Notation by work
Catalan Constant: source and evidence
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423 symbol spellings in 1272 mathematical passages across 11 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 423 entries are available below.
+Definition passage · 201 passages
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Definition passage: Catalan's constant · chapters01_03.tex, line 13
[Catalan's constant] The real number studied in this book is G=∑_k=0^∞frac(-1)^k(2k+1)^2. The series converges absolutely because ∑_k=0^∞frac1(2k+1)^2 ≤∑_j=1^∞frac1j^2<∞.
Exact mathematical passage
G=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^2}. \label{eq:c01-definition}
=Definition passage · 423 passages
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Definition passage: The distinguished rational sequences · chapter03.tex, line 40
[The distinguished rational sequences] Let (u_n)_n≥0 and (v_n)_n≥0 satisfy [source reference], with (u_0,u_1)=(1,frac74), (v_0,v_1)=(0,frac138). Their rational approximants are r_n=fracv_nu_n.
Exact mathematical passage
\label{cb_eq_uv_initial} (u_0,u_1)=\left(1,\frac74\right),\qquad (v_0,v_1)=\left(0,\frac{13}{8}\right).
GDefinition passage · 94 passages
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Definition passage: Catalan's constant · chapters01_03.tex, line 13
[Catalan's constant] The real number studied in this book is G=∑_k=0^∞frac(-1)^k(2k+1)^2. The series converges absolutely because ∑_k=0^∞frac1(2k+1)^2 ≤∑_j=1^∞frac1j^2<∞.
Exact mathematical passage
G=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^2}. \label{eq:c01-definition}
jDefinition passage · 94 passages
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Definition passage: Catalan's constant · chapters01_03.tex, line 18
[Catalan's constant] The real number studied in this book is G=∑_k=0^∞frac(-1)^k(2k+1)^2. The series converges absolutely because ∑_k=0^∞frac1(2k+1)^2 ≤∑_j=1^∞frac1j^2<∞.
Exact mathematical passage
\sum_{k=0}^{\infty}\frac1{(2k+1)^2} \leq\sum_{j=1}^{\infty}\frac1{j^2}<\infty.
j^2Definition passage · 1 passage
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Definition passage: Catalan's constant · chapters01_03.tex, line 18
[Catalan's constant] The real number studied in this book is G=∑_k=0^∞frac(-1)^k(2k+1)^2. The series converges absolutely because ∑_k=0^∞frac1(2k+1)^2 ≤∑_j=1^∞frac1j^2<∞.
Exact mathematical passage
\sum_{k=0}^{\infty}\frac1{(2k+1)^2} \leq\sum_{j=1}^{\infty}\frac1{j^2}<\infty.
kDefinition passage · 47 passages
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Definition passage: Catalan's constant · chapters01_03.tex, line 13
[Catalan's constant] The real number studied in this book is G=∑_k=0^∞frac(-1)^k(2k+1)^2. The series converges absolutely because ∑_k=0^∞frac1(2k+1)^2 ≤∑_j=1^∞frac1j^2<∞.
Exact mathematical passage
G=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^2}. \label{eq:c01-definition}
nDefinition passage · 368 passages
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Definition passage: The distinguished rational sequences · chapter03.tex, line 38
[The distinguished rational sequences] Let (u_n)_n≥0 and (v_n)_n≥0 satisfy [source reference], with (u_0,u_1)=(1,frac74), (v_0,v_1)=(0,frac138). Their rational approximants are r_n=fracv_nu_n.
Exact mathematical passage
(u_n)_{n\geq0}
rDefinition passage · 44 passages
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Definition passage: The distinguished rational sequences · chapter03.tex, line 45
[The distinguished rational sequences] Let (u_n)_n≥0 and (v_n)_n≥0 satisfy [source reference], with (u_0,u_1)=(1,frac74), (v_0,v_1)=(0,frac138). Their rational approximants are r_n=fracv_nu_n.
Exact mathematical passage
\label{cb_eq_r_def} r_n=\frac{v_n}{u_n}.
r_nDefinition passage · 12 passages
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Definition passage: The distinguished rational sequences · chapter03.tex, line 45
[The distinguished rational sequences] Let (u_n)_n≥0 and (v_n)_n≥0 satisfy [source reference], with (u_0,u_1)=(1,frac74), (v_0,v_1)=(0,frac138). Their rational approximants are r_n=fracv_nu_n.
Exact mathematical passage
\label{cb_eq_r_def} r_n=\frac{v_n}{u_n}.
uDefinition passage · 76 passages
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Definition passage: The distinguished rational sequences · chapter03.tex, line 38
[The distinguished rational sequences] Let (u_n)_n≥0 and (v_n)_n≥0 satisfy [source reference], with (u_0,u_1)=(1,frac74), (v_0,v_1)=(0,frac138). Their rational approximants are r_n=fracv_nu_n.
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(u_n)_{n\geq0}
u_0Definition passage · 6 passages
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Definition passage: The distinguished rational sequences · chapter03.tex, line 40
[The distinguished rational sequences] Let (u_n)_n≥0 and (v_n)_n≥0 satisfy [source reference], with (u_0,u_1)=(1,frac74), (v_0,v_1)=(0,frac138). Their rational approximants are r_n=fracv_nu_n.
Exact mathematical passage
\label{cb_eq_uv_initial} (u_0,u_1)=\left(1,\frac74\right),\qquad (v_0,v_1)=\left(0,\frac{13}{8}\right).
u_1Definition passage · 6 passages
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Definition passage: The distinguished rational sequences · chapter03.tex, line 40
[The distinguished rational sequences] Let (u_n)_n≥0 and (v_n)_n≥0 satisfy [source reference], with (u_0,u_1)=(1,frac74), (v_0,v_1)=(0,frac138). Their rational approximants are r_n=fracv_nu_n.
Exact mathematical passage
\label{cb_eq_uv_initial} (u_0,u_1)=\left(1,\frac74\right),\qquad (v_0,v_1)=\left(0,\frac{13}{8}\right).
u_nDefinition passage · 38 passages
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Definition passage: The distinguished rational sequences · chapter03.tex, line 38
[The distinguished rational sequences] Let (u_n)_n≥0 and (v_n)_n≥0 satisfy [source reference], with (u_0,u_1)=(1,frac74), (v_0,v_1)=(0,frac138). Their rational approximants are r_n=fracv_nu_n.
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(u_n)_{n\geq0}
vDefinition passage · 91 passages
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Definition passage: The distinguished rational sequences · chapter03.tex, line 38
[The distinguished rational sequences] Let (u_n)_n≥0 and (v_n)_n≥0 satisfy [source reference], with (u_0,u_1)=(1,frac74), (v_0,v_1)=(0,frac138). Their rational approximants are r_n=fracv_nu_n.
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(v_n)_{n\geq0}
v_0Definition passage · 10 passages
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Definition passage: The distinguished rational sequences · chapter03.tex, line 40
[The distinguished rational sequences] Let (u_n)_n≥0 and (v_n)_n≥0 satisfy [source reference], with (u_0,u_1)=(1,frac74), (v_0,v_1)=(0,frac138). Their rational approximants are r_n=fracv_nu_n.
Exact mathematical passage
\label{cb_eq_uv_initial} (u_0,u_1)=\left(1,\frac74\right),\qquad (v_0,v_1)=\left(0,\frac{13}{8}\right).
v_1Definition passage · 8 passages
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Definition passage: The distinguished rational sequences · chapter03.tex, line 40
[The distinguished rational sequences] Let (u_n)_n≥0 and (v_n)_n≥0 satisfy [source reference], with (u_0,u_1)=(1,frac74), (v_0,v_1)=(0,frac138). Their rational approximants are r_n=fracv_nu_n.
Exact mathematical passage
\label{cb_eq_uv_initial} (u_0,u_1)=\left(1,\frac74\right),\qquad (v_0,v_1)=\left(0,\frac{13}{8}\right).
v_nDefinition passage · 24 passages
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Definition passage: The distinguished rational sequences · chapter03.tex, line 38
[The distinguished rational sequences] Let (u_n)_n≥0 and (v_n)_n≥0 satisfy [source reference], with (u_0,u_1)=(1,frac74), (v_0,v_1)=(0,frac138). Their rational approximants are r_n=fracv_nu_n.
Exact mathematical passage
(v_n)_{n\geq0}
\sumDefinition passage · 32 passages
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Definition passage: Catalan's constant · chapters01_03.tex, line 13
[Catalan's constant] The real number studied in this book is G=∑_k=0^∞frac(-1)^k(2k+1)^2. The series converges absolutely because ∑_k=0^∞frac1(2k+1)^2 ≤∑_j=1^∞frac1j^2<∞.
Exact mathematical passage
G=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^2}. \label{eq:c01-definition}
\sum_{j=1}^{\infty}Definition passage · 1 passage
Search aliases: ∑_j=1^∞ sum infty
Definition passage: Catalan's constant · chapters01_03.tex, line 18
[Catalan's constant] The real number studied in this book is G=∑_k=0^∞frac(-1)^k(2k+1)^2. The series converges absolutely because ∑_k=0^∞frac1(2k+1)^2 ≤∑_j=1^∞frac1j^2<∞.
Exact mathematical passage
\sum_{k=0}^{\infty}\frac1{(2k+1)^2} \leq\sum_{j=1}^{\infty}\frac1{j^2}<\infty.
\sum_{k=0}^{\infty}Definition passage · 7 passages
Search aliases: ∑_k=0^∞ sum infty
Definition passage: Catalan's constant · chapters01_03.tex, line 13
[Catalan's constant] The real number studied in this book is G=∑_k=0^∞frac(-1)^k(2k+1)^2. The series converges absolutely because ∑_k=0^∞frac1(2k+1)^2 ≤∑_j=1^∞frac1j^2<∞.
Exact mathematical passage
G=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^2}. \label{eq:c01-definition}
\inftyDefinition passage · 28 passages
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Definition passage: Catalan's constant · chapters01_03.tex, line 13
[Catalan's constant] The real number studied in this book is G=∑_k=0^∞frac(-1)^k(2k+1)^2. The series converges absolutely because ∑_k=0^∞frac1(2k+1)^2 ≤∑_j=1^∞frac1j^2<∞.
Exact mathematical passage
G=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^2}. \label{eq:c01-definition}
\leqDefinition passage · 70 passages
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Definition passage: Catalan's constant · chapters01_03.tex, line 18
[Catalan's constant] The real number studied in this book is G=∑_k=0^∞frac(-1)^k(2k+1)^2. The series converges absolutely because ∑_k=0^∞frac1(2k+1)^2 ≤∑_j=1^∞frac1j^2<∞.
Exact mathematical passage
\sum_{k=0}^{\infty}\frac1{(2k+1)^2} \leq\sum_{j=1}^{\infty}\frac1{j^2}<\infty.
\geqDefinition passage · 83 passages
Search aliases: ≥ geq
Definition passage: The distinguished rational sequences · chapter03.tex, line 38
[The distinguished rational sequences] Let (u_n)_n≥0 and (v_n)_n≥0 satisfy [source reference], with (u_0,u_1)=(1,frac74), (v_0,v_1)=(0,frac138). Their rational approximants are r_n=fracv_nu_n.
Exact mathematical passage
(u_n)_{n\geq0}
AIntroduction passage · 28 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
aIntroduction passage · 84 passages
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Introduction passage · chapter03.tex, line 50
The notation a_n,c_n in this chapter and the next denotes recurrence coefficients. The later source notation a_n,b_n is translated explicitly when it occurs, and the cleared packet coefficients A_n,m,B_n,m have two indices. These are different objects.
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a_n,c_n
a_nIntroduction passage · 32 passages
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Introduction passage · chapter03.tex, line 50
The notation a_n,c_n in this chapter and the next denotes recurrence coefficients. The later source notation a_n,b_n is translated explicitly when it occurs, and the cleared packet coefficients A_n,m,B_n,m have two indices. These are different objects.
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a_n,c_n
A_{n,m}Introduction passage · 3 passages
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Introduction passage · chapter03.tex, line 52
The notation a_n,c_n in this chapter and the next denotes recurrence coefficients. The later source notation a_n,b_n is translated explicitly when it occurs, and the cleared packet coefficients A_n,m,B_n,m have two indices. These are different objects.
