Notation by work
Paper V: Polyphase Signal Calculus
All notation · Glossary · Source coverage
671 symbol spellings in 2175 mathematical passages across 41 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 671 entries are available below.
*Notation table · 36 passages
Search aliases: *
Notation table · atlas-pages.tex, line 27
W_α^*W_α · A projector on the original signal space · A scalar packet energy
Exact mathematical passage
W_\alpha^*W_\alpha
+Notation table · 300 passages
Search aliases: +
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1710
P_r · the coordinate projector onto the coset r+L
Exact mathematical passage
r+L
+^mNotation table · 5 passages
Search aliases: +^m
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1716
ϕ_m,τ · the truncated power packet (t-τ)_+^m/m!
Exact mathematical passage
(t-\tau)_+^m/m!
\AcalNotation table · 11 passages
Search aliases: Acal A mathcal
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1711
A_L, S_L · lattice analysis and interleaving synthesis
Exact mathematical passage
\Acal_L,\Scal_L
\Acal_LNotation table · 6 passages
Search aliases: Acal_L A_L Acal mathcal
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1711
A_L, S_L · lattice analysis and interleaving synthesis
Exact mathematical passage
\Acal_L,\Scal_L
DNotation table · 47 passages
Search aliases: D
Notation table · atlas-pages.tex, line 24
D · One channel-index delay · Necessarily one original-sample delay
Exact mathematical passage
D
dNotation table · 88 passages
Search aliases: d
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1708
L≤ Z^d · a finite index sublattice
Exact mathematical passage
L\leq\ZZ^d
ENotation table · 70 passages
Search aliases: E
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1712
E,R · filter bank analysis and synthesis symbols
Exact mathematical passage
E,R
fNotation table · 93 passages
Search aliases: f
Notation table · atlas-pages.tex, line 22
P_N,rf · A masked signal on the original lattice · The shorter decimated channel
Exact mathematical passage
P_{N,r}f
f_rNotation table · 2 passages
Search aliases: f_r
Notation table · atlas-pages.tex, line 23
f_r[q] · The value f[Nq+r] · An average of neighboring samples
Exact mathematical passage
f_r[q]
\GamNotation table · 12 passages
Search aliases: Gam Γ Gamma
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1474
Γ · reflection coefficient · Smith-disk coordinate
Exact mathematical passage
\Gam
HNotation table · 37 passages
Search aliases: H
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1715
H_τ · the step packet 1_[τ,1]
Exact mathematical passage
H_\tau
hNotation table · 87 passages
Search aliases: h
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1717
h · the grid spacing in the recovery theorems
Exact mathematical passage
h
H_\tauNotation table · 11 passages
Search aliases: H_τ tau
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1715
H_τ · the step packet 1_[τ,1]
Exact mathematical passage
H_\tau
jNotation table · 132 passages
Search aliases: j
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1714
j,widehat j · the true and estimated singular packets
Exact mathematical passage
j,\widehat j
LNotation table · 122 passages
Search aliases: L
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1708
L≤ Z^d · a finite index sublattice
Exact mathematical passage
L\leq\ZZ^d
mNotation table · 125 passages
Search aliases: m
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1713
Q_m · a bounded linear approximation operator acting on the smooth part
Exact mathematical passage
Q_m
NNotation table · 153 passages
Search aliases: N
Notation table · atlas-pages.tex, line 22
P_N,rf · A masked signal on the original lattice · The shorter decimated channel
Exact mathematical passage
P_{N,r}f
\oneNotation table · 12 passages
Search aliases: one 1 mathbf
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1715
H_τ · the step packet 1_[τ,1]
Exact mathematical passage
\one_{[\tau,1]}
\one_{[\tau,1]}Notation table · 4 passages
Search aliases: one_[τ,1] 1_[τ,1] one tau mathbf tau
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1715
H_τ · the step packet 1_[τ,1]
Exact mathematical passage
\one_{[\tau,1]}
PNotation table · 46 passages
Search aliases: P
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1710
P_r · the coordinate projector onto the coset r+L
Exact mathematical passage
P_r
P_{N,r}Notation table · 11 passages
Search aliases: P_N,r
Notation table · atlas-pages.tex, line 22
P_N,rf · A masked signal on the original lattice · The shorter decimated channel
Exact mathematical passage
P_{N,r}f
P_rNotation table · 4 passages
Search aliases: P_r
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1710
P_r · the coordinate projector onto the coset r+L
Exact mathematical passage
P_r
QNotation table · 92 passages
Search aliases: Q
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1713
Q_m · a bounded linear approximation operator acting on the smooth part
Exact mathematical passage
Q_m
qNotation table · 90 passages
Search aliases: q
Notation table · atlas-pages.tex, line 23
f_r[q] · The value f[Nq+r] · An average of neighboring samples
Exact mathematical passage
f_r[q]
Q_mNotation table · 42 passages
Search aliases: Q_m
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1713
Q_m · a bounded linear approximation operator acting on the smooth part
Exact mathematical passage
Q_m
RNotation table · 76 passages
Search aliases: R
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1709
R · a set of representatives for Z^d/L
Exact mathematical passage
R
rNotation table · 149 passages
Search aliases: r
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1710
P_r · the coordinate projector onto the coset r+L
Exact mathematical passage
P_r
SNotation table · 15 passages
Search aliases: S
Notation table · atlas-pages.tex, line 21
midrule S,δ_0 · Sample shift and unit impulse · A finite truncation of every ℓ^2 signal
Exact mathematical passage
S,\delta_0
\ScalNotation table · 2 passages
Search aliases: Scal S mathcal
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1711
A_L, S_L · lattice analysis and interleaving synthesis
Exact mathematical passage
\Acal_L,\Scal_L
\Scal_LNotation table · 2 passages
Search aliases: Scal_L S_L Scal mathcal
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1711
A_L, S_L · lattice analysis and interleaving synthesis
Exact mathematical passage
\Acal_L,\Scal_L
tNotation table · 57 passages
Search aliases: t
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1716
ϕ_m,τ · the truncated power packet (t-τ)_+^m/m!
Exact mathematical passage
(t-\tau)_+^m/m!
WNotation table · 25 passages
Search aliases: W
Notation table · atlas-pages.tex, line 26
W_α · The addressed Haar analysis operator · An edge detector by itself
Exact mathematical passage
W_\alpha
W_\alphaNotation table · 6 passages
Search aliases: W_α alpha
Notation table · atlas-pages.tex, line 26
W_α · The addressed Haar analysis operator · An edge detector by itself
Exact mathematical passage
W_\alpha
W_\alpha^*Notation table · 4 passages
Search aliases: W_α^* alpha
Notation table · atlas-pages.tex, line 27
W_α^*W_α · A projector on the original signal space · A scalar packet energy
Exact mathematical passage
W_\alpha^*W_\alpha
\widehat jNotation table · 37 passages
Search aliases: widehat j widehat
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1714
j,widehat j · the true and estimated singular packets
Exact mathematical passage
j,\widehat j
zNotation table · 79 passages
Search aliases: z
Notation table · atlas-pages.tex, line 30
Γ,z · Reflection coefficient and normalized impedance · Coefficients that can be linearly interchanged
Exact mathematical passage
\Gamma,z
\ZZNotation table · 40 passages
Search aliases: ZZ Z mathbb
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1708
L≤ Z^d · a finite index sublattice
Exact mathematical passage
L\leq\ZZ^d
\ZZ^dNotation table · 27 passages
Search aliases: ZZ^d Z^d ZZ mathbb
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1708
L≤ Z^d · a finite index sublattice
Exact mathematical passage
L\leq\ZZ^d
\alphaNotation table · 13 passages
Search aliases: α alpha
Notation table · atlas-pages.tex, line 26
W_α · The addressed Haar analysis operator · An edge detector by itself
Exact mathematical passage
W_\alpha
\GammaNotation table · 47 passages
Search aliases: Γ Gamma
Notation table · atlas-pages.tex, line 30
Γ,z · Reflection coefficient and normalized impedance · Coefficients that can be linearly interchanged
Exact mathematical passage
\Gamma,z
\deltaNotation table · 86 passages
Search aliases: δ delta
Notation table · atlas-pages.tex, line 21
midrule S,δ_0 · Sample shift and unit impulse · A finite truncation of every ℓ^2 signal
Exact mathematical passage
S,\delta_0
\delta_0Notation table · 16 passages
Search aliases: δ_0 delta
Notation table · atlas-pages.tex, line 21
midrule S,δ_0 · Sample shift and unit impulse · A finite truncation of every ℓ^2 signal
Exact mathematical passage
S,\delta_0
\LambdaNotation table · 20 passages
Search aliases: Λ Lambda
Notation table · atlas-pages.tex, line 28
Λ · A complete covering antichain of a finite tree · An arbitrary collection of nodes
Exact mathematical passage
\Lambda
\tauNotation table · 61 passages
Search aliases: τ tau
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1715
H_τ · the step packet 1_[τ,1]
Exact mathematical passage
H_\tau
\phiNotation table · 17 passages
Search aliases: ϕ phi
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1716
ϕ_m,τ · the truncated power packet (t-τ)_+^m/m!
Exact mathematical passage
\phi_{m,\tau}
\phi_{m,\tau}Notation table · 8 passages
Search aliases: ϕ_m,τ phi tau
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1716
ϕ_m,τ · the truncated power packet (t-τ)_+^m/m!
Exact mathematical passage
\phi_{m,\tau}
\ellNotation table · 70 passages
Search aliases: ℓ ell
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1502
[source reference] · HilbertPackets · Complex ℓ^2 impulse expansion, finite-support identity, unitary shifts, and closed shift-generated span.
Exact mathematical passage
\ell^2
\ell^2Notation table · 21 passages
Search aliases: ℓ^2 ell
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1502
[source reference] · HilbertPackets · Complex ℓ^2 impulse expansion, finite-support identity, unitary shifts, and closed shift-generated span.
Exact mathematical passage
\ell^2
\leqNotation table · 73 passages
Search aliases: ≤ leq
Notation table · Read passage · Paper_V_Traditional_Final.tex, line 1708
L≤ Z^d · a finite index sublattice
Exact mathematical passage
L\leq\ZZ^d
=Definition passage · 632 passages
Search aliases: =
Definition passage: Complete packet pruning · Read passage · Paper_V_Traditional_Final.tex, line 898
[Complete packet pruning] Let T be a finite rooted refinement tree. Every node v carries a projector P_v, the root satisfies P_∅=I, and every internal node satisfies P_v=∑_u∈ch(v)P_u, P_uP_w=0 (u≠ w in ch(v)). A set Λ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets Λ exactly once.
Exact mathematical passage
P_{\varnothing}=I
aDefinition passage · 123 passages
Search aliases: a
Definition passage: Sobolev step packet class · Read passage · Paper_V_Traditional_Final.tex, line 2521
[Sobolev step packet class] Fix r≥1, 1≤ p≤∞, and an integer Q≥0. The class F_r,p(Q) consists of functions f=s+∑_q=1^Qa_qϕ_0,τ_q with s∈ W^r,p(0,1) and 0<τ_1<⋯<τ_Q<1. No lower bound on the jump amplitudes is required for representation. Recovery results later impose such a bound.
Exact mathematical passage
f=s+\sum_{q=1}^{Q}a_q\phi_{0,\tau_q}
a_qDefinition passage · 18 passages
Search aliases: a_q
Definition passage: Sobolev step packet class · Read passage · Paper_V_Traditional_Final.tex, line 2521
[Sobolev step packet class] Fix r≥1, 1≤ p≤∞, and an integer Q≥0. The class F_r,p(Q) consists of functions f=s+∑_q=1^Qa_qϕ_0,τ_q with s∈ W^r,p(0,1) and 0<τ_1<⋯<τ_Q<1. No lower bound on the jump amplitudes is required for representation. Recovery results later impose such a bound.
Exact mathematical passage
f=s+\sum_{q=1}^{Q}a_q\phi_{0,\tau_q}
CDefinition passage · 30 passages
Search aliases: C
Definition passage: Linear tube bound · Read passage · Paper_V_Traditional_Final.tex, line 2832
[Linear tube bound] A boundary Γ has a linear tube bound if there are constants C_Γ and δ_0>0 such that μ(N_δ(Γ)) ≤ C_Γδ for 0<δ<δ_0.
Exact mathematical passage
C_\Gamma
C_\GammaDefinition passage · 6 passages
Search aliases: C_Γ Gamma
Definition passage: Linear tube bound · Read passage · Paper_V_Traditional_Final.tex, line 2832
[Linear tube bound] A boundary Γ has a linear tube bound if there are constants C_Γ and δ_0>0 such that μ(N_δ(Γ)) ≤ C_Γδ for 0<δ<δ_0.
Exact mathematical passage
C_\Gamma
\operatorname{ch}Definition passage · 3 passages
Search aliases: ch operatorname
Definition passage: Complete packet pruning · Read passage · Paper_V_Traditional_Final.tex, line 899
[Complete packet pruning] Let T be a finite rooted refinement tree. Every node v carries a projector P_v, the root satisfies P_∅=I, and every internal node satisfies P_v=∑_u∈ch(v)P_u, P_uP_w=0 (u≠ w in ch(v)). A set Λ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets Λ exactly once.
Exact mathematical passage
P_v=\sum_{u\in\operatorname{ch}(v)}P_u, \qquad P_uP_w=0\quad(u\ne w\text{ in }\operatorname{ch}(v)).
\operatorname{child}Definition passage · 4 passages
Search aliases: child operatorname
Definition passage: Unitary packet tree · Read passage · Paper_V_Traditional_Final.tex, line 2334
[Unitary packet tree] A finite unitary packet tree consists of Hilbert spaces H_v at its vertices and, at every internal vertex v, a unitary map U_v: H_v⟶⊕_w∈child(v) H_w. An admissible pruning is a set of vertices that meets every root to terminal path exactly once.
Exact mathematical passage
U_v:\Hcal_v\longrightarrow\bigoplus_{w\in\operatorname{child}(v)}\Hcal_w.
\FcalDefinition passage · 1 passage
Search aliases: Fcal F mathcal
Definition passage: Sobolev step packet class · Read passage · Paper_V_Traditional_Final.tex, line 2520
[Sobolev step packet class] Fix r≥1, 1≤ p≤∞, and an integer Q≥0. The class F_r,p(Q) consists of functions f=s+∑_q=1^Qa_qϕ_0,τ_q with s∈ W^r,p(0,1) and 0<τ_1<⋯<τ_Q<1. No lower bound on the jump amplitudes is required for representation. Recovery results later impose such a bound.
Exact mathematical passage
\Fcal_{r,p}(Q)
\Fcal_{r,p}Definition passage · 1 passage
Search aliases: Fcal_r,p F_r,p Fcal mathcal
Definition passage: Sobolev step packet class · Read passage · Paper_V_Traditional_Final.tex, line 2520
[Sobolev step packet class] Fix r≥1, 1≤ p≤∞, and an integer Q≥0. The class F_r,p(Q) consists of functions f=s+∑_q=1^Qa_qϕ_0,τ_q with s∈ W^r,p(0,1) and 0<τ_1<⋯<τ_Q<1. No lower bound on the jump amplitudes is required for representation. Recovery results later impose such a bound.
Exact mathematical passage
\Fcal_{r,p}(Q)
h_{rs}Definition passage · 5 passages
Search aliases: h_rs
Definition passage: Packet transfer kernel · Read passage · Paper_V_Traditional_Final.tex, line 1918
[Packet transfer kernel] For output residue r∈ R and input residue s∈ R, define a kernel on L by h_rs(m)=h(r-s+m), m∈ L.
Exact mathematical passage
h_{rs}(m)=h(r-s+m), \qquad m\in L.
\HcalDefinition passage · 3 passages
Search aliases: Hcal H mathcal
Definition passage: Unitary packet tree · Read passage · Paper_V_Traditional_Final.tex, line 2333
[Unitary packet tree] A finite unitary packet tree consists of Hilbert spaces H_v at its vertices and, at every internal vertex v, a unitary map U_v: H_v⟶⊕_w∈child(v) H_w. An admissible pruning is a set of vertices that meets every root to terminal path exactly once.
Exact mathematical passage
\Hcal_v
\Hcal_vDefinition passage · 3 passages
Search aliases: Hcal_v H_v Hcal mathcal
Definition passage: Unitary packet tree · Read passage · Paper_V_Traditional_Final.tex, line 2333
[Unitary packet tree] A finite unitary packet tree consists of Hilbert spaces H_v at its vertices and, at every internal vertex v, a unitary map U_v: H_v⟶⊕_w∈child(v) H_w. An admissible pruning is a set of vertices that meets every root to terminal path exactly once.
Exact mathematical passage
\Hcal_v
\Hcal_wDefinition passage · 1 passage
Search aliases: Hcal_w H_w Hcal mathcal
Definition passage: Unitary packet tree · Read passage · Paper_V_Traditional_Final.tex, line 2334
[Unitary packet tree] A finite unitary packet tree consists of Hilbert spaces H_v at its vertices and, at every internal vertex v, a unitary map U_v: H_v⟶⊕_w∈child(v) H_w. An admissible pruning is a set of vertices that meets every root to terminal path exactly once.
Exact mathematical passage
U_v:\Hcal_v\longrightarrow\bigoplus_{w\in\operatorname{child}(v)}\Hcal_w.
IDefinition passage · 57 passages
Search aliases: I
Definition passage: Complete packet pruning · Read passage · Paper_V_Traditional_Final.tex, line 898
[Complete packet pruning] Let T be a finite rooted refinement tree. Every node v carries a projector P_v, the root satisfies P_∅=I, and every internal node satisfies P_v=∑_u∈ch(v)P_u, P_uP_w=0 (u≠ w in ch(v)). A set Λ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets Λ exactly once.
Exact mathematical passage
P_{\varnothing}=I
nDefinition passage · 137 passages
Search aliases: n
Definition passage: Lattice packet · Read passage · Paper_V_Traditional_Final.tex, line 1769
[Lattice packet] For r∈ R, define the coordinate projector (P_rx)(n)= 1_r+L(n)x(n). The family (P_r)_r∈ R is called the packet system associated with L and R.
Exact mathematical passage
(P_rx)(n)=\one_{r+L}(n)x(n).
N_\deltaDefinition passage · 6 passages
Search aliases: N_δ delta
Definition passage: Linear tube bound · Read passage · Paper_V_Traditional_Final.tex, line 2833
[Linear tube bound] A boundary Γ has a linear tube bound if there are constants C_Γ and δ_0>0 such that μ(N_δ(Γ)) ≤ C_Γδ for 0<δ<δ_0.
Exact mathematical passage
\mu(N_\delta(\Gamma)) \leq C_\Gamma\delta
\one_{r+L}Definition passage · 1 passage
Search aliases: one_r+L 1_r+L one mathbf
Definition passage: Lattice packet · Read passage · Paper_V_Traditional_Final.tex, line 1769
[Lattice packet] For r∈ R, define the coordinate projector (P_rx)(n)= 1_r+L(n)x(n). The family (P_r)_r∈ R is called the packet system associated with L and R.
Exact mathematical passage
(P_rx)(n)=\one_{r+L}(n)x(n).
pDefinition passage · 94 passages
Search aliases: p
Definition passage: Sobolev step packet class · Read passage · Paper_V_Traditional_Final.tex, line 2520
[Sobolev step packet class] Fix r≥1, 1≤ p≤∞, and an integer Q≥0. The class F_r,p(Q) consists of functions f=s+∑_q=1^Qa_qϕ_0,τ_q with s∈ W^r,p(0,1) and 0<τ_1<⋯<τ_Q<1. No lower bound on the jump amplitudes is required for representation. Recovery results later impose such a bound.
Exact mathematical passage
1\leq p\leq\infty
P_uDefinition passage · 1 passage
Search aliases: P_u
Definition passage: Complete packet pruning · Read passage · Paper_V_Traditional_Final.tex, line 899
[Complete packet pruning] Let T be a finite rooted refinement tree. Every node v carries a projector P_v, the root satisfies P_∅=I, and every internal node satisfies P_v=∑_u∈ch(v)P_u, P_uP_w=0 (u≠ w in ch(v)). A set Λ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets Λ exactly once.
Exact mathematical passage
P_v=\sum_{u\in\operatorname{ch}(v)}P_u, \qquad P_uP_w=0\quad(u\ne w\text{ in }\operatorname{ch}(v)).
P_vDefinition passage · 6 passages
Search aliases: P_v
Definition passage: Complete packet pruning · Read passage · Paper_V_Traditional_Final.tex, line 898
[Complete packet pruning] Let T be a finite rooted refinement tree. Every node v carries a projector P_v, the root satisfies P_∅=I, and every internal node satisfies P_v=∑_u∈ch(v)P_u, P_uP_w=0 (u≠ w in ch(v)). A set Λ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets Λ exactly once.
Exact mathematical passage
P_v
P_wDefinition passage · 1 passage
Search aliases: P_w
Definition passage: Complete packet pruning · Read passage · Paper_V_Traditional_Final.tex, line 899
[Complete packet pruning] Let T be a finite rooted refinement tree. Every node v carries a projector P_v, the root satisfies P_∅=I, and every internal node satisfies P_v=∑_u∈ch(v)P_u, P_uP_w=0 (u≠ w in ch(v)). A set Λ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets Λ exactly once.
Exact mathematical passage
P_v=\sum_{u\in\operatorname{ch}(v)}P_u, \qquad P_uP_w=0\quad(u\ne w\text{ in }\operatorname{ch}(v)).
P_{\varnothing}Definition passage · 2 passages
Search aliases: P_∅ varnothing
Definition passage: Complete packet pruning · Read passage · Paper_V_Traditional_Final.tex, line 898
[Complete packet pruning] Let T be a finite rooted refinement tree. Every node v carries a projector P_v, the root satisfies P_∅=I, and every internal node satisfies P_v=∑_u∈ch(v)P_u, P_uP_w=0 (u≠ w in ch(v)). A set Λ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets Λ exactly once.
Exact mathematical passage
P_{\varnothing}=I
sDefinition passage · 164 passages
Search aliases: s
Definition passage: Packet transfer kernel · Read passage · Paper_V_Traditional_Final.tex, line 1917
[Packet transfer kernel] For output residue r∈ R and input residue s∈ R, define a kernel on L by h_rs(m)=h(r-s+m), m∈ L.
Exact mathematical passage
s\in R
\mathcal TDefinition passage · 9 passages
Search aliases: T mathcal
Definition passage: Complete packet pruning · Read passage · Paper_V_Traditional_Final.tex, line 898
[Complete packet pruning] Let T be a finite rooted refinement tree. Every node v carries a projector P_v, the root satisfies P_∅=I, and every internal node satisfies P_v=∑_u∈ch(v)P_u, P_uP_w=0 (u≠ w in ch(v)). A set Λ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets Λ exactly once.
Exact mathematical passage
\mathcal T
UDefinition passage · 7 passages
Search aliases: U
Definition passage: Unitary packet tree · Read passage · Paper_V_Traditional_Final.tex, line 2334
[Unitary packet tree] A finite unitary packet tree consists of Hilbert spaces H_v at its vertices and, at every internal vertex v, a unitary map U_v: H_v⟶⊕_w∈child(v) H_w. An admissible pruning is a set of vertices that meets every root to terminal path exactly once.
Exact mathematical passage
U_v:\Hcal_v\longrightarrow\bigoplus_{w\in\operatorname{child}(v)}\Hcal_w.
uDefinition passage · 43 passages
Search aliases: u
Definition passage: Complete packet pruning · Read passage · Paper_V_Traditional_Final.tex, line 899
[Complete packet pruning] Let T be a finite rooted refinement tree. Every node v carries a projector P_v, the root satisfies P_∅=I, and every internal node satisfies P_v=∑_u∈ch(v)P_u, P_uP_w=0 (u≠ w in ch(v)). A set Λ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets Λ exactly once.
Exact mathematical passage
P_v=\sum_{u\in\operatorname{ch}(v)}P_u, \qquad P_uP_w=0\quad(u\ne w\text{ in }\operatorname{ch}(v)).
U_vDefinition passage · 4 passages
Search aliases: U_v
Definition passage: Unitary packet tree · Read passage · Paper_V_Traditional_Final.tex, line 2334
[Unitary packet tree] A finite unitary packet tree consists of Hilbert spaces H_v at its vertices and, at every internal vertex v, a unitary map U_v: H_v⟶⊕_w∈child(v) H_w. An admissible pruning is a set of vertices that meets every root to terminal path exactly once.
Exact mathematical passage
U_v:\Hcal_v\longrightarrow\bigoplus_{w\in\operatorname{child}(v)}\Hcal_w.
vDefinition passage · 71 passages
Search aliases: v
Definition passage: Complete packet pruning · Read passage · Paper_V_Traditional_Final.tex, line 898
[Complete packet pruning] Let T be a finite rooted refinement tree. Every node v carries a projector P_v, the root satisfies P_∅=I, and every internal node satisfies P_v=∑_u∈ch(v)P_u, P_uP_w=0 (u≠ w in ch(v)). A set Λ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets Λ exactly once.
Exact mathematical passage
v
wDefinition passage · 25 passages
Search aliases: w
Definition passage: Complete packet pruning · Read passage · Paper_V_Traditional_Final.tex, line 899
[Complete packet pruning] Let T be a finite rooted refinement tree. Every node v carries a projector P_v, the root satisfies P_∅=I, and every internal node satisfies P_v=∑_u∈ch(v)P_u, P_uP_w=0 (u≠ w in ch(v)). A set Λ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets Λ exactly once.
Exact mathematical passage
P_v=\sum_{u\in\operatorname{ch}(v)}P_u, \qquad P_uP_w=0\quad(u\ne w\text{ in }\operatorname{ch}(v)).
W^{r,p}Definition passage · 11 passages
Search aliases: W^r,p
Definition passage: Sobolev step packet class · Read passage · Paper_V_Traditional_Final.tex, line 2524
[Sobolev step packet class] Fix r≥1, 1≤ p≤∞, and an integer Q≥0. The class F_r,p(Q) consists of functions f=s+∑_q=1^Qa_qϕ_0,τ_q with s∈ W^r,p(0,1) and 0<τ_1<⋯<τ_Q<1. No lower bound on the jump amplitudes is required for representation. Recovery results later impose such a bound.
Exact mathematical passage
s\in W^{r,p}(0,1)
xDefinition passage · 130 passages
Search aliases: x
Definition passage: Lattice packet · Read passage · Paper_V_Traditional_Final.tex, line 1769
[Lattice packet] For r∈ R, define the coordinate projector (P_rx)(n)= 1_r+L(n)x(n). The family (P_r)_r∈ R is called the packet system associated with L and R.
Exact mathematical passage
(P_rx)(n)=\one_{r+L}(n)x(n).
\muDefinition passage · 18 passages
Search aliases: μ mu
Definition passage: Linear tube bound · Read passage · Paper_V_Traditional_Final.tex, line 2833
[Linear tube bound] A boundary Γ has a linear tube bound if there are constants C_Γ and δ_0>0 such that μ(N_δ(Γ)) ≤ C_Γδ for 0<δ<δ_0.
Exact mathematical passage
\mu(N_\delta(\Gamma)) \leq C_\Gamma\delta
\tau_1Definition passage · 2 passages
Search aliases: τ_1 tau
Definition passage: Sobolev step packet class · Read passage · Paper_V_Traditional_Final.tex, line 2525
[Sobolev step packet class] Fix r≥1, 1≤ p≤∞, and an integer Q≥0. The class F_r,p(Q) consists of functions f=s+∑_q=1^Qa_qϕ_0,τ_q with s∈ W^r,p(0,1) and 0<τ_1<⋯<τ_Q<1. No lower bound on the jump amplitudes is required for representation. Recovery results later impose such a bound.
Exact mathematical passage
0<\tau_1<\cdots<\tau_Q<1.
\tau_QDefinition passage · 2 passages
Search aliases: τ_Q tau
Definition passage: Sobolev step packet class · Read passage · Paper_V_Traditional_Final.tex, line 2525
[Sobolev step packet class] Fix r≥1, 1≤ p≤∞, and an integer Q≥0. The class F_r,p(Q) consists of functions f=s+∑_q=1^Qa_qϕ_0,τ_q with s∈ W^r,p(0,1) and 0<τ_1<⋯<τ_Q<1. No lower bound on the jump amplitudes is required for representation. Recovery results later impose such a bound.
Exact mathematical passage
0<\tau_1<\cdots<\tau_Q<1.
\tau_qDefinition passage · 15 passages
Search aliases: τ_q tau
Definition passage: Sobolev step packet class · Read passage · Paper_V_Traditional_Final.tex, line 2521
[Sobolev step packet class] Fix r≥1, 1≤ p≤∞, and an integer Q≥0. The class F_r,p(Q) consists of functions f=s+∑_q=1^Qa_qϕ_0,τ_q with s∈ W^r,p(0,1) and 0<τ_1<⋯<τ_Q<1. No lower bound on the jump amplitudes is required for representation. Recovery results later impose such a bound.
Exact mathematical passage
f=s+\sum_{q=1}^{Q}a_q\phi_{0,\tau_q}
\phi_{0,\tau_q}Definition passage · 1 passage
Search aliases: ϕ_0,τ_q phi tau
Definition passage: Sobolev step packet class · Read passage · Paper_V_Traditional_Final.tex, line 2521
[Sobolev step packet class] Fix r≥1, 1≤ p≤∞, and an integer Q≥0. The class F_r,p(Q) consists of functions f=s+∑_q=1^Qa_qϕ_0,τ_q with s∈ W^r,p(0,1) and 0<τ_1<⋯<τ_Q<1. No lower bound on the jump amplitudes is required for representation. Recovery results later impose such a bound.
Exact mathematical passage
f=s+\sum_{q=1}^{Q}a_q\phi_{0,\tau_q}
\varnothingDefinition passage · 10 passages
Search aliases: ∅ varnothing
Definition passage: Complete packet pruning · Read passage · Paper_V_Traditional_Final.tex, line 898
[Complete packet pruning] Let T be a finite rooted refinement tree. Every node v carries a projector P_v, the root satisfies P_∅=I, and every internal node satisfies P_v=∑_u∈ch(v)P_u, P_uP_w=0 (u≠ w in ch(v)). A set Λ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets Λ exactly once.
Exact mathematical passage
P_{\varnothing}=I
\inDefinition passage · 85 passages
Search aliases: ∈ in
Definition passage: Complete packet pruning · Read passage · Paper_V_Traditional_Final.tex, line 899
[Complete packet pruning] Let T be a finite rooted refinement tree. Every node v carries a projector P_v, the root satisfies P_∅=I, and every internal node satisfies P_v=∑_u∈ch(v)P_u, P_uP_w=0 (u≠ w in ch(v)). A set Λ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets Λ exactly once.
Exact mathematical passage
P_v=\sum_{u\in\operatorname{ch}(v)}P_u, \qquad P_uP_w=0\quad(u\ne w\text{ in }\operatorname{ch}(v)).
\sumDefinition passage · 82 passages
Search aliases: ∑ sum
Definition passage: Complete packet pruning · Read passage · Paper_V_Traditional_Final.tex, line 899
[Complete packet pruning] Let T be a finite rooted refinement tree. Every node v carries a projector P_v, the root satisfies P_∅=I, and every internal node satisfies P_v=∑_u∈ch(v)P_u, P_uP_w=0 (u≠ w in ch(v)). A set Λ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets Λ exactly once.
Exact mathematical passage
P_v=\sum_{u\in\operatorname{ch}(v)}P_u, \qquad P_uP_w=0\quad(u\ne w\text{ in }\operatorname{ch}(v)).
\sum_{q=1}^{Q}Definition passage · 6 passages
Search aliases: ∑_q=1^Q sum
Definition passage: Sobolev step packet class · Read passage · Paper_V_Traditional_Final.tex, line 2521
[Sobolev step packet class] Fix r≥1, 1≤ p≤∞, and an integer Q≥0. The class F_r,p(Q) consists of functions f=s+∑_q=1^Qa_qϕ_0,τ_q with s∈ W^r,p(0,1) and 0<τ_1<⋯<τ_Q<1. No lower bound on the jump amplitudes is required for representation. Recovery results later impose such a bound.
Exact mathematical passage
f=s+\sum_{q=1}^{Q}a_q\phi_{0,\tau_q}
\sum_{u\in\operatorname{ch}(v)}Definition passage · 3 passages
Search aliases: ∑_u∈ch(v) sum in operatorname
Definition passage: Complete packet pruning · Read passage · Paper_V_Traditional_Final.tex, line 899
[Complete packet pruning] Let T be a finite rooted refinement tree. Every node v carries a projector P_v, the root satisfies P_∅=I, and every internal node satisfies P_v=∑_u∈ch(v)P_u, P_uP_w=0 (u≠ w in ch(v)). A set Λ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets Λ exactly once.
Exact mathematical passage
P_v=\sum_{u\in\operatorname{ch}(v)}P_u, \qquad P_uP_w=0\quad(u\ne w\text{ in }\operatorname{ch}(v)).
\inftyDefinition passage · 49 passages
Search aliases: ∞ infty
Definition passage: Sobolev step packet class · Read passage · Paper_V_Traditional_Final.tex, line 2520
[Sobolev step packet class] Fix r≥1, 1≤ p≤∞, and an integer Q≥0. The class F_r,p(Q) consists of functions f=s+∑_q=1^Qa_qϕ_0,τ_q with s∈ W^r,p(0,1) and 0<τ_1<⋯<τ_Q<1. No lower bound on the jump amplitudes is required for representation. Recovery results later impose such a bound.
Exact mathematical passage
1\leq p\leq\infty
\neDefinition passage · 16 passages
Search aliases: ≠ ne
Definition passage: Complete packet pruning · Read passage · Paper_V_Traditional_Final.tex, line 899
[Complete packet pruning] Let T be a finite rooted refinement tree. Every node v carries a projector P_v, the root satisfies P_∅=I, and every internal node satisfies P_v=∑_u∈ch(v)P_u, P_uP_w=0 (u≠ w in ch(v)). A set Λ of nodes is a complete pruning if it is an antichain and every root-to-terminal path meets Λ exactly once.
Exact mathematical passage
P_v=\sum_{u\in\operatorname{ch}(v)}P_u, \qquad P_uP_w=0\quad(u\ne w\text{ in }\operatorname{ch}(v)).
\geqDefinition passage · 20 passages
Search aliases: ≥ geq
Definition passage: Sobolev step packet class · Read passage · Paper_V_Traditional_Final.tex, line 2520
[Sobolev step packet class] Fix r≥1, 1≤ p≤∞, and an integer Q≥0. The class F_r,p(Q) consists of functions f=s+∑_q=1^Qa_qϕ_0,τ_q with s∈ W^r,p(0,1) and 0<τ_1<⋯<τ_Q<1. No lower bound on the jump amplitudes is required for representation. Recovery results later impose such a bound.
Exact mathematical passage
r\geq1
\bigoplusDefinition passage · 5 passages
Search aliases: ⊕ bigoplus
Definition passage: Unitary packet tree · Read passage · Paper_V_Traditional_Final.tex, line 2334
[Unitary packet tree] A finite unitary packet tree consists of Hilbert spaces H_v at its vertices and, at every internal vertex v, a unitary map U_v: H_v⟶⊕_w∈child(v) H_w. An admissible pruning is a set of vertices that meets every root to terminal path exactly once.
Exact mathematical passage
U_v:\Hcal_v\longrightarrow\bigoplus_{w\in\operatorname{child}(v)}\Hcal_w.
\bigoplus_{w\in\operatorname{child}(v)}Definition passage · 1 passage
Search aliases: ⊕_w∈child(v) bigoplus in operatorname
Definition passage: Unitary packet tree · Read passage · Paper_V_Traditional_Final.tex, line 2334
[Unitary packet tree] A finite unitary packet tree consists of Hilbert spaces H_v at its vertices and, at every internal vertex v, a unitary map U_v: H_v⟶⊕_w∈child(v) H_w. An admissible pruning is a set of vertices that meets every root to terminal path exactly once.
Exact mathematical passage
U_v:\Hcal_v\longrightarrow\bigoplus_{w\in\operatorname{child}(v)}\Hcal_w.
\cdotsDefinition passage · 3 passages
Search aliases: ⋯ cdots
Definition passage: Sobolev step packet class · Read passage · Paper_V_Traditional_Final.tex, line 2525
[Sobolev step packet class] Fix r≥1, 1≤ p≤∞, and an integer Q≥0. The class F_r,p(Q) consists of functions f=s+∑_q=1^Qa_qϕ_0,τ_q with s∈ W^r,p(0,1) and 0<τ_1<⋯<τ_Q<1. No lower bound on the jump amplitudes is required for representation. Recovery results later impose such a bound.
Exact mathematical passage
0<\tau_1<\cdots<\tau_Q<1.
\longrightarrowDefinition passage · 9 passages
Search aliases: ⟶ longrightarrow
Definition passage: Unitary packet tree · Read passage · Paper_V_Traditional_Final.tex, line 2334
[Unitary packet tree] A finite unitary packet tree consists of Hilbert spaces H_v at its vertices and, at every internal vertex v, a unitary map U_v: H_v⟶⊕_w∈child(v) H_w. An admissible pruning is a set of vertices that meets every root to terminal path exactly once.
Exact mathematical passage
U_v:\Hcal_v\longrightarrow\bigoplus_{w\in\operatorname{child}(v)}\Hcal_w.
AIntroduction passage · 44 passages
Search aliases: A
Introduction passage · atlas-pages.tex, line 422
vpage23A pole is a chart boundary, not a broken group actionHOMOGENEOUS COORDINATES / NO DIVISION BY ZEROTraditional monograph: Theorem 13.2 (p. 26), Corollary 13.3 (p. 26). Represent impedance by a nonzero pair [u:v], where [cu:cv]=[u:v] for c≠0. A finite impedance is [z:1]; infinity is [1:0]. For a nonzero reflection multiplier w, use A(w)=frac121+w · 1-w 1-w · 1+w. vfigprojective_poles headThe action is defined before a ratio is taken The determinant is w, so the matrix cannot send a nonzero …
Exact mathematical passage
A(w)=\frac12\begin{pmatrix}1+w&1-w\\1-w&1+w\end{pmatrix}.
bIntroduction passage · 72 passages
Search aliases: b
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 621
Write H_0,H_1 for the literal low and detail operators. The preceding energy identity implies H_bx _2≤ x _2, so both have bounded Hilbert adjoints. At depth zero the only packet operator is the identity. If the energy identity holds at depth ℓ, apply the one-level identity to each W_ℓ,αx and sum; this proves it at depth ℓ+1.
Exact mathematical passage
\|H_bx\|_2\le\|x\|_2
\bbCIntroduction passage · 10 passages
Search aliases: bbC C mathbb
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 264
Let ℓ^2( Z; C), abbreviated ℓ^2( Z), be the Hilbert space of square-summable complex discrete signals. The inner product is conjugate-linear in its first argument. A superscript * on a bounded operator denotes its Hilbert adjoint. Write δ_n for the unit impulse at address n and let the bilateral shift be (Sx)[n]=x[n-1]. Then S^nδ_0=δ_n. The primitive signal seed is therefore s=δ_0, the address operators are the shift words S^n, and the completion is the ℓ^2 norm.
Exact mathematical passage
\ell^2(\bbZ;\bbC)
\bbZIntroduction passage · 19 passages
Search aliases: bbZ Z mathbb
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 264
Let ℓ^2( Z; C), abbreviated ℓ^2( Z), be the Hilbert space of square-summable complex discrete signals. The inner product is conjugate-linear in its first argument. A superscript * on a bounded operator denotes its Hilbert adjoint. Write δ_n for the unit impulse at address n and let the bilateral shift be (Sx)[n]=x[n-1]. Then S^nδ_0=δ_n. The primitive signal seed is therefore s=δ_0, the address operators are the shift words S^n, and the completion is the ℓ^2 norm.
Exact mathematical passage
\ell^2(\bbZ;\bbC)
cIntroduction passage · 40 passages
Search aliases: c
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 502
[Perfect-reconstruction criterion] For d∈ Z, such a bank reconstructs every input up to a delay of d channel indices exactly when the following Laurent-polynomial identity holds: R(z) E(z)=z^-dI_N. If the right side is c z^-dI_N with c≠0, reconstruction holds up to gain c and channel delay d. After interleaving, this is a delay of Nd original samples. A negative delay denotes an advance.
Exact mathematical passage
c z^{-d}I_N
D^2Introduction passage · 2 passages
Search aliases: D^2
Introduction passage · atlas-pages.tex, line 76
ix of actual delaysFIR TRANSFER / A FOUR-TAP EXAMPLETraditional monograph: Theorem 3.1 (p. 5). Take the causal filter h=(1,2,-1,1), with taps at 0,1,2,3, acting by y[n]=∑_qh[q]f[n-q]. Resolve the input into even and odd channels. Let D denote one delay in the channel index. vfigfir_transfer headReading one matrix entry The odd-input contribution to the even output is (2D+D^2)f_1. The tap at original delay one reaches the preceding odd channel sample; the tap at delay three reaches the odd channe…
Exact mathematical passage
(2D+D^2)f_1
\bm EIntroduction passage · 7 passages
Search aliases: E bm
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 497
[Perfect-reconstruction criterion] For d∈ Z, such a bank reconstructs every input up to a delay of d channel indices exactly when the following Laurent-polynomial identity holds: R(z) E(z)=z^-dI_N. If the right side is c z^-dI_N with c≠0, reconstruction holds up to gain c and channel delay d. After interleaving, this is a delay of Nd original samples. A negative delay denotes an advance.
Exact mathematical passage
\boxed{ \bm R(z)\bm E(z)=z^{-d}I_N. }
f_0Introduction passage · 2 passages
Search aliases: f_0
Introduction passage · atlas-pages.tex, line 62
of residues 0 and 2 modulo four. Therefore the coarse even mask is the sum of two disjoint fine masks: P_2,0=P_4,0+P_4,2. vfigrefinement platecaptionThe common horizontal axis is the original sample index. Empty cells denote masked-out addresses. There are no interpolated samples between the displayed cells. headWhy the coordinate convention matters The decimated channel f_0[q]=f[2q] has its own index q. At modulus four the descendants are f[4q] and f[4q+2]. These shorter arrays must be interlea…
Exact mathematical passage
f_0[q]=f[2q]
f_1Introduction passage · 2 passages
Search aliases: f_1
Introduction passage · atlas-pages.tex, line 76
ix of actual delaysFIR TRANSFER / A FOUR-TAP EXAMPLETraditional monograph: Theorem 3.1 (p. 5). Take the causal filter h=(1,2,-1,1), with taps at 0,1,2,3, acting by y[n]=∑_qh[q]f[n-q]. Resolve the input into even and odd channels. Let D denote one delay in the channel index. vfigfir_transfer headReading one matrix entry The odd-input contribution to the even output is (2D+D^2)f_1. The tap at original delay one reaches the preceding odd channel sample; the tap at delay three reaches the odd channe…
Exact mathematical passage
(2D+D^2)f_1
GIntroduction passage · 20 passages
Search aliases: G
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 1724
Let 1≤ p≤∞. For a countable set G, write ℓ^p(G) for the usual complex sequence space. For an integer d≥1, when G= Z^d, translation by a∈ Z^d is (T_ax)(n)=x(n-a). Let δ_0 be the impulse at the origin.
Exact mathematical passage
G
H_0Introduction passage · 1 passage
Search aliases: H_0
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 621
Write H_0,H_1 for the literal low and detail operators. The preceding energy identity implies H_bx _2≤ x _2, so both have bounded Hilbert adjoints. At depth zero the only packet operator is the identity. If the energy identity holds at depth ℓ, apply the one-level identity to each W_ℓ,αx and sum; this proves it at depth ℓ+1.
Exact mathematical passage
H_0,H_1
H_1Introduction passage · 1 passage
Search aliases: H_1
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 621
Write H_0,H_1 for the literal low and detail operators. The preceding energy identity implies H_bx _2≤ x _2, so both have bounded Hilbert adjoints. At depth zero the only packet operator is the identity. If the energy identity holds at depth ℓ, apply the one-level identity to each W_ℓ,αx and sum; this proves it at depth ℓ+1.
Exact mathematical passage
H_0,H_1
H_bIntroduction passage · 1 passage
Search aliases: H_b
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 621
Write H_0,H_1 for the literal low and detail operators. The preceding energy identity implies H_bx _2≤ x _2, so both have bounded Hilbert adjoints. At depth zero the only packet operator is the identity. If the energy identity holds at depth ℓ, apply the one-level identity to each W_ℓ,αx and sum; this proves it at depth ℓ+1.
Exact mathematical passage
\|H_bx\|_2\le\|x\|_2
iIntroduction passage · 52 passages
Search aliases: i
Introduction passage · atlas-pages.tex, line 431
1-w · 1+w. vfigprojective_poles headThe action is defined before a ratio is taken The determinant is w, so the matrix cannot send a nonzero pair to zero. Direct multiplication gives A(w)A(w')=A(ww'), A(1)=I. For w=i, the orbit through zero is [0:1] ⟼ [-i:1] ⟼ [1:0] ⟼ [i:1] ⟼ [0:1]. The third point is a legitimate projective point, even though a finite impedance formula has a zero denominator there. The second iterate is A(-1), which swaps the two coordinates. Thus a half turn exchanges 0 and ∞ a…
Exact mathematical passage
w=i
I_NIntroduction passage · 10 passages
Search aliases: I_N
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 497
[Perfect-reconstruction criterion] For d∈ Z, such a bank reconstructs every input up to a delay of d channel indices exactly when the following Laurent-polynomial identity holds: R(z) E(z)=z^-dI_N. If the right side is c z^-dI_N with c≠0, reconstruction holds up to gain c and channel delay d. After interleaving, this is a delay of Nd original samples. A negative delay denotes an advance.
Exact mathematical passage
\boxed{ \bm R(z)\bm E(z)=z^{-d}I_N. }
JIntroduction passage · 14 passages
Search aliases: J
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 922
Let y denote the observed data. For a complete pruning Λ, consider any finite criterion J_y(Λ) = R_y(Λ) +η L_y(Λ) +τ K(Λ), where R_y is a representation cost, L_y is a cross-packet leakage cost, and K is a tree-complexity penalty. Here R_y,L_y,K take finite real values, and η,τ are fixed real weights (usually nonnegative in applications). No additivity is needed for the first theorem.
Exact mathematical passage
J_y(\Lambda) = R_y(\Lambda) +\eta L_y(\Lambda) +\tau K(\Lambda),
J_yIntroduction passage · 6 passages
Search aliases: J_y
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 922
Let y denote the observed data. For a complete pruning Λ, consider any finite criterion J_y(Λ) = R_y(Λ) +η L_y(Λ) +τ K(Λ), where R_y is a representation cost, L_y is a cross-packet leakage cost, and K is a tree-complexity penalty. Here R_y,L_y,K take finite real values, and η,τ are fixed real weights (usually nonnegative in applications). No additivity is needed for the first theorem.
Exact mathematical passage
J_y(\Lambda) = R_y(\Lambda) +\eta L_y(\Lambda) +\tau K(\Lambda),
KIntroduction passage · 3 passages
Search aliases: K
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 922
Let y denote the observed data. For a complete pruning Λ, consider any finite criterion J_y(Λ) = R_y(Λ) +η L_y(Λ) +τ K(Λ), where R_y is a representation cost, L_y is a cross-packet leakage cost, and K is a tree-complexity penalty. Here R_y,L_y,K take finite real values, and η,τ are fixed real weights (usually nonnegative in applications). No additivity is needed for the first theorem.
Exact mathematical passage
J_y(\Lambda) = R_y(\Lambda) +\eta L_y(\Lambda) +\tau K(\Lambda),
kIntroduction passage · 92 passages
Search aliases: k
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 474
[Operator-block form] If H denotes convolution by h, the same identity is H=∑_k,sP_N,kHP_N,s. Thus the familiar polyphase matrix is exactly the transfer-block matrix of the seed calculus.
Exact mathematical passage
H=\sum_{k,s}P_{N,k}HP_{N,s}.
L_yIntroduction passage · 3 passages
Search aliases: L_y
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 922
Let y denote the observed data. For a complete pruning Λ, consider any finite criterion J_y(Λ) = R_y(Λ) +η L_y(Λ) +τ K(Λ), where R_y is a representation cost, L_y is a cross-packet leakage cost, and K is a tree-complexity penalty. Here R_y,L_y,K take finite real values, and η,τ are fixed real weights (usually nonnegative in applications). No additivity is needed for the first theorem.
