse is constructed from coercivity. It is not supplied as a certificate premise. Here A^* is the actual Hilbert adjoint.
Exact mathematical passage
N_\lambda=A^*A+\lambda I,\qquad R_\lambda=N_\lambda^{-1},\qquad B_\lambda=R_\lambda A^*.
r_\lambdaUse in source · 11 passages
Search aliases: r_λ lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 2058
Throughout this section A:G→ V is bounded between real Hilbert spaces, N_A=ker A, S_A=N_A^⊥, and λ>0. Write N_λ=A^*A+λ I, x_λ=N_λ^-1A^*v, r_λ(x)=N_λ x-A^*v . The inverse is the actual positive-parameter constructor already established. The vector r_λ(x) is the normal-equation residual of the computed source x. Its size measures source correction and image correction on different scales. A fitting residual nnAx-v alone does not supply either correction.
Exact mathematical passage
N_\lambda=A^*A+\lambda I,\qquad x_\lambda=N_\lambda^{-1}A^*v,\qquad r_\lambda(x)=N_\lambda x-A^*v .
\RanUse in source · 33 passages
Search aliases: Ran operatorname
Use in source · Sufficient_Resolution_Traditional.tex, line 156
E_X^2=E_X, and E_Xv=v exactly when v∈Ran a_X. Moreover D K(T)=T, K D(K)=E_YKE_X. Consequently a compatible K is the encoding of a unique map T:X→ Y. Encoding preserves composition. Its identity on resolved coordinates is E_X.
Exact mathematical passage
v\in\Ran a_X
\rightleftarrowsUse in source · 2 passages
Search aliases: rightleftarrows
Use in source · Sufficient_Resolution_Traditional.tex, line 2789
For split frames (U_E,R_E):Jrightleftarrows K and (U_F,R_F):Krightleftarrows G, the composite is split, with analysis R_ER_F and nnR_ER_F≤nnR_EnnR_F. The corresponding calibration allowance is at most (η_rm cal+f)nnR_EnnR_F.
Exact mathematical passage
(U_E,R_E):J\rightleftarrows K
\rmUse in source · 12 passages
Search aliases: rm
Use in source · Sufficient_Resolution_Traditional.tex, line 2684
The observed response Y:K→ V is an operator, and the assumptions nnY-AU≤η_rm cal and nnY-CU≤ f are operator-norm enclosures. Individual calibration samples do not automatically give these enclosures. A mathematical or experimental justification of η_rm cal remains a separate input.
Exact mathematical passage
\nn{Y-AU}\le\eta_{\rm cal}
\RRUse in source · 7 passages
Search aliases: RR R mathbb
Use in source · Sufficient_Resolution_Traditional.tex, line 186
[Redundant coordinates] For X= R and V_X= R^2, take a(x)=(x,x) and s(u,v)=(u+v)/2. Then E(u,v)=((u+v)/2,(u+v)/2). The encoded scalar transformation x↦ cx has matrix frac c2(1 · 1
Exact mathematical passage
X=\RR
\RR^2Use in source · 4 passages
Search aliases: RR^2 R^2 RR mathbb
Use in source · Sufficient_Resolution_Traditional.tex, line 186
[Redundant coordinates] For X= R and V_X= R^2, take a(x)=(x,x) and s(u,v)=(u+v)/2. Then E(u,v)=((u+v)/2,(u+v)/2). The encoded scalar transformation x↦ cx has matrix frac c2(1 · 1
Exact mathematical passage
V_X=\RR^2
SUse in source · 58 passages
Search aliases: S
Use in source · Sufficient_Resolution_Traditional.tex, line 102
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
STx=S(Tx)
sUse in source · 61 passages
Search aliases: s
Use in source · Sufficient_Resolution_Traditional.tex, line 143
The first contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
s_X:V_X\to X
s_0Use in source · 3 passages
Search aliases: s_0
Use in source · Sufficient_Resolution_Traditional.tex, line 1498
Range enlargement induces a surjective contraction F:K/R_0→ K/R_1. For the old minimum representative s_0(z), nn z^2=nnFz^2+nn(P_R_1-P_R_0)s_0(z)^2. Without a closedness or projection hypothesis, a general quotient instead has nn[v]=0iff v∈cl R, [v]=0iff v∈ R. Thus its seminorm detects zero exactly when R is closed. Nonclosed range can leave a nonzero class of zero seminorm.
Exact mathematical passage
s_0(z)
S_AUse in source · 13 passages
Search aliases: S_A
Use in source · Sufficient_Resolution_Traditional.tex, line 2057
Throughout this section A:G→ V is bounded between real Hilbert spaces, N_A=ker A, S_A=N_A^⊥, and λ>0. Write N_λ=A^*A+λ I, x_λ=N_λ^-1A^*v, r_λ(x)=N_λ x-A^*v . The inverse is the actual positive-parameter constructor already established. The vector r_λ(x) is the normal-equation residual of the computed source x. Its size measures source correction and image correction on different scales. A fitting residual nnAx-v alone does not supply either correction.
Exact mathematical passage
S_A=N_A^\perp
s_iUse in source · 2 passages
Search aliases: s_i
Use in source · Sufficient_Resolution_Traditional.tex, line 1795
If λ_i>0 tends to zero and nne_i/sqrtλ_i→0, then B_λ_i(Ag+e_i)⟶ P_(ker A)^⊥g. The same limit holds for computed x_i when additionally fracnn(A^*A+λ_iI)x_i-A^*(Ag+e_i)λ_i→0. For example, λ_i=s_i^2, s_i>0→0, and nne_i≤ s_i^2 give the exact-solve consistency condition.
Exact mathematical passage
s_i>0\to0
s_i^2Use in source · 2 passages
Search aliases: s_i^2
Use in source · Sufficient_Resolution_Traditional.tex, line 1795
If λ_i>0 tends to zero and nne_i/sqrtλ_i→0, then B_λ_i(Ag+e_i)⟶ P_(ker A)^⊥g. The same limit holds for computed x_i when additionally fracnn(A^*A+λ_iI)x_i-A^*(Ag+e_i)λ_i→0. For example, λ_i=s_i^2, s_i>0→0, and nne_i≤ s_i^2 give the exact-solve consistency condition.
Exact mathematical passage
\lambda_i=s_i^2
s_XUse in source · 9 passages
Search aliases: s_X
Use in source · Sufficient_Resolution_Traditional.tex, line 143
The first contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
s_X:V_X\to X
s_YUse in source · 4 passages
Search aliases: s_Y
Use in source · Sufficient_Resolution_Traditional.tex, line 146
The first contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
\mathcal K(T)=a_YTs_X,\qquad \mathcal D(K)=s_YKa_X.
\spanofUse in source · 1 passage
Search aliases: spanof span operatorname
Use in source · Sufficient_Resolution_Traditional.tex, line 1533
[A metric must survive its coordinate change] For K={(x,y,0)} and R=span{(1,0,0)}, take the compatible basis b_1=(1,0,0), b_2=(1,2,0). Its Gram matrix is (1 · 1
Exact mathematical passage
R=\spanof\{(1,0,0)\}
TUse in source · 88 passages
Search aliases: T
Use in source · Sufficient_Resolution_Traditional.tex, line 102
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
STx=S(Tx)
tUse in source · 71 passages
Search aliases: t
Use in source · Sufficient_Resolution_Traditional.tex, line 759
Let L be a bounded real operator on a Banach space. Its flow is the actual operator exponential e^tL. Boundedness allows differentiation of the norm-convergent exponential series and integration of continuous operator curves. Equalities in this section hold for all real t, using oriented interval integrals when t<0.
Exact mathematical passage
e^{tL}
T'Use in source · 2 passages
Search aliases: T'
Use in source · Sufficient_Resolution_Traditional.tex, line 173
The splitting gives E_X^2=a_X(s_Xa_X)s_X=a_Xs_X. If E_Xv=v, then v=a_X(s_Xv); conversely E_Xa_Xx=a_Xx. Substituting the definitions yields s_Y(a_YTs_X)a_X=T, a_Y(s_YKa_X)s_X=E_YKE_X. For compatible K, take T= D(K). If another T' encodes to K, decoding proves T'= D(K). Finally, an intermediate resolution of Y has s_Ya_Y=I_Y, so K(S) K(T)=a_ZS(s_Ya_Y)Ts_X=a_ZSTs_X. Encoding I_X gives a_Xs_X=E_X, which acts as identity on compatible kernels. No inverse on all of V_X has been assumed.
Exact mathematical passage
T'
T^nUse in source · 4 passages
Search aliases: T^n
Use in source · Sufficient_Resolution_Traditional.tex, line 307
Let V be visible state and H hidden state. Over a commutative ring, specify four maps and form the actual full dynamics T(y,z)=(Ay+Bz,Gy+Dz). Put C(y,z)=y, Ry=(y,0), E_n=CT^nR, and M_k=BD^kG. The return M_k contains entry into H, k hidden transitions, and return to V. Its first effect occurs after k+2 full transitions.
Exact mathematical passage
E_n=CT^nR
t^nUse in source · 1 passage
Search aliases: t^n
Use in source · Sufficient_Resolution_Traditional.tex, line 824
The nth derivative at zero of e^tL is L^n. Bounded linear maps commute with these derivatives, so equality for all times implies equality of every observed power. Conversely bounded linear maps commute with the sum of the norm-convergent exponential series: K(e^tL)=∑_n(t^n/n!)K(L^n). Equality term by term gives equality of the sums. Sandwiching and compression are bounded linear maps on the corresponding operator spaces, so they give the specializations. For the last identity, use the flow law a…
Exact mathematical passage
K(e^{tL})=\sum_n(t^n/n!)K(L^n)
T_{[i_1,\ldots,i_{j-1}]}Use in source · 1 passage
Search aliases: T_[i_1,…,i_j-1] ldots
Use in source · Sufficient_Resolution_Traditional.tex, line 629
One has E_∅=0, E_i::w=E_wT_i+U_wδ_i, E_umathbin+ +v=E_vT_u+U_vE_u. For w=[i_1,…,i_n] this yields E_w=∑_j=1^n U_[i_j+1,…,i_n]δ_i_jT_[i_1,…,i_j-1]. The output defect is (O-Vq)T_w+VE_w.