Exact mathematical passage
A_{n,m},B_{n,m}
BIntroduction passage · 24 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
bIntroduction passage · 53 passages
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Introduction passage · chapter03.tex, line 51
The notation a_n,c_n in this chapter and the next denotes recurrence coefficients. The later source notation a_n,b_n is translated explicitly when it occurs, and the cleared packet coefficients A_n,m,B_n,m have two indices. These are different objects.
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a_n,b_n
b_nIntroduction passage · 8 passages
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Introduction passage · chapter03.tex, line 51
The notation a_n,c_n in this chapter and the next denotes recurrence coefficients. The later source notation a_n,b_n is translated explicitly when it occurs, and the cleared packet coefficients A_n,m,B_n,m have two indices. These are different objects.
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a_n,b_n
B_{n,m}Introduction passage · 3 passages
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Introduction passage · chapter03.tex, line 52
The notation a_n,c_n in this chapter and the next denotes recurrence coefficients. The later source notation a_n,b_n is translated explicitly when it occurs, and the cleared packet coefficients A_n,m,B_n,m have two indices. These are different objects.
Exact mathematical passage
A_{n,m},B_{n,m}
CIntroduction passage · 10 passages
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Introduction passage · chapters08_10.tex, line 246
The prime number theorem supplies the standard asymptotic~[source reference] log L_k=k+o(k). Indeed log L_k=∑_p^a≤ klog p is the Chebyshev function ψ_rm Ch(k), and its ratio to k tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write log L_k=∑_a≥1ϑ(k^1/a), where ϑ(x)=∑_p≤ xlog p. The first term is k+o(k). The remaining terms are O(sqrtklog k)=o(k), using ϑ(x)=O(x) and at most log_2 k nonzero terms. The symbol ψ_rm Ch her…
Exact mathematical passage
\psi_{\rm Ch}(k)
cIntroduction passage · 45 passages
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Introduction passage · chapter03.tex, line 50
The notation a_n,c_n in this chapter and the next denotes recurrence coefficients. The later source notation a_n,b_n is translated explicitly when it occurs, and the cleared packet coefficients A_n,m,B_n,m have two indices. These are different objects.
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a_n,c_n
c_nIntroduction passage · 18 passages
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Introduction passage · chapter03.tex, line 50
The notation a_n,c_n in this chapter and the next denotes recurrence coefficients. The later source notation a_n,b_n is translated explicitly when it occurs, and the cleared packet coefficients A_n,m,B_n,m have two indices. These are different objects.
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a_n,c_n
DIntroduction passage · 71 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
dIntroduction passage · 23 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
d_nIntroduction passage · 8 passages
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Introduction passage · chapters04_07.tex, line 106
Set r_n=fracv_nu_n, d_n=fracp(n+1)2(2n+1)^2(2n+2)^2u_nu_n+1. The symbol d_n denotes a positive rational step size. It is unrelated to the least common multiples L_N used in denominator clearing. Dividing the Wronskian identity by u_nu_n+1 yields r_n+1-r_n=(-1)^n d_n. The movement of the approximants therefore alternates exactly.
Exact mathematical passage
\label{cb_eq_ratio_steps} r_n=\frac{v_n}{u_n},\qquad d_n=\frac{p(n+1)}{2(2n+1)^2(2n+2)^2u_nu_{n+1}}.
EIntroduction passage · 15 passages
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Introduction passage · chapters04_07.tex, line 471
Let E_n denote the well poised _6F_5 series at -1 in Zudilin's displayed Whipple transformation, with a=3n+1, b=c=d=n+1/2, and e=n+1. The transformation states H_n=fracΓ(3n+2)Γ(n+tfrac12) Γ(2n+tfrac32)Γ(2n+1)E_n. Its condition Re(1+a-d-e)>0 becomes n+1/2>0 here. The displayed prefactor of the source hypergeometric form F_n^rm src is F_n^rm src=(-1)^n fracΓ(n+1)^2Γ(3n+2)Γ(n+tfrac12)^3 Γ(2n+tfrac32)^3Γ(2n+1)E_n. Substitution in [source reference] therefore gives F_n^rm src=(-1)^nJ_n. There is no a…
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E_n
eIntroduction passage · 15 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
E_nIntroduction passage · 3 passages
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Introduction passage · chapters04_07.tex, line 471
Let E_n denote the well poised _6F_5 series at -1 in Zudilin's displayed Whipple transformation, with a=3n+1, b=c=d=n+1/2, and e=n+1. The transformation states H_n=fracΓ(3n+2)Γ(n+tfrac12) Γ(2n+tfrac32)Γ(2n+1)E_n. Its condition Re(1+a-d-e)>0 becomes n+1/2>0 here. The displayed prefactor of the source hypergeometric form F_n^rm src is F_n^rm src=(-1)^n fracΓ(n+1)^2Γ(3n+2)Γ(n+tfrac12)^3 Γ(2n+tfrac32)^3Γ(2n+1)E_n. Substitution in [source reference] therefore gives F_n^rm src=(-1)^nJ_n. There is no a…
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E_n
FIntroduction passage · 21 passages
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Introduction passage · chapters04_07.tex, line 471
Let E_n denote the well poised _6F_5 series at -1 in Zudilin's displayed Whipple transformation, with a=3n+1, b=c=d=n+1/2, and e=n+1. The transformation states H_n=fracΓ(3n+2)Γ(n+tfrac12) Γ(2n+tfrac32)Γ(2n+1)E_n. Its condition Re(1+a-d-e)>0 becomes n+1/2>0 here. The displayed prefactor of the source hypergeometric form F_n^rm src is F_n^rm src=(-1)^n fracΓ(n+1)^2Γ(3n+2)Γ(n+tfrac12)^3 Γ(2n+tfrac32)^3Γ(2n+1)E_n. Substitution in [source reference] therefore gives F_n^rm src=(-1)^nJ_n. There is no a…
Exact mathematical passage
{}_6F_5
fIntroduction passage · 1 passage
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Introduction passage · chapters04_07.tex, line 563
If E denotes the shift operator (Ef)_n=f_n+1, the packet is (1+E)^mT at index n. This is an identity about a finite polynomial in one shift. It does not introduce a further assumption about recurrence solutions or convergence. Reduction modulo a power of two commutes with this polynomial, which is the basis of the valuation computation in the next chapter.
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(Ef)_n=f_{n+1}
F_5Introduction passage · 1 passage
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Introduction passage · chapters04_07.tex, line 471
Let E_n denote the well poised _6F_5 series at -1 in Zudilin's displayed Whipple transformation, with a=3n+1, b=c=d=n+1/2, and e=n+1. The transformation states H_n=fracΓ(3n+2)Γ(n+tfrac12) Γ(2n+tfrac32)Γ(2n+1)E_n. Its condition Re(1+a-d-e)>0 becomes n+1/2>0 here. The displayed prefactor of the source hypergeometric form F_n^rm src is F_n^rm src=(-1)^n fracΓ(n+1)^2Γ(3n+2)Γ(n+tfrac12)^3 Γ(2n+tfrac32)^3Γ(2n+1)E_n. Substitution in [source reference] therefore gives F_n^rm src=(-1)^nJ_n. There is no a…
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{}_6F_5
f_{n+1}Introduction passage · 1 passage
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Introduction passage · chapters04_07.tex, line 563
If E denotes the shift operator (Ef)_n=f_n+1, the packet is (1+E)^mT at index n. This is an identity about a finite polynomial in one shift. It does not introduce a further assumption about recurrence solutions or convergence. Reduction modulo a power of two commutes with this polynomial, which is the basis of the valuation computation in the next chapter.
Exact mathematical passage
(Ef)_n=f_{n+1}
F_n^{\rm src}Introduction passage · 4 passages
Search aliases: F_n^rm src rm
Introduction passage · chapters04_07.tex, line 479
Let E_n denote the well poised _6F_5 series at -1 in Zudilin's displayed Whipple transformation, with a=3n+1, b=c=d=n+1/2, and e=n+1. The transformation states H_n=fracΓ(3n+2)Γ(n+tfrac12) Γ(2n+tfrac32)Γ(2n+1)E_n. Its condition Re(1+a-d-e)>0 becomes n+1/2>0 here. The displayed prefactor of the source hypergeometric form F_n^rm src is F_n^rm src=(-1)^n fracΓ(n+1)^2Γ(3n+2)Γ(n+tfrac12)^3 Γ(2n+tfrac32)^3Γ(2n+1)E_n. Substitution in [source reference] therefore gives F_n^rm src=(-1)^nJ_n. There is no a…
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F_n^{\rm src}
gIntroduction passage · 28 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
\gcdIntroduction passage · 10 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
HIntroduction passage · 35 passages
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Introduction passage · appendices.tex, line 93
Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are widetilde U_n,m=8(-1)^n16^-nS_n,m and widetilde V_n,m=8(-1)^n16^-nH_n,m. They must not be mistaken for the fixed-pair aliases U,V above.
Exact mathematical passage
\widetilde V_{n,m}=8(-1)^n16^{-n}H_{n,m}
hIntroduction passage · 50 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
H_nIntroduction passage · 4 passages
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Introduction passage · chapters04_07.tex, line 474
Let E_n denote the well poised _6F_5 series at -1 in Zudilin's displayed Whipple transformation, with a=3n+1, b=c=d=n+1/2, and e=n+1. The transformation states H_n=fracΓ(3n+2)Γ(n+tfrac12) Γ(2n+tfrac32)Γ(2n+1)E_n. Its condition Re(1+a-d-e)>0 becomes n+1/2>0 here. The displayed prefactor of the source hypergeometric form F_n^rm src is F_n^rm src=(-1)^n fracΓ(n+1)^2Γ(3n+2)Γ(n+tfrac12)^3 Γ(2n+tfrac32)^3Γ(2n+1)E_n. Substitution in [source reference] therefore gives F_n^rm src=(-1)^nJ_n. There is no a…
Exact mathematical passage
H_n=\frac{\Gamma(3n+2)\Gamma(n+\tfrac12)} {\Gamma(2n+\tfrac32)\Gamma(2n+1)}E_n.
H_{n,m}Introduction passage · 13 passages
Search aliases: H_n,m
Introduction passage · appendices.tex, line 93
Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are widetilde U_n,m=8(-1)^n16^-nS_n,m and widetilde V_n,m=8(-1)^n16^-nH_n,m. They must not be mistaken for the fixed-pair aliases U,V above.
Exact mathematical passage
\widetilde V_{n,m}=8(-1)^n16^{-n}H_{n,m}
iIntroduction passage · 12 passages
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Introduction passage · chapters04_07.tex, line 525
For integers n,m≥0, define S_n,m · =∑_i=0^mbinom miT_n+i, H_n,m · =∑_i=0^mbinom mi16^n+iv_n+i. The two index symbol H_n,m denotes a rational packet coefficient. It is distinct from the hypergeometric function H_n used in the integral normalization derivation. The theorem on T_n proves S_n,m∈Z_>0 at every pair. The coefficient H_n,m is rational and requires separate denominator clearing.
Exact mathematical passage
S_{n,m}&=\sum_{i=0}^m\binom miT_{n+i},\label{cb_eq_S_def}\\ H_{n,m}&=\sum_{i=0}^m\binom mi16^{n+i}v_{n+i}. \label{cb_eq_H_packet_def}
k^{1/a}Introduction passage · 1 passage
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Introduction passage · chapters08_10.tex, line 246
The prime number theorem supplies the standard asymptotic~[source reference] log L_k=k+o(k). Indeed log L_k=∑_p^a≤ klog p is the Chebyshev function ψ_rm Ch(k), and its ratio to k tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write log L_k=∑_a≥1ϑ(k^1/a), where ϑ(x)=∑_p≤ xlog p. The first term is k+o(k). The remaining terms are O(sqrtklog k)=o(k), using ϑ(x)=O(x) and at most log_2 k nonzero terms. The symbol ψ_rm Ch her…
Exact mathematical passage
\log L_k=\sum_{a\geq1}\vartheta(k^{1/a})
LIntroduction passage · 35 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
L_0Introduction passage · 3 passages
Search aliases: L_0
Introduction passage · chapters04_07.tex, line 371
Put L_N=lcm(1,2,…,N), L_0=1. The source's least common multiple notation is translated to L_N here. The letter D will later denote an actual reduced denominator, which can be much smaller than this universal clearing bound. Krattenthaler and Rivoal's Theorem 2 gives 16^nL_2n^2b_n∈Z for positive integer n [source reference]. In the present normalization it implies 4·16^nL_2n^2v_n∈Z (n≥0). The case n=0 is immediate from v_0=0. In particular the historical script's weaker clearing factor, with 8 in…
Exact mathematical passage
L_N=\operatorname{lcm}(1,2,\ldots,N),\qquad L_0=1.
