Notation by work
Collatz and Riemann: source and evidence
All notation · Glossary · Source coverage
441 symbol spellings in 1129 mathematical passages across 10 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 441 entries are available below.
+Notation table · 218 passages
Search aliases: +
Notation table · body.tex, line 825
ρ=β+iγ · A nontrivial zero, counted with its multiplicity.
Exact mathematical passage
\rho=\beta+i\gamma
=Notation table · 308 passages
Search aliases: =
Notation table · body.tex, line 825
ρ=β+iγ · A nontrivial zero, counted with its multiplicity.
Exact mathematical passage
\rho=\beta+i\gamma
\mathcal ANotation table · 7 passages
Search aliases: A mathcal
Notation table · body.tex, line 836
c_K, A_Q · The signed coefficient and its full absolute majorant.
Exact mathematical passage
c_K,\mathcal A_Q
aNotation table · 56 passages
Search aliases: a
Notation table · body.tex, line 831
g_t,a · A Gaussian of norm one in the physical measure du.
Exact mathematical passage
g_{t,a}
\mathcal A_QNotation table · 7 passages
Search aliases: A_Q mathcal
Notation table · body.tex, line 836
c_K, A_Q · The signed coefficient and its full absolute majorant.
Exact mathematical passage
c_K,\mathcal A_Q
cNotation table · 34 passages
Search aliases: c
Notation table · body.tex, line 836
c_K, A_Q · The signed coefficient and its full absolute majorant.
Exact mathematical passage
c_K,\mathcal A_Q
c_KNotation table · 10 passages
Search aliases: c_K
Notation table · body.tex, line 836
c_K, A_Q · The signed coefficient and its full absolute majorant.
Exact mathematical passage
c_K,\mathcal A_Q
dNotation table · 47 passages
Search aliases: d
Notation table · body.tex, line 831
g_t,a · A Gaussian of norm one in the physical measure du.
Exact mathematical passage
du
gNotation table · 16 passages
Search aliases: g
Notation table · body.tex, line 831
g_t,a · A Gaussian of norm one in the physical measure du.
Exact mathematical passage
g_{t,a}
g_{t,a}Notation table · 5 passages
Search aliases: g_t,a
Notation table · body.tex, line 831
g_t,a · A Gaussian of norm one in the physical measure du.
Exact mathematical passage
g_{t,a}
iNotation table · 23 passages
Search aliases: i
Notation table · body.tex, line 825
ρ=β+iγ · A nontrivial zero, counted with its multiplicity.
Exact mathematical passage
\rho=\beta+i\gamma
JNotation table · 25 passages
Search aliases: J
Notation table · body.tex, line 830
widetilde E_J · Squared packet energy on one unit translation band and one fixed heat strip.
Exact mathematical passage
\widetilde E_J
KNotation table · 42 passages
Search aliases: K
Notation table · body.tex, line 833
K,M · Integer sequence index and its natural scale M=sqrt K.
Exact mathematical passage
K,M
LNotation table · 13 passages
Search aliases: L
Notation table · body.tex, line 829
L · The operator 1/4-∂_u^2 removing the explicit pole channel.
Exact mathematical passage
L
MNotation table · 65 passages
Search aliases: M
Notation table · body.tex, line 833
K,M · Integer sequence index and its natural scale M=sqrt K.
Exact mathematical passage
K,M
QNotation table · 68 passages
Search aliases: Q
Notation table · body.tex, line 834
Q,d · A squarefree product of primes and its number of prime coordinates.
Exact mathematical passage
Q,d
RNotation table · 56 passages
Search aliases: R
Notation table · body.tex, line 832
λ,R · Squared coherent displacement and Hermite cutoff. The cutoff R in this section is separate from the fixed tail multiplier in the barrier.
Exact mathematical passage
\lambda,R
tNotation table · 111 passages
Search aliases: t
Notation table · body.tex, line 827
t,u · Positive heat time and translation, respectively.
Exact mathematical passage
t,u
uNotation table · 81 passages
Search aliases: u
Notation table · body.tex, line 827
t,u · Positive heat time and translation, respectively.
Exact mathematical passage
t,u
wNotation table · 48 passages
Search aliases: w
Notation table · body.tex, line 835
Δ_Qw_K · The alternating divisor cube acting on the positive weight.
Exact mathematical passage
\Delta_Qw_K
w_KNotation table · 9 passages
Search aliases: w_K
Notation table · body.tex, line 835
Δ_Qw_K · The alternating divisor cube acting on the positive weight.
Exact mathematical passage
\Delta_Qw_K
\widetilde ENotation table · 7 passages
Search aliases: widetilde E widetilde
Notation table · body.tex, line 830
widetilde E_J · Squared packet energy on one unit translation band and one fixed heat strip.
Exact mathematical passage
\widetilde E_J
\widetilde E_JNotation table · 6 passages
Search aliases: widetilde E_J widetilde
Notation table · body.tex, line 830
widetilde E_J · Squared packet energy on one unit translation band and one fixed heat strip.
Exact mathematical passage
\widetilde E_J
zNotation table · 45 passages
Search aliases: z
Notation table · body.tex, line 826
z_ρ · The centered coordinate ρ-1/2.
Exact mathematical passage
z_\rho
z_\rhoNotation table · 9 passages
Search aliases: z_ρ rho
Notation table · body.tex, line 826
z_ρ · The centered coordinate ρ-1/2.
Exact mathematical passage
z_\rho
\betaNotation table · 3 passages
Search aliases: β beta
Notation table · body.tex, line 825
ρ=β+iγ · A nontrivial zero, counted with its multiplicity.
Exact mathematical passage
\rho=\beta+i\gamma
\gammaNotation table · 17 passages
Search aliases: γ gamma
Notation table · body.tex, line 825
ρ=β+iγ · A nontrivial zero, counted with its multiplicity.
Exact mathematical passage
\rho=\beta+i\gamma
\DeltaNotation table · 12 passages
Search aliases: Δ Delta
Notation table · body.tex, line 835
Δ_Qw_K · The alternating divisor cube acting on the positive weight.
Exact mathematical passage
\Delta_Qw_K
\Delta_QNotation table · 9 passages
Search aliases: Δ_Q Delta
Notation table · body.tex, line 835
Δ_Qw_K · The alternating divisor cube acting on the positive weight.
Exact mathematical passage
\Delta_Qw_K
\ThetaNotation table · 18 passages
Search aliases: Θ Theta
Notation table · body.tex, line 828
Θ_t · The Gaussian sum over centered zero modes.
Exact mathematical passage
\Theta_t
\Theta_tNotation table · 18 passages
Search aliases: Θ_t Theta
Notation table · body.tex, line 828
Θ_t · The Gaussian sum over centered zero modes.
Exact mathematical passage
\Theta_t
\lambdaNotation table · 18 passages
Search aliases: λ lambda
Notation table · body.tex, line 832
λ,R · Squared coherent displacement and Hermite cutoff. The cutoff R in this section is separate from the fixed tail multiplier in the barrier.
Exact mathematical passage
\lambda,R
\rhoNotation table · 45 passages
Search aliases: ρ rho
Notation table · body.tex, line 825
ρ=β+iγ · A nontrivial zero, counted with its multiplicity.
Exact mathematical passage
\rho=\beta+i\gamma
\partialNotation table · 7 passages
Search aliases: ∂ partial
Notation table · body.tex, line 829
L · The operator 1/4-∂_u^2 removing the explicit pole channel.
Exact mathematical passage
1/4-\partial_u^2
\partial_u^2Notation table · 6 passages
Search aliases: ∂_u^2 partial
Notation table · body.tex, line 829
L · The operator 1/4-∂_u^2 removing the explicit pole channel.
Exact mathematical passage
1/4-\partial_u^2
ADefinition passage · 28 passages
Search aliases: A
Definition passage: Classical, shortcut, and odd maps · body.tex, line 27
n/2, · n even. For a positive odd integer n, let a(n)=ν_2(3n+1), A(n)=frac3n+12^a(n). Here ν_2(m) is the largest exponent a for which 2^a divides the positive integer m. Thus a(n)≥1, and A(n) is positive and odd.
Exact mathematical passage
\label{eq:oddmap} a(n)=\nu_2(3n+1),\qquad A(n)=\frac{3n+1}{2^{a(n)}}.
a_0Definition passage · 7 passages
Search aliases: a_0
Definition passage: Valuation word and offset · body.tex, line 55
[Valuation word and offset] A valuation word is a finite sequence w=(a_0,…,a_k-1) of positive integers. Define its partial depths and offsets by S_0=0, S_j=∑_i<ja_i, C_0=0, C_j+1=3C_j+2^S_j. Its total depth is S=S_k and its terminal offset is C=C_k. The word is realized by n if these are the first k valuations of the actual odd orbit from n.
Exact mathematical passage
w=(a_0,\ldots,a_{k-1})
a_iDefinition passage · 1 passage
Search aliases: a_i
Definition passage: Valuation word and offset · body.tex, line 56
[Valuation word and offset] A valuation word is a finite sequence w=(a_0,…,a_k-1) of positive integers. Define its partial depths and offsets by S_0=0, S_j=∑_i<ja_i, C_0=0, C_j+1=3C_j+2^S_j. Its total depth is S=S_k and its terminal offset is C=C_k. The word is realized by n if these are the first k valuations of the actual odd orbit from n.
Exact mathematical passage
\label{eq:offsetrec} S_0=0,\quad S_j=\sum_{i<j}a_i,\qquad C_0=0,\quad C_{j+1}=3C_j+2^{S_j}.
a_{k-1}Definition passage · 2 passages
Search aliases: a_k-1
Definition passage: Valuation word and offset · body.tex, line 55
[Valuation word and offset] A valuation word is a finite sequence w=(a_0,…,a_k-1) of positive integers. Define its partial depths and offsets by S_0=0, S_j=∑_i<ja_i, C_0=0, C_j+1=3C_j+2^S_j. Its total depth is S=S_k and its terminal offset is C=C_k. The word is realized by n if these are the first k valuations of the actual odd orbit from n.
Exact mathematical passage
w=(a_0,\ldots,a_{k-1})
CDefinition passage · 58 passages
Search aliases: C
Definition passage: Valuation word and offset · body.tex, line 56
[Valuation word and offset] A valuation word is a finite sequence w=(a_0,…,a_k-1) of positive integers. Define its partial depths and offsets by S_0=0, S_j=∑_i<ja_i, C_0=0, C_j+1=3C_j+2^S_j. Its total depth is S=S_k and its terminal offset is C=C_k. The word is realized by n if these are the first k valuations of the actual odd orbit from n.
Exact mathematical passage
\label{eq:offsetrec} S_0=0,\quad S_j=\sum_{i<j}a_i,\qquad C_0=0,\quad C_{j+1}=3C_j+2^{S_j}.
C_0Definition passage · 1 passage
Search aliases: C_0
Definition passage: Valuation word and offset · body.tex, line 56
[Valuation word and offset] A valuation word is a finite sequence w=(a_0,…,a_k-1) of positive integers. Define its partial depths and offsets by S_0=0, S_j=∑_i<ja_i, C_0=0, C_j+1=3C_j+2^S_j. Its total depth is S=S_k and its terminal offset is C=C_k. The word is realized by n if these are the first k valuations of the actual odd orbit from n.
Exact mathematical passage
\label{eq:offsetrec} S_0=0,\quad S_j=\sum_{i<j}a_i,\qquad C_0=0,\quad C_{j+1}=3C_j+2^{S_j}.
C_jDefinition passage · 8 passages
Search aliases: C_j
Definition passage: Valuation word and offset · body.tex, line 56
[Valuation word and offset] A valuation word is a finite sequence w=(a_0,…,a_k-1) of positive integers. Define its partial depths and offsets by S_0=0, S_j=∑_i<ja_i, C_0=0, C_j+1=3C_j+2^S_j. Its total depth is S=S_k and its terminal offset is C=C_k. The word is realized by n if these are the first k valuations of the actual odd orbit from n.
Exact mathematical passage
\label{eq:offsetrec} S_0=0,\quad S_j=\sum_{i<j}a_i,\qquad C_0=0,\quad C_{j+1}=3C_j+2^{S_j}.
C_{j+1}Definition passage · 1 passage
Search aliases: C_j+1
Definition passage: Valuation word and offset · body.tex, line 56
[Valuation word and offset] A valuation word is a finite sequence w=(a_0,…,a_k-1) of positive integers. Define its partial depths and offsets by S_0=0, S_j=∑_i<ja_i, C_0=0, C_j+1=3C_j+2^S_j. Its total depth is S=S_k and its terminal offset is C=C_k. The word is realized by n if these are the first k valuations of the actual odd orbit from n.
Exact mathematical passage
\label{eq:offsetrec} S_0=0,\quad S_j=\sum_{i<j}a_i,\qquad C_0=0,\quad C_{j+1}=3C_j+2^{S_j}.
C_kDefinition passage · 17 passages
Search aliases: C_k
Definition passage: Valuation word and offset · body.tex, line 60
[Valuation word and offset] A valuation word is a finite sequence w=(a_0,…,a_k-1) of positive integers. Define its partial depths and offsets by S_0=0, S_j=∑_i<ja_i, C_0=0, C_j+1=3C_j+2^S_j. Its total depth is S=S_k and its terminal offset is C=C_k. The word is realized by n if these are the first k valuations of the actual odd orbit from n.
Exact mathematical passage
C=C_k
\operatorname{Col}Definition passage · 1 passage
Search aliases: Col operatorname
Definition passage: Classical, shortcut, and odd maps · body.tex, line 19
[Classical, shortcut, and odd maps] For a positive integer n, define the classical map and its shortcut by Col(n)= 3n+1, · n odd, n/2, · n even, T(n)= (3n+1)/2, · n odd, n/2, · n even. For a positive odd integer n, let a(n)=ν_2(3n+1), A(n)=frac3n+12^a(n). Here ν_2(m) is the largest exponent a for which 2^a divides the positive integer m. Thus a(n)≥1, and A(n) is positive and odd.
Exact mathematical passage
\label{eq:classical} \operatorname{Col}(n)= \begin{cases}3n+1,&n\text{ odd},\\n/2,&n\text{ even},\end{cases} \qquad T(n)= \begin{cases}(3n+1)/2,&n\text{ odd},\\n/2,&n\text{ even}.\end{cases}
d_rDefinition passage · 9 passages
Search aliases: d_r
Definition passage: Coefficient ratio and deficit · body.tex, line 499
[Coefficient ratio and deficit] Define ρ_r=frac2^S_r3^r, d_r=lfloor rlog_2 3rfloor-S_r. Permanent coefficient noncrossing means ρ_r≤1 for every r≥0. Under that hypothesis d_r is a nonnegative integer.
Exact mathematical passage
\label{eq:rho} \rho_r=\frac{2^{S_r}}{3^r},\qquad d_r=\lfloor r\log_2 3\rfloor-S_r.
eDefinition passage · 54 passages
Search aliases: e
Definition passage: Pole annihilated packet · body.tex, line 235
[Pole annihilated packet] Let L=1/4-∂_u^2, and define widetildeΘ_t=LΘ_t =∑_ρ∈ Zρ(1-ρ) e^tz_ρ^2+uz_ρ.
Exact mathematical passage
\label{rh:tilde} \widetilde\Theta_t=L\Theta_t =\sum_{\rho\in\mathcal Z}\rho(1-\rho) e^{tz_\rho^2+uz_\rho}.
e^{tz_\rho^2+uz_\rho}Definition passage · 3 passages
Search aliases: e^tz_ρ^2+uz_ρ rho rho
Definition passage: Pole annihilated packet · body.tex, line 235
[Pole annihilated packet] Let L=1/4-∂_u^2, and define widetildeΘ_t=LΘ_t =∑_ρ∈ Zρ(1-ρ) e^tz_ρ^2+uz_ρ.
Exact mathematical passage
\label{rh:tilde} \widetilde\Theta_t=L\Theta_t =\sum_{\rho\in\mathcal Z}\rho(1-\rho) e^{tz_\rho^2+uz_\rho}.
\expDefinition passage · 5 passages
Search aliases: exp
Definition passage: Zero heat packet · body.tex, line 66
[Zero heat packet] For a real heat parameter t>0 and a complex translation u, define Θ_t(u)=∑_ρ∈ Z exp(tz_ρ^2+uz_ρ).
Exact mathematical passage
\label{rh:theta} \Theta_t(u)=\sum_{\rho\in\mathcal Z} \exp\bigl(tz_\rho^2+uz_\rho\bigr).
jDefinition passage · 80 passages
Search aliases: j
Definition passage: Valuation word and offset · body.tex, line 56
[Valuation word and offset] A valuation word is a finite sequence w=(a_0,…,a_k-1) of positive integers. Define its partial depths and offsets by S_0=0, S_j=∑_i<ja_i, C_0=0, C_j+1=3C_j+2^S_j. Its total depth is S=S_k and its terminal offset is C=C_k. The word is realized by n if these are the first k valuations of the actual odd orbit from n.
Exact mathematical passage
\label{eq:offsetrec} S_0=0,\quad S_j=\sum_{i<j}a_i,\qquad C_0=0,\quad C_{j+1}=3C_j+2^{S_j}.
kDefinition passage · 76 passages
Search aliases: k
Definition passage: Valuation word and offset · body.tex, line 55
[Valuation word and offset] A valuation word is a finite sequence w=(a_0,…,a_k-1) of positive integers. Define its partial depths and offsets by S_0=0, S_j=∑_i<ja_i, C_0=0, C_j+1=3C_j+2^S_j. Its total depth is S=S_k and its terminal offset is C=C_k. The word is realized by n if these are the first k valuations of the actual odd orbit from n.
Exact mathematical passage
w=(a_0,\ldots,a_{k-1})
\lfloorDefinition passage · 7 passages
Search aliases: lfloor
Definition passage: Coefficient ratio and deficit · body.tex, line 499
[Coefficient ratio and deficit] Define ρ_r=frac2^S_r3^r, d_r=lfloor rlog_2 3rfloor-S_r. Permanent coefficient noncrossing means ρ_r≤1 for every r≥0. Under that hypothesis d_r is a nonnegative integer.
Exact mathematical passage
\label{eq:rho} \rho_r=\frac{2^{S_r}}{3^r},\qquad d_r=\lfloor r\log_2 3\rfloor-S_r.
\logDefinition passage · 41 passages
Search aliases: log
Definition passage: Coefficient ratio and deficit · body.tex, line 499
[Coefficient ratio and deficit] Define ρ_r=frac2^S_r3^r, d_r=lfloor rlog_2 3rfloor-S_r. Permanent coefficient noncrossing means ρ_r≤1 for every r≥0. Under that hypothesis d_r is a nonnegative integer.
Exact mathematical passage
\label{eq:rho} \rho_r=\frac{2^{S_r}}{3^r},\qquad d_r=\lfloor r\log_2 3\rfloor-S_r.
\log_2Definition passage · 6 passages
Search aliases: log_2 log
Definition passage: Coefficient ratio and deficit · body.tex, line 499
[Coefficient ratio and deficit] Define ρ_r=frac2^S_r3^r, d_r=lfloor rlog_2 3rfloor-S_r. Permanent coefficient noncrossing means ρ_r≤1 for every r≥0. Under that hypothesis d_r is a nonnegative integer.
Exact mathematical passage
\label{eq:rho} \rho_r=\frac{2^{S_r}}{3^r},\qquad d_r=\lfloor r\log_2 3\rfloor-S_r.
mDefinition passage · 69 passages
Search aliases: m
Definition passage: Classical, shortcut, and odd maps · body.tex, line 30
n/2, · n even. For a positive odd integer n, let a(n)=ν_2(3n+1), A(n)=frac3n+12^a(n). Here ν_2(m) is the largest exponent a for which 2^a divides the positive integer m. Thus a(n)≥1, and A(n) is positive and odd.
Exact mathematical passage
\nu_2(m)
nDefinition passage · 141 passages
Search aliases: n
Definition passage: Classical, shortcut, and odd maps · body.tex, line 18
[Classical, shortcut, and odd maps] For a positive integer n, define the classical map and its shortcut by Col(n)= 3n+1, · n odd,
Exact mathematical passage
n
rDefinition passage · 101 passages
Search aliases: r
Definition passage: Coefficient ratio and deficit · body.tex, line 499
[Coefficient ratio and deficit] Define ρ_r=frac2^S_r3^r, d_r=lfloor rlog_2 3rfloor-S_r. Permanent coefficient noncrossing means ρ_r≤1 for every r≥0. Under that hypothesis d_r is a nonnegative integer.
Exact mathematical passage
\label{eq:rho} \rho_r=\frac{2^{S_r}}{3^r},\qquad d_r=\lfloor r\log_2 3\rfloor-S_r.
\rfloorDefinition passage · 7 passages
Search aliases: rfloor
Definition passage: Coefficient ratio and deficit · body.tex, line 499
[Coefficient ratio and deficit] Define ρ_r=frac2^S_r3^r, d_r=lfloor rlog_2 3rfloor-S_r. Permanent coefficient noncrossing means ρ_r≤1 for every r≥0. Under that hypothesis d_r is a nonnegative integer.
Exact mathematical passage
\label{eq:rho} \rho_r=\frac{2^{S_r}}{3^r},\qquad d_r=\lfloor r\log_2 3\rfloor-S_r.
SDefinition passage · 82 passages
Search aliases: S
Definition passage: Valuation word and offset · body.tex, line 56
[Valuation word and offset] A valuation word is a finite sequence w=(a_0,…,a_k-1) of positive integers. Define its partial depths and offsets by S_0=0, S_j=∑_i<ja_i, C_0=0, C_j+1=3C_j+2^S_j. Its total depth is S=S_k and its terminal offset is C=C_k. The word is realized by n if these are the first k valuations of the actual odd orbit from n.
Exact mathematical passage
\label{eq:offsetrec} S_0=0,\quad S_j=\sum_{i<j}a_i,\qquad C_0=0,\quad C_{j+1}=3C_j+2^{S_j}.
S_0Definition passage · 1 passage
Search aliases: S_0
Definition passage: Valuation word and offset · body.tex, line 56
[Valuation word and offset] A valuation word is a finite sequence w=(a_0,…,a_k-1) of positive integers. Define its partial depths and offsets by S_0=0, S_j=∑_i<ja_i, C_0=0, C_j+1=3C_j+2^S_j. Its total depth is S=S_k and its terminal offset is C=C_k. The word is realized by n if these are the first k valuations of the actual odd orbit from n.
Exact mathematical passage
\label{eq:offsetrec} S_0=0,\quad S_j=\sum_{i<j}a_i,\qquad C_0=0,\quad C_{j+1}=3C_j+2^{S_j}.
S_jDefinition passage · 11 passages
Search aliases: S_j
Definition passage: Valuation word and offset · body.tex, line 56
[Valuation word and offset] A valuation word is a finite sequence w=(a_0,…,a_k-1) of positive integers. Define its partial depths and offsets by S_0=0, S_j=∑_i<ja_i, C_0=0, C_j+1=3C_j+2^S_j. Its total depth is S=S_k and its terminal offset is C=C_k. The word is realized by n if these are the first k valuations of the actual odd orbit from n.
Exact mathematical passage
\label{eq:offsetrec} S_0=0,\quad S_j=\sum_{i<j}a_i,\qquad C_0=0,\quad C_{j+1}=3C_j+2^{S_j}.
S_kDefinition passage · 3 passages
Search aliases: S_k
Definition passage: Valuation word and offset · body.tex, line 60
[Valuation word and offset] A valuation word is a finite sequence w=(a_0,…,a_k-1) of positive integers. Define its partial depths and offsets by S_0=0, S_j=∑_i<ja_i, C_0=0, C_j+1=3C_j+2^S_j. Its total depth is S=S_k and its terminal offset is C=C_k. The word is realized by n if these are the first k valuations of the actual odd orbit from n.
Exact mathematical passage
S=S_k
S_rDefinition passage · 14 passages
Search aliases: S_r
Definition passage: Coefficient ratio and deficit · body.tex, line 499
[Coefficient ratio and deficit] Define ρ_r=frac2^S_r3^r, d_r=lfloor rlog_2 3rfloor-S_r. Permanent coefficient noncrossing means ρ_r≤1 for every r≥0. Under that hypothesis d_r is a nonnegative integer.
Exact mathematical passage
\label{eq:rho} \rho_r=\frac{2^{S_r}}{3^r},\qquad d_r=\lfloor r\log_2 3\rfloor-S_r.
TDefinition passage · 12 passages
Search aliases: T
Definition passage: Classical, shortcut, and odd maps · body.tex, line 19
[Classical, shortcut, and odd maps] For a positive integer n, define the classical map and its shortcut by Col(n)= 3n+1, · n odd, n/2, · n even, T(n)= (3n+1)/2, · n odd, n/2, · n even. For a positive odd integer n, let a(n)=ν_2(3n+1), A(n)=frac3n+12^a(n). Here ν_2(m) is the largest exponent a for which 2^a divides the positive integer m. Thus a(n)≥1, and A(n) is positive and odd.
Exact mathematical passage
\label{eq:classical} \operatorname{Col}(n)= \begin{cases}3n+1,&n\text{ odd},\\n/2,&n\text{ even},\end{cases} \qquad T(n)= \begin{cases}(3n+1)/2,&n\text{ odd},\\n/2,&n\text{ even}.\end{cases}
\widetilde\ThetaDefinition passage · 8 passages
Search aliases: widetildeΘ widetilde Theta
Definition passage: Pole annihilated packet · body.tex, line 235
[Pole annihilated packet] Let L=1/4-∂_u^2, and define widetildeΘ_t=LΘ_t =∑_ρ∈ Zρ(1-ρ) e^tz_ρ^2+uz_ρ.
Exact mathematical passage
\label{rh:tilde} \widetilde\Theta_t=L\Theta_t =\sum_{\rho\in\mathcal Z}\rho(1-\rho) e^{tz_\rho^2+uz_\rho}.
\widetilde\Theta_tDefinition passage · 4 passages
Search aliases: widetildeΘ_t widetilde Theta
Definition passage: Pole annihilated packet · body.tex, line 235
[Pole annihilated packet] Let L=1/4-∂_u^2, and define widetildeΘ_t=LΘ_t =∑_ρ∈ Zρ(1-ρ) e^tz_ρ^2+uz_ρ.
Exact mathematical passage
\label{rh:tilde} \widetilde\Theta_t=L\Theta_t =\sum_{\rho\in\mathcal Z}\rho(1-\rho) e^{tz_\rho^2+uz_\rho}.
\mathcal ZDefinition passage · 8 passages
Search aliases: Z mathcal
Definition passage: Zero heat packet · body.tex, line 66
[Zero heat packet] For a real heat parameter t>0 and a complex translation u, define Θ_t(u)=∑_ρ∈ Z exp(tz_ρ^2+uz_ρ).
Exact mathematical passage
\label{rh:theta} \Theta_t(u)=\sum_{\rho\in\mathcal Z} \exp\bigl(tz_\rho^2+uz_\rho\bigr).
z_\rho^2Definition passage · 9 passages
Search aliases: z_ρ^2 rho
Definition passage: Zero heat packet · body.tex, line 66
[Zero heat packet] For a real heat parameter t>0 and a complex translation u, define Θ_t(u)=∑_ρ∈ Z exp(tz_ρ^2+uz_ρ).
Exact mathematical passage
\label{rh:theta} \Theta_t(u)=\sum_{\rho\in\mathcal Z} \exp\bigl(tz_\rho^2+uz_\rho\bigr).
\nuDefinition passage · 5 passages
Search aliases: ν nu
Definition passage: Classical, shortcut, and odd maps · body.tex, line 27
n/2, · n even. For a positive odd integer n, let a(n)=ν_2(3n+1), A(n)=frac3n+12^a(n). Here ν_2(m) is the largest exponent a for which 2^a divides the positive integer m. Thus a(n)≥1, and A(n) is positive and odd.
Exact mathematical passage
\label{eq:oddmap} a(n)=\nu_2(3n+1),\qquad A(n)=\frac{3n+1}{2^{a(n)}}.
\nu_2Definition passage · 2 passages
Search aliases: ν_2 nu
Definition passage: Classical, shortcut, and odd maps · body.tex, line 27
n/2, · n even. For a positive odd integer n, let a(n)=ν_2(3n+1), A(n)=frac3n+12^a(n). Here ν_2(m) is the largest exponent a for which 2^a divides the positive integer m. Thus a(n)≥1, and A(n) is positive and odd.
Exact mathematical passage
\label{eq:oddmap} a(n)=\nu_2(3n+1),\qquad A(n)=\frac{3n+1}{2^{a(n)}}.
\rho_rDefinition passage · 13 passages
Search aliases: ρ_r rho
Definition passage: Coefficient ratio and deficit · body.tex, line 499
[Coefficient ratio and deficit] Define ρ_r=frac2^S_r3^r, d_r=lfloor rlog_2 3rfloor-S_r. Permanent coefficient noncrossing means ρ_r≤1 for every r≥0. Under that hypothesis d_r is a nonnegative integer.
Exact mathematical passage
\label{eq:rho} \rho_r=\frac{2^{S_r}}{3^r},\qquad d_r=\lfloor r\log_2 3\rfloor-S_r.
\ldotsDefinition passage · 4 passages
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Definition passage: Valuation word and offset · body.tex, line 55
[Valuation word and offset] A valuation word is a finite sequence w=(a_0,…,a_k-1) of positive integers. Define its partial depths and offsets by S_0=0, S_j=∑_i<ja_i, C_0=0, C_j+1=3C_j+2^S_j. Its total depth is S=S_k and its terminal offset is C=C_k. The word is realized by n if these are the first k valuations of the actual odd orbit from n.
Exact mathematical passage
w=(a_0,\ldots,a_{k-1})
\inDefinition passage · 13 passages
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Definition passage: Zero heat packet · body.tex, line 66
[Zero heat packet] For a real heat parameter t>0 and a complex translation u, define Θ_t(u)=∑_ρ∈ Z exp(tz_ρ^2+uz_ρ).
Exact mathematical passage
\label{rh:theta} \Theta_t(u)=\sum_{\rho\in\mathcal Z} \exp\bigl(tz_\rho^2+uz_\rho\bigr).
\sumDefinition passage · 42 passages
Search aliases: ∑ sum
Definition passage: Valuation word and offset · body.tex, line 56
[Valuation word and offset] A valuation word is a finite sequence w=(a_0,…,a_k-1) of positive integers. Define its partial depths and offsets by S_0=0, S_j=∑_i<ja_i, C_0=0, C_j+1=3C_j+2^S_j. Its total depth is S=S_k and its terminal offset is C=C_k. The word is realized by n if these are the first k valuations of the actual odd orbit from n.
Exact mathematical passage
\label{eq:offsetrec} S_0=0,\quad S_j=\sum_{i<j}a_i,\qquad C_0=0,\quad C_{j+1}=3C_j+2^{S_j}.
\sum_{i<j}Definition passage · 1 passage
Search aliases: ∑_i<j sum
Definition passage: Valuation word and offset · body.tex, line 56
[Valuation word and offset] A valuation word is a finite sequence w=(a_0,…,a_k-1) of positive integers. Define its partial depths and offsets by S_0=0, S_j=∑_i<ja_i, C_0=0, C_j+1=3C_j+2^S_j. Its total depth is S=S_k and its terminal offset is C=C_k. The word is realized by n if these are the first k valuations of the actual odd orbit from n.
Exact mathematical passage
\label{eq:offsetrec} S_0=0,\quad S_j=\sum_{i<j}a_i,\qquad C_0=0,\quad C_{j+1}=3C_j+2^{S_j}.
\sum_{\rho\in\mathcal Z}Definition passage · 5 passages
Search aliases: ∑_ρ∈ Z sum rho in mathcal
Definition passage: Zero heat packet · body.tex, line 66
[Zero heat packet] For a real heat parameter t>0 and a complex translation u, define Θ_t(u)=∑_ρ∈ Z exp(tz_ρ^2+uz_ρ).
Exact mathematical passage
\label{rh:theta} \Theta_t(u)=\sum_{\rho\in\mathcal Z} \exp\bigl(tz_\rho^2+uz_\rho\bigr).
\leDefinition passage · 75 passages
Search aliases: ≤ le
Definition passage: Coefficient ratio and deficit · body.tex, line 503
[Coefficient ratio and deficit] Define ρ_r=frac2^S_r3^r, d_r=lfloor rlog_2 3rfloor-S_r. Permanent coefficient noncrossing means ρ_r≤1 for every r≥0. Under that hypothesis d_r is a nonnegative integer.
Exact mathematical passage
\rho_r\le1
\geDefinition passage · 79 passages
Search aliases: ≥ ge
Definition passage: Classical, shortcut, and odd maps · body.tex, line 30
n/2, · n even. For a positive odd integer n, let a(n)=ν_2(3n+1), A(n)=frac3n+12^a(n). Here ν_2(m) is the largest exponent a for which 2^a divides the positive integer m. Thus a(n)≥1, and A(n) is positive and odd.
Exact mathematical passage
a(n)\ge1
\mathfrak AIntroduction passage · 2 passages
Search aliases: A mathfrak
Introduction passage · body.tex, line 184
) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv, P_t(u) · =frac12sqrtπ t ∑_n≥2fracΛ(n)sqrt n (e^-(u-log n)^2/(4t)+ e^-(u+log n)^2/(4t)). The Gaussian makes the prime series absolutely convergent. The completed explicit formula gives Θ_t(u)=2e^t/4cos…
Exact mathematical passage
\mathfrak A_t(u)&=\frac1{2\pi}\int_{\mathbb R} e^{-tv^2}\cos(uv)\Psi_\infty(v)\,dv, \label{rh:arch}\\ \mathfrak P_t(u)&=\frac1{2\sqrt{\pi t}} \sum_{n\ge2}\frac{\Lambda(n)}{\sqrt n} \left(e^{-(u-\log n)^2/(4t)}+ e^{-(u+\log n)^2/(4t)}\right). \label{rh:primepacket}
\mathfrak A_tIntroduction passage · 2 passages
Search aliases: A_t mathfrak
Introduction passage · body.tex, line 184
) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv, P_t(u) · =frac12sqrtπ t ∑_n≥2fracΛ(n)sqrt n (e^-(u-log n)^2/(4t)+ e^-(u+log n)^2/(4t)). The Gaussian makes the prime series absolutely convergent. The completed explicit formula gives Θ_t(u)=2e^t/4cos…
Exact mathematical passage
\mathfrak A_t(u)&=\frac1{2\pi}\int_{\mathbb R} e^{-tv^2}\cos(uv)\Psi_\infty(v)\,dv, \label{rh:arch}\\ \mathfrak P_t(u)&=\frac1{2\sqrt{\pi t}} \sum_{n\ge2}\frac{\Lambda(n)}{\sqrt n} \left(e^{-(u-\log n)^2/(4t)}+ e^{-(u+\log n)^2/(4t)}\right). \label{rh:primepacket}
c_wIntroduction passage · 2 passages
Search aliases: c_w
Introduction passage · body.tex, line 171
Composition and the descent budget. A realized word acts by the affine transformation F_w(n)=λ_w n+c_w, λ_w=frac3^k2^S, c_w=fracC_k2^S. The transfer law is evaluated only on the integer cylinder of the word. As a rational affine expression it is defined elsewhere, but there it need not describe an actual Collatz segment.
Exact mathematical passage
\label{eq:transfer} F_w(n)=\lambda_w n+c_w,\qquad \lambda_w=\frac{3^k}{2^S},\qquad c_w=\frac{C_k}{2^S}.