Exact mathematical passage
E_w=\sum_{j=1}^n U_{[i_{j+1},\ldots,i_n]}\delta_{i_j}T_{[i_1,\ldots,i_{j-1}]}.
T_AUse in source · 3 passages
Search aliases: T_A
Use in source · Sufficient_Resolution_Traditional.tex, line 595
[A noncommuting minimal model] On Q^3, let O(x,y,h)=x, T_A(x,y,h)=(y,0,h), and T_B(x,y,h)=(0,x,h). The two operations do not commute: on (1,0,0) their two compositions give (1,0,0) and (0,0,0). The map q(x,y,h)=(x,y) is generated by O and OT_A, intertwines both operations, and factors O. It therefore has a depth-one certificate. The hidden coordinate h is invisible to every future word, and the minimum exact state dimension is two. The current observation alone has dimension one and misses the d…
Exact mathematical passage
T_A(x,y,h)=(y,0,h)
T_BUse in source · 1 passage
Search aliases: T_B
Use in source · Sufficient_Resolution_Traditional.tex, line 595
[A noncommuting minimal model] On Q^3, let O(x,y,h)=x, T_A(x,y,h)=(y,0,h), and T_B(x,y,h)=(0,x,h). The two operations do not commute: on (1,0,0) their two compositions give (1,0,0) and (0,0,0). The map q(x,y,h)=(x,y) is generated by O and OT_A, intertwines both operations, and factors O. It therefore has a depth-one certificate. The hidden coordinate h is invisible to every future word, and the minimum exact state dimension is two. The current observation alone has dimension one and misses the d…
Exact mathematical passage
T_B(x,y,h)=(0,x,h)
T_iUse in source · 19 passages
Search aliases: T_i
Use in source · Sufficient_Resolution_Traditional.tex, line 268
smallHypotheses. Commutative-ring modules; a linear observation O:X→ Y and an arbitrary family of allowed endomorphisms T_i:X→ X; every finite chronological word is included.
Exact mathematical passage
T_i:X\to X
T_{i::w}Use in source · 1 passage
Search aliases: T_i::w
Use in source · Sufficient_Resolution_Traditional.tex, line 284
An intersection of kernels is a submodule. If x∈ N, then OT_wT_ix=OT_i::wx=0, so T_ix∈ N. If an invariant submodule L⊆ker O contains x, induction on word length gives T_wx∈ L; therefore x∈ N. This establishes maximality.
Exact mathematical passage
OT_wT_ix=OT_{i::w}x=0
T_i^{-1}Use in source · 1 passage
Search aliases: T_i^-1
Use in source · Sufficient_Resolution_Traditional.tex, line 475
N_0=ker O and N_r+1=ker O∩⋂_i T_i^-1N_r⊆ N_r. One plateau N_r+1=N_r implies N_r=N. In all cases N_k=N. Thus equality of future observations through word length k is equivalent to equality for every finite word.
Exact mathematical passage
N_{r+1}=\ker O\cap\bigcap_i T_i^{-1}N_r\subseteq N_r.
T_{i_1}Use in source · 1 passage
Search aliases: T_i_1
Use in source · Sufficient_Resolution_Traditional.tex, line 104
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
T_w=T_{i_n}\cdots T_{i_1}
T_{i_n}Use in source · 1 passage
Search aliases: T_i_n
Use in source · Sufficient_Resolution_Traditional.tex, line 104
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
T_w=T_{i_n}\cdots T_{i_1}
T_uUse in source · 4 passages
Search aliases: T_u
Use in source · Sufficient_Resolution_Traditional.tex, line 105
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
T_{u\mathbin{+\!+}v}=T_vT_u
T_{u\mathbin{+\!+}v}Use in source · 2 passages
Search aliases: T_umathbin+ +v mathbin
Use in source · Sufficient_Resolution_Traditional.tex, line 105
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
T_{u\mathbin{+\!+}v}=T_vT_u
T_vUse in source · 2 passages
Search aliases: T_v
Use in source · Sufficient_Resolution_Traditional.tex, line 105
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
T_{u\mathbin{+\!+}v}=T_vT_u
T_wUse in source · 24 passages
Search aliases: T_w
Use in source · Sufficient_Resolution_Traditional.tex, line 104
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
T_w=T_{i_n}\cdots T_{i_1}
T_\varnothingUse in source · 1 passage
Search aliases: T_∅ varnothing
Use in source · Sufficient_Resolution_Traditional.tex, line 104
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
T_\varnothing=I
UUse in source · 67 passages
Search aliases: U
Use in source · Sufficient_Resolution_Traditional.tex, line 195
smallHypotheses. Commutative-ring modules; linear C:X→ Z, R:Z→ X with CR=I_Z; arbitrary endomorphisms S,T,U of X.
Exact mathematical passage
S,T,U
uUse in source · 29 passages
Search aliases: u
Use in source · Sufficient_Resolution_Traditional.tex, line 104
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
u
u^2Use in source · 3 passages
Search aliases: u^2
Use in source · Sufficient_Resolution_Traditional.tex, line 1622
For y=R_λ r, its energy gives λnn y^2≤ip ry≤nn rnn y, hence the resolvent bound. For x_λ, set u=nnAx_λ. The energy gives u^2+λnnx_λ^2≤nn v u. Completing the square in u yields 4λnnx_λ^2≤nn v^2 and also u≤nn v. Comparing the minimum objective with F_λ(v,0)=nn v^2 bounds the fitting residual. Subtract the exact normal equation from the computed one and apply the resolvent for the correction identity.
Exact mathematical passage
u^2+\lambda\nn{x_\lambda}^2\le\nn v\,u
U_{[i_{j+1},\ldots,i_n]}Use in source · 1 passage
Search aliases: U_[i_j+1,…,i_n] ldots
Use in source · Sufficient_Resolution_Traditional.tex, line 629
One has E_∅=0, E_i::w=E_wT_i+U_wδ_i, E_umathbin+ +v=E_vT_u+U_vE_u. For w=[i_1,…,i_n] this yields E_w=∑_j=1^n U_[i_j+1,…,i_n]δ_i_jT_[i_1,…,i_j-1]. The output defect is (O-Vq)T_w+VE_w.
Exact mathematical passage
E_w=\sum_{j=1}^n U_{[i_{j+1},\ldots,i_n]}\delta_{i_j}T_{[i_1,\ldots,i_{j-1}]}.
U_EUse in source · 3 passages
Search aliases: U_E
Use in source · Sufficient_Resolution_Traditional.tex, line 2789
For split frames (U_E,R_E):Jrightleftarrows K and (U_F,R_F):Krightleftarrows G, the composite is split, with analysis R_ER_F and nnR_ER_F≤nnR_EnnR_F. The corresponding calibration allowance is at most (η_rm cal+f)nnR_EnnR_F.
Exact mathematical passage
(U_E,R_E):J\rightleftarrows K
U_FUse in source · 3 passages
Search aliases: U_F
Use in source · Sufficient_Resolution_Traditional.tex, line 2790
For split frames (U_E,R_E):Jrightleftarrows K and (U_F,R_F):Krightleftarrows G, the composite is split, with analysis R_ER_F and nnR_ER_F≤nnR_EnnR_F. The corresponding calibration allowance is at most (η_rm cal+f)nnR_EnnR_F.
Exact mathematical passage
(U_F,R_F):K\rightleftarrows G
U_iUse in source · 11 passages
Search aliases: U_i
Use in source · Sufficient_Resolution_Traditional.tex, line 278
Let N=⋂_wker(OT_w). It is the largest submodule contained in ker O and invariant under every T_i. The operations and observation descend to X/N, preserving every future observation. In particular, [x]=[y]iff OT_wx=OT_wy for every finite word w. If q:X→ Z satisfies qT_i=U_iq and Vq=O, then ker q⊆ N.
Exact mathematical passage
qT_i=U_iq
U_vUse in source · 2 passages
Search aliases: U_v
Use in source · Sufficient_Resolution_Traditional.tex, line 624
One has E_∅=0, E_i::w=E_wT_i+U_wδ_i, E_umathbin+ +v=E_vT_u+U_vE_u. For w=[i_1,…,i_n] this yields E_w=∑_j=1^n U_[i_j+1,…,i_n]δ_i_jT_[i_1,…,i_j-1]. The output defect is (O-Vq)T_w+VE_w.
Exact mathematical passage
E_{i::w}=E_wT_i+U_w\delta_i,\qquad E_{u\mathbin{+\!+}v}=E_vT_u+U_vE_u.
U_wUse in source · 18 passages
Search aliases: U_w
Use in source · Sufficient_Resolution_Traditional.tex, line 292
Invariance defines quotient operations [x]↦[T_ix]; containment in ker O defines [x]↦ Ox. Induction gives their all-word fidelity. Equality [x]=[y] means x-y∈ N, which is precisely equality of the displayed observations by linearity. In an exact competing model, induction gives qT_w=U_wq and hence OT_w=VU_wq. If qx=0, every future observation of x is zero. Thus a competing model can retain extra distinctions, but cannot erase any distinction visible to the specified future task.
Exact mathematical passage
qT_w=U_wq
VUse in source · 43 passages
Search aliases: V
Use in source · Sufficient_Resolution_Traditional.tex, line 143
The first contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
a_X:X\to V_X
v^2Use in source · 6 passages
Search aliases: v^2
Use in source · Sufficient_Resolution_Traditional.tex, line 1080
The components lie in R, K∩ R^⊥, and K^⊥, respectively, are pairwise orthogonal, and give the unique decomposition v=r(v)+o(v)+i(v). Furthermore, nn v^2=nn r^2+nn o^2+nn i^2, dist(v,R)^2=nn o^2+nn i^2. For any u∈ R, nnv-u^2=dist(v,R)^2+nnr(v)-u^2. Thus r(v) is the unique nearest represented vector.
Exact mathematical passage
\nn v^2=\nn r^2+\nn o^2+\nn i^2,\qquad \dist(v,R)^2=\nn o^2+\nn i^2.
v_iUse in source · 8 passages
Search aliases: v_i
Use in source · Sufficient_Resolution_Traditional.tex, line 1972
smallHypotheses. Known bounded A:G→ V between real Hilbert spaces and a fixed true source g; a valid pointwise allowance nnAg-v≤η. Accepted-source consistency uses a fixed τ≥1 and genuine bounds nnAg-v_i≤η_i with η_i→0. Total selector existence additionally requires η>0 and τ>1.