L_{2(n+m)}^2Introduction passage · 1 passage
Search aliases: L_2(n+m)^2
Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
L_{2n}^2Introduction passage · 2 passages
Search aliases: L_2n^2
Introduction passage · chapters04_07.tex, line 378
Put L_N=lcm(1,2,…,N), L_0=1. The source's least common multiple notation is translated to L_N here. The letter D will later denote an actual reduced denominator, which can be much smaller than this universal clearing bound. Krattenthaler and Rivoal's Theorem 2 gives 16^nL_2n^2b_n∈Z for positive integer n [source reference]. In the present normalization it implies 4·16^nL_2n^2v_n∈Z (n≥0). The case n=0 is immediate from v_0=0. In particular the historical script's weaker clearing factor, with 8 in…
Exact mathematical passage
16^nL_{2n}^2b_n\in\Z
L_kIntroduction passage · 4 passages
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Introduction passage · chapters08_10.tex, line 242
The prime number theorem supplies the standard asymptotic~[source reference] log L_k=k+o(k). Indeed log L_k=∑_p^a≤ klog p is the Chebyshev function ψ_rm Ch(k), and its ratio to k tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write log L_k=∑_a≥1ϑ(k^1/a), where ϑ(x)=∑_p≤ xlog p. The first term is k+o(k). The remaining terms are O(sqrtklog k)=o(k), using ϑ(x)=O(x) and at most log_2 k nonzero terms. The symbol ψ_rm Ch her…
Exact mathematical passage
\log L_k=k+o(k). \label{eq:PNT-LCM}
L_NIntroduction passage · 3 passages
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Introduction passage · chapters04_07.tex, line 111
Set r_n=fracv_nu_n, d_n=fracp(n+1)2(2n+1)^2(2n+2)^2u_nu_n+1. The symbol d_n denotes a positive rational step size. It is unrelated to the least common multiples L_N used in denominator clearing. Dividing the Wronskian identity by u_nu_n+1 yields r_n+1-r_n=(-1)^n d_n. The movement of the approximants therefore alternates exactly.
Exact mathematical passage
L_N
\operatorname{lcm}Introduction passage · 3 passages
Search aliases: lcm operatorname
Introduction passage · chapters04_07.tex, line 371
Put L_N=lcm(1,2,…,N), L_0=1. The source's least common multiple notation is translated to L_N here. The letter D will later denote an actual reduced denominator, which can be much smaller than this universal clearing bound. Krattenthaler and Rivoal's Theorem 2 gives 16^nL_2n^2b_n∈Z for positive integer n [source reference]. In the present normalization it implies 4·16^nL_2n^2v_n∈Z (n≥0). The case n=0 is immediate from v_0=0. In particular the historical script's weaker clearing factor, with 8 in…
Exact mathematical passage
L_N=\operatorname{lcm}(1,2,\ldots,N),\qquad L_0=1.
\logIntroduction passage · 49 passages
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Introduction passage · chapters08_10.tex, line 242
The prime number theorem supplies the standard asymptotic~[source reference] log L_k=k+o(k). Indeed log L_k=∑_p^a≤ klog p is the Chebyshev function ψ_rm Ch(k), and its ratio to k tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write log L_k=∑_a≥1ϑ(k^1/a), where ϑ(x)=∑_p≤ xlog p. The first term is k+o(k). The remaining terms are O(sqrtklog k)=o(k), using ϑ(x)=O(x) and at most log_2 k nonzero terms. The symbol ψ_rm Ch her…
Exact mathematical passage
\log L_k=k+o(k). \label{eq:PNT-LCM}
\log_2Introduction passage · 1 passage
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Introduction passage · chapters08_10.tex, line 246
the Chebyshev function ψ_rm Ch(k), and its ratio to k tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write log L_k=∑_a≥1ϑ(k^1/a), where ϑ(x)=∑_p≤ xlog p. The first term is k+o(k). The remaining terms are O(sqrtklog k)=o(k), using ϑ(x)=O(x) and at most log_2 k nonzero terms. The symbol ψ_rm Ch here denotes the arithmetic function, not the two variable kernel in [source reference]. The prime number theorem is an external…
Exact mathematical passage
\log_2 k
mIntroduction passage · 118 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
NIntroduction passage · 73 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
OIntroduction passage · 3 passages
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Introduction passage · chapters08_10.tex, line 246
-LCM Indeed log L_k=∑_p^a≤ klog p is the Chebyshev function ψ_rm Ch(k), and its ratio to k tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write log L_k=∑_a≥1ϑ(k^1/a), where ϑ(x)=∑_p≤ xlog p. The first term is k+o(k). The remaining terms are O(sqrtklog k)=o(k), using ϑ(x)=O(x) and at most log_2 k nonzero terms. The symbol ψ_rm Ch here denotes the arithmetic function, not the two variable kernel in [source reference]. Th…
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O(\sqrt{k}\log k)=o(k)
oIntroduction passage · 9 passages
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Introduction passage · chapters08_10.tex, line 242
The prime number theorem supplies the standard asymptotic~[source reference] log L_k=k+o(k). Indeed log L_k=∑_p^a≤ klog p is the Chebyshev function ψ_rm Ch(k), and its ratio to k tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write log L_k=∑_a≥1ϑ(k^1/a), where ϑ(x)=∑_p≤ xlog p. The first term is k+o(k). The remaining terms are O(sqrtklog k)=o(k), using ϑ(x)=O(x) and at most log_2 k nonzero terms. The symbol ψ_rm Ch her…
Exact mathematical passage
\log L_k=k+o(k). \label{eq:PNT-LCM}
pIntroduction passage · 24 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
p^aIntroduction passage · 1 passage
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Introduction passage · chapters08_10.tex, line 246
The prime number theorem supplies the standard asymptotic~[source reference] log L_k=k+o(k). Indeed log L_k=∑_p^a≤ klog p is the Chebyshev function ψ_rm Ch(k), and its ratio to k tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write log L_k=∑_a≥1ϑ(k^1/a), where ϑ(x)=∑_p≤ xlog p. The first term is k+o(k). The remaining terms are O(sqrtklog k)=o(k), using ϑ(x)=O(x) and at most log_2 k nonzero terms. The symbol ψ_rm Ch her…
Exact mathematical passage
\log L_k=\sum_{p^a\leq k}\log p
p_{\rm red}Introduction passage · 1 passage
Search aliases: p_rm red rm
Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
QIntroduction passage · 75 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
qIntroduction passage · 31 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
q_{\rm red}Introduction passage · 2 passages
Search aliases: q_rm red rm
Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
r_{n+1}Introduction passage · 1 passage
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Introduction passage · chapters04_07.tex, line 113
Set r_n=fracv_nu_n, d_n=fracp(n+1)2(2n+1)^2(2n+2)^2u_nu_n+1. The symbol d_n denotes a positive rational step size. It is unrelated to the least common multiples L_N used in denominator clearing. Dividing the Wronskian identity by u_nu_n+1 yields r_n+1-r_n=(-1)^n d_n. The movement of the approximants therefore alternates exactly.
Exact mathematical passage
\label{cb_eq_alternating_steps} r_{n+1}-r_n=(-1)^n d_n.
\operatorname{Re}Introduction passage · 1 passage
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Introduction passage · chapters04_07.tex, line 478
Let E_n denote the well poised _6F_5 series at -1 in Zudilin's displayed Whipple transformation, with a=3n+1, b=c=d=n+1/2, and e=n+1. The transformation states H_n=fracΓ(3n+2)Γ(n+tfrac12) Γ(2n+tfrac32)Γ(2n+1)E_n. Its condition Re(1+a-d-e)>0 becomes n+1/2>0 here. The displayed prefactor of the source hypergeometric form F_n^rm src is F_n^rm src=(-1)^n fracΓ(n+1)^2Γ(3n+2)Γ(n+tfrac12)^3 Γ(2n+tfrac32)^3Γ(2n+1)E_n. Substitution in [source reference] therefore gives F_n^rm src=(-1)^nJ_n. There is no a…
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\operatorname{Re}(1+a-d-e)>0
\rmIntroduction passage · 8 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
SIntroduction passage · 36 passages
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Introduction passage · appendices.tex, line 93
Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are widetilde U_n,m=8(-1)^n16^-nS_n,m and widetilde V_n,m=8(-1)^n16^-nH_n,m. They must not be mistaken for the fixed-pair aliases U,V above.
Exact mathematical passage
\widetilde U_{n,m}=8(-1)^n16^{-n}S_{n,m}
sIntroduction passage · 36 passages
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Introduction passage · chapters04_07.tex, line 479
Let E_n denote the well poised _6F_5 series at -1 in Zudilin's displayed Whipple transformation, with a=3n+1, b=c=d=n+1/2, and e=n+1. The transformation states H_n=fracΓ(3n+2)Γ(n+tfrac12) Γ(2n+tfrac32)Γ(2n+1)E_n. Its condition Re(1+a-d-e)>0 becomes n+1/2>0 here. The displayed prefactor of the source hypergeometric form F_n^rm src is F_n^rm src=(-1)^n fracΓ(n+1)^2Γ(3n+2)Γ(n+tfrac12)^3 Γ(2n+tfrac32)^3Γ(2n+1)E_n. Substitution in [source reference] therefore gives F_n^rm src=(-1)^nJ_n. There is no a…
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F_n^{\rm src}
S_{n,m}Introduction passage · 19 passages
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Introduction passage · appendices.tex, line 93
Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are widetilde U_n,m=8(-1)^n16^-nS_n,m and widetilde V_n,m=8(-1)^n16^-nH_n,m. They must not be mistaken for the fixed-pair aliases U,V above.
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\widetilde U_{n,m}=8(-1)^n16^{-n}S_{n,m}
TIntroduction passage · 29 passages
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Introduction passage · chapters04_07.tex, line 525
For integers n,m≥0, define S_n,m · =∑_i=0^mbinom miT_n+i, H_n,m · =∑_i=0^mbinom mi16^n+iv_n+i. The two index symbol H_n,m denotes a rational packet coefficient. It is distinct from the hypergeometric function H_n used in the integral normalization derivation. The theorem on T_n proves S_n,m∈Z_>0 at every pair. The coefficient H_n,m is rational and requires separate denominator clearing.
Exact mathematical passage
S_{n,m}&=\sum_{i=0}^m\binom miT_{n+i},\label{cb_eq_S_def}\\ H_{n,m}&=\sum_{i=0}^m\binom mi16^{n+i}v_{n+i}. \label{cb_eq_H_packet_def}
T_nIntroduction passage · 16 passages
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Introduction passage · chapters04_07.tex, line 532
H_n,m · =∑_i=0^mbinom mi16^n+iv_n+i. The two index symbol H_n,m denotes a rational packet coefficient. It is distinct from the hypergeometric function H_n used in the integral normalization derivation. The theorem on T_n proves S_n,m∈Z_>0 at every pair. The coefficient H_n,m is rational and requires separate denominator clearing.
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T_n
T_{n+i}Introduction passage · 4 passages
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Introduction passage · chapters04_07.tex, line 525
For integers n,m≥0, define S_n,m · =∑_i=0^mbinom miT_n+i, H_n,m · =∑_i=0^mbinom mi16^n+iv_n+i. The two index symbol H_n,m denotes a rational packet coefficient. It is distinct from the hypergeometric function H_n used in the integral normalization derivation. The theorem on T_n proves S_n,m∈Z_>0 at every pair. The coefficient H_n,m is rational and requires separate denominator clearing.