\cosIntroduction passage · 4 passages
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Introduction passage · body.tex, line 184
) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv, P_t(u) · =frac12sqrtπ t ∑_n≥2fracΛ(n)sqrt n (e^-(u-log n)^2/(4t)+ e^-(u+log n)^2/(4t)). The Gaussian makes the prime series absolutely convergent. The completed explicit formula gives Θ_t(u)=2e^t/4cos…
Exact mathematical passage
\mathfrak A_t(u)&=\frac1{2\pi}\int_{\mathbb R} e^{-tv^2}\cos(uv)\Psi_\infty(v)\,dv, \label{rh:arch}\\ \mathfrak P_t(u)&=\frac1{2\sqrt{\pi t}} \sum_{n\ge2}\frac{\Lambda(n)}{\sqrt n} \left(e^{-(u-\log n)^2/(4t)}+ e^{-(u+\log n)^2/(4t)}\right). \label{rh:primepacket}
EIntroduction passage · 6 passages
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Introduction passage · body.tex, line 552
There are binoms-15ℓ-1 positive words of total s. If W_ℓ denotes the finite total number of selected words, summing the cylinder counts gives #(E_ℓ∩[1,H])≤ε_ℓ H+W_ℓ, ε_ℓ=frac12∑_s=5ℓ^h binoms-15ℓ-12^-s. This count is arithmetic. It does not assume that orbit valuations are independent random variables.
Exact mathematical passage
\#(E_\ell\cap[1,H])\le\varepsilon_\ell H+W_\ell, \qquad \varepsilon_\ell=\frac12\sum_{s=5\ell}^{h} \binom{s-1}{5\ell-1}2^{-s}.
e^{-(u+\log n)^2/(4t)}Introduction passage · 1 passage
Search aliases: e^-(u+log n)^2/(4t) log
Introduction passage · body.tex, line 184
) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv, P_t(u) · =frac12sqrtπ t ∑_n≥2fracΛ(n)sqrt n (e^-(u-log n)^2/(4t)+ e^-(u+log n)^2/(4t)). The Gaussian makes the prime series absolutely convergent. The completed explicit formula gives Θ_t(u)=2e^t/4cos…
Exact mathematical passage
\mathfrak A_t(u)&=\frac1{2\pi}\int_{\mathbb R} e^{-tv^2}\cos(uv)\Psi_\infty(v)\,dv, \label{rh:arch}\\ \mathfrak P_t(u)&=\frac1{2\sqrt{\pi t}} \sum_{n\ge2}\frac{\Lambda(n)}{\sqrt n} \left(e^{-(u-\log n)^2/(4t)}+ e^{-(u+\log n)^2/(4t)}\right). \label{rh:primepacket}
e^{-(u-\log n)^2/(4t)}Introduction passage · 1 passage
Search aliases: e^-(u-log n)^2/(4t) log
Introduction passage · body.tex, line 184
) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv, P_t(u) · =frac12sqrtπ t ∑_n≥2fracΛ(n)sqrt n (e^-(u-log n)^2/(4t)+ e^-(u+log n)^2/(4t)). The Gaussian makes the prime series absolutely convergent. The completed explicit formula gives Θ_t(u)=2e^t/4cos…
Exact mathematical passage
\mathfrak A_t(u)&=\frac1{2\pi}\int_{\mathbb R} e^{-tv^2}\cos(uv)\Psi_\infty(v)\,dv, \label{rh:arch}\\ \mathfrak P_t(u)&=\frac1{2\sqrt{\pi t}} \sum_{n\ge2}\frac{\Lambda(n)}{\sqrt n} \left(e^{-(u-\log n)^2/(4t)}+ e^{-(u+\log n)^2/(4t)}\right). \label{rh:primepacket}
e^{-1/r^2}Introduction passage · 1 passage
Search aliases: e^-1/r^2
Introduction passage · body.tex, line 525
The active scale follows from the weight. For a real variable x>1, write y=x^-2. Then w_K(x)=y(1-y)^K, and differentiation in y puts its maximum at y=1/(K+1). The maximum of the continuously extended weight is therefore at x=sqrtK+1, close to M=sqrt K. For fixed r>0, as M tends to infinity through square roots of integers, M^2w_M^2(rM) =r^-2(1-frac1r^2M^2)^M^2 ⟶ r^-2e^-1/r^2. This is a scaling description of the positive weight. It says nothing about the sign distribution of μ(n) across its acti…
Exact mathematical passage
M^2w_{M^2}(rM) =r^{-2}\left(1-\frac1{r^2M^2}\right)^{M^2} \longrightarrow r^{-2}e^{-1/r^2}.
e^{-tv^2}Introduction passage · 4 passages
Search aliases: e^-tv^2
Introduction passage · body.tex, line 184
) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv, P_t(u) · =frac12sqrtπ t ∑_n≥2fracΛ(n)sqrt n (e^-(u-log n)^2/(4t)+ e^-(u+log n)^2/(4t)). The Gaussian makes the prime series absolutely convergent. The completed explicit formula gives Θ_t(u)=2e^t/4cos…
Exact mathematical passage
\mathfrak A_t(u)&=\frac1{2\pi}\int_{\mathbb R} e^{-tv^2}\cos(uv)\Psi_\infty(v)\,dv, \label{rh:arch}\\ \mathfrak P_t(u)&=\frac1{2\sqrt{\pi t}} \sum_{n\ge2}\frac{\Lambda(n)}{\sqrt n} \left(e^{-(u-\log n)^2/(4t)}+ e^{-(u+\log n)^2/(4t)}\right). \label{rh:primepacket}
E_\ellIntroduction passage · 3 passages
Search aliases: E_ℓ ell
Introduction passage · body.tex, line 552
There are binoms-15ℓ-1 positive words of total s. If W_ℓ denotes the finite total number of selected words, summing the cylinder counts gives #(E_ℓ∩[1,H])≤ε_ℓ H+W_ℓ, ε_ℓ=frac12∑_s=5ℓ^h binoms-15ℓ-12^-s. This count is arithmetic. It does not assume that orbit valuations are independent random variables.
Exact mathematical passage
\#(E_\ell\cap[1,H])\le\varepsilon_\ell H+W_\ell, \qquad \varepsilon_\ell=\frac12\sum_{s=5\ell}^{h} \binom{s-1}{5\ell-1}2^{-s}.
FIntroduction passage · 9 passages
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Introduction passage · body.tex, line 171
Composition and the descent budget. A realized word acts by the affine transformation F_w(n)=λ_w n+c_w, λ_w=frac3^k2^S, c_w=fracC_k2^S. The transfer law is evaluated only on the integer cylinder of the word. As a rational affine expression it is defined elsewhere, but there it need not describe an actual Collatz segment.
Exact mathematical passage
\label{eq:transfer} F_w(n)=\lambda_w n+c_w,\qquad \lambda_w=\frac{3^k}{2^S},\qquad c_w=\frac{C_k}{2^S}.
F_wIntroduction passage · 3 passages
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Introduction passage · body.tex, line 171
Composition and the descent budget. A realized word acts by the affine transformation F_w(n)=λ_w n+c_w, λ_w=frac3^k2^S, c_w=fracC_k2^S. The transfer law is evaluated only on the integer cylinder of the word. As a rational affine expression it is defined elsewhere, but there it need not describe an actual Collatz segment.
Exact mathematical passage
\label{eq:transfer} F_w(n)=\lambda_w n+c_w,\qquad \lambda_w=\frac{3^k}{2^S},\qquad c_w=\frac{C_k}{2^S}.
HIntroduction passage · 25 passages
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Introduction passage · body.tex, line 552
There are binoms-15ℓ-1 positive words of total s. If W_ℓ denotes the finite total number of selected words, summing the cylinder counts gives #(E_ℓ∩[1,H])≤ε_ℓ H+W_ℓ, ε_ℓ=frac12∑_s=5ℓ^h binoms-15ℓ-12^-s. This count is arithmetic. It does not assume that orbit valuations are independent random variables.
Exact mathematical passage
\#(E_\ell\cap[1,H])\le\varepsilon_\ell H+W_\ell, \qquad \varepsilon_\ell=\frac12\sum_{s=5\ell}^{h} \binom{s-1}{5\ell-1}2^{-s}.
hIntroduction passage · 17 passages
Search aliases: h
Introduction passage · body.tex, line 552
There are binoms-15ℓ-1 positive words of total s. If W_ℓ denotes the finite total number of selected words, summing the cylinder counts gives #(E_ℓ∩[1,H])≤ε_ℓ H+W_ℓ, ε_ℓ=frac12∑_s=5ℓ^h binoms-15ℓ-12^-s. This count is arithmetic. It does not assume that orbit valuations are independent random variables.
Exact mathematical passage
\#(E_\ell\cap[1,H])\le\varepsilon_\ell H+W_\ell, \qquad \varepsilon_\ell=\frac12\sum_{s=5\ell}^{h} \binom{s-1}{5\ell-1}2^{-s}.
M^2Introduction passage · 13 passages
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Introduction passage · body.tex, line 525
The active scale follows from the weight. For a real variable x>1, write y=x^-2. Then w_K(x)=y(1-y)^K, and differentiation in y puts its maximum at y=1/(K+1). The maximum of the continuously extended weight is therefore at x=sqrtK+1, close to M=sqrt K. For fixed r>0, as M tends to infinity through square roots of integers, M^2w_M^2(rM) =r^-2(1-frac1r^2M^2)^M^2 ⟶ r^-2e^-1/r^2. This is a scaling description of the positive weight. It says nothing about the sign distribution of μ(n) across its acti…
Exact mathematical passage
M^2w_{M^2}(rM) =r^{-2}\left(1-\frac1{r^2M^2}\right)^{M^2} \longrightarrow r^{-2}e^{-1/r^2}.
\mathfrak PIntroduction passage · 3 passages
Search aliases: P mathfrak
Introduction passage · body.tex, line 184
) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv, P_t(u) · =frac12sqrtπ t ∑_n≥2fracΛ(n)sqrt n (e^-(u-log n)^2/(4t)+ e^-(u+log n)^2/(4t)). The Gaussian makes the prime series absolutely convergent. The completed explicit formula gives Θ_t(u)=2e^t/4cos…
Exact mathematical passage
\mathfrak A_t(u)&=\frac1{2\pi}\int_{\mathbb R} e^{-tv^2}\cos(uv)\Psi_\infty(v)\,dv, \label{rh:arch}\\ \mathfrak P_t(u)&=\frac1{2\sqrt{\pi t}} \sum_{n\ge2}\frac{\Lambda(n)}{\sqrt n} \left(e^{-(u-\log n)^2/(4t)}+ e^{-(u+\log n)^2/(4t)}\right). \label{rh:primepacket}
pIntroduction passage · 23 passages
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Introduction passage · body.tex, line 177
Let Λ(n) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv,
Exact mathematical passage
\log p
p^{-\rho}Introduction passage · 1 passage
Search aliases: p^-ρ rho
Introduction passage · body.tex, line 756
First, an exact prime high pass followed by exact coprime reconstruction preserves a power mode. If a mode has local multiplier 1-p^-ρ, reconstruction contributes the reciprocal multiplier whenever it is defined. Their product is one. The high pass alone can look smaller, but exact reconstruction restores the removed factor. The historical Lean module HighpassReconstructionNoGo.lean certifies this algebraic cancellation. It does not estimate a prime sum.
Exact mathematical passage
1-p^{-\rho}
p^kIntroduction passage · 3 passages
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Introduction passage · body.tex, line 177
Let Λ(n) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv,
Exact mathematical passage
n=p^k
\mathfrak P_tIntroduction passage · 3 passages
Search aliases: P_t mathfrak
Introduction passage · body.tex, line 184
) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv, P_t(u) · =frac12sqrtπ t ∑_n≥2fracΛ(n)sqrt n (e^-(u-log n)^2/(4t)+ e^-(u+log n)^2/(4t)). The Gaussian makes the prime series absolutely convergent. The completed explicit formula gives Θ_t(u)=2e^t/4cos…
Exact mathematical passage
\mathfrak A_t(u)&=\frac1{2\pi}\int_{\mathbb R} e^{-tv^2}\cos(uv)\Psi_\infty(v)\,dv, \label{rh:arch}\\ \mathfrak P_t(u)&=\frac1{2\sqrt{\pi t}} \sum_{n\ge2}\frac{\Lambda(n)}{\sqrt n} \left(e^{-(u-\log n)^2/(4t)}+ e^{-(u+\log n)^2/(4t)}\right). \label{rh:primepacket}
\mathbb RIntroduction passage · 8 passages
Search aliases: R mathbb
Introduction passage · body.tex, line 178
Let Λ(n) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv,
Exact mathematical passage
\label{rh:psiinf} \Psi_\infty(v)=-\log\pi+ \Re\psi\left(\frac14+\frac{iv}{2}\right) \qquad(v\in\mathbb R).
r^{-2}Introduction passage · 1 passage
Search aliases: r^-2
Introduction passage · body.tex, line 525
The active scale follows from the weight. For a real variable x>1, write y=x^-2. Then w_K(x)=y(1-y)^K, and differentiation in y puts its maximum at y=1/(K+1). The maximum of the continuously extended weight is therefore at x=sqrtK+1, close to M=sqrt K. For fixed r>0, as M tends to infinity through square roots of integers, M^2w_M^2(rM) =r^-2(1-frac1r^2M^2)^M^2 ⟶ r^-2e^-1/r^2. This is a scaling description of the positive weight. It says nothing about the sign distribution of μ(n) across its acti…
Exact mathematical passage
M^2w_{M^2}(rM) =r^{-2}\left(1-\frac1{r^2M^2}\right)^{M^2} \longrightarrow r^{-2}e^{-1/r^2}.
r^2Introduction passage · 1 passage
Search aliases: r^2
Introduction passage · body.tex, line 525
The active scale follows from the weight. For a real variable x>1, write y=x^-2. Then w_K(x)=y(1-y)^K, and differentiation in y puts its maximum at y=1/(K+1). The maximum of the continuously extended weight is therefore at x=sqrtK+1, close to M=sqrt K. For fixed r>0, as M tends to infinity through square roots of integers, M^2w_M^2(rM) =r^-2(1-frac1r^2M^2)^M^2 ⟶ r^-2e^-1/r^2. This is a scaling description of the positive weight. It says nothing about the sign distribution of μ(n) across its acti…
Exact mathematical passage
M^2w_{M^2}(rM) =r^{-2}\left(1-\frac1{r^2M^2}\right)^{M^2} \longrightarrow r^{-2}e^{-1/r^2}.
\ReIntroduction passage · 15 passages
Search aliases: Re
Introduction passage · body.tex, line 178
Let Λ(n) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv,
Exact mathematical passage
\label{rh:psiinf} \Psi_\infty(v)=-\log\pi+ \Re\psi\left(\frac14+\frac{iv}{2}\right) \qquad(v\in\mathbb R).
sIntroduction passage · 58 passages
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Introduction passage · body.tex, line 551
There are binoms-15ℓ-1 positive words of total s. If W_ℓ denotes the finite total number of selected words, summing the cylinder counts gives #(E_ℓ∩[1,H])≤ε_ℓ H+W_ℓ, ε_ℓ=frac12∑_s=5ℓ^h binoms-15ℓ-12^-s. This count is arithmetic. It does not assume that orbit valuations are independent random variables.
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\binom{s-1}{5\ell-1}
vIntroduction passage · 23 passages
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Introduction passage · body.tex, line 178
Let Λ(n) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv,
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\label{rh:psiinf} \Psi_\infty(v)=-\log\pi+ \Re\psi\left(\frac14+\frac{iv}{2}\right) \qquad(v\in\mathbb R).
v^2Introduction passage · 5 passages
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Introduction passage · body.tex, line 184
) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv, P_t(u) · =frac12sqrtπ t ∑_n≥2fracΛ(n)sqrt n (e^-(u-log n)^2/(4t)+ e^-(u+log n)^2/(4t)). The Gaussian makes the prime series absolutely convergent. The completed explicit formula gives Θ_t(u)=2e^t/4cos…
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\mathfrak A_t(u)&=\frac1{2\pi}\int_{\mathbb R} e^{-tv^2}\cos(uv)\Psi_\infty(v)\,dv, \label{rh:arch}\\ \mathfrak P_t(u)&=\frac1{2\sqrt{\pi t}} \sum_{n\ge2}\frac{\Lambda(n)}{\sqrt n} \left(e^{-(u-\log n)^2/(4t)}+ e^{-(u+\log n)^2/(4t)}\right). \label{rh:primepacket}
WIntroduction passage · 6 passages
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Introduction passage · body.tex, line 551
There are binoms-15ℓ-1 positive words of total s. If W_ℓ denotes the finite total number of selected words, summing the cylinder counts gives #(E_ℓ∩[1,H])≤ε_ℓ H+W_ℓ, ε_ℓ=frac12∑_s=5ℓ^h binoms-15ℓ-12^-s. This count is arithmetic. It does not assume that orbit valuations are independent random variables.
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W_\ell
w_{M^2}Introduction passage · 5 passages
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Introduction passage · body.tex, line 525
The active scale follows from the weight. For a real variable x>1, write y=x^-2. Then w_K(x)=y(1-y)^K, and differentiation in y puts its maximum at y=1/(K+1). The maximum of the continuously extended weight is therefore at x=sqrtK+1, close to M=sqrt K. For fixed r>0, as M tends to infinity through square roots of integers, M^2w_M^2(rM) =r^-2(1-frac1r^2M^2)^M^2 ⟶ r^-2e^-1/r^2. This is a scaling description of the positive weight. It says nothing about the sign distribution of μ(n) across its acti…
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M^2w_{M^2}(rM) =r^{-2}\left(1-\frac1{r^2M^2}\right)^{M^2} \longrightarrow r^{-2}e^{-1/r^2}.
W_\ellIntroduction passage · 6 passages
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Introduction passage · body.tex, line 551
There are binoms-15ℓ-1 positive words of total s. If W_ℓ denotes the finite total number of selected words, summing the cylinder counts gives #(E_ℓ∩[1,H])≤ε_ℓ H+W_ℓ, ε_ℓ=frac12∑_s=5ℓ^h binoms-15ℓ-12^-s. This count is arithmetic. It does not assume that orbit valuations are independent random variables.
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W_\ell
xIntroduction passage · 51 passages
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Introduction passage · body.tex, line 524
The active scale follows from the weight. For a real variable x>1, write y=x^-2. Then w_K(x)=y(1-y)^K, and differentiation in y puts its maximum at y=1/(K+1). The maximum of the continuously extended weight is therefore at x=sqrtK+1, close to M=sqrt K. For fixed r>0, as M tends to infinity through square roots of integers, M^2w_M^2(rM) =r^-2(1-frac1r^2M^2)^M^2 ⟶ r^-2e^-1/r^2. This is a scaling description of the positive weight. It says nothing about the sign distribution of μ(n) across its acti…
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x>1
x^{-2}Introduction passage · 1 passage
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Introduction passage · body.tex, line 524
The active scale follows from the weight. For a real variable x>1, write y=x^-2. Then w_K(x)=y(1-y)^K, and differentiation in y puts its maximum at y=1/(K+1). The maximum of the continuously extended weight is therefore at x=sqrtK+1, close to M=sqrt K. For fixed r>0, as M tends to infinity through square roots of integers, M^2w_M^2(rM) =r^-2(1-frac1r^2M^2)^M^2 ⟶ r^-2e^-1/r^2. This is a scaling description of the positive weight. It says nothing about the sign distribution of μ(n) across its acti…
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y=x^{-2}
yIntroduction passage · 11 passages
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Introduction passage · body.tex, line 524
The active scale follows from the weight. For a real variable x>1, write y=x^-2. Then w_K(x)=y(1-y)^K, and differentiation in y puts its maximum at y=1/(K+1). The maximum of the continuously extended weight is therefore at x=sqrtK+1, close to M=sqrt K. For fixed r>0, as M tends to infinity through square roots of integers, M^2w_M^2(rM) =r^-2(1-frac1r^2M^2)^M^2 ⟶ r^-2e^-1/r^2. This is a scaling description of the positive weight. It says nothing about the sign distribution of μ(n) across its acti…
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y=x^{-2}
\GammaIntroduction passage · 3 passages
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Introduction passage · body.tex, line 177
Let Λ(n) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv,
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\psi=\Gamma'/\Gamma
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Introduction passage · body.tex, line 177
Let Λ(n) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv,
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\psi=\Gamma'/\Gamma
\varepsilonIntroduction passage · 30 passages
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Introduction passage · body.tex, line 552
There are binoms-15ℓ-1 positive words of total s. If W_ℓ denotes the finite total number of selected words, summing the cylinder counts gives #(E_ℓ∩[1,H])≤ε_ℓ H+W_ℓ, ε_ℓ=frac12∑_s=5ℓ^h binoms-15ℓ-12^-s. This count is arithmetic. It does not assume that orbit valuations are independent random variables.
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\#(E_\ell\cap[1,H])\le\varepsilon_\ell H+W_\ell, \qquad \varepsilon_\ell=\frac12\sum_{s=5\ell}^{h} \binom{s-1}{5\ell-1}2^{-s}.
\varepsilon_\ellIntroduction passage · 8 passages
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Introduction passage · body.tex, line 552
There are binoms-15ℓ-1 positive words of total s. If W_ℓ denotes the finite total number of selected words, summing the cylinder counts gives #(E_ℓ∩[1,H])≤ε_ℓ H+W_ℓ, ε_ℓ=frac12∑_s=5ℓ^h binoms-15ℓ-12^-s. This count is arithmetic. It does not assume that orbit valuations are independent random variables.
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\#(E_\ell\cap[1,H])\le\varepsilon_\ell H+W_\ell, \qquad \varepsilon_\ell=\frac12\sum_{s=5\ell}^{h} \binom{s-1}{5\ell-1}2^{-s}.
\LambdaIntroduction passage · 12 passages
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Introduction passage · body.tex, line 177
Let Λ(n) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv,
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\Lambda(n)
\lambda_wIntroduction passage · 2 passages
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Introduction passage · body.tex, line 171
Composition and the descent budget. A realized word acts by the affine transformation F_w(n)=λ_w n+c_w, λ_w=frac3^k2^S, c_w=fracC_k2^S. The transfer law is evaluated only on the integer cylinder of the word. As a rational affine expression it is defined elsewhere, but there it need not describe an actual Collatz segment.
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\label{eq:transfer} F_w(n)=\lambda_w n+c_w,\qquad \lambda_w=\frac{3^k}{2^S},\qquad c_w=\frac{C_k}{2^S}.
\piIntroduction passage · 17 passages
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Introduction passage · body.tex, line 178
Let Λ(n) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv,
Exact mathematical passage
\label{rh:psiinf} \Psi_\infty(v)=-\log\pi+ \Re\psi\left(\frac14+\frac{iv}{2}\right) \qquad(v\in\mathbb R).
\PsiIntroduction passage · 5 passages
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Introduction passage · body.tex, line 178
Let Λ(n) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv,
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\label{rh:psiinf} \Psi_\infty(v)=-\log\pi+ \Re\psi\left(\frac14+\frac{iv}{2}\right) \qquad(v\in\mathbb R).
\psiIntroduction passage · 7 passages
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Introduction passage · body.tex, line 177
Let Λ(n) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv,
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\psi=\Gamma'/\Gamma
\Psi_\inftyIntroduction passage · 5 passages
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Introduction passage · body.tex, line 178
Let Λ(n) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv,
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\label{rh:psiinf} \Psi_\infty(v)=-\log\pi+ \Re\psi\left(\frac14+\frac{iv}{2}\right) \qquad(v\in\mathbb R).
\ellIntroduction passage · 43 passages
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Introduction passage · body.tex, line 551
There are binoms-15ℓ-1 positive words of total s. If W_ℓ denotes the finite total number of selected words, summing the cylinder counts gives #(E_ℓ∩[1,H])≤ε_ℓ H+W_ℓ, ε_ℓ=frac12∑_s=5ℓ^h binoms-15ℓ-12^-s. This count is arithmetic. It does not assume that orbit valuations are independent random variables.
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\binom{s-1}{5\ell-1}
\sum_{n\ge2}Introduction passage · 3 passages
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Introduction passage · body.tex, line 184
) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv, P_t(u) · =frac12sqrtπ t ∑_n≥2fracΛ(n)sqrt n (e^-(u-log n)^2/(4t)+ e^-(u+log n)^2/(4t)). The Gaussian makes the prime series absolutely convergent. The completed explicit formula gives Θ_t(u)=2e^t/4cos…
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\mathfrak A_t(u)&=\frac1{2\pi}\int_{\mathbb R} e^{-tv^2}\cos(uv)\Psi_\infty(v)\,dv, \label{rh:arch}\\ \mathfrak P_t(u)&=\frac1{2\sqrt{\pi t}} \sum_{n\ge2}\frac{\Lambda(n)}{\sqrt n} \left(e^{-(u-\log n)^2/(4t)}+ e^{-(u+\log n)^2/(4t)}\right). \label{rh:primepacket}
\sum_{s=5\ell}^{h}Introduction passage · 1 passage
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Introduction passage · body.tex, line 552
There are binoms-15ℓ-1 positive words of total s. If W_ℓ denotes the finite total number of selected words, summing the cylinder counts gives #(E_ℓ∩[1,H])≤ε_ℓ H+W_ℓ, ε_ℓ=frac12∑_s=5ℓ^h binoms-15ℓ-12^-s. This count is arithmetic. It does not assume that orbit valuations are independent random variables.
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\#(E_\ell\cap[1,H])\le\varepsilon_\ell H+W_\ell, \qquad \varepsilon_\ell=\frac12\sum_{s=5\ell}^{h} \binom{s-1}{5\ell-1}2^{-s}.
\inftyIntroduction passage · 17 passages
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Introduction passage · body.tex, line 178
Let Λ(n) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv,
Exact mathematical passage
\label{rh:psiinf} \Psi_\infty(v)=-\log\pi+ \Re\psi\left(\frac14+\frac{iv}{2}\right) \qquad(v\in\mathbb R).
\capIntroduction passage · 1 passage
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Introduction passage · body.tex, line 552
There are binoms-15ℓ-1 positive words of total s. If W_ℓ denotes the finite total number of selected words, summing the cylinder counts gives #(E_ℓ∩[1,H])≤ε_ℓ H+W_ℓ, ε_ℓ=frac12∑_s=5ℓ^h binoms-15ℓ-12^-s. This count is arithmetic. It does not assume that orbit valuations are independent random variables.
Exact mathematical passage
\#(E_\ell\cap[1,H])\le\varepsilon_\ell H+W_\ell, \qquad \varepsilon_\ell=\frac12\sum_{s=5\ell}^{h} \binom{s-1}{5\ell-1}2^{-s}.
\intIntroduction passage · 14 passages
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Introduction passage · body.tex, line 184
) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv, P_t(u) · =frac12sqrtπ t ∑_n≥2fracΛ(n)sqrt n (e^-(u-log n)^2/(4t)+ e^-(u+log n)^2/(4t)). The Gaussian makes the prime series absolutely convergent. The completed explicit formula gives Θ_t(u)=2e^t/4cos…
Exact mathematical passage
\mathfrak A_t(u)&=\frac1{2\pi}\int_{\mathbb R} e^{-tv^2}\cos(uv)\Psi_\infty(v)\,dv, \label{rh:arch}\\ \mathfrak P_t(u)&=\frac1{2\sqrt{\pi t}} \sum_{n\ge2}\frac{\Lambda(n)}{\sqrt n} \left(e^{-(u-\log n)^2/(4t)}+ e^{-(u+\log n)^2/(4t)}\right). \label{rh:primepacket}
\int_{\mathbb R}Introduction passage · 5 passages
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Introduction passage · body.tex, line 184
) denote the von Mangoldt function: it is log p when n=p^k for a prime p and an integer k≥1, and it is zero otherwise. Let ψ=Γ'/Γ be the digamma function, and set Ψ_∞(v)=-logπ+ Reψ(frac14+fraciv2) (v∈ R). Define the archimedean term and prime packet by A_t(u) · =frac12π∫_ R e^-tv^2cos(uv)Ψ_∞(v) dv, P_t(u) · =frac12sqrtπ t ∑_n≥2fracΛ(n)sqrt n (e^-(u-log n)^2/(4t)+ e^-(u+log n)^2/(4t)). The Gaussian makes the prime series absolutely convergent. The completed explicit formula gives Θ_t(u)=2e^t/4cos…
Exact mathematical passage
\mathfrak A_t(u)&=\frac1{2\pi}\int_{\mathbb R} e^{-tv^2}\cos(uv)\Psi_\infty(v)\,dv, \label{rh:arch}\\ \mathfrak P_t(u)&=\frac1{2\sqrt{\pi t}} \sum_{n\ge2}\frac{\Lambda(n)}{\sqrt n} \left(e^{-(u-\log n)^2/(4t)}+ e^{-(u+\log n)^2/(4t)}\right). \label{rh:primepacket}
\longrightarrowIntroduction passage · 1 passage
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Introduction passage · body.tex, line 525
The active scale follows from the weight. For a real variable x>1, write y=x^-2. Then w_K(x)=y(1-y)^K, and differentiation in y puts its maximum at y=1/(K+1). The maximum of the continuously extended weight is therefore at x=sqrtK+1, close to M=sqrt K. For fixed r>0, as M tends to infinity through square roots of integers, M^2w_M^2(rM) =r^-2(1-frac1r^2M^2)^M^2 ⟶ r^-2e^-1/r^2. This is a scaling description of the positive weight. It says nothing about the sign distribution of μ(n) across its acti…
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M^2w_{M^2}(rM) =r^{-2}\left(1-\frac1{r^2M^2}\right)^{M^2} \longrightarrow r^{-2}e^{-1/r^2}.
*Use in source · 3 passages
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Use in source · body.tex, line 216
The threshold as a fixed point. When G=2^S-3^k>0, the affine transfer has one real fixed point n_* = fracC_kG=fracc_w1-λ_w. For a real input above n_*, the affine expression falls. For an input below n_*, it rises. Equality gives a fixed point of the affine expression. These are statements about that expression before its integer cylinder is imposed. An actual periodic state would additionally have to be positive odd and realize the word.
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n_* = \frac{C_k}{G}=\frac{c_w}{1-\lambda_w}.
A^2Use in source · 1 passage
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Use in source · body.tex, line 307
For example, the word (1,2) has S_2=3 and C_2=5, so 8·31=9·27+5. The corresponding cylinder is n≡11pmod16. Writing n=16t+11 gives A(n)=24t+17, A^2(n)=18t+13. Every member of this progression grows over the two steps, even though the second step falls from 24t+17 to 18t+13. For t=1, the local fall is 41 to 31, while both states remain above 27. These identities are present in the recovered Lean source [source reference].
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A(n)=24t+17,\qquad A^2(n)=18t+13.
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Use in source · body.tex, line 481
· ⟼31104t+8261⟼5832t+1549. The valuations are 1,1,1,1,1,4. The next numerator is 3(5832t+1549)+1=2^3(2187t+581). Therefore A^7(n)=oddPart(2187t+581) ≤2187t+581<4096t+1087. The final valuation may exceed 3 for some parameters. The inequality remains valid because removing further factors of two only reduces the positive quotient. This proof is represented in the recovered deep residue certificate source.
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A^7(n)=\operatorname{oddPart}(2187t+581) \le2187t+581<4096t+1087.
A^jUse in source · 1 passage
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Use in source · body.tex, line 363
ntheorem[Compressed run exit] Suppose j≥0, q≥1, b≥0, and m is positive odd, with 3^j+1q=2^b m+1. For n=2^j+1q-1, the exact exit is A^j+1(n)=m. If m<n, then j+1 is the exact first strict descent time. Lemma~[source reference] gives A^j(n)=2·3^j q-1. Its next odd numerator is 3(2·3^j q-1)+1=2(3^j+1q-1)=2^b+1m. Oddness of m proves the exact final valuation b+1 and the exit. Earlier states are at least n by the lemma.
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A^j(n)=2\cdot3^j q-1
A^{j+1}Use in source · 3 passages
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Use in source · body.tex, line 357
[Compressed run exit] Suppose j≥0, q≥1, b≥0, and m is positive odd, with 3^j+1q=2^b m+1. For n=2^j+1q-1, the exact exit is A^j+1(n)=m. If m<n, then j+1 is the exact first strict descent time. Lemma~[source reference] gives A^j(n)=2·3^j q-1. Its next odd numerator is 3(2·3^j q-1)+1=2(3^j+1q-1)=2^b+1m. Oddness of m proves the exact final valuation b+1 and the exit. Earlier states are at least n by the lemma.
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\label{eq:exit} A^{j+1}(n)=m.
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Use in source · body.tex, line 40
[Collatz] For every positive odd integer n, there exists r≥0 such that A^r(n)=1.
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A^r(n)=1
A^{r+s}Use in source · 1 passage
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Use in source · body.tex, line 268
abeleq:universal for every odd n>1 there exists r≥1 with A^r(n)<n. Convergence to 1 implies descent for each n>1. For the converse, use strong induction on positive odd integers. The case n=1 is immediate. For n>1, choose a descendant m=A^r(n)<n. It is positive and odd, so the induction hypothesis gives A^s(m)=1. Hence A^r+s(n)=1.
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A^{r+s}(n)=1
A^sUse in source · 1 passage
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Use in source · body.tex, line 268
for every odd n>1 there exists r≥1 with A^r(n)<n. Convergence to 1 implies descent for each n>1. For the converse, use strong induction on positive odd integers. The case n=1 is immediate. For n>1, choose a descendant m=A^r(n)<n. It is positive and odd, so the induction hypothesis gives A^s(m)=1. Hence A^r+s(n)=1.
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A^s(m)=1
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Use in source · body.tex, line 333
[A run of unit valuations] For integers j≥0, q≥1, and 0≤ t≤ j, put n=2^j+1q-1. Then A^t(n)=2^j+1-t3^t q-1. Every one of the first j valuations is exactly 1. Moreover A^t(n)≥ n for 0≤ t≤ j. For t<j, the displayed state has the form 4u-1 with u>0. Thus 3(4u-1)+1=2(6u-1), whose quotient is odd. This proves the next state and valuation by induction. The comparison with the source follows from 3^t≥2^t.
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\label{eq:run} A^t(n)=2^{j+1-t}3^t q-1.
a_1Use in source · 3 passages
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Use in source · body.tex, line 84
One can see the offset being created before writing the general formula. The first step is 2^a_0x_1=3n+1. Multiplying the second step by 2^a_0 gives 2^a_0+a_1x_2=3(3n+1)+2^a_0 =9n+3+2^a_0. The first addition has been multiplied by 3. The second addition has been multiplied by the dyadic depth already removed. After a third step the offset is 9+3·2^a_0+2^a_0+a_1. Every later addition enters at its own prefix depth and receives one factor of 3 for each later odd multiplication. This explains each …
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2^{a_0+a_1}x_2=3(3n+1)+2^{a_0} =9n+3+2^{a_0}.
a_jUse in source · 2 passages
Search aliases: a_j
Use in source · body.tex, line 117
r of two. Reducing [source reference] modulo 2^S_j+1 gives 3^k-j(3^j n+C_j)+2^S_jD_j,k≡0pmod2^S_j+1. Every odd multiple of 2^S_j has the same residue modulo 2^S_j+1. Since 3^k-j is invertible, it follows that 3^j n+C_j≡2^S_jpmod2^S_j+1. Thus x_j is integral and odd. It is positive because n>0 and C_j≥0. Finally the recurrence gives 3x_j+1=2^a_jx_j+1. Oddness of x_j+1 makes a_j the exact valuation. These candidates are therefore the actual orbit. Invertibility of 3^k yields a unique residue class…
Exact mathematical passage
3x_j+1=2^{a_j}x_{j+1}
a_rUse in source · 5 passages
Search aliases: a_r
Use in source · body.tex, line 49
Exact valuation packets. Fix a positive odd seed n, and write x_r=A^r(n). Each step has the exact equation 2^a_rx_r+1=3x_r+1, a_r=a(x_r). The exponent records how much dyadic information is removed at that step. It is determined by the actual state, rather than chosen independently.