Exact mathematical passage
\nn{Ag-v_i}\le\eta_i
V_XUse in source · 5 passages
Search aliases: V_X
Use in source · Sufficient_Resolution_Traditional.tex, line 143
The first contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
a_X:X\to V_X
\VertUse in source · 19 passages
Search aliases: Vert
Use in source · Sufficient_Resolution_Traditional.tex, line 651
smallHypotheses. Normed spaces over nontrivially normed field; bounded maps; nonnegative letter bounds a_i,b_i,epsilon_i and eta≥VertO-VqVert.
Exact mathematical passage
\ge\Vert
WUse in source · 2 passages
Search aliases: W
Use in source · Sufficient_Resolution_Traditional.tex, line 233
smallHypotheses. Commutative-ring modules and a linear presentation GxrightarrowAVxrightarrowδW with δ A=0; a compatible vector v∈kerδ.
Exact mathematical passage
G\xrightarrow{A}V\xrightarrow{\delta}W
wUse in source · 75 passages
Search aliases: w
Use in source · Sufficient_Resolution_Traditional.tex, line 103
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
w=[i_1,\ldots,i_n]
\widehat rUse in source · 3 passages
Search aliases: widehat r widehat
Use in source · Sufficient_Resolution_Traditional.tex, line 2277
If the normal residual itself is evaluated approximately as widehat r, the distinct assumptions nnr_λ(x)-widehat r≤ν and nnwidehat r≤widehatρ give the allowance widehatρ+ν. Substituting that sum into both correction tests retains the evaluation error rather than silently identifying it with zero.
Exact mathematical passage
\widehat r
\widehat\rhoUse in source · 2 passages
Search aliases: widehatρ widehat rho
Use in source · Sufficient_Resolution_Traditional.tex, line 2280
If the normal residual itself is evaluated approximately as widehat r, the distinct assumptions nnr_λ(x)-widehat r≤ν and nnwidehat r≤widehatρ give the allowance widehatρ+ν. Substituting that sum into both correction tests retains the evaluation error rather than silently identifying it with zero.
Exact mathematical passage
\nn{\widehat r}\le\widehat\rho
XUse in source · 52 passages
Search aliases: X
Use in source · Sufficient_Resolution_Traditional.tex, line 143
The first contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
a_X:X\to V_X
xUse in source · 206 passages
Search aliases: x
Use in source · Sufficient_Resolution_Traditional.tex, line 102
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
STx=S(Tx)
x^2Use in source · 2 passages
Search aliases: x^2
Use in source · Sufficient_Resolution_Traditional.tex, line 1576
The bilinear form b(x,y)=ipAxAy+λip xy is bounded and satisfies b(x,x)≥λnn x^2. Lax--Milgram gives a continuous linear equivalence from the source to itself under the Riesz identification. The adjoint identity identifies its map with A^*A+λ I. The two inverse laws give an actual resolvent. Applying one inverse law to A^*v yields the normal equation. Taking its inner product with x_λ and using ipA^*vx_λ=ip vAx_λ gives the energy identity. Strictly positive λ supplies coercivity even if A has a no…
Exact mathematical passage
b(x,x)\ge\lambda\nn x^2
X^nUse in source · 1 passage
Search aliases: X^n
Use in source · Sufficient_Resolution_Traditional.tex, line 399
Divide X^n by the monic p to obtain a remainder q_n of degree below deg p. Evaluation at D gives D^n=q_n(D) because p(D)=0. Writing q_n(D) as a finite linear combination of the powers D^k below deg p shows BD^nG=0. For finite-dimensional H, Cayley--Hamilton provides the monic characteristic polynomial of D with degree d. SR-06 then yields both displayed equivalences. In dimension zero, all maps through H vanish and the empty list of memory tests is sufficient. If exact closure fails, the contrap…
Exact mathematical passage
X^n
x_iUse in source · 19 passages
Search aliases: x_i
Use in source · Sufficient_Resolution_Traditional.tex, line 1791
If λ_i>0 tends to zero and nne_i/sqrtλ_i→0, then B_λ_i(Ag+e_i)⟶ P_(ker A)^⊥g. The same limit holds for computed x_i when additionally fracnn(A^*A+λ_iI)x_i-A^*(Ag+e_i)λ_i→0. For example, λ_i=s_i^2, s_i>0→0, and nne_i≤ s_i^2 give the exact-solve consistency condition.
Exact mathematical passage
x_i
x_\lambdaUse in source · 36 passages
Search aliases: x_λ lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 1560
N_λ is a bounded bijection with bounded inverse R_λ. For x_λ=B_λ v, N_λ x_λ=A^*v, A^*(Ax_λ-v)+λ x_λ=0, and nnAx_λ^2+λnnx_λ^2 =ip vAx_λ.
Exact mathematical passage
x_\lambda=B_\lambda v
x_\muUse in source · 2 passages
Search aliases: x_μ mu
Use in source · Sufficient_Resolution_Traditional.tex, line 1913
Compare the minimum objectives at 0<λ<μ using each minimizer as a competitor for the other. Adding the two inequalities gives (μ-λ)(nnx_μ^2-nnx_λ^2)≤0, so source norms are antitone. Substitution back into the smaller-parameter comparison shows residual monotonicity. If the two residuals were equal, both norms and objective values would be equal, and unique minimization would give x_λ=x_μ. Subtracting their normal equations gives (μ-λ)x_λ=0, hence x_λ=0 and A^*v=0. This proves strictness and the …
Exact mathematical passage
(\mu-\lambda)(\nn{x_\mu}^2-\nn{x_\lambda}^2)\le0
\xrightarrowUse in source · 1 passage
Search aliases: xrightarrow
Use in source · Sufficient_Resolution_Traditional.tex, line 233
smallHypotheses. Commutative-ring modules and a linear presentation GxrightarrowAVxrightarrowδW with δ A=0; a compatible vector v∈kerδ.
Exact mathematical passage
G\xrightarrow{A}V\xrightarrow{\delta}W
YUse in source · 19 passages
Search aliases: Y
Use in source · Sufficient_Resolution_Traditional.tex, line 145
The first contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
Y
yUse in source · 78 passages
Search aliases: y
Use in source · Sufficient_Resolution_Traditional.tex, line 275
Let N=⋂_wker(OT_w). It is the largest submodule contained in ker O and invariant under every T_i. The operations and observation descend to X/N, preserving every future observation. In particular, [x]=[y]iff OT_wx=OT_wy for every finite word w. If q:X→ Z satisfies qT_i=U_iq and Vq=O, then ker q⊆ N.
Exact mathematical passage
[x]=[y]\iff OT_wx=OT_wy\quad\text{for every finite word }w.
y'Use in source · 2 passages
Search aliases: y'
Use in source · Sufficient_Resolution_Traditional.tex, line 845
For L(y,z)=(Ay+Bz,Gy+Dz) and (y(t),z(t))=e^tL(y_0,z_0), z(t)=e^tDz_0+∫_0^t e^(t-s)DGy(s) ds. Writing M(t)=Be^tDG, the visible dynamics satisfy y'(t)=Ay(t)+Be^tDz_0+∫_0^t M(t-s)y(s) ds. Prepared initialization removes the term involving z_0.
Exact mathematical passage
y'(t)=Ay(t)+Be^{tD}z_0+\int_0^t M(t-s)y(s)\,ds.
y^2Use in source · 2 passages
Search aliases: y^2
Use in source · Sufficient_Resolution_Traditional.tex, line 1620
For y=R_λ r, its energy gives λnn y^2≤ip ry≤nn rnn y, hence the resolvent bound. For x_λ, set u=nnAx_λ. The energy gives u^2+λnnx_λ^2≤nn v u. Completing the square in u yields 4λnnx_λ^2≤nn v^2 and also u≤nn v. Comparing the minimum objective with F_λ(v,0)=nn v^2 bounds the fitting residual. Subtract the exact normal equation from the computed one and apply the resolvent for the correction identity.
Exact mathematical passage
\lambda\nn y^2\le\ip ry\le\nn r\nn y
y_0Use in source · 6 passages
Search aliases: y_0
Use in source · Sufficient_Resolution_Traditional.tex, line 316
For the trajectory (y_n,z_n)=T^n(y_0,z_0), z_n=D^nz_0+∑_j=0^n-1D^n-1-jGy_j, and y_n+1=Ay_n+BD^nz_0+∑_j=0^n-1M_n-1-jy_j. Prepared evolution (z_0=0) satisfies E_0=I, E_1=A, and E_n+1=AE_n+∑_j=0^n-1M_n-1-jE_j.
Exact mathematical passage
(y_n,z_n)=T^n(y_0,z_0)
y_iUse in source · 17 passages
Search aliases: y_i
Use in source · Sufficient_Resolution_Traditional.tex, line 2197
Choose accepted comparison sources y_i from the definition. Exact accepted consistency gives y_i→ p=P_S_Ag, while nnx_i-y_i≤κ_i→0 gives x_i-y_i→0. Adding the two limits proves x_i→ p. Continuity of the norm gives the source-error limit, because p-g=-P_N_Ag.
Exact mathematical passage
y_i
y_jUse in source · 3 passages
Search aliases: y_j
Use in source · Sufficient_Resolution_Traditional.tex, line 317
For the trajectory (y_n,z_n)=T^n(y_0,z_0), z_n=D^nz_0+∑_j=0^n-1D^n-1-jGy_j, and y_n+1=Ay_n+BD^nz_0+∑_j=0^n-1M_n-1-jy_j. Prepared evolution (z_0=0) satisfies E_0=I, E_1=A, and E_n+1=AE_n+∑_j=0^n-1M_n-1-jE_j.