Exact mathematical passage
S_{n,m}&=\sum_{i=0}^m\binom miT_{n+i},\label{cb_eq_S_def}\\ H_{n,m}&=\sum_{i=0}^m\binom mi16^{n+i}v_{n+i}. \label{cb_eq_H_packet_def}
UIntroduction passage · 32 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
u_2Introduction passage · 7 passages
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Introduction passage · chapter03.tex, line 274
For a positive solution, ratios of consecutive values give another finite description. If x_n=w_n/w_n-1 is defined, then x_n+1=α_n+fracβ_nx_n. For u, every such ratio is defined. The first two iterations are fracu_2u_1=α_1+fracβ_17/4, fracu_3u_2= α_2+fracβ_2α_1+β_1/(7/4). Repeated substitution yields a finite generalized continued fraction. For v, one cannot begin with v_1/v_0, because v_0=0. Its ratio iteration instead begins at v_2/v_1=α_1. These formulas are exact finite identities. Establish…
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\frac{u_2}{u_1}=\alpha_1+\frac{\beta_1}{7/4},\qquad \frac{u_3}{u_2}= \alpha_2+\frac{\beta_2}{\alpha_1+\beta_1/(7/4)}.
u_3Introduction passage · 3 passages
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Introduction passage · chapter03.tex, line 274
For a positive solution, ratios of consecutive values give another finite description. If x_n=w_n/w_n-1 is defined, then x_n+1=α_n+fracβ_nx_n. For u, every such ratio is defined. The first two iterations are fracu_2u_1=α_1+fracβ_17/4, fracu_3u_2= α_2+fracβ_2α_1+β_1/(7/4). Repeated substitution yields a finite generalized continued fraction. For v, one cannot begin with v_1/v_0, because v_0=0. Its ratio iteration instead begins at v_2/v_1=α_1. These formulas are exact finite identities. Establish…
Exact mathematical passage
\frac{u_2}{u_1}=\alpha_1+\frac{\beta_1}{7/4},\qquad \frac{u_3}{u_2}= \alpha_2+\frac{\beta_2}{\alpha_1+\beta_1/(7/4)}.
u_{n+1}Introduction passage · 8 passages
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Introduction passage · chapters04_07.tex, line 106
Set r_n=fracv_nu_n, d_n=fracp(n+1)2(2n+1)^2(2n+2)^2u_nu_n+1. The symbol d_n denotes a positive rational step size. It is unrelated to the least common multiples L_N used in denominator clearing. Dividing the Wronskian identity by u_nu_n+1 yields r_n+1-r_n=(-1)^n d_n. The movement of the approximants therefore alternates exactly.
Exact mathematical passage
\label{cb_eq_ratio_steps} r_n=\frac{v_n}{u_n},\qquad d_n=\frac{p(n+1)}{2(2n+1)^2(2n+2)^2u_nu_{n+1}}.
VIntroduction passage · 16 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
v_2Introduction passage · 26 passages
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Introduction passage · chapter03.tex, line 89
The initial u values are positive, and each subsequent value is a positive linear combination of its two predecessors. For v, the initial v_0 is zero and v_1>0. The first forward step gives v_2>0, and the same induction applies thereafter. Consequently the quotient r_n is defined at every index.
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v_2>0
v_{n+i}Introduction passage · 3 passages
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Introduction passage · chapters04_07.tex, line 525
For integers n,m≥0, define S_n,m · =∑_i=0^mbinom miT_n+i, H_n,m · =∑_i=0^mbinom mi16^n+iv_n+i. The two index symbol H_n,m denotes a rational packet coefficient. It is distinct from the hypergeometric function H_n used in the integral normalization derivation. The theorem on T_n proves S_n,m∈Z_>0 at every pair. The coefficient H_n,m is rational and requires separate denominator clearing.
Exact mathematical passage
S_{n,m}&=\sum_{i=0}^m\binom miT_{n+i},\label{cb_eq_S_def}\\ H_{n,m}&=\sum_{i=0}^m\binom mi16^{n+i}v_{n+i}. \label{cb_eq_H_packet_def}
wIntroduction passage · 23 passages
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Introduction passage · chapter03.tex, line 269
For a positive solution, ratios of consecutive values give another finite description. If x_n=w_n/w_n-1 is defined, then x_n+1=α_n+fracβ_nx_n. For u, every such ratio is defined. The first two iterations are fracu_2u_1=α_1+fracβ_17/4, fracu_3u_2= α_2+fracβ_2α_1+β_1/(7/4). Repeated substitution yields a finite generalized continued fraction. For v, one cannot begin with v_1/v_0, because v_0=0. Its ratio iteration instead begins at v_2/v_1=α_1. These formulas are exact finite identities. Establish…
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x_n=w_n/w_{n-1}
w_nIntroduction passage · 9 passages
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Introduction passage · chapter03.tex, line 269
For a positive solution, ratios of consecutive values give another finite description. If x_n=w_n/w_n-1 is defined, then x_n+1=α_n+fracβ_nx_n. For u, every such ratio is defined. The first two iterations are fracu_2u_1=α_1+fracβ_17/4, fracu_3u_2= α_2+fracβ_2α_1+β_1/(7/4). Repeated substitution yields a finite generalized continued fraction. For v, one cannot begin with v_1/v_0, because v_0=0. Its ratio iteration instead begins at v_2/v_1=α_1. These formulas are exact finite identities. Establish…
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x_n=w_n/w_{n-1}
w_{n-1}Introduction passage · 6 passages
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Introduction passage · chapter03.tex, line 269
For a positive solution, ratios of consecutive values give another finite description. If x_n=w_n/w_n-1 is defined, then x_n+1=α_n+fracβ_nx_n. For u, every such ratio is defined. The first two iterations are fracu_2u_1=α_1+fracβ_17/4, fracu_3u_2= α_2+fracβ_2α_1+β_1/(7/4). Repeated substitution yields a finite generalized continued fraction. For v, one cannot begin with v_1/v_0, because v_0=0. Its ratio iteration instead begins at v_2/v_1=α_1. These formulas are exact finite identities. Establish…
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x_n=w_n/w_{n-1}
\widetilde JIntroduction passage · 21 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
\widetilde J_{n,m}Introduction passage · 13 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
\widetilde UIntroduction passage · 4 passages
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Introduction passage · appendices.tex, line 93
Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are widetilde U_n,m=8(-1)^n16^-nS_n,m and widetilde V_n,m=8(-1)^n16^-nH_n,m. They must not be mistaken for the fixed-pair aliases U,V above.
Exact mathematical passage
\widetilde U_{n,m}=8(-1)^n16^{-n}S_{n,m}
\widetilde U_{n,m}Introduction passage · 1 passage
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Introduction passage · appendices.tex, line 93
Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are widetilde U_n,m=8(-1)^n16^-nS_n,m and widetilde V_n,m=8(-1)^n16^-nH_n,m. They must not be mistaken for the fixed-pair aliases U,V above.
Exact mathematical passage
\widetilde U_{n,m}=8(-1)^n16^{-n}S_{n,m}
\widetilde VIntroduction passage · 1 passage
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Introduction passage · appendices.tex, line 93
Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are widetilde U_n,m=8(-1)^n16^-nS_n,m and widetilde V_n,m=8(-1)^n16^-nH_n,m. They must not be mistaken for the fixed-pair aliases U,V above.
Exact mathematical passage
\widetilde V_{n,m}=8(-1)^n16^{-n}H_{n,m}
\widetilde V_{n,m}Introduction passage · 1 passage
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Introduction passage · appendices.tex, line 93
Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are widetilde U_n,m=8(-1)^n16^-nS_n,m and widetilde V_n,m=8(-1)^n16^-nH_n,m. They must not be mistaken for the fixed-pair aliases U,V above.
Exact mathematical passage
\widetilde V_{n,m}=8(-1)^n16^{-n}H_{n,m}
xIntroduction passage · 66 passages
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Introduction passage · chapter03.tex, line 269
For a positive solution, ratios of consecutive values give another finite description. If x_n=w_n/w_n-1 is defined, then x_n+1=α_n+fracβ_nx_n. For u, every such ratio is defined. The first two iterations are fracu_2u_1=α_1+fracβ_17/4, fracu_3u_2= α_2+fracβ_2α_1+β_1/(7/4). Repeated substitution yields a finite generalized continued fraction. For v, one cannot begin with v_1/v_0, because v_0=0. Its ratio iteration instead begins at v_2/v_1=α_1. These formulas are exact finite identities. Establish…
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x_n=w_n/w_{n-1}
x_nIntroduction passage · 5 passages
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Introduction passage · chapter03.tex, line 269
For a positive solution, ratios of consecutive values give another finite description. If x_n=w_n/w_n-1 is defined, then x_n+1=α_n+fracβ_nx_n. For u, every such ratio is defined. The first two iterations are fracu_2u_1=α_1+fracβ_17/4, fracu_3u_2= α_2+fracβ_2α_1+β_1/(7/4). Repeated substitution yields a finite generalized continued fraction. For v, one cannot begin with v_1/v_0, because v_0=0. Its ratio iteration instead begins at v_2/v_1=α_1. These formulas are exact finite identities. Establish…
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x_n=w_n/w_{n-1}
x_{n+1}Introduction passage · 1 passage
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Introduction passage · chapter03.tex, line 270
For a positive solution, ratios of consecutive values give another finite description. If x_n=w_n/w_n-1 is defined, then x_n+1=α_n+fracβ_nx_n. For u, every such ratio is defined. The first two iterations are fracu_2u_1=α_1+fracβ_17/4, fracu_3u_2= α_2+fracβ_2α_1+β_1/(7/4). Repeated substitution yields a finite generalized continued fraction. For v, one cannot begin with v_1/v_0, because v_0=0. Its ratio iteration instead begins at v_2/v_1=α_1. These formulas are exact finite identities. Establish…
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x_{n+1}=\alpha_n+\frac{\beta_n}{x_n}.
\ZIntroduction passage · 18 passages
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Introduction passage · chapters04_07.tex, line 378
Put L_N=lcm(1,2,…,N), L_0=1. The source's least common multiple notation is translated to L_N here. The letter D will later denote an actual reduced denominator, which can be much smaller than this universal clearing bound. Krattenthaler and Rivoal's Theorem 2 gives 16^nL_2n^2b_n∈Z for positive integer n [source reference]. In the present normalization it implies 4·16^nL_2n^2v_n∈Z (n≥0). The case n=0 is immediate from v_0=0. In particular the historical script's weaker clearing factor, with 8 in…
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16^nL_{2n}^2b_n\in\Z
\Z_{>0}Introduction passage · 3 passages
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Introduction passage · chapters04_07.tex, line 533
H_n,m · =∑_i=0^mbinom mi16^n+iv_n+i. The two index symbol H_n,m denotes a rational packet coefficient. It is distinct from the hypergeometric function H_n used in the integral normalization derivation. The theorem on T_n proves S_n,m∈Z_>0 at every pair. The coefficient H_n,m is rational and requires separate denominator clearing.
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S_{n,m}\in\Z_{>0}
|Introduction passage · 15 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
\cdotIntroduction passage · 14 passages
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Introduction passage · chapters04_07.tex, line 380
melcm(1,2,…,N), L_0=1. The source's least common multiple notation is translated to L_N here. The letter D will later denote an actual reduced denominator, which can be much smaller than this universal clearing bound. Krattenthaler and Rivoal's Theorem 2 gives 16^nL_2n^2b_n∈Z for positive integer n [source reference]. In the present normalization it implies 4·16^nL_2n^2v_n∈Z (n≥0). The case n=0 is immediate from v_0=0. In particular the historical script's weaker clearing factor, with 8 in place…
Exact mathematical passage
\label{cb_eq_v_sharp_clear} 4\cdot16^nL_{2n}^2v_n\in\Z\qquad(n\geq0).
\alphaIntroduction passage · 97 passages
Search aliases: α alpha
Introduction passage · chapter03.tex, line 270
For a positive solution, ratios of consecutive values give another finite description. If x_n=w_n/w_n-1 is defined, then x_n+1=α_n+fracβ_nx_n. For u, every such ratio is defined. The first two iterations are fracu_2u_1=α_1+fracβ_17/4, fracu_3u_2= α_2+fracβ_2α_1+β_1/(7/4). Repeated substitution yields a finite generalized continued fraction. For v, one cannot begin with v_1/v_0, because v_0=0. Its ratio iteration instead begins at v_2/v_1=α_1. These formulas are exact finite identities. Establish…
Exact mathematical passage
x_{n+1}=\alpha_n+\frac{\beta_n}{x_n}.