Exact mathematical passage
J_y(\Lambda) = R_y(\Lambda) +\eta L_y(\Lambda) +\tau K(\Lambda),
\one_{[\tau,\infty)}Introduction passage · 1 passage
Search aliases: one_[τ,∞) 1_[τ,∞) one tau infty mathbf tau infty
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 3185
The adjacent difference detector is local in the sample domain. Wavelet coefficients and higher differences give related multiscale packets. In the Haar subsections the signal is defined on R. Here the step convention is H_τ(t)= 1_[τ,∞)(t), whose restriction to (0,1) agrees with the preceding interval step. A chosen extension beyond the observation interval must use this half-line step convention, and the smooth extension must satisfy the stated Lipschitz hypothesis on every displayed wavelet su…
Exact mathematical passage
H_\tau(t)=\one_{[\tau,\infty)}(t)
P_{2,0}Introduction passage · 4 passages
Search aliases: P_2,0
Introduction passage · atlas-pages.tex, line 56
vpage4Refine packets on a common latticeEXACT REFINEMENT / WHAT CAN BE ADDEDTraditional monograph: Theorem 2.3 (p. 5). An even sample belongs to exactly one of residues 0 and 2 modulo four. Therefore the coarse even mask is the sum of two disjoint fine masks: P_2,0=P_4,0+P_4,2. vfigrefinement platecaptionThe common horizontal axis is the original sample index. Empty cells denote masked-out addresses. There are no interpolated samples between the displayed cells. headWhy the coordinate convention…
Exact mathematical passage
P_{2,0}=P_{4,0}+P_{4,2}.
P_{4,0}Introduction passage · 3 passages
Search aliases: P_4,0
Introduction passage · atlas-pages.tex, line 56
vpage4Refine packets on a common latticeEXACT REFINEMENT / WHAT CAN BE ADDEDTraditional monograph: Theorem 2.3 (p. 5). An even sample belongs to exactly one of residues 0 and 2 modulo four. Therefore the coarse even mask is the sum of two disjoint fine masks: P_2,0=P_4,0+P_4,2. vfigrefinement platecaptionThe common horizontal axis is the original sample index. Empty cells denote masked-out addresses. There are no interpolated samples between the displayed cells. headWhy the coordinate convention…
Exact mathematical passage
P_{2,0}=P_{4,0}+P_{4,2}.
P_{4,2}Introduction passage · 3 passages
Search aliases: P_4,2
Introduction passage · atlas-pages.tex, line 56
vpage4Refine packets on a common latticeEXACT REFINEMENT / WHAT CAN BE ADDEDTraditional monograph: Theorem 2.3 (p. 5). An even sample belongs to exactly one of residues 0 and 2 modulo four. Therefore the coarse even mask is the sum of two disjoint fine masks: P_2,0=P_4,0+P_4,2. vfigrefinement platecaptionThe common horizontal axis is the original sample index. Empty cells denote masked-out addresses. There are no interpolated samples between the displayed cells. headWhy the coordinate convention…
Exact mathematical passage
P_{2,0}=P_{4,0}+P_{4,2}.
P_{N,k}Introduction passage · 7 passages
Search aliases: P_N,k
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 474
[Operator-block form] If H denotes convolution by h, the same identity is H=∑_k,sP_N,kHP_N,s. Thus the familiar polyphase matrix is exactly the transfer-block matrix of the seed calculus.
Exact mathematical passage
H=\sum_{k,s}P_{N,k}HP_{N,s}.
P_{N,s}Introduction passage · 5 passages
Search aliases: P_N,s
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 474
[Operator-block form] If H denotes convolution by h, the same identity is H=∑_k,sP_N,kHP_N,s. Thus the familiar polyphase matrix is exactly the transfer-block matrix of the seed calculus.
Exact mathematical passage
H=\sum_{k,s}P_{N,k}HP_{N,s}.
\bm RIntroduction passage · 6 passages
Search aliases: R bm
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 497
[Perfect-reconstruction criterion] For d∈ Z, such a bank reconstructs every input up to a delay of d channel indices exactly when the following Laurent-polynomial identity holds: R(z) E(z)=z^-dI_N. If the right side is c z^-dI_N with c≠0, reconstruction holds up to gain c and channel delay d. After interleaving, this is a delay of Nd original samples. A negative delay denotes an advance.
Exact mathematical passage
\boxed{ \bm R(z)\bm E(z)=z^{-d}I_N. }
R_yIntroduction passage · 3 passages
Search aliases: R_y
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 922
Let y denote the observed data. For a complete pruning Λ, consider any finite criterion J_y(Λ) = R_y(Λ) +η L_y(Λ) +τ K(Λ), where R_y is a representation cost, L_y is a cross-packet leakage cost, and K is a tree-complexity penalty. Here R_y,L_y,K take finite real values, and η,τ are fixed real weights (usually nonnegative in applications). No additivity is needed for the first theorem.
Exact mathematical passage
J_y(\Lambda) = R_y(\Lambda) +\eta L_y(\Lambda) +\tau K(\Lambda),
\RRIntroduction passage · 5 passages
Search aliases: RR R mathbb
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 3185
The adjacent difference detector is local in the sample domain. Wavelet coefficients and higher differences give related multiscale packets. In the Haar subsections the signal is defined on R. Here the step convention is H_τ(t)= 1_[τ,∞)(t), whose restriction to (0,1) agrees with the preceding interval step. A chosen extension beyond the observation interval must use this half-line step convention, and the smooth extension must satisfy the stated Lipschitz hypothesis on every displayed wavelet su…
Exact mathematical passage
\RR
S^nIntroduction passage · 5 passages
Search aliases: S^n
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 269
Let ℓ^2( Z; C), abbreviated ℓ^2( Z), be the Hilbert space of square-summable complex discrete signals. The inner product is conjugate-linear in its first argument. A superscript * on a bounded operator denotes its Hilbert adjoint. Write δ_n for the unit impulse at address n and let the bilateral shift be (Sx)[n]=x[n-1]. Then S^nδ_0=δ_n. The primitive signal seed is therefore s=δ_0, the address operators are the shift words S^n, and the completion is the ℓ^2 norm.
Exact mathematical passage
S^n\delta_0=\delta_n.
TIntroduction passage · 8 passages
Search aliases: T
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 1725
Let 1≤ p≤∞. For a countable set G, write ℓ^p(G) for the usual complex sequence space. For an integer d≥1, when G= Z^d, translation by a∈ Z^d is (T_ax)(n)=x(n-a). Let δ_0 be the impulse at the origin.
Exact mathematical passage
(T_ax)(n)=x(n-a).
T_aIntroduction passage · 3 passages
Search aliases: T_a
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 1725
Let 1≤ p≤∞. For a countable set G, write ℓ^p(G) for the usual complex sequence space. For an integer d≥1, when G= Z^d, translation by a∈ Z^d is (T_ax)(n)=x(n-a). Let δ_0 be the impulse at the origin.
Exact mathematical passage
(T_ax)(n)=x(n-a).
w'Introduction passage · 2 passages
Search aliases: w'
Introduction passage · atlas-pages.tex, line 428
1-w · 1+w. vfigprojective_poles headThe action is defined before a ratio is taken The determinant is w, so the matrix cannot send a nonzero pair to zero. Direct multiplication gives A(w)A(w')=A(ww'), A(1)=I. For w=i, the orbit through zero is [0:1] ⟼ [-i:1] ⟼ [1:0] ⟼ [i:1] ⟼ [0:1]. The third point is a legitimate projective point, even though a finite impedance formula has a zero denominator there. The second iterate is A(-1), which swaps the two coordinates. Thus a half turn exchanges 0 and ∞ a…
Exact mathematical passage
A(w)A(w')=A(ww'),\qquad A(1)=I.
W_{\ell,\alpha}Introduction passage · 5 passages
Search aliases: W_ℓ,α ell alpha
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 621
Write H_0,H_1 for the literal low and detail operators. The preceding energy identity implies H_bx _2≤ x _2, so both have bounded Hilbert adjoints. At depth zero the only packet operator is the identity. If the energy identity holds at depth ℓ, apply the one-level identity to each W_ℓ,αx and sum; this proves it at depth ℓ+1.
Exact mathematical passage
W_{\ell,\alpha}x
yIntroduction passage · 51 passages
Search aliases: y
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 921
Let y denote the observed data. For a complete pruning Λ, consider any finite criterion J_y(Λ) = R_y(Λ) +η L_y(Λ) +τ K(Λ), where R_y is a representation cost, L_y is a cross-packet leakage cost, and K is a tree-complexity penalty. Here R_y,L_y,K take finite real values, and η,τ are fixed real weights (usually nonnegative in applications). No additivity is needed for the first theorem.
Exact mathematical passage
y
z^{-d}Introduction passage · 6 passages
Search aliases: z^-d
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 497
[Perfect-reconstruction criterion] For d∈ Z, such a bank reconstructs every input up to a delay of d channel indices exactly when the following Laurent-polynomial identity holds: R(z) E(z)=z^-dI_N. If the right side is c z^-dI_N with c≠0, reconstruction holds up to gain c and channel delay d. After interleaving, this is a delay of Nd original samples. A negative delay denotes an advance.
Exact mathematical passage
\boxed{ \bm R(z)\bm E(z)=z^{-d}I_N. }
\delta_nIntroduction passage · 5 passages
Search aliases: δ_n delta
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 264
Let ℓ^2( Z; C), abbreviated ℓ^2( Z), be the Hilbert space of square-summable complex discrete signals. The inner product is conjugate-linear in its first argument. A superscript * on a bounded operator denotes its Hilbert adjoint. Write δ_n for the unit impulse at address n and let the bilateral shift be (Sx)[n]=x[n-1]. Then S^nδ_0=δ_n. The primitive signal seed is therefore s=δ_0, the address operators are the shift words S^n, and the completion is the ℓ^2 norm.
Exact mathematical passage
\delta_n
\etaIntroduction passage · 17 passages
Search aliases: η eta
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 922
Let y denote the observed data. For a complete pruning Λ, consider any finite criterion J_y(Λ) = R_y(Λ) +η L_y(Λ) +τ K(Λ), where R_y is a representation cost, L_y is a cross-packet leakage cost, and K is a tree-complexity penalty. Here R_y,L_y,K take finite real values, and η,τ are fixed real weights (usually nonnegative in applications). No additivity is needed for the first theorem.
Exact mathematical passage
J_y(\Lambda) = R_y(\Lambda) +\eta L_y(\Lambda) +\tau K(\Lambda),
\ell^pIntroduction passage · 9 passages
Search aliases: ℓ^p ell
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 1724
Let 1≤ p≤∞. For a countable set G, write ℓ^p(G) for the usual complex sequence space. For an integer d≥1, when G= Z^d, translation by a∈ Z^d is (T_ax)(n)=x(n-a). Let δ_0 be the impulse at the origin.
Exact mathematical passage
\ell^p(G)
\sum_{k,s}Introduction passage · 1 passage
Search aliases: ∑_k,s sum
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 474
[Operator-block form] If H denotes convolution by h, the same identity is H=∑_k,sP_N,kHP_N,s. Thus the familiar polyphase matrix is exactly the transfer-block matrix of the seed calculus.
Exact mathematical passage
H=\sum_{k,s}P_{N,k}HP_{N,s}.
\sum_qIntroduction passage · 8 passages
Search aliases: ∑_q sum
Introduction passage · atlas-pages.tex, line 73
vpage5A filter becomes a matrix of actual delaysFIR TRANSFER / A FOUR-TAP EXAMPLETraditional monograph: Theorem 3.1 (p. 5). Take the causal filter h=(1,2,-1,1), with taps at 0,1,2,3, acting by y[n]=∑_qh[q]f[n-q]. Resolve the input into even and odd channels. Let D denote one delay in the channel index. vfigfir_transfer headReading one matrix entry The odd-input contribution to the even output is (2D+D^2)f_1. The tap at original delay one reaches the preceding odd channel sample; the tap at delay…
Exact mathematical passage
y[n]=\sum_qh[q]f[n-q]
\leIntroduction passage · 32 passages
Search aliases: ≤ le
Introduction passage · Read passage · Paper_V_Traditional_Final.tex, line 621
Write H_0,H_1 for the literal low and detail operators. The preceding energy identity implies H_bx _2≤ x _2, so both have bounded Hilbert adjoints. At depth zero the only packet operator is the identity. If the energy identity holds at depth ℓ, apply the one-level identity to each W_ℓ,αx and sum; this proves it at depth ℓ+1.
Exact mathematical passage
\|H_bx\|_2\le\|x\|_2
\longmapstoIntroduction passage · 3 passages
Search aliases: ⟼ longmapsto
Introduction passage · atlas-pages.tex, line 432
1-w · 1+w. vfigprojective_poles headThe action is defined before a ratio is taken The determinant is w, so the matrix cannot send a nonzero pair to zero. Direct multiplication gives A(w)A(w')=A(ww'), A(1)=I. For w=i, the orbit through zero is [0:1] ⟼ [-i:1] ⟼ [1:0] ⟼ [i:1] ⟼ [0:1]. The third point is a legitimate projective point, even though a finite impedance formula has a zero denominator there. The second iterate is A(-1), which swaps the two coordinates. Thus a half turn exchanges 0 and ∞ a…
Exact mathematical passage
[0:1]\ \longmapsto\ [-i:1]\ \longmapsto\ [1:0] \ \longmapsto\ [i:1]\ \longmapsto\ [0:1].
\operatorname*Use in source · 1 passage
Search aliases: * operatorname
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 936
[label=(roman*)] item A global minimizer Λ_y^*∈*argmin_Λ∈ P( T)J_y(Λ) always exists. item If the minimizer is unique and at least one competing pruning exists, its gap Γ_y = min_Λ≠Λ_y^* (J_y(Λ)-J_y(Λ_y^*)) is attained and strictly positive. If the pruning set is a singleton, its only element minimizes every criterion; no real-valued minimum over an empty competitor set is introduced. item In the case with competitors, if new data y' satisfy sup_Λ∈ P( T) |J_y'(Λ)-J_y(Λ)| <fracΓ_y2, then Λ_y'^*=Λ_…
Exact mathematical passage
\Lambda_y^*\in\operatorname*{argmin}_{\Lambda\in\mathfrak P(\mathcal T)}J_y(\Lambda)
+^{m-\ell}Use in source · 1 passage
Search aliases: +^m-ℓ ell
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2559
For 0≤ ℓ≤ m, the classical derivative of ϕ_m,τ on the two open sides is ϕ_m,τ^(ℓ)(t) = frac(t-τ)_+^m-ℓ(m-ℓ)!. This derivative has zero jump when ℓ<m and a unit jump when ℓ=m. Derivatives of order larger than m vanish on both open sides. It follows that subtracting the term a_q,mϕ_m,τ_q removes exactly the jump in the derivative of order m without changing the other classical trace jumps. Hence s_f has matching traces through order r-1.
Exact mathematical passage
\phi_{m,\tau}^{(\ell)}(t) = \frac{(t-\tau)_+^{m-\ell}}{(m-\ell)!}.
\mathbf1Use in source · 13 passages
Search aliases: 1 mathbf
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 315
Fix an integer N≥1. For 0≤ r<N, define (P_N,rx)[n] = 1_n≡ r (mod N)x[n]. The corresponding decimated channel is x_r[q]=x[Nq+r].
Exact mathematical passage
(P_{N,r}x)[n] = \mathbf 1_{n\equiv r\, (\mathrm{mod}\,N)}x[n].
\mathbf1_{[\tau,1]}Use in source · 1 passage
Search aliases: 1_[τ,1] mathbf tau
Use in source · atlas-pages.tex, line 578
vpage31Derivative jumps have their own packetsTRACE DATA / A STEP AND A RAMPTraditional monograph: Theorem E7.2 (p. 56), Corollary E7.3 (p. 57). On [0,1], let H_τ=1_[τ,1] and take f(t)=t^2+2H_1/3(t)-3(t-2/3)_+. The function has a value jump 2 at 1/3 and a first-derivative jump -3 at 2/3. Its packet and remainder are j_f(t)=2ϕ_0,1/3(t)-3ϕ_1,2/3(t), s_f(t)=t^2. vfig[.92]ext_trace platecaptionThe displayed curves are exact polynomial pieces. The solid jump is not connected across its discontinuity.…
Exact mathematical passage
H_\tau=\mathbf1_{[\tau,1]}
\mathbf1_{[\tau,\infty)}Use in source · 1 passage
Search aliases: 1_[τ,∞) mathbf tau infty
Use in source · atlas-pages.tex, line 698
vpage37Haar sees a jump only through its position in the supportSCALE SEPARATION / A QUANTIFIED THRESHOLDTraditional monograph: Theorem E10.1 (p. 68), Theorem E10.2 (p. 69), Corollary E10.3 (p. 69). Use a normalized Haar wavelet on an available interval of length L on the real line, positive on its left half and negative on its right half. Here H_τ is the half-line step 1_[τ,∞), so there is no artificial second endpoint jump. If u is the relative jump position in the support, then |⟨ aH_τ,ψ⟩| =|…
Exact mathematical passage
\mathbf1_{[\tau,\infty)}
\mathbf1_{n\equiv r\, (\mathrm{mod}\,N)}Use in source · 1 passage
Search aliases: 1_n≡ r (mod N) mathbf equiv mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 315
Fix an integer N≥1. For 0≤ r<N, define (P_N,rx)[n] = 1_n≡ r (mod N)x[n]. The corresponding decimated channel is x_r[q]=x[Nq+r].
Exact mathematical passage
(P_{N,r}x)[n] = \mathbf 1_{n\equiv r\, (\mathrm{mod}\,N)}x[n].
\mathbf1_{\widehat\Omega}Use in source · 6 passages
Search aliases: 1_widehatΩ mathbf widehat Omega
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 706
Let (X,Σ,μ) be a measure space, with measurable masks Ω,widehatΩ⊆ X and scalar amplitudes a,widehat a∈ C (real amplitudes are a special case). Suppose a scalar jump packet is j=a 1_Ω, and an estimated packet is widehat j=widehat a 1_widehatΩ.
Exact mathematical passage
\widehat j=\widehat a\mathbf 1_{\widehat\Omega}.
\mathbf1_{\{|t|<1/4\}}Use in source · 1 passage
Search aliases: 1_{|t|<1/4} mathbf
Use in source · atlas-pages.tex, line 266
vpage16Why a Fourier cutoff cannot reproduce this jump pointwiseA CONTINUOUS PULSE / EXACT ENDPOINT VALUESTraditional monograph: Section 6 (p. 11), Section 8 (p. 13). Use a separate period-one example on [-1/2,1/2): j(t)= 1_{|t|<1/4}, S_mj(t)=tfrac12+frac2π∑_n=1^mfracsin(π n/2)ncos(2π nt). This is the exact finite Fourier formula. The drawn curves sample it numerically; the endpoint calculation below is symbolic. vfigfourier_jump headOne value is fixed at every cutoff At t=±1/4, every summand va…
Exact mathematical passage
j(t)=\mathbf 1_{\{|t|<1/4\}},\qquad S_mj(t)=\tfrac12+\frac2\pi\sum_{n=1}^{m}\frac{\sin(\pi n/2)}{n}\cos(2\pi nt).
\mathbf1_\OmegaUse in source · 3 passages
Search aliases: 1_Ω mathbf Omega
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 741
Write a 1_Ω-widehat a 1_widehatΩ =(a-widehat a) 1_Ω +widehat a( 1_Ω- 1_widehatΩ). Apply the triangle inequality and use 1_Ω- 1_widehatΩ _L^p =μ(ΩtrianglewidehatΩ)^1/p. For the supremum estimate, |1_Ω-1_widehatΩ|≤1 pointwise, and this function vanishes everywhere when the symmetric difference is empty. The same pointwise bound also bounds the essential supremum. Nonempty null symmetric differences need not contribute to the essential norm; the stated bound remains valid in that case and need not …
Exact mathematical passage
|\mathbf1_\Omega-\mathbf1_{\widehat\Omega}|\le1
\mathbf1_{\Omega}Use in source · 3 passages
Search aliases: 1_Ω mathbf Omega
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 702
Let (X,Σ,μ) be a measure space, with measurable masks Ω,widehatΩ⊆ X and scalar amplitudes a,widehat a∈ C (real amplitudes are a special case). Suppose a scalar jump packet is j=a 1_Ω, and an estimated packet is widehat j=widehat a 1_widehatΩ.
Exact mathematical passage
j=a\mathbf 1_{\Omega},
\mathbf1_{\Omega_q}Use in source · 1 passage
Search aliases: 1_Ω_q mathbf Omega
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1294
[Edge-packet approximation class] There should exist natural function spaces in which a signal is represented as f=s+∑_q a_q1_Ω_q, with separate norms for smooth approximation, boundary geometry, packet amplitude, and packet identification. A projector-resolved approximation theory on such spaces should yield rates that do not mix smooth truncation error with discontinuity-location error.
Exact mathematical passage
f=s+\sum_q a_q\mathbf1_{\Omega_q},
\mathbf1_{\Omega\triangle\widehat\Omega\ne\varnothing}Use in source · 1 passage
Search aliases: 1_ΩtrianglewidehatΩ≠∅ mathbf Omega triangle widehat Omega ne varnothing
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 722
orms, j-widehat j _L^p ≤ |a-widehat a| μ(Ω)^1/p + |widehat a| μ(ΩtrianglewidehatΩ)^1/p. If μ(Ω) and μ(ΩtrianglewidehatΩ) are finite, all terms in this displayed jump bound are finite and it is an ordinary real-valued L^p inequality. For the pointwise supremum, and consequently also the essential L^∞ seminorm, j-widehat j _∞ ≤ |a-widehat a|+|widehat a| 1_ΩtrianglewidehatΩ≠∅.
Exact mathematical passage
\|j-\widehat j\|_{\infty} \le |a-\widehat a|+|\widehat a|\, \mathbf 1_{\Omega\triangle\widehat\Omega\ne\varnothing}.
\mathcal AUse in source · 1 passage
Search aliases: A mathcal
Use in source · atlas-pages.tex, line 43
ump packet, not an estimated feature or a frequency band. headWhat has changed? Only where the values are stored The complete tuple of channels contains sixteen values, just as the original vector does. A single decimated channel is shorter, but keeping all four loses no samples. The residue projector P_4,rf is different: it keeps the original lattice and writes zeros outside phase r. S_4 A_4f=f, f _2^2=∑_r=0^3 f_r _2^2=57. The norm on the channel tuple is the sum-of-squares Hilbert norm, not th…
Exact mathematical passage
\mathcal S_4\mathcal A_4f=f, \qquad \|f\|_2^2=\sum_{r=0}^3\|f_r\|_2^2=57.
a^2Use in source · 2 passages
Search aliases: a^2
Use in source · atlas-pages.tex, line 340
Section 9 (p. 15). On the integer 8×8 grid, choose the disk-shaped mask Ω={(i,j):(i-3)^2+(j-3)^2≤5} and translate it one pixel to the right. Each mask contains 21 pixels; their symmetric difference contains exactly 10. vfigmask_geometry headA geometric error calculation is not a segmentation algorithm For a common amplitude a, the squared counting-measure error is 10a^2, entirely on the disagreement pixels. The full-image mean squared error is 10a^2/64. Both are exact, but answer different norma…
Exact mathematical passage
10a^2
A^qUse in source · 1 passage
Search aliases: A^q
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1075
[Cyclic product law] P_N,k(AB) = ∑_i+j≡ k (mod N)A_iB_j. For q∈ N, the qth-power identity is P_N,k(A^q) = ∑_i_1+⋯+i_q≡ k A_i_1⋯ A_i_q.
Exact mathematical passage
P_{N,k}(A^q) = \sum_{i_1+\cdots+i_q\equiv k} A_{i_1}\cdots A_{i_q}.
a_0Use in source · 3 passages
Search aliases: a_0
Use in source · atlas-pages.tex, line 205
nch entirely. For actual Haar packets, set P_v=W_v^*W_v. Parent-to-child conservation gives P_v=P_v0+P_v1. Replacing unselected parents by their children until the complete pruning is reached proves ∑_v∈ΛP_v=I. This is why the coefficients at different depths still form an exact coordinate system. headAn exhaustive finite check At depth three the pruning count obeys a_0=1 and a_d+1=1+a_d^2, giving 1,2,5,26. The supplied suite checks all 26 covering conditions and all 26 actual 16×16 projector-su…
Exact mathematical passage
a_0=1
a_1Use in source · 2 passages
Search aliases: a_1
Use in source · atlas-pages.tex, line 565
STANTS / A MIXED-DEPTH CERTIFICATETraditional monograph: Proposition E6.4 (p. 53). Suppose the root split has nonnegative squared-norm bounds A=1/2 and B=2. Retain the left child. Split the right child with bounds 3/4 and 5/4, then retain both grandchildren. vfig[.85]ext_tree headStart with one at every retained node The left subtree has bounds a_0=b_0=1. The right subtree has a_1=3/4 and b_1=5/4. The recursive certificate gives a_rm root=tfrac12min{1,tfrac34}=tfrac38, b_rm root=2max{1,tfrac54}=…
Exact mathematical passage
a_1=3/4
a_2Use in source · 1 passage
Search aliases: a_2
Use in source · ext_overlap.tex, line 13
le(4.199999999999999,2.0);node at(3.5,1.5) 1 ; node[left]at(0,1.5)a_1=2; filldraw[fill=orange!25,draw=white](0.0,0.0)rectangle(1.4,1.0);node at(0.7,0.5) 1 ; filldraw[fill=black!3,draw=white](1.4,0.0)rectangle(2.8,1.0);node at(2.0999999999999996,0.5) 0 ; filldraw[fill=teal!12,draw=white](2.8,0.0)rectangle(4.199999999999999,1.0);node at(3.5,0.5) 4 ; node[left]at(0,0.5)a_2=5; node[above]at(0.7,3)b_0=0;node[above]at(2.0999999999999996,3)b_1=2;node[above]at(3.5,3)b_2=5;node[anchor=west,align=left] at…
Exact mathematical passage
a_2=5
\mathcal A_4Use in source · 1 passage
Search aliases: A_4 mathcal
Use in source · atlas-pages.tex, line 43
ump packet, not an estimated feature or a frequency band. headWhat has changed? Only where the values are stored The complete tuple of channels contains sixteen values, just as the original vector does. A single decimated channel is shorter, but keeping all four loses no samples. The residue projector P_4,rf is different: it keeps the original lattice and writes zeros outside phase r. S_4 A_4f=f, f _2^2=∑_r=0^3 f_r _2^2=57. The norm on the channel tuple is the sum-of-squares Hilbert norm, not th…
Exact mathematical passage
\mathcal S_4\mathcal A_4f=f, \qquad \|f\|_2^2=\sum_{r=0}^3\|f_r\|_2^2=57.
a_{d+1}Use in source · 1 passage
Search aliases: a_d+1
Use in source · atlas-pages.tex, line 205
. For actual Haar packets, set P_v=W_v^*W_v. Parent-to-child conservation gives P_v=P_v0+P_v1. Replacing unselected parents by their children until the complete pruning is reached proves ∑_v∈ΛP_v=I. This is why the coefficients at different depths still form an exact coordinate system. headAn exhaustive finite check At depth three the pruning count obeys a_0=1 and a_d+1=1+a_d^2, giving 1,2,5,26. The supplied suite checks all 26 covering conditions and all 26 actual 16×16 projector-sum identities…
Exact mathematical passage
a_{d+1}=1+a_d^2
a_d^2Use in source · 1 passage
Search aliases: a_d^2
Use in source · atlas-pages.tex, line 205
. For actual Haar packets, set P_v=W_v^*W_v. Parent-to-child conservation gives P_v=P_v0+P_v1. Replacing unselected parents by their children until the complete pruning is reached proves ∑_v∈ΛP_v=I. This is why the coefficients at different depths still form an exact coordinate system. headAn exhaustive finite check At depth three the pruning count obeys a_0=1 and a_d+1=1+a_d^2, giving 1,2,5,26. The supplied suite checks all 26 covering conditions and all 26 actual 16×16 projector-sum identities…
Exact mathematical passage
a_{d+1}=1+a_d^2
A_iUse in source · 3 passages
Search aliases: A_i
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1059
Linear time invariance is not required for exact packet coordinates. Let A and B be formal power series over a commutative coefficient ring (in particular C). Set A(z)=∑_n≥0a_nz^n, A_i=P_N,iA. All phase indices range from 0 to N-1. For another representation B(z), multiplication gives a nonlinear or bilinear interaction.
Exact mathematical passage
A(z)=\sum_{n\ge0}a_nz^n, \qquad A_i=P_{N,i}A.
a_iUse in source · 6 passages
Search aliases: a_i
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2777
Let (E_i)_i=0^I and (F_j)_j=0^J be measurable partitions of the same domain, up to null sets. Let a_i and b_j be the corresponding values. The index zero may be used for a background region.
Exact mathematical passage
a_i
A_{i_1}Use in source · 1 passage
Search aliases: A_i_1
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1075
[Cyclic product law] P_N,k(AB) = ∑_i+j≡ k (mod N)A_iB_j. For q∈ N, the qth-power identity is P_N,k(A^q) = ∑_i_1+⋯+i_q≡ k A_i_1⋯ A_i_q.
Exact mathematical passage
P_{N,k}(A^q) = \sum_{i_1+\cdots+i_q\equiv k} A_{i_1}\cdots A_{i_q}.
A_{i_q}Use in source · 1 passage
Search aliases: A_i_q
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1075
[Cyclic product law] P_N,k(AB) = ∑_i+j≡ k (mod N)A_iB_j. For q∈ N, the qth-power identity is P_N,k(A^q) = ∑_i_1+⋯+i_q≡ k A_i_1⋯ A_i_q.
Exact mathematical passage
P_{N,k}(A^q) = \sum_{i_1+\cdots+i_q\equiv k} A_{i_1}\cdots A_{i_q}.
a_jUse in source · 3 passages
Search aliases: a_j
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3519
[Shifted pointwise monomial law] Let y(n)=∏_j=1^mx(n-a_j) and write a_j=Nc_j+b_j, 0≤ b_j<N. For output residue r, define r_j≡ r-b_jpmod N, 0≤ r_j<N, and let κ_r,b_j = 1, · r<b_j,
Exact mathematical passage
y(n)=\prod_{j=1}^{m}x(n-a_j)
a_kUse in source · 2 passages
Search aliases: a_k
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2803
[Background and packet error] Suppose the true and estimated packets use the same amplitude list (a_i)_i=0^I, but possibly different regions. Then j-widehat j _p^p = ∑_i,k|a_i-a_k|^pμ(E_i∩ F_k). Only off diagonal overlaps contribute.
Exact mathematical passage
\|j-\widehat j\|_p^p = \sum_{i,k}|a_i-a_k|^p\mu(E_i\cap F_k).
a_{\min}Use in source · 17 passages
Search aliases: a_min min
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1691
The principal recovery theorem uses adjacent sample contrasts. Let b_h=Mh+2ε be the largest contrast that can be produced by a Lipschitz background and bounded noise in one grid cell. If a_min>2b_h, then every threshold between b_h and a_min-b_h detects exactly the cells containing jumps. The jump locations are recovered within h/2, and each amplitude error is at most b_h. The probability theorem replaces the deterministic noise term by 2σsqrtlog(2n/δ). Two half shifted Haar grids provide a seco…
Exact mathematical passage
a_{\min}>2b_h,
A_nUse in source · 1 passage
Search aliases: A_n
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1467
A_n · S^n · address operator
Exact mathematical passage
A_n
a_nUse in source · 1 passage
Search aliases: a_n
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1059
Linear time invariance is not required for exact packet coordinates. Let A and B be formal power series over a commutative coefficient ring (in particular C). Set A(z)=∑_n≥0a_nz^n, A_i=P_N,iA. All phase indices range from 0 to N-1. For another representation B(z), multiplication gives a nonlinear or bilinear interaction.
Exact mathematical passage
A(z)=\sum_{n\ge0}a_nz^n, \qquad A_i=P_{N,i}A.
a_{q,m}Use in source · 3 passages
Search aliases: a_q,m
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2537
egintheorem[Higher order singular packet decomposition] Let 0=τ_0<τ_1<⋯<τ_Q<τ_Q+1=1. Suppose the restriction of f to every interval (τ_q,τ_q+1) belongs to W^r,p, and choose the one dimensional Sobolev representatives whose derivatives through order r-1 have one sided traces at the endpoints. Define a_q,m = jumpf^(m)_τ_q = f^(m)(τ_q+)-f^(m)(τ_q-) and j_f(t) = ∑_q=1^Q∑_m=0^r-1a_q,mϕ_m,τ_q(t). Then s_f=f-j_f has matching one sided derivatives through order r-1 at every breakpoint and belongs to W^r…
Exact mathematical passage
a_{q,m} = \jump{f^{(m)}}_{\tau_q} = f^{(m)}(\tau_q+)-f^{(m)}(\tau_q-)
a_{\rm root}Use in source · 1 passage
Search aliases: a_rm root rm
Use in source · atlas-pages.tex, line 566
Proposition E6.4 (p. 53). Suppose the root split has nonnegative squared-norm bounds A=1/2 and B=2. Retain the left child. Split the right child with bounds 3/4 and 5/4, then retain both grandchildren. vfig[.85]ext_tree headStart with one at every retained node The left subtree has bounds a_0=b_0=1. The right subtree has a_1=3/4 and b_1=5/4. The recursive certificate gives a_rm root=tfrac12min{1,tfrac34}=tfrac38, b_rm root=2max{1,tfrac54}=tfrac52. Consequently the total leaf coefficient energy l…
Exact mathematical passage
a_{\rm root}=\tfrac12\min\{1,\tfrac34\}=\tfrac38, \qquad b_{\rm root}=2\max\{1,\tfrac54\}=\tfrac52.
A_vUse in source · 4 passages
Search aliases: A_v
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2377
[Bounds for nonunitary local splits] Suppose every internal node has an analysis map U_v and constants 0≤ A_v≤ B_v<∞ satisfying A_v x ^2 ≤ ∑_w∈child(v) (U_vx)_w ^2 ≤ B_v x ^2. Define recursively a(v)= 1, · v retained,
Exact mathematical passage
0\leq A_v\leq B_v<\infty
\Acal_EUse in source · 2 passages
Search aliases: Acal_E A_E Acal mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2138
[Frame bounds from the polyphase symbol] Assume that A I_N ≤ E( e^ iω)^*E( e^ iω) ≤ B I_N for almost every ω, where 0<A≤ B<∞. Then the analysis bank satisfies A x _2^2 ≤ A_E x _2^2 ≤ B x _2^2. The pointwise matrix R_rm can( e^ iω) = (E^*E)^-1E^* is a bounded left inverse and gives the canonical dual synthesis bank.
Exact mathematical passage
A\|x\|_2^2 \leq \|\Acal_E x\|_2^2 \leq B\|x\|_2^2.
\Acal_L^{-1}Use in source · 2 passages
Search aliases: Acal_L^-1 A_L^-1 Acal mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1933
[Exact convolution transfer law] Let y=h*x. Then y_r(ℓ) = ∑_s∈ R∑_m∈ Lh_rs(m)x_s(ℓ-m). Equivalently, A_LH A_L^-1 = (H_rs)_r,s∈ R, where H_rs is convolution on L by h_rs.
Exact mathematical passage
\Acal_LH\Acal_L^{-1} = (H_{rs})_{r,s\in R},
\Acal_{\Lcal}Use in source · 2 passages
Search aliases: Acal_Lcal A_ L Acal Lcal mathcal mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2342
[Reconstruction on a pruned packet tree] Every admissible pruning L defines a unitary analysis map A_ L: H_rm root ⟶ ⊕_v∈ L H_v. Its inverse is obtained by applying the adjoints of the local splits from the leaves toward the root. In particular, x ^2 = ∑_v∈ L x_v ^2.
Exact mathematical passage
\Acal_{\Lcal}:\Hcal_{\rm root} \longrightarrow \bigoplus_{v\in\Lcal}\Hcal_v.
\Acal_{\Lcal}^{-1}Use in source · 1 passage
Search aliases: Acal_Lcal^-1 A_ L^-1 Acal Lcal mathcal mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2369
The synthesis map is unitary by [source reference]. Hence widehat x-x = A_ L^-1e has the same norm as e.
Exact mathematical passage
\widehat x-x =\Acal_{\Lcal}^{-1}e
BUse in source · 28 passages
Search aliases: B
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 506
multiplication: composing shifts adds their delays, and distributivity handles finite sums. Thus the matrix identity implies the asserted action. Conversely, let B= R E-cz^-dI_N and suppose its action vanishes on every channel sequence. Insert an impulse at channel s, address zero, with every other channel zero. Output channel k reads the coefficients of the Laurent polynomial B_ks at all addresses. All of these coefficients vanish, so B_ks=0 for every k,s. The representation is faithful, provin…
Exact mathematical passage
B_{ks}
\bm BUse in source · 1 passage
Search aliases: B bm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 506
Finite convolution represents Laurent-polynomial multiplication: composing shifts adds their delays, and distributivity handles finite sums. Thus the matrix identity implies the asserted action. Conversely, let B= R E-cz^-dI_N and suppose its action vanishes on every channel sequence. Insert an impulse at channel s, address zero, with every other channel zero. Output channel k reads the coefficients of the Laurent polynomial B_ks at all addresses. All of these coefficients vanish, so B_ks=0 for …
Exact mathematical passage
\bm B=\bm R\bm E-cz^{-d}I_N
b_0Use in source · 2 passages
Search aliases: b_0
Use in source · atlas-pages.tex, line 565
ate along branchesLOCAL FRAME CONSTANTS / A MIXED-DEPTH CERTIFICATETraditional monograph: Proposition E6.4 (p. 53). Suppose the root split has nonnegative squared-norm bounds A=1/2 and B=2. Retain the left child. Split the right child with bounds 3/4 and 5/4, then retain both grandchildren. vfig[.85]ext_tree headStart with one at every retained node The left subtree has bounds a_0=b_0=1. The right subtree has a_1=3/4 and b_1=5/4. The recursive certificate gives a_rm root=tfrac12min{1,tfrac34}=tf…
Exact mathematical passage
a_0=b_0=1
b_1Use in source · 2 passages
Search aliases: b_1
Use in source · atlas-pages.tex, line 565
ED-DEPTH CERTIFICATETraditional monograph: Proposition E6.4 (p. 53). Suppose the root split has nonnegative squared-norm bounds A=1/2 and B=2. Retain the left child. Split the right child with bounds 3/4 and 5/4, then retain both grandchildren. vfig[.85]ext_tree headStart with one at every retained node The left subtree has bounds a_0=b_0=1. The right subtree has a_1=3/4 and b_1=5/4. The recursive certificate gives a_rm root=tfrac12min{1,tfrac34}=tfrac38, b_rm root=2max{1,tfrac54}=tfrac52. Conse…
Exact mathematical passage
b_1=5/4
b_2Use in source · 1 passage
Search aliases: b_2
Use in source · ext_overlap.tex, line 14
raw=white](0.0,0.0)rectangle(1.4,1.0);node at(0.7,0.5) 1 ; filldraw[fill=black!3,draw=white](1.4,0.0)rectangle(2.8,1.0);node at(2.0999999999999996,0.5) 0 ; filldraw[fill=teal!12,draw=white](2.8,0.0)rectangle(4.199999999999999,1.0);node at(3.5,0.5) 4 ; node[left]at(0,0.5)a_2=5; node[above]at(0.7,3)b_0=0;node[above]at(2.0999999999999996,3)b_1=2;node[above]at(3.5,3)b_2=5;node[anchor=west,align=left] at(5,2)off-diagonal squared error
Exact mathematical passage
b_2=5
b_hUse in source · 31 passages
Search aliases: b_h
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1687
The principal recovery theorem uses adjacent sample contrasts. Let b_h=Mh+2ε be the largest contrast that can be produced by a Lipschitz background and bounded noise in one grid cell. If a_min>2b_h, then every threshold between b_h and a_min-b_h detects exactly the cells containing jumps. The jump locations are recovered within h/2, and each amplitude error is at most b_h. The probability theorem replaces the deterministic noise term by 2σsqrtlog(2n/δ). Two half shifted Haar grids provide a seco…
Exact mathematical passage
b_h=Mh+2\varepsilon
b_{h,\delta}Use in source · 7 passages
Search aliases: b_h,δ delta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3105
[Probability recovery bound] Fix 0<δ<1 and define t_n,δ = 2σ sqrtlog(frac2nδ) and b_h,δ=Mh+t_n,δ. With probability at least 1-δ, every contrast satisfies the conclusions of [source reference] with b_h replaced by b_h,δ.
Exact mathematical passage
b_{h,\delta}=Mh+t_{n,\delta}.
B_jUse in source · 2 passages
Search aliases: B_j
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1067
[Cyclic product law] P_N,k(AB) = ∑_i+j≡ k (mod N)A_iB_j. For q∈ N, the qth-power identity is P_N,k(A^q) = ∑_i_1+⋯+i_q≡ k A_i_1⋯ A_i_q.
Exact mathematical passage
\boxed{ P_{N,k}(AB) = \sum_{i+j\equiv k\, (\mathrm{mod}\,N)}A_iB_j. }
b_jUse in source · 12 passages
Search aliases: b_j
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2777
Let (E_i)_i=0^I and (F_j)_j=0^J be measurable partitions of the same domain, up to null sets. Let a_i and b_j be the corresponding values. The index zero may be used for a background region.
Exact mathematical passage
b_j
b_kUse in source · 1 passage
Search aliases: b_k
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2812
Apply [source reference] with b_k=a_k. The diagonal terms vanish.
Exact mathematical passage
b_k=a_k
B_{ks}Use in source · 2 passages
Search aliases: B_ks
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 506
multiplication: composing shifts adds their delays, and distributivity handles finite sums. Thus the matrix identity implies the asserted action. Conversely, let B= R E-cz^-dI_N and suppose its action vanishes on every channel sequence. Insert an impulse at channel s, address zero, with every other channel zero. Output channel k reads the coefficients of the Laurent polynomial B_ks at all addresses. All of these coefficients vanish, so B_ks=0 for every k,s. The representation is faithful, provin…
Exact mathematical passage
B_{ks}
b_{q,m}Use in source · 3 passages
Search aliases: b_q,m
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2571
[Uniqueness of the trace packet] Fix the breakpoints τ_q. If f=s+∑_q=1^Q∑_m=0^r-1b_q,mϕ_m,τ_q with s∈ W^r,p(0,1), then b_q,m=jumpf^(m)_τ_q. The finite truncated power packet is therefore the unique packet supported at the declared breakpoints that removes all trace jumps through order r-1.
Exact mathematical passage
f=s+\sum_{q=1}^{Q}\sum_{m=0}^{r-1}b_{q,m}\phi_{m,\tau_q}
b_{\rm root}Use in source · 1 passage
Search aliases: b_rm root rm
Use in source · atlas-pages.tex, line 566
Proposition E6.4 (p. 53). Suppose the root split has nonnegative squared-norm bounds A=1/2 and B=2. Retain the left child. Split the right child with bounds 3/4 and 5/4, then retain both grandchildren. vfig[.85]ext_tree headStart with one at every retained node The left subtree has bounds a_0=b_0=1. The right subtree has a_1=3/4 and b_1=5/4. The recursive certificate gives a_rm root=tfrac12min{1,tfrac34}=tfrac38, b_rm root=2max{1,tfrac54}=tfrac52. Consequently the total leaf coefficient energy l…
Exact mathematical passage
a_{\rm root}=\tfrac12\min\{1,\tfrac34\}=\tfrac38, \qquad b_{\rm root}=2\max\{1,\tfrac54\}=\tfrac52.
B_vUse in source · 4 passages
Search aliases: B_v
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2377
[Bounds for nonunitary local splits] Suppose every internal node has an analysis map U_v and constants 0≤ A_v≤ B_v<∞ satisfying A_v x ^2 ≤ ∑_w∈child(v) (U_vx)_w ^2 ≤ B_v x ^2. Define recursively a(v)= 1, · v retained,
Exact mathematical passage
0\leq A_v\leq B_v<\infty
\bbNUse in source · 1 passage
Search aliases: bbN N mathbb
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1074
[Cyclic product law] P_N,k(AB) = ∑_i+j≡ k (mod N)A_iB_j. For q∈ N, the qth-power identity is P_N,k(A^q) = ∑_i_1+⋯+i_q≡ k A_i_1⋯ A_i_q.
Exact mathematical passage
q\in\bbN
\bigsqcupUse in source · 1 passage
Search aliases: bigsqcup
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1759
Let L≤ Z^d be a sublattice of finite index q. Choose a set R⊂ Z^d containing one representative from every coset in Z^d/L. Thus Z^d=bigsqcup_r∈ R(r+L). For x: Z^d→ C, its channel at r is the sequence on L x_r(ℓ)=x(r+ℓ).
Exact mathematical passage
\ZZ^d=\bigsqcup_{r\in R}(r+L).
\bigsqcup_{r\in R}Use in source · 1 passage
Search aliases: bigsqcup_r∈ R bigsqcup in
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1759
Let L≤ Z^d be a sublattice of finite index q. Choose a set R⊂ Z^d containing one representative from every coset in Z^d/L. Thus Z^d=bigsqcup_r∈ R(r+L). For x: Z^d→ C, its channel at r is the sequence on L x_r(ℓ)=x(r+ℓ).
Exact mathematical passage
\ZZ^d=\bigsqcup_{r\in R}(r+L).
\bmodUse in source · 10 passages
Search aliases: bmod
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 371
one also has A_N S_N=I. Equip this finite product with the Hilbert norm (x_0,…,x_N-1) _ch^2 =∑_r=0^N-1 x_r _2^2. Then A_N is a linear isometric equivalence and x _2^2=∑_r=0^N-1 x_r _2^2. Euclidean division gives a unique pair (r,q) with 0≤ r<N and n=Nq+r, including negative n. The address maps n↦(nbmod N,nmathbindivN) and (r,q)↦ Nq+r are inverse. Analysis and synthesis are reindexing by this bijection. Reindexing the nonnegative sum ∑_n|x[n]|^2 proves the norm identity and shows that both maps p…
Exact mathematical passage
n\mapsto(n\bmod N,n\mathbin{\mathrm{div}}N)
\bottomruleUse in source · 3 passages
Search aliases: bottomrule
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 780
[One-dimensional errors] For 1024 samples on [-1,1) and Fourier cutoff m=18, the retained frequency indices are -18,…,18: 37 bins including the zero-frequency bin. The measured values were lrr method · RMSE · maximum absolute error global Fourier truncation · 5.4023×10^-2 · 3.4968×10^-1 known edge packet · 1.3692×10^-4 · 1.4503×10^-3 estimated edge packet · 2.4029×10^-2 · 5.9530×10^-2 The simple estimator recovered the binary mask exactly in this deliberately separated two-class example. That re…
Exact mathematical passage
\begin{array}{lrr} \toprule \text{method}&\RMSE&\text{maximum absolute error}\\ \midrule \text{global Fourier truncation}&5.4023\times10^{-2}&3.4968\times10^{-1}\\ \text{known edge packet}&1.3692\times10^{-4}&1.4503\times10^{-3}\\ \text{estimated edge packet}&2.4029\times10^{-2}&5.9530\times10^{-2}\\ \bottomrule \end{array}
\BVUse in source · 1 passage
Search aliases: BV operatorname
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3755
The higher order singular packet theorem removes finitely many trace jumps. A more complete approximation theory should connect packet norms to geometric measure quantities in BV, piecewise Sobolev spaces, and free discontinuity models. It should also determine when a redundant packet dictionary can identify the derivative order of a singularity rather than merely its location.
Exact mathematical passage
\BV
C^{-1}Use in source · 5 passages
Search aliases: C^-1
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1159
Let C(z)=(z-1)/(z+1). Then T_N,k=C^-1∘ R_N,k∘ C. For a division-free construction, represent a point by a nonzero pair [u:v], identifying [cu:cv]=[u:v] for c≠0. With w=ω_N^k, use the matrix A(w)=frac121+w · 1-w
Exact mathematical passage
T_{N,k}=C^{-1}\circ R_{N,k}\circ C.
c_0Use in source · 5 passages
Search aliases: c_0
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1737
[Impulse generation] For 1≤ p<∞, overlinespan{T_aδ_0:a∈ Z^d}^ℓ^p = ℓ^p( Z^d). The same statement holds in c_0( Z^d) with the supremum norm. In ℓ^∞( Z^d), the closed span of the translation orbit is exactly c_0( Z^d).