Exact mathematical passage
z_n=D^nz_0+\sum_{j=0}^{n-1}D^{n-1-j}Gy_j,
y_nUse in source · 4 passages
Search aliases: y_n
Use in source · Sufficient_Resolution_Traditional.tex, line 316
For the trajectory (y_n,z_n)=T^n(y_0,z_0), z_n=D^nz_0+∑_j=0^n-1D^n-1-jGy_j, and y_n+1=Ay_n+BD^nz_0+∑_j=0^n-1M_n-1-jy_j. Prepared evolution (z_0=0) satisfies E_0=I, E_1=A, and E_n+1=AE_n+∑_j=0^n-1M_n-1-jE_j.
Exact mathematical passage
(y_n,z_n)=T^n(y_0,z_0)
y_{n+1}Use in source · 2 passages
Search aliases: y_n+1
Use in source · Sufficient_Resolution_Traditional.tex, line 321
For the trajectory (y_n,z_n)=T^n(y_0,z_0), z_n=D^nz_0+∑_j=0^n-1D^n-1-jGy_j, and y_n+1=Ay_n+BD^nz_0+∑_j=0^n-1M_n-1-jy_j. Prepared evolution (z_0=0) satisfies E_0=I, E_1=A, and E_n+1=AE_n+∑_j=0^n-1M_n-1-jE_j.
Exact mathematical passage
y_{n+1}=Ay_n+BD^nz_0+\sum_{j=0}^{n-1}M_{n-1-j}y_j.
ZUse in source · 23 passages
Search aliases: Z
Use in source · Sufficient_Resolution_Traditional.tex, line 176
The splitting gives E_X^2=a_X(s_Xa_X)s_X=a_Xs_X. If E_Xv=v, then v=a_X(s_Xv); conversely E_Xa_Xx=a_Xx. Substituting the definitions yields s_Y(a_YTs_X)a_X=T, a_Y(s_YKa_X)s_X=E_YKE_X. For compatible K, take T= D(K). If another T' encodes to K, decoding proves T'= D(K). Finally, an intermediate resolution of Y has s_Ya_Y=I_Y, so K(S) K(T)=a_ZS(s_Ya_Y)Ts_X=a_ZSTs_X. Encoding I_X gives a_Xs_X=E_X, which acts as identity on compatible kernels. No inverse on all of V_X has been assumed.
Exact mathematical passage
\mathcal K(S)\mathcal K(T)=a_ZS(s_Ya_Y)Ts_X=a_ZSTs_X
zUse in source · 64 passages
Search aliases: z
Use in source · Sufficient_Resolution_Traditional.tex, line 304
Let V be visible state and H hidden state. Over a commutative ring, specify four maps and form the actual full dynamics T(y,z)=(Ay+Bz,Gy+Dz). Put C(y,z)=y, Ry=(y,0), E_n=CT^nR, and M_k=BD^kG. The return M_k contains entry into H, k hidden transitions, and return to V. Its first effect occurs after k+2 full transitions.
Exact mathematical passage
T(y,z)=(Ay+Bz,Gy+Dz).
z'Use in source · 2 passages
Search aliases: z'
Use in source · Sufficient_Resolution_Traditional.tex, line 771
frac ddte^tD=De^tD=e^tDD, e^(s+t)D=e^sDe^tD, and e^-tD is the inverse flow. Every actual curve satisfying z'=Dz+f with continuous f obeys z(t)=e^tDz(0)+∫_0^t e^(t-s)Df(s) ds. The initial value uniquely determines the homogeneous solution; two forced solutions with the same forcing and initial value also agree.
Exact mathematical passage
z'=Dz+f
z^2Use in source · 2 passages
Search aliases: z^2
Use in source · Sufficient_Resolution_Traditional.tex, line 1427
Every class z has a unique compatible minimum-norm representative s(z), and every compatible representative v satisfies nn v^2=nn z^2+nnv-s(z)^2. For arbitrary data define c(v)=[P_Kv]. This is nonexpansive and dist(v,R)^2=nnc(v)^2+nni(v)^2. The datum is realizable exactly when both c(v)=0 and δ v=0. When the canonical inverse exists, nnv-Ag^2=nnc(v)^2+nni(v)^2+nnA(g-Bv)^2.
Exact mathematical passage
\nn v^2=\nn z^2+\nn{v-s(z)}^2.
z_0Use in source · 12 passages
Search aliases: z_0
Use in source · Sufficient_Resolution_Traditional.tex, line 316
For the trajectory (y_n,z_n)=T^n(y_0,z_0), z_n=D^nz_0+∑_j=0^n-1D^n-1-jGy_j, and y_n+1=Ay_n+BD^nz_0+∑_j=0^n-1M_n-1-jy_j. Prepared evolution (z_0=0) satisfies E_0=I, E_1=A, and E_n+1=AE_n+∑_j=0^n-1M_n-1-jE_j.
Exact mathematical passage
(y_n,z_n)=T^n(y_0,z_0)
z_1Use in source · 5 passages
Search aliases: z_1
Use in source · Sufficient_Resolution_Traditional.tex, line 450
[Delayed return and disconnected return] On Q^3, set T(y,z_1,z_2)=(z_1,z_2,y) and retain y. Then A=0, M_0=0, M_1=I, and E_1=E_2=0 while E_3=I. Two shallow exact checks cannot certify all depths without the dimension budget. For another system, let A=D=I, B(z_1,z_2)=z_1, and G(y)=(0,y). Both couplings are nonzero but every BD^kG vanishes. Prepared closure is exact, while an arbitrary hidden initial first coordinate affects the next visible state.
Exact mathematical passage
T(y,z_1,z_2)=(z_1,z_2,y)
z_2Use in source · 5 passages
Search aliases: z_2
Use in source · Sufficient_Resolution_Traditional.tex, line 450
[Delayed return and disconnected return] On Q^3, set T(y,z_1,z_2)=(z_1,z_2,y) and retain y. Then A=0, M_0=0, M_1=I, and E_1=E_2=0 while E_3=I. Two shallow exact checks cannot certify all depths without the dimension budget. For another system, let A=D=I, B(z_1,z_2)=z_1, and G(y)=(0,y). Both couplings are nonzero but every BD^kG vanishes. Prepared closure is exact, while an arbitrary hidden initial first coordinate affects the next visible state.
Exact mathematical passage
T(y,z_1,z_2)=(z_1,z_2,y)
Z_kUse in source · 3 passages
Search aliases: Z_k
Use in source · Sufficient_Resolution_Traditional.tex, line 536
Let H_kx=(OT_wx)_|w|≤ k and Z_k=Ran H_k. Then ker H_k=N and X/N≅ Z_k. The formulas U_i(H_kx)=H_kT_ix, V(H_kx)=Ox define actual linear maps on Z_k and preserve every finite future word. This model has minimum exact dimension.
Exact mathematical passage
Z_k=\Ran H_k
z_nUse in source · 4 passages
Search aliases: z_n
Use in source · Sufficient_Resolution_Traditional.tex, line 316
For the trajectory (y_n,z_n)=T^n(y_0,z_0), z_n=D^nz_0+∑_j=0^n-1D^n-1-jGy_j, and y_n+1=Ay_n+BD^nz_0+∑_j=0^n-1M_n-1-jy_j. Prepared evolution (z_0=0) satisfies E_0=I, E_1=A, and E_n+1=AE_n+∑_j=0^n-1M_n-1-jE_j.
Exact mathematical passage
(y_n,z_n)=T^n(y_0,z_0)
z_{n+1}Use in source · 1 passage
Search aliases: z_n+1
Use in source · Sufficient_Resolution_Traditional.tex, line 334
At n=0, the first formula is z_0=z_0 with an empty sum. If it holds at n, the actual recurrence z_n+1=Gy_n+Dz_n gives the same formula at n+1 after distributing D and appending the newest forcing term. Substituting this expression into y_n+1=Ay_n+Bz_n proves the visible memory equation. For a prepared input y_0, each y_j equals E_jy_0. The pointwise memory equation is therefore an operator identity because it holds for every y_0. The equations are derived from powers of T; the trajectory recurre…
Exact mathematical passage
z_{n+1}=Gy_n+Dz_n
|Use in source · 22 passages
Search aliases: |
Use in source · Sufficient_Resolution_Traditional.tex, line 463
Now work over a field and, for dimension statements, assume dim X<∞. The allowed operations need not commute. Define N_r=⋂_|w|≤ rker(OT_w), k=dimker O. A finite horizon means bounded word length. It gives a finite enumeration of constraints only when the operation alphabet is finite.
Exact mathematical passage
N_r=\bigcap_{|w|\le r}\ker(OT_w),\qquad k=\dim\ker O.
\cdotUse in source · 2 passages
Search aliases: · cdot
Use in source · Sufficient_Resolution_Traditional.tex, line 2655
For the scalar boundary, the true data error is zero and nnC-A=1≤δ, so both stated truth allowances are valid. The small-data branch would require 1≤2/10, which is false. For every positive parameter, the actual regularized source of C=0 is zero and its fitting residual is one. Its candidate noise is 1/10+1·0=1/10, so the upper generalized band would again require 1≤2/10. Every branch fails. The relative feasibility condition also fails, since 1<(2-1)/10 is false. Thus the boundary refutes an un…
Exact mathematical passage
1/10+1\cdot0=1/10
\timesUse in source · 1 passage
Search aliases: × times
Use in source · Sufficient_Resolution_Traditional.tex, line 423
For C:X→ Z, R:Z→ X, CR=I, let Q=I-RC and H=ker C. The maps x⟼(Cx,Qx), (y,z)⟼ Ry+z are inverse linear maps between X and Z× H. An endomorphism T is conjugate to block dynamics with A=CTR, B=CTι_H, G=QTR, D=QTι_H. Their prepared maps equal CT^nR. Hence exact compressed powers are equivalent to zero canonical memory. If dimker C=d, complete compressed map tests through d+1 decide all depths.
Exact mathematical passage
Z\times H
\alphaUse in source · 16 passages
Search aliases: α alpha
Use in source · Sufficient_Resolution_Traditional.tex, line 1302
Some α>0 satisfies αnn g≤nnAg on S, and any supplied such constant gives nn B≤1/α. Therefore nnB(Ag+e)-g^2≤(nn e/α)^2+nnP_N_Ag^2. For an arbitrary computed source h, nnh-Bv^2 ≤(fracnnAh-P_Rvα)^2+nnP_N_Ah^2. If also δ A=0, a compatibility constant σ bounds the local error as in SR-27. The constants α and σ have separate roles.