Exact mathematical passage
\label{eq:onestep} 2^{a_r}x_{r+1}=3x_r+1,\qquad a_r=a(x_r).
a_{r-1}Use in source · 1 passage
Search aliases: a_r-1
Use in source · body.tex, line 287
[htbp]centering captionExact odd packet coordinates for the first ten steps from 27. rrrrr r · x_r · a_r-1 · S_r · C_r
Exact mathematical passage
a_{r-1}
\asympUse in source · 1 passage
Search aliases: asymp
Use in source · body.tex, line 605
The cube's shape is also scale dependent. At fixed K, large m makes every factor (1-1/(bm)^2)^K close to one, so its leading coefficient is ∏_pmid Q(1-p^-2). At the active scale masympsqrt K, the different divisors sample different parts of the weight profile. This difference can create local cancellation within a cube. It does not remove the remaining sign μ(m), which is the quantity the signed criterion still asks us to control.
Exact mathematical passage
m\asymp\sqrt K
BUse in source · 13 passages
Search aliases: B
Use in source · body.tex, line 526
[Injectivity] A permanently noncrossing positive odd orbit never repeats a state. Suppose x_u=x_v with v>u. Put k=v-u and B=S_v-S_u. The affine law for this positive length block gives (2^B-3^k)x_u=C_u,v>0. Hence 2^B/3^k>1. Determinism makes the same states and valuations repeat, so ρ_u+hk=ρ_u(2^B/3^k)^h. This exceeds 1 for sufficiently large h, contrary to permanent noncrossing.
Exact mathematical passage
B=S_v-S_u
bUse in source · 30 passages
Search aliases: b
Use in source · body.tex, line 258
One can also retain a weaker certificate when exact final valuation is unnecessary. If the last numerator is known to equal 2^b q with q>0 integral, then the next odd state is at most q. Exact equality requires q odd. A descent proof that only uses an upper bound can accept both parities. Example~[source reference] exploits this distinction. It obtains a uniform falling estimate without claiming one final exact word for every member of its coarse progression.
Exact mathematical passage
2^b q
b^{-4}Use in source · 1 passage
Search aliases: b^-4
Use in source · body.tex, line 651
The limiting divisor sum is ∑_bmid Qfracμ(b)b^2 =∏_pmid Q(1-p^-2)≥ζ(2)^-1. For 0≤ x≤1, Bernoulli's inequality gives 0≤1-(1-x)^K≤ Kx. With K=M^2, the error in replacing each power by one is at most fracM^2m^2∑_bmid Qb^-4 ≤fracζ(4)R^2. Multiplying the resulting lower bound by m^-2 proves the lemma.
Exact mathematical passage
\frac{M^2}{m^2}\sum_{b\mid Q}b^{-4} \le\frac{\zeta(4)}{R^2}.
B^2Use in source · 2 passages
Search aliases: B^2
Use in source · body.tex, line 92
It helps to make the normal convergence argument quantitative enough to see which operations it permits. Write z=x+iγ, where |x|<1/2, and u=a+ib. Then Re(tz^2+uz)=t(x^2-γ^2)+ax-bγ. If t∈[t_0,T] and |a|,|b|≤ B, this is at most T/4+B/2-t_0γ^2+B|γ| ≤ T/4+B/2+fracB^22t_0-fract_02γ^2. The last step is the elementary inequality B|γ|≤(t_0/2)γ^2+B^2/(2t_0). Thus a compact set of heat times and translations has one Gaussian majorant. A finite number of derivatives only inserts a polynomial in z. Such a p…
Exact mathematical passage
T/4+B/2-t_0\gamma^2+B|\gamma| \le T/4+B/2+\frac{B^2}{2t_0}-\frac{t_0}2\gamma^2.
b^2Use in source · 1 passage
Search aliases: b^2
Use in source · body.tex, line 645
The limiting divisor sum is ∑_bmid Qfracμ(b)b^2 =∏_pmid Q(1-p^-2)≥ζ(2)^-1. For 0≤ x≤1, Bernoulli's inequality gives 0≤1-(1-x)^K≤ Kx. With K=M^2, the error in replacing each power by one is at most fracM^2m^2∑_bmid Qb^-4 ≤fracζ(4)R^2. Multiplying the resulting lower bound by m^-2 proves the lemma.
Exact mathematical passage
\sum_{b\mid Q}\frac{\mu(b)}{b^2} =\prod_{p\mid Q}(1-p^{-2})\ge\zeta(2)^{-1}.
\mathbb CUse in source · 1 passage
Search aliases: C mathbb
Use in source · body.tex, line 312
Conversely, regard u↦widetildeΘ_·(u) as a function taking values in H=L^2([t_0,T]; C). Cauchy Schwarz on each unit interval and the assumed energy bound imply ∫_0^∞ widetildeΘ_·(u) _H e^-au du<∞ (a>0). Its H-valued Laplace transform is holomorphic for Re w>0. Initially for Re w>1/2, it also equals the normally convergent meromorphic series with terms fracρ(1-ρ)e^t z_ρ^2w-z_ρ. The residue at a distinct centered zero is a nonzero function of t in H. Indeed, the multiplicity and ρ(1-ρ) are nonzero,…
Exact mathematical passage
H=L^2([t_0,T];\mathbb C)
C'Use in source · 1 passage
Search aliases: C'
Use in source · body.tex, line 670
This is the obstruction behind pulling a smaller endpoint backward through an unchanged late prefix. A useful smaller source merger must eventually change the word. For two different transfers ending at the same y, elimination gives m=frac2^S'3^k n+2^S'C-2^S C'2^S3^k'. The remaining task is to force a realizing alternative with 0<m<n. The formula does not itself supply such an alternative. The saved route notes develop this direction while retaining its unproved universal existence step [source …
Exact mathematical passage
m=\frac{2^{S'}3^k n+2^{S'}C-2^S C'}{2^S3^{k'}}.
C^2Use in source · 1 passage
Search aliases: C^2
Use in source · body.tex, line 431
Suppose a fixed positive width constant C>0 is chosen and the active centers in a shell satisfy |a-c_J|≤ Csqrt J+O(1) and t≥ t_0>0. Put α=C^2/(4t_0). Then λ≤α J+O(sqrt J). For a fixed constant q>α, projection through R_J=lceil qJrceil has norm error bounded by exp(-frac12 I(qmidα)J+o(J)), I(qmidα)=qlog(q/α)-q+α. The coefficient mass of the positive prime packet through X_J=exp(J+O(sqrt J)) is at most ∑_n≤ X_JfracΛ(n)sqrt n ≤2sqrtX_Jlog X_J=exp(J/2+o(J)). Thus I(qmidα)>1 makes the aggregate compr…
Exact mathematical passage
\alpha=C^2/(4t_0)
C_1Use in source · 1 passage
Search aliases: C_1
Use in source · body.tex, line 123
Two ordered words with the same total depth. For w=(1,3), the recurrence gives C_1=1 and C_2=5, with total depth S=4. The affine expression is F_(1,3)(n)=frac9n+516. The exact cylinder requires 9n+5≡16pmod32. The inverse of 9 modulo 32 is 25, so its address is 19 modulo 32. Every positive representative can be written n=32t+19 with an integer t≥0. Direct substitution gives the full orbit segment 32t+19⟼48t+29⟼18t+11. The first numerator is 2(48t+29), with an odd quotient. The next numerator is 8…
Exact mathematical passage
C_1=1
c_1Use in source · 1 passage
Search aliases: c_1
Use in source · body.tex, line 513
For example, c_1=frac1ζ(2)-frac1ζ(4), c_2=frac1ζ(2)-frac2ζ(4)+frac1ζ(6). These formulas are exact. At large K, their alternating binomial terms can be individually much larger than their sum, so straightforward floating point evaluation may lose accuracy. A finite expression is not automatically a stable evaluation scheme. For a direct truncation of the original series at n=N, the elementary certified remainder bound is instead |c_K-∑_n≤ Nμ(n)w_K(n)| ≤∑_n>Nn^-2≤frac1N. This bound is useful for c…
Exact mathematical passage
c_1=\frac1{\zeta(2)}-\frac1{\zeta(4)},\qquad c_2=\frac1{\zeta(2)}-\frac2{\zeta(4)}+\frac1{\zeta(6)}.
C_2Use in source · 2 passages
Search aliases: C_2
Use in source · body.tex, line 123
Two ordered words with the same total depth. For w=(1,3), the recurrence gives C_1=1 and C_2=5, with total depth S=4. The affine expression is F_(1,3)(n)=frac9n+516. The exact cylinder requires 9n+5≡16pmod32. The inverse of 9 modulo 32 is 25, so its address is 19 modulo 32. Every positive representative can be written n=32t+19 with an integer t≥0. Direct substitution gives the full orbit segment 32t+19⟼48t+29⟼18t+11. The first numerator is 2(48t+29), with an odd quotient. The next numerator is 8…
Exact mathematical passage
C_2=5
c_2Use in source · 1 passage
Search aliases: c_2
Use in source · body.tex, line 513
For example, c_1=frac1ζ(2)-frac1ζ(4), c_2=frac1ζ(2)-frac2ζ(4)+frac1ζ(6). These formulas are exact. At large K, their alternating binomial terms can be individually much larger than their sum, so straightforward floating point evaluation may lose accuracy. A finite expression is not automatically a stable evaluation scheme. For a direct truncation of the original series at n=N, the elementary certified remainder bound is instead |c_K-∑_n≤ Nμ(n)w_K(n)| ≤∑_n>Nn^-2≤frac1N. This bound is useful for c…
Exact mathematical passage
c_1=\frac1{\zeta(2)}-\frac1{\zeta(4)},\qquad c_2=\frac1{\zeta(2)}-\frac2{\zeta(4)}+\frac1{\zeta(6)}.
C_{37}Use in source · 1 passage
Search aliases: C_37
Use in source · body.tex, line 313
The fresh finite integer audit for this edition finds the first actual descent at accelerated time 37: x_37=23, S_37=59, C_37=1100931843921811423. Thus 2^59·23=3^37·27+1100931843921811423. The same segment has 59 shortcut steps. The historical serious audit separately records a Lean check of T^59(27)=23 [source reference]. The two statements use different iteration counts and the same endpoint.
Exact mathematical passage
\label{eq:27descent} x_{37}=23,\qquad S_{37}=59,\qquad C_{37}=1100931843921811423.
c_fUse in source · 2 passages
Search aliases: c_f
Use in source · body.tex, line 208
The composition formula also explains why a local fall need not resolve a global obstruction. If an expanding segment has data (λ_g,c_g) and a following falling segment has (λ_f,c_f), the final value is λ_fλ_g n+λ_f c_g+c_f. A comparison with the local peak controls only part of this expression. To prove descent below n, the entire expression must be compared with the same original source.
Exact mathematical passage
(\lambda_f,c_f)
c_gUse in source · 2 passages
Search aliases: c_g
Use in source · body.tex, line 208
The composition formula also explains why a local fall need not resolve a global obstruction. If an expanding segment has data (λ_g,c_g) and a following falling segment has (λ_f,c_f), the final value is λ_fλ_g n+λ_f c_g+c_f. A comparison with the local peak controls only part of this expression. To prove descent below n, the entire expression must be compared with the same original source.
Exact mathematical passage
(\lambda_g,c_g)
c_JUse in source · 1 passage
Search aliases: c_J
Use in source · body.tex, line 430
Suppose a fixed positive width constant C>0 is chosen and the active centers in a shell satisfy |a-c_J|≤ Csqrt J+O(1) and t≥ t_0>0. Put α=C^2/(4t_0). Then λ≤α J+O(sqrt J). For a fixed constant q>α, projection through R_J=lceil qJrceil has norm error bounded by exp(-frac12 I(qmidα)J+o(J)), I(qmidα)=qlog(q/α)-q+α. The coefficient mass of the positive prime packet through X_J=exp(J+O(sqrt J)) is at most ∑_n≤ X_JfracΛ(n)sqrt n ≤2sqrtX_Jlog X_J=exp(J/2+o(J)). Thus I(qmidα)>1 makes the aggregate compr…
Exact mathematical passage
|a-c_J|\le C\sqrt J+O(1)
c_{M^2}Use in source · 5 passages
Search aliases: c_M^2
Use in source · body.tex, line 499
[Báez Duarte criterion] The Riemann hypothesis holds if and only if, for every ε>0, |c_K|=O_ε(K^-3/4+ε). Equivalently, for every ε>0, |c_M^2|=O_ε(M^-3/2+ε), with M restricted to square roots of positive integers.
Exact mathematical passage
|c_{M^2}|=O_\varepsilon(M^{-3/2+\varepsilon})
C_rUse in source · 8 passages
Search aliases: C_r
Use in source · body.tex, line 282
A complete finite example from 27. The seed 27 makes the distinction between the clocks and the floors visible. Its first ten odd accelerated steps are shown in Table~[source reference]. Each row is checked by multiplying the displayed state by 2^S_r and comparing it with 3^r·27+C_r.
Exact mathematical passage
3^r\cdot27+C_r
C_{r+1}Use in source · 1 passage
Search aliases: C_r+1
Use in source · body.tex, line 694
endix Complete odd coordinates for the source 27. Table~[source reference] records the entire accelerated orbit through its first arrival at 1. The column a is the valuation used to arrive at the state in that row. The column S is both cumulative dyadic depth and cumulative shortcut time. The column L=S+r is cumulative classical time. The offset is computed by C_r+1=3C_r+2^S_r, starting from zero. Each row satisfies the exact identity 2^Sx=3^r·27+C.
Exact mathematical passage
C_{r+1}=3C_r+2^{S_r}
C_{u,v}Use in source · 1 passage
Search aliases: C_u,v
Use in source · body.tex, line 527
[Injectivity] A permanently noncrossing positive odd orbit never repeats a state. Suppose x_u=x_v with v>u. Put k=v-u and B=S_v-S_u. The affine law for this positive length block gives (2^B-3^k)x_u=C_u,v>0. Hence 2^B/3^k>1. Determinism makes the same states and valuations repeat, so ρ_u+hk=ρ_u(2^B/3^k)^h. This exceeds 1 for sufficiently large h, contrary to permanent noncrossing.
Exact mathematical passage
(2^B-3^k)x_u=C_{u,v}>0.
C_\varepsilonUse in source · 6 passages
Search aliases: C_ε varepsilon
Use in source · body.tex, line 296
For a nonnegative quantity E_J, call its growth subexponential when lim_J→∞fraclog(1+E_J)J=0. Equivalently, for every ε>0, there is a constant C_ε such that E_J≤ C_ε e^ε J for all positive integers J. This is an upper growth condition. It does not assert a positive lower bound.
Exact mathematical passage
C_\varepsilon
\coshUse in source · 5 passages
Search aliases: cosh
Use in source · body.tex, line 195
P_t(u) · =frac12sqrtπ t ∑_n≥2fracΛ(n)sqrt n (e^-(u-log n)^2/(4t)+ e^-(u+log n)^2/(4t)). The Gaussian makes the prime series absolutely convergent. The completed explicit formula gives Θ_t(u)=2e^t/4cosh(u/2)+ A_t(u)- P_t(u). We use the formula in the recovered normalization [source reference]. With Fourier convention g(x)=(2π)^-1∫ h(v)e^-ivx dv, the even test h(v)=e^-tv^2cos(uv) has transform g(x)=frac14sqrtπ t (e^-(x-u)^2/(4t)+e^-(x+u)^2/(4t)). The zero symmetry identifies its spectral sum with …
Exact mathematical passage
\label{rh:explicit} \Theta_t(u)=2e^{t/4}\cosh(u/2)+\mathfrak A_t(u)-\mathfrak P_t(u).
DUse in source · 34 passages
Search aliases: D
Use in source · body.tex, line 106
Conversely, define the rational candidates x_j=(3^j n+C_j)/2^S_j. We show that every candidate is a positive odd integer. The terminal congruence supplies this for j=k. For 0≤ j<k, split the offset sum at j: C_k=3^k-jC_j+2^S_jD_j,k. The integer D_j,k is odd. Its first summand is 3^k-1-j, while every later summand has an additional factor of two. Reducing [source reference] modulo 2^S_j+1 gives 3^k-j(3^j n+C_j)+2^S_jD_j,k≡0pmod2^S_j+1. Every odd multiple of 2^S_j has the same residue modulo 2^S_j…
Exact mathematical passage
C_k=3^{k-j}C_j+2^{S_j}D_{j,k}.
d^rUse in source · 1 passage
Search aliases: d^r
Use in source · body.tex, line 367
belrh:gaussian g_t,a(u)=(2π t)^-1/4e^-(u-a)^2/(4t). It has norm one in L^2( R,du). For a chosen center c, let ψ_r,t,c, r=0,1,2,…, be the orthonormal scaled Hermite functions ψ_r,t,c(u)= frac(2π t)^-1/4sqrt2^r r! H_r(fracu-csqrt2t)e^-(u-c)^2/(4t). Here H_r(x)=(-1)^r e^x^2(d^r/dx^r)e^-x^2 is the physicists' Hermite polynomial. If P_R is orthogonal projection onto indices 0≤ r≤ R, the coherent expansion gives g_t,a · =e^-λ/2 ∑_r≥0fracσ^rsqrtr!ψ_r,t,c,
Exact mathematical passage
H_r(x)=(-1)^r e^{x^2}(d^r/dx^r)e^{-x^2}
D_{j,k}Use in source · 3 passages
Search aliases: D_j,k
Use in source · body.tex, line 106
Conversely, define the rational candidates x_j=(3^j n+C_j)/2^S_j. We show that every candidate is a positive odd integer. The terminal congruence supplies this for j=k. For 0≤ j<k, split the offset sum at j: C_k=3^k-jC_j+2^S_jD_j,k. The integer D_j,k is odd. Its first summand is 3^k-1-j, while every later summand has an additional factor of two. Reducing [source reference] modulo 2^S_j+1 gives 3^k-j(3^j n+C_j)+2^S_jD_j,k≡0pmod2^S_j+1. Every odd multiple of 2^S_j has the same residue modulo 2^S_j…
Exact mathematical passage
C_k=3^{k-j}C_j+2^{S_j}D_{j,k}.
D_RUse in source · 2 passages
Search aliases: D_R
Use in source · body.tex, line 624
What the proof does not transfer to an unbounded strip. If the deficit bound is replaced by a varying number D_R, the height multiplier 2^D_R+1 varies with the same orbit window. A smaller cylinder density may be offset by a much larger permitted height. The finite family correction may also grow if its block length is allowed to vary. The preceding proof gives no uniform comparison between those competing quantities.
Exact mathematical passage
D_R
\mathbb EUse in source · 1 passage
Search aliases: E mathbb
Use in source · body.tex, line 447
The exponent in [source reference] comes from a standard elementary exponential bound for a Poisson variable X of mean λ. For h>0, Pr(X≥ R)≤ e^-hR E(e^hX) =exp(-hR+λ(e^h-1)). When R>λ>0, choose e^h=R/λ. For λ=0, every positive tail is zero. The positive mean case gives the rate Rlog(R/λ)-R+λ. Passing to the squared norm tail and then taking a square root supplies the factor one half in [source reference]. Changing R by one has no effect on its leading shell exponent.
Exact mathematical passage
\Pr(X\ge R)\le e^{-hR}\mathbb E(e^{hX}) =\exp\bigl(-hR+\lambda(e^h-1)\bigr).
e^{-(a-b)^2/(8t)}Use in source · 3 passages
Search aliases: e^-(a-b)^2/(8t)
Use in source · body.tex, line 383
The word physical refers to the measure du in which the packet is actually evaluated. It does not mean that a separately chosen coefficient norm can be substituted. Completing the square gives the exact overlap ⟨ g_t,a,g_t,b⟩ =e^-(a-b)^2/(8t). For two translates, the Gram matrix has diagonal entries one and off diagonal entry κ=e^-(a-b)^2/(8t). Its eigenvalues are 1+κ and 1-κ. When the centers are close, their difference is almost invisible in the physical norm even if the two raw coefficients a…
Exact mathematical passage
\label{rh:overlap} \langle g_{t,a},g_{t,b}\rangle =e^{-(a-b)^2/(8t)}.
e^{-(\log n)^2/(4t)}Use in source · 1 passage
Search aliases: e^-(log n)^2/(4t) log
Use in source · body.tex, line 222
At u=0, for example, the prime packet is P_t(0)=frac1sqrtπ t ∑_n≥2fracΛ(n)sqrt n e^-(log n)^2/(4t). Every summand is nonnegative. The complete formula is nonetheless a difference involving this positive quantity. At large positive u, the Gaussian centered at log n selects prime powers with logarithms near u, so the active arithmetic scale is around n=e^u. A fixed heat time gives a fixed width in logarithmic coordinates, which corresponds to a multiplicative window in the integers.
Exact mathematical passage
\mathfrak P_t(0)=\frac1{\sqrt{\pi t}} \sum_{n\ge2}\frac{\Lambda(n)}{\sqrt n} e^{-(\log n)^2/(4t)}.
e^{-(u-a)^2/(4t)}Use in source · 2 passages
Search aliases: e^-(u-a)^2/(4t)
Use in source · body.tex, line 249
For a translated Gaussian G_t,a(u)=e^-(u-a)^2/(4t), direct differentiation gives LG_t,a(u)= (frac14+frac12t-frac(u-a)^24t^2) G_t,a(u). The annihilated prime packet therefore carries a polynomial factor with both signs. Positivity of the original Gaussian sum is no longer positivity of every transformed term. On the zero side the same operator is simpler because each exponential is an eigenfunction of ∂_u^2. On the explicit pole channel, ∂_u^2cosh(u/2)=frac14cosh(u/2), so cancellation is exact.
Exact mathematical passage
G_{t,a}(u)=e^{-(u-a)^2/(4t)}
e^{-(u-c)^2/(4s)}Use in source · 1 passage
Search aliases: e^-(u-c)^2/(4s)
Use in source · body.tex, line 475
To verify [source reference], the scalar Gaussian integral is ∫_ R(2π s)^-1/4 e^-(u-c)^2/(4s)e^uz du =2sqrtπ s(2π s)^-1/4e^cz+sz^2. It holds for every complex z, either by completing the square first for real z and continuing analytically, or by the entire Gaussian transform. Multiplying by the mode coefficient at heat time t replaces t by t+s. The annihilator weight is unchanged. Absolute values before integration are bounded using |Re z|<1/2 and Gaussian decay in the ordinate, so the sum and i…
Exact mathematical passage
\int_{\mathbb R}(2\pi s)^{-1/4} e^{-(u-c)^2/(4s)}e^{uz}\,du =2\sqrt{\pi s}(2\pi s)^{-1/4}e^{cz+sz^2}.
e^{-(u-c)^2/(4t)}Use in source · 1 passage
Search aliases: e^-(u-c)^2/(4t)
Use in source · body.tex, line 362
For t>0 and a real center a, define the normalized Gaussian g_t,a(u)=(2π t)^-1/4e^-(u-a)^2/(4t). It has norm one in L^2( R,du). For a chosen center c, let ψ_r,t,c, r=0,1,2,…, be the orthonormal scaled Hermite functions ψ_r,t,c(u)= frac(2π t)^-1/4sqrt2^r r! H_r(fracu-csqrt2t)e^-(u-c)^2/(4t). Here H_r(x)=(-1)^r e^x^2(d^r/dx^r)e^-x^2 is the physicists' Hermite polynomial. If P_R is orthogonal projection onto indices 0≤ r≤ R, the coherent expansion gives g_t,a · =e^-λ/2 ∑_r≥0fracσ^rsqrtr!ψ_r,t,c,
Exact mathematical passage
\label{rh:hermite} \psi_{r,t,c}(u)= \frac{(2\pi t)^{-1/4}}{\sqrt{2^r r!}} H_r\left(\frac{u-c}{\sqrt{2t}}\right)e^{-(u-c)^2/(4t)}.
e^{-(x+u)^2/(4t)}Use in source · 1 passage
Search aliases: e^-(x+u)^2/(4t)
Use in source · body.tex, line 201
ies absolutely convergent. The completed explicit formula gives Θ_t(u)=2e^t/4cosh(u/2)+ A_t(u)- P_t(u). We use the formula in the recovered normalization [source reference]. With Fourier convention g(x)=(2π)^-1∫ h(v)e^-ivx dv, the even test h(v)=e^-tv^2cos(uv) has transform g(x)=frac14sqrtπ t (e^-(x-u)^2/(4t)+e^-(x+u)^2/(4t)). The zero symmetry identifies its spectral sum with Θ_t(u). Its two pole evaluations give 2e^t/4cosh(u/2). The prime contribution uses g(log n)+g(-log n), which fixes the c…
Exact mathematical passage
g(x)=\frac1{4\sqrt{\pi t}} \left(e^{-(x-u)^2/(4t)}+e^{-(x+u)^2/(4t)}\right).
e^{-(x-u)^2/(4t)}Use in source · 1 passage
Search aliases: e^-(x-u)^2/(4t)
Use in source · body.tex, line 201
ies absolutely convergent. The completed explicit formula gives Θ_t(u)=2e^t/4cosh(u/2)+ A_t(u)- P_t(u). We use the formula in the recovered normalization [source reference]. With Fourier convention g(x)=(2π)^-1∫ h(v)e^-ivx dv, the even test h(v)=e^-tv^2cos(uv) has transform g(x)=frac14sqrtπ t (e^-(x-u)^2/(4t)+e^-(x+u)^2/(4t)). The zero symmetry identifies its spectral sum with Θ_t(u). Its two pole evaluations give 2e^t/4cosh(u/2). The prime contribution uses g(log n)+g(-log n), which fixes the c…
Exact mathematical passage
g(x)=\frac1{4\sqrt{\pi t}} \left(e^{-(x-u)^2/(4t)}+e^{-(x+u)^2/(4t)}\right).
e^{-aJ}Use in source · 1 passage
Search aliases: e^-aJ
Use in source · body.tex, line 334
The energy statement is deliberately local in u. It does not assume a global square integrable packet, which would already rule out many possible growth behaviors. The Hilbert space H collects only the fixed positive heat interval. Tonelli's theorem identifies widetilde E_J=∫_J^J+1 widetildeΘ_·(u) _H^2 du. Let a>0. On a unit band, Cauchy Schwarz gives ∫_J^J+1 widetildeΘ_·(u) _H e^-au du ≤ e^-aJwidetilde E_J^1/2. Choose the energy exponent to be ε=a. Subexponential growth then bounds the right si…
Exact mathematical passage
\int_J^{J+1}\|\widetilde\Theta_\cdot(u)\|_H e^{-au}\,du \le e^{-aJ}\widetilde E_J^{1/2}.
e^{-aJ/2}Use in source · 1 passage
Search aliases: e^-aJ/2
Use in source · body.tex, line 338
fixed positive heat interval. Tonelli's theorem identifies widetilde E_J=∫_J^J+1 widetildeΘ_·(u) _H^2 du. Let a>0. On a unit band, Cauchy Schwarz gives ∫_J^J+1 widetildeΘ_·(u) _H e^-au du ≤ e^-aJwidetilde E_J^1/2. Choose the energy exponent to be ε=a. Subexponential growth then bounds the right side by a constant times e^-aJ/2. The geometric sum converges. The compact initial band [0,1] is harmless by normal convergence. This proves the weighted integrability used in the theorem for every a>0.
Exact mathematical passage
e^{-aJ/2}
e^{-au}Use in source · 3 passages
Search aliases: e^-au
Use in source · body.tex, line 142
If Θ_t is bounded, then on a compact subset of Re w>0 the integrand of its Laplace transform is bounded by a constant times e^-au, for some a>0. Differentiation with respect to w inserts powers of u, all of which remain integrable. This proves holomorphy on the claimed half plane.
Exact mathematical passage
e^{-au}
e^{-c\gamma^2}Use in source · 1 passage
Search aliases: e^-cγ^2 gamma
Use in source · body.tex, line 101
For completeness, divide the positive ordinates into intervals [j,j+1). The number in an interval is no greater than N(j+1), which is O((j+1)log(j+2)). Even this deliberately loose bound gives ∑_ρ∈ Z(1+|γ|)^A e^-cγ^2<∞ (c>0, A≥0). The finitely many small ordinates cause no difficulty, and the negative ones follow by conjugation. This establishes uniform convergence on the compact sets just described. It permits differentiation in u, differentiation in a positive heat interval, and multiplication…
Exact mathematical passage
\sum_{\rho\in\mathcal Z}(1+|\gamma|)^A e^{-c\gamma^2}<\infty \qquad(c>0,\ A\ge0).
e^{-hR}Use in source · 1 passage
Search aliases: e^-hR
Use in source · body.tex, line 447
The exponent in [source reference] comes from a standard elementary exponential bound for a Poisson variable X of mean λ. For h>0, Pr(X≥ R)≤ e^-hR E(e^hX) =exp(-hR+λ(e^h-1)). When R>λ>0, choose e^h=R/λ. For λ=0, every positive tail is zero. The positive mean case gives the rate Rlog(R/λ)-R+λ. Passing to the squared norm tail and then taking a square root supplies the factor one half in [source reference]. Changing R by one has no effect on its leading shell exponent.
Exact mathematical passage
\Pr(X\ge R)\le e^{-hR}\mathbb E(e^{hX}) =\exp\bigl(-hR+\lambda(e^h-1)\bigr).
e^{-iuv}Use in source · 1 passage
Search aliases: e^-iuv
Use in source · body.tex, line 217
The normalization can be checked without any arithmetic approximation. For real x, the elementary Gaussian Fourier integral is ∫_ Re^-tv^2e^-ivx dv =sqrtfracπ t e^-x^2/(4t). Write cos(uv)=(e^iuv+e^-iuv)/2. The factor one half from this cosine and the factor (2π)^-1 in the Fourier convention give the coefficient 1/(4sqrtπ t) in g. The function g is even, so the sum g(log n)+g(-log n) doubles it. This produces the coefficient 1/(2sqrtπ t) in the prime packet, not twice that coefficient.
Exact mathematical passage
\cos(uv)=(e^{iuv}+e^{-iuv})/2
e^{-ivx}Use in source · 2 passages
Search aliases: e^-ivx
Use in source · body.tex, line 199
e^-(u+log n)^2/(4t)). The Gaussian makes the prime series absolutely convergent. The completed explicit formula gives Θ_t(u)=2e^t/4cosh(u/2)+ A_t(u)- P_t(u). We use the formula in the recovered normalization [source reference]. With Fourier convention g(x)=(2π)^-1∫ h(v)e^-ivx dv, the even test h(v)=e^-tv^2cos(uv) has transform g(x)=frac14sqrtπ t (e^-(x-u)^2/(4t)+e^-(x+u)^2/(4t)). The zero symmetry identifies its spectral sum with Θ_t(u). Its two pole evaluations give 2e^t/4cosh(u/2). The prime c…
Exact mathematical passage
g(x)=(2\pi)^{-1}\int h(v)e^{-ivx}\,dv
e^{-t\gamma^2}Use in source · 2 passages
Search aliases: e^-tγ^2 gamma
Use in source · body.tex, line 75
The series converges normally on compact sets of translations and on compact intervals of positive heat times. Indeed, for real u, |e^tz_ρ^2+uz_ρ| ≤ e^t/4+|u|/2e^-tγ^2. For complex u, there is an additional exponential factor linear in |γ| on each compact set. Gaussian decay still dominates. The classical zero counting estimate N(Y)=O(Ylog Y), where N(Y) counts zeros with 0<Imρ≤ Y, gives summability [source reference]. The same domination permits differentiation. Consequently ∂_tΘ_t=∂_u^2Θ_t. Th…
Exact mathematical passage
\label{rh:maj} \left|e^{tz_\rho^2+uz_\rho}\right| \le e^{t/4+|u|/2}e^{-t\gamma^2}.
e^{-wu}Use in source · 1 passage
Search aliases: e^-wu
Use in source · body.tex, line 125
Under the hypothesis, [source reference] gives the asserted bound and determinant inequality. Conversely, suppose the packet is bounded on the positive half line. Its Laplace transform F_t(w)=∫_0^∞Θ_t(u)e^-wu du is holomorphic for Re w>0. For Re w>1/2, absolute convergence gives F_t(w)=∑_ρ∈ Z frace^tz_ρ^2w-z_ρ. The series on the right is meromorphic, with normal convergence away from its poles. At each distinct centered zero its residue is its positive integer multiplicity times e^tz_ρ^2, which …
Exact mathematical passage
F_t(w)=\int_0^{\infty}\Theta_t(u)e^{-wu}\,du
e^{-x^2}Use in source · 1 passage
Search aliases: e^-x^2
Use in source · body.tex, line 367
belrh:gaussian g_t,a(u)=(2π t)^-1/4e^-(u-a)^2/(4t). It has norm one in L^2( R,du). For a chosen center c, let ψ_r,t,c, r=0,1,2,…, be the orthonormal scaled Hermite functions ψ_r,t,c(u)= frac(2π t)^-1/4sqrt2^r r! H_r(fracu-csqrt2t)e^-(u-c)^2/(4t). Here H_r(x)=(-1)^r e^x^2(d^r/dx^r)e^-x^2 is the physicists' Hermite polynomial. If P_R is orthogonal projection onto indices 0≤ r≤ R, the coherent expansion gives g_t,a · =e^-λ/2 ∑_r≥0fracσ^rsqrtr!ψ_r,t,c,
Exact mathematical passage
H_r(x)=(-1)^r e^{x^2}(d^r/dx^r)e^{-x^2}
e^{-x^2/(4t)}Use in source · 1 passage
Search aliases: e^-x^2/(4t)
Use in source · body.tex, line 213
The normalization can be checked without any arithmetic approximation. For real x, the elementary Gaussian Fourier integral is ∫_ Re^-tv^2e^-ivx dv =sqrtfracπ t e^-x^2/(4t). Write cos(uv)=(e^iuv+e^-iuv)/2. The factor one half from this cosine and the factor (2π)^-1 in the Fourier convention give the coefficient 1/(4sqrtπ t) in g. The function g is even, so the sum g(log n)+g(-log n) doubles it. This produces the coefficient 1/(2sqrtπ t) in the prime packet, not twice that coefficient.