Exact mathematical passage
c_0(\ZZ^d)
c_jUse in source · 4 passages
Search aliases: c_j
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3523
[Shifted pointwise monomial law] Let y(n)=∏_j=1^mx(n-a_j) and write a_j=Nc_j+b_j, 0≤ b_j<N. For output residue r, define r_j≡ r-b_jpmod N, 0≤ r_j<N, and let κ_r,b_j = 1, · r<b_j,
Exact mathematical passage
a_j=Nc_j+b_j, \qquad 0\leq b_j<N.
C_QUse in source · 4 passages
Search aliases: C_Q
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3426
[Estimated packet Gibbs bound] Let Q_m be linear and uniformly bounded on L^p, with sup_m Q_m ≤ C_Q. For f=s+j and an estimated packet widehat j_m, define widehat f_m =Q_m(f-widehat j_m)+widehat j_m. Then f-widehat f_m _p ≤ s-Q_ms _p +(1+C_Q) j-widehat j_m _p. Consequently, smooth approximation and packet recovery may be improved independently. No claim of Gibbs removal is justified unless both terms are controlled in the norm under consideration.
Exact mathematical passage
\sup_m\|Q_m\|\leq C_Q.
c_vUse in source · 7 passages
Search aliases: c_v
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 977
[Dynamic-programming best packet basis] Suppose J_y(Λ)=∑_v∈Λc_v(y). For a node v, define recursively V_y(v) = c_v(y), · v terminal,
Exact mathematical passage
J_y(\Lambda)=\sum_{v\in\Lambda}c_v(y).
\cAUse in source · 4 passages
Search aliases: cA A mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 350
Define the analysis and synthesis maps A_Nx=(x_0,…,x_N-1), S_N(x_0,…,x_N-1)[Nq+r]=x_r[q].
Exact mathematical passage
\cA_Nx=(x_0,\ldots,x_{N-1}), \qquad \cS_N(x_0,\ldots,x_{N-1})[Nq+r]=x_r[q].
\cA_NUse in source · 4 passages
Search aliases: cA_N A_N cA mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 350
Define the analysis and synthesis maps A_Nx=(x_0,…,x_N-1), S_N(x_0,…,x_N-1)[Nq+r]=x_r[q].
Exact mathematical passage
\cA_Nx=(x_0,\ldots,x_{N-1}), \qquad \cS_N(x_0,\ldots,x_{N-1})[Nq+r]=x_r[q].
\mathrm{can}Use in source · 2 passages
Search aliases: can mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2164
[Coefficient noise under canonical synthesis] Under the hypotheses of [source reference], suppose the observed coefficient vector is c= A_Ex+ν. Canonical synthesis gives widehat x=R_canc and widehat x-x _2 ≤ frac1sqrt A ν _2.
Exact mathematical passage
\widehat x=R_{\mathrm{can}}c
\CCUse in source · 5 passages
Search aliases: CC C mathbb
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1749
[Finite periodic signals] Let G= Z^d/L, where L≤ Z^d has finite index. The translation orbit of the impulse at the identity is a basis of C^G. In particular, one impulse generates every signal on the finite periodic lattice.
Exact mathematical passage
\CC^G
\CC^GUse in source · 2 passages
Search aliases: CC^G C^G CC mathbb
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1749
[Finite periodic signals] Let G= Z^d/L, where L≤ Z^d has finite index. The translation orbit of the impulse at the identity is a basis of C^G. In particular, one impulse generates every signal on the finite periodic lattice.
Exact mathematical passage
\CC^G
\mathrm{ch}Use in source · 1 passage
Search aliases: ch mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 361
[Analysis--synthesis identity] S_N A_N=I. On the product of the channel spaces one also has A_N S_N=I. Equip this finite product with the Hilbert norm (x_0,…,x_N-1) _ch^2 =∑_r=0^N-1 x_r _2^2. Then A_N is a linear isometric equivalence and x _2^2=∑_r=0^N-1 x_r _2^2. Euclidean division gives a unique pair (r,q) with 0≤ r<N and n=Nq+r, including negative n. The address maps n↦(nbmod N,nmathbindivN) and (r,q)↦ Nq+r are inverse. Analysis and synthesis are reindexing by this bijection. Reindexing the …
Exact mathematical passage
\|(x_0,\ldots,x_{N-1})\|_{\mathrm{ch}}^2 =\sum_{r=0}^{N-1}\|x_r\|_2^2.
\operatorname{Child}Use in source · 3 passages
Search aliases: Child operatorname
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 384
Suppose N,M≥1 and Nmid M. Every coarse phase has a disjoint family of fine descendants: Child_M(N,r) = {s:0≤ s<M, s≡ rpmod N}.
Exact mathematical passage
\operatorname{Child}_{M}(N,r) = \{s:0\le s<M,\ s\equiv r\pmod N\}.
\operatorname{Child}_{M}Use in source · 1 passage
Search aliases: Child_M operatorname
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 384
Suppose N,M≥1 and Nmid M. Every coarse phase has a disjoint family of fine descendants: Child_M(N,r) = {s:0≤ s<M, s≡ rpmod N}.
Exact mathematical passage
\operatorname{Child}_{M}(N,r) = \{s:0\le s<M,\ s\equiv r\pmod N\}.
\cosUse in source · 5 passages
Search aliases: cos
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1199
For 0≤ρ<1 and angle θ, Γ=ρ e^ iθ gives z(ρ,θ) =frac1+ρ e^ iθ1-ρ e^ iθ, with r(ρ,θ)=frac1-ρ^21-2ρcosθ+ρ^2, x(ρ,θ)=frac2ρsinθ1-2ρcosθ+ρ^2. A fixed ρ is a constant-mismatch circle. Refining N increases angular resolution without changing ρ.
Exact mathematical passage
r(\rho,\theta)=\frac{1-\rho^2}{1-2\rho\cos\theta+\rho^2}, \qquad x(\rho,\theta)=\frac{2\rho\sin\theta}{1-2\rho\cos\theta+\rho^2}.
\cSUse in source · 3 passages
Search aliases: cS S mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 350
Define the analysis and synthesis maps A_Nx=(x_0,…,x_N-1), S_N(x_0,…,x_N-1)[Nq+r]=x_r[q].
Exact mathematical passage
\cA_Nx=(x_0,\ldots,x_{N-1}), \qquad \cS_N(x_0,\ldots,x_{N-1})[Nq+r]=x_r[q].
\cS_NUse in source · 3 passages
Search aliases: cS_N S_N cS mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 350
Define the analysis and synthesis maps A_Nx=(x_0,…,x_N-1), S_N(x_0,…,x_N-1)[Nq+r]=x_r[q].
Exact mathematical passage
\cA_Nx=(x_0,\ldots,x_{N-1}), \qquad \cS_N(x_0,\ldots,x_{N-1})[Nq+r]=x_r[q].
\mathbb DUse in source · 3 passages
Search aliases: D mathbb
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3657
Let H = {z∈ C:Re z>0} and define C(z)=fracz-1z+1. It maps H conformally onto the unit disk D, with inverse C^-1(w)=frac1+w1-w. For θ∈ R, let R_θ(w)= e^ iθw.
Exact mathematical passage
\mathbb D
D^{2}Use in source · 1 passage
Search aliases: D^2
Use in source · fir_transfer.tex, line 7
node[] at (6.30000,4.60000) h=(1,2,-1,1), D: u[n]↦ u[n-1]; node[anchor=east] at (1.80000,3.10000) output 0; draw[fill=black!0,draw=black!40,line width=.35pt] (2.20000,2.55000) rectangle ++(4.10000,1.05000); node[] at (4.25000,3.07500) 1 - D; draw[fill=black!14,draw=black!40,line width=.35pt] (6.30000,2.55000) rectangle ++(4.10000,1.05000); node[] at (8.35000,3.07500) D^2 + 2 D; node[anchor=east] at (1.80000,1.85000) output 1; draw[fill=black!14,draw=black!40,line width=.35pt] (2.20000,1.30000) r…
Exact mathematical passage
D^{2} + 2 D
D^qUse in source · 1 passage
Search aliases: D^q
Use in source · atlas-pages.tex, line 109
matrix. The same delay algebra explains both the off-diagonal phase swap and the channel powers. headSigned delays need the same care For an original-sample delay d at modulus N, y_k[n]=f_(k-d)bmod N [n+lfloorfrack-dNrfloor]. This includes advances when d<0. The finite verification checks the formula on signed sample addresses. A bank with product D^qI_N delays the original signal by Nq samples, not q samples. takeawayAlways state the unit of a delay. A phase permutation and a channel delay toge…
Exact mathematical passage
D^qI_N
D_kUse in source · 10 passages
Search aliases: D_k
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2920
me f(t) = s(t)+∑_q=1^Qa_qH_τ_q(t), H_τ(t)= 1_[τ,1](t), where s is Lipschitz with constant M, and Q≥1 is finite with nonzero jump amplitudes. The zero-jump case requires no minimum-amplitude condition. Suppose no jump lies at a sample point and no grid cell contains more than one jump. Observe y_k=f(t_k)+η_k. Define the adjacent contrast D_k=y_k+1-y_k.
Exact mathematical passage
D_k=y_{k+1}-y_k.
\ddUse in source · 2 passages
Search aliases: dd d mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2760
Assume τ<widehatτ. Differentiation with respect to the location parameter gives frac∂∂ uϕ_m,u(t) =-ϕ_m-1,u(t) for almost every t. Hence ϕ_m,τ-ϕ_m,widehatτ = ∫_τ^widehatτϕ_m-1,u d u. Minkowski's inequality gives ϕ_m,τ-ϕ_m,widehatτ _p ≤ |τ-widehatτ| sup_u ϕ_m-1,u _p. The supremum is bounded by the norm of t^m-1/(m-1)! on [0,1], which is the stated constant. The supremum norm argument is identical.
Exact mathematical passage
\phi_{m,\tau}-\phi_{m,\widehat\tau} = \int_\tau^{\widehat\tau}\phi_{m-1,u}\dd u.
\operatorname{diag}Use in source · 8 passages
Search aliases: diag operatorname
Use in source · atlas-pages.tex, line 124
finite kernels, composition requires only finitely many operations at each address. The converse uses impulses. Insert an impulse in one input channel. Each output channel lists every coefficient of the corresponding transfer polynomial. If all such responses vanish for RE-I, every coefficient is zero; hence the polynomial matrix itself is zero. headA deliberately misleading test The matrix B(D)=diag(D,1) satisfies B(1)=I_2. At that one transform value it appears perfect. In the time domain, how…
Exact mathematical passage
B(D)=\operatorname{diag}(D,1)
\distUse in source · 1 passage
Search aliases: dist operatorname
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2825
For a closed set Γ, write N_δ(Γ) = {x:dist(x,Γ)<δ}.
Exact mathematical passage
N_\delta(\Gamma) = \{x:\dist(x,\Gamma)<\delta\}.
\mathrm{div}Use in source · 2 passages
Search aliases: div mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 371
one also has A_N S_N=I. Equip this finite product with the Hilbert norm (x_0,…,x_N-1) _ch^2 =∑_r=0^N-1 x_r _2^2. Then A_N is a linear isometric equivalence and x _2^2=∑_r=0^N-1 x_r _2^2. Euclidean division gives a unique pair (r,q) with 0≤ r<N and n=Nq+r, including negative n. The address maps n↦(nbmod N,nmathbindivN) and (r,q)↦ Nq+r are inverse. Analysis and synthesis are reindexing by this bijection. Reindexing the nonnegative sum ∑_n|x[n]|^2 proves the norm identity and shows that both maps p…
Exact mathematical passage
n\mapsto(n\bmod N,n\mathbin{\mathrm{div}}N)
\eUse in source · 16 passages
Search aliases: e mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2052
[Quantitative imperfect reconstruction] Define the delay compensated defect D( e^ iω) = e^ i dωR( e^ iω)E( e^ iω)-I_N. If ε_rec = *ess sup_ω D( e^ iω) _op<∞, then the delay compensated reconstruction widetilde x satisfies widetilde x-x _2 ≤ ε_rec x _2. The constant is sharp as an operator norm.
Exact mathematical passage
D(\e^{\ii\omega}) = \e^{\ii d\omega}R(\e^{\ii\omega})E(\e^{\ii\omega})-I_N.
\mathbb EUse in source · 2 passages
Search aliases: E mathbb
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3090
Assume the noise variables are independent, centered, and σ sub Gaussian: E e^λη_k ≤ e^σ^2λ^2/2.
Exact mathematical passage
\mathbb E\e^{\lambda\eta_k} \leq \e^{\sigma^2\lambda^2/2}.
eUse in source · 46 passages
Search aliases: e
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 570
Split x into even and odd channels e,o. Define d=o-e, s=e+frac12d. The inverse is e=s-frac12d, o=d+e.
Exact mathematical passage
e,o
E^*Use in source · 3 passages
Search aliases: E^*
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2146
[Frame bounds from the polyphase symbol] Assume that A I_N ≤ E( e^ iω)^*E( e^ iω) ≤ B I_N for almost every ω, where 0<A≤ B<∞. Then the analysis bank satisfies A x _2^2 ≤ A_E x _2^2 ≤ B x _2^2. The pointwise matrix R_rm can( e^ iω) = (E^*E)^-1E^* is a bounded left inverse and gives the canonical dual synthesis bank.
Exact mathematical passage
R_{\rm can}(\e^{\ii\omega}) = (E^*E)^{-1}E^*
e^{-2\ii\beta\ell}Use in source · 1 passage
Search aliases: e^-2iiβℓ e^-2 iβℓ ii beta ell mathrm beta ell
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1214
For a lossless transmission line, Γ_rm in=Γ_L e^-2 iβℓ. One N-phase step corresponds to a line-length increment λ/(2N), up to the chosen orientation convention. Hence the phase address has a direct electrical interpretation.
Exact mathematical passage
\Gam_{\rm in}=\Gam_L e^{-2\ii\beta\ell}.
e^{10}Use in source · 1 passage
Search aliases: e^10
Use in source · atlas-pages.tex, line 691
1/2, σ=1/50, δ=1/20, and a_min=2/25. Since log(10240)<10<(16/5)^2, choosing R=16 makes the right-hand side strictly less than frac1256+frac150frac165 =frac217332000<frac225. Thus sixteen independent observations per sample site suffice for this bound. It is a sufficient budget, not a claim of optimal sample complexity. The exact checker certifies e^10>10240 using a rational Taylor partial sum, so the displayed logarithm bound is not based on rounded numerical logarithms. headDependence between a…
Exact mathematical passage
e^{10}>10240
\e^{2\pi\ii k/N}Use in source · 1 passage
Search aliases: e^2πii k/N e^2π i k/N e pi ii mathrm pi mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3691
+1) into C^-1 and simplify. Preservation of H follows because C maps it to D, rotation preserves D, and C^-1 maps back. The group law follows by conjugating the rotation group law. When θ=2π/N, the Nth power is the identity. Conversely, if its kth power is the identity, conjugation by C and evaluation at w=1/2 give e^2π i k/N=1, so N divides k. Hence the order is exactly N.
Exact mathematical passage
\e^{2\pi\ii k/N}=1
e^{2\pi\ii/N}Use in source · 1 passage
Search aliases: e^2πii/N e^2π i/N pi ii pi mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1133
Fix ω_N=e^2π i/N and rotate the reflection coefficient by R_N,k(Γ)=ω_N^kΓ.
Exact mathematical passage
\omega_N=e^{2\pi\ii/N}
\e^{\ii d\omega}Use in source · 1 passage
Search aliases: e^ii dω e^ i dω e ii omega mathrm mathrm omega
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2052
[Quantitative imperfect reconstruction] Define the delay compensated defect D( e^ iω) = e^ i dωR( e^ iω)E( e^ iω)-I_N. If ε_rec = *ess sup_ω D( e^ iω) _op<∞, then the delay compensated reconstruction widetilde x satisfies widetilde x-x _2 ≤ ε_rec x _2. The constant is sharp as an operator norm.
Exact mathematical passage
D(\e^{\ii\omega}) = \e^{\ii d\omega}R(\e^{\ii\omega})E(\e^{\ii\omega})-I_N.
\e^{\ii(\varphi+\theta)}Use in source · 1 passage
Search aliases: e^ii(φ+θ) e^ i(φ+θ) e ii varphi theta mathrm mathrm varphi theta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3714
[Radial and phase coordinates] Write C(z)=ρ e^ iφ, 0≤ρ<1. Then C(Φ_θ(z)) = ρ e^ i(φ+θ). The radial mismatch ρ is invariant and the phase coordinate advances by θ.
Exact mathematical passage
C(\Phi_\theta(z)) = \rho\e^{\ii(\varphi+\theta)}.
\e^{\ii\theta}Use in source · 3 passages
Search aliases: e^iiθ e^ iθ e ii theta mathrm mathrm theta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3662
Let H = {z∈ C:Re z>0} and define C(z)=fracz-1z+1. It maps H conformally onto the unit disk D, with inverse C^-1(w)=frac1+w1-w. For θ∈ R, let R_θ(w)= e^ iθw.
Exact mathematical passage
R_\theta(w)=\e^{\ii\theta}w.
e^{\ii\theta}Use in source · 2 passages
Search aliases: e^iiθ e^ iθ ii theta mathrm theta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1190
For 0≤ρ<1 and angle θ, Γ=ρ e^ iθ gives z(ρ,θ) =frac1+ρ e^ iθ1-ρ e^ iθ, with r(ρ,θ)=frac1-ρ^21-2ρcosθ+ρ^2, x(ρ,θ)=frac2ρsinθ1-2ρcosθ+ρ^2. A fixed ρ is a constant-mismatch circle. Refining N increases angular resolution without changing ρ.
Exact mathematical passage
\Gam=\rho e^{\ii\theta}
\e^{\ii\pi}Use in source · 1 passage
Search aliases: e^iiπ e^ iπ e ii pi mathrm mathrm pi
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3703
Insert e^ iπ=-1 in the formula of [source reference].
Exact mathematical passage
\e^{\ii\pi}=-1
\e^{\ii\varphi}Use in source · 1 passage
Search aliases: e^iiφ e^ iφ e ii varphi mathrm mathrm varphi
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3708
[Radial and phase coordinates] Write C(z)=ρ e^ iφ, 0≤ρ<1. Then C(Φ_θ(z)) = ρ e^ i(φ+θ). The radial mismatch ρ is invariant and the phase coordinate advances by θ.
Exact mathematical passage
C(z)=\rho\e^{\ii\varphi}, \qquad 0\leq\rho<1.
\e^{\ii\omega}Use in source · 8 passages
Search aliases: e^iiω e^ iω e ii omega mathrm mathrm omega
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2052
[Quantitative imperfect reconstruction] Define the delay compensated defect D( e^ iω) = e^ i dωR( e^ iω)E( e^ iω)-I_N. If ε_rec = *ess sup_ω D( e^ iω) _op<∞, then the delay compensated reconstruction widetilde x satisfies widetilde x-x _2 ≤ ε_rec x _2. The constant is sharp as an operator norm.
Exact mathematical passage
D(\e^{\ii\omega}) = \e^{\ii d\omega}R(\e^{\ii\omega})E(\e^{\ii\omega})-I_N.
e^{\lambda\eta}Use in source · 1 passage
Search aliases: e^λη lambda eta
Use in source · atlas-pages.tex, line 679
vpage36Repeated observations shrink the noise side of the gapSUBGAUSSIAN RECOVERY / A CONSERVATIVE SAMPLE BUDGETTraditional monograph: Theorem E9.8 (p. 67), Corollary E9.9 (p. 67), Corollary E9.11 (p. 67). Suppose all noise observations are independent and centered, with moment-generating bound E e^λη≤ e^σ^2λ^2/2. Averaging R observations at each of n+1 sample sites replaces σ by σ/sqrt R. A sufficient recovery condition is a_min>2Mh+frac4σsqrt Rsqrtlog(2n/δ). The probability of simultaneous rec…
Exact mathematical passage
\mathbb E e^{\lambda\eta}\le e^{\sigma^2\lambda^2/2}
\e^{\lambda\eta_k}Use in source · 1 passage
Search aliases: e^λη_k e lambda eta mathrm lambda eta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3090
Assume the noise variables are independent, centered, and σ sub Gaussian: E e^λη_k ≤ e^σ^2λ^2/2.
Exact mathematical passage
\mathbb E\e^{\lambda\eta_k} \leq \e^{\sigma^2\lambda^2/2}.
\e^{\sigma^2\lambda^2/2}Use in source · 1 passage
Search aliases: e^σ^2λ^2/2 e sigma lambda mathrm sigma lambda
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3090
Assume the noise variables are independent, centered, and σ sub Gaussian: E e^λη_k ≤ e^σ^2λ^2/2.
Exact mathematical passage
\mathbb E\e^{\lambda\eta_k} \leq \e^{\sigma^2\lambda^2/2}.
e^{\sigma^2\lambda^2/2}Use in source · 1 passage
Search aliases: e^σ^2λ^2/2 sigma lambda
Use in source · atlas-pages.tex, line 679
vpage36Repeated observations shrink the noise side of the gapSUBGAUSSIAN RECOVERY / A CONSERVATIVE SAMPLE BUDGETTraditional monograph: Theorem E9.8 (p. 67), Corollary E9.9 (p. 67), Corollary E9.11 (p. 67). Suppose all noise observations are independent and centered, with moment-generating bound E e^λη≤ e^σ^2λ^2/2. Averaging R observations at each of n+1 sample sites replaces σ by σ/sqrt R. A sufficient recovery condition is a_min>2Mh+frac4σsqrt Rsqrtlog(2n/δ). The probability of simultaneous rec…
Exact mathematical passage
\mathbb E e^{\lambda\eta}\le e^{\sigma^2\lambda^2/2}
E_0Use in source · 3 passages
Search aliases: E_0
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2285
[Exact lifting update] The lifting matrices are invertible with L_P^-1=L_-P, U_Q^-1=U_-Q. If R_0E_0=I, then every finite product V of lifting matrices gives another exact bank by E=VE_0, R=R_0V^-1. Every lifting step has determinant one.
Exact mathematical passage
R_0E_0=I
e_0Use in source · 2 passages
Search aliases: e_0
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2102
[Stability under filter perturbation] Assume RE=z^-dI_N. Let widetilde E=E+Δ E, widetilde R=R+Δ R. Set e_0 · =*ess sup E _op, · r_0 · =*ess sup R _op, δ_E · =*ess sup Δ E _op, · δ_R · =*ess sup Δ R _op. Then the delay compensated reconstruction defect obeys ε_rec ≤ r_0δ_E+e_0δ_R+δ_Rδ_E.
Exact mathematical passage
\begin{aligned} e_0&=\esssup\|E\|_{\mathrm{op}}, & r_0&=\esssup\|R\|_{\mathrm{op}},\\ \delta_E&=\esssup\|\Delta E\|_{\mathrm{op}}, & \delta_R&=\esssup\|\Delta R\|_{\mathrm{op}}. \end{aligned}
E_HUse in source · 2 passages
Search aliases: E_H
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2232
The Haar matrix is E_H = frac1sqrt2 1 · 1 1 · -1 . For even and odd samples e(n)=x(2n) and o(n)=x(2n+1), the low and high coordinates are a(n)=frace(n)+o(n)sqrt2, d(n)=frace(n)-o(n)sqrt2.
Exact mathematical passage
E_H = \frac1{\sqrt2} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}.
\bm E_{\mathrm{H}}Use in source · 2 passages
Search aliases: E_H bm mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 524
The Haar analysis matrix is E_H = frac1sqrt2 1 · 1 1 · -1 . It is orthogonal, so R_H= E_H^ T and R_H E_H=I_2. The channels are a[n]=fracx[2n]+x[2n+1]sqrt2, d[n]=fracx[2n]-x[2n+1]sqrt2.
Exact mathematical passage
\bm E_{\mathrm{H}} = \frac1{\sqrt2} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}.
\bm E_{\mathrm{H}}^{\mathsf T}Use in source · 1 passage
Search aliases: E_H^ T bm mathrm mathsf
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 533
1 · -1 . It is orthogonal, so R_H= E_H^ T and R_H E_H=I_2. The channels are a[n]=fracx[2n]+x[2n+1]sqrt2, d[n]=fracx[2n]-x[2n+1]sqrt2.
Exact mathematical passage
\bm R_{\mathrm{H}}=\bm E_{\mathrm{H}}^{\mathsf T}
E_iUse in source · 8 passages
Search aliases: E_i
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2777
Let (E_i)_i=0^I and (F_j)_j=0^J be measurable partitions of the same domain, up to null sets. Let a_i and b_j be the corresponding values. The index zero may be used for a background region.
Exact mathematical passage
(E_i)_{i=0}^{I}
\esssupUse in source · 4 passages
Search aliases: esssup *ess sup operatorname
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2058
[Quantitative imperfect reconstruction] Define the delay compensated defect D( e^ iω) = e^ i dωR( e^ iω)E( e^ iω)-I_N. If ε_rec = *ess sup_ω D( e^ iω) _op<∞, then the delay compensated reconstruction widetilde x satisfies widetilde x-x _2 ≤ ε_rec x _2. The constant is sharp as an operator norm.
Exact mathematical passage
\varepsilon_{\mathrm{rec}} = \esssup_{\omega}\|D(\e^{\ii\omega})\|_{\mathrm{op}}<\infty,
\esssup_\omegaUse in source · 2 passages
Search aliases: esssup_ω *ess sup_ω esssup omega operatorname omega
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2087
0 · 0.02sinω . Then *ess sup_ω D( e^ iω) _op =0.02. Every input therefore has relative reconstruction error at most two percent in ℓ^2. The estimate is independent of the input length.
Exact mathematical passage
\esssup_\omega\|D(\e^{\ii\omega})\|_{\mathrm{op}} =0.02.
\esssup_{\omega}Use in source · 1 passage
Search aliases: esssup_ω *ess sup_ω esssup omega operatorname omega
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2058
[Quantitative imperfect reconstruction] Define the delay compensated defect D( e^ iω) = e^ i dωR( e^ iω)E( e^ iω)-I_N. If ε_rec = *ess sup_ω D( e^ iω) _op<∞, then the delay compensated reconstruction widetilde x satisfies widetilde x-x _2 ≤ ε_rec x _2. The constant is sharp as an operator norm.
Exact mathematical passage
\varepsilon_{\mathrm{rec}} = \esssup_{\omega}\|D(\e^{\ii\omega})\|_{\mathrm{op}}<\infty,
\expUse in source · 2 passages
Search aliases: exp
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1177
[Impedance--admittance is the two-phase split] For N=2 and finite z≠0, T_2,1(z)=z^-1. On the full projective line it exchanges 0 and ∞. Thus a half-turn of the reflection disk exchanges normalized impedance and admittance. ω_2=exp(π i)=-1, so A(ω_2)[u:v]=[v:u]. The finite ratio is 1/z when z≠0; the coordinate swap also gives [0:1]↔[1:0]. No ordinary division by zero is used to represent infinity.
Exact mathematical passage
\omega_2=\exp(\pi\ii)=-1
FUse in source · 48 passages
Search aliases: F
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 290
The shift orbit is the standard orthonormal basis {δ_n}_n∈ Z. For x∈ℓ^2, the finite truncations x_F=∑_n∈ Fx[n]δ_n, F⊂ Z finite, satisfy x-x_F _2^2=∑_n∉ F|x[n]|^2→0 as the finite sets exhaust Z. Thus the unordered Hilbert-space sum ∑_nx[n]δ_n converges to x, and the finite-support statement is exact. The shift is a surjective isometry because n↦ n-1 is a bijection of Z, with inverse shift n↦ n+1. This proves the closed-span assertion.
Exact mathematical passage
x_F=\sum_{n\in F}x[n]\delta_n,\qquad F\subset\bbZ\text{ finite},
f^{(m)}Use in source · 3 passages
Search aliases: f^(m)
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1672
The singular packet results begin with trace data. If a piecewise smooth function has a jump in its derivative of order m at τ_q, then the packet jumpf^(m)_τ_q frac(t-τ_q)_+^mm! removes exactly that trace jump. Subtracting all packets through order r-1 leaves a global W^r,p remainder. The packet coefficients are uniquely determined by the one sided trace jumps. This gives a finite and explicit singular part for the class covered by [source reference].
Exact mathematical passage
\jump{f^{(m)}}_{\tau_q} \frac{(t-\tau_q)_+^m}{m!}
f^{(r)}Use in source · 1 passage
Search aliases: f^(r)
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2554
sided traces at the endpoints. Define a_q,m = jumpf^(m)_τ_q = f^(m)(τ_q+)-f^(m)(τ_q-) and j_f(t) = ∑_q=1^Q∑_m=0^r-1a_q,mϕ_m,τ_q(t). Then s_f=f-j_f has matching one sided derivatives through order r-1 at every breakpoint and belongs to W^r,p(0,1). Its weak derivative of order r equals the piecewise classical derivative f^(r) almost everywhere.
Exact mathematical passage
f^{(r)}
f_{(k-d)\bmod N}Use in source · 1 passage
Search aliases: f_(k-d)bmod N bmod
Use in source · atlas-pages.tex, line 105
inal sample δ_1. Under DI_2 it stays even but moves to channel address one: this is δ_2. Thus J(D)^2=DI_2, H(D)=I_2+2J(D)-J(D)^2+J(D)^3 recovers the previous page's four-tap transfer matrix. The same delay algebra explains both the off-diagonal phase swap and the channel powers. headSigned delays need the same care For an original-sample delay d at modulus N, y_k[n]=f_(k-d)bmod N [n+lfloorfrack-dNrfloor]. This includes advances when d<0. The finite verification checks the formula on signed sampl…
Exact mathematical passage
y_k[n]=f_{(k-d)\bmod N} \!\left[n+\left\lfloor\frac{k-d}{N}\right\rfloor\right].
F_*Use in source · 4 passages
Search aliases: F_*
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2466
[Stability of the optimal cost] Let c and widetilde c be two node cost functions. For an admissible pruning L, set ε( L) = ∑_v∈ L|c(v)-widetilde c(v)| and let ε_* = max_ Lε( L). If F_* and widetilde F_* are the two optimal root costs, then |F_*-widetilde F_*| ≤ ε_*.
Exact mathematical passage
F_*
F_0Use in source · 1 passage
Search aliases: F_0
Use in source · nonlinear_shift.tex, line 4
[x=1cm,y=1cm,>=Latex,font=sffamilysmall,line cap=round,line join=round] node[] at (6.50000,4.60000) F(x)[n]=x[n]+tfrac12x[n]x[n-1]; draw[fill=black!10,draw=black!40,line width=.35pt] (0.60000,2.70000) rectangle ++(5.70000,1.00000); node[] at (3.45000,3.20000) F_0[n]=e[n]+tfrac12e[n]o[n-1]; draw[fill=black!10,draw=black!40,line width=.35pt] (6.80000,2.70000) rectangle ++(5.70000,1.00000); node[] at (9.65000,3.20000) F_1[n]=o[n]+tfrac12o[n]e[n]; node[] at (3.45000,1.65000) e[1]=1/2, o[0]=1/4; node…
Exact mathematical passage
F_0[n]=e[n]+\tfrac12e[n]o[n-1]
F_1Use in source · 1 passage
Search aliases: F_1
Use in source · nonlinear_shift.tex, line 6
all,line cap=round,line join=round] node[] at (6.50000,4.60000) F(x)[n]=x[n]+tfrac12x[n]x[n-1]; draw[fill=black!10,draw=black!40,line width=.35pt] (0.60000,2.70000) rectangle ++(5.70000,1.00000); node[] at (3.45000,3.20000) F_0[n]=e[n]+tfrac12e[n]o[n-1]; draw[fill=black!10,draw=black!40,line width=.35pt] (6.80000,2.70000) rectangle ++(5.70000,1.00000); node[] at (9.65000,3.20000) F_1[n]=o[n]+tfrac12o[n]e[n]; node[] at (3.45000,1.65000) e[1]=1/2, o[0]=1/4; node[] at (9.65000,1.65000) o[1]=1/4, e[…
Exact mathematical passage
F_1[n]=o[n]+\tfrac12o[n]e[n]
F_jUse in source · 7 passages
Search aliases: F_j
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2777
Let (E_i)_i=0^I and (F_j)_j=0^J be measurable partitions of the same domain, up to null sets. Let a_i and b_j be the corresponding values. The index zero may be used for a background region.
Exact mathematical passage
(F_j)_{j=0}^{J}
F_kUse in source · 1 passage
Search aliases: F_k
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2803
[Background and packet error] Suppose the true and estimated packets use the same amplitude list (a_i)_i=0^I, but possibly different regions. Then j-widehat j _p^p = ∑_i,k|a_i-a_k|^pμ(E_i∩ F_k). Only off diagonal overlaps contribute.
Exact mathematical passage
\|j-\widehat j\|_p^p = \sum_{i,k}|a_i-a_k|^p\mu(E_i\cap F_k).
f_nUse in source · 1 passage
Search aliases: f_n
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3492
[Residue law for Cauchy products] Let F(z)=∑_n≥0f_nz^n, G(z)=∑_n≥0g_nz^n be formal power series. The same statement holds for power series that converge on a common disk. Decompose both series into residue packets modulo N: F=∑_r=0^N-1F_r, G=∑_s=0^N-1G_s. If H=FG, then its residue packet H_t is H_t = ∑_r+s≡ t pmod NF_rG_s.
Exact mathematical passage
F(z)=\sum_{n\geq0}f_nz^n, \qquad G(z)=\sum_{n\geq0}g_nz^n
F_rUse in source · 4 passages
Search aliases: F_r
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1099
For a finite nonlinear shift operator F(x)[n]=x[n]+α x[n]x[n-1], the exact packet coordinates are simply F_r((x_s)_s)=P_N,rF (∑_sP_N,sx). Analysis followed by this packet system and synthesis reproduces the direct nonlinear output exactly. The numerical residual in the companion implementation was zero to machine representation.
Exact mathematical passage
F_r((x_s)_s)=P_{N,r}F\!\left(\sum_sP_{N,s}x\right).
gUse in source · 4 passages
Search aliases: g
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 936
[label=(roman*)] item A global minimizer Λ_y^*∈*argmin_Λ∈ P( T)J_y(Λ) always exists. item If the minimizer is unique and at least one competing pruning exists, its gap Γ_y = min_Λ≠Λ_y^* (J_y(Λ)-J_y(Λ_y^*)) is attained and strictly positive. If the pruning set is a singleton, its only element minimizes every criterion; no real-valued minimum over an empty competitor set is introduced. item In the case with competitors, if new data y' satisfy sup_Λ∈ P( T) |J_y'(Λ)-J_y(Λ)| <fracΓ_y2, then Λ_y'^*=Λ_…
Exact mathematical passage
\Lambda_y^*\in\operatorname*{argmin}_{\Lambda\in\mathfrak P(\mathcal T)}J_y(\Lambda)
g_nUse in source · 1 passage
Search aliases: g_n
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3492
[Residue law for Cauchy products] Let F(z)=∑_n≥0f_nz^n, G(z)=∑_n≥0g_nz^n be formal power series. The same statement holds for power series that converge on a common disk. Decompose both series into residue packets modulo N: F=∑_r=0^N-1F_r, G=∑_s=0^N-1G_s. If H=FG, then its residue packet H_t is H_t = ∑_r+s≡ t pmod NF_rG_s.
Exact mathematical passage
F(z)=\sum_{n\geq0}f_nz^n, \qquad G(z)=\sum_{n\geq0}g_nz^n
G_{rs}Use in source · 2 passages
Search aliases: G_rs
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1965
[Composition of transfer matrices] If G and H are convolution operators with packet matrices (G_rs) and (H_rs), then the packet matrix of GH is (GH)_rt = ∑_s∈ RG_rsH_st. In one dimensional transform notation, the corresponding matrix symbols multiply pointwise.
Exact mathematical passage
(G_{rs})
G_sUse in source · 3 passages
Search aliases: G_s
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3498
[Residue law for Cauchy products] Let F(z)=∑_n≥0f_nz^n, G(z)=∑_n≥0g_nz^n be formal power series. The same statement holds for power series that converge on a common disk. Decompose both series into residue packets modulo N: F=∑_r=0^N-1F_r, G=∑_s=0^N-1G_s. If H=FG, then its residue packet H_t is H_t = ∑_r+s≡ t pmod NF_rG_s.
Exact mathematical passage
F=\sum_{r=0}^{N-1}F_r, \qquad G=\sum_{s=0}^{N-1}G_s.
\Gam_LUse in source · 1 passage
Search aliases: Gam_L Γ_L Gam Gamma
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1214
For a lossless transmission line, Γ_rm in=Γ_L e^-2 iβℓ. One N-phase step corresponds to a line-length increment λ/(2N), up to the chosen orientation convention. Hence the phase address has a direct electrical interpretation.
Exact mathematical passage
\Gam_{\rm in}=\Gam_L e^{-2\ii\beta\ell}.
\Gam_{\rm in}Use in source · 1 passage
Search aliases: Gam_rm in Γ_rm in Gam rm Gamma rm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1214
For a lossless transmission line, Γ_rm in=Γ_L e^-2 iβℓ. One N-phase step corresponds to a line-length increment λ/(2N), up to the chosen orientation convention. Hence the phase address has a direct electrical interpretation.
Exact mathematical passage
\Gam_{\rm in}=\Gam_L e^{-2\ii\beta\ell}.
\bm HUse in source · 3 passages
Search aliases: H bm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 438
[Exact FIR polyphase transfer law] The output channels satisfy y_k[n] = ∑_s=0^N-1∑_m∈ Z h_ks[m]x_s[n-m]. For the numerical channel transform, take z∈ C∖{0} and assume ∑_n∈ Z|x_s[n]z^-n|<∞ (0≤ s<N). Define X_s(z)=∑_nx_s[n]z^-n. The output series then also converge absolutely, and Y(z)= H^(N)(z) X(z), H^(N)_ks(z)=∑_m h[Nm+k-s]z^-m.
Exact mathematical passage
\bm Y(z)=\bm H^{(N)}(z)\bm X(z), \qquad H^{(N)}_{ks}(z)=\sum_m h[Nm+k-s]z^{-m}.
\mathbb HUse in source · 4 passages
Search aliases: H mathbb
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3648
Let H = {z∈ C:Re z>0} and define C(z)=fracz-1z+1. It maps H conformally onto the unit disk D, with inverse C^-1(w)=frac1+w1-w. For θ∈ R, let R_θ(w)= e^ iθw.
Exact mathematical passage
\mathbb H = \{z\in\CC:\Re z>0\}
\mathrm{H}Use in source · 3 passages
Search aliases: H mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 524
The Haar analysis matrix is E_H = frac1sqrt2 1 · 1 1 · -1 . It is orthogonal, so R_H= E_H^ T and R_H E_H=I_2. The channels are a[n]=fracx[2n]+x[2n+1]sqrt2, d[n]=fracx[2n]-x[2n+1]sqrt2.
Exact mathematical passage
\bm E_{\mathrm{H}} = \frac1{\sqrt2} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}.
\bm H^{(N)}Use in source · 1 passage
Search aliases: H^(N) bm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 438
[Exact FIR polyphase transfer law] The output channels satisfy y_k[n] = ∑_s=0^N-1∑_m∈ Z h_ks[m]x_s[n-m]. For the numerical channel transform, take z∈ C∖{0} and assume ∑_n∈ Z|x_s[n]z^-n|<∞ (0≤ s<N). Define X_s(z)=∑_nx_s[n]z^-n. The output series then also converge absolutely, and Y(z)= H^(N)(z) X(z), H^(N)_ks(z)=∑_m h[Nm+k-s]z^-m.
Exact mathematical passage
\bm Y(z)=\bm H^{(N)}(z)\bm X(z), \qquad H^{(N)}_{ks}(z)=\sum_m h[Nm+k-s]z^{-m}.
H^{(N)}_{ks}Use in source · 1 passage
Search aliases: H^(N)_ks
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 438
[Exact FIR polyphase transfer law] The output channels satisfy y_k[n] = ∑_s=0^N-1∑_m∈ Z h_ks[m]x_s[n-m]. For the numerical channel transform, take z∈ C∖{0} and assume ∑_n∈ Z|x_s[n]z^-n|<∞ (0≤ s<N). Define X_s(z)=∑_nx_s[n]z^-n. The output series then also converge absolutely, and Y(z)= H^(N)(z) X(z), H^(N)_ks(z)=∑_m h[Nm+k-s]z^-m.
Exact mathematical passage
\bm Y(z)=\bm H^{(N)}(z)\bm X(z), \qquad H^{(N)}_{ks}(z)=\sum_m h[Nm+k-s]z^{-m}.
H_{1/3}Use in source · 1 passage
Search aliases: H_1/3
Use in source · atlas-pages.tex, line 579
vpage31Derivative jumps have their own packetsTRACE DATA / A STEP AND A RAMPTraditional monograph: Theorem E7.2 (p. 56), Corollary E7.3 (p. 57). On [0,1], let H_τ=1_[τ,1] and take f(t)=t^2+2H_1/3(t)-3(t-2/3)_+. The function has a value jump 2 at 1/3 and a first-derivative jump -3 at 2/3. Its packet and remainder are j_f(t)=2ϕ_0,1/3(t)-3ϕ_1,2/3(t), s_f(t)=t^2. vfig[.92]ext_trace platecaptionThe displayed curves are exact polynomial pieces. The solid jump is not connected across its discontinuity.…
Exact mathematical passage
f(t)=t^2+2H_{1/3}(t)-3(t-2/3)_+.
\bm H_4Use in source · 1 passage
Search aliases: H_4 bm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2014
[A four phase finite impulse response operator] Let h(0)=1, h(1)=2, h(2)=3, h(3)=4, h(4)=5, with all other taps zero. For four residue channels, the transfer symbol is H_4(z) = 1+5z^-1 · 4z^-1 · 3z^-1 · 2z^-1 2 · 1+5z^-1 · 4z^-1 · 3z^-1 3 · 2 · 1+5z^-1 · 4z^-1 4 · 3 · 2 · 1+5z^-1 . The delay term records the tap that crosses a coarse cell boundary. No approximation is introduced by the matrix representation.
Exact mathematical passage
\bm H_4(z) = \begin{pmatrix} 1+5z^{-1}&4z^{-1}&3z^{-1}&2z^{-1}\\ 2&1+5z^{-1}&4z^{-1}&3z^{-1}\\ 3&2&1+5z^{-1}&4z^{-1}\\ 4&3&2&1+5z^{-1} \end{pmatrix}.
H_{ks}Use in source · 1 passage
Search aliases: H_ks
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1987
For L=N Z with N≥1, first take |z|=1. The assumption h∈ℓ^1( Z) makes the following symbol series absolutely convergent. Define H_ks(z)=∑_m∈ Zh(Nm+k-s)z^-m. For x∈ℓ^1( Z), the channel transforms converge absolutely on the unit circle and satisfy Y(z)= H_N(z) X(z). For x∈ℓ^2( Z), the same identity holds almost everywhere in the L^2 Fourier sense. For finite impulse-response kernels it also holds at nonzero z wherever the input channel transforms are weighted absolutely summable. This is the classi…
Exact mathematical passage
H_{ks}(z)=\sum_{m\in\ZZ}h(Nm+k-s)z^{-m}.
h_{ks}Use in source · 6 passages
Search aliases: h_ks
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 416
Let h be a finite impulse response and y=h*x, y[n]=∑_q∈ Zh[q]x[n-q]. Resolve both input and output into N phases. For output phase k and input phase s, define the transfer kernel h_ks[m]=h[Nm+k-s]. Out-of-range taps are understood to be zero.
Exact mathematical passage
\boxed{ h_{ks}[m]=h[Nm+k-s]. }
\bm H_NUse in source · 1 passage
Search aliases: H_N bm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1991
For L=N Z with N≥1, first take |z|=1. The assumption h∈ℓ^1( Z) makes the following symbol series absolutely convergent. Define H_ks(z)=∑_m∈ Zh(Nm+k-s)z^-m. For x∈ℓ^1( Z), the channel transforms converge absolutely on the unit circle and satisfy Y(z)= H_N(z) X(z). For x∈ℓ^2( Z), the same identity holds almost everywhere in the L^2 Fourier sense. For finite impulse-response kernels it also holds at nonzero z wherever the input channel transforms are weighted absolutely summable. This is the classi…
Exact mathematical passage
\bm Y(z)=\bm H_N(z)\bm X(z).
H_{rs}Use in source · 3 passages
Search aliases: H_rs
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1933
[Exact convolution transfer law] Let y=h*x. Then y_r(ℓ) = ∑_s∈ R∑_m∈ Lh_rs(m)x_s(ℓ-m). Equivalently, A_LH A_L^-1 = (H_rs)_r,s∈ R, where H_rs is convolution on L by h_rs.
Exact mathematical passage
\Acal_LH\Acal_L^{-1} = (H_{rs})_{r,s\in R},
H_{st}Use in source · 1 passage
Search aliases: H_st
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1966
[Composition of transfer matrices] If G and H are convolution operators with packet matrices (G_rs) and (H_rs), then the packet matrix of GH is (GH)_rt = ∑_s∈ RG_rsH_st. In one dimensional transform notation, the corresponding matrix symbols multiply pointwise.
Exact mathematical passage
(GH)_{rt} = \sum_{s\in R}G_{rs}H_{st}.
H_tUse in source · 2 passages
Search aliases: H_t
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3503
[Residue law for Cauchy products] Let F(z)=∑_n≥0f_nz^n, G(z)=∑_n≥0g_nz^n be formal power series. The same statement holds for power series that converge on a common disk. Decompose both series into residue packets modulo N: F=∑_r=0^N-1F_r, G=∑_s=0^N-1G_s. If H=FG, then its residue packet H_t is H_t = ∑_r+s≡ t pmod NF_rG_s.
Exact mathematical passage
H_t
H_{\widehat\tau}Use in source · 2 passages
Search aliases: H_widehatτ widehat tau
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2724
[Location error for one step] Let 0≤τ,widehatτ≤1 and H_τ= 1_[τ,1]. For 1≤ p<∞, aH_τ-aH_widehatτ _p = |a||τ-widehatτ|^1/p. In the supremum norm, the error equals |a| whenever τ≠widehatτ.
Exact mathematical passage
\|aH_\tau-aH_{\widehat\tau}\|_p = |a||\tau-\widehat\tau|^{1/p}.
H_{\widehat\tau_q}Use in source · 2 passages
Search aliases: H_widehatτ_q widehat tau
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3006
[Packet reconstruction after detection] Let j=∑_q=1^Qa_qH_τ_q, widehat j=∑_q=1^Qwidehat a_qH_widehatτ_q. Under the hypotheses of [source reference], for 1≤ p<∞, j-widehat j _p ≤ Qb_h + ( frac h2∑_q=1^Q|a_q|^p )^1/p. If Q_m is bounded and linear and widehat f_m=Q_m(f-widehat j)+widehat j, then f-widehat f_m _p ≤ · s-Q_ms _p
Exact mathematical passage
j=\sum_{q=1}^{Q}a_qH_{\tau_q}, \qquad \widehat j=\sum_{q=1}^{Q}\widehat a_qH_{\widehat\tau_q}.
H_{\tau_q}Use in source · 3 passages
Search aliases: H_τ_q tau
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2908
Let t_k=kh, k=0,…,n, h=frac1n. Assume f(t) = s(t)+∑_q=1^Qa_qH_τ_q(t), H_τ(t)= 1_[τ,1](t), where s is Lipschitz with constant M, and Q≥1 is finite with nonzero jump amplitudes. The zero-jump case requires no minimum-amplitude condition. Suppose no jump lies at a sample point and no grid cell contains more than one jump. Observe y_k=f(t_k)+η_k. Define the adjacent contrast D_k=y_k+1-y_k.
Exact mathematical passage
f(t) = s(t)+\sum_{q=1}^{Q}a_qH_{\tau_q}(t), \qquad H_\tau(t)=\one_{[\tau,1]}(t),
\Hcal_{\rm root}Use in source · 1 passage
Search aliases: Hcal_rm root H_rm root Hcal rm mathcal rm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2342
[Reconstruction on a pruned packet tree] Every admissible pruning L defines a unitary analysis map A_ L: H_rm root ⟶ ⊕_v∈ L H_v. Its inverse is obtained by applying the adjoints of the local splits from the leaves toward the root. In particular, x ^2 = ∑_v∈ L x_v ^2.
Exact mathematical passage
\Acal_{\Lcal}:\Hcal_{\rm root} \longrightarrow \bigoplus_{v\in\Lcal}\Hcal_v.