Exact mathematical passage
\alpha>0
\alpha^2Use in source · 3 passages
Search aliases: α^2 alpha
Use in source · Sufficient_Resolution_Traditional.tex, line 1648
Write p=P_(ker A)^⊥g, k=P_ker Ag. Then B_λ Ag+λ R_λ g=g, B_λ(Ag+e)-g=-λ R_λ g+B_λ e, and nnB_λ(Ag+e)-g^2 =nn-λ R_λ p+B_λ e^2+nn k^2. If A has source coercivity α>0, this is at most (fracλα^2+λnn p +fracnn e2sqrtλ)^2+nn k^2. For a supplied source condition g=A^*Aw, nnB_λ Ag-g≤λnn w without requiring closed range.
Exact mathematical passage
\left(\frac\lambda{\alpha^2+\lambda}\nn p +\frac{\nn e}{2\sqrt\lambda}\right)^2+\nn k^2.
\DeltaUse in source · 5 passages
Search aliases: Δ Delta
Use in source · Sufficient_Resolution_Traditional.tex, line 200
Set P=RC, Q=I-P, Φ(T)=CTR, and Δ(S,T)=Φ(ST)-Φ(S)Φ(T). Then Δ(S,T)=CSQTR, and associativity gives Δ(S,T)Φ(U)+Δ(ST,U) =Φ(S)Δ(T,U)+Δ(S,TU). For two successive split compressions, Δ_21(S,T)=Φ_2(Δ_1(S,T)) +Δ_2(Φ_1(S),Φ_1(T)).
Exact mathematical passage
\Delta(S,T)=\Phi(ST)-\Phi(S)\Phi(T)
\delta'Use in source · 2 passages
Search aliases: δ' delta
Use in source · Sufficient_Resolution_Traditional.tex, line 242
Commuting presentation maps LA=A'F and δ'L=Mδ induce a map overline L: H→ H' by [v]↦[Lv]. These maps compose and preserve zero classes. Reflection of zero follows when overline L is injective.
Exact mathematical passage
\delta'L=M\delta
\Delta_1Use in source · 2 passages
Search aliases: Δ_1 Delta
Use in source · Sufficient_Resolution_Traditional.tex, line 210
Set P=RC, Q=I-P, Φ(T)=CTR, and Δ(S,T)=Φ(ST)-Φ(S)Φ(T). Then Δ(S,T)=CSQTR, and associativity gives Δ(S,T)Φ(U)+Δ(ST,U) =Φ(S)Δ(T,U)+Δ(S,TU). For two successive split compressions, Δ_21(S,T)=Φ_2(Δ_1(S,T)) +Δ_2(Φ_1(S),Φ_1(T)).
Exact mathematical passage
\Delta_{21}(S,T)=\Phi_2(\Delta_1(S,T)) +\Delta_2(\Phi_1(S),\Phi_1(T)).
\Delta_2Use in source · 1 passage
Search aliases: Δ_2 Delta
Use in source · Sufficient_Resolution_Traditional.tex, line 210
Set P=RC, Q=I-P, Φ(T)=CTR, and Δ(S,T)=Φ(ST)-Φ(S)Φ(T). Then Δ(S,T)=CSQTR, and associativity gives Δ(S,T)Φ(U)+Δ(ST,U) =Φ(S)Δ(T,U)+Δ(S,TU). For two successive split compressions, Δ_21(S,T)=Φ_2(Δ_1(S,T)) +Δ_2(Φ_1(S),Φ_1(T)).
Exact mathematical passage
\Delta_{21}(S,T)=\Phi_2(\Delta_1(S,T)) +\Delta_2(\Phi_1(S),\Phi_1(T)).
\Delta_{21}Use in source · 1 passage
Search aliases: Δ_21 Delta
Use in source · Sufficient_Resolution_Traditional.tex, line 210
Set P=RC, Q=I-P, Φ(T)=CTR, and Δ(S,T)=Φ(ST)-Φ(S)Φ(T). Then Δ(S,T)=CSQTR, and associativity gives Δ(S,T)Φ(U)+Δ(ST,U) =Φ(S)Δ(T,U)+Δ(S,TU). For two successive split compressions, Δ_21(S,T)=Φ_2(Δ_1(S,T)) +Δ_2(Φ_1(S),Φ_1(T)).
Exact mathematical passage
\Delta_{21}(S,T)=\Phi_2(\Delta_1(S,T)) +\Delta_2(\Phi_1(S),\Phi_1(T)).
\delta_iUse in source · 14 passages
Search aliases: δ_i delta
Use in source · Sufficient_Resolution_Traditional.tex, line 611
Let q:X→ Z be a proposed representation, with true operations T_i, reduced operations U_i, and readouts O:X→ Y, V:Z→ Y. Define the step and word defects δ_i=qT_i-U_iq, E_w=qT_w-U_wq. Algebraic identities require only the stated commutative-ring structure. Quantitative estimates below use continuous linear maps over a nontrivially normed field and their whole-map operator norms.
Exact mathematical passage
\delta_i=qT_i-U_iq,\qquad E_w=qT_w-U_wq.
\delta_{i_j}Use in source · 1 passage
Search aliases: δ_i_j delta
Use in source · Sufficient_Resolution_Traditional.tex, line 629
One has E_∅=0, E_i::w=E_wT_i+U_wδ_i, E_umathbin+ +v=E_vT_u+U_vE_u. For w=[i_1,…,i_n] this yields E_w=∑_j=1^n U_[i_j+1,…,i_n]δ_i_jT_[i_1,…,i_j-1]. The output defect is (O-Vq)T_w+VE_w.
Exact mathematical passage
E_w=\sum_{j=1}^n U_{[i_{j+1},\ldots,i_n]}\delta_{i_j}T_{[i_1,\ldots,i_{j-1}]}.
\delta_pUse in source · 1 passage
Search aliases: δ_p delta
Use in source · Sufficient_Resolution_Traditional.tex, line 936
One has the operator identity E(t)=∫_0^t e^(t-s)Aδ e^sL ds, and δ=0 exactly when E(t)=0 for every t. Also E(s+t)=E(s)e^tL+e^sAE(t). For a second representation p:Z→ W with generator B, δ_pq=pδ_q+δ_pq, E_pq(t)=pE_q(t)+E_p(t)q.
Exact mathematical passage
\delta_{pq}=p\delta_q+\delta_pq,\qquad E_{pq}(t)=pE_q(t)+E_p(t)q.
\delta_{pq}Use in source · 1 passage
Search aliases: δ_pq delta
Use in source · Sufficient_Resolution_Traditional.tex, line 936
One has the operator identity E(t)=∫_0^t e^(t-s)Aδ e^sL ds, and δ=0 exactly when E(t)=0 for every t. Also E(s+t)=E(s)e^tL+e^sAE(t). For a second representation p:Z→ W with generator B, δ_pq=pδ_q+δ_pq, E_pq(t)=pE_q(t)+E_p(t)q.
Exact mathematical passage
\delta_{pq}=p\delta_q+\delta_pq,\qquad E_{pq}(t)=pE_q(t)+E_p(t)q.
\delta_qUse in source · 1 passage
Search aliases: δ_q delta
Use in source · Sufficient_Resolution_Traditional.tex, line 936
One has the operator identity E(t)=∫_0^t e^(t-s)Aδ e^sL ds, and δ=0 exactly when E(t)=0 for every t. Also E(s+t)=E(s)e^tL+e^sAE(t). For a second representation p:Z→ W with generator B, δ_pq=pδ_q+δ_pq, E_pq(t)=pE_q(t)+E_p(t)q.
Exact mathematical passage
\delta_{pq}=p\delta_q+\delta_pq,\qquad E_{pq}(t)=pE_q(t)+E_p(t)q.
\delta_sUse in source · 1 passage
Search aliases: δ_s delta
Use in source · Sufficient_Resolution_Traditional.tex, line 1203
Nested orthogonal projections split o_0(v)=o_1(v)+g(v) into orthogonal sectors, and the inconsistency component stays fixed. Pythagoras gives both refinement identities. For the scalar family δ_s(x)=sx, s>0, take R=0 and v=1. The reconstruction distance is one and the residual is s. For any proposed common c, choosing s sufficiently small violates 1≤ cs. The theorem guarantees a constant per presentation, not a common constant across degenerating presentations.
Exact mathematical passage
\delta_s(x)=sx
\epsilonUse in source · 37 passages
Search aliases: ϵ epsilon
Use in source · Sufficient_Resolution_Traditional.tex, line 655
Suppose nnT_i≤ a_i, nnU_i≤ b_i, nnδ_i≤ϵ_i. Let A_w,B_w be the products of the respective letter bounds and define D_∅=0, D_i::w=D_wa_i+B_wϵ_i. Then nnT_w≤ A_w, nnU_w≤ B_w, nnE_w≤ D_w, and D_umathbin+ +v=D_vA_u+B_vD_u. If nnO-Vq≤η, then nnOT_w-VU_wq≤η A_w+nn V D_w. For arbitrary initial x∈ X, z∈ Z, nnqT_wx-U_wz≤ D_wnn x+B_wnnqx-z.
Exact mathematical passage
\nn{\delta_i}\le\epsilon_i
\epsilon_iUse in source · 7 passages
Search aliases: ϵ_i epsilon
Use in source · Sufficient_Resolution_Traditional.tex, line 655
Suppose nnT_i≤ a_i, nnU_i≤ b_i, nnδ_i≤ϵ_i. Let A_w,B_w be the products of the respective letter bounds and define D_∅=0, D_i::w=D_wa_i+B_wϵ_i. Then nnT_w≤ A_w, nnU_w≤ B_w, nnE_w≤ D_w, and D_umathbin+ +v=D_vA_u+B_vD_u. If nnO-Vq≤η, then nnOT_w-VU_wq≤η A_w+nn V D_w. For arbitrary initial x∈ X, z∈ Z, nnqT_wx-U_wz≤ D_wnn x+B_wnnqx-z.