\alpha_1Introduction passage · 4 passages
Search aliases: α_1 alpha
Introduction passage · chapter03.tex, line 274
For a positive solution, ratios of consecutive values give another finite description. If x_n=w_n/w_n-1 is defined, then x_n+1=α_n+fracβ_nx_n. For u, every such ratio is defined. The first two iterations are fracu_2u_1=α_1+fracβ_17/4, fracu_3u_2= α_2+fracβ_2α_1+β_1/(7/4). Repeated substitution yields a finite generalized continued fraction. For v, one cannot begin with v_1/v_0, because v_0=0. Its ratio iteration instead begins at v_2/v_1=α_1. These formulas are exact finite identities. Establish…
Exact mathematical passage
\frac{u_2}{u_1}=\alpha_1+\frac{\beta_1}{7/4},\qquad \frac{u_3}{u_2}= \alpha_2+\frac{\beta_2}{\alpha_1+\beta_1/(7/4)}.
\alpha_2Introduction passage · 2 passages
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Introduction passage · chapter03.tex, line 274
For a positive solution, ratios of consecutive values give another finite description. If x_n=w_n/w_n-1 is defined, then x_n+1=α_n+fracβ_nx_n. For u, every such ratio is defined. The first two iterations are fracu_2u_1=α_1+fracβ_17/4, fracu_3u_2= α_2+fracβ_2α_1+β_1/(7/4). Repeated substitution yields a finite generalized continued fraction. For v, one cannot begin with v_1/v_0, because v_0=0. Its ratio iteration instead begins at v_2/v_1=α_1. These formulas are exact finite identities. Establish…
Exact mathematical passage
\frac{u_2}{u_1}=\alpha_1+\frac{\beta_1}{7/4},\qquad \frac{u_3}{u_2}= \alpha_2+\frac{\beta_2}{\alpha_1+\beta_1/(7/4)}.
\alpha_nIntroduction passage · 7 passages
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Introduction passage · chapter03.tex, line 270
For a positive solution, ratios of consecutive values give another finite description. If x_n=w_n/w_n-1 is defined, then x_n+1=α_n+fracβ_nx_n. For u, every such ratio is defined. The first two iterations are fracu_2u_1=α_1+fracβ_17/4, fracu_3u_2= α_2+fracβ_2α_1+β_1/(7/4). Repeated substitution yields a finite generalized continued fraction. For v, one cannot begin with v_1/v_0, because v_0=0. Its ratio iteration instead begins at v_2/v_1=α_1. These formulas are exact finite identities. Establish…
Exact mathematical passage
x_{n+1}=\alpha_n+\frac{\beta_n}{x_n}.
\betaIntroduction passage · 21 passages
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Introduction passage · chapter03.tex, line 270
For a positive solution, ratios of consecutive values give another finite description. If x_n=w_n/w_n-1 is defined, then x_n+1=α_n+fracβ_nx_n. For u, every such ratio is defined. The first two iterations are fracu_2u_1=α_1+fracβ_17/4, fracu_3u_2= α_2+fracβ_2α_1+β_1/(7/4). Repeated substitution yields a finite generalized continued fraction. For v, one cannot begin with v_1/v_0, because v_0=0. Its ratio iteration instead begins at v_2/v_1=α_1. These formulas are exact finite identities. Establish…
Exact mathematical passage
x_{n+1}=\alpha_n+\frac{\beta_n}{x_n}.
\beta_1Introduction passage · 4 passages
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Introduction passage · chapter03.tex, line 274
For a positive solution, ratios of consecutive values give another finite description. If x_n=w_n/w_n-1 is defined, then x_n+1=α_n+fracβ_nx_n. For u, every such ratio is defined. The first two iterations are fracu_2u_1=α_1+fracβ_17/4, fracu_3u_2= α_2+fracβ_2α_1+β_1/(7/4). Repeated substitution yields a finite generalized continued fraction. For v, one cannot begin with v_1/v_0, because v_0=0. Its ratio iteration instead begins at v_2/v_1=α_1. These formulas are exact finite identities. Establish…
Exact mathematical passage
\frac{u_2}{u_1}=\alpha_1+\frac{\beta_1}{7/4},\qquad \frac{u_3}{u_2}= \alpha_2+\frac{\beta_2}{\alpha_1+\beta_1/(7/4)}.
\beta_2Introduction passage · 2 passages
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Introduction passage · chapter03.tex, line 274
For a positive solution, ratios of consecutive values give another finite description. If x_n=w_n/w_n-1 is defined, then x_n+1=α_n+fracβ_nx_n. For u, every such ratio is defined. The first two iterations are fracu_2u_1=α_1+fracβ_17/4, fracu_3u_2= α_2+fracβ_2α_1+β_1/(7/4). Repeated substitution yields a finite generalized continued fraction. For v, one cannot begin with v_1/v_0, because v_0=0. Its ratio iteration instead begins at v_2/v_1=α_1. These formulas are exact finite identities. Establish…
Exact mathematical passage
\frac{u_2}{u_1}=\alpha_1+\frac{\beta_1}{7/4},\qquad \frac{u_3}{u_2}= \alpha_2+\frac{\beta_2}{\alpha_1+\beta_1/(7/4)}.
\beta_nIntroduction passage · 7 passages
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Introduction passage · chapter03.tex, line 270
For a positive solution, ratios of consecutive values give another finite description. If x_n=w_n/w_n-1 is defined, then x_n+1=α_n+fracβ_nx_n. For u, every such ratio is defined. The first two iterations are fracu_2u_1=α_1+fracβ_17/4, fracu_3u_2= α_2+fracβ_2α_1+β_1/(7/4). Repeated substitution yields a finite generalized continued fraction. For v, one cannot begin with v_1/v_0, because v_0=0. Its ratio iteration instead begins at v_2/v_1=α_1. These formulas are exact finite identities. Establish…
Exact mathematical passage
x_{n+1}=\alpha_n+\frac{\beta_n}{x_n}.
\GammaIntroduction passage · 3 passages
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Introduction passage · chapters04_07.tex, line 474
Let E_n denote the well poised _6F_5 series at -1 in Zudilin's displayed Whipple transformation, with a=3n+1, b=c=d=n+1/2, and e=n+1. The transformation states H_n=fracΓ(3n+2)Γ(n+tfrac12) Γ(2n+tfrac32)Γ(2n+1)E_n. Its condition Re(1+a-d-e)>0 becomes n+1/2>0 here. The displayed prefactor of the source hypergeometric form F_n^rm src is F_n^rm src=(-1)^n fracΓ(n+1)^2Γ(3n+2)Γ(n+tfrac12)^3 Γ(2n+tfrac32)^3Γ(2n+1)E_n. Substitution in [source reference] therefore gives F_n^rm src=(-1)^nJ_n. There is no a…
Exact mathematical passage
H_n=\frac{\Gamma(3n+2)\Gamma(n+\tfrac12)} {\Gamma(2n+\tfrac32)\Gamma(2n+1)}E_n.
\varthetaIntroduction passage · 3 passages
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Introduction passage · chapters08_10.tex, line 246
The prime number theorem supplies the standard asymptotic~[source reference] log L_k=k+o(k). Indeed log L_k=∑_p^a≤ klog p is the Chebyshev function ψ_rm Ch(k), and its ratio to k tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write log L_k=∑_a≥1ϑ(k^1/a), where ϑ(x)=∑_p≤ xlog p. The first term is k+o(k). The remaining terms are O(sqrtklog k)=o(k), using ϑ(x)=O(x) and at most log_2 k nonzero terms. The symbol ψ_rm Ch her…
Exact mathematical passage
\log L_k=\sum_{a\geq1}\vartheta(k^{1/a})
\LambdaIntroduction passage · 11 passages
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Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
\Lambda_{n,m}Introduction passage · 2 passages
Search aliases: Λ_n,m Lambda
Introduction passage · appendices.tex, line 83
For a fixed pair, put U=S_n,m and write V=H_n,m=N/D in reduced form. The exact dictionary is Q · =8L_2(n+m)^2,notag A · =(-1)^nQU, B=(-1)^nQV,notag h · =gcd(DU,N)=gcd(U,N),notag g · =gcd(|A|,|B|)=frac QD h,notag q_rm red · =fracDUh, p_rm red=frac Nh,notag Λ_n,m · =frac AgG-frac Bg =frac16^nL_2(n+m)^2gwidetilde J_n,m. The symbol q_rm red denotes a reduced denominator. It is unrelated to the polynomial q(n) in the recurrence. The rational coefficients of Eskandari's integral normalization are wide…
Exact mathematical passage
Q&=8L_{2(n+m)}^2,\notag\\ A&=(-1)^nQU,\qquad B=(-1)^nQV,\notag\\ h&=\gcd(DU,N)=\gcd(U,N),\notag\\ g&=\gcd(|A|,|B|)=\frac QD\,h,\notag\\ q_{\rm red}&=\frac{DU}{h},\qquad p_{\rm red}=\frac Nh,\notag\\ \Lambda_{n,m}&=\frac AgG-\frac Bg =\frac{16^nL_{2(n+m)}^2}{g}\widetilde J_{n,m}. \label{eq:app-reduction-dictionary}
\psiIntroduction passage · 18 passages
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Introduction passage · chapters08_10.tex, line 246
The prime number theorem supplies the standard asymptotic~[source reference] log L_k=k+o(k). Indeed log L_k=∑_p^a≤ klog p is the Chebyshev function ψ_rm Ch(k), and its ratio to k tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write log L_k=∑_a≥1ϑ(k^1/a), where ϑ(x)=∑_p≤ xlog p. The first term is k+o(k). The remaining terms are O(sqrtklog k)=o(k), using ϑ(x)=O(x) and at most log_2 k nonzero terms. The symbol ψ_rm Ch her…
Exact mathematical passage
\psi_{\rm Ch}(k)
\psi_{\rm Ch}Introduction passage · 2 passages
Search aliases: ψ_rm Ch psi rm
Introduction passage · chapters08_10.tex, line 246
The prime number theorem supplies the standard asymptotic~[source reference] log L_k=k+o(k). Indeed log L_k=∑_p^a≤ klog p is the Chebyshev function ψ_rm Ch(k), and its ratio to k tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write log L_k=∑_a≥1ϑ(k^1/a), where ϑ(x)=∑_p≤ xlog p. The first term is k+o(k). The remaining terms are O(sqrtklog k)=o(k), using ϑ(x)=O(x) and at most log_2 k nonzero terms. The symbol ψ_rm Ch her…
Exact mathematical passage
\psi_{\rm Ch}(k)
\ldotsIntroduction passage · 10 passages
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Introduction passage · chapters04_07.tex, line 371
Put L_N=lcm(1,2,…,N), L_0=1. The source's least common multiple notation is translated to L_N here. The letter D will later denote an actual reduced denominator, which can be much smaller than this universal clearing bound. Krattenthaler and Rivoal's Theorem 2 gives 16^nL_2n^2b_n∈Z for positive integer n [source reference]. In the present normalization it implies 4·16^nL_2n^2v_n∈Z (n≥0). The case n=0 is immediate from v_0=0. In particular the historical script's weaker clearing factor, with 8 in…
Exact mathematical passage
L_N=\operatorname{lcm}(1,2,\ldots,N),\qquad L_0=1.