Exact mathematical passage
\int_{\mathbb R}e^{-tv^2}e^{-ivx}\,dv =\sqrt{\frac\pi t}\,e^{-x^2/(4t)}.
e^{-\lambda}Use in source · 2 passages
Search aliases: e^-λ lambda
Use in source · body.tex, line 368
e the orthonormal scaled Hermite functions ψ_r,t,c(u)= frac(2π t)^-1/4sqrt2^r r! H_r(fracu-csqrt2t)e^-(u-c)^2/(4t). Here H_r(x)=(-1)^r e^x^2(d^r/dx^r)e^-x^2 is the physicists' Hermite polynomial. If P_R is orthogonal projection onto indices 0≤ r≤ R, the coherent expansion gives g_t,a · =e^-λ/2 ∑_r≥0fracσ^rsqrtr!ψ_r,t,c, σ · =fraca-c2sqrt t, λ=σ^2, notag (I-P_R)g_t,a _2^2 · =e^-λ∑_r>Rfracλ^rr!. The last expression is a Poisson upper tail. These formulas and the compression estimates are recovered…
Exact mathematical passage
g_{t,a}&=e^{-\lambda/2} \sum_{r\ge0}\frac{\sigma^r}{\sqrt{r!}}\psi_{r,t,c}, \label{rh:coherent}\\ \sigma&=\frac{a-c}{2\sqrt t},\qquad \lambda=\sigma^2, \notag\\ \|(I-P_R)g_{t,a}\|_2^2 &=e^{-\lambda}\sum_{r>R}\frac{\lambda^r}{r!}. \label{rh:poisson}
e^{-\lambda/2}Use in source · 1 passage
Search aliases: e^-λ/2 lambda
Use in source · body.tex, line 368
e the orthonormal scaled Hermite functions ψ_r,t,c(u)= frac(2π t)^-1/4sqrt2^r r! H_r(fracu-csqrt2t)e^-(u-c)^2/(4t). Here H_r(x)=(-1)^r e^x^2(d^r/dx^r)e^-x^2 is the physicists' Hermite polynomial. If P_R is orthogonal projection onto indices 0≤ r≤ R, the coherent expansion gives g_t,a · =e^-λ/2 ∑_r≥0fracσ^rsqrtr!ψ_r,t,c, σ · =fraca-c2sqrt t, λ=σ^2, notag (I-P_R)g_t,a _2^2 · =e^-λ∑_r>Rfracλ^rr!. The last expression is a Poisson upper tail. These formulas and the compression estimates are recovered…
Exact mathematical passage
g_{t,a}&=e^{-\lambda/2} \sum_{r\ge0}\frac{\sigma^r}{\sqrt{r!}}\psi_{r,t,c}, \label{rh:coherent}\\ \sigma&=\frac{a-c}{2\sqrt t},\qquad \lambda=\sigma^2, \notag\\ \|(I-P_R)g_{t,a}\|_2^2 &=e^{-\lambda}\sum_{r>R}\frac{\lambda^r}{r!}. \label{rh:poisson}
e^{-\sigma^2/2}Use in source · 1 passage
Search aliases: e^-σ^2/2 sigma
Use in source · body.tex, line 408
Put x=(u-c)/sqrt2t. The ratio of a translated Gaussian to the centered one is fracg_t,a(u)g_t,c(u) =e^sqrt2σ x-σ^2. The Hermite generating identity e^2xv-v^2=∑_r≥0H_r(x)v^r/r!, with v=σ/sqrt2, turns this ratio into e^-σ^2/2∑_r≥0 fracH_r(x)σ^r2^r/2r!. Substitution of the normalized Hermite functions gives [source reference]. Squaring its orthogonal coefficients gives e^-λλ^r/r!. Their total is one because the exponential series sums to e^λ. The discarded squared norm is therefore exactly a Poisso…
Exact mathematical passage
e^{-\sigma^2/2}\sum_{r\ge0} \frac{H_r(x)\sigma^r}{2^{r/2}r!}.
e^{2xv-v^2}Use in source · 1 passage
Search aliases: e^2xv-v^2
Use in source · body.tex, line 406
Put x=(u-c)/sqrt2t. The ratio of a translated Gaussian to the centered one is fracg_t,a(u)g_t,c(u) =e^sqrt2σ x-σ^2. The Hermite generating identity e^2xv-v^2=∑_r≥0H_r(x)v^r/r!, with v=σ/sqrt2, turns this ratio into e^-σ^2/2∑_r≥0 fracH_r(x)σ^r2^r/2r!. Substitution of the normalized Hermite functions gives [source reference]. Squaring its orthogonal coefficients gives e^-λλ^r/r!. Their total is one because the exponential series sums to e^λ. The discarded squared norm is therefore exactly a Poisso…
Exact mathematical passage
e^{2xv-v^2}=\sum_{r\ge0}H_r(x)v^r/r!
e^{au}Use in source · 1 passage
Search aliases: e^au
Use in source · body.tex, line 171
To spell out the last implication, suppose the limit superior in [source reference] were below some centered zero's real part. Choose a number a strictly between them. Beyond a finite initial interval the packet would then satisfy |Θ_t(u)|≤ e^au, after slightly enlarging a if necessary. Its Laplace transform would be holomorphic for Re w>a, but the chosen zero would still place a pole there. This proves the lower bound without requiring that the supremum δ_ζ be attained. When δ_ζ=0, the definiti…
Exact mathematical passage
|\Theta_t(u)|\le e^{au}
e^{cz+sz^2}Use in source · 1 passage
Search aliases: e^cz+sz^2
Use in source · body.tex, line 475
To verify [source reference], the scalar Gaussian integral is ∫_ R(2π s)^-1/4 e^-(u-c)^2/(4s)e^uz du =2sqrtπ s(2π s)^-1/4e^cz+sz^2. It holds for every complex z, either by completing the square first for real z and continuing analytically, or by the entire Gaussian transform. Multiplying by the mode coefficient at heat time t replaces t by t+s. The annihilator weight is unchanged. Absolute values before integration are bounded using |Re z|<1/2 and Gaussian decay in the ordinate, so the sum and i…
Exact mathematical passage
\int_{\mathbb R}(2\pi s)^{-1/4} e^{-(u-c)^2/(4s)}e^{uz}\,du =2\sqrt{\pi s}(2\pi s)^{-1/4}e^{cz+sz^2}.
e^hUse in source · 2 passages
Search aliases: e^h
Use in source · body.tex, line 447
The exponent in [source reference] comes from a standard elementary exponential bound for a Poisson variable X of mean λ. For h>0, Pr(X≥ R)≤ e^-hR E(e^hX) =exp(-hR+λ(e^h-1)). When R>λ>0, choose e^h=R/λ. For λ=0, every positive tail is zero. The positive mean case gives the rate Rlog(R/λ)-R+λ. Passing to the squared norm tail and then taking a square root supplies the factor one half in [source reference]. Changing R by one has no effect on its leading shell exponent.
Exact mathematical passage
\Pr(X\ge R)\le e^{-hR}\mathbb E(e^{hX}) =\exp\bigl(-hR+\lambda(e^h-1)\bigr).
e^{hX}Use in source · 1 passage
Search aliases: e^hX
Use in source · body.tex, line 447
The exponent in [source reference] comes from a standard elementary exponential bound for a Poisson variable X of mean λ. For h>0, Pr(X≥ R)≤ e^-hR E(e^hX) =exp(-hR+λ(e^h-1)). When R>λ>0, choose e^h=R/λ. For λ=0, every positive tail is zero. The positive mean case gives the rate Rlog(R/λ)-R+λ. Passing to the squared norm tail and then taking a square root supplies the factor one half in [source reference]. Changing R by one has no effect on its leading shell exponent.
Exact mathematical passage
\Pr(X\ge R)\le e^{-hR}\mathbb E(e^{hX}) =\exp\bigl(-hR+\lambda(e^h-1)\bigr).
e^{iuv}Use in source · 1 passage
Search aliases: e^iuv
Use in source · body.tex, line 217
The normalization can be checked without any arithmetic approximation. For real x, the elementary Gaussian Fourier integral is ∫_ Re^-tv^2e^-ivx dv =sqrtfracπ t e^-x^2/(4t). Write cos(uv)=(e^iuv+e^-iuv)/2. The factor one half from this cosine and the factor (2π)^-1 in the Fourier convention give the coefficient 1/(4sqrtπ t) in g. The function g is even, so the sum g(log n)+g(-log n) doubles it. This produces the coefficient 1/(2sqrtπ t) in the prime packet, not twice that coefficient.
Exact mathematical passage
\cos(uv)=(e^{iuv}+e^{-iuv})/2
e^{iu\gamma}Use in source · 1 passage
Search aliases: e^iuγ gamma
Use in source · body.tex, line 110
Under the Riemann hypothesis the packet is Θ_t(u)=∑_ρ∈ Ze^-tγ^2e^iuγ. It is then the Fourier transform of a finite positive discrete measure. In particular, its magnitude is bounded by its value at the origin. The converse can be proved without estimating cancellation among different zeros.
Exact mathematical passage
\label{rh:criticalpacket} \Theta_t(u)=\sum_{\rho\in\mathcal Z}e^{-t\gamma^2}e^{iu\gamma}.
e^{j+1}Use in source · 1 passage
Search aliases: e^j+1
Use in source · body.tex, line 229
Absolute convergence of the prime series requires no prime number theorem. Use the elementary bound Λ(n)≤log n and partition the integers by j≤log n<j+1. There are at most e^j+1 terms in such a group. The Gaussian in j dominates the resulting exponential factor. The same reasoning covers differentiated packets because their derivatives insert only polynomials in u-log n. This supplies a direct analytic justification for the operations applied to the prime side.
Exact mathematical passage
e^{j+1}
e^{\sqrt2\sigma x-\sigma^2}Use in source · 1 passage
Search aliases: e^sqrt2σ x-σ^2 sqrt sigma sigma
Use in source · body.tex, line 401
Put x=(u-c)/sqrt2t. The ratio of a translated Gaussian to the centered one is fracg_t,a(u)g_t,c(u) =e^sqrt2σ x-σ^2. The Hermite generating identity e^2xv-v^2=∑_r≥0H_r(x)v^r/r!, with v=σ/sqrt2, turns this ratio into e^-σ^2/2∑_r≥0 fracH_r(x)σ^r2^r/2r!. Substitution of the normalized Hermite functions gives [source reference]. Squaring its orthogonal coefficients gives e^-λλ^r/r!. Their total is one because the exponential series sums to e^λ. The discarded squared norm is therefore exactly a Poisso…
Exact mathematical passage
\frac{g_{t,a}(u)}{g_{t,c}(u)} =e^{\sqrt2\sigma x-\sigma^2}.
e^{t z_\rho^2}Use in source · 1 passage
Search aliases: e^t z_ρ^2 rho
Use in source · body.tex, line 318
Theta_·(u) as a function taking values in H=L^2([t_0,T]; C). Cauchy Schwarz on each unit interval and the assumed energy bound imply ∫_0^∞ widetildeΘ_·(u) _H e^-au du<∞ (a>0). Its H-valued Laplace transform is holomorphic for Re w>0. Initially for Re w>1/2, it also equals the normally convergent meromorphic series with terms fracρ(1-ρ)e^t z_ρ^2w-z_ρ. The residue at a distinct centered zero is a nonzero function of t in H. Indeed, the multiplicity and ρ(1-ρ) are nonzero, and the exponential never…
Exact mathematical passage
\frac{\rho(1-\rho)e^{t z_\rho^2}}{w-z_\rho}.
e^{t/4}Use in source · 3 passages
Search aliases: e^t/4
Use in source · body.tex, line 195
P_t(u) · =frac12sqrtπ t ∑_n≥2fracΛ(n)sqrt n (e^-(u-log n)^2/(4t)+ e^-(u+log n)^2/(4t)). The Gaussian makes the prime series absolutely convergent. The completed explicit formula gives Θ_t(u)=2e^t/4cosh(u/2)+ A_t(u)- P_t(u). We use the formula in the recovered normalization [source reference]. With Fourier convention g(x)=(2π)^-1∫ h(v)e^-ivx dv, the even test h(v)=e^-tv^2cos(uv) has transform g(x)=frac14sqrtπ t (e^-(x-u)^2/(4t)+e^-(x+u)^2/(4t)). The zero symmetry identifies its spectral sum with …
Exact mathematical passage
\label{rh:explicit} \Theta_t(u)=2e^{t/4}\cosh(u/2)+\mathfrak A_t(u)-\mathfrak P_t(u).
e^{t/4+|u|/2}Use in source · 1 passage
Search aliases: e^t/4+|u|/2
Use in source · body.tex, line 75
The series converges normally on compact sets of translations and on compact intervals of positive heat times. Indeed, for real u, |e^tz_ρ^2+uz_ρ| ≤ e^t/4+|u|/2e^-tγ^2. For complex u, there is an additional exponential factor linear in |γ| on each compact set. Gaussian decay still dominates. The classical zero counting estimate N(Y)=O(Ylog Y), where N(Y) counts zeros with 0<Imρ≤ Y, gives summability [source reference]. The same domination permits differentiation. Consequently ∂_tΘ_t=∂_u^2Θ_t. Th…
Exact mathematical passage
\label{rh:maj} \left|e^{tz_\rho^2+uz_\rho}\right| \le e^{t/4+|u|/2}e^{-t\gamma^2}.
e^{tz^2}Use in source · 1 passage
Search aliases: e^tz^2
Use in source · body.tex, line 145
In the smaller initial domain Re w>1/2, absolute summation justifies integration of each zero term: ∫_0^∞ e^tz^2+uz-wu du =frace^tz^2w-z. For a compact set of w avoiding all centered zeros, large ordinates make |w-z| large. The Gaussian coefficient then gives a summable bound for the meromorphic series. Near one distinct centered zero z_0, isolate all copies of that zero. If its multiplicity is m_0, their sum is fracm_0e^tz_0^2w-z_0. Every other term is holomorphic in a sufficiently small disk. …
Exact mathematical passage
\int_0^\infty e^{tz^2+uz-wu}\,du =\frac{e^{tz^2}}{w-z}.
e^{tz^2+uz}Use in source · 2 passages
Search aliases: e^tz^2+uz
Use in source · body.tex, line 107
For a single centered mode, ∂_t e^tz^2+uz=z^2e^tz^2+uz and ∂_u^2e^tz^2+uz=z^2e^tz^2+uz. Normal convergence transfers this equality to the entire packet. The heat equation here is an exact identity for the chosen series, rather than an empirical fit to a finite list of zeros.
Exact mathematical passage
\partial_t e^{tz^2+uz}=z^2e^{tz^2+uz}
e^{tz^2+uz-wu}Use in source · 1 passage
Search aliases: e^tz^2+uz-wu
Use in source · body.tex, line 145
In the smaller initial domain Re w>1/2, absolute summation justifies integration of each zero term: ∫_0^∞ e^tz^2+uz-wu du =frace^tz^2w-z. For a compact set of w avoiding all centered zeros, large ordinates make |w-z| large. The Gaussian coefficient then gives a summable bound for the meromorphic series. Near one distinct centered zero z_0, isolate all copies of that zero. If its multiplicity is m_0, their sum is fracm_0e^tz_0^2w-z_0. Every other term is holomorphic in a sufficiently small disk. …
Exact mathematical passage
\int_0^\infty e^{tz^2+uz-wu}\,du =\frac{e^{tz^2}}{w-z}.
e^{tz_0^2}Use in source · 2 passages
Search aliases: e^tz_0^2
Use in source · body.tex, line 150
justifies integration of each zero term: ∫_0^∞ e^tz^2+uz-wu du =frace^tz^2w-z. For a compact set of w avoiding all centered zeros, large ordinates make |w-z| large. The Gaussian coefficient then gives a summable bound for the meromorphic series. Near one distinct centered zero z_0, isolate all copies of that zero. If its multiplicity is m_0, their sum is fracm_0e^tz_0^2w-z_0. Every other term is holomorphic in a sufficiently small disk. The residue cannot vanish because m_0≥1 and an exponential …
Exact mathematical passage
\frac{m_0e^{tz_0^2}}{w-z_0}.
e^{tz_\rho^2}Use in source · 3 passages
Search aliases: e^tz_ρ^2 rho
Use in source · body.tex, line 129
Under the hypothesis, [source reference] gives the asserted bound and determinant inequality. Conversely, suppose the packet is bounded on the positive half line. Its Laplace transform F_t(w)=∫_0^∞Θ_t(u)e^-wu du is holomorphic for Re w>0. For Re w>1/2, absolute convergence gives F_t(w)=∑_ρ∈ Z frace^tz_ρ^2w-z_ρ. The series on the right is meromorphic, with normal convergence away from its poles. At each distinct centered zero its residue is its positive integer multiplicity times e^tz_ρ^2, which …
Exact mathematical passage
\label{rh:laplace} F_t(w)=\sum_{\rho\in\mathcal Z} \frac{e^{tz_\rho^2}}{w-z_\rho}.
e^uUse in source · 1 passage
Search aliases: e^u
Use in source · body.tex, line 227
example, the prime packet is P_t(0)=frac1sqrtπ t ∑_n≥2fracΛ(n)sqrt n e^-(log n)^2/(4t). Every summand is nonnegative. The complete formula is nonetheless a difference involving this positive quantity. At large positive u, the Gaussian centered at log n selects prime powers with logarithms near u, so the active arithmetic scale is around n=e^u. A fixed heat time gives a fixed width in logarithmic coordinates, which corresponds to a multiplicative window in the integers.
Exact mathematical passage
n=e^u
e^{uz}Use in source · 1 passage
Search aliases: e^uz
Use in source · body.tex, line 475
To verify [source reference], the scalar Gaussian integral is ∫_ R(2π s)^-1/4 e^-(u-c)^2/(4s)e^uz du =2sqrtπ s(2π s)^-1/4e^cz+sz^2. It holds for every complex z, either by completing the square first for real z and continuing analytically, or by the entire Gaussian transform. Multiplying by the mode coefficient at heat time t replaces t by t+s. The annihilator weight is unchanged. Absolute values before integration are bounded using |Re z|<1/2 and Gaussian decay in the ordinate, so the sum and i…
Exact mathematical passage
\int_{\mathbb R}(2\pi s)^{-1/4} e^{-(u-c)^2/(4s)}e^{uz}\,du =2\sqrt{\pi s}(2\pi s)^{-1/4}e^{cz+sz^2}.
e^{x^2}Use in source · 1 passage
Search aliases: e^x^2
Use in source · body.tex, line 367
belrh:gaussian g_t,a(u)=(2π t)^-1/4e^-(u-a)^2/(4t). It has norm one in L^2( R,du). For a chosen center c, let ψ_r,t,c, r=0,1,2,…, be the orthonormal scaled Hermite functions ψ_r,t,c(u)= frac(2π t)^-1/4sqrt2^r r! H_r(fracu-csqrt2t)e^-(u-c)^2/(4t). Here H_r(x)=(-1)^r e^x^2(d^r/dx^r)e^-x^2 is the physicists' Hermite polynomial. If P_R is orthogonal projection onto indices 0≤ r≤ R, the coherent expansion gives g_t,a · =e^-λ/2 ∑_r≥0fracσ^rsqrtr!ψ_r,t,c,
Exact mathematical passage
H_r(x)=(-1)^r e^{x^2}(d^r/dx^r)e^{-x^2}
e^{\delta u}Use in source · 1 passage
Search aliases: e^δ u delta
Use in source · body.tex, line 173
figplaceholderrh_heatfrontAmplitude envelopes of hypothetical centered modes. A displacement δ gives amplitude growth e^δ u, with a fixed nonzero heat multiplier omitted in the illustration. These model curves illustrate the theorem's exponent and are not sampled off line zeta zeros or a complete symmetric zero packet.
Exact mathematical passage
e^{\delta u}
e^{\varepsilon J}Use in source · 2 passages
Search aliases: e^ε J varepsilon
Use in source · body.tex, line 296
For a nonnegative quantity E_J, call its growth subexponential when lim_J→∞fraclog(1+E_J)J=0. Equivalently, for every ε>0, there is a constant C_ε such that E_J≤ C_ε e^ε J for all positive integers J. This is an upper growth condition. It does not assert a positive lower bound.
Exact mathematical passage
E_J\le C_\varepsilon e^{\varepsilon J}
e^\lambdaUse in source · 1 passage
Search aliases: e^λ lambda
Use in source · body.tex, line 413
The Hermite generating identity e^2xv-v^2=∑_r≥0H_r(x)v^r/r!, with v=σ/sqrt2, turns this ratio into e^-σ^2/2∑_r≥0 fracH_r(x)σ^r2^r/2r!. Substitution of the normalized Hermite functions gives [source reference]. Squaring its orthogonal coefficients gives e^-λλ^r/r!. Their total is one because the exponential series sums to e^λ. The discarded squared norm is therefore exactly a Poisson probability, rather than a bound inferred from a plot.
Exact mathematical passage
e^\lambda
e^{\pi t/4}Use in source · 1 passage
Search aliases: e^π t/4 pi
Use in source · body.tex, line 60
figplaceholderrh_xiA finite evaluation of the completed function on the critical line. The lower panel multiplies the normalized value by e^π t/4, where t is the plotted height. The symmetry and normalization are exact classical facts. The plotted interval is a finite numerical illustration and supplies no conclusion about all zeros.
Exact mathematical passage
e^{\pi t/4}
E_JUse in source · 3 passages
Search aliases: E_J
Use in source · body.tex, line 291
For a nonnegative quantity E_J, call its growth subexponential when lim_J→∞fraclog(1+E_J)J=0. Equivalently, for every ε>0, there is a constant C_ε such that E_J≤ C_ε e^ε J for all positive integers J. This is an upper growth condition. It does not assert a positive lower bound.
Exact mathematical passage
E_J
fUse in source · 6 passages
Search aliases: f
Use in source · body.tex, line 208
The composition formula also explains why a local fall need not resolve a global obstruction. If an expanding segment has data (λ_g,c_g) and a following falling segment has (λ_f,c_f), the final value is λ_fλ_g n+λ_f c_g+c_f. A comparison with the local peak controls only part of this expression. To prove descent below n, the entire expression must be compared with the same original source.
Exact mathematical passage
(\lambda_f,c_f)
F_{(1,3)}Use in source · 1 passage
Search aliases: F_(1,3)
Use in source · body.tex, line 124
Two ordered words with the same total depth. For w=(1,3), the recurrence gives C_1=1 and C_2=5, with total depth S=4. The affine expression is F_(1,3)(n)=frac9n+516. The exact cylinder requires 9n+5≡16pmod32. The inverse of 9 modulo 32 is 25, so its address is 19 modulo 32. Every positive representative can be written n=32t+19 with an integer t≥0. Direct substitution gives the full orbit segment 32t+19⟼48t+29⟼18t+11. The first numerator is 2(48t+29), with an odd quotient. The next numerator is 8…
Exact mathematical passage
F_{(1,3)}(n)=\frac{9n+5}{16}.
F_{(3,1)}Use in source · 1 passage
Search aliases: F_(3,1)
Use in source · body.tex, line 134
For v=(3,1), the first offset is still 1, but the second is 3+2^3=11. Its affine expression and cylinder are F_(3,1)(n)=frac9n+1116, n≡29pmod32. The realized segment is 32t+29⟼12t+11⟼18t+17, t≥0. Both words have two odd multiplications and four divisions by two. Their coefficients are the same. Their offsets, residue addresses, and intermediate states differ. The first word grows before falling. The second falls before growing. Both endpoints lie below their sources for every displayed parameter…
Exact mathematical passage
F_{(3,1)}(n)=\frac{9n+11}{16},\qquad n\equiv29\pmod{32}.
F_tUse in source · 2 passages
Search aliases: F_t
Use in source · body.tex, line 125
Under the hypothesis, [source reference] gives the asserted bound and determinant inequality. Conversely, suppose the packet is bounded on the positive half line. Its Laplace transform F_t(w)=∫_0^∞Θ_t(u)e^-wu du is holomorphic for Re w>0. For Re w>1/2, absolute convergence gives F_t(w)=∑_ρ∈ Z frace^tz_ρ^2w-z_ρ. The series on the right is meromorphic, with normal convergence away from its poles. At each distinct centered zero its residue is its positive integer multiplicity times e^tz_ρ^2, which …
Exact mathematical passage
F_t(w)=\int_0^{\infty}\Theta_t(u)e^{-wu}\,du
F_vUse in source · 2 passages
Search aliases: F_v
Use in source · body.tex, line 182
[Concatenation] Let a word w have data (k,S,C) and a following word v have data (ℓ,U,D). Their concatenation has data (k+ℓ,S+U,3^ℓ C+2^S D). On every realizing source, its transfer is F_v∘ F_w. Substitute 2^S y=3^k n+C into 2^U z=3^ℓ y+D. Multiplication by 2^S gives 2^S+Uz=3^k+ℓn+3^ℓ C+2^S D. The exact cylinder theorem identifies the admissible integer sources.
Exact mathematical passage
F_v\circ F_w
GUse in source · 14 passages
Search aliases: G
Use in source · body.tex, line 197
[Exact descent criterion] A realized word of positive length ends below its source if and only if (2^S-3^k)n>C_k. If 2^S≤3^k, strict descent is impossible. If 2^S>3^k, write G=2^S-3^k. The endpoint is at least the source if and only if n≤lfloorfracC_kGrfloor. The affine identity gives 2^S(x_k-n)=C_k-(2^S-3^k)n. Its sign proves the assertions. Since n is an integer, the inequality Gn≤ C_k is equivalent to [source reference].
Exact mathematical passage
G=2^S-3^k
g_{s,c}Use in source · 1 passage
Search aliases: g_s,c
Use in source · body.tex, line 465
The ground channel shows why the distinction matters. Direct integration gives ∫_ Rg_s,c(u)widetildeΘ_t(u) du =K_swidetildeΘ_t+s(c), K_s=2sqrtπ s(2π s)^-1/4. The Gaussian test makes this integral absolutely convergent even without the conjecture. It is a Gaussian pairing, with no assertion that the unweighted zero packet belongs to L^2( R). Completing the square proves the identity term by term. Normal convergence and the Gaussian majorant justify summation [source reference].
Exact mathematical passage
\label{rh:ground} \int_{\mathbb R}g_{s,c}(u)\widetilde\Theta_t(u)\,du =K_s\widetilde\Theta_{t+s}(c), \qquad K_s=2\sqrt{\pi s}(2\pi s)^{-1/4}.
G_{t,a}Use in source · 2 passages
Search aliases: G_t,a
Use in source · body.tex, line 249
For a translated Gaussian G_t,a(u)=e^-(u-a)^2/(4t), direct differentiation gives LG_t,a(u)= (frac14+frac12t-frac(u-a)^24t^2) G_t,a(u). The annihilated prime packet therefore carries a polynomial factor with both signs. Positivity of the original Gaussian sum is no longer positivity of every transformed term. On the zero side the same operator is simpler because each exponential is an eigenfunction of ∂_u^2. On the explicit pole channel, ∂_u^2cosh(u/2)=frac14cosh(u/2), so cancellation is exact.
Exact mathematical passage
G_{t,a}(u)=e^{-(u-a)^2/(4t)}
g_{t,b}Use in source · 1 passage
Search aliases: g_t,b
Use in source · body.tex, line 383
The word physical refers to the measure du in which the packet is actually evaluated. It does not mean that a separately chosen coefficient norm can be substituted. Completing the square gives the exact overlap ⟨ g_t,a,g_t,b⟩ =e^-(a-b)^2/(8t). For two translates, the Gram matrix has diagonal entries one and off diagonal entry κ=e^-(a-b)^2/(8t). Its eigenvalues are 1+κ and 1-κ. When the centers are close, their difference is almost invisible in the physical norm even if the two raw coefficients a…
Exact mathematical passage
\label{rh:overlap} \langle g_{t,a},g_{t,b}\rangle =e^{-(a-b)^2/(8t)}.
g_{t,c}Use in source · 1 passage
Search aliases: g_t,c
Use in source · body.tex, line 401
Put x=(u-c)/sqrt2t. The ratio of a translated Gaussian to the centered one is fracg_t,a(u)g_t,c(u) =e^sqrt2σ x-σ^2. The Hermite generating identity e^2xv-v^2=∑_r≥0H_r(x)v^r/r!, with v=σ/sqrt2, turns this ratio into e^-σ^2/2∑_r≥0 fracH_r(x)σ^r2^r/2r!. Substitution of the normalized Hermite functions gives [source reference]. Squaring its orthogonal coefficients gives e^-λλ^r/r!. Their total is one because the exponential series sums to e^λ. The discarded squared norm is therefore exactly a Poisso…
Exact mathematical passage
\frac{g_{t,a}(u)}{g_{t,c}(u)} =e^{\sqrt2\sigma x-\sigma^2}.
H^2Use in source · 2 passages
Search aliases: H^2
Use in source · body.tex, line 329
The energy statement is deliberately local in u. It does not assume a global square integrable packet, which would already rule out many possible growth behaviors. The Hilbert space H collects only the fixed positive heat interval. Tonelli's theorem identifies widetilde E_J=∫_J^J+1 widetildeΘ_·(u) _H^2 du. Let a>0. On a unit band, Cauchy Schwarz gives ∫_J^J+1 widetildeΘ_·(u) _H e^-au du ≤ e^-aJwidetilde E_J^1/2. Choose the energy exponent to be ε=a. Subexponential growth then bounds the right si…
Exact mathematical passage
\widetilde E_J=\int_J^{J+1} \|\widetilde\Theta_\cdot(u)\|_H^2\,du.
H_RUse in source · 4 passages
Search aliases: H_R
Use in source · body.tex, line 570
For R>N, Lemmas~[source reference] and [source reference] place the R-N distinct states x_N,…,x_R-1 in E_ℓ below H_R=lceil2^D+1(n+R/3)rceil. Thus R-N≤ε_ℓ H_R+W_ℓ. Fix ℓ and let R tend to infinity. Division by R yields 1≤ε_ℓ2^D+1/3. Choosing ℓ sufficiently large contradicts this inequality. No eventual bounded deficit strip exists.
Exact mathematical passage
H_R=\lceil2^{D+1}(n+R/3)\rceil
H_rUse in source · 6 passages
Search aliases: H_r
Use in source · body.tex, line 628
The deficit itself can be handled entirely with integers. Let H_r be the least nonnegative integer h such that 3^r<2^h. Then lfloor rlog_2 3rfloor=H_r-1, including r=0. For r>0, equality of a positive power of 3 with a power of 2 is impossible, so the strict threshold identifies the floor exactly. The displayed logarithm is a convenient notation for this integer comparison. A certificate implementation need not approximate an irrational logarithm to evaluate the deficit.
Exact mathematical passage
H_r
IUse in source · 8 passages
Search aliases: I
Use in source · body.tex, line 368
e the orthonormal scaled Hermite functions ψ_r,t,c(u)= frac(2π t)^-1/4sqrt2^r r! H_r(fracu-csqrt2t)e^-(u-c)^2/(4t). Here H_r(x)=(-1)^r e^x^2(d^r/dx^r)e^-x^2 is the physicists' Hermite polynomial. If P_R is orthogonal projection onto indices 0≤ r≤ R, the coherent expansion gives g_t,a · =e^-λ/2 ∑_r≥0fracσ^rsqrtr!ψ_r,t,c, σ · =fraca-c2sqrt t, λ=σ^2, notag (I-P_R)g_t,a _2^2 · =e^-λ∑_r>Rfracλ^rr!. The last expression is a Poisson upper tail. These formulas and the compression estimates are recovered…
Exact mathematical passage
g_{t,a}&=e^{-\lambda/2} \sum_{r\ge0}\frac{\sigma^r}{\sqrt{r!}}\psi_{r,t,c}, \label{rh:coherent}\\ \sigma&=\frac{a-c}{2\sqrt t},\qquad \lambda=\sigma^2, \notag\\ \|(I-P_R)g_{t,a}\|_2^2 &=e^{-\lambda}\sum_{r>R}\frac{\lambda^r}{r!}. \label{rh:poisson}
\ImUse in source · 1 passage
Search aliases: Im
Use in source · body.tex, line 79
tervals of positive heat times. Indeed, for real u, |e^tz_ρ^2+uz_ρ| ≤ e^t/4+|u|/2e^-tγ^2. For complex u, there is an additional exponential factor linear in |γ| on each compact set. Gaussian decay still dominates. The classical zero counting estimate N(Y)=O(Ylog Y), where N(Y) counts zeros with 0<Imρ≤ Y, gives summability [source reference]. The same domination permits differentiation. Consequently ∂_tΘ_t=∂_u^2Θ_t. The zero symmetries make Θ_t(u) real and even when u is real.
Exact mathematical passage
0<\Im\rho\le Y
k'Use in source · 1 passage
Search aliases: k'
Use in source · body.tex, line 670
This is the obstruction behind pulling a smaller endpoint backward through an unchanged late prefix. A useful smaller source merger must eventually change the word. For two different transfers ending at the same y, elimination gives m=frac2^S'3^k n+2^S'C-2^S C'2^S3^k'. The remaining task is to force a realizing alternative with 0<m<n. The formula does not itself supply such an alternative. The saved route notes develop this direction while retaining its unproved universal existence step [source …
Exact mathematical passage
m=\frac{2^{S'}3^k n+2^{S'}C-2^S C'}{2^S3^{k'}}.
K^{-3/4+\varepsilon}Use in source · 2 passages
Search aliases: K^-3/4+ε varepsilon
Use in source · body.tex, line 495
[Báez Duarte criterion] The Riemann hypothesis holds if and only if, for every ε>0, |c_K|=O_ε(K^-3/4+ε). Equivalently, for every ε>0, |c_M^2|=O_ε(M^-3/2+ε), with M restricted to square roots of positive integers.
Exact mathematical passage
\label{rh:bdcriterion} |c_K|=O_\varepsilon(K^{-3/4+\varepsilon}).
K_sUse in source · 1 passage
Search aliases: K_s
Use in source · body.tex, line 465
The ground channel shows why the distinction matters. Direct integration gives ∫_ Rg_s,c(u)widetildeΘ_t(u) du =K_swidetildeΘ_t+s(c), K_s=2sqrtπ s(2π s)^-1/4. The Gaussian test makes this integral absolutely convergent even without the conjecture. It is a Gaussian pairing, with no assertion that the unweighted zero packet belongs to L^2( R). Completing the square proves the identity term by term. Normal convergence and the Gaussian majorant justify summation [source reference].
Exact mathematical passage
\label{rh:ground} \int_{\mathbb R}g_{s,c}(u)\widetilde\Theta_t(u)\,du =K_s\widetilde\Theta_{t+s}(c), \qquad K_s=2\sqrt{\pi s}(2\pi s)^{-1/4}.
L^2Use in source · 3 passages
Search aliases: L^2
Use in source · body.tex, line 312
Conversely, regard u↦widetildeΘ_·(u) as a function taking values in H=L^2([t_0,T]; C). Cauchy Schwarz on each unit interval and the assumed energy bound imply ∫_0^∞ widetildeΘ_·(u) _H e^-au du<∞ (a>0). Its H-valued Laplace transform is holomorphic for Re w>0. Initially for Re w>1/2, it also equals the normally convergent meromorphic series with terms fracρ(1-ρ)e^t z_ρ^2w-z_ρ. The residue at a distinct centered zero is a nonzero function of t in H. Indeed, the multiplicity and ρ(1-ρ) are nonzero,…
Exact mathematical passage
H=L^2([t_0,T];\mathbb C)
L^4Use in source · 1 passage
Search aliases: L^4
Use in source · body.tex, line 762
Fourth, a raw positive fourth moment cannot lie below the square of its quadratic energy in a normalized probability space. If ν is its probability measure and F∈ L^4(ν) is the packet, Cauchy Schwarz gives ∫ |F|^4 dν≥(∫ |F|^2 dν)^2. This is the precise obstruction to a strict saving below that scale for the raw moment. It does not assert positivity of every separately grouped collision term for arbitrary signed coefficients. A viable replacement would require a signed expression, a justified quo…
Exact mathematical passage
F\in L^4(\nu)
L^qUse in source · 1 passage
Search aliases: L^q
Use in source · body.tex, line 760
Third, global Hardy norms on a prime torus are blind to the squarefree Möbius sign. For a finite prime set, a squarefree monomial is ∏_p∈ Sz_p. Replacing every coordinate by -z_p multiplies it by (-1)^|S|, exactly its Möbius coefficient. The substitution preserves Haar measure on the torus. Every global L^q norm formed with that measure is consequently unchanged by this sign operation. A bound in that invariant norm cannot distinguish the signed packet from the corresponding unsigned squarefree …
Exact mathematical passage
L^q
\lceilUse in source · 2 passages
Search aliases: lceil
Use in source · body.tex, line 570
For R>N, Lemmas~[source reference] and [source reference] place the R-N distinct states x_N,…,x_R-1 in E_ℓ below H_R=lceil2^D+1(n+R/3)rceil. Thus R-N≤ε_ℓ H_R+W_ℓ. Fix ℓ and let R tend to infinity. Division by R yields 1≤ε_ℓ2^D+1/3. Choosing ℓ sufficiently large contradicts this inequality. No eventual bounded deficit strip exists.