Exact mathematical passage
\nn{\delta_i}\le\epsilon_i
\eta_iUse in source · 15 passages
Search aliases: η_i eta
Use in source · Sufficient_Resolution_Traditional.tex, line 1972
smallHypotheses. Known bounded A:G→ V between real Hilbert spaces and a fixed true source g; a valid pointwise allowance nnAg-v≤η. Accepted-source consistency uses a fixed τ≥1 and genuine bounds nnAg-v_i≤η_i with η_i→0. Total selector existence additionally requires η>0 and τ>1.
Exact mathematical passage
\nn{Ag-v_i}\le\eta_i
\eta_{\rm cal}Use in source · 11 passages
Search aliases: η_rm cal eta rm
Use in source · Sufficient_Resolution_Traditional.tex, line 2684
The observed response Y:K→ V is an operator, and the assumptions nnY-AU≤η_rm cal and nnY-CU≤ f are operator-norm enclosures. Individual calibration samples do not automatically give these enclosures. A mathematical or experimental justification of η_rm cal remains a separate input.
Exact mathematical passage
\nn{Y-AU}\le\eta_{\rm cal}
\eta_{{\rm cal},i}Use in source · 1 passage
Search aliases: η_rm cal,i eta rm
Use in source · Sufficient_Resolution_Traditional.tex, line 2739
For recovery assume G,V are real Hilbert spaces, A is a fixed bounded true operator, τ≥0 is fixed, and p∈(ker A)^⊥ is fixed. Retain the whole calibration bounds and the true data bounds nnAp-v_i≤η_i. Put δ_i=(η_rm cal,i+f_i)nn R. If the true data, calibration-truth and model-fit allowances tend to zero, then δ_i→0. Generalized accepted sources using this generated allowance converge to a fixed identifiable true source p. For an arbitrary true source g, apply this with p=P_S_Ag and genuine data b…
Exact mathematical passage
\delta_i=(\eta_{{\rm cal},i}+f_i)\nn R
\iotaUse in source · 2 passages
Search aliases: ι iota
Use in source · Sufficient_Resolution_Traditional.tex, line 425
For C:X→ Z, R:Z→ X, CR=I, let Q=I-RC and H=ker C. The maps x⟼(Cx,Qx), (y,z)⟼ Ry+z are inverse linear maps between X and Z× H. An endomorphism T is conjugate to block dynamics with A=CTR, B=CTι_H, G=QTR, D=QTι_H. Their prepared maps equal CT^nR. Hence exact compressed powers are equivalent to zero canonical memory. If dimker C=d, complete compressed map tests through d+1 decide all depths.
Exact mathematical passage
A=CTR,\quad B=CT\iota_H,\quad G=QTR,\quad D=QT\iota_H.
\iota_HUse in source · 1 passage
Search aliases: ι_H iota
Use in source · Sufficient_Resolution_Traditional.tex, line 425
For C:X→ Z, R:Z→ X, CR=I, let Q=I-RC and H=ker C. The maps x⟼(Cx,Qx), (y,z)⟼ Ry+z are inverse linear maps between X and Z× H. An endomorphism T is conjugate to block dynamics with A=CTR, B=CTι_H, G=QTR, D=QTι_H. Their prepared maps equal CT^nR. Hence exact compressed powers are equivalent to zero canonical memory. If dimker C=d, complete compressed map tests through d+1 decide all depths.
Exact mathematical passage
A=CTR,\quad B=CT\iota_H,\quad G=QTR,\quad D=QT\iota_H.
\iota_SUse in source · 1 passage
Search aliases: ι_S iota
Use in source · Sufficient_Resolution_Traditional.tex, line 1237
Restriction A_S:S→ R is a bounded linear bijection with bounded inverse. Define B=ι_S A_S^-1P_R. Then AB=P_R, BA=P_S, ABA=A, and BAB=B. Its two projections are symmetric, ker B=R^⊥, and Ran B=S. The best-image identity AB=P_R and the condition Ran B⊆ S uniquely determine B.
Exact mathematical passage
B=\iota_S A_S^{-1}P_R.
\kappaUse in source · 26 passages
Search aliases: κ kappa
Use in source · Sufficient_Resolution_Traditional.tex, line 2112
The distinction matters when λ is small. A normal residual sufficient for a prescribed image tolerance may be insufficient for a prescribed source tolerance. If source correction must be at most κ and image correction at most ξ, a sound pair of conditions is nnr_λ(x)≤λκ and nnr_λ(x)≤2sqrtλ ξ. These are parameter-dependent requirements, not a uniform inverse bound as λ tends to zero.
Exact mathematical passage
\kappa
\kappa^2Use in source · 1 passage
Search aliases: κ^2 kappa
Use in source · Sufficient_Resolution_Traditional.tex, line 2542
A generalized accepted x obeys nnAx-Ap≤(τ+1)c_η,δ(p). A source within κ of such an x obeys the same bound with nn A κ added. For p∈ S_A, vanishing data/operator allowances and near-source budgets imply strong convergence to p. If p=A^*w, an inexact accepted candidate also satisfies nnx-p^2≤ 2κ^2+4nn w(τ+1)c_η,δ(p).
Exact mathematical passage
\nn{x-p}^2\le 2\kappa^2+4\nn w(\tau+1)c_{\eta,\delta}(p).
\kappa_iUse in source · 6 passages
Search aliases: κ_i kappa
Use in source · Sufficient_Resolution_Traditional.tex, line 2186
Call x near accepted with budget κ when some exact accepted source y satisfies nnx-y≤κ. If nnAg-v_i≤η_i, η_i→0, and near-accepted x_i have κ_i→0, then x_i⟶ P_S_Ag, nnx_i-g⟶nnP_N_Ag. At every index nnP_N_Ax_i≤κ_i; therefore the computed true-kernel component tends to zero. No separate parameter schedule is assumed.
Exact mathematical passage
\kappa_i\to0
\lambdaUse in source · 143 passages
Search aliases: λ lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 1546
Fix a bounded operator A:G→ V between real Hilbert spaces, without assuming closed range. For λ>0, put N_λ=A^*A+λ I, R_λ=N_λ^-1, B_λ=R_λ A^*. The inverse is constructed from coercivity. It is not supplied as a certificate premise. Here A^* is the actual Hilbert adjoint.
Exact mathematical passage
\lambda>0
\lambda^2Use in source · 1 passage
Search aliases: λ^2 lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 1688
For two operators and parameters, R_A,λ-R_C,μ =R_A,λ(N_C,μ-N_A,λ)R_C,μ. At fixed positive λ, with ϵ=nnA-C, nnB_A,λ-B_C,λ ≤fracϵλ+ frac(nn A+nn C)ϵnn Cλ^2. If A has closed range and canonical inverse B, then B_λ v-Bv=-λ R_λ Bv. For δ A=0, nnv-AB_λ v^2 =nnc(v)^2+nni(v)^2+nnA(B_λ v-Bv)^2.
Exact mathematical passage
\nn{B_{A,\lambda}-B_{C,\lambda}} \le\frac\epsilon\lambda+ \frac{(\nn A+\nn C)\epsilon\nn C}{\lambda^2}.
\lambda_iUse in source · 7 passages
Search aliases: λ_i lambda
Use in source · Sufficient_Resolution_Traditional.tex, line 1786
If λ_i>0 tends to zero and nne_i/sqrtλ_i→0, then B_λ_i(Ag+e_i)⟶ P_(ker A)^⊥g. The same limit holds for computed x_i when additionally fracnn(A^*A+λ_iI)x_i-A^*(Ag+e_i)λ_i→0. For example, λ_i=s_i^2, s_i>0→0, and nne_i≤ s_i^2 give the exact-solve consistency condition.
Exact mathematical passage
\lambda_i>0
\muUse in source · 6 passages
Search aliases: μ mu
Use in source · Sufficient_Resolution_Traditional.tex, line 1683
For two operators and parameters, R_A,λ-R_C,μ =R_A,λ(N_C,μ-N_A,λ)R_C,μ. At fixed positive λ, with ϵ=nnA-C, nnB_A,λ-B_C,λ ≤fracϵλ+ frac(nn A+nn C)ϵnn Cλ^2. If A has closed range and canonical inverse B, then B_λ v-Bv=-λ R_λ Bv. For δ A=0, nnv-AB_λ v^2 =nnc(v)^2+nni(v)^2+nnA(B_λ v-Bv)^2.
Exact mathematical passage
R_{A,\lambda}-R_{C,\mu} =R_{A,\lambda}(N_{C,\mu}-N_{A,\lambda})R_{C,\mu}.
\nuUse in source · 2 passages
Search aliases: ν nu
Use in source · Sufficient_Resolution_Traditional.tex, line 2279
If the normal residual itself is evaluated approximately as widehat r, the distinct assumptions nnr_λ(x)-widehat r≤ν and nnwidehat r≤widehatρ give the allowance widehatρ+ν. Substituting that sum into both correction tests retains the evaluation error rather than silently identifying it with zero.
Exact mathematical passage
\nn{r_\lambda(x)-\widehat r}\le\nu
\xiUse in source · 12 passages
Search aliases: ξ xi
Use in source · Sufficient_Resolution_Traditional.tex, line 2113
The distinction matters when λ is small. A normal residual sufficient for a prescribed image tolerance may be insufficient for a prescribed source tolerance. If source correction must be at most κ and image correction at most ξ, a sound pair of conditions is nnr_λ(x)≤λκ and nnr_λ(x)≤2sqrtλ ξ. These are parameter-dependent requirements, not a uniform inverse bound as λ tends to zero.
Exact mathematical passage
\xi
\xi^2Use in source · 2 passages
Search aliases: ξ^2 xi
Use in source · Sufficient_Resolution_Traditional.tex, line 2238
Suppose a record supplies λ>0, nonnegative ρ,ξ,κ,m,ω, and nnr_λ(x)^2≤ρ^2, ρ^2≤(λκ)^2, ρ^2≤4λξ^2, max(0,m-ω)^2≤nnAx-v^2≤(m+ω)^2, η+ω+ξ≤ m≤τη-ω-ξ . Then x is near accepted with source budget κ. Families of such records give SR-51 when the genuine data allowances and source budgets tend to zero.