\inIntroduction passage · 24 passages
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Introduction passage · chapters04_07.tex, line 378
Put L_N=lcm(1,2,…,N), L_0=1. The source's least common multiple notation is translated to L_N here. The letter D will later denote an actual reduced denominator, which can be much smaller than this universal clearing bound. Krattenthaler and Rivoal's Theorem 2 gives 16^nL_2n^2b_n∈Z for positive integer n [source reference]. In the present normalization it implies 4·16^nL_2n^2v_n∈Z (n≥0). The case n=0 is immediate from v_0=0. In particular the historical script's weaker clearing factor, with 8 in…
Exact mathematical passage
16^nL_{2n}^2b_n\in\Z
\sum_{a\geq1}Introduction passage · 1 passage
Search aliases: ∑_a≥1 sum geq
Introduction passage · chapters08_10.tex, line 246
The prime number theorem supplies the standard asymptotic~[source reference] log L_k=k+o(k). Indeed log L_k=∑_p^a≤ klog p is the Chebyshev function ψ_rm Ch(k), and its ratio to k tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write log L_k=∑_a≥1ϑ(k^1/a), where ϑ(x)=∑_p≤ xlog p. The first term is k+o(k). The remaining terms are O(sqrtklog k)=o(k), using ϑ(x)=O(x) and at most log_2 k nonzero terms. The symbol ψ_rm Ch her…
Exact mathematical passage
\log L_k=\sum_{a\geq1}\vartheta(k^{1/a})
\sum_{i=0}^mIntroduction passage · 5 passages
Search aliases: ∑_i=0^m sum
Introduction passage · chapters04_07.tex, line 525
For integers n,m≥0, define S_n,m · =∑_i=0^mbinom miT_n+i, H_n,m · =∑_i=0^mbinom mi16^n+iv_n+i. The two index symbol H_n,m denotes a rational packet coefficient. It is distinct from the hypergeometric function H_n used in the integral normalization derivation. The theorem on T_n proves S_n,m∈Z_>0 at every pair. The coefficient H_n,m is rational and requires separate denominator clearing.
Exact mathematical passage
S_{n,m}&=\sum_{i=0}^m\binom miT_{n+i},\label{cb_eq_S_def}\\ H_{n,m}&=\sum_{i=0}^m\binom mi16^{n+i}v_{n+i}. \label{cb_eq_H_packet_def}
\sum_{p^a\leq k}Introduction passage · 1 passage
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Introduction passage · chapters08_10.tex, line 246
The prime number theorem supplies the standard asymptotic~[source reference] log L_k=k+o(k). Indeed log L_k=∑_p^a≤ klog p is the Chebyshev function ψ_rm Ch(k), and its ratio to k tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write log L_k=∑_a≥1ϑ(k^1/a), where ϑ(x)=∑_p≤ xlog p. The first term is k+o(k). The remaining terms are O(sqrtklog k)=o(k), using ϑ(x)=O(x) and at most log_2 k nonzero terms. The symbol ψ_rm Ch her…
Exact mathematical passage
\log L_k=\sum_{p^a\leq k}\log p
\sum_{p\leq x}Introduction passage · 1 passage
Search aliases: ∑_p≤ x sum leq
Introduction passage · chapters08_10.tex, line 246
e standard asymptotic~[source reference] log L_k=k+o(k). Indeed log L_k=∑_p^a≤ klog p is the Chebyshev function ψ_rm Ch(k), and its ratio to k tends to one. Source contract LC10 records the prime number theorem and the elementary prime power conversion. Equivalently, write log L_k=∑_a≥1ϑ(k^1/a), where ϑ(x)=∑_p≤ xlog p. The first term is k+o(k). The remaining terms are O(sqrtklog k)=o(k), using ϑ(x)=O(x) and at most log_2 k nonzero terms. The symbol ψ_rm Ch here denotes the arithmetic function, n…
Exact mathematical passage
\vartheta(x)=\sum_{p\leq x}\log p
a_0Use in source · 9 passages
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Use in source · appendices.tex, line 236
The Euler interval uses a_k=(2k+1)^-2 and the positive difference convention Δ a_k=a_k-a_k+1. The identity Δ^ja_0=∫_0^1(1-x^2)^j(-log x) dd x gives 0≤Δ^ja_0≤1. The positive Euler series for G is ∑_j≥02^-j-1Δ^ja_0. Keeping the first 1001 terms therefore gives a lower endpoint and an upper endpoint obtained by adding 2^-1001. All retained terms are rational. The analytic tail bound is a written argument. The computed interval is its finite arithmetic application.
Exact mathematical passage
\Delta^ja_0=\int_0^1(1-x^2)^j(-\log x)\,\dd x
a_1Use in source · 2 passages
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Use in source · appendices.tex, line 130
At n=1, direct substitution in [source reference] gives p(1)=13, p(2)=65, q(1)=10699, a_1=1872, c_1=260. The recurrence therefore yields u_2 · =frac10699·(7/4)+2601872 =frac64964,
Exact mathematical passage
p(1)=13,\quad p(2)=65,\quad q(1)=10699, \quad a_1=1872,\quad c_1=260.
a_2Use in source · 1 passage
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Use in source · chapter03.tex, line 144
At n=2 the coefficients become p(2)=65, p(3)=157, q(2)=434871, a_2=58500, c_2=22608. Using the fractions just obtained, u_3 · =frac434871(649/64)+22608(7/4)58500
Exact mathematical passage
p(2)=65,\quad p(3)=157,\quad q(2)=434871,\qquad a_2=58500,\quad c_2=22608.
A_jUse in source · 9 passages
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Use in source · chapters01_03.tex, line 545
Suppose a cleared form is A_jα-B_j with A_j,B_j∈Z and (A_j,B_j)≠(0,0). Define its common content by h_j=gcd(absA_j,absB_j)>0. The primitive pair is (fracA_jh_j,fracB_jh_j), and the associated form has size fracabsA_jα-B_jh_j. Both coefficients remain integers. Thus a proved lower bound on common content may improve an irrationality argument. The word ``proved'' is essential when the improvement is used at infinitely many indices.
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A_j\alpha-B_j
a_jUse in source · 2 passages
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Use in source · chapters01_03.tex, line 534
Such a D_j always exists for one finite rational pair. Write u_j=a_j/b_j and v_j=c_j/d_j in lowest terms with positive denominators. The smallest positive common clearing factor is D_j^min=lcm(b_j,d_j). Indeed, integrality of D_ja_j/b_j forces b_jmid D_j by coprimality, and integrality of D_jc_j/d_j forces d_jmid D_j. Conversely, their least common multiple clears both denominators.
Exact mathematical passage
u_j=a_j/b_j
a_kUse in source · 4 passages
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Use in source · appendices.tex, line 235
The Euler interval uses a_k=(2k+1)^-2 and the positive difference convention Δ a_k=a_k-a_k+1. The identity Δ^ja_0=∫_0^1(1-x^2)^j(-log x) dd x gives 0≤Δ^ja_0≤1. The positive Euler series for G is ∑_j≥02^-j-1Δ^ja_0. Keeping the first 1001 terms therefore gives a lower endpoint and an upper endpoint obtained by adding 2^-1001. All retained terms are rational. The analytic tail bound is a written argument. The computed interval is its finite arithmetic application.
Exact mathematical passage
a_k=(2k+1)^{-2}
a_{k+1}Use in source · 2 passages
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Use in source · appendices.tex, line 235
The Euler interval uses a_k=(2k+1)^-2 and the positive difference convention Δ a_k=a_k-a_k+1. The identity Δ^ja_0=∫_0^1(1-x^2)^j(-log x) dd x gives 0≤Δ^ja_0≤1. The positive Euler series for G is ∑_j≥02^-j-1Δ^ja_0. Keeping the first 1001 terms therefore gives a lower endpoint and an upper endpoint obtained by adding 2^-1001. All retained terms are rational. The analytic tail bound is a written argument. The computed interval is its finite arithmetic application.
Exact mathematical passage
\Delta a_k=a_k-a_{k+1}
A_nUse in source · 1 passage
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Use in source · chapter03.tex, line 117
[tbp] centeringincludegraphics[width=0.93linewidth]../visual/figures/09_recurrence_structure.pdf captionTwo preceding values determine the next rational value. The forward rule begins at n=1 and uses the stated initial pair. Here and in subsequent recurrence diagrams, A_n=a_n, Q_n=q(n) and C_n=c_n.
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A_n=a_n
\absUse in source · 22 passages
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Use in source · chapters01_03.tex, line 188
For 0<δ<1, integration by parts gives ∫_δ^1fracarctan xx dd x =-arctan(δ)logδ -∫_δ^1fraclog x1+x^2 dd x. The boundary expression tends to zero as δdownarrow0, since 0≤arctanδ≤δ and δabslogδ→0. The first equality follows from Proposition~[source reference].
Exact mathematical passage
\delta\abs{\log\delta}\to0
\arcsinUse in source · 1 passage
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Use in source · chapters04_07.tex, line 428
The initial integral supplies a short diagnostic. Substitute x=t^2 and y=1-s^2. Then J_0 · =4∫_0^1 ∫_0^1fracdd s dd t1-t^2+t^2s^2 · =4∫_0^1fracarcsin ttsqrt1-t^2 dd t =4∫_0^π/2θcscθ ddθ. Since an antiderivative of cscθ is logtan(θ/2), integration by parts yields J_0=-4∫_0^π/2logtan(θ/2) ddθ =-8∫_0^π/4logtan z dd z=8G. Both boundary terms vanish. At zero this uses θlogtan(θ/2)→0, and at π/2 the logarithm is zero.
Exact mathematical passage
J_0 &=4\int_0^1\!\int_0^1\frac{\dd s\,\dd t}{1-t^2+t^2s^2}\\ &=4\int_0^1\frac{\arcsin t}{t\sqrt{1-t^2}}\,\dd t =4\int_0^{\pi/2}\theta\csc\theta\,\dd\theta.
\arctanUse in source · 5 passages
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Use in source · chapters01_03.tex, line 173
[Arctangent and tangent forms] The following equalities hold: G=∫_0^1fracarctan xx dd x =-∫_0^π/4log(tanθ) ddθ. The integrand arctan x/x is assigned its continuous value 1 at x=0.
Exact mathematical passage
G=\int_0^1\frac{\arctan x}{x}\,\dd x =-\int_0^{\pi/4}\log(\tan\theta)\,\dd\theta. \label{eq:c01-tangent-forms}
b^{-1}Use in source · 1 passage
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Use in source · chapters01_03.tex, line 376
For a fixed rational α, every integer form Qα-P lies in the discrete set b^-1Z. Its nonzero values cannot enter the open interval (-1/b,1/b). The denominator b need not be known to use this obstruction. If a constructed infinite family has nonzero values tending to zero, it eventually violates the gap for any proposed b.
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b^{-1}\Z
B_jUse in source · 7 passages
Search aliases: B_j
Use in source · chapters01_03.tex, line 545
Suppose a cleared form is A_jα-B_j with A_j,B_j∈Z and (A_j,B_j)≠(0,0). Define its common content by h_j=gcd(absA_j,absB_j)>0. The primitive pair is (fracA_jh_j,fracB_jh_j), and the associated form has size fracabsA_jα-B_jh_j. Both coefficients remain integers. Thus a proved lower bound on common content may improve an irrationality argument. The word ``proved'' is essential when the improvement is used at infinitely many indices.
Exact mathematical passage
A_j\alpha-B_j
b_jUse in source · 4 passages
Search aliases: b_j
Use in source · chapters01_03.tex, line 534
Such a D_j always exists for one finite rational pair. Write u_j=a_j/b_j and v_j=c_j/d_j in lowest terms with positive denominators. The smallest positive common clearing factor is D_j^min=lcm(b_j,d_j). Indeed, integrality of D_ja_j/b_j forces b_jmid D_j by coprimality, and integrality of D_jc_j/d_j forces d_jmid D_j. Conversely, their least common multiple clears both denominators.