Exact mathematical passage
H_R=\lceil2^{D+1}(n+R/3)\rceil
\limUse in source · 1 passage
Search aliases: lim
Use in source · body.tex, line 292
For a nonnegative quantity E_J, call its growth subexponential when lim_J→∞fraclog(1+E_J)J=0. Equivalently, for every ε>0, there is a constant C_ε such that E_J≤ C_ε e^ε J for all positive integers J. This is an upper growth condition. It does not assert a positive lower bound.
Exact mathematical passage
\label{rh:subexp} \lim_{J\to\infty}\frac{\log(1+E_J)}J=0.
\lim_{J\to\infty}Use in source · 1 passage
Search aliases: lim_J→∞ lim to infty
Use in source · body.tex, line 292
For a nonnegative quantity E_J, call its growth subexponential when lim_J→∞fraclog(1+E_J)J=0. Equivalently, for every ε>0, there is a constant C_ε such that E_J≤ C_ε e^ε J for all positive integers J. This is an upper growth condition. It does not assert a positive lower bound.
Exact mathematical passage
\label{rh:subexp} \lim_{J\to\infty}\frac{\log(1+E_J)}J=0.
\limsupUse in source · 1 passage
Search aliases: limsup
Use in source · body.tex, line 166
The packet also makes the obstruction quantitative. Define δ_ζ=sup_ρ∈ Z(Reρ-1/2), M_t(U)=max(1,sup_0≤ u≤ U|Θ_t(u)|). Then 0≤δ_ζ≤1/2. The same Laplace argument proves limsup_U→∞fraclog M_t(U)U=δ_ζ. The upper bound follows from summing absolute values. If a smaller exponential rate held, the Laplace transform would be holomorphic beyond that rate, contradicting a pole at a zero further to the right. An off line zero may have a very large ordinate and a very small Gaussian coefficient. Its coeffici…
Exact mathematical passage
\label{rh:front} \limsup_{U\to\infty}\frac{\log M_t(U)}{U}=\delta_\zeta.
\limsup_{U\to\infty}Use in source · 1 passage
Search aliases: limsup_U→∞ limsup to infty
Use in source · body.tex, line 166
The packet also makes the obstruction quantitative. Define δ_ζ=sup_ρ∈ Z(Reρ-1/2), M_t(U)=max(1,sup_0≤ u≤ U|Θ_t(u)|). Then 0≤δ_ζ≤1/2. The same Laplace argument proves limsup_U→∞fraclog M_t(U)U=δ_ζ. The upper bound follows from summing absolute values. If a smaller exponential rate held, the Laplace transform would be holomorphic beyond that rate, contradicting a pole at a zero further to the right. An off line zero may have a very large ordinate and a very small Gaussian coefficient. Its coeffici…
Exact mathematical passage
\label{rh:front} \limsup_{U\to\infty}\frac{\log M_t(U)}{U}=\delta_\zeta.
M^{-1}Use in source · 1 passage
Search aliases: M^-1
Use in source · body.tex, line 532
The size difference behind the criterion is now visible. There are on the order of M integers in a window of length proportional to M, and a typical weight there has order M^-2. An unsigned mass therefore naturally has order M^-1. The desired signed amplitude has the smaller scale M^-3/2+ε. This count is only orientation. The barrier proof below supplies a rigorous lower bound for the actual full absolute majorant, with its squarefree support and cube dependence included.
Exact mathematical passage
M^{-1}
M^{-1-o(1)}Use in source · 4 passages
Search aliases: M^-1-o(1)
Use in source · body.tex, line 713
[Absolute cube barrier] Fix R with κ_R>0. Suppose squarefree Q=Q(M) has d(M)=o(log M) distinct prime divisors. For all sufficiently large M, A_Q(M)≥ fracκ_R8Rζ(2) fracφ(Q)QM ≥fracκ_R8Rζ(2)M(d+1). Consequently A_Q(M)≥ M^-1-o(1). It cannot itself satisfy the target bound in Theorem~[source reference].
Exact mathematical passage
\mathcal A_Q(M)\ge M^{-1-o(1)}
M^{-1/2+o(1)}Use in source · 1 passage
Search aliases: M^-1/2+o(1)
Use in source · body.tex, line 732
The dimension condition means that the cube size 2^d is M^o(1). In this regime the target amplitude is M^-3/2+o(1), whereas the triangle majorant is at least M^-1-o(1). The missing factor is M^-1/2+o(1). It must arise from cancellation or another operation which retains and controls the signs.
Exact mathematical passage
M^{-1/2+o(1)}
M^{-2}Use in source · 1 passage
Search aliases: M^-2
Use in source · body.tex, line 532
The size difference behind the criterion is now visible. There are on the order of M integers in a window of length proportional to M, and a typical weight there has order M^-2. An unsigned mass therefore naturally has order M^-1. The desired signed amplitude has the smaller scale M^-3/2+ε. This count is only orientation. The barrier proof below supplies a rigorous lower bound for the actual full absolute majorant, with its squarefree support and cube dependence included.
Exact mathematical passage
M^{-2}
m^{-2}Use in source · 1 passage
Search aliases: m^-2
Use in source · body.tex, line 655
The limiting divisor sum is ∑_bmid Qfracμ(b)b^2 =∏_pmid Q(1-p^-2)≥ζ(2)^-1. For 0≤ x≤1, Bernoulli's inequality gives 0≤1-(1-x)^K≤ Kx. With K=M^2, the error in replacing each power by one is at most fracM^2m^2∑_bmid Qb^-4 ≤fracζ(4)R^2. Multiplying the resulting lower bound by m^-2 proves the lemma.
Exact mathematical passage
m^{-2}
M^{-3/2+o(1)}Use in source · 2 passages
Search aliases: M^-3/2+o(1)
Use in source · body.tex, line 732
The dimension condition means that the cube size 2^d is M^o(1). In this regime the target amplitude is M^-3/2+o(1), whereas the triangle majorant is at least M^-1-o(1). The missing factor is M^-1/2+o(1). It must arise from cancellation or another operation which retains and controls the signs.
Exact mathematical passage
M^{-3/2+o(1)}
M^{-3/2+\varepsilon}Use in source · 3 passages
Search aliases: M^-3/2+ε varepsilon
Use in source · body.tex, line 499
[Báez Duarte criterion] The Riemann hypothesis holds if and only if, for every ε>0, |c_K|=O_ε(K^-3/4+ε). Equivalently, for every ε>0, |c_M^2|=O_ε(M^-3/2+ε), with M restricted to square roots of positive integers.
Exact mathematical passage
|c_{M^2}|=O_\varepsilon(M^{-3/2+\varepsilon})
M^{1/2+o(1)}Use in source · 1 passage
Search aliases: M^1/2+o(1)
Use in source · body.tex, line 723
Order the prime divisors as p_1<⋯<p_d. Since p_i≥ i+1, fracφ(Q)Q=∏_i=1^d(1-p_i^-1) ≥∏_i=1^dfrac ii+1=frac1d+1. Also ∏_pnmid Q(1-p^-2)≥1/ζ(2). The error in [source reference] is M^1/2+o(1), while its main term is at least RM/[ζ(2)(d+1)]. Therefore eventually N_Q,R(M)≥fracRM2ζ(2)fracφ(Q)Q. On every counted base, |μ(m)|=1, and Lemma~[source reference] gives |Δ_Qw_M^2(m)|≥κ_R/(4R^2M^2). Their contribution is a lower bound for the full majorant. Multiplication gives [source reference].
Exact mathematical passage
M^{1/2+o(1)}
M^{1/2-\varepsilon-o(1)}Use in source · 1 passage
Search aliases: M^1/2-ε-o(1) varepsilon
Use in source · body.tex, line 748
To see the incompatibility with the target, choose any fixed amplitude exponent loss ε<1/2 in the M formulation of the criterion. A quantity bounded below by M^-1-o(1) cannot also be bounded above by a constant times M^-3/2+ε. Their ratio grows as M^1/2-ε-o(1). The criterion requires every positive ε, so a failure for one such exponent already shows that this majorant cannot supply it. The signed coefficient can still be much smaller than the majorant.
Exact mathematical passage
M^{1/2-\varepsilon-o(1)}
m^2Use in source · 3 passages
Search aliases: m^2
Use in source · body.tex, line 579
At K=1, the weights are rational, since w_1(n)=n^-2-n^-4. Factoring the divisor sums gives Δ_6w_1(m)=frac23m^2-frac2527m^4, Δ_30w_1(m)=frac1625m^2-frac208225m^4. The following entries are exact. Every listed base is coprime to the Q of its row. The final column multiplies the cube value by the remaining Möbius sign. rrrrr Q · m · μ(m) · Δ_Qw_1(m) · Contribution
Exact mathematical passage
\label{rh:smallcubes} \Delta_6w_1(m)=\frac2{3m^2}-\frac{25}{27m^4}, \qquad \Delta_{30}w_1(m)=\frac{16}{25m^2}-\frac{208}{225m^4}.
m^4Use in source · 1 passage
Search aliases: m^4
Use in source · body.tex, line 579
At K=1, the weights are rational, since w_1(n)=n^-2-n^-4. Factoring the divisor sums gives Δ_6w_1(m)=frac23m^2-frac2527m^4, Δ_30w_1(m)=frac1625m^2-frac208225m^4. The following entries are exact. Every listed base is coprime to the Q of its row. The final column multiplies the cube value by the remaining Möbius sign. rrrrr Q · m · μ(m) · Δ_Qw_1(m) · Contribution
Exact mathematical passage
\label{rh:smallcubes} \Delta_6w_1(m)=\frac2{3m^2}-\frac{25}{27m^4}, \qquad \Delta_{30}w_1(m)=\frac{16}{25m^2}-\frac{208}{225m^4}.
M^{o(1)}Use in source · 3 passages
Search aliases: M^o(1)
Use in source · body.tex, line 732
The dimension condition means that the cube size 2^d is M^o(1). In this regime the target amplitude is M^-3/2+o(1), whereas the triangle majorant is at least M^-1-o(1). The missing factor is M^-1/2+o(1). It must arise from cancellation or another operation which retains and controls the signs.
Exact mathematical passage
M^{o(1)}
m_0Use in source · 5 passages
Search aliases: m_0
Use in source · body.tex, line 149
, absolute summation justifies integration of each zero term: ∫_0^∞ e^tz^2+uz-wu du =frace^tz^2w-z. For a compact set of w avoiding all centered zeros, large ordinates make |w-z| large. The Gaussian coefficient then gives a summable bound for the meromorphic series. Near one distinct centered zero z_0, isolate all copies of that zero. If its multiplicity is m_0, their sum is fracm_0e^tz_0^2w-z_0. Every other term is holomorphic in a sufficiently small disk. The residue cannot vanish because m_0≥…
Exact mathematical passage
m_0
M_tUse in source · 3 passages
Search aliases: M_t
Use in source · body.tex, line 160
The packet also makes the obstruction quantitative. Define δ_ζ=sup_ρ∈ Z(Reρ-1/2), M_t(U)=max(1,sup_0≤ u≤ U|Θ_t(u)|). Then 0≤δ_ζ≤1/2. The same Laplace argument proves limsup_U→∞fraclog M_t(U)U=δ_ζ. The upper bound follows from summing absolute values. If a smaller exponential rate held, the Laplace transform would be holomorphic beyond that rate, contradicting a pole at a zero further to the right. An off line zero may have a very large ordinate and a very small Gaussian coefficient. Its coeffici…
Exact mathematical passage
\label{rh:delta} \delta_\zeta=\sup_{\rho\in\mathcal Z}(\Re\rho-1/2), \qquad M_t(U)=\max\left(1,\sup_{0\le u\le U}|\Theta_t(u)|\right).
\maxUse in source · 1 passage
Search aliases: max
Use in source · body.tex, line 160
The packet also makes the obstruction quantitative. Define δ_ζ=sup_ρ∈ Z(Reρ-1/2), M_t(U)=max(1,sup_0≤ u≤ U|Θ_t(u)|). Then 0≤δ_ζ≤1/2. The same Laplace argument proves limsup_U→∞fraclog M_t(U)U=δ_ζ. The upper bound follows from summing absolute values. If a smaller exponential rate held, the Laplace transform would be holomorphic beyond that rate, contradicting a pole at a zero further to the right. An off line zero may have a very large ordinate and a very small Gaussian coefficient. Its coeffici…
Exact mathematical passage
\label{rh:delta} \delta_\zeta=\sup_{\rho\in\mathcal Z}(\Re\rho-1/2), \qquad M_t(U)=\max\left(1,\sup_{0\le u\le U}|\Theta_t(u)|\right).
\midUse in source · 11 passages
Search aliases: mid
Use in source · body.tex, line 666
merger obstruction] Suppose two positive sources n>m realize the same affine word with data (k,S,C) and endpoints y>x. Then n-m is a positive multiple of 2^S. In particular, if 2^S≥ n, no smaller positive source can share that word. Subtract the endpoint equations to obtain 3^k(n-m)=2^S(y-x). Coprimality of 3^k and 2^S forces 2^Smid n-m. A positive multiple of 2^S is at least 2^S, so m=n-(n-m)≤0 when 2^S≥ n.
Exact mathematical passage
2^S\mid n-m
NUse in source · 22 passages
Search aliases: N
Use in source · body.tex, line 537
[No eventually bounded noncrossing deficit] For a positive odd actual orbit with ρ_r≤1 at every time, the deficit is unbounded on every tail. More precisely, for every pair of integers D,N≥0 there exists r≥ N with 2^D+12^S_r≤3^r. Assume instead that d_r≤ D for every r≥ N. At any such starting time, a local block of k valuations has sum s=S_r+k-S_r satisfying s<klog_2 3+D+1. This follows from ρ_r+k≤1 and ρ_r>2^-D-1. Take k=5ℓ, where ℓ≥1 is an integer. Since 3^5<2^8, every local word has total val…
Exact mathematical passage
D,N\ge0
n^{-1/2}Use in source · 1 passage
Search aliases: n^-1/2
Use in source · body.tex, line 460
The simple coefficient mass bound also has an elementary proof. Since Λ(n)≤log X for n≤ X, and ∑_n≤ Xn^-1/2≤2sqrt X, their product gives the stated estimate. Compression may therefore be arranged without first proving any delicate prime cancellation theorem. The remaining arithmetic estimate lies in the compressed packet, not in the omitted Gaussian tail.
Exact mathematical passage
\sum_{n\le X}n^{-1/2}\le2\sqrt X
n^{-2}Use in source · 2 passages
Search aliases: n^-2
Use in source · body.tex, line 518
ζ(4)+frac1ζ(6). These formulas are exact. At large K, their alternating binomial terms can be individually much larger than their sum, so straightforward floating point evaluation may lose accuracy. A finite expression is not automatically a stable evaluation scheme. For a direct truncation of the original series at n=N, the elementary certified remainder bound is instead |c_K-∑_n≤ Nμ(n)w_K(n)| ≤∑_n>Nn^-2≤frac1N. This bound is useful for checking a finite computation, but it is much too coarse t…
Exact mathematical passage
\left|c_K-\sum_{n\le N}\mu(n)w_K(n)\right| \le\sum_{n>N}n^{-2}\le\frac1N.
n^{-2j-2}Use in source · 1 passage
Search aliases: n^-2j-2
Use in source · body.tex, line 510
Although c_K is written as an infinite arithmetic series, each fixed K has an exact finite expression in even zeta values. Expand its power by the binomial theorem: c_K=∑_j=0^K(-1)^jbinom Kjfrac1ζ(2j+2). To justify this identity, note first that the expansion has only K+1 terms. Each inner series ∑μ(n)n^-2j-2 is absolutely convergent. Its Euler product is ∏_p(1-p^-2j-2), which equals the reciprocal zeta value in the same half plane of absolute convergence. Thus no limiting interchange near the c…
Exact mathematical passage
\sum\mu(n)n^{-2j-2}
n^{-4}Use in source · 1 passage
Search aliases: n^-4
Use in source · body.tex, line 578
At K=1, the weights are rational, since w_1(n)=n^-2-n^-4. Factoring the divisor sums gives Δ_6w_1(m)=frac23m^2-frac2527m^4, Δ_30w_1(m)=frac1625m^2-frac208225m^4. The following entries are exact. Every listed base is coprime to the Q of its row. The final column multiplies the cube value by the remaining Möbius sign. rrrrr Q · m · μ(m) · Δ_Qw_1(m) · Contribution
Exact mathematical passage
w_1(n)=n^{-2}-n^{-4}
n^{-s}Use in source · 1 passage
Search aliases: n^-s
Use in source · body.tex, line 16
For a complex number s with real part greater than one, the zeta function is ζ(s)=∑_n=1^∞n^-s =∏_p(1-p^-s)^-1, where n ranges over positive integers and p ranges over primes. Analytic continuation extends it to a meromorphic function on the complex plane with a simple pole at s=1. The completed function ξ(s)=frac12s(s-1)π^-s/2Γ(s/2)ζ(s) is entire and satisfies ξ(s)=ξ(1-s). Here Γ is the gamma function. These are classical facts, with the normalization in [source reference] fixed throughout this …
Exact mathematical passage
\label{rh:zeta} \zeta(s)=\sum_{n=1}^{\infty}n^{-s} =\prod_{p}(1-p^{-s})^{-1},
n^{-\sigma}Use in source · 1 passage
Search aliases: n^-σ sigma
Use in source · body.tex, line 271
. Differentiating gives -fracζ'(s)ζ(s) =∑_p∑_k≥1(log p)p^-ks =∑_n≥2fracΛ(n)n^s. The factor k from differentiation cancels the denominator k, which is why the coefficient of p^k is log p, rather than log(p^k). Absolute convergence of these derivative series holds for Re s>1. For instance, Λ(n)≤log n, and ∑(log n)n^-σ converges for σ>1.
Exact mathematical passage
\sum(\log n)n^{-\sigma}
n^2Use in source · 2 passages
Search aliases: n^2
Use in source · body.tex, line 394
s approximately 0.999999875, and the Gram condition number (1+κ)/(1-κ) is about 1.60×10^7. These values diagnose a coordinate system, not the size of a zeta coefficient. The asymptotic explanation is exact: 1-e^-(a-b)^2/(8t)∼frac(a-b)^28t, log(n+1)-log n∼frac1n. Consecutive integer centers therefore produce a Gram condition number of order n^2 at fixed heat width.
Exact mathematical passage
n^2
n^sUse in source · 1 passage
Search aliases: n^s
Use in source · body.tex, line 266
The appearance of Λ can be traced directly to the Euler product in its first convergence half plane. There, logζ(s)=∑_p∑_k≥1fracp^-ksk. Differentiating gives -fracζ'(s)ζ(s) =∑_p∑_k≥1(log p)p^-ks =∑_n≥2fracΛ(n)n^s. The factor k from differentiation cancels the denominator k, which is why the coefficient of p^k is log p, rather than log(p^k). Absolute convergence of these derivative series holds for Re s>1. For instance, Λ(n)≤log n, and ∑(log n)n^-σ converges for σ>1.
Exact mathematical passage
-\frac{\zeta'(s)}{\zeta(s)} =\sum_p\sum_{k\ge1}(\log p)p^{-ks} =\sum_{n\ge2}\frac{\Lambda(n)}{n^s}.
n_*Use in source · 3 passages
Search aliases: n_*
Use in source · body.tex, line 216
The threshold as a fixed point. When G=2^S-3^k>0, the affine transfer has one real fixed point n_* = fracC_kG=fracc_w1-λ_w. For a real input above n_*, the affine expression falls. For an input below n_*, it rises. Equality gives a fixed point of the affine expression. These are statements about that expression before its integer cylinder is imposed. An actual periodic state would additionally have to be positive odd and realize the word.
Exact mathematical passage
n_* = \frac{C_k}{G}=\frac{c_w}{1-\lambda_w}.
n_1Use in source · 1 passage
Search aliases: n_1
Use in source · body.tex, line 401
The first examples are n_1=3 and n_2=151. The latter follows the exact odd orbit 151,227,341,1. The family does more than show that some prefixes can grow. It certifies arbitrarily large first descent times while retaining termination. Such delays must therefore be allowed in any proposed global proof.
Exact mathematical passage
n_1=3
n_2Use in source · 2 passages
Search aliases: n_2
Use in source · body.tex, line 401
The first examples are n_1=3 and n_2=151. The latter follows the exact odd orbit 151,227,341,1. The family does more than show that some prefixes can grow. It certifies arbitrarily large first descent times while retaining termination. Such delays must therefore be allowed in any proposed global proof.
Exact mathematical passage
n_2=151
n_jUse in source · 4 passages
Search aliases: n_j
Use in source · body.tex, line 374
Explicit terminating seeds with unbounded delay. The program also constructs a closed family of sources whose first descent occurs arbitrarily late, yet whose first lower value is exactly 1 [source reference]. Define q_j=frac2^3^j+13^j+1, n_j=2^j+1q_j-1.
Exact mathematical passage
\label{eq:seeds} q_j=\frac{2^{3^j}+1}{3^{j+1}},\qquad n_j=2^{j+1}q_j-1.
N_{Q,R}Use in source · 3 passages
Search aliases: N_Q,R
Use in source · body.tex, line 667
[Squarefree coprime bases] For fixed R>0, the number N_Q,R(M) of squarefree integers m in [RM,2RM] with (m,Q)=1 satisfies, uniformly in squarefree Q, N_Q,R(M)=RMfracφ(Q)Q ∏_pnmid Q(1-p^-2)+O_R(2^dsqrt M). Here φ(Q) is Euler's totient function, which counts the positive residues coprime to Q.
Exact mathematical passage
N_{Q,R}(M)
\nmidUse in source · 4 passages
Search aliases: nmid
Use in source · body.tex, line 668
[Squarefree coprime bases] For fixed R>0, the number N_Q,R(M) of squarefree integers m in [RM,2RM] with (m,Q)=1 satisfies, uniformly in squarefree Q, N_Q,R(M)=RMfracφ(Q)Q ∏_pnmid Q(1-p^-2)+O_R(2^dsqrt M). Here φ(Q) is Euler's totient function, which counts the positive residues coprime to Q.
Exact mathematical passage
\label{rh:countformula} N_{Q,R}(M)=RM\frac{\varphi(Q)}Q \prod_{p\nmid Q}(1-p^{-2})+O_R(2^d\sqrt M).
OUse in source · 15 passages
Search aliases: O
Use in source · body.tex, line 79
ally on compact sets of translations and on compact intervals of positive heat times. Indeed, for real u, |e^tz_ρ^2+uz_ρ| ≤ e^t/4+|u|/2e^-tγ^2. For complex u, there is an additional exponential factor linear in |γ| on each compact set. Gaussian decay still dominates. The classical zero counting estimate N(Y)=O(Ylog Y), where N(Y) counts zeros with 0<Imρ≤ Y, gives summability [source reference]. The same domination permits differentiation. Consequently ∂_tΘ_t=∂_u^2Θ_t. The zero symmetries make Θ_…
Exact mathematical passage
N(Y)=O(Y\log Y)
oUse in source · 18 passages
Search aliases: o
Use in source · body.tex, line 432
Suppose a fixed positive width constant C>0 is chosen and the active centers in a shell satisfy |a-c_J|≤ Csqrt J+O(1) and t≥ t_0>0. Put α=C^2/(4t_0). Then λ≤α J+O(sqrt J). For a fixed constant q>α, projection through R_J=lceil qJrceil has norm error bounded by exp(-frac12 I(qmidα)J+o(J)), I(qmidα)=qlog(q/α)-q+α. The coefficient mass of the positive prime packet through X_J=exp(J+O(sqrt J)) is at most ∑_n≤ X_JfracΛ(n)sqrt n ≤2sqrtX_Jlog X_J=exp(J/2+o(J)). Thus I(qmidα)>1 makes the aggregate compr…
Exact mathematical passage
\label{rh:rate} \exp\left(-\frac12 I(q\mid\alpha)J+o(J)\right), \qquad I(q\mid\alpha)=q\log(q/\alpha)-q+\alpha.
O_RUse in source · 4 passages
Search aliases: O_R
Use in source · body.tex, line 668
[Squarefree coprime bases] For fixed R>0, the number N_Q,R(M) of squarefree integers m in [RM,2RM] with (m,Q)=1 satisfies, uniformly in squarefree Q, N_Q,R(M)=RMfracφ(Q)Q ∏_pnmid Q(1-p^-2)+O_R(2^dsqrt M). Here φ(Q) is Euler's totient function, which counts the positive residues coprime to Q.
Exact mathematical passage
\label{rh:countformula} N_{Q,R}(M)=RM\frac{\varphi(Q)}Q \prod_{p\nmid Q}(1-p^{-2})+O_R(2^d\sqrt M).
O_\varepsilonUse in source · 2 passages
Search aliases: O_ε varepsilon
Use in source · body.tex, line 495
[Báez Duarte criterion] The Riemann hypothesis holds if and only if, for every ε>0, |c_K|=O_ε(K^-3/4+ε). Equivalently, for every ε>0, |c_M^2|=O_ε(M^-3/2+ε), with M restricted to square roots of positive integers.
Exact mathematical passage
\label{rh:bdcriterion} |c_K|=O_\varepsilon(K^{-3/4+\varepsilon}).
\operatorname{oddPart}Use in source · 1 passage
Search aliases: oddPart operatorname
Use in source · body.tex, line 481
· ⟼31104t+8261⟼5832t+1549. The valuations are 1,1,1,1,1,4. The next numerator is 3(5832t+1549)+1=2^3(2187t+581). Therefore A^7(n)=oddPart(2187t+581) ≤2187t+581<4096t+1087. The final valuation may exceed 3 for some parameters. The inequality remains valid because removing further factors of two only reduces the positive quotient. This proof is represented in the recovered deep residue certificate source.
Exact mathematical passage
A^7(n)=\operatorname{oddPart}(2187t+581) \le2187t+581<4096t+1087.
\overline zUse in source · 1 passage
Search aliases: overline z overline
Use in source · body.tex, line 36
Write ρ=β+iγ, with real numbers β and γ, and define its centered coordinate z_ρ=ρ-frac12. The symmetries become z↦-z and z↦overline z. Under the conjecture, every centered coordinate is purely imaginary. Away from the line, the real part of z becomes an exponential growth parameter in the packet below.
Exact mathematical passage
z\mapsto\overline z
\overline{1/2+i\gamma}Use in source · 1 passage
Search aliases: overline1/2+iγ overline gamma
Use in source · body.tex, line 56
On the critical line, conjugation and reflection identify the same point: overline1/2+iγ=1-(1/2+iγ). Therefore ξ(1/2+iγ) is real. Real values on that line explain why one can track sign changes in a one dimensional plot. They do not explain why every zero must lie there. A real analytic function restricted to one line can have additional zeros elsewhere in the complex plane.
Exact mathematical passage
\overline{1/2+i\gamma}=1-(1/2+i\gamma)
\overline\rhoUse in source · 2 passages
Search aliases: overlineρ overline rho
Use in source · body.tex, line 26
Let Z be the multiset of nontrivial zeros, with every zero repeated according to its multiplicity. They lie in the strip 0<Re s<1. Conjugation and the functional equation imply that a zero ρ is accompanied by overlineρ, 1-ρ, and 1-overlineρ, allowing coincidences on the critical line [source reference].
Exact mathematical passage
\overline\rho
PUse in source · 2 passages
Search aliases: P
Use in source · body.tex, line 367
rm one in L^2( R,du). For a chosen center c, let ψ_r,t,c, r=0,1,2,…, be the orthonormal scaled Hermite functions ψ_r,t,c(u)= frac(2π t)^-1/4sqrt2^r r! H_r(fracu-csqrt2t)e^-(u-c)^2/(4t). Here H_r(x)=(-1)^r e^x^2(d^r/dx^r)e^-x^2 is the physicists' Hermite polynomial. If P_R is orthogonal projection onto indices 0≤ r≤ R, the coherent expansion gives g_t,a · =e^-λ/2 ∑_r≥0fracσ^rsqrtr!ψ_r,t,c,
Exact mathematical passage
P_R
p^{-2}Use in source · 6 passages
Search aliases: p^-2
Use in source · body.tex, line 605
The cube's shape is also scale dependent. At fixed K, large m makes every factor (1-1/(bm)^2)^K close to one, so its leading coefficient is ∏_pmid Q(1-p^-2). At the active scale masympsqrt K, the different divisors sample different parts of the weight profile. This difference can create local cancellation within a cube. It does not remove the remaining sign μ(m), which is the quantity the signed criterion still asks us to control.
Exact mathematical passage
\prod_{p\mid Q}(1-p^{-2})
p^{-2j-2}Use in source · 1 passage
Search aliases: p^-2j-2
Use in source · body.tex, line 510
hmetic series, each fixed K has an exact finite expression in even zeta values. Expand its power by the binomial theorem: c_K=∑_j=0^K(-1)^jbinom Kjfrac1ζ(2j+2). To justify this identity, note first that the expansion has only K+1 terms. Each inner series ∑μ(n)n^-2j-2 is absolutely convergent. Its Euler product is ∏_p(1-p^-2j-2), which equals the reciprocal zeta value in the same half plane of absolute convergence. Thus no limiting interchange near the critical line is involved.
Exact mathematical passage
\prod_p(1-p^{-2j-2})
p^{-ks}Use in source · 2 passages
Search aliases: p^-ks
Use in source · body.tex, line 262
The appearance of Λ can be traced directly to the Euler product in its first convergence half plane. There, logζ(s)=∑_p∑_k≥1fracp^-ksk. Differentiating gives -fracζ'(s)ζ(s) =∑_p∑_k≥1(log p)p^-ks =∑_n≥2fracΛ(n)n^s. The factor k from differentiation cancels the denominator k, which is why the coefficient of p^k is log p, rather than log(p^k). Absolute convergence of these derivative series holds for Re s>1. For instance, Λ(n)≤log n, and ∑(log n)n^-σ converges for σ>1.
Exact mathematical passage
\log\zeta(s)=\sum_p\sum_{k\ge1}\frac{p^{-ks}}k.
p^{-s}Use in source · 1 passage
Search aliases: p^-s
Use in source · body.tex, line 16
For a complex number s with real part greater than one, the zeta function is ζ(s)=∑_n=1^∞n^-s =∏_p(1-p^-s)^-1, where n ranges over positive integers and p ranges over primes. Analytic continuation extends it to a meromorphic function on the complex plane with a simple pole at s=1. The completed function ξ(s)=frac12s(s-1)π^-s/2Γ(s/2)ζ(s) is entire and satisfies ξ(s)=ξ(1-s). Here Γ is the gamma function. These are classical facts, with the normalization in [source reference] fixed throughout this …
Exact mathematical passage
\label{rh:zeta} \zeta(s)=\sum_{n=1}^{\infty}n^{-s} =\prod_{p}(1-p^{-s})^{-1},
p_1Use in source · 1 passage
Search aliases: p_1
Use in source · body.tex, line 717
Order the prime divisors as p_1<⋯<p_d. Since p_i≥ i+1, fracφ(Q)Q=∏_i=1^d(1-p_i^-1) ≥∏_i=1^dfrac ii+1=frac1d+1. Also ∏_pnmid Q(1-p^-2)≥1/ζ(2). The error in [source reference] is M^1/2+o(1), while its main term is at least RM/[ζ(2)(d+1)]. Therefore eventually N_Q,R(M)≥fracRM2ζ(2)fracφ(Q)Q. On every counted base, |μ(m)|=1, and Lemma~[source reference] gives |Δ_Qw_M^2(m)|≥κ_R/(4R^2M^2). Their contribution is a lower bound for the full majorant. Multiplication gives [source reference].
Exact mathematical passage
p_1<\cdots<p_d
p_dUse in source · 1 passage
Search aliases: p_d
Use in source · body.tex, line 717
Order the prime divisors as p_1<⋯<p_d. Since p_i≥ i+1, fracφ(Q)Q=∏_i=1^d(1-p_i^-1) ≥∏_i=1^dfrac ii+1=frac1d+1. Also ∏_pnmid Q(1-p^-2)≥1/ζ(2). The error in [source reference] is M^1/2+o(1), while its main term is at least RM/[ζ(2)(d+1)]. Therefore eventually N_Q,R(M)≥fracRM2ζ(2)fracφ(Q)Q. On every counted base, |μ(m)|=1, and Lemma~[source reference] gives |Δ_Qw_M^2(m)|≥κ_R/(4R^2M^2). Their contribution is a lower bound for the full majorant. Multiplication gives [source reference].
Exact mathematical passage
p_1<\cdots<p_d
p_iUse in source · 1 passage
Search aliases: p_i
Use in source · body.tex, line 717
Order the prime divisors as p_1<⋯<p_d. Since p_i≥ i+1, fracφ(Q)Q=∏_i=1^d(1-p_i^-1) ≥∏_i=1^dfrac ii+1=frac1d+1. Also ∏_pnmid Q(1-p^-2)≥1/ζ(2). The error in [source reference] is M^1/2+o(1), while its main term is at least RM/[ζ(2)(d+1)]. Therefore eventually N_Q,R(M)≥fracRM2ζ(2)fracφ(Q)Q. On every counted base, |μ(m)|=1, and Lemma~[source reference] gives |Δ_Qw_M^2(m)|≥κ_R/(4R^2M^2). Their contribution is a lower bound for the full majorant. Multiplication gives [source reference].
Exact mathematical passage
p_i\ge i+1
p_i^{-1}Use in source · 1 passage
Search aliases: p_i^-1
Use in source · body.tex, line 718
Order the prime divisors as p_1<⋯<p_d. Since p_i≥ i+1, fracφ(Q)Q=∏_i=1^d(1-p_i^-1) ≥∏_i=1^dfrac ii+1=frac1d+1. Also ∏_pnmid Q(1-p^-2)≥1/ζ(2). The error in [source reference] is M^1/2+o(1), while its main term is at least RM/[ζ(2)(d+1)]. Therefore eventually N_Q,R(M)≥fracRM2ζ(2)fracφ(Q)Q. On every counted base, |μ(m)|=1, and Lemma~[source reference] gives |Δ_Qw_M^2(m)|≥κ_R/(4R^2M^2). Their contribution is a lower bound for the full majorant. Multiplication gives [source reference].
Exact mathematical passage
\frac{\varphi(Q)}Q=\prod_{i=1}^d(1-p_i^{-1}) \ge\prod_{i=1}^d\frac i{i+1}=\frac1{d+1}.
P_RUse in source · 2 passages
Search aliases: P_R
Use in source · body.tex, line 367
rm one in L^2( R,du). For a chosen center c, let ψ_r,t,c, r=0,1,2,…, be the orthonormal scaled Hermite functions ψ_r,t,c(u)= frac(2π t)^-1/4sqrt2^r r! H_r(fracu-csqrt2t)e^-(u-c)^2/(4t). Here H_r(x)=(-1)^r e^x^2(d^r/dx^r)e^-x^2 is the physicists' Hermite polynomial. If P_R is orthogonal projection onto indices 0≤ r≤ R, the coherent expansion gives g_t,a · =e^-λ/2 ∑_r≥0fracσ^rsqrtr!ψ_r,t,c,
Exact mathematical passage
P_R
\pmodUse in source · 10 passages
Search aliases: pmod
Use in source · body.tex, line 94
[Exact integer cylinder] Let w be a nonempty valuation word of length k and total depth S. A positive odd integer n realizes w if and only if 3^k n+C_k≡ 2^Spmod2^S+1. Consequently the realizing integers form one odd residue class modulo 2^S+1, whose residue is u_w≡(2^S-C_k)(3^k)^-1pmod2^S+1. If the word is realized, [source reference] has an odd terminal quotient x_k. Multiplying an odd integer by 2^S gives exactly the congruence [source reference].
Exact mathematical passage
\label{eq:cylinder} 3^k n+C_k\equiv 2^S\pmod{2^{S+1}}.