Exact mathematical passage
\nn{r_\lambda(x)}^2\le\rho^2,\quad \rho^2\le(\lambda\kappa)^2,\quad \rho^2\le4\lambda\xi^2,
\rhoUse in source · 18 passages
Search aliases: ρ rho
Use in source · Sufficient_Resolution_Traditional.tex, line 1901
The residual is continuous and nondecreasing on positive parameters; the source norm is nonincreasing. It is strictly increasing when A^*v≠0, and constant at nn v when A^*v=0. For dist(v,Ran A)<ρ<nn v, there is exactly one λ>0 with r_v(λ)=ρ. The resulting exact real parameter is a noncomputable choice with a proved specification.
Exact mathematical passage
\dist(v,\Ran A)<\rho<\nn v,
\rho^2Use in source · 1 passage
Search aliases: ρ^2 rho
Use in source · Sufficient_Resolution_Traditional.tex, line 2238
Suppose a record supplies λ>0, nonnegative ρ,ξ,κ,m,ω, and nnr_λ(x)^2≤ρ^2, ρ^2≤(λκ)^2, ρ^2≤4λξ^2, max(0,m-ω)^2≤nnAx-v^2≤(m+ω)^2, η+ω+ξ≤ m≤τη-ω-ξ . Then x is near accepted with source budget κ. Families of such records give SR-51 when the genuine data allowances and source budgets tend to zero.
Exact mathematical passage
\nn{r_\lambda(x)}^2\le\rho^2,\quad \rho^2\le(\lambda\kappa)^2,\quad \rho^2\le4\lambda\xi^2,
\sigmaUse in source · 11 passages
Search aliases: σ sigma
Use in source · Sufficient_Resolution_Traditional.tex, line 1144
If σnn x≤nnδ x for x∈ K^⊥ and σ>0, then δ i(v)=δ v, nni(v)≤fracnnδ vσ, and nno(v)≤dist(v,R), dist(v,R)^2≤nno(v)^2+(fracnnδ vσ)^2. For compatible data, dist(v,R)=nno(v).
Exact mathematical passage
\sigma\nn x\le\nn{\delta x}
\PhiUse in source · 9 passages
Search aliases: Φ Phi
Use in source · Sufficient_Resolution_Traditional.tex, line 199
Set P=RC, Q=I-P, Φ(T)=CTR, and Δ(S,T)=Φ(ST)-Φ(S)Φ(T). Then Δ(S,T)=CSQTR, and associativity gives Δ(S,T)Φ(U)+Δ(ST,U) =Φ(S)Δ(T,U)+Δ(S,TU). For two successive split compressions, Δ_21(S,T)=Φ_2(Δ_1(S,T)) +Δ_2(Φ_1(S),Φ_1(T)).
Exact mathematical passage
\Phi(T)=CTR
\Phi_1Use in source · 2 passages
Search aliases: Φ_1 Phi
Use in source · Sufficient_Resolution_Traditional.tex, line 210
Set P=RC, Q=I-P, Φ(T)=CTR, and Δ(S,T)=Φ(ST)-Φ(S)Φ(T). Then Δ(S,T)=CSQTR, and associativity gives Δ(S,T)Φ(U)+Δ(ST,U) =Φ(S)Δ(T,U)+Δ(S,TU). For two successive split compressions, Δ_21(S,T)=Φ_2(Δ_1(S,T)) +Δ_2(Φ_1(S),Φ_1(T)).
Exact mathematical passage
\Delta_{21}(S,T)=\Phi_2(\Delta_1(S,T)) +\Delta_2(\Phi_1(S),\Phi_1(T)).
\Phi_2Use in source · 4 passages
Search aliases: Φ_2 Phi
Use in source · Sufficient_Resolution_Traditional.tex, line 210
Set P=RC, Q=I-P, Φ(T)=CTR, and Δ(S,T)=Φ(ST)-Φ(S)Φ(T). Then Δ(S,T)=CSQTR, and associativity gives Δ(S,T)Φ(U)+Δ(ST,U) =Φ(S)Δ(T,U)+Δ(S,TU). For two successive split compressions, Δ_21(S,T)=Φ_2(Δ_1(S,T)) +Δ_2(Φ_1(S),Φ_1(T)).
Exact mathematical passage
\Delta_{21}(S,T)=\Phi_2(\Delta_1(S,T)) +\Delta_2(\Phi_1(S),\Phi_1(T)).
\omegaUse in source · 12 passages
Search aliases: ω omega
Use in source · Sufficient_Resolution_Traditional.tex, line 2128
Suppose m is a measured fitting norm with |m-nnAx-v|≤ω, nnr_λ(x)≤λκ, nnr_λ(x)≤2sqrtλ ξ . If η+ω+ξ≤ m≤τη-ω-ξ, then the actual exact source x_λ has discrepancy in [η,τη], and nnx-x_λ≤κ. A protected scalar window exists exactly when 2(ω+ξ)≤(τ-1)η . Its midpoint (τ+1)η/2 is then admissible.
Exact mathematical passage
|m-\nn{Ax-v}|\le\omega,\quad \nn{r_\lambda(x)}\le\lambda\kappa,\quad \nn{r_\lambda(x)}\le2\sqrt\lambda\,\xi .
\ldotsUse in source · 3 passages
Search aliases: … ldots
Use in source · Sufficient_Resolution_Traditional.tex, line 103
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
w=[i_1,\ldots,i_n]
\toUse in source · 60 passages
Search aliases: → to
Use in source · Sufficient_Resolution_Traditional.tex, line 143
The first contract supplies two maps rather than identifying a coordinate space with the object by convention. Let a_X:X→ V_X and s_X:V_X→ X be linear with s_Xa_X=I_X. The ambient space V_X may have redundant coordinates. Define E_X=a_Xs_X. For another such resolution of Y, put K(T)=a_YTs_X, D(K)=s_YKa_X. We call a proposed coordinate operator compatible when E_YKE_X=K.
Exact mathematical passage
a_X:X\to V_X
\mapstoUse in source · 6 passages
Search aliases: ↦ mapsto
Use in source · Sufficient_Resolution_Traditional.tex, line 188
[Redundant coordinates] For X= R and V_X= R^2, take a(x)=(x,x) and s(u,v)=(u+v)/2. Then E(u,v)=((u+v)/2,(u+v)/2). The encoded scalar transformation x↦ cx has matrix frac c2(1 · 1
Exact mathematical passage
x\mapsto cx
\forallUse in source · 6 passages
Search aliases: ∀ forall
Use in source · Sufficient_Resolution_Traditional.tex, line 351
The following conditions are equivalent: E_n=A^n (∀ n), M_k=0 (∀ k), E_m+n=E_mE_n (∀ m,n). More precisely, E_n=A^n (n≤ N+1)iff M_k=0 (k<N). If M_j=0 for j<k, then E_k+2-A^k+2=M_k. Closure for every initial hidden state is equivalent to B=0.
Exact mathematical passage
E_n=A^n\ (\forall n),\qquad M_k=0\ (\forall k),\qquad E_{m+n}=E_mE_n\ (\forall m,n).
\existsUse in source · 3 passages
Search aliases: ∃ exists
Use in source · Sufficient_Resolution_Traditional.tex, line 252
The chain condition puts the actual range of A inside K. The zero class criterion for a quotient is membership in the submodule divided out, giving [v]=0iff v∈ Riff∃ g,Ag=v. Every compatible class is zero exactly when K⊆ R; the reverse inclusion is already known. Injectivity of A gives uniqueness of any preimage.
Exact mathematical passage
[v]=0\iff v\in R\iff\exists g,Ag=v
\varnothingUse in source · 3 passages
Search aliases: ∅ varnothing
Use in source · Sufficient_Resolution_Traditional.tex, line 104
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
T_\varnothing=I
\notinUse in source · 1 passage
Search aliases: ∉ notin
Use in source · Sufficient_Resolution_Traditional.tex, line 1877
iable space is (ker A^*)^⊥=clRan A, yielding the fitted limit. SR-38 shows AB_λ is contractive, so changing data by e_i changes the fitted value by at most nne_i. If reconstructed sources converged in norm to some g, continuity of A would make their images converge to Ag. On a nontrivial filter, the already established image limit v would force Ag=v, contrary to v∉Ran A. This proves absence of a source norm limit without adding an unproved assertion that source norms diverge.
Exact mathematical passage
v\notin\Ran A
\sumUse in source · 6 passages
Search aliases: ∑ sum
Use in source · Sufficient_Resolution_Traditional.tex, line 317
For the trajectory (y_n,z_n)=T^n(y_0,z_0), z_n=D^nz_0+∑_j=0^n-1D^n-1-jGy_j, and y_n+1=Ay_n+BD^nz_0+∑_j=0^n-1M_n-1-jy_j. Prepared evolution (z_0=0) satisfies E_0=I, E_1=A, and E_n+1=AE_n+∑_j=0^n-1M_n-1-jE_j.
Exact mathematical passage
z_n=D^nz_0+\sum_{j=0}^{n-1}D^{n-1-j}Gy_j,
\sum_{j=0}^{n-1}Use in source · 4 passages
Search aliases: ∑_j=0^n-1 sum
Use in source · Sufficient_Resolution_Traditional.tex, line 317
For the trajectory (y_n,z_n)=T^n(y_0,z_0), z_n=D^nz_0+∑_j=0^n-1D^n-1-jGy_j, and y_n+1=Ay_n+BD^nz_0+∑_j=0^n-1M_n-1-jy_j. Prepared evolution (z_0=0) satisfies E_0=I, E_1=A, and E_n+1=AE_n+∑_j=0^n-1M_n-1-jE_j.
Exact mathematical passage
z_n=D^nz_0+\sum_{j=0}^{n-1}D^{n-1-j}Gy_j,
\sum_{j=1}^nUse in source · 1 passage
Search aliases: ∑_j=1^n sum
Use in source · Sufficient_Resolution_Traditional.tex, line 629
One has E_∅=0, E_i::w=E_wT_i+U_wδ_i, E_umathbin+ +v=E_vT_u+U_vE_u. For w=[i_1,…,i_n] this yields E_w=∑_j=1^n U_[i_j+1,…,i_n]δ_i_jT_[i_1,…,i_j-1]. The output defect is (O-Vq)T_w+VE_w.