\operatorname{Hol}Use in source · 1 passage
Search aliases: Hol operatorname
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1473
Hol(γ) · closed phase transfer · cycle consistency in a network
Exact mathematical passage
\operatorname{Hol}(\gamma)
i^kUse in source · 2 passages
Search aliases: i^k
Use in source · atlas-pages.tex, line 13
se alternatives can all be seen The full binary tree of depth three has only 26 complete prunings. We enumerate all of them, compute a unique mixed-depth optimum, and certify its stability with rational costs. The displayed objective is deliberately explicit, not claimed to be a universal signal-quality score. headA coordinate orbit with no hidden numerical poles The four reflection points Γ=i^k/2 map to four exact impedances. A second orbit passes through infinity; homogeneous coordinates make …
Exact mathematical passage
\Gamma=i^k/2
i_1Use in source · 1 passage
Search aliases: i_1
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1075
[Cyclic product law] P_N,k(AB) = ∑_i+j≡ k (mod N)A_iB_j. For q∈ N, the qth-power identity is P_N,k(A^q) = ∑_i_1+⋯+i_q≡ k A_i_1⋯ A_i_q.
Exact mathematical passage
P_{N,k}(A^q) = \sum_{i_1+\cdots+i_q\equiv k} A_{i_1}\cdots A_{i_q}.
I_{16}Use in source · 1 passage
Search aliases: I_16
Use in source · depth_two.tex, line 44
00,0.65000); node[] at (5.90000,1.04500) - frac14; draw[fill=black!18,draw=black!40,line width=.35pt] (6.80000,0.72000) rectangle ++(1.80000,0.65000); node[] at (7.70000,1.04500) frac14; node[] at (10.90000,1.05000) frac14; node[] at (5.00000,4.60000) coefficient positions within each packet; node[] at (10.90000,4.60000) energy; node[] at (6.25000,-0.30000) ∑_α∈{0,1}^2W_α^*W_α=I_16;
Exact mathematical passage
\sum_{\alpha\in\{0,1\}^2}W_\alpha^*W_\alpha=I_{16}
I_2Use in source · 7 passages
Search aliases: I_2
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 519
1 · 0, whereas z^-1I_2 delays the original signal by two samples. The general address identity is valid for negative sample addresses as well.
Exact mathematical passage
z^{-1}I_2
I_{j,k}Use in source · 4 passages
Search aliases: I_j,k
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3190
Let I_j,k = [k2^-j,(k+1)2^-j) and let ψ_j,k be the normalized Haar wavelet supported on I_j,k, positive on the left half and negative on the right half.
Exact mathematical passage
I_{j,k} = [k2^{-j},(k+1)2^{-j})
i_qUse in source · 1 passage
Search aliases: i_q
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1075
[Cyclic product law] P_N,k(AB) = ∑_i+j≡ k (mod N)A_iB_j. For q∈ N, the qth-power identity is P_N,k(A^q) = ∑_i_1+⋯+i_q≡ k A_i_1⋯ A_i_q.
Exact mathematical passage
P_{N,k}(A^q) = \sum_{i_1+\cdots+i_q\equiv k} A_{i_1}\cdots A_{i_q}.
\iiUse in source · 23 passages
Search aliases: ii i mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 623
Set A=∑_α W_ℓ,α^*W_ℓ,α. This is a finite sum of bounded operators. The energy result says ⟨ x,Ax⟩=⟨ x,x⟩ for every x. Complex polarization, applied also to x+y and x+ i y, gives ⟨ y,(A-I)x⟩=0 for all x,y, hence A=I. Applying the operator identity to x proves synthesis. This argument uses the actual Hilbert adjoints of the decimation operators, not a separately postulated inverse symbol.
Exact mathematical passage
x+\ii y
\operatorname{Im}Use in source · 1 passage
Search aliases: Im operatorname
Use in source · atlas-pages.tex, line 415
headThe coordinates preserve structure, not Euclidean distances A quarter turn is a rotation in the reflection disk. After pulling back, it is a M obius map, not a Euclidean rotation of the impedance plane. Equal reflection magnitudes need not give equal impedance magnitudes. The opposite points 1/2 and -1/2 become 3 and 1/3. The continuous impedance locus is the circle (Rez-5/3)^2+(Imz)^2=16/9, also drawn in the plate. Finite passive impedances with Rez≥0 map to the closed disk except Γ=1. That…
Exact mathematical passage
(\operatorname{Re}z-5/3)^2+(\operatorname{Im}z)^2=16/9
J'Use in source · 1 passage
Search aliases: J'
Use in source · atlas-pages.tex, line 234
e every node cost changes by at most ε_v=1/32. A pruning may select as many as eight leaves, so its total cost can change by at most 8/32=1/4. The recursive envelope proves this without guessing which pruning will be selected: E(v)=max{ε_v, E(v0)+E(v1)}, E(∅)=tfrac14. For any competitor Λ, the changed cost difference is bounded below by J'(Λ)-J'(Λ_*)≥ Γ-2E(∅)=tfrac12>0. Thus the optimum is unchanged for every permitted simultaneous perturbation, not just a sampled noise realization. The envelope…
Exact mathematical passage
J'(\Lambda)-J'(\Lambda_*)\ge \Gamma-2E(\varnothing)=\tfrac12>0.
j_2Use in source · 1 passage
Search aliases: j_2
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1235
[Dyadic--odd packet factorization] After identifying the residue labels by the Chinese remainder map, the phase projector factors as P_N,j=P_2^s,j_2P_M,j_o.
Exact mathematical passage
\boxed{ P_{N,j}=P_{2^s,j_2}P_{M,j_o}. }
j_fUse in source · 4 passages
Search aliases: j_f
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2545
_packet_decomposition Let 0=τ_0<τ_1<⋯<τ_Q<τ_Q+1=1. Suppose the restriction of f to every interval (τ_q,τ_q+1) belongs to W^r,p, and choose the one dimensional Sobolev representatives whose derivatives through order r-1 have one sided traces at the endpoints. Define a_q,m = jumpf^(m)_τ_q = f^(m)(τ_q+)-f^(m)(τ_q-) and j_f(t) = ∑_q=1^Q∑_m=0^r-1a_q,mϕ_m,τ_q(t). Then s_f=f-j_f has matching one sided derivatives through order r-1 at every breakpoint and belongs to W^r,p(0,1). Its weak derivative of or…
Exact mathematical passage
j_f(t) = \sum_{q=1}^{Q}\sum_{m=0}^{r-1}a_{q,m}\phi_{m,\tau_q}(t).
j_oUse in source · 1 passage
Search aliases: j_o
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1235
[Dyadic--odd packet factorization] After identifying the residue labels by the Chinese remainder map, the phase projector factors as P_N,j=P_2^s,j_2P_M,j_o.
Exact mathematical passage
\boxed{ P_{N,j}=P_{2^s,j_2}P_{M,j_o}. }
j_qUse in source · 4 passages
Search aliases: j_q
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 681
[Several exact edge packets] If f=s+∑_q=1^Mj_q and every j_q is retained exactly, then f-(Q_ms+∑_qj_q) = s-Q_ms . Curved edges, corners, and different jump amplitudes do not alter the identity. The finite sum j=∑_q=1^M j_q is a single retained packet. Applying the preceding cancellation identity gives the result; the empty family M=0 is included with sum zero.
Exact mathematical passage
f=s+\sum_{q=1}^{M}j_q
J_{y'}Use in source · 2 passages
Search aliases: J_y'
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 949
imizer is unique and at least one competing pruning exists, its gap Γ_y = min_Λ≠Λ_y^* (J_y(Λ)-J_y(Λ_y^*)) is attained and strictly positive. If the pruning set is a singleton, its only element minimizes every criterion; no real-valued minimum over an empty competitor set is introduced. item In the case with competitors, if new data y' satisfy sup_Λ∈ P( T) |J_y'(Λ)-J_y(Λ)| <fracΓ_y2, then Λ_y'^*=Λ_y^*. item Every selected minimizer preserves the exact synthesis identity of the preceding theorem.
Exact mathematical passage
\sup_{\Lambda\in\mathfrak P(\mathcal T)} |J_{y'}(\Lambda)-J_y(\Lambda)| <\frac{\Gamma_y}{2},
\jumpUse in source · 3 passages
Search aliases: jump
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1672
The singular packet results begin with trace data. If a piecewise smooth function has a jump in its derivative of order m at τ_q, then the packet jumpf^(m)_τ_q frac(t-τ_q)_+^mm! removes exactly that trace jump. Subtracting all packets through order r-1 leaves a global W^r,p remainder. The packet coefficients are uniquely determined by the one sided trace jumps. This gives a finite and explicit singular part for the class covered by [source reference].
Exact mathematical passage
\jump{f^{(m)}}_{\tau_q} \frac{(t-\tau_q)_+^m}{m!}
k_xUse in source · 1 passage
Search aliases: k_x
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 820
[Two-dimensional errors] For the 192×192 disk experiment, a square Fourier mask retains indices |k_x|,|k_y|≤16, or 33^2=1089 bins. The reported errors are lrr method · full-image RMSE · boundary-band RMSE
Exact mathematical passage
|k_x|,|k_y|\le16
k_yUse in source · 1 passage
Search aliases: k_y
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 820
[Two-dimensional errors] For the 192×192 disk experiment, a square Fourier mask retains indices |k_x|,|k_y|≤16, or 33^2=1089 bins. The reported errors are lrr method · full-image RMSE · boundary-band RMSE
Exact mathematical passage
|k_x|,|k_y|\le16
\mathcal LUse in source · 3 passages
Search aliases: L mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 490
Work with finite impulse-response banks. Let L= C[z,z^-1] be the Laurent-polynomial algebra, with the monomial z^-m acting on a bilateral sequence by x[n]↦ x[n-m]. Let E∈Mat_N× N( L) and R∈Mat_N× N( L) be analysis and synthesis matrices. Their composition acts on all bilateral channel sequences by finite convolutions. At a nonzero z where the input transforms are absolutely convergent, this gives widehat X(z)= R(z) E(z) X(z).
Exact mathematical passage
\mathcal L=\bbC[z,z^{-1}]
lUse in source · 3 passages
Search aliases: l
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 780
[One-dimensional errors] For 1024 samples on [-1,1) and Fourier cutoff m=18, the retained frequency indices are -18,…,18: 37 bins including the zero-frequency bin. The measured values were lrr method · RMSE · maximum absolute error global Fourier truncation · 5.4023×10^-2 · 3.4968×10^-1 known edge packet · 1.3692×10^-4 · 1.4503×10^-3 estimated edge packet · 2.4029×10^-2 · 5.9530×10^-2 The simple estimator recovered the binary mask exactly in this deliberately separated two-class example. That re…
Exact mathematical passage
\begin{array}{lrr} \toprule \text{method}&\RMSE&\text{maximum absolute error}\\ \midrule \text{global Fourier truncation}&5.4023\times10^{-2}&3.4968\times10^{-1}\\ \text{known edge packet}&1.3692\times10^{-4}&1.4503\times10^{-3}\\ \text{estimated edge packet}&2.4029\times10^{-2}&5.9530\times10^{-2}\\ \bottomrule \end{array}
L'Use in source · 7 passages
Search aliases: L'
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1648
a finite index sublattice of Z^d, and let R contain one representative from every coset. The analysis map x⟼ (x_r)_r∈ R, x_r(ℓ)=x(r+ℓ), is an isometric isomorphism on every ℓ^p space. This statement includes the usual one dimensional polyphase decomposition, but it also applies to arbitrary finite index lattice packets in several dimensions. If L'≤ L, the coarse packets split exactly into their fine descendants. A convolution operator becomes a matrix of convolution operators on L, with entries …
Exact mathematical passage
L'\leq L
L^{-1/2}Use in source · 1 passage
Search aliases: L^-1/2
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3208
Let L=2^-j and let a be the left endpoint of the support. Pair every point t∈[a,a+L/2] with t+L/2. Then ⟨ s,ψ_j,k⟩ = L^-1/2 ∫_a^a+L/2 overlines(t)-s(t+L/2) d t. The integrand has magnitude at most ML/2, and the integration interval has length L/2. The result is ML^3/2/4.
Exact mathematical passage
\langle s,\psi_{j,k}\rangle = L^{-1/2} \int_a^{a+L/2} \overline{s(t)-s(t+L/2)}\,\dd t.
L^1Use in source · 2 passages
Search aliases: L^1
Use in source · atlas-pages.tex, line 298
ehatΩ=[tfrac516,tfrac1316], a=widehat a=3. The two intervals have the same length. Their disagreement consists of one missed strip and one extra strip, each of width 1/16. vfigmask_translation |ΩmathbintrianglewidehatΩ|=tfrac18, j-widehat j=3(1_Ω-1_widehatΩ). c|ccc error measure · L^1 norm · squared L^2 norm · L^∞ norm
Exact mathematical passage
L^1
L^{1/2}Use in source · 7 passages
Search aliases: L^1/2
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3224
[Exact Haar coefficient of a step] Let τ∈ I_j,k, let L=2^-j, and write u=fracτ-kLL∈[0,1]. Then | ⟨ aH_τ,ψ_j,k ⟩ | = |a|L^1/2min{u,1-u}. Every Haar coefficient whose support does not cross τ is zero.
Exact mathematical passage
\left| \left\langle aH_\tau,\psi_{j,k} \right\rangle \right| = |a|L^{1/2}\min\{u,1-u\}.
L^2Use in source · 7 passages
Search aliases: L^2
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1994
ZZ with N≥1, first take |z|=1. The assumption h∈ℓ^1( Z) makes the following symbol series absolutely convergent. Define H_ks(z)=∑_m∈ Zh(Nm+k-s)z^-m. For x∈ℓ^1( Z), the channel transforms converge absolutely on the unit circle and satisfy Y(z)= H_N(z) X(z). For x∈ℓ^2( Z), the same identity holds almost everywhere in the L^2 Fourier sense. For finite impulse-response kernels it also holds at nonzero z wherever the input channel transforms are weighted absolutely summable. This is the classical pol…
Exact mathematical passage
L^2
L^{3/2}Use in source · 3 passages
Search aliases: L^3/2
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3215
Let L=2^-j and let a be the left endpoint of the support. Pair every point t∈[a,a+L/2] with t+L/2. Then ⟨ s,ψ_j,k⟩ = L^-1/2 ∫_a^a+L/2 overlines(t)-s(t+L/2) d t. The integrand has magnitude at most ML/2, and the integration interval has length L/2. The result is ML^3/2/4.
Exact mathematical passage
ML^{3/2}/4
L^pUse in source · 19 passages
Search aliases: L^p
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 711
[Packet perturbation bound] For 1≤ p<∞, interpreted first as an inequality of extended L^p seminorms, j-widehat j _L^p ≤ |a-widehat a| μ(Ω)^1/p + |widehat a| μ(ΩtrianglewidehatΩ)^1/p. If μ(Ω) and μ(ΩtrianglewidehatΩ) are finite, all terms in this displayed jump bound are finite and it is an ordinary real-valued L^p inequality. For the pointwise supremum, and consequently also the essential L^∞ seminorm, j-widehat j _∞ ≤ |a-widehat a|+|widehat a| 1_ΩtrianglewidehatΩ≠∅.
Exact mathematical passage
L^p
L^\inftyUse in source · 2 passages
Search aliases: L^∞ infty
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 721
f extended L^p seminorms, j-widehat j _L^p ≤ |a-widehat a| μ(Ω)^1/p + |widehat a| μ(ΩtrianglewidehatΩ)^1/p. If μ(Ω) and μ(ΩtrianglewidehatΩ) are finite, all terms in this displayed jump bound are finite and it is an ordinary real-valued L^p inequality. For the pointwise supremum, and consequently also the essential L^∞ seminorm, j-widehat j _∞ ≤ |a-widehat a|+|widehat a| 1_ΩtrianglewidehatΩ≠∅.
Exact mathematical passage
L^\infty
L_{-P}Use in source · 1 passage
Search aliases: L_-P
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2280
[Exact lifting update] The lifting matrices are invertible with L_P^-1=L_-P, U_Q^-1=U_-Q. If R_0E_0=I, then every finite product V of lifting matrices gives another exact bank by E=VE_0, R=R_0V^-1. Every lifting step has determinant one.
Exact mathematical passage
L_P^{-1}=L_{-P}, \qquad U_Q^{-1}=U_{-Q}.
L_PUse in source · 1 passage
Search aliases: L_P
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2262
Let P(z) and Q(z) be Laurent polynomials. Define L_P(z) = 1 · 0 P(z) · 1 , U_Q(z) = 1 · Q(z) 0 · 1 .
Exact mathematical passage
L_P(z) = \begin{pmatrix} 1&0\\ P(z)&1 \end{pmatrix}, \qquad U_Q(z) = \begin{pmatrix} 1&Q(z)\\ 0&1 \end{pmatrix}.
L_P^{-1}Use in source · 1 passage
Search aliases: L_P^-1
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2280
[Exact lifting update] The lifting matrices are invertible with L_P^-1=L_-P, U_Q^-1=U_-Q. If R_0E_0=I, then every finite product V of lifting matrices gives another exact bank by E=VE_0, R=R_0V^-1. Every lifting step has determinant one.
Exact mathematical passage
L_P^{-1}=L_{-P}, \qquad U_Q^{-1}=U_{-Q}.
\LcalUse in source · 14 passages
Search aliases: Lcal L mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2341
[Reconstruction on a pruned packet tree] Every admissible pruning L defines a unitary analysis map A_ L: H_rm root ⟶ ⊕_v∈ L H_v. Its inverse is obtained by applying the adjoints of the local splits from the leaves toward the root. In particular, x ^2 = ∑_v∈ L x_v ^2.
Exact mathematical passage
\Lcal
\Lcal_*Use in source · 2 passages
Search aliases: Lcal_* L_* Lcal mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2485
[Stability of the selected basis] Assume the optimal pruning L_* for c is unique and at least one competing pruning exists. Define g=min_ L≠ L_*(C( L)-C( L_*))>0. If 2ε_*<g, then the optimal pruning for widetilde c is the same. If only one admissible pruning exists, it remains selected under every cost change without a gap condition.
Exact mathematical passage
\Lcal_*
\Lcal_wUse in source · 1 passage
Search aliases: Lcal_w L_w Lcal mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2406
Assume the child subtree bounds have been proved. Then ∑_w∑_ℓ∈ L_w x_ℓ ^2 ≥ min_w a(w) ∑_w (U_vx)_w ^2 ≥ A_vmin_w a(w) x ^2. The upper bound follows in the same way with the maximum and B_v. Induction completes the proof.
Exact mathematical passage
\sum_{w}\sum_{\ell\in\Lcal_w}\|x_\ell\|^2 \geq \min_w a(w) \sum_w\|(U_vx)_w\|^2 \geq A_v\min_w a(w)\|x\|^2.
\lfloorUse in source · 1 passage
Search aliases: lfloor
Use in source · atlas-pages.tex, line 105
inal sample δ_1. Under DI_2 it stays even but moves to channel address one: this is δ_2. Thus J(D)^2=DI_2, H(D)=I_2+2J(D)-J(D)^2+J(D)^3 recovers the previous page's four-tap transfer matrix. The same delay algebra explains both the off-diagonal phase swap and the channel powers. headSigned delays need the same care For an original-sample delay d at modulus N, y_k[n]=f_(k-d)bmod N [n+lfloorfrack-dNrfloor]. This includes advances when d<0. The finite verification checks the formula on signed sampl…
Exact mathematical passage
y_k[n]=f_{(k-d)\bmod N} \!\left[n+\left\lfloor\frac{k-d}{N}\right\rfloor\right].
\logUse in source · 8 passages
Search aliases: log
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1695
on be the largest contrast that can be produced by a Lipschitz background and bounded noise in one grid cell. If a_min>2b_h, then every threshold between b_h and a_min-b_h detects exactly the cells containing jumps. The jump locations are recovered within h/2, and each amplitude error is at most b_h. The probability theorem replaces the deterministic noise term by 2σsqrtlog(2n/δ). Two half shifted Haar grids provide a second recovery mechanism. Every point is at least one quarter of a cell away …
Exact mathematical passage
2\sigma\sqrt{\log(2n/\delta)}.
MUse in source · 65 passages
Search aliases: M
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 383
Suppose N,M≥1 and Nmid M. Every coarse phase has a disjoint family of fine descendants: Child_M(N,r) = {s:0≤ s<M, s≡ rpmod N}.
Exact mathematical passage
N,M\ge1
M'Use in source · 3 passages
Search aliases: M'
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1895
[Rectangular image phases] For L=M Z× N Z⊂ Z^2, one may take R={0,…,M-1}×{0,…,N-1}. The analysis map separates an image into MN exact phases. Refining (M,N) to (M',N'), where Mmid M' and Nmid N', splits every coarse phase into fracM'MfracN'N fine phases.
Exact mathematical passage
(M',N')
M_{ij}Use in source · 4 passages
Search aliases: M_ij
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2781
[Overlap matrix formula] Define M_ij=μ(E_i∩ F_j). For 1≤ p<∞, ∑_i=0^Ia_i 1_E_i - ∑_j=0^Jb_j 1_F_j _p^p = ∑_i=0^I∑_j=0^J|a_i-b_j|^pM_ij. The row sums of M are μ(E_i), and the column sums are μ(F_j).
Exact mathematical passage
M_{ij}=\mu(E_i\cap F_j).
\matUse in source · 2 passages
Search aliases: mat
Use in source · atlas-pages.tex, line 114
vpage7Perfect reconstruction is an operator identityFILTER BANKS / WHY A FREQUENCY SPOT-CHECK IS INSUFFICIENTTraditional monograph: Theorem 4.1 (p. 7). Consider the finite Laurent bank E(D)=mat1 · D 0 · 1, R(D)=mat1 · -D 0 · 1. Its synthesis cancels the exact delayed cross-channel contribution introduced by analysis. vfigperfect_reconstruction headTwo inverses, all inputs Direct polynomial multiplication gives RE=ER=I_2. Since D acts as a shift, these identities imply perfect reconstruction on e…
Exact mathematical passage
E(D)=\mat{1&D\\0&1},\qquad R(D)=\mat{1&-D\\0&1}.
\operatorname{Mat}Use in source · 2 passages
Search aliases: Mat operatorname
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 490
Work with finite impulse-response banks. Let L= C[z,z^-1] be the Laurent-polynomial algebra, with the monomial z^-m acting on a bilateral sequence by x[n]↦ x[n-m]. Let E∈Mat_N× N( L) and R∈Mat_N× N( L) be analysis and synthesis matrices. Their composition acts on all bilateral channel sequences by finite convolutions. At a nonzero z where the input transforms are absolutely convergent, this gives widehat X(z)= R(z) E(z) X(z).
Exact mathematical passage
\bm E\in\operatorname{Mat}_{N\times N}(\mathcal L)
\operatorname{Mat}_{N\times N}Use in source · 2 passages
Search aliases: Mat_N× N operatorname times
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 490
Work with finite impulse-response banks. Let L= C[z,z^-1] be the Laurent-polynomial algebra, with the monomial z^-m acting on a bilateral sequence by x[n]↦ x[n-m]. Let E∈Mat_N× N( L) and R∈Mat_N× N( L) be analysis and synthesis matrices. Their composition acts on all bilateral channel sequences by finite convolutions. At a nonzero z where the input transforms are absolutely convergent, this gives widehat X(z)= R(z) E(z) X(z).
Exact mathematical passage
\bm E\in\operatorname{Mat}_{N\times N}(\mathcal L)
\mathbinUse in source · 6 passages
Search aliases: mathbin
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 371
one also has A_N S_N=I. Equip this finite product with the Hilbert norm (x_0,…,x_N-1) _ch^2 =∑_r=0^N-1 x_r _2^2. Then A_N is a linear isometric equivalence and x _2^2=∑_r=0^N-1 x_r _2^2. Euclidean division gives a unique pair (r,q) with 0≤ r<N and n=Nq+r, including negative n. The address maps n↦(nbmod N,nmathbindivN) and (r,q)↦ Nq+r are inverse. Analysis and synthesis are reindexing by this bijection. Reindexing the nonnegative sum ∑_n|x[n]|^2 proves the norm identity and shows that both maps p…
Exact mathematical passage
n\mapsto(n\bmod N,n\mathbin{\mathrm{div}}N)
\maxUse in source · 7 passages
Search aliases: max
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1004
[Local perturbation certificate] Assume |c_v(y')-c_v(y)|≤ε_v. Define E(v) = ε_v, · v terminal, [1mm] displaystyle max{ε_v, ∑_u∈ch(v)E(u)}, · v internal. For the complete prunings P_v below v, the envelope is exactly E(v)=max_Λ∈ P_v∑_a∈Λε_a. The maximum is attained. Every complete-pruning cost changes by at most E(∅), and the two globally optimal values differ by at most this quantity. Consequently, for a unique optimum with competitors, Γ_y>2E(∅) is a deterministic certificate that the globally …
Exact mathematical passage
E(v) = \begin{cases} \varepsilon_v,&v\text{ terminal},\\[1mm] \displaystyle \max\left\{\varepsilon_v, \sum_{u\in\operatorname{ch}(v)}E(u)\right\},&v\text{ internal}. \end{cases}
\max_kUse in source · 1 passage
Search aliases: max_k max
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3125
inproof If σ=0, every centered zero-sub-Gaussian noise variable vanishes almost surely, and the noise-free deterministic conclusion applies. For σ>0, the difference η_k+1-η_k is sub Gaussian with variance proxy 2σ^2, so P(|η_k+1-η_k|>t) ≤ 2exp(-fract^24σ^2). A union bound over the n cells gives P( max_k|η_k+1-η_k|>t_n,δ ) ≤ δ. On the complementary event, repeat the deterministic proof.
Exact mathematical passage
\mathbb P\left( \max_k|\eta_{k+1}-\eta_k|>t_{n,\delta} \right) \leq \delta.
\max_{\Lcal}Use in source · 1 passage
Search aliases: max_Lcal max_ L max Lcal max mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2461
[Stability of the optimal cost] Let c and widetilde c be two node cost functions. For an admissible pruning L, set ε( L) = ∑_v∈ L|c(v)-widetilde c(v)| and let ε_* = max_ Lε( L). If F_* and widetilde F_* are the two optimal root costs, then |F_*-widetilde F_*| ≤ ε_*.
Exact mathematical passage
\varepsilon_* = \max_{\Lcal}\varepsilon(\Lcal).
\max_wUse in source · 1 passage
Search aliases: max_w max
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2394
A_vmin_w a(w), · v split, and b(v)= 1, · v retained, B_vmax_w b(w), · v split. Then the leaf analysis of the pruned tree has bounds a(rm root) and b(rm root).
Exact mathematical passage
b(v)= \begin{cases} 1,&v\text{ retained},\\ B_v\max_w b(w),&v\text{ split}. \end{cases}
\max_{\Lambda\in\mathfrak P_v}Use in source · 1 passage
Search aliases: max_Λ∈ P_v max Lambda in mathfrak
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1015
[1mm] displaystyle max{ε_v, ∑_u∈ch(v)E(u)}, · v internal. For the complete prunings P_v below v, the envelope is exactly E(v)=max_Λ∈ P_v∑_a∈Λε_a. The maximum is attained. Every complete-pruning cost changes by at most E(∅), and the two globally optimal values differ by at most this quantity. Consequently, for a unique optimum with competitors, Γ_y>2E(∅) is a deterministic certificate that the globally optimal pruning is unchanged. A singleton pruning set is stable without a gap condition. Assume…
Exact mathematical passage
E(v)=\max_{\Lambda\in\mathfrak P_v}\sum_{a\in\Lambda}\varepsilon_a.
\midUse in source · 4 passages
Search aliases: mid
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 383
Suppose N,M≥1 and Nmid M. Every coarse phase has a disjoint family of fine descendants: Child_M(N,r) = {s:0≤ s<M, s≡ rpmod N}.
Exact mathematical passage
N\mid M
\midruleUse in source · 3 passages
Search aliases: midrule
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 780
[One-dimensional errors] For 1024 samples on [-1,1) and Fourier cutoff m=18, the retained frequency indices are -18,…,18: 37 bins including the zero-frequency bin. The measured values were lrr method · RMSE · maximum absolute error global Fourier truncation · 5.4023×10^-2 · 3.4968×10^-1 known edge packet · 1.3692×10^-4 · 1.4503×10^-3 estimated edge packet · 2.4029×10^-2 · 5.9530×10^-2 The simple estimator recovered the binary mask exactly in this deliberately separated two-class example. That re…
Exact mathematical passage
\begin{array}{lrr} \toprule \text{method}&\RMSE&\text{maximum absolute error}\\ \midrule \text{global Fourier truncation}&5.4023\times10^{-2}&3.4968\times10^{-1}\\ \text{known edge packet}&1.3692\times10^{-4}&1.4503\times10^{-3}\\ \text{estimated edge packet}&2.4029\times10^{-2}&5.9530\times10^{-2}\\ \bottomrule \end{array}
\minUse in source · 28 passages
Search aliases: min
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 941
[label=(roman*)] item A global minimizer Λ_y^*∈*argmin_Λ∈ P( T)J_y(Λ) always exists. item If the minimizer is unique and at least one competing pruning exists, its gap Γ_y = min_Λ≠Λ_y^* (J_y(Λ)-J_y(Λ_y^*)) is attained and strictly positive. If the pruning set is a singleton, its only element minimizes every criterion; no real-valued minimum over an empty competitor set is introduced. item In the case with competitors, if new data y' satisfy sup_Λ∈ P( T) |J_y'(Λ)-J_y(Λ)| <fracΓ_y2, then Λ_y'^*=Λ_…
Exact mathematical passage
\Gamma_y = \min_{\Lambda\ne\Lambda_y^*} \bigl(J_y(\Lambda)-J_y(\Lambda_y^*)\bigr)
\min_{\Lcal\ne\Lcal_*}Use in source · 1 passage
Search aliases: min_Lcal≠Lcal_* min_ L≠ L_* min Lcal ne Lcal min mathcal ne mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2485
[Stability of the selected basis] Assume the optimal pruning L_* for c is unique and at least one competing pruning exists. Define g=min_ L≠ L_*(C( L)-C( L_*))>0. If 2ε_*<g, then the optimal pruning for widetilde c is the same. If only one admissible pruning exists, it remains selected under every cost change without a gap condition.
Exact mathematical passage
g=\min_{\Lcal\ne\Lcal_*}(C(\Lcal)-C(\Lcal_*))>0
\min_qUse in source · 1 passage
Search aliases: min_q min
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2953
[Deterministic jump cell recovery] Let a_min=min_q|a_q| and assume a_min>2b_h. Choose a threshold θ satisfying b_h<θ<a_min-b_h. Then widehat K = {k:|D_k|>θ} is exactly the set of grid cells containing jumps. If widehatτ_q is the midpoint of the detected cell and widehat a_q=D_k, then after matching by order, |widehatτ_q-τ_q| ≤ frac h2 and |widehat a_q-a_q| ≤ b_h.
Exact mathematical passage
a_{\min}=\min_q|a_q|
\min_wUse in source · 2 passages
Search aliases: min_w min
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2386
[Bounds for nonunitary local splits] Suppose every internal node has an analysis map U_v and constants 0≤ A_v≤ B_v<∞ satisfying A_v x ^2 ≤ ∑_w∈child(v) (U_vx)_w ^2 ≤ B_v x ^2. Define recursively a(v)= 1, · v retained, A_vmin_w a(w), · v split, and b(v)= 1, · v retained,
Exact mathematical passage
a(v)= \begin{cases} 1,&v\text{ retained},\\ A_v\min_w a(w),&v\text{ split}, \end{cases}
\min_{\Lambda\ne\Lambda_y^*}Use in source · 1 passage
Search aliases: min_Λ≠Λ_y^* min Lambda ne Lambda
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 941
[label=(roman*)] item A global minimizer Λ_y^*∈*argmin_Λ∈ P( T)J_y(Λ) always exists. item If the minimizer is unique and at least one competing pruning exists, its gap Γ_y = min_Λ≠Λ_y^* (J_y(Λ)-J_y(Λ_y^*)) is attained and strictly positive. If the pruning set is a singleton, its only element minimizes every criterion; no real-valued minimum over an empty competitor set is introduced. item In the case with competitors, if new data y' satisfy sup_Λ∈ P( T) |J_y'(Λ)-J_y(Λ)| <fracΓ_y2, then Λ_y'^*=Λ_…
Exact mathematical passage
\Gamma_y = \min_{\Lambda\ne\Lambda_y^*} \bigl(J_y(\Lambda)-J_y(\Lambda_y^*)\bigr)
\mathrm{mod}Use in source · 5 passages
Search aliases: mod mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 315
Fix an integer N≥1. For 0≤ r<N, define (P_N,rx)[n] = 1_n≡ r (mod N)x[n]. The corresponding decimated channel is x_r[q]=x[Nq+r].
Exact mathematical passage
(P_{N,r}x)[n] = \mathbf 1_{n\equiv r\, (\mathrm{mod}\,N)}x[n].
N'Use in source · 3 passages
Search aliases: N'
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1895
[Rectangular image phases] For L=M Z× N Z⊂ Z^2, one may take R={0,…,M-1}×{0,…,N-1}. The analysis map separates an image into MN exact phases. Refining (M,N) to (M',N'), where Mmid M' and Nmid N', splits every coarse phase into fracM'MfracN'N fine phases.
Exact mathematical passage
(M',N')
oUse in source · 26 passages
Search aliases: o
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 570
Split x into even and odd channels e,o. Define d=o-e, s=e+frac12d. The inverse is e=s-frac12d, o=d+e.
Exact mathematical passage
e,o
\one_{[-a,a]}Use in source · 1 passage
Search aliases: one_[-a,a] 1_[-a,a] one mathbf
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3468
[Square wave Fourier tail] Let j(t)= 1_[-a,a](t) on the circle and let s be smooth. If Q_m is a Fourier partial sum, then Q_m(s+j) has the ordinary oscillatory boundary layer. In contrast, Q_ms+j has error exactly s-Q_ms. The visual comparison in [source reference] illustrates this change in approximation target. It is not a new Fourier summability theorem.
Exact mathematical passage
j(t)=\one_{[-a,a]}(t)
\one_EUse in source · 4 passages
Search aliases: one_E 1_E one mathbf
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2701
[Exact scalar region formula] Let E,F be measurable subsets of a measure space, let a,b∈ C, and let 1≤ p<∞. Then a 1_E-b 1_F _p^p = · |a-b|^pμ(E∩ F) +|a|^pμ(E∖ F) · +|b|^pμ(F∖ E). For equal amplitudes, a 1_E-a 1_F _p = |a|μ(Emathbintriangle F)^1/p.
Exact mathematical passage
\begin{aligned} \|a\one_E-b\one_F\|_p^p ={}& |a-b|^p\mu(E\cap F) +|a|^p\mu(E\setminus F)\\ &+|b|^p\mu(F\setminus E). \end{aligned}
\one_{E_i}Use in source · 1 passage
Search aliases: one_E_i 1_E_i one mathbf
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2785
[Overlap matrix formula] Define M_ij=μ(E_i∩ F_j). For 1≤ p<∞, ∑_i=0^Ia_i 1_E_i - ∑_j=0^Jb_j 1_F_j _p^p = ∑_i=0^I∑_j=0^J|a_i-b_j|^pM_ij. The row sums of M are μ(E_i), and the column sums are μ(F_j).
Exact mathematical passage
\left\| \sum_{i=0}^{I}a_i\one_{E_i} - \sum_{j=0}^{J}b_j\one_{F_j} \right\|_p^p = \sum_{i=0}^{I}\sum_{j=0}^{J}|a_i-b_j|^pM_{ij}.
\one_FUse in source · 4 passages
Search aliases: one_F 1_F one mathbf
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2701
[Exact scalar region formula] Let E,F be measurable subsets of a measure space, let a,b∈ C, and let 1≤ p<∞. Then a 1_E-b 1_F _p^p = · |a-b|^pμ(E∩ F) +|a|^pμ(E∖ F) · +|b|^pμ(F∖ E). For equal amplitudes, a 1_E-a 1_F _p = |a|μ(Emathbintriangle F)^1/p.
Exact mathematical passage
\begin{aligned} \|a\one_E-b\one_F\|_p^p ={}& |a-b|^p\mu(E\cap F) +|a|^p\mu(E\setminus F)\\ &+|b|^p\mu(F\setminus E). \end{aligned}
\one_{F_j}Use in source · 1 passage
Search aliases: one_F_j 1_F_j one mathbf
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2785
[Overlap matrix formula] Define M_ij=μ(E_i∩ F_j). For 1≤ p<∞, ∑_i=0^Ia_i 1_E_i - ∑_j=0^Jb_j 1_F_j _p^p = ∑_i=0^I∑_j=0^J|a_i-b_j|^pM_ij. The row sums of M are μ(E_i), and the column sums are μ(F_j).
Exact mathematical passage
\left\| \sum_{i=0}^{I}a_i\one_{E_i} - \sum_{j=0}^{J}b_j\one_{F_j} \right\|_p^p = \sum_{i=0}^{I}\sum_{j=0}^{J}|a_i-b_j|^pM_{ij}.
\mathrm{op}Use in source · 4 passages
Search aliases: op mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2058
[Quantitative imperfect reconstruction] Define the delay compensated defect D( e^ iω) = e^ i dωR( e^ iω)E( e^ iω)-I_N. If ε_rec = *ess sup_ω D( e^ iω) _op<∞, then the delay compensated reconstruction widetilde x satisfies widetilde x-x _2 ≤ ε_rec x _2. The constant is sharp as an operator norm.
Exact mathematical passage
\varepsilon_{\mathrm{rec}} = \esssup_{\omega}\|D(\e^{\ii\omega})\|_{\mathrm{op}}<\infty,
\overline{s(t)-s(t+L/2)}Use in source · 1 passage
Search aliases: overlines(t)-s(t+L/2) overline
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3208
Let L=2^-j and let a be the left endpoint of the support. Pair every point t∈[a,a+L/2] with t+L/2. Then ⟨ s,ψ_j,k⟩ = L^-1/2 ∫_a^a+L/2 overlines(t)-s(t+L/2) d t. The integrand has magnitude at most ML/2, and the integration interval has length L/2. The result is ML^3/2/4.
Exact mathematical passage
\langle s,\psi_{j,k}\rangle = L^{-1/2} \int_a^{a+L/2} \overline{s(t)-s(t+L/2)}\,\dd t.
\overline{\operatorname{span}\{S^n\delta_0:n\in\bbZ\}}Use in source · 1 passage
Search aliases: overlinespan{S^nδ_0:n∈bbZ} overlinespan{S^nδ_0:n∈ Z} overline operatorname delta in bbZ overline operatorname delta in mathbb
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 276
[Impulse seed theorem] The closed shift-generated hull of δ_0 is all of ℓ^2( Z): overlinespan{S^nδ_0:n∈ Z}^ ℓ^2 =ℓ^2( Z). Every finitely supported signal has the exact finite expression x=∑_n∈ Zx[n]S^nδ_0.
Exact mathematical passage
\boxed{ \overline{\operatorname{span}\{S^n\delta_0:n\in\bbZ\}}^{\,\ell^2} =\ell^2(\bbZ). }
\overline{\operatorname{span}\{S^n\delta_0:n\in\bbZ\}}^{\,\ell^2}Use in source · 1 passage
Search aliases: overlinespan{S^nδ_0:n∈bbZ}^ ℓ^2 overlinespan{S^nδ_0:n∈ Z}^ ℓ^2 overline operatorname delta in bbZ ell overline operatorname delta in mathbb ell
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 276
[Impulse seed theorem] The closed shift-generated hull of δ_0 is all of ℓ^2( Z): overlinespan{S^nδ_0:n∈ Z}^ ℓ^2 =ℓ^2( Z). Every finitely supported signal has the exact finite expression x=∑_n∈ Zx[n]S^nδ_0.
Exact mathematical passage
\boxed{ \overline{\operatorname{span}\{S^n\delta_0:n\in\bbZ\}}^{\,\ell^2} =\ell^2(\bbZ). }
\overline{\Span\{T_a\delta_0:a\in\ZZ^d\}}Use in source · 1 passage
Search aliases: overlineSpan{T_aδ_0:a∈ZZ^d} overlinespan{T_aδ_0:a∈ Z^d} overline Span delta in ZZ overline operatorname delta in mathbb
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1732
[Impulse generation] For 1≤ p<∞, overlinespan{T_aδ_0:a∈ Z^d}^ℓ^p = ℓ^p( Z^d). The same statement holds in c_0( Z^d) with the supremum norm. In ℓ^∞( Z^d), the closed span of the translation orbit is exactly c_0( Z^d).
Exact mathematical passage
\overline{\Span\{T_a\delta_0:a\in\ZZ^d\}}^{\ell^p} = \ell^p(\ZZ^d).
\overline{\Span\{T_a\delta_0:a\in\ZZ^d\}}^{\ell^p}Use in source · 1 passage
Search aliases: overlineSpan{T_aδ_0:a∈ZZ^d}^ℓ^p overlinespan{T_aδ_0:a∈ Z^d}^ℓ^p overline Span delta in ZZ ell overline operatorname delta in mathbb ell
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1732
[Impulse generation] For 1≤ p<∞, overlinespan{T_aδ_0:a∈ Z^d}^ℓ^p = ℓ^p( Z^d). The same statement holds in c_0( Z^d) with the supremum norm. In ℓ^∞( Z^d), the closed span of the translation orbit is exactly c_0( Z^d).
Exact mathematical passage
\overline{\Span\{T_a\delta_0:a\in\ZZ^d\}}^{\ell^p} = \ell^p(\ZZ^d).
\mathbb PUse in source · 4 passages
Search aliases: P mathbb
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1142
[Polyphase M obius action] Regard normalized impedance as the projective line P^1( C)= C∪{∞}. For N≥1, the pullback is a globally defined projective action. In the finite chart, wherever the denominator is nonzero, it is T_N,k(z) = frac(1+ω_N^k)z+(1-ω_N^k) (1-ω_N^k)z+(1+ω_N^k). As maps on P^1( C) it satisfies T_N,k∘ T_N,ℓ=T_N,k+ℓ (mod N).
Exact mathematical passage
\mathbb P^1(\bbC)=\bbC\cup\{\infty\}
\mathfrak PUse in source · 5 passages
Search aliases: P mathfrak
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 932
[Finite globally optimal stable selector] Let P( T) be the finite set of complete prunings of T.
Exact mathematical passage
\mathfrak P(\mathcal T)
\mathbb P^1Use in source · 2 passages
Search aliases: P^1 mathbb
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1142
[Polyphase M obius action] Regard normalized impedance as the projective line P^1( C)= C∪{∞}. For N≥1, the pullback is a globally defined projective action. In the finite chart, wherever the denominator is nonzero, it is T_N,k(z) = frac(1+ω_N^k)z+(1-ω_N^k) (1-ω_N^k)z+(1+ω_N^k). As maps on P^1( C) it satisfies T_N,k∘ T_N,ℓ=T_N,k+ℓ (mod N).
Exact mathematical passage
\mathbb P^1(\bbC)=\bbC\cup\{\infty\}
p^pUse in source · 6 passages
Search aliases: p^p
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1776
For finite p, equip ⊕_r∈ Rℓ^p(L) with (u_r) ^p=∑_r∈ R u_r _p^p. For p=∞, use the maximum norm.
Exact mathematical passage
\|(u_r)\|^p=\sum_{r\in R}\|u_r\|_p^p.
P_{12,k}Use in source · 1 passage
Search aliases: P_12,k
Use in source · atlas-pages.tex, line 449
VE ADDRESSESTraditional monograph: Proposition 14.1 (p. 29). At phase order 12=4·3, the map k⟼(kbmod4, kbmod3) is a bijection from twelve residue addresses to the 4×3 grid. Every cell below contains its unique original address. vfigcrt_refinement headThe factorization is an intersection of conditions For integer sample addresses, including negative ones, P_12,k=P_4,kbmod4 P_3,kbmod3. The product on the right is composition of two residue masks: both congruences must hold. Coprimality makes their…
Exact mathematical passage
P_{12,k}=P_{4,k\bmod4}\,P_{3,k\bmod3}.
P_{2,1}Use in source · 1 passage
Search aliases: P_2,1
Use in source · product_conventions.tex, line 22
[fill=black!20,draw=black!40,line width=.35pt] (2.92000,2.35000) rectangle ++(0.84000,0.60000); node[] at (3.34000,2.65000) 3; draw[fill=black!0,draw=black!40,line width=.35pt] (3.76000,2.35000) rectangle ++(0.84000,0.60000); draw[fill=black!20,draw=black!40,line width=.35pt] (4.60000,2.35000) rectangle ++(0.84000,0.60000); node[] at (5.02000,2.65000) 5; node[] at (2.90000,1.60000) (P_2,0x) (P_2,1y)=0; draw[fill=black!22,draw=black!40,line width=.35pt] (7.30000,3.60000) rectangle ++(1.50000,0.78…
Exact mathematical passage
(P_{2,0}x)\,(P_{2,1}y)=0
P_{2^s,j_2}Use in source · 1 passage
Search aliases: P_2^s,j_2
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1235
[Dyadic--odd packet factorization] After identifying the residue labels by the Chinese remainder map, the phase projector factors as P_N,j=P_2^s,j_2P_M,j_o.
Exact mathematical passage
\boxed{ P_{N,j}=P_{2^s,j_2}P_{M,j_o}. }
P_{3,0}Use in source · 2 passages
Search aliases: P_3,0
Use in source · atlas-pages.tex, line 369
)·(P_N,ky). Here the dot means multiplication at the same sample, not convolution. headCauchy multiplication adds degrees For formal series A(t),B(t), the coefficient of t^n is a sum over a+b=n. Consequently the output residue is i+jpmod N: P_N,k(AB)=∑_i+j≡ k (N)(P_N,iA)(P_N,jB). For N=3, take A=1+2t+3t^2 and B=2+t^2. Direct multiplication gives AB=2+4t+7t^2+2t^3+3t^4, P_3,0(AB)=2+2t^3. The 2t^3 term comes from residues 1+2, not 0+0. At power zero, the empty product contributes 1 only in residue…
Exact mathematical passage
AB=2+4t+7t^2+2t^3+3t^4, \qquad P_{3,0}(AB)=2+2t^3.
P_{3,k\bmod3}Use in source · 1 passage
Search aliases: P_3,kbmod3 bmod
Use in source · atlas-pages.tex, line 449
VE ADDRESSESTraditional monograph: Proposition 14.1 (p. 29). At phase order 12=4·3, the map k⟼(kbmod4, kbmod3) is a bijection from twelve residue addresses to the 4×3 grid. Every cell below contains its unique original address. vfigcrt_refinement headThe factorization is an intersection of conditions For integer sample addresses, including negative ones, P_12,k=P_4,kbmod4 P_3,kbmod3. The product on the right is composition of two residue masks: both congruences must hold. Coprimality makes their…
Exact mathematical passage
P_{12,k}=P_{4,k\bmod4}\,P_{3,k\bmod3}.
P_{4,k\bmod4}Use in source · 1 passage
Search aliases: P_4,kbmod4 bmod
Use in source · atlas-pages.tex, line 449
VE ADDRESSESTraditional monograph: Proposition 14.1 (p. 29). At phase order 12=4·3, the map k⟼(kbmod4, kbmod3) is a bijection from twelve residue addresses to the 4×3 grid. Every cell below contains its unique original address. vfigcrt_refinement headThe factorization is an intersection of conditions For integer sample addresses, including negative ones, P_12,k=P_4,kbmod4 P_3,kbmod3. The product on the right is composition of two residue masks: both congruences must hold. Coprimality makes their…
Exact mathematical passage
P_{12,k}=P_{4,k\bmod4}\,P_{3,k\bmod3}.
P_{4,r}Use in source · 1 passage
Search aliases: P_4,r
Use in source · atlas-pages.tex, line 42
y cell contains an original address and its rational sample value. Shading marks the known jump packet, not an estimated feature or a frequency band. headWhat has changed? Only where the values are stored The complete tuple of channels contains sixteen values, just as the original vector does. A single decimated channel is shorter, but keeping all four loses no samples. The residue projector P_4,rf is different: it keeps the original lattice and writes zeros outside phase r. S_4 A_4f=f, f _2^2=∑…
Exact mathematical passage
P_{4,r}f
P_{M,a}Use in source · 1 passage
Search aliases: P_M,a
Use in source · atlas-pages.tex, line 66
leaved to recover the coarse channel. Adding them entrywise instead produces sums of different original samples. For example, at the first descendant address the values f[0]=0 and f[2]=1/2 are two separate values. Their sum 1/2 cannot tell us both of them. Exact refinement is a disjoint partition, not an averaging rule or a new measurement. headThe general formula When N divides M, P_N,r=∑_substack0≤ a<M a≡ rpmod NP_M,a. The divisibility hypothesis ensures that every fine residue class lies enti…
Exact mathematical passage
P_{N,r}=\sum_{\substack{0\le a<M\\a\equiv r\pmod N}}P_{M,a}.