\PrUse in source · 1 passage
Search aliases: Pr
Use in source · body.tex, line 447
The exponent in [source reference] comes from a standard elementary exponential bound for a Poisson variable X of mean λ. For h>0, Pr(X≥ R)≤ e^-hR E(e^hX) =exp(-hR+λ(e^h-1)). When R>λ>0, choose e^h=R/λ. For λ=0, every positive tail is zero. The positive mean case gives the rate Rlog(R/λ)-R+λ. Passing to the squared norm tail and then taking a square root supplies the factor one half in [source reference]. Changing R by one has no effect on its leading shell exponent.
Exact mathematical passage
\Pr(X\ge R)\le e^{-hR}\mathbb E(e^{hX}) =\exp\bigl(-hR+\lambda(e^h-1)\bigr).
qUse in source · 30 passages
Search aliases: q
Use in source · body.tex, line 258
One can also retain a weaker certificate when exact final valuation is unnecessary. If the last numerator is known to equal 2^b q with q>0 integral, then the next odd state is at most q. Exact equality requires q odd. A descent proof that only uses an upper bound can accept both parities. Example~[source reference] exploits this distinction. It obtains a uniform falling estimate without claiming one final exact word for every member of its coarse progression.
Exact mathematical passage
2^b q
q_2Use in source · 1 passage
Search aliases: q_2
Use in source · body.tex, line 438
The full progression through 151. The parameter j=2 gives 3^j=9 and q_2=frac2^9+13^3=19, n_2=8·19-1=151. The exit equation is 27·19=2^9·1+1. Thus the first two valuations are 1 and the third is 10. Their sum is 12. The cylinder modulus is 2^13=8192, rather than 2^12. That last factor of two keeps the exiting quotient odd.
Exact mathematical passage
q_2=\frac{2^9+1}{3^3}=19,\qquad n_2=8\cdot19-1=151.
q_jUse in source · 3 passages
Search aliases: q_j
Use in source · body.tex, line 374
Explicit terminating seeds with unbounded delay. The program also constructs a closed family of sources whose first descent occurs arbitrarily late, yet whose first lower value is exactly 1 [source reference]. Define q_j=frac2^3^j+13^j+1, n_j=2^j+1q_j-1.
Exact mathematical passage
\label{eq:seeds} q_j=\frac{2^{3^j}+1}{3^{j+1}},\qquad n_j=2^{j+1}q_j-1.
R^2Use in source · 4 passages
Search aliases: R^2
Use in source · body.tex, line 633
[Uniform positive cube tail] For a fixed R>0, put κ_R=frac1ζ(2)-fracζ(4)R^2. For every squarefree Q and every integer m≥ RM, Δ_Qw_M^2(m)≥fracκ_Rm^2. In particular, R=3 gives κ_R>0.
Exact mathematical passage
\label{rh:cr} \kappa_R=\frac1{\zeta(2)}-\frac{\zeta(4)}{R^2}.
r_1Use in source · 1 passage
Search aliases: r_1
Use in source · body.tex, line 167
figplaceholdercollatz_residue_treeValuation words give exact dyadic cylinders. The diagram labels the first two valuations by r_1=a_0 and r_2=a_1. Refinement reveals further valuation information. A bounded diagram displays finitely many branches of this infinite arithmetic structure.
Exact mathematical passage
r_1=a_0
r_2Use in source · 1 passage
Search aliases: r_2
Use in source · body.tex, line 167
figplaceholdercollatz_residue_treeValuation words give exact dyadic cylinders. The diagram labels the first two valuations by r_1=a_0 and r_2=a_1. Refinement reveals further valuation information. A bounded diagram displays finitely many branches of this infinite arithmetic structure.
Exact mathematical passage
r_2=a_1
R_JUse in source · 1 passage
Search aliases: R_J
Use in source · body.tex, line 431
Suppose a fixed positive width constant C>0 is chosen and the active centers in a shell satisfy |a-c_J|≤ Csqrt J+O(1) and t≥ t_0>0. Put α=C^2/(4t_0). Then λ≤α J+O(sqrt J). For a fixed constant q>α, projection through R_J=lceil qJrceil has norm error bounded by exp(-frac12 I(qmidα)J+o(J)), I(qmidα)=qlog(q/α)-q+α. The coefficient mass of the positive prime packet through X_J=exp(J+O(sqrt J)) is at most ∑_n≤ X_JfracΛ(n)sqrt n ≤2sqrtX_Jlog X_J=exp(J/2+o(J)). Thus I(qmidα)>1 makes the aggregate compr…
Exact mathematical passage
R_J=\lceil qJ\rceil
r_{z_0}Use in source · 2 passages
Search aliases: r_z_0
Use in source · body.tex, line 343
Near a distinct centered zero z_0 of multiplicity m_0, the residue vector is r_z_0(t)=m_0ρ_0(1-ρ_0)e^tz_0^2. It is continuous and never zero on [t_0,T], so its H norm is positive. To reduce the contradiction to ordinary complex analysis, choose the continuous linear functional given by pairing with this residue vector. The scalar transform then has nonzero residue r_z_0 _H^2, whereas the Laplace integral remains holomorphic. This makes explicit why integrating over heat times cannot average the …
Exact mathematical passage
r_{z_0}(t)=m_0\rho_0(1-\rho_0)e^{tz_0^2}.
\rceilUse in source · 2 passages
Search aliases: rceil
Use in source · body.tex, line 570
For R>N, Lemmas~[source reference] and [source reference] place the R-N distinct states x_N,…,x_R-1 in E_ℓ below H_R=lceil2^D+1(n+R/3)rceil. Thus R-N≤ε_ℓ H_R+W_ℓ. Fix ℓ and let R tend to infinity. Division by R yields 1≤ε_ℓ2^D+1/3. Choosing ℓ sufficiently large contradicts this inequality. No eventual bounded deficit strip exists.
Exact mathematical passage
H_R=\lceil2^{D+1}(n+R/3)\rceil
S'Use in source · 1 passage
Search aliases: S'
Use in source · body.tex, line 670
This is the obstruction behind pulling a smaller endpoint backward through an unchanged late prefix. A useful smaller source merger must eventually change the word. For two different transfers ending at the same y, elimination gives m=frac2^S'3^k n+2^S'C-2^S C'2^S3^k'. The remaining task is to force a realizing alternative with 0<m<n. The formula does not itself supply such an alternative. The saved route notes develop this direction while retaining its unproved universal existence step [source …
Exact mathematical passage
m=\frac{2^{S'}3^k n+2^{S'}C-2^S C'}{2^S3^{k'}}.
S_2Use in source · 1 passage
Search aliases: S_2
Use in source · body.tex, line 302
For example, the word (1,2) has S_2=3 and C_2=5, so 8·31=9·27+5. The corresponding cylinder is n≡11pmod16. Writing n=16t+11 gives A(n)=24t+17, A^2(n)=18t+13. Every member of this progression grows over the two steps, even though the second step falls from 24t+17 to 18t+13. For t=1, the local fall is 41 to 31, while both states remain above 27. These identities are present in the recovered Lean source [source reference].
Exact mathematical passage
S_2=3
S_{37}Use in source · 1 passage
Search aliases: S_37
Use in source · body.tex, line 313
The fresh finite integer audit for this edition finds the first actual descent at accelerated time 37: x_37=23, S_37=59, C_37=1100931843921811423. Thus 2^59·23=3^37·27+1100931843921811423. The same segment has 59 shortcut steps. The historical serious audit separately records a Lean check of T^59(27)=23 [source reference]. The two statements use different iteration counts and the same endpoint.
Exact mathematical passage
\label{eq:27descent} x_{37}=23,\qquad S_{37}=59,\qquad C_{37}=1100931843921811423.
S_IUse in source · 2 passages
Search aliases: S_I
Use in source · body.tex, line 619
For any finite interval I of remaining bases, distinguish its signed contribution from its absolute mass: S_I=∑_substackm∈ I (m,Q)=1μ(m)Δ_Qw_M^2(m), V_I=∑_substackm∈ I (m,Q)=1 |μ(m)| |Δ_Qw_M^2(m)|. The valid inequalities are |S_I|≤ V_I≤ A_Q(M) and |c_M^2|≤ A_Q(M). There is no general inequality |c_M^2|≤ V_I. Terms outside the chosen interval remain in the full signed sum. Even a two term toy sum with values 3 inside the interval and 10 outside it shows why its interval mass cannot bound the full…
Exact mathematical passage
S_I=\sum_{\substack{m\in I\\(m,Q)=1}}\mu(m)\Delta_Qw_{M^2}(m), \qquad V_I=\sum_{\substack{m\in I\\(m,Q)=1}} |\mu(m)|\,|\Delta_Qw_{M^2}(m)|.
S_iUse in source · 1 passage
Search aliases: S_i
Use in source · body.tex, line 69
[Affine packet law] If n realizes w, then every prefix satisfies 2^S_jx_j=3^j n+C_j. The terminal offset also has the explicit expression C_k=∑_i=0^k-13^k-1-i2^S_i. At j=0 the identity is x_0=n. If it holds at j, multiply [source reference] by 2^S_j to obtain 2^S_j+1x_j+1=3^j+1n+3C_j+2^S_j. This is the next identity by [source reference]. Iterating the same offset recurrence gives [source reference].
Exact mathematical passage
\label{eq:offsetsum} C_k=\sum_{i=0}^{k-1}3^{k-1-i}2^{S_i}.
S_{j+1}Use in source · 1 passage
Search aliases: S_j+1
Use in source · body.tex, line 75
If n realizes w, then every prefix satisfies 2^S_jx_j=3^j n+C_j. The terminal offset also has the explicit expression C_k=∑_i=0^k-13^k-1-i2^S_i. At j=0 the identity is x_0=n. If it holds at j, multiply [source reference] by 2^S_j to obtain 2^S_j+1x_j+1=3^j+1n+3C_j+2^S_j. This is the next identity by [source reference]. Iterating the same offset recurrence gives [source reference].
Exact mathematical passage
2^{S_{j+1}}x_{j+1}=3^{j+1}n+3C_j+2^{S_j}.
S_{r+k}Use in source · 1 passage
Search aliases: S_r+k
Use in source · body.tex, line 543
e odd actual orbit with ρ_r≤1 at every time, the deficit is unbounded on every tail. More precisely, for every pair of integers D,N≥0 there exists r≥ N with 2^D+12^S_r≤3^r. Assume instead that d_r≤ D for every r≥ N. At any such starting time, a local block of k valuations has sum s=S_r+k-S_r satisfying s<klog_2 3+D+1. This follows from ρ_r+k≤1 and ρ_r>2^-D-1. Take k=5ℓ, where ℓ≥1 is an integer. Since 3^5<2^8, every local word has total valuation at most h=8ℓ+D+1.
Exact mathematical passage
s=S_{r+k}-S_r
S_uUse in source · 1 passage
Search aliases: S_u
Use in source · body.tex, line 526
[Injectivity] A permanently noncrossing positive odd orbit never repeats a state. Suppose x_u=x_v with v>u. Put k=v-u and B=S_v-S_u. The affine law for this positive length block gives (2^B-3^k)x_u=C_u,v>0. Hence 2^B/3^k>1. Determinism makes the same states and valuations repeat, so ρ_u+hk=ρ_u(2^B/3^k)^h. This exceeds 1 for sufficiently large h, contrary to permanent noncrossing.
Exact mathematical passage
B=S_v-S_u
S_vUse in source · 1 passage
Search aliases: S_v
Use in source · body.tex, line 526
[Injectivity] A permanently noncrossing positive odd orbit never repeats a state. Suppose x_u=x_v with v>u. Put k=v-u and B=S_v-S_u. The affine law for this positive length block gives (2^B-3^k)x_u=C_u,v>0. Hence 2^B/3^k>1. Determinism makes the same states and valuations repeat, so ρ_u+hk=ρ_u(2^B/3^k)^h. This exceeds 1 for sufficiently large h, contrary to permanent noncrossing.
Exact mathematical passage
B=S_v-S_u
\supUse in source · 1 passage
Search aliases: sup
Use in source · body.tex, line 160
The packet also makes the obstruction quantitative. Define δ_ζ=sup_ρ∈ Z(Reρ-1/2), M_t(U)=max(1,sup_0≤ u≤ U|Θ_t(u)|). Then 0≤δ_ζ≤1/2. The same Laplace argument proves limsup_U→∞fraclog M_t(U)U=δ_ζ. The upper bound follows from summing absolute values. If a smaller exponential rate held, the Laplace transform would be holomorphic beyond that rate, contradicting a pole at a zero further to the right. An off line zero may have a very large ordinate and a very small Gaussian coefficient. Its coeffici…
Exact mathematical passage
\label{rh:delta} \delta_\zeta=\sup_{\rho\in\mathcal Z}(\Re\rho-1/2), \qquad M_t(U)=\max\left(1,\sup_{0\le u\le U}|\Theta_t(u)|\right).
\sup_{0\le u\le U}Use in source · 1 passage
Search aliases: sup_0≤ u≤ U sup le le
Use in source · body.tex, line 160
The packet also makes the obstruction quantitative. Define δ_ζ=sup_ρ∈ Z(Reρ-1/2), M_t(U)=max(1,sup_0≤ u≤ U|Θ_t(u)|). Then 0≤δ_ζ≤1/2. The same Laplace argument proves limsup_U→∞fraclog M_t(U)U=δ_ζ. The upper bound follows from summing absolute values. If a smaller exponential rate held, the Laplace transform would be holomorphic beyond that rate, contradicting a pole at a zero further to the right. An off line zero may have a very large ordinate and a very small Gaussian coefficient. Its coeffici…
Exact mathematical passage
\label{rh:delta} \delta_\zeta=\sup_{\rho\in\mathcal Z}(\Re\rho-1/2), \qquad M_t(U)=\max\left(1,\sup_{0\le u\le U}|\Theta_t(u)|\right).
\sup_{\rho\in\mathcal Z}Use in source · 1 passage
Search aliases: sup_ρ∈ Z sup rho in mathcal
Use in source · body.tex, line 160
The packet also makes the obstruction quantitative. Define δ_ζ=sup_ρ∈ Z(Reρ-1/2), M_t(U)=max(1,sup_0≤ u≤ U|Θ_t(u)|). Then 0≤δ_ζ≤1/2. The same Laplace argument proves limsup_U→∞fraclog M_t(U)U=δ_ζ. The upper bound follows from summing absolute values. If a smaller exponential rate held, the Laplace transform would be holomorphic beyond that rate, contradicting a pole at a zero further to the right. An off line zero may have a very large ordinate and a very small Gaussian coefficient. Its coeffici…
Exact mathematical passage
\label{rh:delta} \delta_\zeta=\sup_{\rho\in\mathcal Z}(\Re\rho-1/2), \qquad M_t(U)=\max\left(1,\sup_{0\le u\le U}|\Theta_t(u)|\right).
t^2Use in source · 1 passage
Search aliases: t^2
Use in source · body.tex, line 250
For a translated Gaussian G_t,a(u)=e^-(u-a)^2/(4t), direct differentiation gives LG_t,a(u)= (frac14+frac12t-frac(u-a)^24t^2) G_t,a(u). The annihilated prime packet therefore carries a polynomial factor with both signs. Positivity of the original Gaussian sum is no longer positivity of every transformed term. On the zero side the same operator is simpler because each exponential is an eigenfunction of ∂_u^2. On the explicit pole channel, ∂_u^2cosh(u/2)=frac14cosh(u/2), so cancellation is exact.
Exact mathematical passage
LG_{t,a}(u)= \left(\frac14+\frac1{2t}-\frac{(u-a)^2}{4t^2}\right) G_{t,a}(u).
T^{59}Use in source · 1 passage
Search aliases: T^59
Use in source · body.tex, line 321
The fresh finite integer audit for this edition finds the first actual descent at accelerated time 37: x_37=23, S_37=59, C_37=1100931843921811423. Thus 2^59·23=3^37·27+1100931843921811423. The same segment has 59 shortcut steps. The historical serious audit separately records a Lean check of T^59(27)=23 [source reference]. The two statements use different iteration counts and the same endpoint.
Exact mathematical passage
T^{59}(27)=23
t_0Use in source · 13 passages
Search aliases: t_0
Use in source · body.tex, line 91
It helps to make the normal convergence argument quantitative enough to see which operations it permits. Write z=x+iγ, where |x|<1/2, and u=a+ib. Then Re(tz^2+uz)=t(x^2-γ^2)+ax-bγ. If t∈[t_0,T] and |a|,|b|≤ B, this is at most T/4+B/2-t_0γ^2+B|γ| ≤ T/4+B/2+fracB^22t_0-fract_02γ^2. The last step is the elementary inequality B|γ|≤(t_0/2)γ^2+B^2/(2t_0). Thus a compact set of heat times and translations has one Gaussian majorant. A finite number of derivatives only inserts a polynomial in z. Such a p…
Exact mathematical passage
t\in[t_0,T]
UUse in source · 8 passages
Search aliases: U
Use in source · body.tex, line 178
[Concatenation] Let a word w have data (k,S,C) and a following word v have data (ℓ,U,D). Their concatenation has data (k+ℓ,S+U,3^ℓ C+2^S D). On every realizing source, its transfer is F_v∘ F_w. Substitute 2^S y=3^k n+C into 2^U z=3^ℓ y+D. Multiplication by 2^S gives 2^S+Uz=3^k+ℓn+3^ℓ C+2^S D. The exact cylinder theorem identifies the admissible integer sources.
Exact mathematical passage
(\ell,U,D)
u^2Use in source · 1 passage
Search aliases: u^2
Use in source · body.tex, line 383
For every j≥0, q_j is a positive integer. At j=0 the numerator is 3. If 3^j+1 divides u+1 for u=2^3^j, then 2^3^j+1+1=u^3+1=(u+1)(u^2-u+1). Since u≡-1pmod3, the second factor is divisible by 3. Induction proves divisibility by 3^j+2. Positivity of the quotient is immediate.
Exact mathematical passage
2^{3^{j+1}}+1=u^3+1=(u+1)(u^2-u+1).
u^3Use in source · 1 passage
Search aliases: u^3
Use in source · body.tex, line 383
For every j≥0, q_j is a positive integer. At j=0 the numerator is 3. If 3^j+1 divides u+1 for u=2^3^j, then 2^3^j+1+1=u^3+1=(u+1)(u^2-u+1). Since u≡-1pmod3, the second factor is divisible by 3. Induction proves divisibility by 3^j+2. Positivity of the quotient is immediate.
Exact mathematical passage
2^{3^{j+1}}+1=u^3+1=(u+1)(u^2-u+1).
u_wUse in source · 1 passage
Search aliases: u_w
Use in source · body.tex, line 98
[Exact integer cylinder] Let w be a nonempty valuation word of length k and total depth S. A positive odd integer n realizes w if and only if 3^k n+C_k≡ 2^Spmod2^S+1. Consequently the realizing integers form one odd residue class modulo 2^S+1, whose residue is u_w≡(2^S-C_k)(3^k)^-1pmod2^S+1. If the word is realized, [source reference] has an odd terminal quotient x_k. Multiplying an odd integer by 2^S gives exactly the congruence [source reference].
Exact mathematical passage
\label{eq:address} u_w\equiv(2^S-C_k)(3^k)^{-1}\pmod{2^{S+1}}.
VUse in source · 3 passages
Search aliases: V
Use in source · body.tex, line 619
For any finite interval I of remaining bases, distinguish its signed contribution from its absolute mass: S_I=∑_substackm∈ I (m,Q)=1μ(m)Δ_Qw_M^2(m), V_I=∑_substackm∈ I (m,Q)=1 |μ(m)| |Δ_Qw_M^2(m)|. The valid inequalities are |S_I|≤ V_I≤ A_Q(M) and |c_M^2|≤ A_Q(M). There is no general inequality |c_M^2|≤ V_I. Terms outside the chosen interval remain in the full signed sum. Even a two term toy sum with values 3 inside the interval and 10 outside it shows why its interval mass cannot bound the full…
Exact mathematical passage
S_I=\sum_{\substack{m\in I\\(m,Q)=1}}\mu(m)\Delta_Qw_{M^2}(m), \qquad V_I=\sum_{\substack{m\in I\\(m,Q)=1}} |\mu(m)|\,|\Delta_Qw_{M^2}(m)|.
v^rUse in source · 1 passage
Search aliases: v^r
Use in source · body.tex, line 406
Put x=(u-c)/sqrt2t. The ratio of a translated Gaussian to the centered one is fracg_t,a(u)g_t,c(u) =e^sqrt2σ x-σ^2. The Hermite generating identity e^2xv-v^2=∑_r≥0H_r(x)v^r/r!, with v=σ/sqrt2, turns this ratio into e^-σ^2/2∑_r≥0 fracH_r(x)σ^r2^r/2r!. Substitution of the normalized Hermite functions gives [source reference]. Squaring its orthogonal coefficients gives e^-λλ^r/r!. Their total is one because the exponential series sums to e^λ. The discarded squared norm is therefore exactly a Poisso…
Exact mathematical passage
e^{2xv-v^2}=\sum_{r\ge0}H_r(x)v^r/r!
V_IUse in source · 3 passages
Search aliases: V_I
Use in source · body.tex, line 619
For any finite interval I of remaining bases, distinguish its signed contribution from its absolute mass: S_I=∑_substackm∈ I (m,Q)=1μ(m)Δ_Qw_M^2(m), V_I=∑_substackm∈ I (m,Q)=1 |μ(m)| |Δ_Qw_M^2(m)|. The valid inequalities are |S_I|≤ V_I≤ A_Q(M) and |c_M^2|≤ A_Q(M). There is no general inequality |c_M^2|≤ V_I. Terms outside the chosen interval remain in the full signed sum. Even a two term toy sum with values 3 inside the interval and 10 outside it shows why its interval mass cannot bound the full…
Exact mathematical passage
S_I=\sum_{\substack{m\in I\\(m,Q)=1}}\mu(m)\Delta_Qw_{M^2}(m), \qquad V_I=\sum_{\substack{m\in I\\(m,Q)=1}} |\mu(m)|\,|\Delta_Qw_{M^2}(m)|.
w_1Use in source · 3 passages
Search aliases: w_1
Use in source · body.tex, line 578
At K=1, the weights are rational, since w_1(n)=n^-2-n^-4. Factoring the divisor sums gives Δ_6w_1(m)=frac23m^2-frac2527m^4, Δ_30w_1(m)=frac1625m^2-frac208225m^4. The following entries are exact. Every listed base is coprime to the Q of its row. The final column multiplies the cube value by the remaining Möbius sign. rrrrr Q · m · μ(m) · Δ_Qw_1(m) · Contribution
Exact mathematical passage
w_1(n)=n^{-2}-n^{-4}
\widetilde{\mathfrak A}Use in source · 3 passages
Search aliases: widetilde A widetilde mathfrak
Use in source · body.tex, line 243
The identity 1/4-z_ρ^2=ρ(1-ρ) explains the weight. It is nonzero for every nontrivial zero. Thus the operator kills the explicit pole contribution in [source reference] while retaining every zero mode. Differentiated normal convergence gives widetildeΘ_t=widetilde A_t- widetilde P_t, where tildes on the right mean application of L.
Exact mathematical passage
\label{rh:tildeexplicit} \widetilde\Theta_t=\widetilde{\mathfrak A}_t- \widetilde{\mathfrak P}_t,
\widetilde{\mathfrak A}_tUse in source · 3 passages
Search aliases: widetilde A_t widetilde mathfrak
Use in source · body.tex, line 243
The identity 1/4-z_ρ^2=ρ(1-ρ) explains the weight. It is nonzero for every nontrivial zero. Thus the operator kills the explicit pole contribution in [source reference] while retaining every zero mode. Differentiated normal convergence gives widetildeΘ_t=widetilde A_t- widetilde P_t, where tildes on the right mean application of L.
Exact mathematical passage
\label{rh:tildeexplicit} \widetilde\Theta_t=\widetilde{\mathfrak A}_t- \widetilde{\mathfrak P}_t,
\widetilde E_J^{1/2}Use in source · 1 passage
Search aliases: widetilde E_J^1/2 widetilde
Use in source · body.tex, line 334
The energy statement is deliberately local in u. It does not assume a global square integrable packet, which would already rule out many possible growth behaviors. The Hilbert space H collects only the fixed positive heat interval. Tonelli's theorem identifies widetilde E_J=∫_J^J+1 widetildeΘ_·(u) _H^2 du. Let a>0. On a unit band, Cauchy Schwarz gives ∫_J^J+1 widetildeΘ_·(u) _H e^-au du ≤ e^-aJwidetilde E_J^1/2. Choose the energy exponent to be ε=a. Subexponential growth then bounds the right si…
Exact mathematical passage
\int_J^{J+1}\|\widetilde\Theta_\cdot(u)\|_H e^{-au}\,du \le e^{-aJ}\widetilde E_J^{1/2}.
\widetilde{\mathfrak P}Use in source · 2 passages
Search aliases: widetilde P widetilde mathfrak
Use in source · body.tex, line 243
The identity 1/4-z_ρ^2=ρ(1-ρ) explains the weight. It is nonzero for every nontrivial zero. Thus the operator kills the explicit pole contribution in [source reference] while retaining every zero mode. Differentiated normal convergence gives widetildeΘ_t=widetilde A_t- widetilde P_t, where tildes on the right mean application of L.
Exact mathematical passage
\label{rh:tildeexplicit} \widetilde\Theta_t=\widetilde{\mathfrak A}_t- \widetilde{\mathfrak P}_t,
\widetilde{\mathfrak P}_tUse in source · 2 passages
Search aliases: widetilde P_t widetilde mathfrak
Use in source · body.tex, line 243
The identity 1/4-z_ρ^2=ρ(1-ρ) explains the weight. It is nonzero for every nontrivial zero. Thus the operator kills the explicit pole contribution in [source reference] while retaining every zero mode. Differentiated normal convergence gives widetildeΘ_t=widetilde A_t- widetilde P_t, where tildes on the right mean application of L.
Exact mathematical passage
\label{rh:tildeexplicit} \widetilde\Theta_t=\widetilde{\mathfrak A}_t- \widetilde{\mathfrak P}_t,
\widetilde\Theta_{t+s}Use in source · 1 passage
Search aliases: widetildeΘ_t+s widetilde Theta
Use in source · body.tex, line 465
The ground channel shows why the distinction matters. Direct integration gives ∫_ Rg_s,c(u)widetildeΘ_t(u) du =K_swidetildeΘ_t+s(c), K_s=2sqrtπ s(2π s)^-1/4. The Gaussian test makes this integral absolutely convergent even without the conjecture. It is a Gaussian pairing, with no assertion that the unweighted zero packet belongs to L^2( R). Completing the square proves the identity term by term. Normal convergence and the Gaussian majorant justify summation [source reference].
Exact mathematical passage
\label{rh:ground} \int_{\mathbb R}g_{s,c}(u)\widetilde\Theta_t(u)\,du =K_s\widetilde\Theta_{t+s}(c), \qquad K_s=2\sqrt{\pi s}(2\pi s)^{-1/4}.
\widetilde\Theta_\cdotUse in source · 4 passages
Search aliases: widetildeΘ_· widetilde Theta cdot
Use in source · body.tex, line 312
Conversely, regard u↦widetildeΘ_·(u) as a function taking values in H=L^2([t_0,T]; C). Cauchy Schwarz on each unit interval and the assumed energy bound imply ∫_0^∞ widetildeΘ_·(u) _H e^-au du<∞ (a>0). Its H-valued Laplace transform is holomorphic for Re w>0. Initially for Re w>1/2, it also equals the normally convergent meromorphic series with terms fracρ(1-ρ)e^t z_ρ^2w-z_ρ. The residue at a distinct centered zero is a nonzero function of t in H. Indeed, the multiplicity and ρ(1-ρ) are nonzero,…
Exact mathematical passage
u\mapsto\widetilde\Theta_\cdot(u)
XUse in source · 7 passages
Search aliases: X
Use in source · body.tex, line 437
(1) and t≥ t_0>0. Put α=C^2/(4t_0). Then λ≤α J+O(sqrt J). For a fixed constant q>α, projection through R_J=lceil qJrceil has norm error bounded by exp(-frac12 I(qmidα)J+o(J)), I(qmidα)=qlog(q/α)-q+α. The coefficient mass of the positive prime packet through X_J=exp(J+O(sqrt J)) is at most ∑_n≤ X_JfracΛ(n)sqrt n ≤2sqrtX_Jlog X_J=exp(J/2+o(J)). Thus I(qmidα)>1 makes the aggregate compression error exponentially small. The pole annihilator adds only polynomial factors to the Hermite ladder and leav…
Exact mathematical passage
X_J=\exp(J+O(\sqrt J))
x'Use in source · 1 passage
Search aliases: x'
Use in source · body.tex, line 761
This appendix is finite evidence with an unusually transparent verification rule. A reader can check any row using integer multiplication and exponentiation, and can check consecutive rows using 3x+1=2^a x'. The large offsets are retained because deleting them would obscure the exact descent budget. Their size has no bearing on a universal theorem beyond this certified source.
Exact mathematical passage
3x+1=2^a x'
x^2Use in source · 3 passages
Search aliases: x^2
Use in source · body.tex, line 88
It helps to make the normal convergence argument quantitative enough to see which operations it permits. Write z=x+iγ, where |x|<1/2, and u=a+ib. Then Re(tz^2+uz)=t(x^2-γ^2)+ax-bγ. If t∈[t_0,T] and |a|,|b|≤ B, this is at most T/4+B/2-t_0γ^2+B|γ| ≤ T/4+B/2+fracB^22t_0-fract_02γ^2. The last step is the elementary inequality B|γ|≤(t_0/2)γ^2+B^2/(2t_0). Thus a compact set of heat times and translations has one Gaussian majorant. A finite number of derivatives only inserts a polynomial in z. Such a p…
Exact mathematical passage
\Re(tz^2+uz)=t(x^2-\gamma^2)+ax-b\gamma.
x^rUse in source · 1 passage
Search aliases: x^r
Use in source · body.tex, line 367
belrh:gaussian g_t,a(u)=(2π t)^-1/4e^-(u-a)^2/(4t). It has norm one in L^2( R,du). For a chosen center c, let ψ_r,t,c, r=0,1,2,…, be the orthonormal scaled Hermite functions ψ_r,t,c(u)= frac(2π t)^-1/4sqrt2^r r! H_r(fracu-csqrt2t)e^-(u-c)^2/(4t). Here H_r(x)=(-1)^r e^x^2(d^r/dx^r)e^-x^2 is the physicists' Hermite polynomial. If P_R is orthogonal projection onto indices 0≤ r≤ R, the coherent expansion gives g_t,a · =e^-λ/2 ∑_r≥0fracσ^rsqrtr!ψ_r,t,c,
Exact mathematical passage
H_r(x)=(-1)^r e^{x^2}(d^r/dx^r)e^{-x^2}
x_0Use in source · 1 passage
Search aliases: x_0
Use in source · body.tex, line 74
[Affine packet law] If n realizes w, then every prefix satisfies 2^S_jx_j=3^j n+C_j. The terminal offset also has the explicit expression C_k=∑_i=0^k-13^k-1-i2^S_i. At j=0 the identity is x_0=n. If it holds at j, multiply [source reference] by 2^S_j to obtain 2^S_j+1x_j+1=3^j+1n+3C_j+2^S_j. This is the next identity by [source reference]. Iterating the same offset recurrence gives [source reference].
Exact mathematical passage
x_0=n
x_1Use in source · 1 passage
Search aliases: x_1
Use in source · body.tex, line 83
One can see the offset being created before writing the general formula. The first step is 2^a_0x_1=3n+1. Multiplying the second step by 2^a_0 gives 2^a_0+a_1x_2=3(3n+1)+2^a_0 =9n+3+2^a_0. The first addition has been multiplied by 3. The second addition has been multiplied by the dyadic depth already removed. After a third step the offset is 9+3·2^a_0+2^a_0+a_1. Every later addition enters at its own prefix depth and receives one factor of 3 for each later odd multiplication. This explains each …
Exact mathematical passage
2^{a_0}x_1=3n+1
x_2Use in source · 1 passage
Search aliases: x_2
Use in source · body.tex, line 84
One can see the offset being created before writing the general formula. The first step is 2^a_0x_1=3n+1. Multiplying the second step by 2^a_0 gives 2^a_0+a_1x_2=3(3n+1)+2^a_0 =9n+3+2^a_0. The first addition has been multiplied by 3. The second addition has been multiplied by the dyadic depth already removed. After a third step the offset is 9+3·2^a_0+2^a_0+a_1. Every later addition enters at its own prefix depth and receives one factor of 3 for each later odd multiplication. This explains each …
Exact mathematical passage
2^{a_0+a_1}x_2=3(3n+1)+2^{a_0} =9n+3+2^{a_0}.
x_{37}Use in source · 1 passage
Search aliases: x_37
Use in source · body.tex, line 313
The fresh finite integer audit for this edition finds the first actual descent at accelerated time 37: x_37=23, S_37=59, C_37=1100931843921811423. Thus 2^59·23=3^37·27+1100931843921811423. The same segment has 59 shortcut steps. The historical serious audit separately records a Lean check of T^59(27)=23 [source reference]. The two statements use different iteration counts and the same endpoint.
Exact mathematical passage
\label{eq:27descent} x_{37}=23,\qquad S_{37}=59,\qquad C_{37}=1100931843921811423.
X_JUse in source · 2 passages
Search aliases: X_J
Use in source · body.tex, line 437
(1) and t≥ t_0>0. Put α=C^2/(4t_0). Then λ≤α J+O(sqrt J). For a fixed constant q>α, projection through R_J=lceil qJrceil has norm error bounded by exp(-frac12 I(qmidα)J+o(J)), I(qmidα)=qlog(q/α)-q+α. The coefficient mass of the positive prime packet through X_J=exp(J+O(sqrt J)) is at most ∑_n≤ X_JfracΛ(n)sqrt n ≤2sqrtX_Jlog X_J=exp(J/2+o(J)). Thus I(qmidα)>1 makes the aggregate compression error exponentially small. The pole annihilator adds only polynomial factors to the Hermite ladder and leav…
Exact mathematical passage
X_J=\exp(J+O(\sqrt J))
x_jUse in source · 4 passages
Search aliases: x_j
Use in source · body.tex, line 65
[Affine packet law] If n realizes w, then every prefix satisfies 2^S_jx_j=3^j n+C_j. The terminal offset also has the explicit expression C_k=∑_i=0^k-13^k-1-i2^S_i. At j=0 the identity is x_0=n. If it holds at j, multiply [source reference] by 2^S_j to obtain 2^S_j+1x_j+1=3^j+1n+3C_j+2^S_j. This is the next identity by [source reference]. Iterating the same offset recurrence gives [source reference].
Exact mathematical passage
\label{eq:affine} 2^{S_j}x_j=3^j n+C_j.
x_{j+1}Use in source · 3 passages
Search aliases: x_j+1
Use in source · body.tex, line 75
If n realizes w, then every prefix satisfies 2^S_jx_j=3^j n+C_j. The terminal offset also has the explicit expression C_k=∑_i=0^k-13^k-1-i2^S_i. At j=0 the identity is x_0=n. If it holds at j, multiply [source reference] by 2^S_j to obtain 2^S_j+1x_j+1=3^j+1n+3C_j+2^S_j. This is the next identity by [source reference]. Iterating the same offset recurrence gives [source reference].
Exact mathematical passage
2^{S_{j+1}}x_{j+1}=3^{j+1}n+3C_j+2^{S_j}.
x_kUse in source · 2 passages
Search aliases: x_k
Use in source · body.tex, line 103
s w if and only if 3^k n+C_k≡ 2^Spmod2^S+1. Consequently the realizing integers form one odd residue class modulo 2^S+1, whose residue is u_w≡(2^S-C_k)(3^k)^-1pmod2^S+1. If the word is realized, [source reference] has an odd terminal quotient x_k. Multiplying an odd integer by 2^S gives exactly the congruence [source reference].