Exact mathematical passage
E_w=\sum_{j=1}^n U_{[i_{j+1},\ldots,i_n]}\delta_{i_j}T_{[i_1,\ldots,i_{j-1}]}.
\sum_nUse in source · 1 passage
Search aliases: ∑_n sum
Use in source · Sufficient_Resolution_Traditional.tex, line 824
The nth derivative at zero of e^tL is L^n. Bounded linear maps commute with these derivatives, so equality for all times implies equality of every observed power. Conversely bounded linear maps commute with the sum of the norm-convergent exponential series: K(e^tL)=∑_n(t^n/n!)K(L^n). Equality term by term gives equality of the sums. Sandwiching and compression are bounded linear maps on the corresponding operator spaces, so they give the specializations. For the last identity, use the flow law a…
Exact mathematical passage
K(e^{tL})=\sum_n(t^n/n!)K(L^n)
\setminusUse in source · 2 passages
Search aliases: ∖ setminus
Use in source · Sufficient_Resolution_Traditional.tex, line 1522
For any submodule, the native quotient seminorm equals distance to that submodule. Zero distance is equivalent to membership in its closure by the metric closure criterion; algebraic quotient zero remains actual membership. If R is nonclosed, choose v∈cl R∖ R. Its class is nonzero with zero seminorm. Conversely closedness makes zero distance imply membership. This is why the projection hypotheses in the preceding metric theory cannot be discarded silently.
Exact mathematical passage
v\in\cl R\setminus R
\inftyUse in source · 2 passages
Search aliases: ∞ infty
Use in source · Sufficient_Resolution_Traditional.tex, line 388
If a monic polynomial p annihilates D, vanishing of M_k for k<deg p implies vanishing for all k. If dim H=d<∞, then E_n=A^n (∀ n) iff M_k=0 (0≤ k<d) iff E_n=A^n (0≤ n≤ d+1). The visible space need not be finite-dimensional. The case d=0 is included.
Exact mathematical passage
\dim H=d<\infty
\capUse in source · 7 passages
Search aliases: ∩ cap
Use in source · Sufficient_Resolution_Traditional.tex, line 475
N_0=ker O and N_r+1=ker O∩⋂_i T_i^-1N_r⊆ N_r. One plateau N_r+1=N_r implies N_r=N. In all cases N_k=N. Thus equality of future observations through word length k is equivalent to equality for every finite word.
Exact mathematical passage
N_{r+1}=\ker O\cap\bigcap_i T_i^{-1}N_r\subseteq N_r.
\intUse in source · 6 passages
Search aliases: ∫ int
Use in source · Sufficient_Resolution_Traditional.tex, line 772
frac ddte^tD=De^tD=e^tDD, e^(s+t)D=e^sDe^tD, and e^-tD is the inverse flow. Every actual curve satisfying z'=Dz+f with continuous f obeys z(t)=e^tDz(0)+∫_0^t e^(t-s)Df(s) ds. The initial value uniquely determines the homogeneous solution; two forced solutions with the same forcing and initial value also agree.
Exact mathematical passage
z(t)=e^{tD}z(0)+\int_0^t e^{(t-s)D}f(s)\,ds.
\int_0^tUse in source · 6 passages
Search aliases: ∫_0^t int
Use in source · Sufficient_Resolution_Traditional.tex, line 772
frac ddte^tD=De^tD=e^tDD, e^(s+t)D=e^sDe^tD, and e^-tD is the inverse flow. Every actual curve satisfying z'=Dz+f with continuous f obeys z(t)=e^tDz(0)+∫_0^t e^(t-s)Df(s) ds. The initial value uniquely determines the homogeneous solution; two forced solutions with the same forcing and initial value also agree.
Exact mathematical passage
z(t)=e^{tD}z(0)+\int_0^t e^{(t-s)D}f(s)\,ds.
\congUse in source · 5 passages
Search aliases: ≅ cong
Use in source · Sufficient_Resolution_Traditional.tex, line 511
The model X/N is observable. Every exact competitor induces a unique surjection Ran q→ X/N, taking qx to [x]. Therefore dim(X/N)≤dimRan q, with equality exactly when ker q=N. In that case Ran q≅ X/N. If Z is finite-dimensional, also dimRan q≤dim Z. Surjectivity of q onto Z is not required.
Exact mathematical passage
\Ran q\cong X/N
\neUse in source · 4 passages
Search aliases: ≠ ne
Use in source · Sufficient_Resolution_Traditional.tex, line 698
For H_0(a,b)=0 and H_n+1(a,b)=aH_n(a,b)+b^n, H_n(a,b)=∑_j=0^n-1a^jb^n-1-j, (a-b)H_n=a^n-b^n. If a≠ b, H_n=(a^n-b^n)/(a-b); for equal rates, H_n+1(a,a)=(n+1)a^n. Uniform step bounds give nnE_w≤ϵ H_n(a,b), nnOT_w-VU_wq≤η a^n+nn Vϵ H_n(a,b).
Exact mathematical passage
a\ne b
\subseteqUse in source · 14 passages
Search aliases: ⊆ subseteq
Use in source · Sufficient_Resolution_Traditional.tex, line 237
For δ A=0, define K=kerδ, R=Ran A⊆ K, and H=K/R. A compatible v has class zero exactly when v=Ag for some g. All classes vanish exactly when R=K. If A is injective, a zero class has a unique source.
Exact mathematical passage
R=\Ran A\subseteq K
\perpUse in source · 31 passages
Search aliases: ⊥ perp
Use in source · Sufficient_Resolution_Traditional.tex, line 1077
The components lie in R, K∩ R^⊥, and K^⊥, respectively, are pairwise orthogonal, and give the unique decomposition v=r(v)+o(v)+i(v). Furthermore, nn v^2=nn r^2+nn o^2+nn i^2, dist(v,R)^2=nn o^2+nn i^2. For any u∈ R, nnv-u^2=dist(v,R)^2+nnr(v)-u^2. Thus r(v) is the unique nearest represented vector.
Exact mathematical passage
K\cap R^\perp
\bigcapUse in source · 3 passages
Search aliases: ⋂ bigcap
Use in source · Sufficient_Resolution_Traditional.tex, line 272
Let N=⋂_wker(OT_w). It is the largest submodule contained in ker O and invariant under every T_i. The operations and observation descend to X/N, preserving every future observation. In particular, [x]=[y]iff OT_wx=OT_wy for every finite word w. If q:X→ Z satisfies qT_i=U_iq and Vq=O, then ker q⊆ N.
Exact mathematical passage
N=\bigcap_w\ker(OT_w)
\bigcap_iUse in source · 1 passage
Search aliases: ⋂_i bigcap
Use in source · Sufficient_Resolution_Traditional.tex, line 475
N_0=ker O and N_r+1=ker O∩⋂_i T_i^-1N_r⊆ N_r. One plateau N_r+1=N_r implies N_r=N. In all cases N_k=N. Thus equality of future observations through word length k is equivalent to equality for every finite word.
Exact mathematical passage
N_{r+1}=\ker O\cap\bigcap_i T_i^{-1}N_r\subseteq N_r.
\bigcap_wUse in source · 1 passage
Search aliases: ⋂_w bigcap
Use in source · Sufficient_Resolution_Traditional.tex, line 272
Let N=⋂_wker(OT_w). It is the largest submodule contained in ker O and invariant under every T_i. The operations and observation descend to X/N, preserving every future observation. In particular, [x]=[y]iff OT_wx=OT_wy for every finite word w. If q:X→ Z satisfies qT_i=U_iq and Vq=O, then ker q⊆ N.
Exact mathematical passage
N=\bigcap_w\ker(OT_w)
\bigcap_{|w|\le r}Use in source · 1 passage
Search aliases: ⋂_|w|≤ r bigcap le
Use in source · Sufficient_Resolution_Traditional.tex, line 463
Now work over a field and, for dimension statements, assume dim X<∞. The allowed operations need not commute. Define N_r=⋂_|w|≤ rker(OT_w), k=dimker O. A finite horizon means bounded word length. It gives a finite enumeration of constraints only when the operation alphabet is finite.
Exact mathematical passage
N_r=\bigcap_{|w|\le r}\ker(OT_w),\qquad k=\dim\ker O.
\cdotsUse in source · 2 passages
Search aliases: ⋯ cdots
Use in source · Sufficient_Resolution_Traditional.tex, line 104
Conventions and domains. Operator multiplication applies the right factor first: STx=S(Tx). A word w=[i_1,…,i_n] is chronological, so T_w=T_i_n⋯ T_i_1; T_∅=I. Concatenating u then v gives T_umathbin+ +v=T_vT_u. Empty sums are zero and empty products are one. All equalities of operators are equalities on the entire stated domain.
Exact mathematical passage
T_w=T_{i_n}\cdots T_{i_1}
\longrightarrowUse in source · 4 passages
Search aliases: ⟶ longrightarrow
Use in source · Sufficient_Resolution_Traditional.tex, line 1752
One has clRan(A^*A)=(ker A)^⊥, and for every g,w, nnB_λ Ag-g≤nng-A^*Aw+λnn w. As λdownarrow0, B_λ Ag⟶ P_(ker A)^⊥g, nnB_λ Ag-g⟶nnP_ker Ag. For a nontrivial parameter filter, convergence to g itself holds exactly when g∈(ker A)^⊥.
Exact mathematical passage
B_\lambda Ag\longrightarrow P_{(\ker A)^\perp}g, \qquad \nn{B_\lambda Ag-g}\longrightarrow\nn{P_{\ker A}g}.
\longmapstoUse in source · 2 passages
Search aliases: ⟼ longmapsto
Use in source · Sufficient_Resolution_Traditional.tex, line 420
For C:X→ Z, R:Z→ X, CR=I, let Q=I-RC and H=ker C. The maps x⟼(Cx,Qx), (y,z)⟼ Ry+z are inverse linear maps between X and Z× H. An endomorphism T is conjugate to block dynamics with A=CTR, B=CTι_H, G=QTR, D=QTι_H. Their prepared maps equal CT^nR. Hence exact compressed powers are equivalent to zero canonical memory. If dimker C=d, complete compressed map tests through d+1 decide all depths.
Exact mathematical passage
x\longmapsto(Cx,Qx),\qquad (y,z)\longmapsto Ry+z
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