P_{M,j_o}Use in source · 1 passage
Search aliases: P_M,j_o
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1235
[Dyadic--odd packet factorization] After identifying the residue labels by the Chinese remainder map, the phase projector factors as P_N,j=P_2^s,j_2P_M,j_o.
Exact mathematical passage
\boxed{ P_{N,j}=P_{2^s,j_2}P_{M,j_o}. }
P_{M,s}Use in source · 2 passages
Search aliases: P_M,s
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 392
[Exact nested refinement] For N,M≥1 and Nmid M, P_N,r = ∑_substack0≤ s<M s≡ r (mod N)P_M,s.
Exact mathematical passage
\boxed{ P_{N,r} = \sum_{\substack{0\le s<M\\s\equiv r\, (\mathrm{mod}\,N)}}P_{M,s}. }
P_{N,i}Use in source · 3 passages
Search aliases: P_N,i
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1059
Linear time invariance is not required for exact packet coordinates. Let A and B be formal power series over a commutative coefficient ring (in particular C). Set A(z)=∑_n≥0a_nz^n, A_i=P_N,iA. All phase indices range from 0 to N-1. For another representation B(z), multiplication gives a nonlinear or bilinear interaction.
Exact mathematical passage
A(z)=\sum_{n\ge0}a_nz^n, \qquad A_i=P_{N,i}A.
P_{N,j}Use in source · 3 passages
Search aliases: P_N,j
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1235
[Dyadic--odd packet factorization] After identifying the residue labels by the Chinese remainder map, the phase projector factors as P_N,j=P_2^s,j_2P_M,j_o.
Exact mathematical passage
\boxed{ P_{N,j}=P_{2^s,j_2}P_{M,j_o}. }
P_r^{L}Use in source · 1 passage
Search aliases: P_r^L
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1874
[Exact lattice refinement] For every coarse packet, P_r^L = ∑_s∈Child(r)P_s^L'. The fine packets in the sum are pairwise disjoint. Refinement through several nested sublattices is transitive.
Exact mathematical passage
P_r^{L} = \sum_{s\in\operatorname{Child}(r)}P_s^{L'}.
P_sUse in source · 1 passage
Search aliases: P_s
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1793
[Exact lattice polyphase decomposition] For every 1≤ p≤∞, the analysis map A_L:ℓ^p( Z^d)⟶⊕_r∈ Rℓ^p(L), A_Lx=(x_r)_r∈ R, is an isometric isomorphism. Its inverse is the interleaving map S_L(u)(r+ℓ)=u_r(ℓ). The projectors satisfy P_rP_s=δ_rsP_r, ∑_r∈ RP_r=I.
Exact mathematical passage
P_rP_s=\delta_{rs}P_r, \qquad \sum_{r\in R}P_r=I.
P_s^{L'}Use in source · 1 passage
Search aliases: P_s^L'
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1874
[Exact lattice refinement] For every coarse packet, P_r^L = ∑_s∈Child(r)P_s^L'. The fine packets in the sum are pairwise disjoint. Refinement through several nested sublattices is transitive.
Exact mathematical passage
P_r^{L} = \sum_{s\in\operatorname{Child}(r)}P_s^{L'}.
\mathfrak P_vUse in source · 2 passages
Search aliases: P_v mathfrak
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1014
[1mm] displaystyle max{ε_v, ∑_u∈ch(v)E(u)}, · v internal. For the complete prunings P_v below v, the envelope is exactly E(v)=max_Λ∈ P_v∑_a∈Λε_a. The maximum is attained. Every complete-pruning cost changes by at most E(∅), and the two globally optimal values differ by at most this quantity. Consequently, for a unique optimum with competitors, Γ_y>2E(∅) is a deterministic certificate that the globally optimal pruning is unchanged. A singleton pruning set is stable without a gap condition. Assume…
Exact mathematical passage
\mathfrak P_v
P_{v0}Use in source · 1 passage
Search aliases: P_v0
Use in source · atlas-pages.tex, line 200
es, not time intervals in the original signal. vfigcomplete_cuts headThree candidate selections The complete selection {0,10,11} covers every terminal branch once. The selection {0,00,1} covers the 00 branch twice: it is not an antichain. The selection {00,01,10} misses the 11 branch entirely. For actual Haar packets, set P_v=W_v^*W_v. Parent-to-child conservation gives P_v=P_v0+P_v1. Replacing unselected parents by their children until the complete pruning is reached proves ∑_v∈ΛP_v=I. This is …
Exact mathematical passage
P_v=P_{v0}+P_{v1}.
P_{v1}Use in source · 1 passage
Search aliases: P_v1
Use in source · atlas-pages.tex, line 200
es, not time intervals in the original signal. vfigcomplete_cuts headThree candidate selections The complete selection {0,10,11} covers every terminal branch once. The selection {0,00,1} covers the 00 branch twice: it is not an antichain. The selection {00,01,10} misses the 11 branch entirely. For actual Haar packets, set P_v=W_v^*W_v. Parent-to-child conservation gives P_v=P_v0+P_v1. Replacing unselected parents by their children until the complete pruning is reached proves ∑_v∈ΛP_v=I. This is …
Exact mathematical passage
P_v=P_{v0}+P_{v1}.
\pmodUse in source · 7 passages
Search aliases: pmod
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 384
Suppose N,M≥1 and Nmid M. Every coarse phase has a disjoint family of fine descendants: Child_M(N,r) = {s:0≤ s<M, s≡ rpmod N}.
Exact mathematical passage
\operatorname{Child}_{M}(N,r) = \{s:0\le s<M,\ s\equiv r\pmod N\}.
Q^2Use in source · 1 passage
Search aliases: Q^2
Use in source · atlas-pages.tex, line 608
0 · 0, s=(1,0), e=(0,1). Then Q^2=Q and Qs=s, but (I-Q)e=(-2,1). The smooth error is zero while the unit packet error becomes a reconstruction error of norm sqrt5. For a fully specified realization, take j=e and widehat j=0. Then f=(1,1), widehat f=Qf=(3,0), and their difference is exactly (-2,1). headUse the actual operator bound The always-valid estimate in this setting is f-widehat f ≤ s-Qs +(1+ Q ) e . Here Q =sqrt5, so the bound is conservative. For an orthogonal projection, I-Q ≤1 gives a …
Exact mathematical passage
Q^2=Q
q_1Use in source · 3 passages
Search aliases: q_1
Use in source · atlas-pages.tex, line 528
vpage28Six image phases are six exact address classesLATTICE COORDINATES / A RECTANGULAR EXAMPLETraditional monograph: Theorem E3.2 (p. 43), Theorem E3.5 (p. 44), Remark E2.2 (p. 43). Take L=2 Z×3 Z inside Z^2, with representatives R={0,1}×{0,1,2}. Every address has a unique expression (i,j)=(r_1,r_2)+(2q_1,3q_2). The six channels are x_r(2q_1,3q_2)=x(r_1+2q_1,r_2+3q_2). vfig[.80]ext_lattice platecaptionEach cell shows its residue label. Colors identify six disjoint classes; they do not average …
Exact mathematical passage
(i,j)=(r_1,r_2)+(2q_1,3q_2)
q_2Use in source · 3 passages
Search aliases: q_2
Use in source · atlas-pages.tex, line 528
vpage28Six image phases are six exact address classesLATTICE COORDINATES / A RECTANGULAR EXAMPLETraditional monograph: Theorem E3.2 (p. 43), Theorem E3.5 (p. 44), Remark E2.2 (p. 43). Take L=2 Z×3 Z inside Z^2, with representatives R={0,1}×{0,1,2}. Every address has a unique expression (i,j)=(r_1,r_2)+(2q_1,3q_2). The six channels are x_r(2q_1,3q_2)=x(r_1+2q_1,r_2+3q_2). vfig[.80]ext_lattice platecaptionEach cell shows its residue label. Colors identify six disjoint classes; they do not average …
Exact mathematical passage
(i,j)=(r_1,r_2)+(2q_1,3q_2)
q_kUse in source · 2 passages
Search aliases: q_k
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 511
Original-sample delay. For an arbitrary original-sample delay D∈ Z, the output phase k has y_k[n]=x_r_k[n+q_k], r_k=(k-D)bmod N, q_k=(k-D)mathbindivN. Its channel matrix has the sole nonzero entry z^q_k in row k, column r_k. In particular, a one-sample delay at N=2 has matrix 0 · z^-1
Exact mathematical passage
y_k[n]=x_{r_k}[n+q_k],\qquad r_k=(k-D)\bmod N,\qquad q_k=(k-D)\mathbin{\mathrm{div}}N.
Q_nUse in source · 3 passages
Search aliases: Q_n
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3382
[Classical square-wave Gibbs behavior] Let q(x)=tfrac12sign(sin x), whose jump at zero has size one. For n≥1, its partial sum through the first n positive odd frequencies is Q_n(x)=frac2π∑_k=0^n-1fracsin((2k+1)x)2k+1. At x_n=π/(2n), x_n⟶0, Q_n(x_n)⟶fracSi(π)π =frac12+G, G≈0.089489872236, where Si(z)=∫_0^zsin(t)/t dt and G=Si(π)/π-1/2>0. Thus a nonzero overshoot persists along points approaching the jump. Scaling by a real amplitude a gives an overshoot or undershoot of magnitude |a|G.
Exact mathematical passage
Q_n(x)=\frac{2}{\pi}\sum_{k=0}^{n-1}\frac{\sin((2k+1)x)}{2k+1}.
\QQUse in source · 2 passages
Search aliases: QQ
Use in source · atlas-pages.tex, line 5
Section 13 (p. 24). A packet calculus can reorganize a signal exactly, choose among finitely many representations, or describe a nonlinear coordinate geometry. These are not the same task. This guide keeps one concrete object in view for each, so that the reader can compute rather than only follow arrows. vfigthree_worlds headA signal that can be checked by hand Our main finite signal is f=s+j∈QQ^16. The background repeats tfrac14(0,1,2,1,0,-1,-2,-1), and j[n]=3 at 6≤ n<12, with j[n]=0 elsewhere…
Exact mathematical passage
f=s+j\in\QQ^{16}
\QQ^{16}Use in source · 2 passages
Search aliases: QQ^16 QQ
Use in source · atlas-pages.tex, line 5
Section 13 (p. 24). A packet calculus can reorganize a signal exactly, choose among finitely many representations, or describe a nonlinear coordinate geometry. These are not the same task. This guide keeps one concrete object in view for each, so that the reader can compute rather than only follow arrows. vfigthree_worlds headA signal that can be checked by hand Our main finite signal is f=s+j∈QQ^16. The background repeats tfrac14(0,1,2,1,0,-1,-2,-1), and j[n]=3 at 6≤ n<12, with j[n]=0 elsewhere…
Exact mathematical passage
f=s+j\in\QQ^{16}
R'Use in source · 2 passages
Search aliases: R'
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1865
Let L'≤ L≤ Z^d, with both indices finite. Choose representative systems R for Z^d/L and R' for Z^d/L'. For r∈ R, define Child(r) = {s∈ R':s+L'⊂ r+L}.
Exact mathematical passage
R'
r^2Use in source · 1 passage
Search aliases: r^2
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2872
[A perturbed disk] Let E and F be concentric disks of radii r and r+δ in R^2, with δ>0. Then μ(Emathbintriangle F) = π((r+δ)^2-r^2) = 2π rδ+πδ^2. For equal amplitude a, the exact L^2 packet error is |a|sqrt2π rδ+πδ^2. This gives the expected square root dependence on boundary displacement.
Exact mathematical passage
\mu(E\mathbin\triangle F) = \pi\bigl((r+\delta)^2-r^2\bigr) = 2\pi r\delta+\pi\delta^2.
R_0Use in source · 3 passages
Search aliases: R_0
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2285
[Exact lifting update] The lifting matrices are invertible with L_P^-1=L_-P, U_Q^-1=U_-Q. If R_0E_0=I, then every finite product V of lifting matrices gives another exact bank by E=VE_0, R=R_0V^-1. Every lifting step has determinant one.
Exact mathematical passage
R_0E_0=I
r_0Use in source · 2 passages
Search aliases: r_0
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2102
[Stability under filter perturbation] Assume RE=z^-dI_N. Let widetilde E=E+Δ E, widetilde R=R+Δ R. Set e_0 · =*ess sup E _op, · r_0 · =*ess sup R _op, δ_E · =*ess sup Δ E _op, · δ_R · =*ess sup Δ R _op. Then the delay compensated reconstruction defect obeys ε_rec ≤ r_0δ_E+e_0δ_R+δ_Rδ_E.
Exact mathematical passage
\begin{aligned} e_0&=\esssup\|E\|_{\mathrm{op}}, & r_0&=\esssup\|R\|_{\mathrm{op}},\\ \delta_E&=\esssup\|\Delta E\|_{\mathrm{op}}, & \delta_R&=\esssup\|\Delta R\|_{\mathrm{op}}. \end{aligned}
r_1Use in source · 3 passages
Search aliases: r_1
Use in source · atlas-pages.tex, line 528
vpage28Six image phases are six exact address classesLATTICE COORDINATES / A RECTANGULAR EXAMPLETraditional monograph: Theorem E3.2 (p. 43), Theorem E3.5 (p. 44), Remark E2.2 (p. 43). Take L=2 Z×3 Z inside Z^2, with representatives R={0,1}×{0,1,2}. Every address has a unique expression (i,j)=(r_1,r_2)+(2q_1,3q_2). The six channels are x_r(2q_1,3q_2)=x(r_1+2q_1,r_2+3q_2). vfig[.80]ext_lattice platecaptionEach cell shows its residue label. Colors identify six disjoint classes; they do not average …
Exact mathematical passage
(i,j)=(r_1,r_2)+(2q_1,3q_2)
r_2Use in source · 3 passages
Search aliases: r_2
Use in source · atlas-pages.tex, line 528
vpage28Six image phases are six exact address classesLATTICE COORDINATES / A RECTANGULAR EXAMPLETraditional monograph: Theorem E3.2 (p. 43), Theorem E3.5 (p. 44), Remark E2.2 (p. 43). Take L=2 Z×3 Z inside Z^2, with representatives R={0,1}×{0,1,2}. Every address has a unique expression (i,j)=(r_1,r_2)+(2q_1,3q_2). The six channels are x_r(2q_1,3q_2)=x(r_1+2q_1,r_2+3q_2). vfig[.80]ext_lattice platecaptionEach cell shows its residue label. Colors identify six disjoint classes; they do not average …
Exact mathematical passage
(i,j)=(r_1,r_2)+(2q_1,3q_2)
r_aUse in source · 3 passages
Search aliases: r_a
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1830
[Translation covariance] Fix a∈ Z^d. For every r∈ R, write uniquely r-a=r_a+λ_a,r, r_a∈ R, λ_a,r∈ L. Then ( A_LT_ax)_r(ℓ) = x_r_a(ℓ+λ_a,r). Thus a fine translation permutes the lattice packets and applies an internal translation on L.
Exact mathematical passage
r-a=r_a+\lambda_{a,r}, \qquad r_a\in R, \quad \lambda_{a,r}\in L.
R_{\mathrm{can}}Use in source · 2 passages
Search aliases: R_can mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2164
[Coefficient noise under canonical synthesis] Under the hypotheses of [source reference], suppose the observed coefficient vector is c= A_Ex+ν. Canonical synthesis gives widehat x=R_canc and widehat x-x _2 ≤ frac1sqrt A ν _2.
Exact mathematical passage
\widehat x=R_{\mathrm{can}}c
\bm R_{\mathrm{H}}Use in source · 2 passages
Search aliases: R_H bm mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 533
1 · -1 . It is orthogonal, so R_H= E_H^ T and R_H E_H=I_2. The channels are a[n]=fracx[2n]+x[2n+1]sqrt2, d[n]=fracx[2n]-x[2n+1]sqrt2.
Exact mathematical passage
\bm R_{\mathrm{H}}=\bm E_{\mathrm{H}}^{\mathsf T}
r_jUse in source · 2 passages
Search aliases: r_j
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3529
[Shifted pointwise monomial law] Let y(n)=∏_j=1^mx(n-a_j) and write a_j=Nc_j+b_j, 0≤ b_j<N. For output residue r, define r_j≡ r-b_jpmod N, 0≤ r_j<N, and let κ_r,b_j = 1, · r<b_j,
Exact mathematical passage
r_j\equiv r-b_j\pmod N, \qquad 0\leq r_j<N,
r_kUse in source · 2 passages
Search aliases: r_k
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 511
Original-sample delay. For an arbitrary original-sample delay D∈ Z, the output phase k has y_k[n]=x_r_k[n+q_k], r_k=(k-D)bmod N, q_k=(k-D)mathbindivN. Its channel matrix has the sole nonzero entry z^q_k in row k, column r_k. In particular, a one-sample delay at N=2 has matrix 0 · z^-1
Exact mathematical passage
y_k[n]=x_{r_k}[n+q_k],\qquad r_k=(k-D)\bmod N,\qquad q_k=(k-D)\mathbin{\mathrm{div}}N.
R_{N,k}Use in source · 2 passages
Search aliases: R_N,k
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1137
Fix ω_N=e^2π i/N and rotate the reflection coefficient by R_N,k(Γ)=ω_N^kΓ.
Exact mathematical passage
R_{N,k}(\Gam)=\omega_N^k\Gam.
R_{\rm can}Use in source · 2 passages
Search aliases: R_rm can rm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2146
[Frame bounds from the polyphase symbol] Assume that A I_N ≤ E( e^ iω)^*E( e^ iω) ≤ B I_N for almost every ω, where 0<A≤ B<∞. Then the analysis bank satisfies A x _2^2 ≤ A_E x _2^2 ≤ B x _2^2. The pointwise matrix R_rm can( e^ iω) = (E^*E)^-1E^* is a bounded left inverse and gives the canonical dual synthesis bank.
Exact mathematical passage
R_{\rm can}(\e^{\ii\omega}) = (E^*E)^{-1}E^*
r_{\widehat\tau}Use in source · 1 passage
Search aliases: r_widehatτ widehat tau
Use in source · atlas-pages.tex, line 649
y on the interval between its two locations: H_τ-H_widehatτ _2=sqrtδ=tfrac14, H_τ-H_widehatτ _∞=1. vfig[.89]ext_location headFor a ramp, the error height shrinks too Write r_τ(t)=(t-τ)_+. The difference grows linearly from zero to δ on [τ,widehatτ] and stays at δ on [widehatτ,1]. Direct integration gives r_τ-r_widehatτ _2^2 =fracδ^33+(1-widehatτ)δ^2 =frac176144<frac1256=δ^2. Thus the L^2 ramp error is at most δ, and its supremum error is exactly δ. Positive-order truncated powers are Lipschitz i…
Exact mathematical passage
\|r_\tau-r_{\widehat\tau}\|_2^2 =\frac{\delta^3}{3}+(1-\widehat\tau)\delta^2 =\frac{17}{6144}<\frac1{256}=\delta^2.
R_\thetaUse in source · 3 passages
Search aliases: R_θ theta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3662
Let H = {z∈ C:Re z>0} and define C(z)=fracz-1z+1. It maps H conformally onto the unit disk D, with inverse C^-1(w)=frac1+w1-w. For θ∈ R, let R_θ(w)= e^ iθw.
Exact mathematical passage
R_\theta(w)=\e^{\ii\theta}w.
r_\tauUse in source · 2 passages
Search aliases: r_τ tau
Use in source · atlas-pages.tex, line 648
Traditional monograph: Corollary E8.2 (p. 60), Theorem E8.3 (p. 60), Theorem E8.7 (p. 62). Set τ=1/4, widehatτ=5/16, and δ=1/16. A unit step differs only on the interval between its two locations: H_τ-H_widehatτ _2=sqrtδ=tfrac14, H_τ-H_widehatτ _∞=1. vfig[.89]ext_location headFor a ramp, the error height shrinks too Write r_τ(t)=(t-τ)_+. The difference grows linearly from zero to δ on [τ,widehatτ] and stays at δ on [widehatτ,1]. Direct integration gives r_τ-r_widehatτ _2^2 =fracδ^33+(1-widehatτ)…
Exact mathematical passage
r_\tau(t)=(t-\tau)_+
\ReUse in source · 1 passage
Search aliases: Re
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3648
Let H = {z∈ C:Re z>0} and define C(z)=fracz-1z+1. It maps H conformally onto the unit disk D, with inverse C^-1(w)=frac1+w1-w. For θ∈ R, let R_θ(w)= e^ iθw.
Exact mathematical passage
\mathbb H = \{z\in\CC:\Re z>0\}
\operatorname{Re}Use in source · 2 passages
Search aliases: Re operatorname
Use in source · atlas-pages.tex, line 415
headThe coordinates preserve structure, not Euclidean distances A quarter turn is a rotation in the reflection disk. After pulling back, it is a M obius map, not a Euclidean rotation of the impedance plane. Equal reflection magnitudes need not give equal impedance magnitudes. The opposite points 1/2 and -1/2 become 3 and 1/3. The continuous impedance locus is the circle (Rez-5/3)^2+(Imz)^2=16/9, also drawn in the plate. Finite passive impedances with Rez≥0 map to the closed disk except Γ=1. That…
Exact mathematical passage
(\operatorname{Re}z-5/3)^2+(\operatorname{Im}z)^2=16/9
\mathrm{rec}Use in source · 3 passages
Search aliases: rec mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2058
[Quantitative imperfect reconstruction] Define the delay compensated defect D( e^ iω) = e^ i dωR( e^ iω)E( e^ iω)-I_N. If ε_rec = *ess sup_ω D( e^ iω) _op<∞, then the delay compensated reconstruction widetilde x satisfies widetilde x-x _2 ≤ ε_rec x _2. The constant is sharp as an operator norm.
Exact mathematical passage
\varepsilon_{\mathrm{rec}} = \esssup_{\omega}\|D(\e^{\ii\omega})\|_{\mathrm{op}}<\infty,
\rfloorUse in source · 1 passage
Search aliases: rfloor
Use in source · atlas-pages.tex, line 105
inal sample δ_1. Under DI_2 it stays even but moves to channel address one: this is δ_2. Thus J(D)^2=DI_2, H(D)=I_2+2J(D)-J(D)^2+J(D)^3 recovers the previous page's four-tap transfer matrix. The same delay algebra explains both the off-diagonal phase swap and the channel powers. headSigned delays need the same care For an original-sample delay d at modulus N, y_k[n]=f_(k-d)bmod N [n+lfloorfrack-dNrfloor]. This includes advances when d<0. The finite verification checks the formula on signed sampl…
Exact mathematical passage
y_k[n]=f_{(k-d)\bmod N} \!\left[n+\left\lfloor\frac{k-d}{N}\right\rfloor\right].
\rmUse in source · 10 passages
Search aliases: rm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1214
For a lossless transmission line, Γ_rm in=Γ_L e^-2 iβℓ. One N-phase step corresponds to a line-length increment λ/(2N), up to the chosen orientation convention. Hence the phase address has a direct electrical interpretation.
Exact mathematical passage
\Gam_{\rm in}=\Gam_L e^{-2\ii\beta\ell}.
\RMSEUse in source · 1 passage
Search aliases: RMSE operatorname
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 780
[One-dimensional errors] For 1024 samples on [-1,1) and Fourier cutoff m=18, the retained frequency indices are -18,…,18: 37 bins including the zero-frequency bin. The measured values were lrr method · RMSE · maximum absolute error global Fourier truncation · 5.4023×10^-2 · 3.4968×10^-1 known edge packet · 1.3692×10^-4 · 1.4503×10^-3 estimated edge packet · 2.4029×10^-2 · 5.9530×10^-2 The simple estimator recovered the binary mask exactly in this deliberately separated two-class example. That re…
Exact mathematical passage
\begin{array}{lrr} \toprule \text{method}&\RMSE&\text{maximum absolute error}\\ \midrule \text{global Fourier truncation}&5.4023\times10^{-2}&3.4968\times10^{-1}\\ \text{known edge packet}&1.3692\times10^{-4}&1.4503\times10^{-3}\\ \text{estimated edge packet}&2.4029\times10^{-2}&5.9530\times10^{-2}\\ \bottomrule \end{array}
\RR^2Use in source · 2 passages
Search aliases: RR^2 R^2 RR mathbb
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2871
[A perturbed disk] Let E and F be concentric disks of radii r and r+δ in R^2, with δ>0. Then μ(Emathbintriangle F) = π((r+δ)^2-r^2) = 2π rδ+πδ^2. For equal amplitude a, the exact L^2 packet error is |a|sqrt2π rδ+πδ^2. This gives the expected square root dependence on boundary displacement.
Exact mathematical passage
\RR^2
\mathcal SUse in source · 1 passage
Search aliases: S mathcal
Use in source · atlas-pages.tex, line 43
ump packet, not an estimated feature or a frequency band. headWhat has changed? Only where the values are stored The complete tuple of channels contains sixteen values, just as the original vector does. A single decimated channel is shorter, but keeping all four loses no samples. The residue projector P_4,rf is different: it keeps the original lattice and writes zeros outside phase r. S_4 A_4f=f, f _2^2=∑_r=0^3 f_r _2^2=57. The norm on the channel tuple is the sum-of-squares Hilbert norm, not th…
Exact mathematical passage
\mathcal S_4\mathcal A_4f=f, \qquad \|f\|_2^2=\sum_{r=0}^3\|f_r\|_2^2=57.
\sUse in source · 1 passage
Search aliases: s
Use in source · ext_frame.tex, line 7
draw[->] (0,0)--(9.4,0) node[right]singular value; draw[line width=6pt,teal!35] (3,1)--(9,1); draw[line width=6pt,black!35] (4,2)--(8,2); node[anchor=east]at(2.7,2)original;node[anchor=east]at(2.7,1)perturbed; foreach x/s in 3/3/4,4/1,8/2,9/9/4 draw(x,-.12)--(x,.12);node[below]at(x,-.15)s; fill (4,2)circle(2pt);fill(8,2)circle(2pt);fill[teal](3,1)circle(2pt);fill[teal](9,1)circle(2pt); draw[<->] (3,2.5)--(4,2.5)node[midway,above]δ; node at (6,-1.0)squared bounds: 9/16 ≤ widetilde Ex ^2/ x ^2 ≤ 8…
Exact mathematical passage
\s
S^{-1}Use in source · 1 passage
Search aliases: S^-1
Use in source · known_edge.tex, line 57
h=.45pt] (11.31329,1.52632) -- (11.31329,1.25000); draw[fill=white,line width=.7pt] (11.31329,1.25000) circle (0.045); draw[black!45,line width=.45pt] (12.05380,1.52632) -- (12.05380,1.34211); draw[fill=white,line width=.7pt] (12.05380,1.34211) circle (0.045); node[] at (6.50000,4.80000) solid: f dashed: Qf open: Qs+j; node[] at (6.50000,0.10000) periodic 16-sample model; Q=(S^-1+2I+S)/4;
Exact mathematical passage
Q=(S^{-1}+2I+S)/4
\mathcal S_4Use in source · 1 passage
Search aliases: S_4 mathcal
Use in source · atlas-pages.tex, line 43
ump packet, not an estimated feature or a frequency band. headWhat has changed? Only where the values are stored The complete tuple of channels contains sixteen values, just as the original vector does. A single decimated channel is shorter, but keeping all four loses no samples. The residue projector P_4,rf is different: it keeps the original lattice and writes zeros outside phase r. S_4 A_4f=f, f _2^2=∑_r=0^3 f_r _2^2=57. The norm on the channel tuple is the sum-of-squares Hilbert norm, not th…
Exact mathematical passage
\mathcal S_4\mathcal A_4f=f, \qquad \|f\|_2^2=\sum_{r=0}^3\|f_r\|_2^2=57.
s_fUse in source · 3 passages
Search aliases: s_f
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2551
Suppose the restriction of f to every interval (τ_q,τ_q+1) belongs to W^r,p, and choose the one dimensional Sobolev representatives whose derivatives through order r-1 have one sided traces at the endpoints. Define a_q,m = jumpf^(m)_τ_q = f^(m)(τ_q+)-f^(m)(τ_q-) and j_f(t) = ∑_q=1^Q∑_m=0^r-1a_q,mϕ_m,τ_q(t). Then s_f=f-j_f has matching one sided derivatives through order r-1 at every breakpoint and belongs to W^r,p(0,1). Its weak derivative of order r equals the piecewise classical derivative f^(…
Exact mathematical passage
s_f=f-j_f
S_mUse in source · 5 passages
Search aliases: S_m
Use in source · atlas-pages.tex, line 266
vpage16Why a Fourier cutoff cannot reproduce this jump pointwiseA CONTINUOUS PULSE / EXACT ENDPOINT VALUESTraditional monograph: Section 6 (p. 11), Section 8 (p. 13). Use a separate period-one example on [-1/2,1/2): j(t)= 1_{|t|<1/4}, S_mj(t)=tfrac12+frac2π∑_n=1^mfracsin(π n/2)ncos(2π nt). This is the exact finite Fourier formula. The drawn curves sample it numerically; the endpoint calculation below is symbolic. vfigfourier_jump headOne value is fixed at every cutoff At t=±1/4, every summand va…
Exact mathematical passage
j(t)=\mathbf 1_{\{|t|<1/4\}},\qquad S_mj(t)=\tfrac12+\frac2\pi\sum_{n=1}^{m}\frac{\sin(\pi n/2)}{n}\cos(2\pi nt).
\sgnUse in source · 1 passage
Search aliases: sgn operatorname
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2990
[Jump count and sign recovery] Under the hypotheses of [source reference], the number of detected cells equals Q, and sgn(widehat a_q)=sgn(a_q) for every real valued jump.
Exact mathematical passage
\sgn(\widehat a_q)=\sgn(a_q)
\operatorname{Si}Use in source · 4 passages
Search aliases: Si operatorname
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3386
[Classical square-wave Gibbs behavior] Let q(x)=tfrac12sign(sin x), whose jump at zero has size one. For n≥1, its partial sum through the first n positive odd frequencies is Q_n(x)=frac2π∑_k=0^n-1fracsin((2k+1)x)2k+1. At x_n=π/(2n), x_n⟶0, Q_n(x_n)⟶fracSi(π)π =frac12+G, G≈0.089489872236, where Si(z)=∫_0^zsin(t)/t dt and G=Si(π)/π-1/2>0. Thus a nonzero overshoot persists along points approaching the jump. Scaling by a real amplitude a gives an overshoot or undershoot of magnitude |a|G.
Exact mathematical passage
x_n\longrightarrow0,\qquad Q_n(x_n)\longrightarrow\frac{\operatorname{Si}(\pi)}{\pi} =\frac12+G,\qquad G\approx0.089489872236,
\operatorname{sign}Use in source · 1 passage
Search aliases: sign operatorname
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3381
[Classical square-wave Gibbs behavior] Let q(x)=tfrac12sign(sin x), whose jump at zero has size one. For n≥1, its partial sum through the first n positive odd frequencies is Q_n(x)=frac2π∑_k=0^n-1fracsin((2k+1)x)2k+1. At x_n=π/(2n), x_n⟶0, Q_n(x_n)⟶fracSi(π)π =frac12+G, G≈0.089489872236, where Si(z)=∫_0^zsin(t)/t dt and G=Si(π)/π-1/2>0. Thus a nonzero overshoot persists along points approaching the jump. Scaling by a real amplitude a gives an overshoot or undershoot of magnitude |a|G.
Exact mathematical passage
q(x)=\tfrac12\operatorname{sign}(\sin x)
\sinUse in source · 8 passages
Search aliases: sin
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1199
For 0≤ρ<1 and angle θ, Γ=ρ e^ iθ gives z(ρ,θ) =frac1+ρ e^ iθ1-ρ e^ iθ, with r(ρ,θ)=frac1-ρ^21-2ρcosθ+ρ^2, x(ρ,θ)=frac2ρsinθ1-2ρcosθ+ρ^2. A fixed ρ is a constant-mismatch circle. Refining N increases angular resolution without changing ρ.
Exact mathematical passage
r(\rho,\theta)=\frac{1-\rho^2}{1-2\rho\cos\theta+\rho^2}, \qquad x(\rho,\theta)=\frac{2\rho\sin\theta}{1-2\rho\cos\theta+\rho^2}.
\succUse in source · 1 passage
Search aliases: succ
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1358
The signal realization makes the seed, address, packet, transfer, refinement, and synthesis operations concrete. The objects and laws can be read directly: seed · =δ_0, address · =S^n, packet · =P_N,rx, transfer · =P_N,kHP_N,s, refinement · =P_N,r=∑_ssucc rP_M,s, synthesis · =∑_rP_N,rx, nonlinearity · =P_N,rF(∑_sP_N,sx). Filter banks, wavelet packets, adaptive complete prunings, edge packets, and Smith coordinates inhabit the same operator grammar, but their claims differ. Polyphase and filter-b…
Exact mathematical passage
\boxed{ \begin{aligned} \text{seed}&=\delta_0,\\ \text{address}&=S^n,\\ \text{packet}&=P_{N,r}x,\\ \text{transfer}&=P_{N,k}HP_{N,s},\\ \text{refinement}&=P_{N,r}=\sum_{s\succ r}P_{M,s},\\ \text{synthesis}&=\sum_rP_{N,r}x,\\ \text{nonlinearity}&=P_{N,r}F\left(\sum_sP_{N,s}x\right). \end{aligned} }
\supUse in source · 5 passages
Search aliases: sup
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 949
imizer is unique and at least one competing pruning exists, its gap Γ_y = min_Λ≠Λ_y^* (J_y(Λ)-J_y(Λ_y^*)) is attained and strictly positive. If the pruning set is a singleton, its only element minimizes every criterion; no real-valued minimum over an empty competitor set is introduced. item In the case with competitors, if new data y' satisfy sup_Λ∈ P( T) |J_y'(Λ)-J_y(Λ)| <fracΓ_y2, then Λ_y'^*=Λ_y^*. item Every selected minimizer preserves the exact synthesis identity of the preceding theorem.
Exact mathematical passage
\sup_{\Lambda\in\mathfrak P(\mathcal T)} |J_{y'}(\Lambda)-J_y(\Lambda)| <\frac{\Gamma_y}{2},
\sup_mUse in source · 2 passages
Search aliases: sup_m sup
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3426
[Estimated packet Gibbs bound] Let Q_m be linear and uniformly bounded on L^p, with sup_m Q_m ≤ C_Q. For f=s+j and an estimated packet widehat j_m, define widehat f_m =Q_m(f-widehat j_m)+widehat j_m. Then f-widehat f_m _p ≤ s-Q_ms _p +(1+C_Q) j-widehat j_m _p. Consequently, smooth approximation and packet recovery may be improved independently. No claim of Gibbs removal is justified unless both terms are controlled in the norm under consideration.
Exact mathematical passage
\sup_m\|Q_m\|\leq C_Q.
\sup_uUse in source · 1 passage
Search aliases: sup_u sup
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2766
Assume τ<widehatτ. Differentiation with respect to the location parameter gives frac∂∂ uϕ_m,u(t) =-ϕ_m-1,u(t) for almost every t. Hence ϕ_m,τ-ϕ_m,widehatτ = ∫_τ^widehatτϕ_m-1,u d u. Minkowski's inequality gives ϕ_m,τ-ϕ_m,widehatτ _p ≤ |τ-widehatτ| sup_u ϕ_m-1,u _p. The supremum is bounded by the norm of t^m-1/(m-1)! on [0,1], which is the stated constant. The supremum norm argument is identical.
Exact mathematical passage
\|\phi_{m,\tau}-\phi_{m,\widehat\tau}\|_p \leq |\tau-\widehat\tau| \sup_u\|\phi_{m-1,u}\|_p.
\sup_{\Lambda\in\mathfrak P(\mathcal T)}Use in source · 1 passage
Search aliases: sup_Λ∈ P( T) sup Lambda in mathfrak mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 949
imizer is unique and at least one competing pruning exists, its gap Γ_y = min_Λ≠Λ_y^* (J_y(Λ)-J_y(Λ_y^*)) is attained and strictly positive. If the pruning set is a singleton, its only element minimizes every criterion; no real-valued minimum over an empty competitor set is introduced. item In the case with competitors, if new data y' satisfy sup_Λ∈ P( T) |J_y'(Λ)-J_y(Λ)| <fracΓ_y2, then Λ_y'^*=Λ_y^*. item Every selected minimizer preserves the exact synthesis identity of the preceding theorem.
Exact mathematical passage
\sup_{\Lambda\in\mathfrak P(\mathcal T)} |J_{y'}(\Lambda)-J_y(\Lambda)| <\frac{\Gamma_y}{2},
\sup_{\Xi}Use in source · 1 passage
Search aliases: sup_Ξ sup Xi
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 961
set, so the feasible set is nonempty and finite. Thus the minimum is attained. If competitors exist, their finitely many differences from a unique optimum are strictly positive, so their minimum is positive and is achieved by a competitor. In the singleton case there is no alternative selector under any objective change. In the remaining case, for any competing pruning Λ≠Λ_y^*, J_y'(Λ)-J_y'(Λ_y^*) · ≥ J_y(Λ)-J_y(Λ_y^*) -2sup_Ξ|J_y'(Ξ)-J_y(Ξ)| · >Γ_y-Γ_y=0. Hence the same pruning remains uniquely…
Exact mathematical passage
\begin{aligned} J_{y'}(\Lambda)-J_{y'}(\Lambda_y^*) &\ge J_y(\Lambda)-J_y(\Lambda_y^*) -2\sup_{\Xi}|J_{y'}(\Xi)-J_y(\Xi)|\\ &>\Gamma_y-\Gamma_y=0. \end{aligned}
\mathsf TUse in source · 1 passage
Search aliases: T mathsf
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 533
1 · -1 . It is orthogonal, so R_H= E_H^ T and R_H E_H=I_2. The channels are a[n]=fracx[2n]+x[2n+1]sqrt2, d[n]=fracx[2n]-x[2n+1]sqrt2.
Exact mathematical passage
\bm R_{\mathrm{H}}=\bm E_{\mathrm{H}}^{\mathsf T}
t^2Use in source · 8 passages
Search aliases: t^2
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3119
If σ=0, every centered zero-sub-Gaussian noise variable vanishes almost surely, and the noise-free deterministic conclusion applies. For σ>0, the difference η_k+1-η_k is sub Gaussian with variance proxy 2σ^2, so P(|η_k+1-η_k|>t) ≤ 2exp(-fract^24σ^2). A union bound over the n cells gives P( max_k|η_k+1-η_k|>t_n,δ ) ≤ δ. On the complementary event, repeat the deterministic proof.
Exact mathematical passage
\mathbb P\bigl(|\eta_{k+1}-\eta_k|>t\bigr) \leq 2\exp\left(-\frac{t^2}{4\sigma^2}\right).
t^3Use in source · 3 passages
Search aliases: t^3
Use in source · atlas-pages.tex, line 369
)·(P_N,ky). Here the dot means multiplication at the same sample, not convolution. headCauchy multiplication adds degrees For formal series A(t),B(t), the coefficient of t^n is a sum over a+b=n. Consequently the output residue is i+jpmod N: P_N,k(AB)=∑_i+j≡ k (N)(P_N,iA)(P_N,jB). For N=3, take A=1+2t+3t^2 and B=2+t^2. Direct multiplication gives AB=2+4t+7t^2+2t^3+3t^4, P_3,0(AB)=2+2t^3. The 2t^3 term comes from residues 1+2, not 0+0. At power zero, the empty product contributes 1 only in residue…
Exact mathematical passage
AB=2+4t+7t^2+2t^3+3t^4, \qquad P_{3,0}(AB)=2+2t^3.
t^4Use in source · 1 passage
Search aliases: t^4
Use in source · atlas-pages.tex, line 369
)·(P_N,ky). Here the dot means multiplication at the same sample, not convolution. headCauchy multiplication adds degrees For formal series A(t),B(t), the coefficient of t^n is a sum over a+b=n. Consequently the output residue is i+jpmod N: P_N,k(AB)=∑_i+j≡ k (N)(P_N,iA)(P_N,jB). For N=3, take A=1+2t+3t^2 and B=2+t^2. Direct multiplication gives AB=2+4t+7t^2+2t^3+3t^4, P_3,0(AB)=2+2t^3. The 2t^3 term comes from residues 1+2, not 0+0. At power zero, the empty product contributes 1 only in residue…
Exact mathematical passage
AB=2+4t+7t^2+2t^3+3t^4, \qquad P_{3,0}(AB)=2+2t^3.
t^{m-1}Use in source · 1 passage
Search aliases: t^m-1
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2772
ion with respect to the location parameter gives frac∂∂ uϕ_m,u(t) =-ϕ_m-1,u(t) for almost every t. Hence ϕ_m,τ-ϕ_m,widehatτ = ∫_τ^widehatτϕ_m-1,u d u. Minkowski's inequality gives ϕ_m,τ-ϕ_m,widehatτ _p ≤ |τ-widehatτ| sup_u ϕ_m-1,u _p. The supremum is bounded by the norm of t^m-1/(m-1)! on [0,1], which is the stated constant. The supremum norm argument is identical.
Exact mathematical passage
t^{m-1}/(m-1)!
t^nUse in source · 1 passage
Search aliases: t^n
Use in source · atlas-pages.tex, line 364
kets can interact. vfigproduct_conventions headPointwise multiplication preserves the same address For time-residue masks, (P_N,ix)[n](P_N,jy)[n]=0 when i≠ j. Therefore P_N,k(x· y)=(P_N,kx)·(P_N,ky). Here the dot means multiplication at the same sample, not convolution. headCauchy multiplication adds degrees For formal series A(t),B(t), the coefficient of t^n is a sum over a+b=n. Consequently the output residue is i+jpmod N: P_N,k(AB)=∑_i+j≡ k (N)(P_N,iA)(P_N,jB). For N=3, take A=1+2t+3t^2 and B…
Exact mathematical passage
t^n
T_{2,1}Use in source · 1 passage
Search aliases: T_2,1
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1171
[Impedance--admittance is the two-phase split] For N=2 and finite z≠0, T_2,1(z)=z^-1. On the full projective line it exchanges 0 and ∞. Thus a half-turn of the reflection disk exchanges normalized impedance and admittance. ω_2=exp(π i)=-1, so A(ω_2)[u:v]=[v:u]. The finite ratio is 1/z when z≠0; the coordinate swap also gives [0:1]↔[1:0]. No ordinary division by zero is used to represent infinity.
Exact mathematical passage
\boxed{T_{2,1}(z)=z^{-1}.}
t_kUse in source · 4 passages
Search aliases: t_k
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2900
Let t_k=kh, k=0,…,n, h=frac1n. Assume f(t) = s(t)+∑_q=1^Qa_qH_τ_q(t), H_τ(t)= 1_[τ,1](t), where s is Lipschitz with constant M, and Q≥1 is finite with nonzero jump amplitudes. The zero-jump case requires no minimum-amplitude condition. Suppose no jump lies at a sample point and no grid cell contains more than one jump. Observe y_k=f(t_k)+η_k. Define the adjacent contrast D_k=y_k+1-y_k.
Exact mathematical passage
t_k=kh, \qquad k=0,\ldots,n, \qquad h=\frac1n.
t_{k+1}Use in source · 2 passages
Search aliases: t_k+1
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2929
[Contrast gap] Assume |η_k|≤ε and set b_h=Mh+2ε. If the cell (t_k,t_k+1) contains no jump, then |D_k|≤ b_h. If it contains one jump of amplitude a_q, then |D_k-a_q|≤ b_h and therefore |D_k|≥ |a_q|-b_h.
Exact mathematical passage
(t_k,t_{k+1})
T_{ks}Use in source · 1 passage
Search aliases: T_ks
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1469
T_ks · P_N,kHP_N,s · input-to-output phase transfer
Exact mathematical passage
T_{ks}
T_{N,k}Use in source · 3 passages
Search aliases: T_N,k
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1143
[Polyphase M obius action] Regard normalized impedance as the projective line P^1( C)= C∪{∞}. For N≥1, the pullback is a globally defined projective action. In the finite chart, wherever the denominator is nonzero, it is T_N,k(z) = frac(1+ω_N^k)z+(1-ω_N^k) (1-ω_N^k)z+(1+ω_N^k). As maps on P^1( C) it satisfies T_N,k∘ T_N,ℓ=T_N,k+ℓ (mod N).
Exact mathematical passage
\boxed{ T_{N,k}(z) = \frac{(1+\omega_N^k)z+(1-\omega_N^k)} {(1-\omega_N^k)z+(1+\omega_N^k)}. }
T_{N,k+\ell\, (\mathrm{mod}\,N)}Use in source · 1 passage
Search aliases: T_N,k+ℓ (mod N) ell mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1152
obius action] Regard normalized impedance as the projective line P^1( C)= C∪{∞}. For N≥1, the pullback is a globally defined projective action. In the finite chart, wherever the denominator is nonzero, it is T_N,k(z) = frac(1+ω_N^k)z+(1-ω_N^k) (1-ω_N^k)z+(1+ω_N^k). As maps on P^1( C) it satisfies T_N,k∘ T_N,ℓ=T_N,k+ℓ (mod N).
Exact mathematical passage
T_{N,k}\circ T_{N,\ell}=T_{N,k+\ell\, (\mathrm{mod}\,N)}.
t_{n,\delta}Use in source · 3 passages
Search aliases: t_n,δ delta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3098
[Probability recovery bound] Fix 0<δ<1 and define t_n,δ = 2σ sqrtlog(frac2nδ) and b_h,δ=Mh+t_n,δ. With probability at least 1-δ, every contrast satisfies the conclusions of [source reference] with b_h replaced by b_h,δ.
Exact mathematical passage
t_{n,\delta} = 2\sigma \sqrt{\log\left(\frac{2n}{\delta}\right)}
T_{N,\ell}Use in source · 1 passage
Search aliases: T_N,ℓ ell
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1152
obius action] Regard normalized impedance as the projective line P^1( C)= C∪{∞}. For N≥1, the pullback is a globally defined projective action. In the finite chart, wherever the denominator is nonzero, it is T_N,k(z) = frac(1+ω_N^k)z+(1-ω_N^k) (1-ω_N^k)z+(1+ω_N^k). As maps on P^1( C) it satisfies T_N,k∘ T_N,ℓ=T_N,k+ℓ (mod N).
Exact mathematical passage
T_{N,k}\circ T_{N,\ell}=T_{N,k+\ell\, (\mathrm{mod}\,N)}.
\topruleUse in source · 3 passages
Search aliases: toprule
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 780
[One-dimensional errors] For 1024 samples on [-1,1) and Fourier cutoff m=18, the retained frequency indices are -18,…,18: 37 bins including the zero-frequency bin. The measured values were lrr method · RMSE · maximum absolute error global Fourier truncation · 5.4023×10^-2 · 3.4968×10^-1 known edge packet · 1.3692×10^-4 · 1.4503×10^-3 estimated edge packet · 2.4029×10^-2 · 5.9530×10^-2 The simple estimator recovered the binary mask exactly in this deliberately separated two-class example. That re…
Exact mathematical passage
\begin{array}{lrr} \toprule \text{method}&\RMSE&\text{maximum absolute error}\\ \midrule \text{global Fourier truncation}&5.4023\times10^{-2}&3.4968\times10^{-1}\\ \text{known edge packet}&1.3692\times10^{-4}&1.4503\times10^{-3}\\ \text{estimated edge packet}&2.4029\times10^{-2}&5.9530\times10^{-2}\\ \bottomrule \end{array}
\triangleUse in source · 15 passages
Search aliases: triangle
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 712
[Packet perturbation bound] For 1≤ p<∞, interpreted first as an inequality of extended L^p seminorms, j-widehat j _L^p ≤ |a-widehat a| μ(Ω)^1/p + |widehat a| μ(ΩtrianglewidehatΩ)^1/p. If μ(Ω) and μ(ΩtrianglewidehatΩ) are finite, all terms in this displayed jump bound are finite and it is an ordinary real-valued L^p inequality. For the pointwise supremum, and consequently also the essential L^∞ seminorm, j-widehat j _∞ ≤ |a-widehat a|+|widehat a| 1_ΩtrianglewidehatΩ≠∅.