Exact mathematical passage
x_k
x_NUse in source · 1 passage
Search aliases: x_N
Use in source · body.tex, line 569
For R>N, Lemmas~[source reference] and [source reference] place the R-N distinct states x_N,…,x_R-1 in E_ℓ below H_R=lceil2^D+1(n+R/3)rceil. Thus R-N≤ε_ℓ H_R+W_ℓ. Fix ℓ and let R tend to infinity. Division by R yields 1≤ε_ℓ2^D+1/3. Choosing ℓ sufficiently large contradicts this inequality. No eventual bounded deficit strip exists.
Exact mathematical passage
x_N,\ldots,x_{R-1}
x_rUse in source · 14 passages
Search aliases: x_r
Use in source · body.tex, line 48
Exact valuation packets. Fix a positive odd seed n, and write x_r=A^r(n). Each step has the exact equation 2^a_rx_r+1=3x_r+1, a_r=a(x_r). The exponent records how much dyadic information is removed at that step. It is determined by the actual state, rather than chosen independently.
Exact mathematical passage
x_r=A^r(n)
x_{r+1}Use in source · 1 passage
Search aliases: x_r+1
Use in source · body.tex, line 49
Exact valuation packets. Fix a positive odd seed n, and write x_r=A^r(n). Each step has the exact equation 2^a_rx_r+1=3x_r+1, a_r=a(x_r). The exponent records how much dyadic information is removed at that step. It is determined by the actual state, rather than chosen independently.
Exact mathematical passage
\label{eq:onestep} 2^{a_r}x_{r+1}=3x_r+1,\qquad a_r=a(x_r).
x_{R-1}Use in source · 1 passage
Search aliases: x_R-1
Use in source · body.tex, line 569
For R>N, Lemmas~[source reference] and [source reference] place the R-N distinct states x_N,…,x_R-1 in E_ℓ below H_R=lceil2^D+1(n+R/3)rceil. Thus R-N≤ε_ℓ H_R+W_ℓ. Fix ℓ and let R tend to infinity. Division by R yields 1≤ε_ℓ2^D+1/3. Choosing ℓ sufficiently large contradicts this inequality. No eventual bounded deficit strip exists.
Exact mathematical passage
x_N,\ldots,x_{R-1}
x_uUse in source · 2 passages
Search aliases: x_u
Use in source · body.tex, line 526
[Injectivity] A permanently noncrossing positive odd orbit never repeats a state. Suppose x_u=x_v with v>u. Put k=v-u and B=S_v-S_u. The affine law for this positive length block gives (2^B-3^k)x_u=C_u,v>0. Hence 2^B/3^k>1. Determinism makes the same states and valuations repeat, so ρ_u+hk=ρ_u(2^B/3^k)^h. This exceeds 1 for sufficiently large h, contrary to permanent noncrossing.
Exact mathematical passage
x_u=x_v
x_vUse in source · 1 passage
Search aliases: x_v
Use in source · body.tex, line 526
[Injectivity] A permanently noncrossing positive odd orbit never repeats a state. Suppose x_u=x_v with v>u. Put k=v-u and B=S_v-S_u. The affine law for this positive length block gives (2^B-3^k)x_u=C_u,v>0. Hence 2^B/3^k>1. Determinism makes the same states and valuations repeat, so ρ_u+hk=ρ_u(2^B/3^k)^h. This exceeds 1 for sufficiently large h, contrary to permanent noncrossing.
Exact mathematical passage
x_u=x_v
YUse in source · 3 passages
Search aliases: Y
Use in source · body.tex, line 79
ally on compact sets of translations and on compact intervals of positive heat times. Indeed, for real u, |e^tz_ρ^2+uz_ρ| ≤ e^t/4+|u|/2e^-tγ^2. For complex u, there is an additional exponential factor linear in |γ| on each compact set. Gaussian decay still dominates. The classical zero counting estimate N(Y)=O(Ylog Y), where N(Y) counts zeros with 0<Imρ≤ Y, gives summability [source reference]. The same domination permits differentiation. Consequently ∂_tΘ_t=∂_u^2Θ_t. The zero symmetries make Θ_…
Exact mathematical passage
N(Y)=O(Y\log Y)
z^{-8\ell}Use in source · 1 passage
Search aliases: z^-8ℓ ell
Use in source · body.tex, line 606
The weight z=3/4 tilts the word sum toward the low total valuations permitted by the hypothetical strip. Since 0<z<1 and s≤ h, one has z^-s≤ z^-h. The identity ∑_a≥1(z/2)^a=fracz2-z=frac35 then accounts for all possible positive valuations in each of the 5ℓ positions. Combining this factor with z^-8ℓ gives (frac43)^8ℓ(frac35)^5ℓ =(frac6553684375)^ℓ. This exact ratio is the source of rarity. It is a bound for a finite union of residue classes, expressed through an infinite convergent generating s…
Exact mathematical passage
z^{-8\ell}
z^{-h}Use in source · 3 passages
Search aliases: z^-h
Use in source · body.tex, line 560
Set z=3/4. For s≤ h, 2^-s≤ z^-h(z/2)^s. Summing over all positive words bounds the truncated sum by ε_ℓ≤frac12z^-h (∑_a≥1(z/2)^a)^5ℓ =frac12(frac43)^D+1 (frac6553684375)^ℓ. The exact rational factor is less than 1, so ε_ℓ tends to zero.
Exact mathematical passage
2^{-s}\le z^{-h}(z/2)^s
z^{-s}Use in source · 1 passage
Search aliases: z^-s
Use in source · body.tex, line 602
The weight z=3/4 tilts the word sum toward the low total valuations permitted by the hypothetical strip. Since 0<z<1 and s≤ h, one has z^-s≤ z^-h. The identity ∑_a≥1(z/2)^a=fracz2-z=frac35 then accounts for all possible positive valuations in each of the 5ℓ positions. Combining this factor with z^-8ℓ gives (frac43)^8ℓ(frac35)^5ℓ =(frac6553684375)^ℓ. This exact ratio is the source of rarity. It is a bound for a finite union of residue classes, expressed through an infinite convergent generating s…
Exact mathematical passage
z^{-s}\le z^{-h}
z^2Use in source · 5 passages
Search aliases: z^2
Use in source · body.tex, line 88
It helps to make the normal convergence argument quantitative enough to see which operations it permits. Write z=x+iγ, where |x|<1/2, and u=a+ib. Then Re(tz^2+uz)=t(x^2-γ^2)+ax-bγ. If t∈[t_0,T] and |a|,|b|≤ B, this is at most T/4+B/2-t_0γ^2+B|γ| ≤ T/4+B/2+fracB^22t_0-fract_02γ^2. The last step is the elementary inequality B|γ|≤(t_0/2)γ^2+B^2/(2t_0). Thus a compact set of heat times and translations has one Gaussian majorant. A finite number of derivatives only inserts a polynomial in z. Such a p…
Exact mathematical passage
\Re(tz^2+uz)=t(x^2-\gamma^2)+ax-b\gamma.
z_0Use in source · 7 passages
Search aliases: z_0
Use in source · body.tex, line 149
In the smaller initial domain Re w>1/2, absolute summation justifies integration of each zero term: ∫_0^∞ e^tz^2+uz-wu du =frace^tz^2w-z. For a compact set of w avoiding all centered zeros, large ordinates make |w-z| large. The Gaussian coefficient then gives a summable bound for the meromorphic series. Near one distinct centered zero z_0, isolate all copies of that zero. If its multiplicity is m_0, their sum is fracm_0e^tz_0^2w-z_0. Every other term is holomorphic in a sufficiently small disk. …
Exact mathematical passage
z_0
z_0^2Use in source · 2 passages
Search aliases: z_0^2
Use in source · body.tex, line 150
justifies integration of each zero term: ∫_0^∞ e^tz^2+uz-wu du =frace^tz^2w-z. For a compact set of w avoiding all centered zeros, large ordinates make |w-z| large. The Gaussian coefficient then gives a summable bound for the meromorphic series. Near one distinct centered zero z_0, isolate all copies of that zero. If its multiplicity is m_0, their sum is fracm_0e^tz_0^2w-z_0. Every other term is holomorphic in a sufficiently small disk. The residue cannot vanish because m_0≥1 and an exponential …
Exact mathematical passage
\frac{m_0e^{tz_0^2}}{w-z_0}.
z_pUse in source · 2 passages
Search aliases: z_p
Use in source · body.tex, line 760
Third, global Hardy norms on a prime torus are blind to the squarefree Möbius sign. For a finite prime set, a squarefree monomial is ∏_p∈ Sz_p. Replacing every coordinate by -z_p multiplies it by (-1)^|S|, exactly its Möbius coefficient. The substitution preserves Haar measure on the torus. Every global L^q norm formed with that measure is consequently unchanged by this sign operation. A bound in that invariant norm cannot distinguish the signed packet from the corresponding unsigned squarefree …
Exact mathematical passage
\prod_{p\in S}z_p
|Use in source · 36 passages
Search aliases: |
Use in source · body.tex, line 75
The series converges normally on compact sets of translations and on compact intervals of positive heat times. Indeed, for real u, |e^tz_ρ^2+uz_ρ| ≤ e^t/4+|u|/2e^-tγ^2. For complex u, there is an additional exponential factor linear in |γ| on each compact set. Gaussian decay still dominates. The classical zero counting estimate N(Y)=O(Ylog Y), where N(Y) counts zeros with 0<Imρ≤ Y, gives summability [source reference]. The same domination permits differentiation. Consequently ∂_tΘ_t=∂_u^2Θ_t. Th…
Exact mathematical passage
\label{rh:maj} \left|e^{tz_\rho^2+uz_\rho}\right| \le e^{t/4+|u|/2}e^{-t\gamma^2}.
|^2Use in source · 4 passages
Search aliases: |^2
Use in source · body.tex, line 302
[Local energy criterion] Fix 0<t_0<T<∞. For positive integers J, set widetilde E_J=∫_t_0^T∫_J^J+1 |widetildeΘ_t(u)|^2 du dt. The Riemann hypothesis is equivalent to subexponential growth of widetilde E_J.
Exact mathematical passage
\label{rh:energydef} \widetilde E_J=\int_{t_0}^T\int_J^{J+1} |\widetilde\Theta_t(u)|^2\,du\,dt.
|^4Use in source · 1 passage
Search aliases: |^4
Use in source · body.tex, line 763
Fourth, a raw positive fourth moment cannot lie below the square of its quadratic energy in a normalized probability space. If ν is its probability measure and F∈ L^4(ν) is the packet, Cauchy Schwarz gives ∫ |F|^4 dν≥(∫ |F|^2 dν)^2. This is the precise obstruction to a strict saving below that scale for the raw moment. It does not assert positivity of every separately grouped collision term for arbitrary signed coefficients. A viable replacement would require a signed expression, a justified quo…
Exact mathematical passage
\int |F|^4\,d\nu\ge\left(\int |F|^2\,d\nu\right)^2.
\pmUse in source · 1 passage
Search aliases: ± pm
Use in source · body.tex, line 257
The choice of operator is tied to a declared removal target. Its two scalar roots are the centered coordinates ±1/2 of the exceptional points s=0,1. Actual nontrivial zeros are strictly inside the critical strip, so none is removed. A different operator which vanished at an unknown nontrivial zero would no longer preserve the obstruction required by the criterion.
Exact mathematical passage
\pm1/2
\cdotUse in source · 24 passages
Search aliases: · cdot
Use in source · body.tex, line 88
One can see the offset being created before writing the general formula. The first step is 2^a_0x_1=3n+1. Multiplying the second step by 2^a_0 gives 2^a_0+a_1x_2=3(3n+1)+2^a_0 =9n+3+2^a_0. The first addition has been multiplied by 3. The second addition has been multiplied by the dyadic depth already removed. After a third step the offset is 9+3·2^a_0+2^a_0+a_1. Every later addition enters at its own prefix depth and receives one factor of 3 for each later odd multiplication. This explains each …
Exact mathematical passage
9+3\cdot2^{a_0}+2^{a_0+a_1}
\timesUse in source · 7 passages
Search aliases: × times
Use in source · body.tex, line 389
For example, take t=1, a=log1000, and b=log1001. The separation is approximately 0.00099950, the overlap is approximately 0.999999875, and the Gram condition number (1+κ)/(1-κ) is about 1.60×10^7. These values diagnose a coordinate system, not the size of a zeta coefficient. The asymptotic explanation is exact: 1-e^-(a-b)^2/(8t)∼frac(a-b)^28t, log(n+1)-log n∼frac1n. Consecutive integer centers therefore produce a Gram condition number of order n^2 at fixed heat width.
Exact mathematical passage
1.60\times10^7
\alphaUse in source · 10 passages
Search aliases: α alpha
Use in source · body.tex, line 613
The contradiction can also be written without a limit. Set B=2^D+1 and choose ℓ so that α=ε_ℓ B/3<1. Since H_R≤ B(n+R/3)+1, the packing inequality would force (1-α)R≤ N+ε_ℓ(Bn+1)+W_ℓ for every integer R>N. The right side is fixed after D,N,n,ℓ are fixed. Selecting an integer R above that finite bound contradicts [source reference]. The limit in the theorem's proof is a compact way of expressing the same comparison.
Exact mathematical passage
\alpha=\varepsilon_\ell B/3<1
\gamma^2Use in source · 7 passages
Search aliases: γ^2 gamma
Use in source · body.tex, line 75
The series converges normally on compact sets of translations and on compact intervals of positive heat times. Indeed, for real u, |e^tz_ρ^2+uz_ρ| ≤ e^t/4+|u|/2e^-tγ^2. For complex u, there is an additional exponential factor linear in |γ| on each compact set. Gaussian decay still dominates. The classical zero counting estimate N(Y)=O(Ylog Y), where N(Y) counts zeros with 0<Imρ≤ Y, gives summability [source reference]. The same domination permits differentiation. Consequently ∂_tΘ_t=∂_u^2Θ_t. Th…
Exact mathematical passage
\label{rh:maj} \left|e^{tz_\rho^2+uz_\rho}\right| \le e^{t/4+|u|/2}e^{-t\gamma^2}.
\deltaUse in source · 10 passages
Search aliases: δ delta
Use in source · body.tex, line 45
The reflection is s↦1-s, not ordinary complex conjugation. These two operations commute, and together they generate the symmetric zero pattern. For example, a hypothetical zero ρ=frac12+δ+iγ, δ>0, would bring centered coordinates δ+iγ, δ-iγ, -δ+iγ, -δ-iγ. The quartet may collapse when δ=0 or γ=0. We sum the actual multiset, so coincidences are handled by multiplicity rather than by pretending that four distinct zeros are always present.
Exact mathematical passage
\rho=\frac12+\delta+i\gamma,\qquad \delta>0,
\Delta_{30}Use in source · 2 passages
Search aliases: Δ_30 Delta
Use in source · body.tex, line 562
For Q=6, the divisors correspond to the subsets of {2,3}, and the exact packet is Δ_6w_K(m)=w_K(m)-w_K(2m)-w_K(3m)+w_K(6m). For Q=30, one adds a prime coordinate: Δ_30w_K(m)= · w_K(m)-w_K(2m)-w_K(3m)-w_K(5m) · +w_K(6m)+w_K(10m)+w_K(15m)-w_K(30m). The complete divisor signs are shown here to make the grouping unambiguous. rrrrrrrrr b · 1 · 2 · 3 · 5 · 6 · 10 · 15 · 30
Exact mathematical passage
\Delta_{30}w_K(m)={}&w_K(m)-w_K(2m)-w_K(3m)-w_K(5m)\\ &+w_K(6m)+w_K(10m)+w_K(15m)-w_K(30m).
\Delta_6Use in source · 2 passages
Search aliases: Δ_6 Delta
Use in source · body.tex, line 558
For Q=6, the divisors correspond to the subsets of {2,3}, and the exact packet is Δ_6w_K(m)=w_K(m)-w_K(2m)-w_K(3m)+w_K(6m). For Q=30, one adds a prime coordinate: Δ_30w_K(m)= · w_K(m)-w_K(2m)-w_K(3m)-w_K(5m)
Exact mathematical passage
\Delta_6w_K(m)=w_K(m)-w_K(2m)-w_K(3m)+w_K(6m).
\delta_\zetaUse in source · 5 passages
Search aliases: δ_ζ delta zeta
Use in source · body.tex, line 160
The packet also makes the obstruction quantitative. Define δ_ζ=sup_ρ∈ Z(Reρ-1/2), M_t(U)=max(1,sup_0≤ u≤ U|Θ_t(u)|). Then 0≤δ_ζ≤1/2. The same Laplace argument proves limsup_U→∞fraclog M_t(U)U=δ_ζ. The upper bound follows from summing absolute values. If a smaller exponential rate held, the Laplace transform would be holomorphic beyond that rate, contradicting a pole at a zero further to the right. An off line zero may have a very large ordinate and a very small Gaussian coefficient. Its coeffici…
Exact mathematical passage
\label{rh:delta} \delta_\zeta=\sup_{\rho\in\mathcal Z}(\Re\rho-1/2), \qquad M_t(U)=\max\left(1,\sup_{0\le u\le U}|\Theta_t(u)|\right).
\zetaUse in source · 22 passages
Search aliases: ζ zeta
Use in source · body.tex, line 16
For a complex number s with real part greater than one, the zeta function is ζ(s)=∑_n=1^∞n^-s =∏_p(1-p^-s)^-1, where n ranges over positive integers and p ranges over primes. Analytic continuation extends it to a meromorphic function on the complex plane with a simple pole at s=1. The completed function ξ(s)=frac12s(s-1)π^-s/2Γ(s/2)ζ(s) is entire and satisfies ξ(s)=ξ(1-s). Here Γ is the gamma function. These are classical facts, with the normalization in [source reference] fixed throughout this …
Exact mathematical passage
\label{rh:zeta} \zeta(s)=\sum_{n=1}^{\infty}n^{-s} =\prod_{p}(1-p^{-s})^{-1},
\zeta'Use in source · 1 passage
Search aliases: ζ' zeta
Use in source · body.tex, line 266
The appearance of Λ can be traced directly to the Euler product in its first convergence half plane. There, logζ(s)=∑_p∑_k≥1fracp^-ksk. Differentiating gives -fracζ'(s)ζ(s) =∑_p∑_k≥1(log p)p^-ks =∑_n≥2fracΛ(n)n^s. The factor k from differentiation cancels the denominator k, which is why the coefficient of p^k is log p, rather than log(p^k). Absolute convergence of these derivative series holds for Re s>1. For instance, Λ(n)≤log n, and ∑(log n)n^-σ converges for σ>1.
Exact mathematical passage
-\frac{\zeta'(s)}{\zeta(s)} =\sum_p\sum_{k\ge1}(\log p)p^{-ks} =\sum_{n\ge2}\frac{\Lambda(n)}{n^s}.
\kappaUse in source · 12 passages
Search aliases: κ kappa
Use in source · body.tex, line 387
e word physical refers to the measure du in which the packet is actually evaluated. It does not mean that a separately chosen coefficient norm can be substituted. Completing the square gives the exact overlap ⟨ g_t,a,g_t,b⟩ =e^-(a-b)^2/(8t). For two translates, the Gram matrix has diagonal entries one and off diagonal entry κ=e^-(a-b)^2/(8t). Its eigenvalues are 1+κ and 1-κ. When the centers are close, their difference is almost invisible in the physical norm even if the two raw coefficients are…
Exact mathematical passage
\kappa=e^{-(a-b)^2/(8t)}
\kappa_3Use in source · 1 passage
Search aliases: κ_3 kappa
Use in source · body.tex, line 659
The bound is uniform in the chosen primes. The finite product cannot fall below its product over all primes, while the error sum cannot exceed its sum over all positive integers. Taking R=3 gives κ_3=frac6π^2-fracπ^4810>0. The first term is approximately 0.6079 and the subtracted term approximately 0.1203. Thus the selected tail lies beyond the central region where the cube's own signs may fluctuate. Its positivity is a useful rigorous fact, but it still leaves the external signs μ(m).
Exact mathematical passage
\kappa_3=\frac6{\pi^2}-\frac{\pi^4}{810}>0.
\kappa_RUse in source · 7 passages
Search aliases: κ_R kappa
Use in source · body.tex, line 633
[Uniform positive cube tail] For a fixed R>0, put κ_R=frac1ζ(2)-fracζ(4)R^2. For every squarefree Q and every integer m≥ RM, Δ_Qw_M^2(m)≥fracκ_Rm^2. In particular, R=3 gives κ_R>0.
Exact mathematical passage
\label{rh:cr} \kappa_R=\frac1{\zeta(2)}-\frac{\zeta(4)}{R^2}.
\lambda^rUse in source · 2 passages
Search aliases: λ^r lambda
Use in source · body.tex, line 368
e the orthonormal scaled Hermite functions ψ_r,t,c(u)= frac(2π t)^-1/4sqrt2^r r! H_r(fracu-csqrt2t)e^-(u-c)^2/(4t). Here H_r(x)=(-1)^r e^x^2(d^r/dx^r)e^-x^2 is the physicists' Hermite polynomial. If P_R is orthogonal projection onto indices 0≤ r≤ R, the coherent expansion gives g_t,a · =e^-λ/2 ∑_r≥0fracσ^rsqrtr!ψ_r,t,c, σ · =fraca-c2sqrt t, λ=σ^2, notag (I-P_R)g_t,a _2^2 · =e^-λ∑_r>Rfracλ^rr!. The last expression is a Poisson upper tail. These formulas and the compression estimates are recovered…
Exact mathematical passage
g_{t,a}&=e^{-\lambda/2} \sum_{r\ge0}\frac{\sigma^r}{\sqrt{r!}}\psi_{r,t,c}, \label{rh:coherent}\\ \sigma&=\frac{a-c}{2\sqrt t},\qquad \lambda=\sigma^2, \notag\\ \|(I-P_R)g_{t,a}\|_2^2 &=e^{-\lambda}\sum_{r>R}\frac{\lambda^r}{r!}. \label{rh:poisson}
\lambda_fUse in source · 2 passages
Search aliases: λ_f lambda
Use in source · body.tex, line 208
The composition formula also explains why a local fall need not resolve a global obstruction. If an expanding segment has data (λ_g,c_g) and a following falling segment has (λ_f,c_f), the final value is λ_fλ_g n+λ_f c_g+c_f. A comparison with the local peak controls only part of this expression. To prove descent below n, the entire expression must be compared with the same original source.
Exact mathematical passage
(\lambda_f,c_f)
\lambda_gUse in source · 2 passages
Search aliases: λ_g lambda
Use in source · body.tex, line 208
The composition formula also explains why a local fall need not resolve a global obstruction. If an expanding segment has data (λ_g,c_g) and a following falling segment has (λ_f,c_f), the final value is λ_fλ_g n+λ_f c_g+c_f. A comparison with the local peak controls only part of this expression. To prove descent below n, the entire expression must be compared with the same original source.
Exact mathematical passage
(\lambda_g,c_g)
\muUse in source · 23 passages
Search aliases: μ mu
Use in source · body.tex, line 486
The Möbius function μ(n) is zero when a square of a prime divides n. Otherwise it equals (-1)^k, where k is the number of distinct prime divisors of n, and μ(1)=1. For an integer K≥1, put w_K(n)=frac1n^2(1-frac1n^2)^K, c_K=∑_n≥1μ(n)w_K(n). The series is absolutely convergent. Its natural active scale is n of order M=sqrt K. The coefficient c_K is classical. Báez Duarte proved the following exact criterion [source reference].
Exact mathematical passage
\mu(n)
\xiUse in source · 4 passages
Search aliases: ξ xi
Use in source · body.tex, line 21
x number s with real part greater than one, the zeta function is ζ(s)=∑_n=1^∞n^-s =∏_p(1-p^-s)^-1, where n ranges over positive integers and p ranges over primes. Analytic continuation extends it to a meromorphic function on the complex plane with a simple pole at s=1. The completed function ξ(s)=frac12s(s-1)π^-s/2Γ(s/2)ζ(s) is entire and satisfies ξ(s)=ξ(1-s). Here Γ is the gamma function. These are classical facts, with the normalization in [source reference] fixed throughout this article [sou…
Exact mathematical passage
\label{rh:xi} \xi(s)=\frac12s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)
\pi^{-s/2}Use in source · 1 passage
Search aliases: π^-s/2 pi
Use in source · body.tex, line 21
x number s with real part greater than one, the zeta function is ζ(s)=∑_n=1^∞n^-s =∏_p(1-p^-s)^-1, where n ranges over positive integers and p ranges over primes. Analytic continuation extends it to a meromorphic function on the complex plane with a simple pole at s=1. The completed function ξ(s)=frac12s(s-1)π^-s/2Γ(s/2)ζ(s) is entire and satisfies ξ(s)=ξ(1-s). Here Γ is the gamma function. These are classical facts, with the normalization in [source reference] fixed throughout this article [sou…
Exact mathematical passage
\label{rh:xi} \xi(s)=\frac12s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s)
\pi^2Use in source · 1 passage
Search aliases: π^2 pi
Use in source · body.tex, line 659
The bound is uniform in the chosen primes. The finite product cannot fall below its product over all primes, while the error sum cannot exceed its sum over all positive integers. Taking R=3 gives κ_3=frac6π^2-fracπ^4810>0. The first term is approximately 0.6079 and the subtracted term approximately 0.1203. Thus the selected tail lies beyond the central region where the cube's own signs may fluctuate. Its positivity is a useful rigorous fact, but it still leaves the external signs μ(m).
Exact mathematical passage
\kappa_3=\frac6{\pi^2}-\frac{\pi^4}{810}>0.
\pi^4Use in source · 1 passage
Search aliases: π^4 pi
Use in source · body.tex, line 659
The bound is uniform in the chosen primes. The finite product cannot fall below its product over all primes, while the error sum cannot exceed its sum over all positive integers. Taking R=3 gives κ_3=frac6π^2-fracπ^4810>0. The first term is approximately 0.6079 and the subtracted term approximately 0.1203. Thus the selected tail lies beyond the central region where the cube's own signs may fluctuate. Its positivity is a useful rigorous fact, but it still leaves the external signs μ(m).
Exact mathematical passage
\kappa_3=\frac6{\pi^2}-\frac{\pi^4}{810}>0.
\rho_0Use in source · 1 passage
Search aliases: ρ_0 rho
Use in source · body.tex, line 343
Near a distinct centered zero z_0 of multiplicity m_0, the residue vector is r_z_0(t)=m_0ρ_0(1-ρ_0)e^tz_0^2. It is continuous and never zero on [t_0,T], so its H norm is positive. To reduce the contradiction to ordinary complex analysis, choose the continuous linear functional given by pairing with this residue vector. The scalar transform then has nonzero residue r_z_0 _H^2, whereas the Laplace integral remains holomorphic. This makes explicit why integrating over heat times cannot average the …
Exact mathematical passage
r_{z_0}(t)=m_0\rho_0(1-\rho_0)e^{tz_0^2}.
\rho_iUse in source · 2 passages
Search aliases: ρ_i rho
Use in source · body.tex, line 510
[Normalized offset and height] Under permanent noncrossing, ρ_r x_r=n+frac13∑_i<rρ_i≤ n+frac r3. At every time with d_r≤ D, where D≥0 is integral, x_r<2^D+1(n+frac r3). Divide the affine law by 3^r and use the offset sum. Noncrossing gives ρ_i≤1 for every preceding time, which is essential for the upper bound. From d_r≤ D we have S_r≥lfloor rlog_2 3rfloor-D>rlog_2 3-D-1. Therefore ρ_r>2^-D-1, proving the height estimate.
Exact mathematical passage
\label{eq:normalized} \rho_r x_r=n+\frac13\sum_{i<r}\rho_i\le n+\frac r3.
\rho_{r+k}Use in source · 2 passages
Search aliases: ρ_r+k rho
Use in source · body.tex, line 547
nbounded on every tail. More precisely, for every pair of integers D,N≥0 there exists r≥ N with 2^D+12^S_r≤3^r. Assume instead that d_r≤ D for every r≥ N. At any such starting time, a local block of k valuations has sum s=S_r+k-S_r satisfying s<klog_2 3+D+1. This follows from ρ_r+k≤1 and ρ_r>2^-D-1. Take k=5ℓ, where ℓ≥1 is an integer. Since 3^5<2^8, every local word has total valuation at most h=8ℓ+D+1.
Exact mathematical passage
\rho_{r+k}\le1
\rho_uUse in source · 1 passage
Search aliases: ρ_u rho
Use in source · body.tex, line 531
[Injectivity] A permanently noncrossing positive odd orbit never repeats a state. Suppose x_u=x_v with v>u. Put k=v-u and B=S_v-S_u. The affine law for this positive length block gives (2^B-3^k)x_u=C_u,v>0. Hence 2^B/3^k>1. Determinism makes the same states and valuations repeat, so ρ_u+hk=ρ_u(2^B/3^k)^h. This exceeds 1 for sufficiently large h, contrary to permanent noncrossing.
Exact mathematical passage
\rho_{u+hk}=\rho_u(2^B/3^k)^h
\rho_{u+hk}Use in source · 1 passage
Search aliases: ρ_u+hk rho
Use in source · body.tex, line 531
[Injectivity] A permanently noncrossing positive odd orbit never repeats a state. Suppose x_u=x_v with v>u. Put k=v-u and B=S_v-S_u. The affine law for this positive length block gives (2^B-3^k)x_u=C_u,v>0. Hence 2^B/3^k>1. Determinism makes the same states and valuations repeat, so ρ_u+hk=ρ_u(2^B/3^k)^h. This exceeds 1 for sufficiently large h, contrary to permanent noncrossing.
Exact mathematical passage
\rho_{u+hk}=\rho_u(2^B/3^k)^h
\sigmaUse in source · 6 passages
Search aliases: σ sigma
Use in source · body.tex, line 271
. Differentiating gives -fracζ'(s)ζ(s) =∑_p∑_k≥1(log p)p^-ks =∑_n≥2fracΛ(n)n^s. The factor k from differentiation cancels the denominator k, which is why the coefficient of p^k is log p, rather than log(p^k). Absolute convergence of these derivative series holds for Re s>1. For instance, Λ(n)≤log n, and ∑(log n)n^-σ converges for σ>1.
Exact mathematical passage
\sum(\log n)n^{-\sigma}
\sigma^2Use in source · 3 passages
Search aliases: σ^2 sigma
Use in source · body.tex, line 368
e the orthonormal scaled Hermite functions ψ_r,t,c(u)= frac(2π t)^-1/4sqrt2^r r! H_r(fracu-csqrt2t)e^-(u-c)^2/(4t). Here H_r(x)=(-1)^r e^x^2(d^r/dx^r)e^-x^2 is the physicists' Hermite polynomial. If P_R is orthogonal projection onto indices 0≤ r≤ R, the coherent expansion gives g_t,a · =e^-λ/2 ∑_r≥0fracσ^rsqrtr!ψ_r,t,c, σ · =fraca-c2sqrt t, λ=σ^2, notag (I-P_R)g_t,a _2^2 · =e^-λ∑_r>Rfracλ^rr!. The last expression is a Poisson upper tail. These formulas and the compression estimates are recovered…
Exact mathematical passage
g_{t,a}&=e^{-\lambda/2} \sum_{r\ge0}\frac{\sigma^r}{\sqrt{r!}}\psi_{r,t,c}, \label{rh:coherent}\\ \sigma&=\frac{a-c}{2\sqrt t},\qquad \lambda=\sigma^2, \notag\\ \|(I-P_R)g_{t,a}\|_2^2 &=e^{-\lambda}\sum_{r>R}\frac{\lambda^r}{r!}. \label{rh:poisson}
\sigma^rUse in source · 2 passages
Search aliases: σ^r sigma
Use in source · body.tex, line 368
e the orthonormal scaled Hermite functions ψ_r,t,c(u)= frac(2π t)^-1/4sqrt2^r r! H_r(fracu-csqrt2t)e^-(u-c)^2/(4t). Here H_r(x)=(-1)^r e^x^2(d^r/dx^r)e^-x^2 is the physicists' Hermite polynomial. If P_R is orthogonal projection onto indices 0≤ r≤ R, the coherent expansion gives g_t,a · =e^-λ/2 ∑_r≥0fracσ^rsqrtr!ψ_r,t,c, σ · =fraca-c2sqrt t, λ=σ^2, notag (I-P_R)g_t,a _2^2 · =e^-λ∑_r>Rfracλ^rr!. The last expression is a Poisson upper tail. These formulas and the compression estimates are recovered…
Exact mathematical passage
g_{t,a}&=e^{-\lambda/2} \sum_{r\ge0}\frac{\sigma^r}{\sqrt{r!}}\psi_{r,t,c}, \label{rh:coherent}\\ \sigma&=\frac{a-c}{2\sqrt t},\qquad \lambda=\sigma^2, \notag\\ \|(I-P_R)g_{t,a}\|_2^2 &=e^{-\lambda}\sum_{r>R}\frac{\lambda^r}{r!}. \label{rh:poisson}
\varphiUse in source · 7 passages
Search aliases: φ varphi
Use in source · body.tex, line 668
[Squarefree coprime bases] For fixed R>0, the number N_Q,R(M) of squarefree integers m in [RM,2RM] with (m,Q)=1 satisfies, uniformly in squarefree Q, N_Q,R(M)=RMfracφ(Q)Q ∏_pnmid Q(1-p^-2)+O_R(2^dsqrt M). Here φ(Q) is Euler's totient function, which counts the positive residues coprime to Q.
Exact mathematical passage
\label{rh:countformula} N_{Q,R}(M)=RM\frac{\varphi(Q)}Q \prod_{p\nmid Q}(1-p^{-2})+O_R(2^d\sqrt M).
\psi_{r,t,c}Use in source · 4 passages
Search aliases: ψ_r,t,c psi
Use in source · body.tex, line 361
For t>0 and a real center a, define the normalized Gaussian g_t,a(u)=(2π t)^-1/4e^-(u-a)^2/(4t). It has norm one in L^2( R,du). For a chosen center c, let ψ_r,t,c, r=0,1,2,…, be the orthonormal scaled Hermite functions ψ_r,t,c(u)= frac(2π t)^-1/4sqrt2^r r! H_r(fracu-csqrt2t)e^-(u-c)^2/(4t). Here H_r(x)=(-1)^r e^x^2(d^r/dx^r)e^-x^2 is the physicists' Hermite polynomial. If P_R is orthogonal projection onto indices 0≤ r≤ R, the coherent expansion gives g_t,a · =e^-λ/2 ∑_r≥0fracσ^rsqrtr!ψ_r,t,c,
Exact mathematical passage
\psi_{r,t,c}
\omegaUse in source · 1 passage
Search aliases: ω omega
Use in source · body.tex, line 534
Let Q be a squarefree positive integer, and let d=ω(Q) be its number of distinct prime divisors. Its alternating prime cube acts on the weight by Δ_Qw_K(m)=∑_bmid Qμ(b)w_K(bm). Here b ranges over all positive divisors of Q. The result is a sum of 2^d terms, one for each subset of the prime coordinates.