Exact mathematical passage
u_j=a_j/b_j
c_1Use in source · 2 passages
Search aliases: c_1
Use in source · appendices.tex, line 130
At n=1, direct substitution in [source reference] gives p(1)=13, p(2)=65, q(1)=10699, a_1=1872, c_1=260. The recurrence therefore yields u_2 · =frac10699·(7/4)+2601872 =frac64964,
Exact mathematical passage
p(1)=13,\quad p(2)=65,\quad q(1)=10699, \quad a_1=1872,\quad c_1=260.
c_2Use in source · 1 passage
Search aliases: c_2
Use in source · chapter03.tex, line 144
At n=2 the coefficients become p(2)=65, p(3)=157, q(2)=434871, a_2=58500, c_2=22608. Using the fractions just obtained, u_3 · =frac434871(649/64)+22608(7/4)58500
Exact mathematical passage
p(2)=65,\quad p(3)=157,\quad q(2)=434871,\qquad a_2=58500,\quad c_2=22608.
c_AUse in source · 2 passages
Search aliases: c_A
Use in source · chapters08_10.tex, line 328
sion of the finite recurrence audit. Together with the filter bound it controls the approximation error |G-frac VU| =frac|widetilde J||widetilde U|. Before any additional quantified gcd saving, the reduced denominator q is at most the absolute cleared coefficient |A|. Its available exponent is c_A(α)=4(1+α)+log R+αlog(1+R). The source's resulting guaranteed approximation exponent is η(α)= fraclog R+αlog(1+R)-log16-log M(α) 4(1+α)+log R+αlog(1+R). The source reports a maximum near 0.62335 at α≈3.…
Exact mathematical passage
c_A(\alpha)=4(1+\alpha)+\log R+\alpha\log(1+R). \label{eq:cA-alpha}
c_jUse in source · 2 passages
Search aliases: c_j
Use in source · chapters01_03.tex, line 534
Such a D_j always exists for one finite rational pair. Write u_j=a_j/b_j and v_j=c_j/d_j in lowest terms with positive denominators. The smallest positive common clearing factor is D_j^min=lcm(b_j,d_j). Indeed, integrality of D_ja_j/b_j forces b_jmid D_j by coprimality, and integrality of D_jc_j/d_j forces d_jmid D_j. Conversely, their least common multiple clears both denominators.
Exact mathematical passage
v_j=c_j/d_j
C_nUse in source · 3 passages
Search aliases: C_n
Use in source · chapter03.tex, line 117
[tbp] centeringincludegraphics[width=0.93linewidth]../visual/figures/09_recurrence_structure.pdf captionTwo preceding values determine the next rational value. The forward rule begins at n=1 and uses the stated initial pair. Here and in subsequent recurrence diagrams, A_n=a_n, Q_n=q(n) and C_n=c_n.
Exact mathematical passage
C_n=c_n
C_{n,m}Use in source · 3 passages
Search aliases: C_n,m
Use in source · chapters04_07.tex, line 712
Let N=n+m and put C_n,m=L_2N^2. The pair produced by the historical script is A_n,m · =8(-1)^nC_n,mS_n,m,
Exact mathematical passage
\label{cb_eq_C_clear} C_{n,m}=L_{2N}^2.
\cscUse in source · 2 passages
Search aliases: csc
Use in source · chapters04_07.tex, line 428
The initial integral supplies a short diagnostic. Substitute x=t^2 and y=1-s^2. Then J_0 · =4∫_0^1 ∫_0^1fracdd s dd t1-t^2+t^2s^2 · =4∫_0^1fracarcsin ttsqrt1-t^2 dd t =4∫_0^π/2θcscθ ddθ. Since an antiderivative of cscθ is logtan(θ/2), integration by parts yields J_0=-4∫_0^π/2logtan(θ/2) ddθ =-8∫_0^π/4logtan z dd z=8G. Both boundary terms vanish. At zero this uses θlogtan(θ/2)→0, and at π/2 the logarithm is zero.
Exact mathematical passage
J_0 &=4\int_0^1\!\int_0^1\frac{\dd s\,\dd t}{1-t^2+t^2s^2}\\ &=4\int_0^1\frac{\arcsin t}{t\sqrt{1-t^2}}\,\dd t =4\int_0^{\pi/2}\theta\csc\theta\,\dd\theta.
\mathcal DUse in source · 4 passages
Search aliases: D mathcal
Use in source · chapters04_07.tex, line 802
For a positive integer N, the scanned domain is D_N={(n,m):1≤ n≤ N, 0≤ m≤4n}. The number of scanned pairs is ∑_n=1^N(4n+1)=2N^2+3N. The upper slope 4 is part of the experiment's definition. The scan supplies no information about pairs above this boundary or with n>N.
Exact mathematical passage
\label{cb_eq_scan_domain} \mathcal D_N=\{(n,m):1\leq n\leq N,\ 0\leq m\leq4n\}.
\mathcal D_{100}Use in source · 1 passage
Search aliases: D_100 mathcal
Use in source · chapters04_07.tex, line 870
The historical smaller crosscheck reports the same maximum from direct sums through N=100. Matching only two summary maxima would be a weak comparison: different entry values can share the same maximum. The release therefore compares every pair in D_100.
Exact mathematical passage
\mathcal D_{100}
\mathcal D_{2000}Use in source · 2 passages
Search aliases: D_2000 mathcal
Use in source · chapters04_07.tex, line 842
The historical run takes N=2000 and K=128. The release reran the unchanged historical algorithm and compared its result with the recovered output. The finite arithmetic results agree. In the domain D_2000 there are exactly 8,006,000 pairs. Every observed residue is nonzero, so the residue lemma applies at every pair. The exact maximum valuation in this finite domain is max_(n,m)∈ D_2000v_2(S_n,m)=31. The recorded maximizing pair is (1466,2113). The record reports the first maximum encountered in…
Exact mathematical passage
\mathcal D_{2000}
d_{2k}Use in source · 2 passages
Search aliases: d_2k
Use in source · chapters04_07.tex, line 167
Equation [source reference] and the preceding lemma identify r_n with partial sums of a convergent alternating series: r_n=∑_j=0^n-1(-1)^j d_j, L=∑_j=0^∞(-1)^j d_j. For the even subsequence, r_2k+2-r_2k=d_2k-d_2k+1>0. For the odd subsequence, r_2k+3-r_2k+1=-d_2k+1+d_2k+2<0. Their adjacent separation d_2k tends to zero, so they have the same limit. Strictness of both subsequences places that limit strictly between every displayed pair. The alternating tail beginning with d_n gives the sign and up…
Exact mathematical passage
r_{2k+2}-r_{2k}=d_{2k}-d_{2k+1}>0
d_{2k+1}Use in source · 2 passages
Search aliases: d_2k+1
Use in source · chapters04_07.tex, line 167
Equation [source reference] and the preceding lemma identify r_n with partial sums of a convergent alternating series: r_n=∑_j=0^n-1(-1)^j d_j, L=∑_j=0^∞(-1)^j d_j. For the even subsequence, r_2k+2-r_2k=d_2k-d_2k+1>0. For the odd subsequence, r_2k+3-r_2k+1=-d_2k+1+d_2k+2<0. Their adjacent separation d_2k tends to zero, so they have the same limit. Strictness of both subsequences places that limit strictly between every displayed pair. The alternating tail beginning with d_n gives the sign and up…
Exact mathematical passage
r_{2k+2}-r_{2k}=d_{2k}-d_{2k+1}>0
d_{2k+2}Use in source · 1 passage
Search aliases: d_2k+2
Use in source · chapters04_07.tex, line 169
Equation [source reference] and the preceding lemma identify r_n with partial sums of a convergent alternating series: r_n=∑_j=0^n-1(-1)^j d_j, L=∑_j=0^∞(-1)^j d_j. For the even subsequence, r_2k+2-r_2k=d_2k-d_2k+1>0. For the odd subsequence, r_2k+3-r_2k+1=-d_2k+1+d_2k+2<0. Their adjacent separation d_2k tends to zero, so they have the same limit. Strictness of both subsequences places that limit strictly between every displayed pair. The alternating tail beginning with d_n gives the sign and up…
Exact mathematical passage
r_{2k+3}-r_{2k+1}=-d_{2k+1}+d_{2k+2}<0
D_jUse in source · 21 passages
Search aliases: D_j
Use in source · chapters01_03.tex, line 527
In the later chapters the starting form is ℓ_j=u_jα-v_j, u_j,v_j∈Q. Choose a positive integer D_j for which D_ju_j and D_jv_j are integers. Then L_j=D_jℓ_j=(D_ju_j)α-D_jv_j has integer coefficients. Its size is D_jabsℓ_j, so the clearance factor must be included in any decay estimate.
Exact mathematical passage
D_j
d_jUse in source · 5 passages
Search aliases: d_j
Use in source · chapters01_03.tex, line 534
Such a D_j always exists for one finite rational pair. Write u_j=a_j/b_j and v_j=c_j/d_j in lowest terms with positive denominators. The smallest positive common clearing factor is D_j^min=lcm(b_j,d_j). Indeed, integrality of D_ja_j/b_j forces b_jmid D_j by coprimality, and integrality of D_jc_j/d_j forces d_jmid D_j. Conversely, their least common multiple clears both denominators.
Exact mathematical passage
v_j=c_j/d_j
D_j^{1/j}Use in source · 1 passage
Search aliases: D_j^1/j
Use in source · chapters01_03.tex, line 572
[Exponential clearance criterion] Let α∈R, and let u_j,v_j∈Q for j≥1. Let D_j be positive integers such that D_ju_j,D_jv_j∈Z. Assume u_jα-v_j≠0 for all sufficiently large j. Suppose limsup_j→∞absu_jα-v_j^1/j≤λ, limsup_j→∞D_j^1/j≤δ, where λ≥0, δ<∞ and δλ<1. Then α is irrational.
Exact mathematical passage
\limsup_{j\to\infty}\abs{u_j\alpha-v_j}^{1/j}\leq\lambda, \qquad \limsup_{j\to\infty}D_j^{1/j}\leq\delta, \label{eq:c02-root-bounds}
D_j^{\min}Use in source · 1 passage
Search aliases: D_j^min min
Use in source · chapters01_03.tex, line 535
Such a D_j always exists for one finite rational pair. Write u_j=a_j/b_j and v_j=c_j/d_j in lowest terms with positive denominators. The smallest positive common clearing factor is D_j^min=lcm(b_j,d_j). Indeed, integrality of D_ja_j/b_j forces b_jmid D_j by coprimality, and integrality of D_jc_j/d_j forces d_jmid D_j. Conversely, their least common multiple clears both denominators.
Exact mathematical passage
D_j^{\min}=\operatorname{lcm}(b_j,d_j). \label{eq:c02-minimal-clearance}
D_nUse in source · 3 passages
Search aliases: D_n
Use in source · chapters08_10.tex, line 379
There is also a formulation independent of the universal clearing choice. With D_n,h_n as in Chapter~[source reference], the primitive error is (-1)^n(D_n/h_n)(S_n,m(n)G-H_n,m(n)). An estimate directly on that expression would suffice. If denominator reduction yields a provably smaller exponential cost, the excess factor Q_n/D_n should simply be removed from the accounting. The irrationality criterion itself is unchanged.
Exact mathematical passage
D_n,h_n
\mathcal D_NUse in source · 1 passage
Search aliases: D_N mathcal
Use in source · chapters04_07.tex, line 802
For a positive integer N, the scanned domain is D_N={(n,m):1≤ n≤ N, 0≤ m≤4n}. The number of scanned pairs is ∑_n=1^N(4n+1)=2N^2+3N. The upper slope 4 is part of the experiment's definition. The scan supplies no information about pairs above this boundary or with n>N.
Exact mathematical passage
\label{cb_eq_scan_domain} \mathcal D_N=\{(n,m):1\leq n\leq N,\ 0\leq m\leq4n\}.
d_{n+1}Use in source · 1 passage
Search aliases: d_n+1
Use in source · chapters04_07.tex, line 153
[Alternating brackets] There is a real number L such that r_n→ L. For every k≥0, r_2k<L<r_2k+1. The even subsequence increases strictly and the odd subsequence decreases strictly. For every n≥0, d_n-d_n+1<|L-r_n|<d_n, sign(L-r_n)=(-1)^n.
Exact mathematical passage
\label{cb_eq_error_bounds} d_n-d_{n+1}<|L-r_n|<d_n, \qquad \operatorname{sign}(L-r_n)=(-1)^n.
d_{n-1}Use in source · 1 passage
Search aliases: d_n-1
Use in source · chapters04_07.tex, line 125
Positivity is immediate. For n≥1, the determinant recurrence gives fracd_nd_n-1 =fracc_na_nfracu_n-1u_n+1<1. The strict inequality follows from a_nu_n+1=q(n)u_n+c_nu_n-1>c_nu_n-1. Finally u_nu_n+1≥1, so 0<d_n≤ fracp(n+1)2(2n+1)^2(2n+2)^2⟶0. No limiting characteristic root is needed for this conclusion.