Exact mathematical passage
\boxed{ \|j-\widehat j\|_{L^p} \le |a-\widehat a|\,\mu(\Omega)^{1/p} + |\widehat a|\,\mu(\Omega\triangle\widehat\Omega)^{1/p}. }
U_{-Q}Use in source · 1 passage
Search aliases: U_-Q
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2280
[Exact lifting update] The lifting matrices are invertible with L_P^-1=L_-P, U_Q^-1=U_-Q. If R_0E_0=I, then every finite product V of lifting matrices gives another exact bank by E=VE_0, R=R_0V^-1. Every lifting step has determinant one.
Exact mathematical passage
L_P^{-1}=L_{-P}, \qquad U_Q^{-1}=U_{-Q}.
u_kUse in source · 2 passages
Search aliases: u_k
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3393
For the limit, put u_k=(k+tfrac12)/n. The identity Q_n(x_n)=frac1π n∑_k=0^n-1fracsin(π u_k)u_k ⟶frac1π∫_0^1fracsin(π u)u du =fracSi(π)π is a midpoint Riemann sum, since the integrand has continuous value π at zero. This is the normalized classical example in [source reference].
Exact mathematical passage
u_k=(k+\tfrac12)/n
U_QUse in source · 1 passage
Search aliases: U_Q
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2262
Let P(z) and Q(z) be Laurent polynomials. Define L_P(z) = 1 · 0 P(z) · 1 , U_Q(z) = 1 · Q(z) 0 · 1 .
Exact mathematical passage
L_P(z) = \begin{pmatrix} 1&0\\ P(z)&1 \end{pmatrix}, \qquad U_Q(z) = \begin{pmatrix} 1&Q(z)\\ 0&1 \end{pmatrix}.
U_Q^{-1}Use in source · 1 passage
Search aliases: U_Q^-1
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2280
[Exact lifting update] The lifting matrices are invertible with L_P^-1=L_-P, U_Q^-1=U_-Q. If R_0E_0=I, then every finite product V of lifting matrices gives another exact bank by E=VE_0, R=R_0V^-1. Every lifting step has determinant one.
Exact mathematical passage
L_P^{-1}=L_{-P}, \qquad U_Q^{-1}=U_{-Q}.
u_rUse in source · 2 passages
Search aliases: u_r
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1776
For finite p, equip ⊕_r∈ Rℓ^p(L) with (u_r) ^p=∑_r∈ R u_r _p^p. For p=∞, use the maximum norm.
Exact mathematical passage
\|(u_r)\|^p=\sum_{r\in R}\|u_r\|_p^p.
U_{\rm root}Use in source · 1 passage
Search aliases: U_rm root rm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2356
Proceed by induction on the depth of the retained subtree. If the root is retained, the statement is immediate. If the root is split, apply the unitary map U_rm root, then apply the induction hypothesis independently in each child subtree. A direct sum of unitary maps is unitary. Reversing the maps with adjoints gives synthesis.
Exact mathematical passage
U_{\rm root}
VUse in source · 11 passages
Search aliases: V
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 981
[Dynamic-programming best packet basis] Suppose J_y(Λ)=∑_v∈Λc_v(y). For a node v, define recursively V_y(v) = c_v(y), · v terminal, [2mm] displaystyle min{c_v(y),∑_u∈ch(v)V_y(u)}, · v internal. Stop at v when c_v(y)≤∑_uV_y(u), including equality; refine otherwise. This deterministic rule produces a globally optimal complete pruning. With the node costs supplied and child values cached, the evaluator uses one comparison per internal node, one visit per node, and at most | T|-1 additions. Its arit…
Exact mathematical passage
V_y(v) = \begin{cases} c_v(y),&v\text{ terminal},\\[2mm] \displaystyle \min\left\{c_v(y),\sum_{u\in\operatorname{ch}(v)}V_y(u)\right\}, &v\text{ internal}. \end{cases}
V^{-1}Use in source · 2 passages
Search aliases: V^-1
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2286
[Exact lifting update] The lifting matrices are invertible with L_P^-1=L_-P, U_Q^-1=U_-Q. If R_0E_0=I, then every finite product V of lifting matrices gives another exact bank by E=VE_0, R=R_0V^-1. Every lifting step has determinant one.
Exact mathematical passage
E=VE_0, \qquad R=R_0V^{-1}.
v^2Use in source · 1 passage
Search aliases: v^2
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1124
Writing Γ=u+ i v, the grid equations are (u-fracr1+r)^2+v^2=frac1(1+r)^2, and, for x≠0, (u-1)^2+(v-frac1x)^2=frac1x^2.
Exact mathematical passage
\left(u-\frac{r}{1+r}\right)^2+v^2=\frac1{(1+r)^2},
V_yUse in source · 2 passages
Search aliases: V_y
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 981
[Dynamic-programming best packet basis] Suppose J_y(Λ)=∑_v∈Λc_v(y). For a node v, define recursively V_y(v) = c_v(y), · v terminal, [2mm] displaystyle min{c_v(y),∑_u∈ch(v)V_y(u)}, · v internal. Stop at v when c_v(y)≤∑_uV_y(u), including equality; refine otherwise. This deterministic rule produces a globally optimal complete pruning. With the node costs supplied and child values cached, the evaluator uses one comparison per internal node, one visit per node, and at most | T|-1 additions. Its arit…
Exact mathematical passage
V_y(v) = \begin{cases} c_v(y),&v\text{ terminal},\\[2mm] \displaystyle \min\left\{c_v(y),\sum_{u\in\operatorname{ch}(v)}V_y(u)\right\}, &v\text{ internal}. \end{cases}
W^{2,p}Use in source · 1 passage
Search aliases: W^2,p
Use in source · atlas-pages.tex, line 590
l smooth remainder. headThe normalization makes the trace coefficient literal The general packet is ϕ_m,τ(t)=(t-τ)_+^m/m!. Its derivative of order m has a unit trace jump, while its other classical trace jumps vanish. Thus its multiplier is exactly the jump being removed, with no hidden factorial. Here the remainder has matching value and first-derivative traces and belongs to W^2,p(0,1) for every 1≤ p≤∞. Its second weak derivative is 2. The singular packet accounts for the distributional bounda…
Exact mathematical passage
W^{2,p}(0,1)
w_{\rm Smith}Use in source · 2 passages
Search aliases: w_rm Smith rm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1253
Away from its pole Γ=1, the inverse Cayley map has derivative fracdzdΓ=frac2(1-Γ)^2, while normalized admittance has derivative proportional to (1+Γ)^-2. Away from Γ=±1, this motivates the symmetric sensitivity w_rm Smith(Γ) = frac1|1-Γ|^2 + frac1|1+Γ|^2.
Exact mathematical passage
\boxed{ w_{\rm Smith}(\Gam) = \frac1{|1-\Gam|^2} + \frac1{|1+\Gam|^2}. }
W_vUse in source · 1 passage
Search aliases: W_v
Use in source · atlas-pages.tex, line 199
al positions below are terminal packet addresses, not time intervals in the original signal. vfigcomplete_cuts headThree candidate selections The complete selection {0,10,11} covers every terminal branch once. The selection {0,00,1} covers the 00 branch twice: it is not an antichain. The selection {00,01,10} misses the 11 branch entirely. For actual Haar packets, set P_v=W_v^*W_v. Parent-to-child conservation gives P_v=P_v0+P_v1. Replacing unselected parents by their children until the complete …
Exact mathematical passage
P_v=W_v^*W_v
W_v^*Use in source · 1 passage
Search aliases: W_v^*
Use in source · atlas-pages.tex, line 199
al positions below are terminal packet addresses, not time intervals in the original signal. vfigcomplete_cuts headThree candidate selections The complete selection {0,10,11} covers every terminal branch once. The selection {0,00,1} covers the 00 branch twice: it is not an antichain. The selection {00,01,10} misses the 11 branch entirely. For actual Haar packets, set P_v=W_v^*W_v. Parent-to-child conservation gives P_v=P_v0+P_v1. Replacing unselected parents by their children until the complete …
Exact mathematical passage
P_v=W_v^*W_v
W_{\ell,\alpha}^*Use in source · 1 passage
Search aliases: W_ℓ,α^* ell alpha
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 623
Set A=∑_α W_ℓ,α^*W_ℓ,α. This is a finite sum of bounded operators. The energy result says ⟨ x,Ax⟩=⟨ x,x⟩ for every x. Complex polarization, applied also to x+y and x+ i y, gives ⟨ y,(A-I)x⟩=0 for all x,y, hence A=I. Applying the operator identity to x proves synthesis. This argument uses the actual Hilbert adjoints of the decimation operators, not a separately postulated inverse symbol.
Exact mathematical passage
A=\sum_\alpha W_{\ell,\alpha}^*W_{\ell,\alpha}
W_{\ell,\alpha}^{*}Use in source · 1 passage
Search aliases: W_ℓ,α^* ell alpha
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 604
[Wavelet-packet reconstruction and energy] At every finite depth ℓ, x=∑_α∈{0,1}^ℓW_ℓ,α^*W_ℓ,αx, and x _2^2 = ∑_α∈{0,1}^ℓ W_ℓ,αx _2^2.
Exact mathematical passage
\boxed{ x=\sum_{\alpha\in\{0,1\}^{\ell}}W_{\ell,\alpha}^{*}W_{\ell,\alpha}x, }
\widehat aUse in source · 16 passages
Search aliases: widehat a widehat
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 701
Let (X,Σ,μ) be a measure space, with measurable masks Ω,widehatΩ⊆ X and scalar amplitudes a,widehat a∈ C (real amplitudes are a special case). Suppose a scalar jump packet is j=a 1_Ω, and an estimated packet is widehat j=widehat a 1_widehatΩ.
Exact mathematical passage
a,\widehat a\in\bbC
\widehat a_qUse in source · 7 passages
Search aliases: widehat a_q widehat
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2970
[Deterministic jump cell recovery] Let a_min=min_q|a_q| and assume a_min>2b_h. Choose a threshold θ satisfying b_h<θ<a_min-b_h. Then widehat K = {k:|D_k|>θ} is exactly the set of grid cells containing jumps. If widehatτ_q is the midpoint of the detected cell and widehat a_q=D_k, then after matching by order, |widehatτ_q-τ_q| ≤ frac h2 and |widehat a_q-a_q| ≤ b_h.
Exact mathematical passage
\widehat a_q=D_k
\widehat cUse in source · 2 passages
Search aliases: widehat c widehat
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2360
[Coefficient perturbations in a unitary tree] Let c= A_ Lx be the leaf coefficients of an admissible unitary pruning, and let widehat c=c+e. If widehat x is synthesized from widehat c, then widehat x-x = e . Thus coefficient quantization and coefficient noise are neither amplified nor attenuated by exact unitary synthesis.
Exact mathematical passage
\widehat c=c+e
\widehat fUse in source · 14 passages
Search aliases: widehat f widehat
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1679
The reconstruction formula with an estimated packet is f-widehat f_m = (I-Q_m)s+(I-Q_m)(j-widehat j). This identity identifies the two terms that must be controlled. The exact packet formulas in [source reference] then evaluate or bound j-widehat j from amplitude error, region overlap, boundary displacement, and the order of a derivative jump. A value jump has location modulus |τ-widehatτ|^1/p in finite L^p, while every positive order truncated power packet is Lipschitz in its location.
Exact mathematical passage
f-\widehat f_m = (I-Q_m)s+(I-Q_m)(j-\widehat j).
\widehat f_mUse in source · 10 passages
Search aliases: widehat f_m widehat
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1679
The reconstruction formula with an estimated packet is f-widehat f_m = (I-Q_m)s+(I-Q_m)(j-widehat j). This identity identifies the two terms that must be controlled. The exact packet formulas in [source reference] then evaluate or bound j-widehat j from amplitude error, region overlap, boundary displacement, and the order of a derivative jump. A value jump has location modulus |τ-widehatτ|^1/p in finite L^p, while every positive order truncated power packet is Lipschitz in its location.
Exact mathematical passage
f-\widehat f_m = (I-Q_m)s+(I-Q_m)(j-\widehat j).
\widehat j_mUse in source · 4 passages
Search aliases: widehat j_m widehat
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3429
[Estimated packet Gibbs bound] Let Q_m be linear and uniformly bounded on L^p, with sup_m Q_m ≤ C_Q. For f=s+j and an estimated packet widehat j_m, define widehat f_m =Q_m(f-widehat j_m)+widehat j_m. Then f-widehat f_m _p ≤ s-Q_ms _p +(1+C_Q) j-widehat j_m _p. Consequently, smooth approximation and packet recovery may be improved independently. No claim of Gibbs removal is justified unless both terms are controlled in the norm under consideration.
Exact mathematical passage
\widehat j_m
\widehat sUse in source · 3 passages
Search aliases: widehat s widehat
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 744
If s,widehat s are measurable smooth-background models, the same triangle inequality gives, for finite p, f-(widehat s+widehat j) _p ≤ s-widehat s _p +|a-widehat a|μ(Ω)^1/p +|widehat a|μ(ΩtrianglewidehatΩ)^1/p. This is the correct separation of errors. Detection error is not hidden inside a generic reconstruction score.
Exact mathematical passage
s,\widehat s
\widehat xUse in source · 5 passages
Search aliases: widehat x widehat
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2164
[Coefficient noise under canonical synthesis] Under the hypotheses of [source reference], suppose the observed coefficient vector is c= A_Ex+ν. Canonical synthesis gives widehat x=R_canc and widehat x-x _2 ≤ frac1sqrt A ν _2.
Exact mathematical passage
\widehat x=R_{\mathrm{can}}c
\widehat{\bm X}Use in source · 1 passage
Search aliases: widehat X widehat bm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 491
with the monomial z^-m acting on a bilateral sequence by x[n]↦ x[n-m]. Let E∈Mat_N× N( L) and R∈Mat_N× N( L) be analysis and synthesis matrices. Their composition acts on all bilateral channel sequences by finite convolutions. At a nonzero z where the input transforms are absolutely convergent, this gives widehat X(z)= R(z) E(z) X(z).
Exact mathematical passage
\widehat{\bm X}(z)=\bm R(z)\bm E(z)\bm X(z).
\widehat\KcalUse in source · 1 passage
Search aliases: widehatKcal widehat K widehat Kcal widehat mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2965
[Deterministic jump cell recovery] Let a_min=min_q|a_q| and assume a_min>2b_h. Choose a threshold θ satisfying b_h<θ<a_min-b_h. Then widehat K = {k:|D_k|>θ} is exactly the set of grid cells containing jumps. If widehatτ_q is the midpoint of the detected cell and widehat a_q=D_k, then after matching by order, |widehatτ_q-τ_q| ≤ frac h2 and |widehat a_q-a_q| ≤ b_h.
Exact mathematical passage
\widehat\Kcal = \{k:|D_k|>\theta\}
\widehat\tauUse in source · 20 passages
Search aliases: widehatτ widehat tau
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1684
reconstruction formula with an estimated packet is f-widehat f_m = (I-Q_m)s+(I-Q_m)(j-widehat j). This identity identifies the two terms that must be controlled. The exact packet formulas in [source reference] then evaluate or bound j-widehat j from amplitude error, region overlap, boundary displacement, and the order of a derivative jump. A value jump has location modulus |τ-widehatτ|^1/p in finite L^p, while every positive order truncated power packet is Lipschitz in its location.
Exact mathematical passage
|\tau-\widehat\tau|^{1/p}
\widehat\tau_qUse in source · 4 passages
Search aliases: widehatτ_q widehat tau
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2970
[Deterministic jump cell recovery] Let a_min=min_q|a_q| and assume a_min>2b_h. Choose a threshold θ satisfying b_h<θ<a_min-b_h. Then widehat K = {k:|D_k|>θ} is exactly the set of grid cells containing jumps. If widehatτ_q is the midpoint of the detected cell and widehat a_q=D_k, then after matching by order, |widehatτ_q-τ_q| ≤ frac h2 and |widehat a_q-a_q| ≤ b_h.
Exact mathematical passage
\widehat\tau_q
\widehat\OmegaUse in source · 20 passages
Search aliases: widehatΩ widehat Omega
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 701
Let (X,Σ,μ) be a measure space, with measurable masks Ω,widehatΩ⊆ X and scalar amplitudes a,widehat a∈ C (real amplitudes are a special case). Suppose a scalar jump packet is j=a 1_Ω, and an estimated packet is widehat j=widehat a 1_widehatΩ.
Exact mathematical passage
\Omega,\widehat\Omega\subseteq X
\widetilde CUse in source · 2 passages
Search aliases: widetilde C widetilde
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2476
For every admissible pruning, |C( L)-widetilde C( L)| ≤ ε_*. Evaluate widetilde C on a minimizer for C to obtain widetilde F_*≤ F_*+ε_*. Reverse the roles to obtain F_*≤widetilde F_*+ε_*.
Exact mathematical passage
|C(\Lcal)-\widetilde C(\Lcal)| \leq \varepsilon_*.
\widetilde cUse in source · 3 passages
Search aliases: widetilde c widetilde
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2454
[Stability of the optimal cost] Let c and widetilde c be two node cost functions. For an admissible pruning L, set ε( L) = ∑_v∈ L|c(v)-widetilde c(v)| and let ε_* = max_ Lε( L). If F_* and widetilde F_* are the two optimal root costs, then |F_*-widetilde F_*| ≤ ε_*.
Exact mathematical passage
\widetilde c
\widetilde EUse in source · 9 passages
Search aliases: widetilde E widetilde
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2096
[Stability under filter perturbation] Assume RE=z^-dI_N. Let widetilde E=E+Δ E, widetilde R=R+Δ R. Set e_0 · =*ess sup E _op, · r_0 · =*ess sup R _op,
Exact mathematical passage
\widetilde E=E+\Delta E, \qquad \widetilde R=R+\Delta R.
\widetilde FUse in source · 4 passages
Search aliases: widetilde F widetilde
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2466
[Stability of the optimal cost] Let c and widetilde c be two node cost functions. For an admissible pruning L, set ε( L) = ∑_v∈ L|c(v)-widetilde c(v)| and let ε_* = max_ Lε( L). If F_* and widetilde F_* are the two optimal root costs, then |F_*-widetilde F_*| ≤ ε_*.
Exact mathematical passage
\widetilde F_*
\widetilde fUse in source · 15 passages
Search aliases: widetilde f widetilde
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 655
Let f=s+j where s belongs to a smooth approximation class and j is a known discontinuity packet. Let Q_m be any approximation operator applied only to s, and define widetilde f_m=Q_ms+j.
Exact mathematical passage
\widetilde f_m=Q_ms+j.
\widetilde F_*Use in source · 4 passages
Search aliases: widetilde F_* widetilde
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2466
[Stability of the optimal cost] Let c and widetilde c be two node cost functions. For an admissible pruning L, set ε( L) = ∑_v∈ L|c(v)-widetilde c(v)| and let ε_* = max_ Lε( L). If F_* and widetilde F_* are the two optimal root costs, then |F_*-widetilde F_*| ≤ ε_*.
Exact mathematical passage
\widetilde F_*
\widetilde f_mUse in source · 7 passages
Search aliases: widetilde f_m widetilde
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 655
Let f=s+j where s belongs to a smooth approximation class and j is a known discontinuity packet. Let Q_m be any approximation operator applied only to s, and define widetilde f_m=Q_ms+j.
Exact mathematical passage
\widetilde f_m=Q_ms+j.
\widetilde RUse in source · 2 passages
Search aliases: widetilde R widetilde
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2096
[Stability under filter perturbation] Assume RE=z^-dI_N. Let widetilde E=E+Δ E, widetilde R=R+Δ R. Set e_0 · =*ess sup E _op, · r_0 · =*ess sup R _op,
Exact mathematical passage
\widetilde E=E+\Delta E, \qquad \widetilde R=R+\Delta R.
\widetilde xUse in source · 2 passages
Search aliases: widetilde x widetilde
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2063
[Quantitative imperfect reconstruction] Define the delay compensated defect D( e^ iω) = e^ i dωR( e^ iω)E( e^ iω)-I_N. If ε_rec = *ess sup_ω D( e^ iω) _op<∞, then the delay compensated reconstruction widetilde x satisfies widetilde x-x _2 ≤ ε_rec x _2. The constant is sharp as an operator norm.
Exact mathematical passage
\widetilde x
XUse in source · 7 passages
Search aliases: X
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 437
[Exact FIR polyphase transfer law] The output channels satisfy y_k[n] = ∑_s=0^N-1∑_m∈ Z h_ks[m]x_s[n-m]. For the numerical channel transform, take z∈ C∖{0} and assume ∑_n∈ Z|x_s[n]z^-n|<∞ (0≤ s<N). Define X_s(z)=∑_nx_s[n]z^-n. The output series then also converge absolutely, and Y(z)= H^(N)(z) X(z), H^(N)_ks(z)=∑_m h[Nm+k-s]z^-m.
Exact mathematical passage
X_s(z)=\sum_nx_s[n]z^{-n}
\bm XUse in source · 4 passages
Search aliases: X bm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 438
[Exact FIR polyphase transfer law] The output channels satisfy y_k[n] = ∑_s=0^N-1∑_m∈ Z h_ks[m]x_s[n-m]. For the numerical channel transform, take z∈ C∖{0} and assume ∑_n∈ Z|x_s[n]z^-n|<∞ (0≤ s<N). Define X_s(z)=∑_nx_s[n]z^-n. The output series then also converge absolutely, and Y(z)= H^(N)(z) X(z), H^(N)_ks(z)=∑_m h[Nm+k-s]z^-m.
Exact mathematical passage
\bm Y(z)=\bm H^{(N)}(z)\bm X(z), \qquad H^{(N)}_{ks}(z)=\sum_m h[Nm+k-s]z^{-m}.
x^2Use in source · 1 passage
Search aliases: x^2
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1128
Writing Γ=u+ i v, the grid equations are (u-fracr1+r)^2+v^2=frac1(1+r)^2, and, for x≠0, (u-1)^2+(v-frac1x)^2=frac1x^2.
Exact mathematical passage
(u-1)^2+\left(v-\frac1x\right)^2=\frac1{x^2}.
x_0Use in source · 5 passages
Search aliases: x_0
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 350
Define the analysis and synthesis maps A_Nx=(x_0,…,x_N-1), S_N(x_0,…,x_N-1)[Nq+r]=x_r[q].
Exact mathematical passage
\cA_Nx=(x_0,\ldots,x_{N-1}), \qquad \cS_N(x_0,\ldots,x_{N-1})[Nq+r]=x_r[q].
x_1Use in source · 2 passages
Search aliases: x_1
Use in source · fir_transfer.tex, line 15
put 1; draw[fill=black!14,draw=black!40,line width=.35pt] (2.20000,1.30000) rectangle ++(4.10000,1.05000); node[] at (4.25000,1.82500) D + 2; draw[fill=black!0,draw=black!40,line width=.35pt] (6.30000,1.30000) rectangle ++(4.10000,1.05000); node[] at (8.35000,1.82500) 1 - D; node[] at (4.25000,3.95000) input 0; node[] at (8.35000,3.95000) input 1; node[] at (6.40000,0.15000) y_0=(1-D)x_0+(2D+D^2)x_1;
Exact mathematical passage
y_0=(1-D)x_0+(2D+D^2)x_1
x_FUse in source · 2 passages
Search aliases: x_F
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 290
The shift orbit is the standard orthonormal basis {δ_n}_n∈ Z. For x∈ℓ^2, the finite truncations x_F=∑_n∈ Fx[n]δ_n, F⊂ Z finite, satisfy x-x_F _2^2=∑_n∉ F|x[n]|^2→0 as the finite sets exhaust Z. Thus the unordered Hilbert-space sum ∑_nx[n]δ_n converges to x, and the finite-support statement is exact. The shift is a surjective isometry because n↦ n-1 is a bijection of Z, with inverse shift n↦ n+1. This proves the closed-span assertion.
Exact mathematical passage
x_F=\sum_{n\in F}x[n]\delta_n,\qquad F\subset\bbZ\text{ finite},
x_nUse in source · 3 passages
Search aliases: x_n
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3385
[Classical square-wave Gibbs behavior] Let q(x)=tfrac12sign(sin x), whose jump at zero has size one. For n≥1, its partial sum through the first n positive odd frequencies is Q_n(x)=frac2π∑_k=0^n-1fracsin((2k+1)x)2k+1. At x_n=π/(2n), x_n⟶0, Q_n(x_n)⟶fracSi(π)π =frac12+G, G≈0.089489872236, where Si(z)=∫_0^zsin(t)/t dt and G=Si(π)/π-1/2>0. Thus a nonzero overshoot persists along points approaching the jump. Scaling by a real amplitude a gives an overshoot or undershoot of magnitude |a|G.
Exact mathematical passage
x_n=\pi/(2n)
x_{N-1}Use in source · 3 passages
Search aliases: x_N-1
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 350
Define the analysis and synthesis maps A_Nx=(x_0,…,x_N-1), S_N(x_0,…,x_N-1)[Nq+r]=x_r[q].
Exact mathematical passage
\cA_Nx=(x_0,\ldots,x_{N-1}), \qquad \cS_N(x_0,\ldots,x_{N-1})[Nq+r]=x_r[q].
x_rUse in source · 11 passages
Search aliases: x_r
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 321
Fix an integer N≥1. For 0≤ r<N, define (P_N,rx)[n] = 1_n≡ r (mod N)x[n]. The corresponding decimated channel is x_r[q]=x[Nq+r].
Exact mathematical passage
x_r[q]=x[Nq+r].
x_{r-1}Use in source · 1 passage
Search aliases: x_r-1
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3578
[Quadratic nearest neighbor interaction] For y(n)=x(n)x(n-1) and modulus N, y_r(q) = x_r(q)x_r-1(q), · r≥1, x_0(q)x_N-1(q-1), · r=0. The carry at residue zero is the nonlinear analogue of a transfer crossing a coarse cell boundary.
Exact mathematical passage
y_r(q) = \begin{cases} x_r(q)x_{r-1}(q),&r\geq1,\\ x_0(q)x_{N-1}(q-1),&r=0. \end{cases}
x_{r_a}Use in source · 1 passage
Search aliases: x_r_a
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1838
[Translation covariance] Fix a∈ Z^d. For every r∈ R, write uniquely r-a=r_a+λ_a,r, r_a∈ R, λ_a,r∈ L. Then ( A_LT_ax)_r(ℓ) = x_r_a(ℓ+λ_a,r). Thus a fine translation permutes the lattice packets and applies an internal translation on L.
Exact mathematical passage
(\Acal_LT_ax)_r(\ell) = x_{r_a}(\ell+\lambda_{a,r}).
x_{r_j}Use in source · 1 passage
Search aliases: x_r_j
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3544
0, · r≥ b_j. Then the output channel is y_r(q) = ∏_j=1^m x_r_j (q-c_j-κ_r,b_j).
Exact mathematical passage
\boxed{ y_r(q) = \prod_{j=1}^{m} x_{r_j} \bigl(q-c_j-\kappa_{r,b_j}\bigr). }
x_{r_k}Use in source · 1 passage
Search aliases: x_r_k
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 511
Original-sample delay. For an arbitrary original-sample delay D∈ Z, the output phase k has y_k[n]=x_r_k[n+q_k], r_k=(k-D)bmod N, q_k=(k-D)mathbindivN. Its channel matrix has the sole nonzero entry z^q_k in row k, column r_k. In particular, a one-sample delay at N=2 has matrix 0 · z^-1
Exact mathematical passage
y_k[n]=x_{r_k}[n+q_k],\qquad r_k=(k-D)\bmod N,\qquad q_k=(k-D)\mathbin{\mathrm{div}}N.
X_sUse in source · 1 passage
Search aliases: X_s
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 437
[Exact FIR polyphase transfer law] The output channels satisfy y_k[n] = ∑_s=0^N-1∑_m∈ Z h_ks[m]x_s[n-m]. For the numerical channel transform, take z∈ C∖{0} and assume ∑_n∈ Z|x_s[n]z^-n|<∞ (0≤ s<N). Define X_s(z)=∑_nx_s[n]z^-n. The output series then also converge absolutely, and Y(z)= H^(N)(z) X(z), H^(N)_ks(z)=∑_m h[Nm+k-s]z^-m.
Exact mathematical passage
X_s(z)=\sum_nx_s[n]z^{-n}
x_sUse in source · 9 passages
Search aliases: x_s
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 425
[Exact FIR polyphase transfer law] The output channels satisfy y_k[n] = ∑_s=0^N-1∑_m∈ Z h_ks[m]x_s[n-m]. For the numerical channel transform, take z∈ C∖{0} and assume ∑_n∈ Z|x_s[n]z^-n|<∞ (0≤ s<N). Define X_s(z)=∑_nx_s[n]z^-n. The output series then also converge absolutely, and Y(z)= H^(N)(z) X(z), H^(N)_ks(z)=∑_m h[Nm+k-s]z^-m.
Exact mathematical passage
\boxed{ y_k[n] = \sum_{s=0}^{N-1}\sum_{m\in\bbZ} h_{ks}[m]x_s[n-m]. }
x_vUse in source · 1 passage
Search aliases: x_v
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2348
[Reconstruction on a pruned packet tree] Every admissible pruning L defines a unitary analysis map A_ L: H_rm root ⟶ ⊕_v∈ L H_v. Its inverse is obtained by applying the adjoints of the local splits from the leaves toward the root. In particular, x ^2 = ∑_v∈ L x_v ^2.
Exact mathematical passage
\|x\|^2 = \sum_{v\in\Lcal}\|x_v\|^2.
x_\ellUse in source · 1 passage
Search aliases: x_ℓ ell
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2406
Assume the child subtree bounds have been proved. Then ∑_w∑_ℓ∈ L_w x_ℓ ^2 ≥ min_w a(w) ∑_w (U_vx)_w ^2 ≥ A_vmin_w a(w) x ^2. The upper bound follows in the same way with the maximum and B_v. Induction completes the proof.
Exact mathematical passage
\sum_{w}\sum_{\ell\in\Lcal_w}\|x_\ell\|^2 \geq \min_w a(w) \sum_w\|(U_vx)_w\|^2 \geq A_v\min_w a(w)\|x\|^2.
\bm YUse in source · 2 passages
Search aliases: Y bm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 438
[Exact FIR polyphase transfer law] The output channels satisfy y_k[n] = ∑_s=0^N-1∑_m∈ Z h_ks[m]x_s[n-m]. For the numerical channel transform, take z∈ C∖{0} and assume ∑_n∈ Z|x_s[n]z^-n|<∞ (0≤ s<N). Define X_s(z)=∑_nx_s[n]z^-n. The output series then also converge absolutely, and Y(z)= H^(N)(z) X(z), H^(N)_ks(z)=∑_m h[Nm+k-s]z^-m.
Exact mathematical passage
\bm Y(z)=\bm H^{(N)}(z)\bm X(z), \qquad H^{(N)}_{ks}(z)=\sum_m h[Nm+k-s]z^{-m}.
y'Use in source · 5 passages
Search aliases: y'
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 948
tem If the minimizer is unique and at least one competing pruning exists, its gap Γ_y = min_Λ≠Λ_y^* (J_y(Λ)-J_y(Λ_y^*)) is attained and strictly positive. If the pruning set is a singleton, its only element minimizes every criterion; no real-valued minimum over an empty competitor set is introduced. item In the case with competitors, if new data y' satisfy sup_Λ∈ P( T) |J_y'(Λ)-J_y(Λ)| <fracΓ_y2, then Λ_y'^*=Λ_y^*. item Every selected minimizer preserves the exact synthesis identity of the prece…
Exact mathematical passage
y'
y_0Use in source · 3 passages
Search aliases: y_0
Use in source · atlas-pages.tex, line 78
note one delay in the channel index. vfigfir_transfer headReading one matrix entry The odd-input contribution to the even output is (2D+D^2)f_1. The tap at original delay one reaches the preceding odd channel sample; the tap at delay three reaches the odd channel two positions back. These are different delays despite coming from the same input phase. The two channel equations are y_0[n] · =f_0[n]-f_0[n-1]+2f_1[n-1]+f_1[n-2], y_1[n] · =2f_0[n]+f_0[n-1]+f_1[n]-f_1[n-1]. Interleaving gives exactly …
Exact mathematical passage
\begin{aligned} y_0[n]&=f_0[n]-f_0[n-1]+2f_1[n-1]+f_1[n-2],\\ y_1[n]&=2f_0[n]+f_0[n-1]+f_1[n]-f_1[n-1]. \end{aligned}
y_1Use in source · 2 passages
Search aliases: y_1
Use in source · atlas-pages.tex, line 78
note one delay in the channel index. vfigfir_transfer headReading one matrix entry The odd-input contribution to the even output is (2D+D^2)f_1. The tap at original delay one reaches the preceding odd channel sample; the tap at delay three reaches the odd channel two positions back. These are different delays despite coming from the same input phase. The two channel equations are y_0[n] · =f_0[n]-f_0[n-1]+2f_1[n-1]+f_1[n-2], y_1[n] · =2f_0[n]+f_0[n-1]+f_1[n]-f_1[n-1]. Interleaving gives exactly …
Exact mathematical passage
\begin{aligned} y_0[n]&=f_0[n]-f_0[n-1]+2f_1[n-1]+f_1[n-2],\\ y_1[n]&=2f_0[n]+f_0[n-1]+f_1[n]-f_1[n-1]. \end{aligned}
y_kUse in source · 8 passages
Search aliases: y_k
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 425
[Exact FIR polyphase transfer law] The output channels satisfy y_k[n] = ∑_s=0^N-1∑_m∈ Z h_ks[m]x_s[n-m]. For the numerical channel transform, take z∈ C∖{0} and assume ∑_n∈ Z|x_s[n]z^-n|<∞ (0≤ s<N). Define X_s(z)=∑_nx_s[n]z^-n. The output series then also converge absolutely, and Y(z)= H^(N)(z) X(z), H^(N)_ks(z)=∑_m h[Nm+k-s]z^-m.
Exact mathematical passage
\boxed{ y_k[n] = \sum_{s=0}^{N-1}\sum_{m\in\bbZ} h_{ks}[m]x_s[n-m]. }
y_{k+1}Use in source · 2 passages
Search aliases: y_k+1
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2920
me f(t) = s(t)+∑_q=1^Qa_qH_τ_q(t), H_τ(t)= 1_[τ,1](t), where s is Lipschitz with constant M, and Q≥1 is finite with nonzero jump amplitudes. The zero-jump case requires no minimum-amplitude condition. Suppose no jump lies at a sample point and no grid cell contains more than one jump. Observe y_k=f(t_k)+η_k. Define the adjacent contrast D_k=y_k+1-y_k.
Exact mathematical passage
D_k=y_{k+1}-y_k.
y_rUse in source · 3 passages
Search aliases: y_r
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1927
[Exact convolution transfer law] Let y=h*x. Then y_r(ℓ) = ∑_s∈ R∑_m∈ Lh_rs(m)x_s(ℓ-m). Equivalently, A_LH A_L^-1 = (H_rs)_r,s∈ R, where H_rs is convolution on L by h_rs.
Exact mathematical passage
y_r(\ell) = \sum_{s\in R}\sum_{m\in L}h_{rs}(m)x_s(\ell-m).
ZUse in source · 2 passages
Search aliases: Z
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1106
Let Z_0>0 be a reference impedance and z=fracZZ_0=r+ i x. The reflection coefficient is Γ=fracz-1z+1, z=frac1+Γ1-Γ.
Exact mathematical passage
Z_0>0
\mathbb ZUse in source · 1 passage
Search aliases: Z mathbb
Use in source · atlas-pages.tex, line 244
vpage15Retain the jump and approximate the other packetAN EXACT IDENTITY / A CONDITIONAL INPUTTraditional monograph: Theorem 6.1 (p. 12), Corollary 6.2 (p. 12). Return to the running signal f=s+j. For a concrete finite calculation, use the periodic three-tap smoother (Qx)[n]=tfrac14x[n-1]+tfrac12x[n]+tfrac14x[n+1], n∈ Z/16 Z. The plots compare smoothing the full signal with retaining its supplied jump packet. Every plotted sample is a rational number computed from the displayed formula. vfigknow…
Exact mathematical passage
(Qx)[n]=\tfrac14x[n-1]+\tfrac12x[n]+\tfrac14x[n+1],\qquad n\in\mathbb Z/16\mathbb Z.
z^{-1}Use in source · 6 passages
Search aliases: z^-1
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 490
Work with finite impulse-response banks. Let L= C[z,z^-1] be the Laurent-polynomial algebra, with the monomial z^-m acting on a bilateral sequence by x[n]↦ x[n-m]. Let E∈Mat_N× N( L) and R∈Mat_N× N( L) be analysis and synthesis matrices. Their composition acts on all bilateral channel sequences by finite convolutions. At a nonzero z where the input transforms are absolutely convergent, this gives widehat X(z)= R(z) E(z) X(z).
Exact mathematical passage
\mathcal L=\bbC[z,z^{-1}]
z^{-j}Use in source · 1 passage
Search aliases: z^-j
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 458
The last factor is x_s[n-m], giving the time-domain formula. Under the additional transform hypothesis, each delayed channel has transform ∑_nx_s[n-m]z^-n =z^-m∑_jx_s[j]z^-j. Absolute summability permits this bijective reindexing, and only finitely many delayed channels are combined. Taking their finite linear combination gives the matrix formula. The time-domain theorem therefore remains valid even when no numerical bilateral transform exists.
Exact mathematical passage
\sum_nx_s[n-m]z^{-n} =z^{-m}\sum_jx_s[j]z^{-j}.
z^{-m}Use in source · 4 passages
Search aliases: z^-m
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 438
[Exact FIR polyphase transfer law] The output channels satisfy y_k[n] = ∑_s=0^N-1∑_m∈ Z h_ks[m]x_s[n-m]. For the numerical channel transform, take z∈ C∖{0} and assume ∑_n∈ Z|x_s[n]z^-n|<∞ (0≤ s<N). Define X_s(z)=∑_nx_s[n]z^-n. The output series then also converge absolutely, and Y(z)= H^(N)(z) X(z), H^(N)_ks(z)=∑_m h[Nm+k-s]z^-m.
Exact mathematical passage
\bm Y(z)=\bm H^{(N)}(z)\bm X(z), \qquad H^{(N)}_{ks}(z)=\sum_m h[Nm+k-s]z^{-m}.
z^{-n}Use in source · 3 passages
Search aliases: z^-n
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 434
[Exact FIR polyphase transfer law] The output channels satisfy y_k[n] = ∑_s=0^N-1∑_m∈ Z h_ks[m]x_s[n-m]. For the numerical channel transform, take z∈ C∖{0} and assume ∑_n∈ Z|x_s[n]z^-n|<∞ (0≤ s<N). Define X_s(z)=∑_nx_s[n]z^-n. The output series then also converge absolutely, and Y(z)= H^(N)(z) X(z), H^(N)_ks(z)=∑_m h[Nm+k-s]z^-m.
Exact mathematical passage
\sum_{n\in\bbZ}|x_s[n]z^{-n}|<\infty\quad(0\le s<N).
z^nUse in source · 2 passages
Search aliases: z^n
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1059
Linear time invariance is not required for exact packet coordinates. Let A and B be formal power series over a commutative coefficient ring (in particular C). Set A(z)=∑_n≥0a_nz^n, A_i=P_N,iA. All phase indices range from 0 to N-1. For another representation B(z), multiplication gives a nonlinear or bilinear interaction.
Exact mathematical passage
A(z)=\sum_{n\ge0}a_nz^n, \qquad A_i=P_{N,i}A.
z^{q_k}Use in source · 1 passage
Search aliases: z^q_k
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 515
Original-sample delay. For an arbitrary original-sample delay D∈ Z, the output phase k has y_k[n]=x_r_k[n+q_k], r_k=(k-D)bmod N, q_k=(k-D)mathbindivN. Its channel matrix has the sole nonzero entry z^q_k in row k, column r_k. In particular, a one-sample delay at N=2 has matrix 0 · z^-1
Exact mathematical passage
z^{q_k}
Z_0Use in source · 2 passages
Search aliases: Z_0
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1106
Let Z_0>0 be a reference impedance and z=fracZZ_0=r+ i x. The reflection coefficient is Γ=fracz-1z+1, z=frac1+Γ1-Γ.
Exact mathematical passage
Z_0>0
\ZZ^2Use in source · 2 passages
Search aliases: ZZ^2 Z^2 ZZ mathbb
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1888
[Rectangular image phases] For L=M Z× N Z⊂ Z^2, one may take R={0,…,M-1}×{0,…,N-1}. The analysis map separates an image into MN exact phases. Refining (M,N) to (M',N'), where Mmid M' and Nmid N', splits every coarse phase into fracM'MfracN'N fine phases.
Exact mathematical passage
L=M\ZZ\times N\ZZ\subset\ZZ^2,
|Use in source · 98 passages
Search aliases: |
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 293
The shift orbit is the standard orthonormal basis {δ_n}_n∈ Z. For x∈ℓ^2, the finite truncations x_F=∑_n∈ Fx[n]δ_n, F⊂ Z finite, satisfy x-x_F _2^2=∑_n∉ F|x[n]|^2→0 as the finite sets exhaust Z. Thus the unordered Hilbert-space sum ∑_nx[n]δ_n converges to x, and the finite-support statement is exact. The shift is a surjective isometry because n↦ n-1 is a bijection of Z, with inverse shift n↦ n+1. This proves the closed-span assertion.
Exact mathematical passage
\|x-x_F\|_2^2=\sum_{n\notin F}|x[n]|^2\to0
|^{-2}Use in source · 1 passage
Search aliases: |^-2
Use in source · atlas-pages.tex, line 479
arepsilon=1/100, the actual increment is 20/9, while derivative times increment is 2. Both are calculated exactly. Their difference comes from using a differential in place of a finite increment. The derivative gives a first order approximation. headWhat a proposed weight can and cannot certify Impedance is sensitive near Γ=1, and admittance near Γ=-1. The symmetric weight w(Γ)=|1-Γ|^-2+|1+Γ|^-2 reflects both chart sensitivities. Its growth motivates denser sampling near the endpoints, but does …
Exact mathematical passage
w(\Gamma)=|1-\Gamma|^{-2}+|1+\Gamma|^{-2}
|^{1/p}Use in source · 2 passages
Search aliases: |^1/p
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1684
reconstruction formula with an estimated packet is f-widehat f_m = (I-Q_m)s+(I-Q_m)(j-widehat j). This identity identifies the two terms that must be controlled. The exact packet formulas in [source reference] then evaluate or bound j-widehat j from amplitude error, region overlap, boundary displacement, and the order of a derivative jump. A value jump has location modulus |τ-widehatτ|^1/p in finite L^p, while every positive order truncated power packet is Lipschitz in its location.
Exact mathematical passage
|\tau-\widehat\tau|^{1/p}
|^2Use in source · 5 passages
Search aliases: |^2
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 293
The shift orbit is the standard orthonormal basis {δ_n}_n∈ Z. For x∈ℓ^2, the finite truncations x_F=∑_n∈ Fx[n]δ_n, F⊂ Z finite, satisfy x-x_F _2^2=∑_n∉ F|x[n]|^2→0 as the finite sets exhaust Z. Thus the unordered Hilbert-space sum ∑_nx[n]δ_n converges to x, and the finite-support statement is exact. The shift is a surjective isometry because n↦ n-1 is a bijection of Z, with inverse shift n↦ n+1. This proves the closed-span assertion.
Exact mathematical passage
\|x-x_F\|_2^2=\sum_{n\notin F}|x[n]|^2\to0
|^pUse in source · 8 passages
Search aliases: |^p
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1802
Every address has a unique representation r+ℓ. For finite p, x _p^p = ∑_r∈ R∑_ℓ∈ L|x(r+ℓ)|^p = ∑_r∈ R x_r _p^p. The corresponding supremum identity is immediate when p=∞. Unique address representation proves that analysis and interleaving are inverse maps. The projector identities follow from the disjoint coset partition.
Exact mathematical passage
\|x\|_p^p = \sum_{r\in R}\sum_{\ell\in L}|x(r+\ell)|^p = \sum_{r\in R}\|x_r\|_p^p.
\pmUse in source · 3 passages
Search aliases: ± pm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1252
Away from its pole Γ=1, the inverse Cayley map has derivative fracdzdΓ=frac2(1-Γ)^2, while normalized admittance has derivative proportional to (1+Γ)^-2. Away from Γ=±1, this motivates the symmetric sensitivity w_rm Smith(Γ) = frac1|1-Γ|^2 + frac1|1+Γ|^2.
Exact mathematical passage
\Gam=\pm1
\cdotUse in source · 4 passages
Search aliases: · cdot
Use in source · atlas-pages.tex, line 327
tfrac52(1_Ω-1_widehatΩ). The missed sample has error 3, the five shared samples have error 1/2, and the extra sample has error -5/2. With counting measure, j-widehat j _1=8, j-widehat j _2^2=tfrac332, j-widehat j _∞=3. headRead the theorem as a budget, not an equality claim Here the L^1 bound is attained: |3-tfrac52| |Ω|+tfrac52|ΩtrianglewidehatΩ| =tfrac12·6+tfrac52·2=8. The L^2 triangle bound is frac12sqrt6+frac52sqrt2, and need not be sharp. A background estimate adds another term s-widehat s …
Exact mathematical passage
|3-\tfrac52|\,|\Omega|+\tfrac52|\Omega\triangle\widehat\Omega| =\tfrac12\cdot6+\tfrac52\cdot2=8.
\timesUse in source · 60 passages
Search aliases: × times
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 371
n division gives a unique pair (r,q) with 0≤ r<N and n=Nq+r, including negative n. The address maps n↦(nbmod N,nmathbindivN) and (r,q)↦ Nq+r are inverse. Analysis and synthesis are reindexing by this bijection. Reindexing the nonnegative sum ∑_n|x[n]|^2 proves the norm identity and shows that both maps preserve square summability. In Lean the target is ℓ^2({0,…,N-1}× Z), which realizes exactly this Hilbert norm.
Exact mathematical passage
\ell^2(\{0,\ldots,N-1\}\times\bbZ)
\betaUse in source · 1 passage
Search aliases: β beta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1214
For a lossless transmission line, Γ_rm in=Γ_L e^-2 iβℓ. One N-phase step corresponds to a line-length increment λ/(2N), up to the chosen orientation convention. Hence the phase address has a direct electrical interpretation.
Exact mathematical passage
\Gam_{\rm in}=\Gam_L e^{-2\ii\beta\ell}.
\gammaUse in source · 2 passages
Search aliases: γ gamma
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1471
W(γ) · filter or path weight · refinement or implementation cost
Exact mathematical passage
W(\gamma)
\Gamma_yUse in source · 4 passages
Search aliases: Γ_y Gamma
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 941
[label=(roman*)] item A global minimizer Λ_y^*∈*argmin_Λ∈ P( T)J_y(Λ) always exists. item If the minimizer is unique and at least one competing pruning exists, its gap Γ_y = min_Λ≠Λ_y^* (J_y(Λ)-J_y(Λ_y^*)) is attained and strictly positive. If the pruning set is a singleton, its only element minimizes every criterion; no real-valued minimum over an empty competitor set is introduced. item In the case with competitors, if new data y' satisfy sup_Λ∈ P( T) |J_y'(Λ)-J_y(Λ)| <fracΓ_y2, then Λ_y'^*=Λ_…
Exact mathematical passage
\Gamma_y = \min_{\Lambda\ne\Lambda_y^*} \bigl(J_y(\Lambda)-J_y(\Lambda_y^*)\bigr)
\DeltaUse in source · 19 passages
Search aliases: Δ Delta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1472
Δ(F) · coarse/fine signal defect · aliasing, leakage, or estimator inconsistency
Exact mathematical passage
\Delta(F)
\Delta^2Use in source · 2 passages
Search aliases: Δ^2 Delta
Use in source · atlas-pages.tex, line 734
vpage39Higher differences annihilate polynomial interiorsBOUNDARY STENCILS / SIGNAL RESPONSE AND NOISE GAINTraditional monograph: Theorem E10.6 (p. 71), Proposition E10.7 (p. 71), Corollary E10.8 (p. 71). Let x(n)=n+2(n-3)_+ on the integer line. Its slope changes from one to three at the breakpoint 3. The second forward difference is Δ^2x(n)=x(n+2)-2x(n+1)+x(n). vfig[.87]ext_differences headAn exact small stencil calculation The values at n=0,…,6 are (0,1,2,3,6,9,12). The second differences at n…
Exact mathematical passage
\Delta^2x(n)=x(n+2)-2x(n+1)+x(n).