Exact mathematical passage
d=\omega(Q)
\ell^{-2}Use in source · 1 passage
Search aliases: ℓ^-2 ell
Use in source · body.tex, line 677
Use the exact identity μ(m)^2=∑_ℓ^2mid mμ(ℓ). In the coprime sector only (ℓ,Q)=1 contributes. Inclusion and exclusion over the prime factors of Q counts the possible multiples of ℓ^2 in the interval with main term RMℓ^-2φ(Q)/Q and error O(2^d). There are O_R(sqrt M) values of ℓ, so the summed error is O_R(2^dsqrt M). Extending the main sum from ℓ≤sqrt2RM to all positive integers contributes only O_R(sqrt M). Absolute convergence identifies the extended sum as ∏_pnmid Q(1-p^-2), giving [source re…
Exact mathematical passage
RM\ell^{-2}\varphi(Q)/Q
\ell^2Use in source · 3 passages
Search aliases: ℓ^2 ell
Use in source · body.tex, line 676
Use the exact identity μ(m)^2=∑_ℓ^2mid mμ(ℓ). In the coprime sector only (ℓ,Q)=1 contributes. Inclusion and exclusion over the prime factors of Q counts the possible multiples of ℓ^2 in the interval with main term RMℓ^-2φ(Q)/Q and error O(2^d). There are O_R(sqrt M) values of ℓ, so the summed error is O_R(2^dsqrt M). Extending the main sum from ℓ≤sqrt2RM to all positive integers contributes only O_R(sqrt M). Absolute convergence identifies the extended sum as ∏_pnmid Q(1-p^-2), giving [source re…
Exact mathematical passage
\mu(m)^2=\sum_{\ell^2\mid m}\mu(\ell)
\toUse in source · 5 passages
Search aliases: → to
Use in source · body.tex, line 582
The quantifier on the conclusion matters. Unbounded deficit on every tail does not imply d_r→∞. The deficit could return to small values between increasingly deep excursions. Even convergence d_r→∞ would only imply ρ_r→0, which is compatible with growing positive states. The present theorem closes every bounded strip, rather than the entire noncrossing branch.
Exact mathematical passage
d_r\to\infty
\mapstoUse in source · 5 passages
Search aliases: ↦ mapsto
Use in source · body.tex, line 35
t uses S_r shortcut steps and S_r+r classical steps. In the shortcut map the division immediately after an odd multiplication is included in that odd step. In the classical map it is a separate step, as are all the other divisions. For example, A(27)=41 is realized classically by 27,82,41. Since its valuation is 1, this has one shortcut step and two classical steps. The next acceleration 41↦31 has valuation 2 and uses two shortcut steps or three classical steps. The combined prefix therefore has…
Exact mathematical passage
41\mapsto31
\partial_tUse in source · 2 passages
Search aliases: ∂_t partial
Use in source · body.tex, line 80
r complex u, there is an additional exponential factor linear in |γ| on each compact set. Gaussian decay still dominates. The classical zero counting estimate N(Y)=O(Ylog Y), where N(Y) counts zeros with 0<Imρ≤ Y, gives summability [source reference]. The same domination permits differentiation. Consequently ∂_tΘ_t=∂_u^2Θ_t. The zero symmetries make Θ_t(u) real and even when u is real.
Exact mathematical passage
\label{rh:heatflow} \partial_t\Theta_t=\partial_u^2\Theta_t.
\prodUse in source · 10 passages
Search aliases: ∏ prod
Use in source · body.tex, line 16
For a complex number s with real part greater than one, the zeta function is ζ(s)=∑_n=1^∞n^-s =∏_p(1-p^-s)^-1, where n ranges over positive integers and p ranges over primes. Analytic continuation extends it to a meromorphic function on the complex plane with a simple pole at s=1. The completed function ξ(s)=frac12s(s-1)π^-s/2Γ(s/2)ζ(s) is entire and satisfies ξ(s)=ξ(1-s). Here Γ is the gamma function. These are classical facts, with the normalization in [source reference] fixed throughout this …
Exact mathematical passage
\label{rh:zeta} \zeta(s)=\sum_{n=1}^{\infty}n^{-s} =\prod_{p}(1-p^{-s})^{-1},
\prod_{i=1}^dUse in source · 1 passage
Search aliases: ∏_i=1^d prod
Use in source · body.tex, line 718
Order the prime divisors as p_1<⋯<p_d. Since p_i≥ i+1, fracφ(Q)Q=∏_i=1^d(1-p_i^-1) ≥∏_i=1^dfrac ii+1=frac1d+1. Also ∏_pnmid Q(1-p^-2)≥1/ζ(2). The error in [source reference] is M^1/2+o(1), while its main term is at least RM/[ζ(2)(d+1)]. Therefore eventually N_Q,R(M)≥fracRM2ζ(2)fracφ(Q)Q. On every counted base, |μ(m)|=1, and Lemma~[source reference] gives |Δ_Qw_M^2(m)|≥κ_R/(4R^2M^2). Their contribution is a lower bound for the full majorant. Multiplication gives [source reference].
Exact mathematical passage
\frac{\varphi(Q)}Q=\prod_{i=1}^d(1-p_i^{-1}) \ge\prod_{i=1}^d\frac i{i+1}=\frac1{d+1}.
\prod_pUse in source · 1 passage
Search aliases: ∏_p prod
Use in source · body.tex, line 510
hmetic series, each fixed K has an exact finite expression in even zeta values. Expand its power by the binomial theorem: c_K=∑_j=0^K(-1)^jbinom Kjfrac1ζ(2j+2). To justify this identity, note first that the expansion has only K+1 terms. Each inner series ∑μ(n)n^-2j-2 is absolutely convergent. Its Euler product is ∏_p(1-p^-2j-2), which equals the reciprocal zeta value in the same half plane of absolute convergence. Thus no limiting interchange near the critical line is involved.
Exact mathematical passage
\prod_p(1-p^{-2j-2})
\prod_{p}Use in source · 1 passage
Search aliases: ∏_p prod
Use in source · body.tex, line 16
For a complex number s with real part greater than one, the zeta function is ζ(s)=∑_n=1^∞n^-s =∏_p(1-p^-s)^-1, where n ranges over positive integers and p ranges over primes. Analytic continuation extends it to a meromorphic function on the complex plane with a simple pole at s=1. The completed function ξ(s)=frac12s(s-1)π^-s/2Γ(s/2)ζ(s) is entire and satisfies ξ(s)=ξ(1-s). Here Γ is the gamma function. These are classical facts, with the normalization in [source reference] fixed throughout this …
Exact mathematical passage
\label{rh:zeta} \zeta(s)=\sum_{n=1}^{\infty}n^{-s} =\prod_{p}(1-p^{-s})^{-1},
\prod_{p\mid Q}Use in source · 2 passages
Search aliases: ∏_pmid Q prod mid
Use in source · body.tex, line 605
The cube's shape is also scale dependent. At fixed K, large m makes every factor (1-1/(bm)^2)^K close to one, so its leading coefficient is ∏_pmid Q(1-p^-2). At the active scale masympsqrt K, the different divisors sample different parts of the weight profile. This difference can create local cancellation within a cube. It does not remove the remaining sign μ(m), which is the quantity the signed criterion still asks us to control.
Exact mathematical passage
\prod_{p\mid Q}(1-p^{-2})
\prod_{p\nmid Q}Use in source · 3 passages
Search aliases: ∏_pnmid Q prod nmid
Use in source · body.tex, line 668
[Squarefree coprime bases] For fixed R>0, the number N_Q,R(M) of squarefree integers m in [RM,2RM] with (m,Q)=1 satisfies, uniformly in squarefree Q, N_Q,R(M)=RMfracφ(Q)Q ∏_pnmid Q(1-p^-2)+O_R(2^dsqrt M). Here φ(Q) is Euler's totient function, which counts the positive residues coprime to Q.
Exact mathematical passage
\label{rh:countformula} N_{Q,R}(M)=RM\frac{\varphi(Q)}Q \prod_{p\nmid Q}(1-p^{-2})+O_R(2^d\sqrt M).
\prod_{p\nmid6}Use in source · 1 passage
Search aliases: ∏_pnmid6 prod nmid
Use in source · body.tex, line 695
For Q=6, the density constant simplifies to fracφ(6)6∏_pnmid6(1-p^-2) =frac12ζ(2). For Q=30, it is 5/(12ζ(2)). As a finite illustration, take M=10 and R=3. The interval is [30,60]. Its Q=6 squarefree coprime bases are 31,35,37,41,43,47,53,55,59. For Q=30, remove 35 and 55, leaving seven bases. In both coprime sectors, 49 would be counted if squarefree support were forgotten. The main term predicts approximately 9.12 bases for Q=6 and 7.60 for Q=30. These small examples illustrate the meaning of …
Exact mathematical passage
\frac{\varphi(6)}6\prod_{p\nmid6}(1-p^{-2}) =\frac1{2\zeta(2)}.
\prod_{p\in S}Use in source · 1 passage
Search aliases: ∏_p∈ S prod in
Use in source · body.tex, line 760
Third, global Hardy norms on a prime torus are blind to the squarefree Möbius sign. For a finite prime set, a squarefree monomial is ∏_p∈ Sz_p. Replacing every coordinate by -z_p multiplies it by (-1)^|S|, exactly its Möbius coefficient. The substitution preserves Haar measure on the torus. Every global L^q norm formed with that measure is consequently unchanged by this sign operation. A bound in that invariant norm cannot distinguish the signed packet from the corresponding unsigned squarefree …
Exact mathematical passage
\prod_{p\in S}z_p
\sum_{a_0,\ldots,a_{k-1}\ge1}Use in source · 1 passage
Search aliases: ∑_a_0,…,a_k-1≥1 sum ldots ge
Use in source · body.tex, line 161
For fixed k≥1, the sum of all relative odd cylinder densities is ∑_a_0,…,a_k-1≥12^-(a_0+⋯+a_k-1) =(∑_a≥12^-a)^k=1. This identity reflects the exact partition of positive odd integers by their first k valuations. It does not assert that valuations along one particular infinite orbit are independent samples. A density calculation describes how sources are distributed among finite cylinders. An orbit theorem describes the behavior of every state of a chosen trajectory. Moving between those statemen…
Exact mathematical passage
\sum_{a_0,\ldots,a_{k-1}\ge1}2^{-(a_0+\cdots+a_{k-1})} =\left(\sum_{a\ge1}2^{-a}\right)^k=1.
\sum_{a\ge1}Use in source · 3 passages
Search aliases: ∑_a≥1 sum ge
Use in source · body.tex, line 161
For fixed k≥1, the sum of all relative odd cylinder densities is ∑_a_0,…,a_k-1≥12^-(a_0+⋯+a_k-1) =(∑_a≥12^-a)^k=1. This identity reflects the exact partition of positive odd integers by their first k valuations. It does not assert that valuations along one particular infinite orbit are independent samples. A density calculation describes how sources are distributed among finite cylinders. An orbit theorem describes the behavior of every state of a chosen trajectory. Moving between those statemen…
Exact mathematical passage
\sum_{a_0,\ldots,a_{k-1}\ge1}2^{-(a_0+\cdots+a_{k-1})} =\left(\sum_{a\ge1}2^{-a}\right)^k=1.
\sum_{b\mid Q}Use in source · 4 passages
Search aliases: ∑_bmid Q sum mid
Use in source · body.tex, line 535
Let Q be a squarefree positive integer, and let d=ω(Q) be its number of distinct prime divisors. Its alternating prime cube acts on the weight by Δ_Qw_K(m)=∑_bmid Qμ(b)w_K(bm). Here b ranges over all positive divisors of Q. The result is a sum of 2^d terms, one for each subset of the prime coordinates.
Exact mathematical passage
\label{rh:deltaq} \Delta_Qw_K(m)=\sum_{b\mid Q}\mu(b)w_K(bm).
\sum_{i<r}Use in source · 1 passage
Search aliases: ∑_i<r sum
Use in source · body.tex, line 510
[Normalized offset and height] Under permanent noncrossing, ρ_r x_r=n+frac13∑_i<rρ_i≤ n+frac r3. At every time with d_r≤ D, where D≥0 is integral, x_r<2^D+1(n+frac r3). Divide the affine law by 3^r and use the offset sum. Noncrossing gives ρ_i≤1 for every preceding time, which is essential for the upper bound. From d_r≤ D we have S_r≥lfloor rlog_2 3rfloor-D>rlog_2 3-D-1. Therefore ρ_r>2^-D-1, proving the height estimate.
Exact mathematical passage
\label{eq:normalized} \rho_r x_r=n+\frac13\sum_{i<r}\rho_i\le n+\frac r3.
\sum_{i=0}^{k-1}Use in source · 1 passage
Search aliases: ∑_i=0^k-1 sum
Use in source · body.tex, line 69
[Affine packet law] If n realizes w, then every prefix satisfies 2^S_jx_j=3^j n+C_j. The terminal offset also has the explicit expression C_k=∑_i=0^k-13^k-1-i2^S_i. At j=0 the identity is x_0=n. If it holds at j, multiply [source reference] by 2^S_j to obtain 2^S_j+1x_j+1=3^j+1n+3C_j+2^S_j. This is the next identity by [source reference]. Iterating the same offset recurrence gives [source reference].
Exact mathematical passage
\label{eq:offsetsum} C_k=\sum_{i=0}^{k-1}3^{k-1-i}2^{S_i}.
\sum_{j=0}^KUse in source · 1 passage
Search aliases: ∑_j=0^K sum
Use in source · body.tex, line 507
Although c_K is written as an infinite arithmetic series, each fixed K has an exact finite expression in even zeta values. Expand its power by the binomial theorem: c_K=∑_j=0^K(-1)^jbinom Kjfrac1ζ(2j+2). To justify this identity, note first that the expansion has only K+1 terms. Each inner series ∑μ(n)n^-2j-2 is absolutely convergent. Its Euler product is ∏_p(1-p^-2j-2), which equals the reciprocal zeta value in the same half plane of absolute convergence. Thus no limiting interchange near the c…
Exact mathematical passage
\label{rh:binomial} c_K=\sum_{j=0}^K(-1)^j\binom Kj\frac1{\zeta(2j+2)}.
\sum_{k\ge0}Use in source · 2 passages
Search aliases: ∑_k≥0 sum ge
Use in source · body.tex, line 278
The archimedean term has a uniform bound on every positive heat strip. A direct proof uses the classical digamma series [source reference]: ψ(a+iy)-ψ(a)= ∑_k≥0fraciy(k+a)(k+a+iy), a>0. Its absolute value is at most |y|∑_k≥0(k+a)^-2. Therefore |Ψ_∞(v)|≤ C(1+|v|) for a fixed constant C. Since widetilde A_t(u)=frac12π ∫_ R(1/4+v^2)e^-tv^2cos(uv) Ψ_∞(v) dv, its magnitude is bounded independently of u and t∈[t_0,T], whenever 0<t_0<T<∞. Its energy on a unit translation band is consequently O(1). Remov…
Exact mathematical passage
\psi(a+iy)-\psi(a)= \sum_{k\ge0}\frac{iy}{(k+a)(k+a+iy)}, \qquad a>0.
\sum_{k\ge1}Use in source · 2 passages
Search aliases: ∑_k≥1 sum ge
Use in source · body.tex, line 262
The appearance of Λ can be traced directly to the Euler product in its first convergence half plane. There, logζ(s)=∑_p∑_k≥1fracp^-ksk. Differentiating gives -fracζ'(s)ζ(s) =∑_p∑_k≥1(log p)p^-ks =∑_n≥2fracΛ(n)n^s. The factor k from differentiation cancels the denominator k, which is why the coefficient of p^k is log p, rather than log(p^k). Absolute convergence of these derivative series holds for Re s>1. For instance, Λ(n)≤log n, and ∑(log n)n^-σ converges for σ>1.
Exact mathematical passage
\log\zeta(s)=\sum_p\sum_{k\ge1}\frac{p^{-ks}}k.
\sum_{n=1}^{\infty}Use in source · 1 passage
Search aliases: ∑_n=1^∞ sum infty
Use in source · body.tex, line 16
For a complex number s with real part greater than one, the zeta function is ζ(s)=∑_n=1^∞n^-s =∏_p(1-p^-s)^-1, where n ranges over positive integers and p ranges over primes. Analytic continuation extends it to a meromorphic function on the complex plane with a simple pole at s=1. The completed function ξ(s)=frac12s(s-1)π^-s/2Γ(s/2)ζ(s) is entire and satisfies ξ(s)=ξ(1-s). Here Γ is the gamma function. These are classical facts, with the normalization in [source reference] fixed throughout this …
Exact mathematical passage
\label{rh:zeta} \zeta(s)=\sum_{n=1}^{\infty}n^{-s} =\prod_{p}(1-p^{-s})^{-1},
\sum_{n>N}Use in source · 1 passage
Search aliases: ∑_n>N sum
Use in source · body.tex, line 518
ζ(4)+frac1ζ(6). These formulas are exact. At large K, their alternating binomial terms can be individually much larger than their sum, so straightforward floating point evaluation may lose accuracy. A finite expression is not automatically a stable evaluation scheme. For a direct truncation of the original series at n=N, the elementary certified remainder bound is instead |c_K-∑_n≤ Nμ(n)w_K(n)| ≤∑_n>Nn^-2≤frac1N. This bound is useful for checking a finite computation, but it is much too coarse t…
Exact mathematical passage
\left|c_K-\sum_{n\le N}\mu(n)w_K(n)\right| \le\sum_{n>N}n^{-2}\le\frac1N.
\sum_{n\le N}Use in source · 1 passage
Search aliases: ∑_n≤ N sum le
Use in source · body.tex, line 518
ζ(4)+frac1ζ(6). These formulas are exact. At large K, their alternating binomial terms can be individually much larger than their sum, so straightforward floating point evaluation may lose accuracy. A finite expression is not automatically a stable evaluation scheme. For a direct truncation of the original series at n=N, the elementary certified remainder bound is instead |c_K-∑_n≤ Nμ(n)w_K(n)| ≤∑_n>Nn^-2≤frac1N. This bound is useful for checking a finite computation, but it is much too coarse t…
Exact mathematical passage
\left|c_K-\sum_{n\le N}\mu(n)w_K(n)\right| \le\sum_{n>N}n^{-2}\le\frac1N.
\sum_{n\le X}Use in source · 1 passage
Search aliases: ∑_n≤ X sum le
Use in source · body.tex, line 460
The simple coefficient mass bound also has an elementary proof. Since Λ(n)≤log X for n≤ X, and ∑_n≤ Xn^-1/2≤2sqrt X, their product gives the stated estimate. Compression may therefore be arranged without first proving any delicate prime cancellation theorem. The remaining arithmetic estimate lies in the compressed packet, not in the omitted Gaussian tail.
Exact mathematical passage
\sum_{n\le X}n^{-1/2}\le2\sqrt X
\sum_{n\le X_J}Use in source · 1 passage
Search aliases: ∑_n≤ X_J sum le
Use in source · body.tex, line 438
=C^2/(4t_0). Then λ≤α J+O(sqrt J). For a fixed constant q>α, projection through R_J=lceil qJrceil has norm error bounded by exp(-frac12 I(qmidα)J+o(J)), I(qmidα)=qlog(q/α)-q+α. The coefficient mass of the positive prime packet through X_J=exp(J+O(sqrt J)) is at most ∑_n≤ X_JfracΛ(n)sqrt n ≤2sqrtX_Jlog X_J=exp(J/2+o(J)). Thus I(qmidα)>1 makes the aggregate compression error exponentially small. The pole annihilator adds only polynomial factors to the Hermite ladder and leaves this exponential con…
Exact mathematical passage
\sum_{n\le X_J}\frac{\Lambda(n)}{\sqrt n} \le2\sqrt{X_J}\log X_J=\exp(J/2+o(J)).
\sum_{n\ge1}Use in source · 1 passage
Search aliases: ∑_n≥1 sum ge
Use in source · body.tex, line 487
The Möbius function μ(n) is zero when a square of a prime divides n. Otherwise it equals (-1)^k, where k is the number of distinct prime divisors of n, and μ(1)=1. For an integer K≥1, put w_K(n)=frac1n^2(1-frac1n^2)^K, c_K=∑_n≥1μ(n)w_K(n). The series is absolutely convergent. Its natural active scale is n of order M=sqrt K. The coefficient c_K is classical. Báez Duarte proved the following exact criterion [source reference].
Exact mathematical passage
\label{rh:w} w_K(n)=\frac1{n^2}\left(1-\frac1{n^2}\right)^K, \qquad c_K=\sum_{n\ge1}\mu(n)w_K(n).
\sum_pUse in source · 2 passages
Search aliases: ∑_p sum
Use in source · body.tex, line 262
The appearance of Λ can be traced directly to the Euler product in its first convergence half plane. There, logζ(s)=∑_p∑_k≥1fracp^-ksk. Differentiating gives -fracζ'(s)ζ(s) =∑_p∑_k≥1(log p)p^-ks =∑_n≥2fracΛ(n)n^s. The factor k from differentiation cancels the denominator k, which is why the coefficient of p^k is log p, rather than log(p^k). Absolute convergence of these derivative series holds for Re s>1. For instance, Λ(n)≤log n, and ∑(log n)n^-σ converges for σ>1.
Exact mathematical passage
\log\zeta(s)=\sum_p\sum_{k\ge1}\frac{p^{-ks}}k.
\sum_rUse in source · 2 passages
Search aliases: ∑_r sum
Use in source · body.tex, line 396
An orthonormal Hermite basis replaces this almost singular coordinate system. If a finite packet is f=∑_r a_rψ_r,t,c, Parseval's identity gives f _2^2=∑_r|a_r|^2 with no unknown Gram loss. This exact equality is the repair. It remains possible for the arithmetic coefficients a_r to be large, or to contain a conjecture level obstruction in one of the first modes.
Exact mathematical passage
f=\sum_r a_r\psi_{r,t,c}
\sum_{r>R}Use in source · 1 passage
Search aliases: ∑_r>R sum
Use in source · body.tex, line 368
e the orthonormal scaled Hermite functions ψ_r,t,c(u)= frac(2π t)^-1/4sqrt2^r r! H_r(fracu-csqrt2t)e^-(u-c)^2/(4t). Here H_r(x)=(-1)^r e^x^2(d^r/dx^r)e^-x^2 is the physicists' Hermite polynomial. If P_R is orthogonal projection onto indices 0≤ r≤ R, the coherent expansion gives g_t,a · =e^-λ/2 ∑_r≥0fracσ^rsqrtr!ψ_r,t,c, σ · =fraca-c2sqrt t, λ=σ^2, notag (I-P_R)g_t,a _2^2 · =e^-λ∑_r>Rfracλ^rr!. The last expression is a Poisson upper tail. These formulas and the compression estimates are recovered…
Exact mathematical passage
g_{t,a}&=e^{-\lambda/2} \sum_{r\ge0}\frac{\sigma^r}{\sqrt{r!}}\psi_{r,t,c}, \label{rh:coherent}\\ \sigma&=\frac{a-c}{2\sqrt t},\qquad \lambda=\sigma^2, \notag\\ \|(I-P_R)g_{t,a}\|_2^2 &=e^{-\lambda}\sum_{r>R}\frac{\lambda^r}{r!}. \label{rh:poisson}
\sum_{r\ge0}Use in source · 3 passages
Search aliases: ∑_r≥0 sum ge
Use in source · body.tex, line 368
e the orthonormal scaled Hermite functions ψ_r,t,c(u)= frac(2π t)^-1/4sqrt2^r r! H_r(fracu-csqrt2t)e^-(u-c)^2/(4t). Here H_r(x)=(-1)^r e^x^2(d^r/dx^r)e^-x^2 is the physicists' Hermite polynomial. If P_R is orthogonal projection onto indices 0≤ r≤ R, the coherent expansion gives g_t,a · =e^-λ/2 ∑_r≥0fracσ^rsqrtr!ψ_r,t,c, σ · =fraca-c2sqrt t, λ=σ^2, notag (I-P_R)g_t,a _2^2 · =e^-λ∑_r>Rfracλ^rr!. The last expression is a Poisson upper tail. These formulas and the compression estimates are recovered…
Exact mathematical passage
g_{t,a}&=e^{-\lambda/2} \sum_{r\ge0}\frac{\sigma^r}{\sqrt{r!}}\psi_{r,t,c}, \label{rh:coherent}\\ \sigma&=\frac{a-c}{2\sqrt t},\qquad \lambda=\sigma^2, \notag\\ \|(I-P_R)g_{t,a}\|_2^2 &=e^{-\lambda}\sum_{r>R}\frac{\lambda^r}{r!}. \label{rh:poisson}
\sum_{s=5\ell}^{8\ell+D+1}Use in source · 1 passage
Search aliases: ∑_s=5ℓ^8ℓ+D+1 sum ell ell
Use in source · body.tex, line 596
For a modulus M, any one residue class contributes at most H/M+1 positive integers up to height H. The +1 accounts for an incomplete terminal period. Applying that estimate separately to every selected word yields the finite term W_ℓ. One may write it exactly as W_ℓ=∑_s=5ℓ^8ℓ+D+1binoms-15ℓ-1 =binom8ℓ+D+15ℓ. It can be enormous. Its size does not invalidate the argument, because ℓ is fixed before the number of orbit states tends to infinity. A growing finite constant is not being silently discarde…
Exact mathematical passage
W_\ell=\sum_{s=5\ell}^{8\ell+D+1}\binom{s-1}{5\ell-1} =\binom{8\ell+D+1}{5\ell}.
\sum_{\substack{m\in I\\(m,Q)=1}}Use in source · 1 passage
Search aliases: ∑_substackm∈ I (m,Q)=1 sum substack in
Use in source · body.tex, line 619
For any finite interval I of remaining bases, distinguish its signed contribution from its absolute mass: S_I=∑_substackm∈ I (m,Q)=1μ(m)Δ_Qw_M^2(m), V_I=∑_substackm∈ I (m,Q)=1 |μ(m)| |Δ_Qw_M^2(m)|. The valid inequalities are |S_I|≤ V_I≤ A_Q(M) and |c_M^2|≤ A_Q(M). There is no general inequality |c_M^2|≤ V_I. Terms outside the chosen interval remain in the full signed sum. Even a two term toy sum with values 3 inside the interval and 10 outside it shows why its interval mass cannot bound the full…
Exact mathematical passage
S_I=\sum_{\substack{m\in I\\(m,Q)=1}}\mu(m)\Delta_Qw_{M^2}(m), \qquad V_I=\sum_{\substack{m\in I\\(m,Q)=1}} |\mu(m)|\,|\Delta_Qw_{M^2}(m)|.
\sum_{\substack{m\ge1\\(m,Q)=1}}Use in source · 3 passages
Search aliases: ∑_substackm≥1 (m,Q)=1 sum substack ge
Use in source · body.tex, line 542
[Exact cube decomposition] For every K≥1 and squarefree Q, c_K=∑_substackm≥1 (m,Q)=1 μ(m)Δ_Qw_K(m). The condition (m,Q)=1 means that m and Q have greatest common divisor one.
Exact mathematical passage
\label{rh:cubeidentity} c_K=\sum_{\substack{m\ge1\\(m,Q)=1}} \mu(m)\Delta_Qw_K(m).
\sum_{\ell^2\mid m}Use in source · 1 passage
Search aliases: ∑_ℓ^2mid m sum ell mid
Use in source · body.tex, line 676
Use the exact identity μ(m)^2=∑_ℓ^2mid mμ(ℓ). In the coprime sector only (ℓ,Q)=1 contributes. Inclusion and exclusion over the prime factors of Q counts the possible multiples of ℓ^2 in the interval with main term RMℓ^-2φ(Q)/Q and error O(2^d). There are O_R(sqrt M) values of ℓ, so the summed error is O_R(2^dsqrt M). Extending the main sum from ℓ≤sqrt2RM to all positive integers contributes only O_R(sqrt M). Absolute convergence identifies the extended sum as ∏_pnmid Q(1-p^-2), giving [source re…
Exact mathematical passage
\mu(m)^2=\sum_{\ell^2\mid m}\mu(\ell)
\circUse in source · 1 passage
Search aliases: ∘ circ
Use in source · body.tex, line 182
[Concatenation] Let a word w have data (k,S,C) and a following word v have data (ℓ,U,D). Their concatenation has data (k+ℓ,S+U,3^ℓ C+2^S D). On every realizing source, its transfer is F_v∘ F_w. Substitute 2^S y=3^k n+C into 2^U z=3^ℓ y+D. Multiplication by 2^S gives 2^S+Uz=3^k+ℓn+3^ℓ C+2^S D. The exact cylinder theorem identifies the admissible integer sources.
Exact mathematical passage
F_v\circ F_w
\int_0^\inftyUse in source · 2 passages
Search aliases: ∫_0^∞ int infty
Use in source · body.tex, line 145
In the smaller initial domain Re w>1/2, absolute summation justifies integration of each zero term: ∫_0^∞ e^tz^2+uz-wu du =frace^tz^2w-z. For a compact set of w avoiding all centered zeros, large ordinates make |w-z| large. The Gaussian coefficient then gives a summable bound for the meromorphic series. Near one distinct centered zero z_0, isolate all copies of that zero. If its multiplicity is m_0, their sum is fracm_0e^tz_0^2w-z_0. Every other term is holomorphic in a sufficiently small disk. …
Exact mathematical passage
\int_0^\infty e^{tz^2+uz-wu}\,du =\frac{e^{tz^2}}{w-z}.
\int_0^{\infty}Use in source · 1 passage
Search aliases: ∫_0^∞ int infty
Use in source · body.tex, line 125
Under the hypothesis, [source reference] gives the asserted bound and determinant inequality. Conversely, suppose the packet is bounded on the positive half line. Its Laplace transform F_t(w)=∫_0^∞Θ_t(u)e^-wu du is holomorphic for Re w>0. For Re w>1/2, absolute convergence gives F_t(w)=∑_ρ∈ Z frace^tz_ρ^2w-z_ρ. The series on the right is meromorphic, with normal convergence away from its poles. At each distinct centered zero its residue is its positive integer multiplicity times e^tz_ρ^2, which …
Exact mathematical passage
F_t(w)=\int_0^{\infty}\Theta_t(u)e^{-wu}\,du
\int_J^{J+1}Use in source · 4 passages
Search aliases: ∫_J^J+1 int
Use in source · body.tex, line 302
[Local energy criterion] Fix 0<t_0<T<∞. For positive integers J, set widetilde E_J=∫_t_0^T∫_J^J+1 |widetildeΘ_t(u)|^2 du dt. The Riemann hypothesis is equivalent to subexponential growth of widetilde E_J.
Exact mathematical passage
\label{rh:energydef} \widetilde E_J=\int_{t_0}^T\int_J^{J+1} |\widetilde\Theta_t(u)|^2\,du\,dt.
\int_{t_0}^TUse in source · 2 passages
Search aliases: ∫_t_0^T int
Use in source · body.tex, line 302
[Local energy criterion] Fix 0<t_0<T<∞. For positive integers J, set widetilde E_J=∫_t_0^T∫_J^J+1 |widetildeΘ_t(u)|^2 du dt. The Riemann hypothesis is equivalent to subexponential growth of widetilde E_J.
Exact mathematical passage
\label{rh:energydef} \widetilde E_J=\int_{t_0}^T\int_J^{J+1} |\widetilde\Theta_t(u)|^2\,du\,dt.
\simUse in source · 1 passage
Search aliases: ∼ sim
Use in source · body.tex, line 390
For example, take t=1, a=log1000, and b=log1001. The separation is approximately 0.00099950, the overlap is approximately 0.999999875, and the Gram condition number (1+κ)/(1-κ) is about 1.60×10^7. These values diagnose a coordinate system, not the size of a zeta coefficient. The asymptotic explanation is exact: 1-e^-(a-b)^2/(8t)∼frac(a-b)^28t, log(n+1)-log n∼frac1n. Consecutive integer centers therefore produce a Gram condition number of order n^2 at fixed heat width.
Exact mathematical passage
1-e^{-(a-b)^2/(8t)}\sim\frac{(a-b)^2}{8t}, \qquad \log(n+1)-\log n\sim\frac1n.
\approxUse in source · 1 passage
Search aliases: ≈ approx
Use in source · body.tex, line 455
For a concrete choice, let C=2sqrtt_0, so α=1, and take q=4. Then I(4mid1)=4log4-3≈2.54518>1. The individual norm error decays with exponent approximately -1.27259J. Multiplication by a coefficient mass with exponent at most J/2+o(J) still leaves decay with exponent approximately -0.77259J. This example verifies the scale balance in the compression argument. It does not give a bound on the retained coordinates. For different shell widths or heat minima, one must recompute α and choose q accordin…
Exact mathematical passage
I(4\mid1)=4\log4-3\approx2.54518>1.
\equivUse in source · 10 passages
Search aliases: ≡ equiv
Use in source · body.tex, line 94
[Exact integer cylinder] Let w be a nonempty valuation word of length k and total depth S. A positive odd integer n realizes w if and only if 3^k n+C_k≡ 2^Spmod2^S+1. Consequently the realizing integers form one odd residue class modulo 2^S+1, whose residue is u_w≡(2^S-C_k)(3^k)^-1pmod2^S+1. If the word is realized, [source reference] has an odd terminal quotient x_k. Multiplying an odd integer by 2^S gives exactly the congruence [source reference].
Exact mathematical passage
\label{eq:cylinder} 3^k n+C_k\equiv 2^S\pmod{2^{S+1}}.
\cdotsUse in source · 2 passages
Search aliases: ⋯ cdots
Use in source · body.tex, line 161
For fixed k≥1, the sum of all relative odd cylinder densities is ∑_a_0,…,a_k-1≥12^-(a_0+⋯+a_k-1) =(∑_a≥12^-a)^k=1. This identity reflects the exact partition of positive odd integers by their first k valuations. It does not assert that valuations along one particular infinite orbit are independent samples. A density calculation describes how sources are distributed among finite cylinders. An orbit theorem describes the behavior of every state of a chosen trajectory. Moving between those statemen…
Exact mathematical passage
\sum_{a_0,\ldots,a_{k-1}\ge1}2^{-(a_0+\cdots+a_{k-1})} =\left(\sum_{a\ge1}2^{-a}\right)^k=1.
\langleUse in source · 1 passage
Search aliases: ⟨ langle
Use in source · body.tex, line 383
The word physical refers to the measure du in which the packet is actually evaluated. It does not mean that a separately chosen coefficient norm can be substituted. Completing the square gives the exact overlap ⟨ g_t,a,g_t,b⟩ =e^-(a-b)^2/(8t). For two translates, the Gram matrix has diagonal entries one and off diagonal entry κ=e^-(a-b)^2/(8t). Its eigenvalues are 1+κ and 1-κ. When the centers are close, their difference is almost invisible in the physical norm even if the two raw coefficients a…
Exact mathematical passage
\label{rh:overlap} \langle g_{t,a},g_{t,b}\rangle =e^{-(a-b)^2/(8t)}.
\rangleUse in source · 1 passage
Search aliases: ⟩ rangle
Use in source · body.tex, line 383
The word physical refers to the measure du in which the packet is actually evaluated. It does not mean that a separately chosen coefficient norm can be substituted. Completing the square gives the exact overlap ⟨ g_t,a,g_t,b⟩ =e^-(a-b)^2/(8t). For two translates, the Gram matrix has diagonal entries one and off diagonal entry κ=e^-(a-b)^2/(8t). Its eigenvalues are 1+κ and 1-κ. When the centers are close, their difference is almost invisible in the physical norm even if the two raw coefficients a…
Exact mathematical passage
\label{rh:overlap} \langle g_{t,a},g_{t,b}\rangle =e^{-(a-b)^2/(8t)}.
\longmapstoUse in source · 6 passages
Search aliases: ⟼ longmapsto
Use in source · body.tex, line 128
,3), the recurrence gives C_1=1 and C_2=5, with total depth S=4. The affine expression is F_(1,3)(n)=frac9n+516. The exact cylinder requires 9n+5≡16pmod32. The inverse of 9 modulo 32 is 25, so its address is 19 modulo 32. Every positive representative can be written n=32t+19 with an integer t≥0. Direct substitution gives the full orbit segment 32t+19⟼48t+29⟼18t+11. The first numerator is 2(48t+29), with an odd quotient. The next numerator is 8(18t+11), again with an odd quotient. Thus the valuat…
Exact mathematical passage
\label{eq:word13} 32t+19\longmapsto48t+29\longmapsto18t+11.
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