\delta^2Use in source · 3 passages
Search aliases: δ^2 delta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2872
[A perturbed disk] Let E and F be concentric disks of radii r and r+δ in R^2, with δ>0. Then μ(Emathbintriangle F) = π((r+δ)^2-r^2) = 2π rδ+πδ^2. For equal amplitude a, the exact L^2 packet error is |a|sqrt2π rδ+πδ^2. This gives the expected square root dependence on boundary displacement.
Exact mathematical passage
\mu(E\mathbin\triangle F) = \pi\bigl((r+\delta)^2-r^2\bigr) = 2\pi r\delta+\pi\delta^2.
\delta^3Use in source · 1 passage
Search aliases: δ^3 delta
Use in source · atlas-pages.tex, line 649
y on the interval between its two locations: H_τ-H_widehatτ _2=sqrtδ=tfrac14, H_τ-H_widehatτ _∞=1. vfig[.89]ext_location headFor a ramp, the error height shrinks too Write r_τ(t)=(t-τ)_+. The difference grows linearly from zero to δ on [τ,widehatτ] and stays at δ on [widehatτ,1]. Direct integration gives r_τ-r_widehatτ _2^2 =fracδ^33+(1-widehatτ)δ^2 =frac176144<frac1256=δ^2. Thus the L^2 ramp error is at most δ, and its supremum error is exactly δ. Positive-order truncated powers are Lipschitz i…
Exact mathematical passage
\|r_\tau-r_{\widehat\tau}\|_2^2 =\frac{\delta^3}{3}+(1-\widehat\tau)\delta^2 =\frac{17}{6144}<\frac1{256}=\delta^2.
\Delta^{d+1}Use in source · 3 passages
Search aliases: Δ^d+1 Delta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3336
[Polynomial annihilation and boundary support] Let x be a sequence that agrees with a polynomial of degree at most d on each interval of a finite partition of Z. Then Δ^d+1x(n)=0 whenever the complete stencil {n,n+1,…,n+d+1} lies inside one polynomial piece. Hence Δ^d+1x is supported in a band of at most d+1 addresses before each breakpoint.
Exact mathematical passage
\Delta^{d+1}x(n)=0
\Delta^mUse in source · 4 passages
Search aliases: Δ^m Delta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3327
Let (Δ x)(n)=x(n+1)-x(n). Then Δ^m x(n) = ∑_r=0^m (-1)^m-rbinom mr x(n+r).
Exact mathematical passage
\Delta^m x(n) = \sum_{r=0}^{m} (-1)^{m-r}\binom mr x(n+r).
\delta_1Use in source · 2 passages
Search aliases: δ_1 delta
Use in source · atlas-pages.tex, line 95
DIFFERENCE VISIBLETraditional monograph: Theorem 4.1 (p. 7). A one-sample delay swaps the two phases and wraps one of them into the preceding channel index. Its matrix is J(D), not DI_2. vfigsample_delay headTrace one impulse The input δ_0 has a single nonzero value in even channel address zero. Under J(D) it appears in odd channel address zero: this is the original sample δ_1. Under DI_2 it stays even but moves to channel address one: this is δ_2. Thus J(D)^2=DI_2, H(D)=I_2+2J(D)-J(D)^2+J(D)^3 …
Exact mathematical passage
\delta_1
\delta_2Use in source · 2 passages
Search aliases: δ_2 delta
Use in source · atlas-pages.tex, line 95
ay swaps the two phases and wraps one of them into the preceding channel index. Its matrix is J(D), not DI_2. vfigsample_delay headTrace one impulse The input δ_0 has a single nonzero value in even channel address zero. Under J(D) it appears in odd channel address zero: this is the original sample δ_1. Under DI_2 it stays even but moves to channel address one: this is δ_2. Thus J(D)^2=DI_2, H(D)=I_2+2J(D)-J(D)^2+J(D)^3 recovers the previous page's four-tap transfer matrix. The same delay algebra…
Exact mathematical passage
\delta_2
\delta_aUse in source · 1 passage
Search aliases: δ_a delta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1741
The translation orbit is the standard coordinate family (δ_a)_a∈ Z^d. Its finite span is the space of finitely supported sequences. Finitely supported sequences are dense in ℓ^p for finite p and in c_0. Their closure in the supremum norm is c_0, which is a proper closed subspace of ℓ^∞.
Exact mathematical passage
(\delta_a)_{a\in\ZZ^d}
\delta_EUse in source · 2 passages
Search aliases: δ_E delta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2102
[Stability under filter perturbation] Assume RE=z^-dI_N. Let widetilde E=E+Δ E, widetilde R=R+Δ R. Set e_0 · =*ess sup E _op, · r_0 · =*ess sup R _op, δ_E · =*ess sup Δ E _op, · δ_R · =*ess sup Δ R _op. Then the delay compensated reconstruction defect obeys ε_rec ≤ r_0δ_E+e_0δ_R+δ_Rδ_E.
Exact mathematical passage
\begin{aligned} e_0&=\esssup\|E\|_{\mathrm{op}}, & r_0&=\esssup\|R\|_{\mathrm{op}},\\ \delta_E&=\esssup\|\Delta E\|_{\mathrm{op}}, & \delta_R&=\esssup\|\Delta R\|_{\mathrm{op}}. \end{aligned}
\delta_RUse in source · 2 passages
Search aliases: δ_R delta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2102
[Stability under filter perturbation] Assume RE=z^-dI_N. Let widetilde E=E+Δ E, widetilde R=R+Δ R. Set e_0 · =*ess sup E _op, · r_0 · =*ess sup R _op, δ_E · =*ess sup Δ E _op, · δ_R · =*ess sup Δ R _op. Then the delay compensated reconstruction defect obeys ε_rec ≤ r_0δ_E+e_0δ_R+δ_Rδ_E.
Exact mathematical passage
\begin{aligned} e_0&=\esssup\|E\|_{\mathrm{op}}, & r_0&=\esssup\|R\|_{\mathrm{op}},\\ \delta_E&=\esssup\|\Delta E\|_{\mathrm{op}}, & \delta_R&=\esssup\|\Delta R\|_{\mathrm{op}}. \end{aligned}
\delta_{rs}Use in source · 2 passages
Search aliases: δ_rs delta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 327
[Polyphase projector theorem] The maps P_N,r are bounded orthogonal projectors on ℓ^2, with operator norm at most one, and satisfy P_N,rP_N,s=δ_rsP_N,r, ∑_r=0^N-1P_N,r=I. Consequently, x=∑_r=0^N-1P_N,rx with no approximation.
Exact mathematical passage
P_{N,r}P_{N,s}=\delta_{rs}P_{N,r}, \qquad \sum_{r=0}^{N-1}P_{N,r}=I.
\epsilonUse in source · 3 passages
Search aliases: ϵ epsilon
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3159
For each fixed δ>0, b_h,δ→0. Apply [source reference]. For any error tolerance ϵ>0, the high-probability bound is eventually smaller than ϵ, so the limit superior of the probability that the packet error exceeds ϵ is at most δ. Since δ is arbitrary, convergence is in probability.
Exact mathematical passage
\epsilon>0
\varepsilonUse in source · 49 passages
Search aliases: ε varepsilon
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1000
[Local perturbation certificate] Assume |c_v(y')-c_v(y)|≤ε_v. Define E(v) = ε_v, · v terminal,
Exact mathematical passage
|c_v(y')-c_v(y)|\le\varepsilon_v.
\varepsilon_*Use in source · 8 passages
Search aliases: ε_* varepsilon
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2461
[Stability of the optimal cost] Let c and widetilde c be two node cost functions. For an admissible pruning L, set ε( L) = ∑_v∈ L|c(v)-widetilde c(v)| and let ε_* = max_ Lε( L). If F_* and widetilde F_* are the two optimal root costs, then |F_*-widetilde F_*| ≤ ε_*.
Exact mathematical passage
\varepsilon_* = \max_{\Lcal}\varepsilon(\Lcal).
\varepsilon_aUse in source · 2 passages
Search aliases: ε_a varepsilon
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1015
[1mm] displaystyle max{ε_v, ∑_u∈ch(v)E(u)}, · v internal. For the complete prunings P_v below v, the envelope is exactly E(v)=max_Λ∈ P_v∑_a∈Λε_a. The maximum is attained. Every complete-pruning cost changes by at most E(∅), and the two globally optimal values differ by at most this quantity. Consequently, for a unique optimum with competitors, Γ_y>2E(∅) is a deterministic certificate that the globally optimal pruning is unchanged. A singleton pruning set is stable without a gap condition. Assume…
Exact mathematical passage
E(v)=\max_{\Lambda\in\mathfrak P_v}\sum_{a\in\Lambda}\varepsilon_a.
\varepsilon_hUse in source · 1 passage
Search aliases: ε_h varepsilon
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3059
[Deterministic consistency] Assume h→0 along grids satisfying the preceding sampling assumptions, ε_h→0, the number of jumps is fixed, and the nonzero jump amplitudes and positive minimum separation remain fixed. Then the detected packet converges to the true packet in every finite L^p norm. If Q_m has a vanishing smooth approximation error and uniformly bounded norm, the complete reconstruction also converges in L^p.
Exact mathematical passage
\varepsilon_h\to0
\varepsilon_{\mathrm{rec}}Use in source · 3 passages
Search aliases: ε_rec varepsilon mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2058
[Quantitative imperfect reconstruction] Define the delay compensated defect D( e^ iω) = e^ i dωR( e^ iω)E( e^ iω)-I_N. If ε_rec = *ess sup_ω D( e^ iω) _op<∞, then the delay compensated reconstruction widetilde x satisfies widetilde x-x _2 ≤ ε_rec x _2. The constant is sharp as an operator norm.
Exact mathematical passage
\varepsilon_{\mathrm{rec}} = \esssup_{\omega}\|D(\e^{\ii\omega})\|_{\mathrm{op}}<\infty,
\varepsilon_vUse in source · 5 passages
Search aliases: ε_v varepsilon
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1000
[Local perturbation certificate] Assume |c_v(y')-c_v(y)|≤ε_v. Define E(v) = ε_v, · v terminal,
Exact mathematical passage
|c_v(y')-c_v(y)|\le\varepsilon_v.
\eta_kUse in source · 8 passages
Search aliases: η_k eta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2916
t_k=kh, k=0,…,n, h=frac1n. Assume f(t) = s(t)+∑_q=1^Qa_qH_τ_q(t), H_τ(t)= 1_[τ,1](t), where s is Lipschitz with constant M, and Q≥1 is finite with nonzero jump amplitudes. The zero-jump case requires no minimum-amplitude condition. Suppose no jump lies at a sample point and no grid cell contains more than one jump. Observe y_k=f(t_k)+η_k. Define the adjacent contrast D_k=y_k+1-y_k.
Exact mathematical passage
y_k=f(t_k)+\eta_k.
\eta_{k+1}Use in source · 4 passages
Search aliases: η_k+1 eta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2945
In a cell without a jump, D_k=s(t_k+1)-s(t_k)+η_k+1-η_k. The Lipschitz and noise bounds give |D_k|≤ Mh+2ε. In a cell with one jump, the same formula has the additional term a_q, which proves the second estimate.
Exact mathematical passage
D_k=s(t_{k+1})-s(t_k)+\eta_{k+1}-\eta_k.
\thetaUse in source · 23 passages
Search aliases: θ theta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1189
For 0≤ρ<1 and angle θ, Γ=ρ e^ iθ gives z(ρ,θ) =frac1+ρ e^ iθ1-ρ e^ iθ, with r(ρ,θ)=frac1-ρ^21-2ρcosθ+ρ^2, x(ρ,θ)=frac2ρsinθ1-2ρcosθ+ρ^2. A fixed ρ is a constant-mismatch circle. Refining N increases angular resolution without changing ρ.
Exact mathematical passage
\theta
\kappaUse in source · 2 passages
Search aliases: κ kappa
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3535
[Shifted pointwise monomial law] Let y(n)=∏_j=1^mx(n-a_j) and write a_j=Nc_j+b_j, 0≤ b_j<N. For output residue r, define r_j≡ r-b_jpmod N, 0≤ r_j<N, and let κ_r,b_j = 1, · r<b_j, 0, · r≥ b_j. Then the output channel is y_r(q) = ∏_j=1^m x_r_j (q-c_j-κ_r,b_j).
Exact mathematical passage
\kappa_{r,b_j} = \begin{cases} 1,&r<b_j,\\ 0,&r\geq b_j. \end{cases}
\kappa_{r,b_j}Use in source · 2 passages
Search aliases: κ_r,b_j kappa
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3535
[Shifted pointwise monomial law] Let y(n)=∏_j=1^mx(n-a_j) and write a_j=Nc_j+b_j, 0≤ b_j<N. For output residue r, define r_j≡ r-b_jpmod N, 0≤ r_j<N, and let κ_r,b_j = 1, · r<b_j, 0, · r≥ b_j. Then the output channel is y_r(q) = ∏_j=1^m x_r_j (q-c_j-κ_r,b_j).
Exact mathematical passage
\kappa_{r,b_j} = \begin{cases} 1,&r<b_j,\\ 0,&r\geq b_j. \end{cases}
\lambdaUse in source · 6 passages
Search aliases: λ lambda
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1217
For a lossless transmission line, Γ_rm in=Γ_L e^-2 iβℓ. One N-phase step corresponds to a line-length increment λ/(2N), up to the chosen orientation convention. Hence the phase address has a direct electrical interpretation.
Exact mathematical passage
\lambda/(2N)
\lambda^2Use in source · 2 passages
Search aliases: λ^2 lambda
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3090
Assume the noise variables are independent, centered, and σ sub Gaussian: E e^λη_k ≤ e^σ^2λ^2/2.
Exact mathematical passage
\mathbb E\e^{\lambda\eta_k} \leq \e^{\sigma^2\lambda^2/2}.
\Lambda_*Use in source · 2 passages
Search aliases: Λ_* Lambda
Use in source · atlas-pages.tex, line 218
)+V(v1)}, V(v)=c_v at a terminal node. headRead the calculation from the bottom upward At 10, the children cost 1+1=2, better than stopping for 7. At 11, stopping costs 4, better than 2+3=5. The right subtree therefore costs 6, better than its stop cost 8. The left subtree costs 1+2=3, better than 5. At the root, 3+6=9 beats 20. The unique optimum is Λ_* = {00,01,100,101,11}, J(Λ_*)=9. All 26 complete prunings are separately enumerated and give the same minimum. At equality, the rule stops, maki…
Exact mathematical passage
\Lambda_* = \{00,01,100,101,11\},\qquad J(\Lambda_*)=9.
\lambda_{a,r}Use in source · 3 passages
Search aliases: λ_a,r lambda
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1830
[Translation covariance] Fix a∈ Z^d. For every r∈ R, write uniquely r-a=r_a+λ_a,r, r_a∈ R, λ_a,r∈ L. Then ( A_LT_ax)_r(ℓ) = x_r_a(ℓ+λ_a,r). Thus a fine translation permutes the lattice packets and applies an internal translation on L.
Exact mathematical passage
r-a=r_a+\lambda_{a,r}, \qquad r_a\in R, \quad \lambda_{a,r}\in L.
\Lambda_{y'}^*Use in source · 1 passage
Search aliases: Λ_y'^* Lambda
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 954
da_y^* (J_y(Λ)-J_y(Λ_y^*)) is attained and strictly positive. If the pruning set is a singleton, its only element minimizes every criterion; no real-valued minimum over an empty competitor set is introduced. item In the case with competitors, if new data y' satisfy sup_Λ∈ P( T) |J_y'(Λ)-J_y(Λ)| <fracΓ_y2, then Λ_y'^*=Λ_y^*. item Every selected minimizer preserves the exact synthesis identity of the preceding theorem.
Exact mathematical passage
\Lambda_{y'}^*=\Lambda_y^*
\Lambda_y^*Use in source · 5 passages
Search aliases: Λ_y^* Lambda
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 936
[label=(roman*)] item A global minimizer Λ_y^*∈*argmin_Λ∈ P( T)J_y(Λ) always exists. item If the minimizer is unique and at least one competing pruning exists, its gap Γ_y = min_Λ≠Λ_y^* (J_y(Λ)-J_y(Λ_y^*)) is attained and strictly positive. If the pruning set is a singleton, its only element minimizes every criterion; no real-valued minimum over an empty competitor set is introduced. item In the case with competitors, if new data y' satisfy sup_Λ∈ P( T) |J_y'(Λ)-J_y(Λ)| <fracΓ_y2, then Λ_y'^*=Λ_…
Exact mathematical passage
\Lambda_y^*\in\operatorname*{argmin}_{\Lambda\in\mathfrak P(\mathcal T)}J_y(\Lambda)
\nuUse in source · 4 passages
Search aliases: ν nu
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2160
[Coefficient noise under canonical synthesis] Under the hypotheses of [source reference], suppose the observed coefficient vector is c= A_Ex+ν. Canonical synthesis gives widehat x=R_canc and widehat x-x _2 ≤ frac1sqrt A ν _2.
Exact mathematical passage
c=\Acal_Ex+\nu.
\XiUse in source · 1 passage
Search aliases: Ξ Xi
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 961
set, so the feasible set is nonempty and finite. Thus the minimum is attained. If competitors exist, their finitely many differences from a unique optimum are strictly positive, so their minimum is positive and is achieved by a competitor. In the singleton case there is no alternative selector under any objective change. In the remaining case, for any competing pruning Λ≠Λ_y^*, J_y'(Λ)-J_y'(Λ_y^*) · ≥ J_y(Λ)-J_y(Λ_y^*) -2sup_Ξ|J_y'(Ξ)-J_y(Ξ)| · >Γ_y-Γ_y=0. Hence the same pruning remains uniquely…
Exact mathematical passage
\begin{aligned} J_{y'}(\Lambda)-J_{y'}(\Lambda_y^*) &\ge J_y(\Lambda)-J_y(\Lambda_y^*) -2\sup_{\Xi}|J_{y'}(\Xi)-J_y(\Xi)|\\ &>\Gamma_y-\Gamma_y=0. \end{aligned}
\piUse in source · 20 passages
Search aliases: π pi
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1133
Fix ω_N=e^2π i/N and rotate the reflection coefficient by R_N,k(Γ)=ω_N^kΓ.
Exact mathematical passage
\omega_N=e^{2\pi\ii/N}
\rhoUse in source · 21 passages
Search aliases: ρ rho
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1189
For 0≤ρ<1 and angle θ, Γ=ρ e^ iθ gives z(ρ,θ) =frac1+ρ e^ iθ1-ρ e^ iθ, with r(ρ,θ)=frac1-ρ^21-2ρcosθ+ρ^2, x(ρ,θ)=frac2ρsinθ1-2ρcosθ+ρ^2. A fixed ρ is a constant-mismatch circle. Refining N increases angular resolution without changing ρ.
Exact mathematical passage
0\le\rho<1
\rho^2Use in source · 1 passage
Search aliases: ρ^2 rho
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1199
For 0≤ρ<1 and angle θ, Γ=ρ e^ iθ gives z(ρ,θ) =frac1+ρ e^ iθ1-ρ e^ iθ, with r(ρ,θ)=frac1-ρ^21-2ρcosθ+ρ^2, x(ρ,θ)=frac2ρsinθ1-2ρcosθ+ρ^2. A fixed ρ is a constant-mismatch circle. Refining N increases angular resolution without changing ρ.
Exact mathematical passage
r(\rho,\theta)=\frac{1-\rho^2}{1-2\rho\cos\theta+\rho^2}, \qquad x(\rho,\theta)=\frac{2\rho\sin\theta}{1-2\rho\cos\theta+\rho^2}.
\SigmaUse in source · 1 passage
Search aliases: Σ Sigma
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 701
Let (X,Σ,μ) be a measure space, with measurable masks Ω,widehatΩ⊆ X and scalar amplitudes a,widehat a∈ C (real amplitudes are a special case). Suppose a scalar jump packet is j=a 1_Ω, and an estimated packet is widehat j=widehat a 1_widehatΩ.
Exact mathematical passage
(X,\Sigma,\mu)
\sigmaUse in source · 23 passages
Search aliases: σ sigma
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1695
on be the largest contrast that can be produced by a Lipschitz background and bounded noise in one grid cell. If a_min>2b_h, then every threshold between b_h and a_min-b_h detects exactly the cells containing jumps. The jump locations are recovered within h/2, and each amplitude error is at most b_h. The probability theorem replaces the deterministic noise term by 2σsqrtlog(2n/δ). Two half shifted Haar grids provide a second recovery mechanism. Every point is at least one quarter of a cell away …
Exact mathematical passage
2\sigma\sqrt{\log(2n/\delta)}.
\sigma^2Use in source · 4 passages
Search aliases: σ^2 sigma
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3090
Assume the noise variables are independent, centered, and σ sub Gaussian: E e^λη_k ≤ e^σ^2λ^2/2.
Exact mathematical passage
\mathbb E\e^{\lambda\eta_k} \leq \e^{\sigma^2\lambda^2/2}.
\sigma_nUse in source · 2 passages
Search aliases: σ_n sigma
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3155
[Probability consistency] Let σ_nsqrtlog n→0 and h=1/n. For every fixed 0<δ<1, along grids satisfying the preceding sampling assumptions, all jump cells are recovered with probability at least 1-δ for sufficiently large n. For the fixed finite jump model above, along grids satisfying the preceding sampling assumptions, the packet error converges to zero in probability in every finite L^p norm.
Exact mathematical passage
\sigma_n\sqrt{\log n}\to0
\tau_0Use in source · 1 passage
Search aliases: τ_0 tau
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2533
[Higher order singular packet decomposition] Let 0=τ_0<τ_1<⋯<τ_Q<τ_Q+1=1. Suppose the restriction of f to every interval (τ_q,τ_q+1) belongs to W^r,p, and choose the one dimensional Sobolev representatives whose derivatives through order r-1 have one sided traces at the endpoints. Define a_q,m = jumpf^(m)_τ_q = f^(m)(τ_q+)-f^(m)(τ_q-) and j_f(t) = ∑_q=1^Q∑_m=0^r-1a_q,mϕ_m,τ_q(t). Then s_f=f-j_f has matching one sided derivatives through order r-1 at every breakpoint and belongs to W^r,p(0,1). It…
Exact mathematical passage
0=\tau_0<\tau_1<\cdots<\tau_Q<\tau_{Q+1}=1.
\tau_{Q+1}Use in source · 1 passage
Search aliases: τ_Q+1 tau
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2533
[Higher order singular packet decomposition] Let 0=τ_0<τ_1<⋯<τ_Q<τ_Q+1=1. Suppose the restriction of f to every interval (τ_q,τ_q+1) belongs to W^r,p, and choose the one dimensional Sobolev representatives whose derivatives through order r-1 have one sided traces at the endpoints. Define a_q,m = jumpf^(m)_τ_q = f^(m)(τ_q+)-f^(m)(τ_q-) and j_f(t) = ∑_q=1^Q∑_m=0^r-1a_q,mϕ_m,τ_q(t). Then s_f=f-j_f has matching one sided derivatives through order r-1 at every breakpoint and belongs to W^r,p(0,1). It…
Exact mathematical passage
0=\tau_0<\tau_1<\cdots<\tau_Q<\tau_{Q+1}=1.
\tau_{q+1}Use in source · 1 passage
Search aliases: τ_q+1 tau
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2536
[Higher order singular packet decomposition] Let 0=τ_0<τ_1<⋯<τ_Q<τ_Q+1=1. Suppose the restriction of f to every interval (τ_q,τ_q+1) belongs to W^r,p, and choose the one dimensional Sobolev representatives whose derivatives through order r-1 have one sided traces at the endpoints. Define a_q,m = jumpf^(m)_τ_q = f^(m)(τ_q+)-f^(m)(τ_q-) and j_f(t) = ∑_q=1^Q∑_m=0^r-1a_q,mϕ_m,τ_q(t). Then s_f=f-j_f has matching one sided derivatives through order r-1 at every breakpoint and belongs to W^r,p(0,1). It…
Exact mathematical passage
(\tau_q,\tau_{q+1})
\PhiUse in source · 6 passages
Search aliases: Φ Phi
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3668
[Cyclic Mobius action] The pullback Φ_θ =C^-1R_θ C is the fractional linear map Φ_θ(z) = frac(1+ e^ iθ)z+(1- e^ iθ) (1- e^ iθ)z+(1+ e^ iθ). It preserves H and satisfies Φ_θΦ_φ = Φ_θ+φ. For an integer N≥1 and θ=2π/N, the map has order N.
Exact mathematical passage
\Phi_\theta =C^{-1}R_\theta C
\varphiUse in source · 3 passages
Search aliases: φ varphi
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3682
[Cyclic Mobius action] The pullback Φ_θ =C^-1R_θ C is the fractional linear map Φ_θ(z) = frac(1+ e^ iθ)z+(1- e^ iθ) (1- e^ iθ)z+(1+ e^ iθ). It preserves H and satisfies Φ_θΦ_φ = Φ_θ+φ. For an integer N≥1 and θ=2π/N, the map has order N.
Exact mathematical passage
\Phi_\theta\Phi_\varphi = \Phi_{\theta+\varphi}.
\phi_{0,1/3}Use in source · 1 passage
Search aliases: ϕ_0,1/3 phi
Use in source · atlas-pages.tex, line 583
vpage31Derivative jumps have their own packetsTRACE DATA / A STEP AND A RAMPTraditional monograph: Theorem E7.2 (p. 56), Corollary E7.3 (p. 57). On [0,1], let H_τ=1_[τ,1] and take f(t)=t^2+2H_1/3(t)-3(t-2/3)_+. The function has a value jump 2 at 1/3 and a first-derivative jump -3 at 2/3. Its packet and remainder are j_f(t)=2ϕ_0,1/3(t)-3ϕ_1,2/3(t), s_f(t)=t^2. vfig[.92]ext_trace platecaptionThe displayed curves are exact polynomial pieces. The solid jump is not connected across its discontinuity.…
Exact mathematical passage
j_f(t)=2\phi_{0,1/3}(t)-3\phi_{1,2/3}(t),\qquad s_f(t)=t^2.
\phi_{0,\tau}Use in source · 1 passage
Search aliases: ϕ_0,τ phi tau
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2517
For m≥0 and 0<τ<1, define ϕ_m,τ(t) = frac(t-τ)_+^mm!. Thus ϕ_0,τ= 1_[τ,1]. On each side of τ, the packet is a polynomial. Its derivative of order m has a unit jump at τ.
Exact mathematical passage
\phi_{0,\tau}=\one_{[\tau,1]}
\phi_{1,2/3}Use in source · 1 passage
Search aliases: ϕ_1,2/3 phi
Use in source · atlas-pages.tex, line 583
vpage31Derivative jumps have their own packetsTRACE DATA / A STEP AND A RAMPTraditional monograph: Theorem E7.2 (p. 56), Corollary E7.3 (p. 57). On [0,1], let H_τ=1_[τ,1] and take f(t)=t^2+2H_1/3(t)-3(t-2/3)_+. The function has a value jump 2 at 1/3 and a first-derivative jump -3 at 2/3. Its packet and remainder are j_f(t)=2ϕ_0,1/3(t)-3ϕ_1,2/3(t), s_f(t)=t^2. vfig[.92]ext_trace platecaptionThe displayed curves are exact polynomial pieces. The solid jump is not connected across its discontinuity.…
Exact mathematical passage
j_f(t)=2\phi_{0,1/3}(t)-3\phi_{1,2/3}(t),\qquad s_f(t)=t^2.
\phi_{m,u}Use in source · 1 passage
Search aliases: ϕ_m,u phi
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2755
Assume τ<widehatτ. Differentiation with respect to the location parameter gives frac∂∂ uϕ_m,u(t) =-ϕ_m-1,u(t) for almost every t. Hence ϕ_m,τ-ϕ_m,widehatτ = ∫_τ^widehatτϕ_m-1,u d u. Minkowski's inequality gives ϕ_m,τ-ϕ_m,widehatτ _p ≤ |τ-widehatτ| sup_u ϕ_m-1,u _p. The supremum is bounded by the norm of t^m-1/(m-1)! on [0,1], which is the stated constant. The supremum norm argument is identical.
Exact mathematical passage
\frac{\partial}{\partial u}\phi_{m,u}(t) =-\phi_{m-1,u}(t)
\phi_{m,\widehat\tau}Use in source · 4 passages
Search aliases: ϕ_m,widehatτ phi widehat tau
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2738
[Location stability for higher order packets] Let m≥1, 1≤ p<∞, and 0≤τ,widehatτ≤1. Then ϕ_m,τ-ϕ_m,widehatτ _L^p(0,1) ≤ frac|τ-widehatτ| (m-1)!((m-1)p+1)^1/p. For p=∞, ϕ_m,τ-ϕ_m,widehatτ _∞ ≤ frac|τ-widehatτ|(m-1)!. Thus location error is Lipschitz for every derivative jump packet of positive order, while a value jump has only the finite L^p modulus in [source reference].
Exact mathematical passage
\|\phi_{m,\tau}-\phi_{m,\widehat\tau}\|_{L^p(0,1)} \leq \frac{|\tau-\widehat\tau|} {(m-1)!((m-1)p+1)^{1/p}}.
\phi_{m,\tau}^{(\ell)}Use in source · 1 passage
Search aliases: ϕ_m,τ^(ℓ) phi tau ell
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2559
For 0≤ ℓ≤ m, the classical derivative of ϕ_m,τ on the two open sides is ϕ_m,τ^(ℓ)(t) = frac(t-τ)_+^m-ℓ(m-ℓ)!. This derivative has zero jump when ℓ<m and a unit jump when ℓ=m. Derivatives of order larger than m vanish on both open sides. It follows that subtracting the term a_q,mϕ_m,τ_q removes exactly the jump in the derivative of order m without changing the other classical trace jumps. Hence s_f has matching traces through order r-1.
Exact mathematical passage
\phi_{m,\tau}^{(\ell)}(t) = \frac{(t-\tau)_+^{m-\ell}}{(m-\ell)!}.
\phi_{m,\tau_q}Use in source · 4 passages
Search aliases: ϕ_m,τ_q phi tau
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2545
_packet_decomposition Let 0=τ_0<τ_1<⋯<τ_Q<τ_Q+1=1. Suppose the restriction of f to every interval (τ_q,τ_q+1) belongs to W^r,p, and choose the one dimensional Sobolev representatives whose derivatives through order r-1 have one sided traces at the endpoints. Define a_q,m = jumpf^(m)_τ_q = f^(m)(τ_q+)-f^(m)(τ_q-) and j_f(t) = ∑_q=1^Q∑_m=0^r-1a_q,mϕ_m,τ_q(t). Then s_f=f-j_f has matching one sided derivatives through order r-1 at every breakpoint and belongs to W^r,p(0,1). Its weak derivative of or…
Exact mathematical passage
j_f(t) = \sum_{q=1}^{Q}\sum_{m=0}^{r-1}a_{q,m}\phi_{m,\tau_q}(t).
\phi_{m-1,u}Use in source · 3 passages
Search aliases: ϕ_m-1,u phi
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2755
Assume τ<widehatτ. Differentiation with respect to the location parameter gives frac∂∂ uϕ_m,u(t) =-ϕ_m-1,u(t) for almost every t. Hence ϕ_m,τ-ϕ_m,widehatτ = ∫_τ^widehatτϕ_m-1,u d u. Minkowski's inequality gives ϕ_m,τ-ϕ_m,widehatτ _p ≤ |τ-widehatτ| sup_u ϕ_m-1,u _p. The supremum is bounded by the norm of t^m-1/(m-1)! on [0,1], which is the stated constant. The supremum norm argument is identical.
Exact mathematical passage
\frac{\partial}{\partial u}\phi_{m,u}(t) =-\phi_{m-1,u}(t)
\Phi_\thetaUse in source · 5 passages
Search aliases: Φ_θ Phi theta
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3668
[Cyclic Mobius action] The pullback Φ_θ =C^-1R_θ C is the fractional linear map Φ_θ(z) = frac(1+ e^ iθ)z+(1- e^ iθ) (1- e^ iθ)z+(1+ e^ iθ). It preserves H and satisfies Φ_θΦ_φ = Φ_θ+φ. For an integer N≥1 and θ=2π/N, the map has order N.
Exact mathematical passage
\Phi_\theta =C^{-1}R_\theta C
\Phi_{\theta+\varphi}Use in source · 1 passage
Search aliases: Φ_θ+φ Phi theta varphi
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3682
[Cyclic Mobius action] The pullback Φ_θ =C^-1R_θ C is the fractional linear map Φ_θ(z) = frac(1+ e^ iθ)z+(1- e^ iθ) (1- e^ iθ)z+(1+ e^ iθ). It preserves H and satisfies Φ_θΦ_φ = Φ_θ+φ. For an integer N≥1 and θ=2π/N, the map has order N.
Exact mathematical passage
\Phi_\theta\Phi_\varphi = \Phi_{\theta+\varphi}.
\Phi_\piUse in source · 1 passage
Search aliases: Φ_π Phi pi
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3696
[Impedance and admittance] For N=2, θ=π and Φ_π(z)=frac1z. Thus the two phase action is normalized impedance and admittance inversion.
Exact mathematical passage
\Phi_\pi(z)=\frac1z.
\Phi_\varphiUse in source · 1 passage
Search aliases: Φ_φ Phi varphi
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3682
[Cyclic Mobius action] The pullback Φ_θ =C^-1R_θ C is the fractional linear map Φ_θ(z) = frac(1+ e^ iθ)z+(1- e^ iθ) (1- e^ iθ)z+(1+ e^ iθ). It preserves H and satisfies Φ_θΦ_φ = Φ_θ+φ. For an integer N≥1 and θ=2π/N, the map has order N.
Exact mathematical passage
\Phi_\theta\Phi_\varphi = \Phi_{\theta+\varphi}.
\psiUse in source · 8 passages
Search aliases: ψ psi
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3195
Let I_j,k = [k2^-j,(k+1)2^-j) and let ψ_j,k be the normalized Haar wavelet supported on I_j,k, positive on the left half and negative on the right half.
Exact mathematical passage
\psi_{j,k}
\psi_{j,k}Use in source · 7 passages
Search aliases: ψ_j,k psi
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3195
Let I_j,k = [k2^-j,(k+1)2^-j) and let ψ_j,k be the normalized Haar wavelet supported on I_j,k, positive on the left half and negative on the right half.
Exact mathematical passage
\psi_{j,k}
\OmegaUse in source · 23 passages
Search aliases: Ω Omega
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 701
Let (X,Σ,μ) be a measure space, with measurable masks Ω,widehatΩ⊆ X and scalar amplitudes a,widehat a∈ C (real amplitudes are a special case). Suppose a scalar jump packet is j=a 1_Ω, and an estimated packet is widehat j=widehat a 1_widehatΩ.
Exact mathematical passage
\Omega,\widehat\Omega\subseteq X
\omegaUse in source · 16 passages
Search aliases: ω omega
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1133
Fix ω_N=e^2π i/N and rotate the reflection coefficient by R_N,k(Γ)=ω_N^kΓ.
Exact mathematical passage
\omega_N=e^{2\pi\ii/N}
\omega_2Use in source · 2 passages
Search aliases: ω_2 omega
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1177
[Impedance--admittance is the two-phase split] For N=2 and finite z≠0, T_2,1(z)=z^-1. On the full projective line it exchanges 0 and ∞. Thus a half-turn of the reflection disk exchanges normalized impedance and admittance. ω_2=exp(π i)=-1, so A(ω_2)[u:v]=[v:u]. The finite ratio is 1/z when z≠0; the coordinate swap also gives [0:1]↔[1:0]. No ordinary division by zero is used to represent infinity.
Exact mathematical passage
\omega_2=\exp(\pi\ii)=-1
\omega_NUse in source · 1 passage
Search aliases: ω_N omega
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1133
Fix ω_N=e^2π i/N and rotate the reflection coefficient by R_N,k(Γ)=ω_N^kΓ.
Exact mathematical passage
\omega_N=e^{2\pi\ii/N}
\omega_N^kUse in source · 3 passages
Search aliases: ω_N^k omega
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1137
Fix ω_N=e^2π i/N and rotate the reflection coefficient by R_N,k(Γ)=ω_N^kΓ.
Exact mathematical passage
R_{N,k}(\Gam)=\omega_N^k\Gam.
\omega_N^NUse in source · 1 passage
Search aliases: ω_N^N omega
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1166
1-w · 1+w. It has determinant w≠0, preserves nonzero pairs, and commutes with rescaling, so it induces a projective map. Direct multiplication gives A(w)A(w')=A(ww') and A(1)=I. Since ω_N^N=1, this proves the cyclic composition law modulo N. The Cayley coordinate of [u:v] is [u-v:u+v], and applying A(w) changes this pair to [w(u-v):u+v]. This proves the conjugacy also at poles. For [z:1], taking the coordinate ratio gives the displayed finite-chart formula.
Exact mathematical passage
\omega_N^N=1
\Omega_qUse in source · 1 passage
Search aliases: Ω_q Omega
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1294
[Edge-packet approximation class] There should exist natural function spaces in which a signal is represented as f=s+∑_q a_q1_Ω_q, with separate norms for smooth approximation, boundary geometry, packet amplitude, and packet identification. A projector-resolved approximation theory on such spaces should yield rates that do not mix smooth truncation error with discontinuity-location error.
Exact mathematical passage
f=s+\sum_q a_q\mathbf1_{\Omega_q},
\ldotsUse in source · 13 passages
Search aliases: … ldots
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 350
Define the analysis and synthesis maps A_Nx=(x_0,…,x_N-1), S_N(x_0,…,x_N-1)[Nq+r]=x_r[q].
Exact mathematical passage
\cA_Nx=(x_0,\ldots,x_{N-1}), \qquad \cS_N(x_0,\ldots,x_{N-1})[Nq+r]=x_r[q].
\ell'Use in source · 3 passages
Search aliases: ℓ' ell
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1942
Evaluate the output at r+ℓ. Write every input address as s+ℓ'. The convolution offset is (r+ℓ)-(s+ℓ') =(r-s)+(ℓ-ℓ'). Setting m=ℓ-ℓ' gives the formula.
Exact mathematical passage
s+\ell'
\ell^1Use in source · 5 passages
Search aliases: ℓ^1 ell
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 305
[Several completions] The same algebraic seed description applies to finite sequences, ℓ^1, ℓ^2, periodic sequences, tempered distributions, and multidimensional lattices. The topology decides which infinite reconstructions are legitimate. Finite-dimensional experiments in this paper use ordinary Euclidean norms; theorem statements specify the ambient norm when it matters.
Exact mathematical passage
\ell^1
\ell^\inftyUse in source · 3 passages
Search aliases: ℓ^∞ ell infty
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1737
[Impulse generation] For 1≤ p<∞, overlinespan{T_aδ_0:a∈ Z^d}^ℓ^p = ℓ^p( Z^d). The same statement holds in c_0( Z^d) with the supremum norm. In ℓ^∞( Z^d), the closed span of the translation orbit is exactly c_0( Z^d).
Exact mathematical passage
\ell^\infty(\ZZ^d)
\toUse in source · 13 passages
Search aliases: → to
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 293
The shift orbit is the standard orthonormal basis {δ_n}_n∈ Z. For x∈ℓ^2, the finite truncations x_F=∑_n∈ Fx[n]δ_n, F⊂ Z finite, satisfy x-x_F _2^2=∑_n∉ F|x[n]|^2→0 as the finite sets exhaust Z. Thus the unordered Hilbert-space sum ∑_nx[n]δ_n converges to x, and the finite-support statement is exact. The shift is a surjective isometry because n↦ n-1 is a bijection of Z, with inverse shift n↦ n+1. This proves the closed-span assertion.
Exact mathematical passage
\|x-x_F\|_2^2=\sum_{n\notin F}|x[n]|^2\to0
\leftrightarrowUse in source · 2 passages
Search aliases: ↔ leftrightarrow
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1177
o-phase split] For N=2 and finite z≠0, T_2,1(z)=z^-1. On the full projective line it exchanges 0 and ∞. Thus a half-turn of the reflection disk exchanges normalized impedance and admittance. ω_2=exp(π i)=-1, so A(ω_2)[u:v]=[v:u]. The finite ratio is 1/z when z≠0; the coordinate swap also gives [0:1]↔[1:0]. No ordinary division by zero is used to represent infinity.
Exact mathematical passage
[0:1]\leftrightarrow[1:0]
\mapstoUse in source · 10 passages
Search aliases: ↦ mapsto
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 293
mal basis {δ_n}_n∈ Z. For x∈ℓ^2, the finite truncations x_F=∑_n∈ Fx[n]δ_n, F⊂ Z finite, satisfy x-x_F _2^2=∑_n∉ F|x[n]|^2→0 as the finite sets exhaust Z. Thus the unordered Hilbert-space sum ∑_nx[n]δ_n converges to x, and the finite-support statement is exact. The shift is a surjective isometry because n↦ n-1 is a bijection of Z, with inverse shift n↦ n+1. This proves the closed-span assertion.
Exact mathematical passage
n\mapsto n-1
\partialUse in source · 2 passages
Search aliases: ∂ partial
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2755
Assume τ<widehatτ. Differentiation with respect to the location parameter gives frac∂∂ uϕ_m,u(t) =-ϕ_m-1,u(t) for almost every t. Hence ϕ_m,τ-ϕ_m,widehatτ = ∫_τ^widehatτϕ_m-1,u d u. Minkowski's inequality gives ϕ_m,τ-ϕ_m,widehatτ _p ≤ |τ-widehatτ| sup_u ϕ_m-1,u _p. The supremum is bounded by the norm of t^m-1/(m-1)! on [0,1], which is the stated constant. The supremum norm argument is identical.
Exact mathematical passage
\frac{\partial}{\partial u}\phi_{m,u}(t) =-\phi_{m-1,u}(t)
\notinUse in source · 1 passage
Search aliases: ∉ notin
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 293
The shift orbit is the standard orthonormal basis {δ_n}_n∈ Z. For x∈ℓ^2, the finite truncations x_F=∑_n∈ Fx[n]δ_n, F⊂ Z finite, satisfy x-x_F _2^2=∑_n∉ F|x[n]|^2→0 as the finite sets exhaust Z. Thus the unordered Hilbert-space sum ∑_nx[n]δ_n converges to x, and the finite-support statement is exact. The shift is a surjective isometry because n↦ n-1 is a bijection of Z, with inverse shift n↦ n+1. This proves the closed-span assertion.
Exact mathematical passage
\|x-x_F\|_2^2=\sum_{n\notin F}|x[n]|^2\to0
\prodUse in source · 2 passages
Search aliases: ∏ prod
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3519
[Shifted pointwise monomial law] Let y(n)=∏_j=1^mx(n-a_j) and write a_j=Nc_j+b_j, 0≤ b_j<N. For output residue r, define r_j≡ r-b_jpmod N, 0≤ r_j<N, and let κ_r,b_j = 1, · r<b_j,
Exact mathematical passage
y(n)=\prod_{j=1}^{m}x(n-a_j)
\prod_{j=1}^{m}Use in source · 2 passages
Search aliases: ∏_j=1^m prod
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3519
[Shifted pointwise monomial law] Let y(n)=∏_j=1^mx(n-a_j) and write a_j=Nc_j+b_j, 0≤ b_j<N. For output residue r, define r_j≡ r-b_jpmod N, 0≤ r_j<N, and let κ_r,b_j = 1, · r<b_j,
Exact mathematical passage
y(n)=\prod_{j=1}^{m}x(n-a_j)
\sum_{a\in\Lambda}Use in source · 1 passage
Search aliases: ∑_a∈Λ sum in Lambda
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1015
[1mm] displaystyle max{ε_v, ∑_u∈ch(v)E(u)}, · v internal. For the complete prunings P_v below v, the envelope is exactly E(v)=max_Λ∈ P_v∑_a∈Λε_a. The maximum is attained. Every complete-pruning cost changes by at most E(∅), and the two globally optimal values differ by at most this quantity. Consequently, for a unique optimum with competitors, Γ_y>2E(∅) is a deterministic certificate that the globally optimal pruning is unchanged. A singleton pruning set is stable without a gap condition. Assume…
Exact mathematical passage
E(v)=\max_{\Lambda\in\mathfrak P_v}\sum_{a\in\Lambda}\varepsilon_a.
\sum_{i+j\equiv k\, (\mathrm{mod}\,N)}Use in source · 1 passage
Search aliases: ∑_i+j≡ k (mod N) sum equiv mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1067
[Cyclic product law] P_N,k(AB) = ∑_i+j≡ k (mod N)A_iB_j. For q∈ N, the qth-power identity is P_N,k(A^q) = ∑_i_1+⋯+i_q≡ k A_i_1⋯ A_i_q.
Exact mathematical passage
\boxed{ P_{N,k}(AB) = \sum_{i+j\equiv k\, (\mathrm{mod}\,N)}A_iB_j. }
\sum_{i+j\equiv k\ (N)}Use in source · 1 passage
Search aliases: ∑_i+j≡ k (N) sum equiv
Use in source · atlas-pages.tex, line 365
eserves the same address For time-residue masks, (P_N,ix)[n](P_N,jy)[n]=0 when i≠ j. Therefore P_N,k(x· y)=(P_N,kx)·(P_N,ky). Here the dot means multiplication at the same sample, not convolution. headCauchy multiplication adds degrees For formal series A(t),B(t), the coefficient of t^n is a sum over a+b=n. Consequently the output residue is i+jpmod N: P_N,k(AB)=∑_i+j≡ k (N)(P_N,iA)(P_N,jB). For N=3, take A=1+2t+3t^2 and B=2+t^2. Direct multiplication gives AB=2+4t+7t^2+2t^3+3t^4, P_3,0(AB)=2+2t…
Exact mathematical passage
P_{N,k}(AB)=\sum_{i+j\equiv k\ (N)}(P_{N,i}A)(P_{N,j}B).
\sum_{i,j}Use in source · 1 passage
Search aliases: ∑_i,j sum
Use in source · atlas-pages.tex, line 634
and 5, so j-widehat j _1=2+3+5=10, j-widehat j _2^2=4+9+25=38. The counting-measure supremum error is 5. The full-domain mean squared error is 38/12=19/6. A mismatch fraction of 3/12 reports a different statistic and omits all three amplitude weights. headThe formula requires partitions For measurable partitions (E_i) and (F_j), with M_ij=μ(E_i∩ F_j), j-widehat j _p^p=∑_i,j|a_i-b_j|^pM_ij, 1≤ p<∞. Each location belongs to exactly one intersection. Overlapping indicator packets that are summed at…
Exact mathematical passage
\|j-\widehat j\|_p^p=\sum_{i,j}|a_i-b_j|^pM_{ij},\qquad 1\le p<\infty.
\sum_{i,k}Use in source · 1 passage
Search aliases: ∑_i,k sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2803
[Background and packet error] Suppose the true and estimated packets use the same amplitude list (a_i)_i=0^I, but possibly different regions. Then j-widehat j _p^p = ∑_i,k|a_i-a_k|^pμ(E_i∩ F_k). Only off diagonal overlaps contribute.
Exact mathematical passage
\|j-\widehat j\|_p^p = \sum_{i,k}|a_i-a_k|^p\mu(E_i\cap F_k).
\sum_{i=0}^{I}Use in source · 1 passage
Search aliases: ∑_i=0^I sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2785
[Overlap matrix formula] Define M_ij=μ(E_i∩ F_j). For 1≤ p<∞, ∑_i=0^Ia_i 1_E_i - ∑_j=0^Jb_j 1_F_j _p^p = ∑_i=0^I∑_j=0^J|a_i-b_j|^pM_ij. The row sums of M are μ(E_i), and the column sums are μ(F_j).
Exact mathematical passage
\left\| \sum_{i=0}^{I}a_i\one_{E_i} - \sum_{j=0}^{J}b_j\one_{F_j} \right\|_p^p = \sum_{i=0}^{I}\sum_{j=0}^{J}|a_i-b_j|^pM_{ij}.
\sum_{i_1+\cdots+i_q\equiv k}Use in source · 1 passage
Search aliases: ∑_i_1+⋯+i_q≡ k sum cdots equiv
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1075
[Cyclic product law] P_N,k(AB) = ∑_i+j≡ k (mod N)A_iB_j. For q∈ N, the qth-power identity is P_N,k(A^q) = ∑_i_1+⋯+i_q≡ k A_i_1⋯ A_i_q.
Exact mathematical passage
P_{N,k}(A^q) = \sum_{i_1+\cdots+i_q\equiv k} A_{i_1}\cdots A_{i_q}.
\sum_jUse in source · 1 passage
Search aliases: ∑_j sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 458
The last factor is x_s[n-m], giving the time-domain formula. Under the additional transform hypothesis, each delayed channel has transform ∑_nx_s[n-m]z^-n =z^-m∑_jx_s[j]z^-j. Absolute summability permits this bijective reindexing, and only finitely many delayed channels are combined. Taking their finite linear combination gives the matrix formula. The time-domain theorem therefore remains valid even when no numerical bilateral transform exists.
Exact mathematical passage
\sum_nx_s[n-m]z^{-n} =z^{-m}\sum_jx_s[j]z^{-j}.
\sum_{j=0}^{J}Use in source · 1 passage
Search aliases: ∑_j=0^J sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2785
[Overlap matrix formula] Define M_ij=μ(E_i∩ F_j). For 1≤ p<∞, ∑_i=0^Ia_i 1_E_i - ∑_j=0^Jb_j 1_F_j _p^p = ∑_i=0^I∑_j=0^J|a_i-b_j|^pM_ij. The row sums of M are μ(E_i), and the column sums are μ(F_j).
Exact mathematical passage
\left\| \sum_{i=0}^{I}a_i\one_{E_i} - \sum_{j=0}^{J}b_j\one_{F_j} \right\|_p^p = \sum_{i=0}^{I}\sum_{j=0}^{J}|a_i-b_j|^pM_{ij}.
\sum_{k=0}^{n-1}Use in source · 2 passages
Search aliases: ∑_k=0^n-1 sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3382
[Classical square-wave Gibbs behavior] Let q(x)=tfrac12sign(sin x), whose jump at zero has size one. For n≥1, its partial sum through the first n positive odd frequencies is Q_n(x)=frac2π∑_k=0^n-1fracsin((2k+1)x)2k+1. At x_n=π/(2n), x_n⟶0, Q_n(x_n)⟶fracSi(π)π =frac12+G, G≈0.089489872236, where Si(z)=∫_0^zsin(t)/t dt and G=Si(π)/π-1/2>0. Thus a nonzero overshoot persists along points approaching the jump. Scaling by a real amplitude a gives an overshoot or undershoot of magnitude |a|G.
Exact mathematical passage
Q_n(x)=\frac{2}{\pi}\sum_{k=0}^{n-1}\frac{\sin((2k+1)x)}{2k+1}.
\sum_mUse in source · 3 passages
Search aliases: ∑_m sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 438
[Exact FIR polyphase transfer law] The output channels satisfy y_k[n] = ∑_s=0^N-1∑_m∈ Z h_ks[m]x_s[n-m]. For the numerical channel transform, take z∈ C∖{0} and assume ∑_n∈ Z|x_s[n]z^-n|<∞ (0≤ s<N). Define X_s(z)=∑_nx_s[n]z^-n. The output series then also converge absolutely, and Y(z)= H^(N)(z) X(z), H^(N)_ks(z)=∑_m h[Nm+k-s]z^-m.
Exact mathematical passage
\bm Y(z)=\bm H^{(N)}(z)\bm X(z), \qquad H^{(N)}_{ks}(z)=\sum_m h[Nm+k-s]z^{-m}.
\sum_{m=0}^{r-1}Use in source · 2 passages
Search aliases: ∑_m=0^r-1 sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2545
_packet_decomposition Let 0=τ_0<τ_1<⋯<τ_Q<τ_Q+1=1. Suppose the restriction of f to every interval (τ_q,τ_q+1) belongs to W^r,p, and choose the one dimensional Sobolev representatives whose derivatives through order r-1 have one sided traces at the endpoints. Define a_q,m = jumpf^(m)_τ_q = f^(m)(τ_q+)-f^(m)(τ_q-) and j_f(t) = ∑_q=1^Q∑_m=0^r-1a_q,mϕ_m,τ_q(t). Then s_f=f-j_f has matching one sided derivatives through order r-1 at every breakpoint and belongs to W^r,p(0,1). Its weak derivative of or…
Exact mathematical passage
j_f(t) = \sum_{q=1}^{Q}\sum_{m=0}^{r-1}a_{q,m}\phi_{m,\tau_q}(t).
\sum_{m\in L}Use in source · 1 passage
Search aliases: ∑_m∈ L sum in
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1927
[Exact convolution transfer law] Let y=h*x. Then y_r(ℓ) = ∑_s∈ R∑_m∈ Lh_rs(m)x_s(ℓ-m). Equivalently, A_LH A_L^-1 = (H_rs)_r,s∈ R, where H_rs is convolution on L by h_rs.
Exact mathematical passage
y_r(\ell) = \sum_{s\in R}\sum_{m\in L}h_{rs}(m)x_s(\ell-m).
\sum_{m\in\bbZ}Use in source · 1 passage
Search aliases: ∑_m∈bbZ ∑_m∈ Z sum in bbZ sum in mathbb
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 425
[Exact FIR polyphase transfer law] The output channels satisfy y_k[n] = ∑_s=0^N-1∑_m∈ Z h_ks[m]x_s[n-m]. For the numerical channel transform, take z∈ C∖{0} and assume ∑_n∈ Z|x_s[n]z^-n|<∞ (0≤ s<N). Define X_s(z)=∑_nx_s[n]z^-n. The output series then also converge absolutely, and Y(z)= H^(N)(z) X(z), H^(N)_ks(z)=∑_m h[Nm+k-s]z^-m.
Exact mathematical passage
\boxed{ y_k[n] = \sum_{s=0}^{N-1}\sum_{m\in\bbZ} h_{ks}[m]x_s[n-m]. }
\sum_{m\in\ZZ}Use in source · 1 passage
Search aliases: ∑_m∈ZZ ∑_m∈ Z sum in ZZ sum in mathbb
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1987
For L=N Z with N≥1, first take |z|=1. The assumption h∈ℓ^1( Z) makes the following symbol series absolutely convergent. Define H_ks(z)=∑_m∈ Zh(Nm+k-s)z^-m. For x∈ℓ^1( Z), the channel transforms converge absolutely on the unit circle and satisfy Y(z)= H_N(z) X(z). For x∈ℓ^2( Z), the same identity holds almost everywhere in the L^2 Fourier sense. For finite impulse-response kernels it also holds at nonzero z wherever the input channel transforms are weighted absolutely summable. This is the classi…
Exact mathematical passage
H_{ks}(z)=\sum_{m\in\ZZ}h(Nm+k-s)z^{-m}.
\sum_nUse in source · 4 passages
Search aliases: ∑_n sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 293
The shift orbit is the standard orthonormal basis {δ_n}_n∈ Z. For x∈ℓ^2, the finite truncations x_F=∑_n∈ Fx[n]δ_n, F⊂ Z finite, satisfy x-x_F _2^2=∑_n∉ F|x[n]|^2→0 as the finite sets exhaust Z. Thus the unordered Hilbert-space sum ∑_nx[n]δ_n converges to x, and the finite-support statement is exact. The shift is a surjective isometry because n↦ n-1 is a bijection of Z, with inverse shift n↦ n+1. This proves the closed-span assertion.
Exact mathematical passage
\sum_nx[n]\delta_n
\sum_{n=1}^{m}Use in source · 1 passage
Search aliases: ∑_n=1^m sum
Use in source · atlas-pages.tex, line 266
vpage16Why a Fourier cutoff cannot reproduce this jump pointwiseA CONTINUOUS PULSE / EXACT ENDPOINT VALUESTraditional monograph: Section 6 (p. 11), Section 8 (p. 13). Use a separate period-one example on [-1/2,1/2): j(t)= 1_{|t|<1/4}, S_mj(t)=tfrac12+frac2π∑_n=1^mfracsin(π n/2)ncos(2π nt). This is the exact finite Fourier formula. The drawn curves sample it numerically; the endpoint calculation below is symbolic. vfigfourier_jump headOne value is fixed at every cutoff At t=±1/4, every summand va…
Exact mathematical passage
j(t)=\mathbf 1_{\{|t|<1/4\}},\qquad S_mj(t)=\tfrac12+\frac2\pi\sum_{n=1}^{m}\frac{\sin(\pi n/2)}{n}\cos(2\pi nt).
\sum_{n\in F}Use in source · 1 passage
Search aliases: ∑_n∈ F sum in
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 290
The shift orbit is the standard orthonormal basis {δ_n}_n∈ Z. For x∈ℓ^2, the finite truncations x_F=∑_n∈ Fx[n]δ_n, F⊂ Z finite, satisfy x-x_F _2^2=∑_n∉ F|x[n]|^2→0 as the finite sets exhaust Z. Thus the unordered Hilbert-space sum ∑_nx[n]δ_n converges to x, and the finite-support statement is exact. The shift is a surjective isometry because n↦ n-1 is a bijection of Z, with inverse shift n↦ n+1. This proves the closed-span assertion.
Exact mathematical passage
x_F=\sum_{n\in F}x[n]\delta_n,\qquad F\subset\bbZ\text{ finite},
\sum_{n\in\bbZ}Use in source · 2 passages
Search aliases: ∑_n∈bbZ ∑_n∈ Z sum in bbZ sum in mathbb
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 283
[Impulse seed theorem] The closed shift-generated hull of δ_0 is all of ℓ^2( Z): overlinespan{S^nδ_0:n∈ Z}^ ℓ^2 =ℓ^2( Z). Every finitely supported signal has the exact finite expression x=∑_n∈ Zx[n]S^nδ_0.
Exact mathematical passage
x=\sum_{n\in\bbZ}x[n]S^n\delta_0.
\sum_{n\notin F}Use in source · 1 passage
Search aliases: ∑_n∉ F sum notin
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 293
The shift orbit is the standard orthonormal basis {δ_n}_n∈ Z. For x∈ℓ^2, the finite truncations x_F=∑_n∈ Fx[n]δ_n, F⊂ Z finite, satisfy x-x_F _2^2=∑_n∉ F|x[n]|^2→0 as the finite sets exhaust Z. Thus the unordered Hilbert-space sum ∑_nx[n]δ_n converges to x, and the finite-support statement is exact. The shift is a surjective isometry because n↦ n-1 is a bijection of Z, with inverse shift n↦ n+1. This proves the closed-span assertion.
Exact mathematical passage
\|x-x_F\|_2^2=\sum_{n\notin F}|x[n]|^2\to0
\sum_{n\equiv r\ (\mathrm{mod}\,N)}Use in source · 1 passage
Search aliases: ∑_n≡ r (mod N) sum equiv mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 341
Each integer belongs to exactly one residue class modulo N. Multiplying the two indicator functions gives zero for distinct classes and the same indicator for equal classes. Summing all residue indicators gives one at every address. Moreover, P_N,rx _2^2=∑_n≡ r (mod N)|x[n]|^2≤ x _2^2, ⟨ P_N,rx,y⟩=⟨ x,P_N,ry⟩. The mask is real-valued, so the inner-product identity follows term by term. Hence each projector is self-adjoint, and the ranges of different residue projectors are orthogonal.
Exact mathematical passage
\|P_{N,r}x\|_2^2=\sum_{n\equiv r\ (\mathrm{mod}\,N)}|x[n]|^2\le\|x\|_2^2, \qquad \langle P_{N,r}x,y\rangle=\langle x,P_{N,r}y\rangle.
\sum_{n\ge0}Use in source · 1 passage
Search aliases: ∑_n≥0 sum ge
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1059
Linear time invariance is not required for exact packet coordinates. Let A and B be formal power series over a commutative coefficient ring (in particular C). Set A(z)=∑_n≥0a_nz^n, A_i=P_N,iA. All phase indices range from 0 to N-1. For another representation B(z), multiplication gives a nonlinear or bilinear interaction.
Exact mathematical passage
A(z)=\sum_{n\ge0}a_nz^n, \qquad A_i=P_{N,i}A.
\sum_{n\geq0}Use in source · 1 passage
Search aliases: ∑_n≥0 sum geq
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3492
[Residue law for Cauchy products] Let F(z)=∑_n≥0f_nz^n, G(z)=∑_n≥0g_nz^n be formal power series. The same statement holds for power series that converge on a common disk. Decompose both series into residue packets modulo N: F=∑_r=0^N-1F_r, G=∑_s=0^N-1G_s. If H=FG, then its residue packet H_t is H_t = ∑_r+s≡ t pmod NF_rG_s.
Exact mathematical passage
F(z)=\sum_{n\geq0}f_nz^n, \qquad G(z)=\sum_{n\geq0}g_nz^n
\sum_{q=1}^MUse in source · 1 passage
Search aliases: ∑_q=1^M sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 692
[Several exact edge packets] If f=s+∑_q=1^Mj_q and every j_q is retained exactly, then f-(Q_ms+∑_qj_q) = s-Q_ms . Curved edges, corners, and different jump amplitudes do not alter the identity. The finite sum j=∑_q=1^M j_q is a single retained packet. Applying the preceding cancellation identity gives the result; the empty family M=0 is included with sum zero.
Exact mathematical passage
j=\sum_{q=1}^M j_q
\sum_{q=1}^{M}Use in source · 1 passage
Search aliases: ∑_q=1^M sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 681
[Several exact edge packets] If f=s+∑_q=1^Mj_q and every j_q is retained exactly, then f-(Q_ms+∑_qj_q) = s-Q_ms . Curved edges, corners, and different jump amplitudes do not alter the identity. The finite sum j=∑_q=1^M j_q is a single retained packet. Applying the preceding cancellation identity gives the result; the empty family M=0 is included with sum zero.
Exact mathematical passage
f=s+\sum_{q=1}^{M}j_q
\sum_{q\in\bbZ}Use in source · 1 passage
Search aliases: ∑_q∈bbZ ∑_q∈ Z sum in bbZ sum in mathbb
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 410
Let h be a finite impulse response and y=h*x, y[n]=∑_q∈ Zh[q]x[n-q]. Resolve both input and output into N phases. For output phase k and input phase s, define the transfer kernel h_ks[m]=h[Nm+k-s]. Out-of-range taps are understood to be zero.
Exact mathematical passage
y=h*x, \qquad y[n]=\sum_{q\in\bbZ}h[q]x[n-q].
\sum_{q\in\ZZ^d}Use in source · 1 passage
Search aliases: ∑_q∈ZZ^d ∑_q∈ Z^d sum in ZZ sum in mathbb
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1905
Let h∈ℓ^1( Z^d) and x∈ℓ^p( Z^d), with 1≤ p≤∞, and define (Hx)(n) =(h*x)(n) =∑_q∈ Z^dh(q)x(n-q). Young's inequality gives Hx _p≤ h _1 x _p for every 1≤ p≤∞.
Exact mathematical passage
(Hx)(n) =(h*x)(n) =\sum_{q\in\ZZ^d}h(q)x(n-q).
\sum_rUse in source · 1 passage
Search aliases: ∑_r sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1358
The signal realization makes the seed, address, packet, transfer, refinement, and synthesis operations concrete. The objects and laws can be read directly: seed · =δ_0, address · =S^n, packet · =P_N,rx, transfer · =P_N,kHP_N,s, refinement · =P_N,r=∑_ssucc rP_M,s, synthesis · =∑_rP_N,rx, nonlinearity · =P_N,rF(∑_sP_N,sx). Filter banks, wavelet packets, adaptive complete prunings, edge packets, and Smith coordinates inhabit the same operator grammar, but their claims differ. Polyphase and filter-b…
Exact mathematical passage
\boxed{ \begin{aligned} \text{seed}&=\delta_0,\\ \text{address}&=S^n,\\ \text{packet}&=P_{N,r}x,\\ \text{transfer}&=P_{N,k}HP_{N,s},\\ \text{refinement}&=P_{N,r}=\sum_{s\succ r}P_{M,s},\\ \text{synthesis}&=\sum_rP_{N,r}x,\\ \text{nonlinearity}&=P_{N,r}F\left(\sum_sP_{N,s}x\right). \end{aligned} }
\sum_{r+s\equiv t\!\!\pmod N}Use in source · 1 passage
Search aliases: ∑_r+s≡ t pmod N sum equiv pmod
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3504
rem[Residue law for Cauchy products] Let F(z)=∑_n≥0f_nz^n, G(z)=∑_n≥0g_nz^n be formal power series. The same statement holds for power series that converge on a common disk. Decompose both series into residue packets modulo N: F=∑_r=0^N-1F_r, G=∑_s=0^N-1G_s. If H=FG, then its residue packet H_t is H_t = ∑_r+s≡ t pmod NF_rG_s.
Exact mathematical passage
\boxed{ H_t = \sum_{r+s\equiv t\!\!\pmod N}F_rG_s. }
\sum_{r=0}^3Use in source · 1 passage
Search aliases: ∑_r=0^3 sum
Use in source · atlas-pages.tex, line 43
ump packet, not an estimated feature or a frequency band. headWhat has changed? Only where the values are stored The complete tuple of channels contains sixteen values, just as the original vector does. A single decimated channel is shorter, but keeping all four loses no samples. The residue projector P_4,rf is different: it keeps the original lattice and writes zeros outside phase r. S_4 A_4f=f, f _2^2=∑_r=0^3 f_r _2^2=57. The norm on the channel tuple is the sum-of-squares Hilbert norm, not th…
Exact mathematical passage
\mathcal S_4\mathcal A_4f=f, \qquad \|f\|_2^2=\sum_{r=0}^3\|f_r\|_2^2=57.
\sum_{r=0}^{m}Use in source · 1 passage
Search aliases: ∑_r=0^m sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3327
Let (Δ x)(n)=x(n+1)-x(n). Then Δ^m x(n) = ∑_r=0^m (-1)^m-rbinom mr x(n+r).
Exact mathematical passage
\Delta^m x(n) = \sum_{r=0}^{m} (-1)^{m-r}\binom mr x(n+r).
\sum_{r=0}^{N-1}Use in source · 5 passages
Search aliases: ∑_r=0^N-1 sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 327
[Polyphase projector theorem] The maps P_N,r are bounded orthogonal projectors on ℓ^2, with operator norm at most one, and satisfy P_N,rP_N,s=δ_rsP_N,r, ∑_r=0^N-1P_N,r=I. Consequently, x=∑_r=0^N-1P_N,rx with no approximation.
Exact mathematical passage
P_{N,r}P_{N,s}=\delta_{rs}P_{N,r}, \qquad \sum_{r=0}^{N-1}P_{N,r}=I.
\sum_{r\in R}Use in source · 3 passages
Search aliases: ∑_r∈ R sum in
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1776
For finite p, equip ⊕_r∈ Rℓ^p(L) with (u_r) ^p=∑_r∈ R u_r _p^p. For p=∞, use the maximum norm.
Exact mathematical passage
\|(u_r)\|^p=\sum_{r\in R}\|u_r\|_p^p.
\sum_sUse in source · 3 passages
Search aliases: ∑_s sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1099
For a finite nonlinear shift operator F(x)[n]=x[n]+α x[n]x[n-1], the exact packet coordinates are simply F_r((x_s)_s)=P_N,rF (∑_sP_N,sx). Analysis followed by this packet system and synthesis reproduces the direct nonlinear output exactly. The numerical residual in the companion implementation was zero to machine representation.
Exact mathematical passage
F_r((x_s)_s)=P_{N,r}F\!\left(\sum_sP_{N,s}x\right).
\sum_{s=0}^{N-1}Use in source · 3 passages
Search aliases: ∑_s=0^N-1 sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 425
[Exact FIR polyphase transfer law] The output channels satisfy y_k[n] = ∑_s=0^N-1∑_m∈ Z h_ks[m]x_s[n-m]. For the numerical channel transform, take z∈ C∖{0} and assume ∑_n∈ Z|x_s[n]z^-n|<∞ (0≤ s<N). Define X_s(z)=∑_nx_s[n]z^-n. The output series then also converge absolutely, and Y(z)= H^(N)(z) X(z), H^(N)_ks(z)=∑_m h[Nm+k-s]z^-m.
Exact mathematical passage
\boxed{ y_k[n] = \sum_{s=0}^{N-1}\sum_{m\in\bbZ} h_{ks}[m]x_s[n-m]. }
\sum_{s\succ r}Use in source · 1 passage
Search aliases: ∑_ssucc r sum succ
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1358
The signal realization makes the seed, address, packet, transfer, refinement, and synthesis operations concrete. The objects and laws can be read directly: seed · =δ_0, address · =S^n, packet · =P_N,rx, transfer · =P_N,kHP_N,s, refinement · =P_N,r=∑_ssucc rP_M,s, synthesis · =∑_rP_N,rx, nonlinearity · =P_N,rF(∑_sP_N,sx). Filter banks, wavelet packets, adaptive complete prunings, edge packets, and Smith coordinates inhabit the same operator grammar, but their claims differ. Polyphase and filter-b…
Exact mathematical passage
\boxed{ \begin{aligned} \text{seed}&=\delta_0,\\ \text{address}&=S^n,\\ \text{packet}&=P_{N,r}x,\\ \text{transfer}&=P_{N,k}HP_{N,s},\\ \text{refinement}&=P_{N,r}=\sum_{s\succ r}P_{M,s},\\ \text{synthesis}&=\sum_rP_{N,r}x,\\ \text{nonlinearity}&=P_{N,r}F\left(\sum_sP_{N,s}x\right). \end{aligned} }
\sum_{\substack{0\le a<M\\a\equiv r\pmod N}}Use in source · 1 passage
Search aliases: ∑_substack0≤ a<M a≡ rpmod N sum substack le a equiv pmod
Use in source · atlas-pages.tex, line 66
leaved to recover the coarse channel. Adding them entrywise instead produces sums of different original samples. For example, at the first descendant address the values f[0]=0 and f[2]=1/2 are two separate values. Their sum 1/2 cannot tell us both of them. Exact refinement is a disjoint partition, not an averaging rule or a new measurement. headThe general formula When N divides M, P_N,r=∑_substack0≤ a<M a≡ rpmod NP_M,a. The divisibility hypothesis ensures that every fine residue class lies enti…
Exact mathematical passage
P_{N,r}=\sum_{\substack{0\le a<M\\a\equiv r\pmod N}}P_{M,a}.
\sum_{\substack{0\le s<M\\s\equiv r\, (\mathrm{mod}\,N)}}Use in source · 1 passage
Search aliases: ∑_substack0≤ s<M s≡ r (mod N) sum substack le s equiv mathrm
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 392
[Exact nested refinement] For N,M≥1 and Nmid M, P_N,r = ∑_substack0≤ s<M s≡ r (mod N)P_M,s.
Exact mathematical passage
\boxed{ P_{N,r} = \sum_{\substack{0\le s<M\\s\equiv r\, (\mathrm{mod}\,N)}}P_{M,s}. }
\sum_{s\in R}Use in source · 3 passages
Search aliases: ∑_s∈ R sum in
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1927
[Exact convolution transfer law] Let y=h*x. Then y_r(ℓ) = ∑_s∈ R∑_m∈ Lh_rs(m)x_s(ℓ-m). Equivalently, A_LH A_L^-1 = (H_rs)_r,s∈ R, where H_rs is convolution on L by h_rs.
Exact mathematical passage
y_r(\ell) = \sum_{s\in R}\sum_{m\in L}h_{rs}(m)x_s(\ell-m).
\sum_{s\in\operatorname{Child}(r)}Use in source · 1 passage
Search aliases: ∑_s∈Child(r) sum in operatorname
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1874
[Exact lattice refinement] For every coarse packet, P_r^L = ∑_s∈Child(r)P_s^L'. The fine packets in the sum are pairwise disjoint. Refinement through several nested sublattices is transitive.
Exact mathematical passage
P_r^{L} = \sum_{s\in\operatorname{Child}(r)}P_s^{L'}.
\sum_uUse in source · 1 passage
Search aliases: ∑_u sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 991
[2mm] displaystyle min{c_v(y),∑_u∈ch(v)V_y(u)}, · v internal. Stop at v when c_v(y)≤∑_uV_y(u), including equality; refine otherwise. This deterministic rule produces a globally optimal complete pruning. With the node costs supplied and child values cached, the evaluator uses one comparison per internal node, one visit per node, and at most | T|-1 additions. Its arithmetic-and-visit count is therefore at most 3| T|. This is a unit-cost statement about the specified evaluator; it excludes evaluati…
Exact mathematical passage
c_v(y)\le\sum_uV_y(u)
\sum_{v\in\Lcal}Use in source · 3 passages
Search aliases: ∑_v∈Lcal ∑_v∈ L sum in Lcal sum in mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2348
[Reconstruction on a pruned packet tree] Every admissible pruning L defines a unitary analysis map A_ L: H_rm root ⟶ ⊕_v∈ L H_v. Its inverse is obtained by applying the adjoints of the local splits from the leaves toward the root. In particular, x ^2 = ∑_v∈ L x_v ^2.
Exact mathematical passage
\|x\|^2 = \sum_{v\in\Lcal}\|x_v\|^2.
\sum_{v\in\Lambda}Use in source · 3 passages
Search aliases: ∑_v∈Λ sum in Lambda
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 909
[Exact synthesis from every complete pruning] For every complete pruning Λ, ∑_v∈ΛP_v=I, x=∑_v∈ΛP_vx. Thus changing the selected tree changes the coordinates but does not introduce reconstruction error.
Exact mathematical passage
\boxed{\sum_{v\in\Lambda}P_v=I,} \qquad \boxed{x=\sum_{v\in\Lambda}P_vx.}
\sum_wUse in source · 1 passage
Search aliases: ∑_w sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2406
Assume the child subtree bounds have been proved. Then ∑_w∑_ℓ∈ L_w x_ℓ ^2 ≥ min_w a(w) ∑_w (U_vx)_w ^2 ≥ A_vmin_w a(w) x ^2. The upper bound follows in the same way with the maximum and B_v. Induction completes the proof.
Exact mathematical passage
\sum_{w}\sum_{\ell\in\Lcal_w}\|x_\ell\|^2 \geq \min_w a(w) \sum_w\|(U_vx)_w\|^2 \geq A_v\min_w a(w)\|x\|^2.
\sum_{w}Use in source · 1 passage
Search aliases: ∑_w sum
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2406
Assume the child subtree bounds have been proved. Then ∑_w∑_ℓ∈ L_w x_ℓ ^2 ≥ min_w a(w) ∑_w (U_vx)_w ^2 ≥ A_vmin_w a(w) x ^2. The upper bound follows in the same way with the maximum and B_v. Induction completes the proof.
Exact mathematical passage
\sum_{w}\sum_{\ell\in\Lcal_w}\|x_\ell\|^2 \geq \min_w a(w) \sum_w\|(U_vx)_w\|^2 \geq A_v\min_w a(w)\|x\|^2.
\sum_{w\in\operatorname{child}(v)}Use in source · 3 passages
Search aliases: ∑_w∈child(v) sum in operatorname
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1661
The packet tree results treat every admissible pruning, not only complete trees of a fixed depth. Unitary local splits give a unitary global transform. Local frame bounds propagate recursively in the nonunitary case. If the cost of a pruning is additive over retained nodes, then F(v) = min{ c(v), ∑_w∈child(v)F(w) } gives the exact optimal tree. The cost and the selected basis are stable when the node costs change by less than the gap separating the optimal pruning from its competitors.
Exact mathematical passage
F(v) = \min\left\{ c(v), \sum_{w\in\operatorname{child}(v)}F(w) \right\}
\sum_{|\alpha|=2}Use in source · 1 passage
Search aliases: ∑_|α|=2 sum alpha
Use in source · atlas-pages.tex, line 171
/4, 0, 1/4, which again sum to 57. Two factors of 1/sqrt2 explain why the depth-two coefficients are rational halves and quarters. headCoefficients and projectors are different objects W_α f is the short coefficient vector. The original-space component is W_α^*W_α f. Summing these synthesized components, not the short arrays entrywise, reconstructs f: f=∑_|α|=2W_α^*W_α f. The finite verification constructs the actual analysis matrices and checks their product with their adjoints. The paper's ℓ^2…
Exact mathematical passage
f=\sum_{|\alpha|=2}W_\alpha^*W_\alpha f.
\sum_\alphaUse in source · 1 passage
Search aliases: ∑_α sum alpha
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 623
Set A=∑_α W_ℓ,α^*W_ℓ,α. This is a finite sum of bounded operators. The energy result says ⟨ x,Ax⟩=⟨ x,x⟩ for every x. Complex polarization, applied also to x+y and x+ i y, gives ⟨ y,(A-I)x⟩=0 for all x,y, hence A=I. Applying the operator identity to x proves synthesis. This argument uses the actual Hilbert adjoints of the decimation operators, not a separately postulated inverse symbol.
Exact mathematical passage
A=\sum_\alpha W_{\ell,\alpha}^*W_{\ell,\alpha}
\sum_{\alpha\in\{0,1\}^2}Use in source · 1 passage
Search aliases: ∑_α∈{0,1}^2 sum alpha in
Use in source · depth_two.tex, line 44
00,0.65000); node[] at (5.90000,1.04500) - frac14; draw[fill=black!18,draw=black!40,line width=.35pt] (6.80000,0.72000) rectangle ++(1.80000,0.65000); node[] at (7.70000,1.04500) frac14; node[] at (10.90000,1.05000) frac14; node[] at (5.00000,4.60000) coefficient positions within each packet; node[] at (10.90000,4.60000) energy; node[] at (6.25000,-0.30000) ∑_α∈{0,1}^2W_α^*W_α=I_16;
Exact mathematical passage
\sum_{\alpha\in\{0,1\}^2}W_\alpha^*W_\alpha=I_{16}
\sum_{\alpha\in\{0,1\}^{\ell}}Use in source · 2 passages
Search aliases: ∑_α∈{0,1}^ℓ sum alpha in ell
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 604
[Wavelet-packet reconstruction and energy] At every finite depth ℓ, x=∑_α∈{0,1}^ℓW_ℓ,α^*W_ℓ,αx, and x _2^2 = ∑_α∈{0,1}^ℓ W_ℓ,αx _2^2.
Exact mathematical passage
\boxed{ x=\sum_{\alpha\in\{0,1\}^{\ell}}W_{\ell,\alpha}^{*}W_{\ell,\alpha}x, }
\sum_{\ell\in L}Use in source · 1 passage
Search aliases: ∑_ℓ∈ L sum ell in
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1802
Every address has a unique representation r+ℓ. For finite p, x _p^p = ∑_r∈ R∑_ℓ∈ L|x(r+ℓ)|^p = ∑_r∈ R x_r _p^p. The corresponding supremum identity is immediate when p=∞. Unique address representation proves that analysis and interleaving are inverse maps. The projector identities follow from the disjoint coset partition.
Exact mathematical passage
\|x\|_p^p = \sum_{r\in R}\sum_{\ell\in L}|x(r+\ell)|^p = \sum_{r\in R}\|x_r\|_p^p.
\sum_{\ell\in\Lcal_w}Use in source · 1 passage
Search aliases: ∑_ℓ∈Lcal_w ∑_ℓ∈ L_w sum ell in Lcal sum ell in mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2406
Assume the child subtree bounds have been proved. Then ∑_w∑_ℓ∈ L_w x_ℓ ^2 ≥ min_w a(w) ∑_w (U_vx)_w ^2 ≥ A_vmin_w a(w) x ^2. The upper bound follows in the same way with the maximum and B_v. Induction completes the proof.
Exact mathematical passage
\sum_{w}\sum_{\ell\in\Lcal_w}\|x_\ell\|^2 \geq \min_w a(w) \sum_w\|(U_vx)_w\|^2 \geq A_v\min_w a(w)\|x\|^2.
\setminusUse in source · 4 passages
Search aliases: ∖ setminus
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 433
[Exact FIR polyphase transfer law] The output channels satisfy y_k[n] = ∑_s=0^N-1∑_m∈ Z h_ks[m]x_s[n-m]. For the numerical channel transform, take z∈ C∖{0} and assume ∑_n∈ Z|x_s[n]z^-n|<∞ (0≤ s<N). Define X_s(z)=∑_nx_s[n]z^-n. The output series then also converge absolutely, and Y(z)= H^(N)(z) X(z), H^(N)_ks(z)=∑_m h[Nm+k-s]z^-m.
Exact mathematical passage
z\in\bbC\setminus\{0\}
\circUse in source · 2 passages
Search aliases: ∘ circ
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1152
obius action] Regard normalized impedance as the projective line P^1( C)= C∪{∞}. For N≥1, the pullback is a globally defined projective action. In the finite chart, wherever the denominator is nonzero, it is T_N,k(z) = frac(1+ω_N^k)z+(1-ω_N^k) (1-ω_N^k)z+(1+ω_N^k). As maps on P^1( C) it satisfies T_N,k∘ T_N,ℓ=T_N,k+ℓ (mod N).
Exact mathematical passage
T_{N,k}\circ T_{N,\ell}=T_{N,k+\ell\, (\mathrm{mod}\,N)}.
\capUse in source · 8 passages
Search aliases: ∩ cap
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2701
[Exact scalar region formula] Let E,F be measurable subsets of a measure space, let a,b∈ C, and let 1≤ p<∞. Then a 1_E-b 1_F _p^p = · |a-b|^pμ(E∩ F) +|a|^pμ(E∖ F) · +|b|^pμ(F∖ E). For equal amplitudes, a 1_E-a 1_F _p = |a|μ(Emathbintriangle F)^1/p.
Exact mathematical passage
\begin{aligned} \|a\one_E-b\one_F\|_p^p ={}& |a-b|^p\mu(E\cap F) +|a|^p\mu(E\setminus F)\\ &+|b|^p\mu(F\setminus E). \end{aligned}
\cupUse in source · 1 passage
Search aliases: ∪ cup
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1142
[Polyphase M obius action] Regard normalized impedance as the projective line P^1( C)= C∪{∞}. For N≥1, the pullback is a globally defined projective action. In the finite chart, wherever the denominator is nonzero, it is T_N,k(z) = frac(1+ω_N^k)z+(1-ω_N^k) (1-ω_N^k)z+(1+ω_N^k). As maps on P^1( C) it satisfies T_N,k∘ T_N,ℓ=T_N,k+ℓ (mod N).
Exact mathematical passage
\mathbb P^1(\bbC)=\bbC\cup\{\infty\}
\intUse in source · 4 passages
Search aliases: ∫ int
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2760
Assume τ<widehatτ. Differentiation with respect to the location parameter gives frac∂∂ uϕ_m,u(t) =-ϕ_m-1,u(t) for almost every t. Hence ϕ_m,τ-ϕ_m,widehatτ = ∫_τ^widehatτϕ_m-1,u d u. Minkowski's inequality gives ϕ_m,τ-ϕ_m,widehatτ _p ≤ |τ-widehatτ| sup_u ϕ_m-1,u _p. The supremum is bounded by the norm of t^m-1/(m-1)! on [0,1], which is the stated constant. The supremum norm argument is identical.
Exact mathematical passage
\phi_{m,\tau}-\phi_{m,\widehat\tau} = \int_\tau^{\widehat\tau}\phi_{m-1,u}\dd u.
\int_0^1Use in source · 1 passage
Search aliases: ∫_0^1 int
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3394
For the limit, put u_k=(k+tfrac12)/n. The identity Q_n(x_n)=frac1π n∑_k=0^n-1fracsin(π u_k)u_k ⟶frac1π∫_0^1fracsin(π u)u du =fracSi(π)π is a midpoint Riemann sum, since the integrand has continuous value π at zero. This is the normalized classical example in [source reference].
Exact mathematical passage
Q_n(x_n)=\frac{1}{\pi n}\sum_{k=0}^{n-1}\frac{\sin(\pi u_k)}{u_k} \longrightarrow\frac1\pi\int_0^1\frac{\sin(\pi u)}u\,du =\frac{\operatorname{Si}(\pi)}{\pi}
\int_0^zUse in source · 1 passage
Search aliases: ∫_0^z int
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3390
Let q(x)=tfrac12sign(sin x), whose jump at zero has size one. For n≥1, its partial sum through the first n positive odd frequencies is Q_n(x)=frac2π∑_k=0^n-1fracsin((2k+1)x)2k+1. At x_n=π/(2n), x_n⟶0, Q_n(x_n)⟶fracSi(π)π =frac12+G, G≈0.089489872236, where Si(z)=∫_0^zsin(t)/t dt and G=Si(π)/π-1/2>0. Thus a nonzero overshoot persists along points approaching the jump. Scaling by a real amplitude a gives an overshoot or undershoot of magnitude |a|G.
Exact mathematical passage
\operatorname{Si}(z)=\int_0^z\sin(t)/t\,dt
\int_a^{a+L/2}Use in source · 1 passage
Search aliases: ∫_a^a+L/2 int
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3208
Let L=2^-j and let a be the left endpoint of the support. Pair every point t∈[a,a+L/2] with t+L/2. Then ⟨ s,ψ_j,k⟩ = L^-1/2 ∫_a^a+L/2 overlines(t)-s(t+L/2) d t. The integrand has magnitude at most ML/2, and the integration interval has length L/2. The result is ML^3/2/4.
Exact mathematical passage
\langle s,\psi_{j,k}\rangle = L^{-1/2} \int_a^{a+L/2} \overline{s(t)-s(t+L/2)}\,\dd t.
\int_\tau^{\widehat\tau}Use in source · 1 passage
Search aliases: ∫_τ^widehatτ int tau widehat tau
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2760
Assume τ<widehatτ. Differentiation with respect to the location parameter gives frac∂∂ uϕ_m,u(t) =-ϕ_m-1,u(t) for almost every t. Hence ϕ_m,τ-ϕ_m,widehatτ = ∫_τ^widehatτϕ_m-1,u d u. Minkowski's inequality gives ϕ_m,τ-ϕ_m,widehatτ _p ≤ |τ-widehatτ| sup_u ϕ_m-1,u _p. The supremum is bounded by the norm of t^m-1/(m-1)! on [0,1], which is the stated constant. The supremum norm argument is identical.
Exact mathematical passage
\phi_{m,\tau}-\phi_{m,\widehat\tau} = \int_\tau^{\widehat\tau}\phi_{m-1,u}\dd u.
\congUse in source · 3 passages
Search aliases: ≅ cong
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1227
Write N=2^sM, M odd. The cyclic index splits by the Chinese remainder theorem: Z/N Z ≅ Z/2^s Z× Z/M Z.
Exact mathematical passage
\bbZ/N\bbZ \cong \bbZ/2^s\bbZ\times\bbZ/M\bbZ.
\approxUse in source · 1 passage
Search aliases: ≈ approx
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 3386
[Classical square-wave Gibbs behavior] Let q(x)=tfrac12sign(sin x), whose jump at zero has size one. For n≥1, its partial sum through the first n positive odd frequencies is Q_n(x)=frac2π∑_k=0^n-1fracsin((2k+1)x)2k+1. At x_n=π/(2n), x_n⟶0, Q_n(x_n)⟶fracSi(π)π =frac12+G, G≈0.089489872236, where Si(z)=∫_0^zsin(t)/t dt and G=Si(π)/π-1/2>0. Thus a nonzero overshoot persists along points approaching the jump. Scaling by a real amplitude a gives an overshoot or undershoot of magnitude |a|G.
Exact mathematical passage
x_n\longrightarrow0,\qquad Q_n(x_n)\longrightarrow\frac{\operatorname{Si}(\pi)}{\pi} =\frac12+G,\qquad G\approx0.089489872236,
\neqUse in source · 1 passage
Search aliases: ≠ neq
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2729
[Location error for one step] Let 0≤τ,widehatτ≤1 and H_τ= 1_[τ,1]. For 1≤ p<∞, aH_τ-aH_widehatτ _p = |a||τ-widehatτ|^1/p. In the supremum norm, the error equals |a| whenever τ≠widehatτ.
Exact mathematical passage
\tau\neq\widehat\tau
\equivUse in source · 12 passages
Search aliases: ≡ equiv
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 315
Fix an integer N≥1. For 0≤ r<N, define (P_N,rx)[n] = 1_n≡ r (mod N)x[n]. The corresponding decimated channel is x_r[q]=x[Nq+r].
Exact mathematical passage
(P_{N,r}x)[n] = \mathbf 1_{n\equiv r\, (\mathrm{mod}\,N)}x[n].
\geUse in source · 11 passages
Search aliases: ≥ ge
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 314
Fix an integer N≥1. For 0≤ r<N, define (P_N,rx)[n] = 1_n≡ r (mod N)x[n]. The corresponding decimated channel is x_r[q]=x[Nq+r].
Exact mathematical passage
N\ge1
\subsetUse in source · 5 passages
Search aliases: ⊂ subset
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 290
The shift orbit is the standard orthonormal basis {δ_n}_n∈ Z. For x∈ℓ^2, the finite truncations x_F=∑_n∈ Fx[n]δ_n, F⊂ Z finite, satisfy x-x_F _2^2=∑_n∉ F|x[n]|^2→0 as the finite sets exhaust Z. Thus the unordered Hilbert-space sum ∑_nx[n]δ_n converges to x, and the finite-support statement is exact. The shift is a surjective isometry because n↦ n-1 is a bijection of Z, with inverse shift n↦ n+1. This proves the closed-span assertion.
Exact mathematical passage
x_F=\sum_{n\in F}x[n]\delta_n,\qquad F\subset\bbZ\text{ finite},
\subseteqUse in source · 3 passages
Search aliases: ⊆ subseteq
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 701
Let (X,Σ,μ) be a measure space, with measurable masks Ω,widehatΩ⊆ X and scalar amplitudes a,widehat a∈ C (real amplitudes are a special case). Suppose a scalar jump packet is j=a 1_Ω, and an estimated packet is widehat j=widehat a 1_widehatΩ.
Exact mathematical passage
\Omega,\widehat\Omega\subseteq X
\bigoplus_{r=0}^{N-1}Use in source · 1 passage
Search aliases: ⊕_r=0^N-1 bigoplus
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1818
[One dimensional residue coordinates] For L=N Z and R={0,…,N-1}, x_r(k)=x(Nk+r) and ℓ^p( Z)≅⊕_r=0^N-1ℓ^p( Z) isometrically.
Exact mathematical passage
\ell^p(\ZZ)\cong\bigoplus_{r=0}^{N-1}\ell^p(\ZZ)
\bigoplus_{r\in R}Use in source · 2 passages
Search aliases: ⊕_r∈ R bigoplus in
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 1775
For finite p, equip ⊕_r∈ Rℓ^p(L) with (u_r) ^p=∑_r∈ R u_r _p^p. For p=∞, use the maximum norm.
Exact mathematical passage
\bigoplus_{r\in R}\ell^p(L)
\bigoplus_{v\in\Lcal}Use in source · 1 passage
Search aliases: ⊕_v∈Lcal ⊕_v∈ L bigoplus in Lcal bigoplus in mathcal
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 2342
[Reconstruction on a pruned packet tree] Every admissible pruning L defines a unitary analysis map A_ L: H_rm root ⟶ ⊕_v∈ L H_v. Its inverse is obtained by applying the adjoints of the local splits from the leaves toward the root. In particular, x ^2 = ∑_v∈ L x_v ^2.
Exact mathematical passage
\Acal_{\Lcal}:\Hcal_{\rm root} \longrightarrow \bigoplus_{v\in\Lcal}\Hcal_v.
\langleUse in source · 9 passages
Search aliases: ⟨ langle
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 341
Each integer belongs to exactly one residue class modulo N. Multiplying the two indicator functions gives zero for distinct classes and the same indicator for equal classes. Summing all residue indicators gives one at every address. Moreover, P_N,rx _2^2=∑_n≡ r (mod N)|x[n]|^2≤ x _2^2, ⟨ P_N,rx,y⟩=⟨ x,P_N,ry⟩. The mask is real-valued, so the inner-product identity follows term by term. Hence each projector is self-adjoint, and the ranges of different residue projectors are orthogonal.
Exact mathematical passage
\|P_{N,r}x\|_2^2=\sum_{n\equiv r\ (\mathrm{mod}\,N)}|x[n]|^2\le\|x\|_2^2, \qquad \langle P_{N,r}x,y\rangle=\langle x,P_{N,r}y\rangle.
\rangleUse in source · 9 passages
Search aliases: ⟩ rangle
Use in source · Read passage · Paper_V_Traditional_Final.tex, line 341
Each integer belongs to exactly one residue class modulo N. Multiplying the two indicator functions gives zero for distinct classes and the same indicator for equal classes. Summing all residue indicators gives one at every address. Moreover, P_N,rx _2^2=∑_n≡ r (mod N)|x[n]|^2≤ x _2^2, ⟨ P_N,rx,y⟩=⟨ x,P_N,ry⟩. The mask is real-valued, so the inner-product identity follows term by term. Hence each projector is self-adjoint, and the ranges of different residue projectors are orthogonal.
Exact mathematical passage
\|P_{N,r}x\|_2^2=\sum_{n\equiv r\ (\mathrm{mod}\,N)}|x[n]|^2\le\|x\|_2^2, \qquad \langle P_{N,r}x,y\rangle=\langle x,P_{N,r}y\rangle.
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- papers/paper-v/current/README.md · 0 mathematical passages
- papers/paper-v/current/README.txt · 0 mathematical passages
- papers/paper-v/current/editable/traditional/Paper_V_Traditional_Final.tex · 1219 mathematical passages
- papers/paper-v/current/editable/traditional/atlas-definitions.tex · 0 mathematical passages
- papers/paper-v/current/editable/traditional/atlas-pages.tex · 535 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/addresses.tex · 27 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/cayley_orbit.tex · 9 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/complete_cuts.tex · 17 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/crt_refinement.tex · 2 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/depth_two.tex · 25 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/dynamic_program.tex · 33 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/error_budget.tex · 38 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/ext_amplification.tex · 5 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/ext_differences.tex · 10 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/ext_frame.tex · 3 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/ext_haar_margin.tex · 3 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/ext_lattice.tex · 40 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/ext_location.tex · 7 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/ext_overlap.tex · 17 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/ext_repetition.tex · 10 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/ext_threshold.tex · 4 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/ext_trace.tex · 6 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/ext_tree.tex · 3 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/ext_two_grids.tex · 7 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/fir_transfer.tex · 6 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/fourier_jump.tex · 10 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/haar_alignment.tex · 3 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/known_edge.tex · 4 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/lifting_geometry.tex · 4 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/mask_geometry.tex · 3 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/mask_translation.tex · 11 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/nonlinear_shift.tex · 7 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/perfect_reconstruction.tex · 12 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/product_conventions.tex · 18 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/projective_poles.tex · 15 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/refinement.tex · 17 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/sample_delay.tex · 15 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/selector_margin.tex · 3 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/sensitivity.tex · 16 mathematical passages
- papers/paper-v/current/editable/traditional/visual_graphics/sparse_tail.tex · 1 mathematical passage
- papers/paper-v/current/editable/traditional/visual_graphics/three_worlds.tex · 10 mathematical passages
PDF reading editions (39)
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- Paper_V_Traditional_Final.pdf
- Paper_V_Visual_Guide_Final.pdf
- Paper_V_Lean_Guide_Final.pdf
- 01_polyphase_analysis.pdf
- 02_filter_transfer_matrix.pdf
- delay_coordinates.pdf
- 03_haar_edge_packets.pdf
- 04_wavelet_packet_energy.pdf
- retained_packet_cancellation.pdf
- 05_edge_reconstruction_1d.pdf
- 06_edge_error_vs_resolution.pdf
- 07_known_edge_packet_2d.pdf
- 08_estimated_edge_packet_2d.pdf
- 09_multiedge_stress_test.pdf
- 10_sparse_wavelet_fourier.pdf
- 17_adaptive_packet_tree.pdf
- 18_adaptive_tree_stability.pdf
- 14_nonlinear_phase_coupling.pdf
- 11_polyphase_smith_chart.pdf
- 12_smith_nested_refinement.pdf
- 13_smith_sensitivity.pdf
- 15_boundary_leakage.pdf
- 16_claim_status.pdf
- 01_polyphase_coordinates.pdf
- 02_transfer_blocks.pdf
- 03_frame_stability.pdf
- 04_packet_tree.pdf
- 05_best_basis.pdf
- 06_higher_order_packets.pdf
- 07_known_packet_reconstruction.pdf
- 08_overlap_matrix.pdf
- jump_contrast_gap.pdf
- 09_jump_detection.pdf
- 10_haar_scale_separation.pdf
- 11_shifted_grids.pdf
- 12_gibbs_comparison.pdf
- 13_sparse_comparison.pdf
- 14_smith_coordinates.pdf
- annals_gibbs_comparison.pdf