Exact mathematical passage
\frac{d_n}{d_{n-1}} =\frac{c_n}{a_n}\frac{u_{n-1}}{u_{n+1}}<1.
\ddUse in source · 25 passages
Search aliases: dd
Use in source · appendices.tex, line 71
widetilde J_n,m · =∑_i=0^m(-16)^ibinom miJ_n+i =16^-nF_n,m. The corrected unfiltered integral is J_n=∫_0^1 ∫_0^1 fracx^n-1/2(1-x)^ny^n(1-y)^n-1/2 (1-xy)^n+1 dd x dd y =8(-1)^n(u_nG-v_n). These are fixed normalization identities. Their factors should be checked together when translating a differently normalized source.
Exact mathematical passage
J_n=\int_0^1\!\int_0^1 \frac{x^{n-1/2}(1-x)^ny^n(1-y)^{n-1/2}} {(1-xy)^{n+1}}\,\dd x\,\dd y =8(-1)^n(u_nG-v_n). \label{eq:app-J}
\operatorname{den}Use in source · 1 passage
Search aliases: den operatorname
Use in source · chapters08_10.tex, line 128
The first row illustrates the exact check without printing its very large numerator and denominator. The independently reduced rational packet has denominator D=den(∑_i=0^66binom66i16^20+iv_20+i). Its recorded quotient Q/D is 1 703 936, and the minimal pair has gcd 8. Multiplication gives 1 703 936·8=13 631 488, exactly the historical value. The complete decimal integers U,N,D,Q are retained in the diagnostic JSON. They are inputs to an exact equality check, rather than values inferred from roun…
Exact mathematical passage
D=\operatorname{den}\left(\sum_{i=0}^{66}\binom{66}{i}16^{20+i}v_{20+i}\right).
\detUse in source · 4 passages
Search aliases: det
Use in source · chapter03.tex, line 188
The determinant is det M_n=-β_n=-fracc_na_n<0. Every transfer is invertible, with M_n^-1=0 · 1
Exact mathematical passage
\det M_n=-\beta_n=-\frac{c_n}{a_n}<0.
\downarrowUse in source · 1 passage
Search aliases: downarrow
Use in source · chapters01_03.tex, line 188
For 0<δ<1, integration by parts gives ∫_δ^1fracarctan xx dd x =-arctan(δ)logδ -∫_δ^1fraclog x1+x^2 dd x. The boundary expression tends to zero as δdownarrow0, since 0≤arctanδ≤δ and δabslogδ→0. The first equality follows from Proposition~[source reference].
Exact mathematical passage
\delta\downarrow0
e^{(-\tau+\eps)j}Use in source · 1 passage
Search aliases: e^(-τ+eps)j tau eps
Use in source · chapters01_03.tex, line 612
Choose eps>0 with σ-τ+2eps<0. Eventually D_j≤ e^(σ+eps)j and absu_jα-v_j≤ e^(-τ+eps)j. Their product tends to zero. Apply the integer form criterion.
Exact mathematical passage
\abs{u_j\alpha-v_j}\leq e^{(-\tau+\eps)j}
e^{(\sigma+\eps)j}Use in source · 1 passage
Search aliases: e^(σ+eps)j sigma eps
Use in source · chapters01_03.tex, line 612
Choose eps>0 with σ-τ+2eps<0. Eventually D_j≤ e^(σ+eps)j and absu_jα-v_j≤ e^(-τ+eps)j. Their product tends to zero. Apply the integer form criterion.
Exact mathematical passage
D_j\leq e^{(\sigma+\eps)j}
e^{-(a+1)t}Use in source · 1 passage
Search aliases: e^-(a+1)t
Use in source · chapters01_03.tex, line 138
Set x=e^-t. Then -log x=t, dd x=-e^-tdd t, and the integral becomes ∫_0^∞t e^-(a+1)t dd t. Integration by parts gives (a+1)^-2. The boundary term at infinity is zero because an exponential with positive decay rate dominates t. The boundary term at zero is also zero.
Exact mathematical passage
\int_0^{\infty}t e^{-(a+1)t}\,\dd t.
e^{-n}Use in source · 2 passages
Search aliases: e^-n
Use in source · chapters08_10.tex, line 389
There is a simple logical reason to resist extrapolation. Given any finite list of positive values x_1,…,x_N, it has a positive extension that tends to zero and another positive extension that tends to infinity. Keep the list fixed and set the tail to e^-n in the first case or e^n in the second. Similarly, any finite list of positive integer gcds can be extended by 1 forever or by integers with an arbitrarily prescribed positive exponential rate.
Exact mathematical passage
e^{-n}
e^{-t}Use in source · 2 passages
Search aliases: e^-t
Use in source · chapters01_03.tex, line 137
Set x=e^-t. Then -log x=t, dd x=-e^-tdd t, and the integral becomes ∫_0^∞t e^-(a+1)t dd t. Integration by parts gives (a+1)^-2. The boundary term at infinity is zero because an exponential with positive decay rate dominates t. The boundary term at zero is also zero.
Exact mathematical passage
x=e^{-t}
e^{cn}Use in source · 1 passage
Search aliases: e^cn
Use in source · chapters08_10.tex, line 393
A positive exponent in an upper bound has the same logical limitation. If x_n≤ e^cn with c>0, both x_n=e^-n and x_n=e^cn/2 satisfy the bound eventually. One sequence decays and the other grows. Thus the failure of the bound to force decay is a proved limitation of the estimate, not proof of divergence of the object it bounds.
Exact mathematical passage
x_n\leq e^{cn}
e^{cn/2}Use in source · 1 passage
Search aliases: e^cn/2
Use in source · chapters08_10.tex, line 393
A positive exponent in an upper bound has the same logical limitation. If x_n≤ e^cn with c>0, both x_n=e^-n and x_n=e^cn/2 satisfy the bound eventually. One sequence decays and the other grows. Thus the failure of the bound to force decay is a proved limitation of the estimate, not proof of divergence of the object it bounds.
Exact mathematical passage
x_n=e^{cn/2}
e^nUse in source · 1 passage
Search aliases: e^n
Use in source · chapters08_10.tex, line 389
There is a simple logical reason to resist extrapolation. Given any finite list of positive values x_1,…,x_N, it has a positive extension that tends to zero and another positive extension that tends to infinity. Keep the list fixed and set the tail to e^-n in the first case or e^n in the second. Similarly, any finite list of positive integer gcds can be extended by 1 forever or by integers with an arbitrarily prescribed positive exponential rate.
Exact mathematical passage
e^n
E_0Use in source · 1 passage
Search aliases: E_0
Use in source · chapters01_03.tex, line 321
[Small Euler intervals] The first difference values are Δ^0a_0=1, Δ a_0=frac89, Δ^2a_0=frac184225. They give E_0=frac12, E_1=frac1318, E_2=frac371450. Thus, for example, frac371450<G<frac17091800. These short transformed sums are useful for checking conventions. Their principal advantage appears as N grows, because the certified width decreases geometrically.
Exact mathematical passage
E_0=\frac12,\qquad E_1=\frac{13}{18},\qquad E_2=\frac{371}{450}.
E_1Use in source · 1 passage
Search aliases: E_1
Use in source · chapters01_03.tex, line 321
[Small Euler intervals] The first difference values are Δ^0a_0=1, Δ a_0=frac89, Δ^2a_0=frac184225. They give E_0=frac12, E_1=frac1318, E_2=frac371450. Thus, for example, frac371450<G<frac17091800. These short transformed sums are useful for checking conventions. Their principal advantage appears as N grows, because the certified width decreases geometrically.
Exact mathematical passage
E_0=\frac12,\qquad E_1=\frac{13}{18},\qquad E_2=\frac{371}{450}.
E_2Use in source · 1 passage
Search aliases: E_2
Use in source · chapters01_03.tex, line 321
[Small Euler intervals] The first difference values are Δ^0a_0=1, Δ a_0=frac89, Δ^2a_0=frac184225. They give E_0=frac12, E_1=frac1318, E_2=frac371450. Thus, for example, frac371450<G<frac17091800. These short transformed sums are useful for checking conventions. Their principal advantage appears as N grows, because the certified width decreases geometrically.
Exact mathematical passage
E_0=\frac12,\qquad E_1=\frac{13}{18},\qquad E_2=\frac{371}{450}.
E_NUse in source · 7 passages
Search aliases: E_N
Use in source · chapters01_03.tex, line 267
[Certified Euler enclosure] Define E_N=∑_j=0^NfracΔ^j a_02^j+1, N≥0. Then G=∑_j=0^∞fracΔ^j a_02^j+1, E_N<G<E_N+2^-N-1. More precisely, G-E_N =∫_0^1frac-log x1+x^2 (frac1-x^22)^N+1 dd x.
Exact mathematical passage
E_N=\sum_{j=0}^{N}\frac{\Delta^j a_0}{2^{j+1}},\qquad N\geq0. \label{eq:c01-E}
\epsUse in source · 7 passages
Search aliases: eps
Use in source · chapters01_03.tex, line 612
Choose eps>0 with σ-τ+2eps<0. Eventually D_j≤ e^(σ+eps)j and absu_jα-v_j≤ e^(-τ+eps)j. Their product tends to zero. Apply the integer form criterion.
Exact mathematical passage
\eps>0
\expUse in source · 5 passages
Search aliases: exp
Use in source · chapters01_03.tex, line 632
The primitive form D_j(u_jα-v_j)/h_j has integer coefficients and remains nonzero. Choose eps>0 so that σ-τ-κ+3eps<0. The upper limits for D_j and the error, and the lower limit for h_j, give eventually fracD_jabsu_jα-v_jh_j ≤exp((σ-τ-κ+3eps)j). The right side tends to zero. Apply Theorem~[source reference] to the primitive pair.
Exact mathematical passage
\frac{D_j\abs{u_j\alpha-v_j}}{h_j} \leq\exp\bigl((\sigma-\tau-\kappa+3\eps)j\bigr).
\mathcal FUse in source · 4 passages
Search aliases: F mathcal
Use in source · chapter03.tex, line 201
The two distinguished solutions can be collected into the fundamental matrix F_n= u_n · v_n u_n-1 · v_n-1, n≥1. It satisfies F_n+1=M_n F_n. Its initial determinant is det F_1= det7/4 · 13/8
Exact mathematical passage
\mathcal F_n= \begin{pmatrix}u_n&v_n\\u_{n-1}&v_{n-1}\end{pmatrix},\qquad n\geq1.
\mathcal F_1Use in source · 1 passage
Search aliases: F_1 mathcal
Use in source · chapter03.tex, line 206
u_n-1 · v_n-1, n≥1. It satisfies F_n+1=M_n F_n. Its initial determinant is det F_1= det7/4 · 13/8 1 · 0=-frac138. Multiplicativity of determinants now gives det F_N+1 =-frac138∏_n=1^N(-β_n)≠0. Thus adjacent values of the two solutions remain independent at every index. Each transfer reverses the determinant sign. The closed evaluation of this product is the Wronskian theorem in Chapter~[source reference], where the factors telescope and yield exact quotient increments. Here the matrix descriptio…
Exact mathematical passage
\det\mathcal F_1= \det\begin{pmatrix}7/4&13/8\\1&0\end{pmatrix}=-\frac{13}{8}.
F_{1,1}Use in source · 5 passages
Search aliases: F_1,1
Use in source · appendices.tex, line 175
The associated rational packet is H_1,1=16v_1+16^2v_2 =26+frac213989 =frac216329. Therefore the raw form is F_1,1=-8(2624G-frac216329) =frac1730569-20992G. A certified interval for G makes this value positive. This is the smallest counterexample in the finite audit to the proposed sign rule signF_n,m=(-1)^m.
Exact mathematical passage
F_{1,1}=-8\left(2624G-\frac{21632}{9}\right) =\frac{173056}{9}-20992G.
F_2Use in source · 1 passage
Search aliases: F_2
Use in source · chapters04_07.tex, line 460
Expanding (1-xy)^-n-1 as its nonnegative binomial series and applying Tonelli's theorem gives J_n=C_nH_n, where C_n=fracΓ(n+1)^2Γ(n+tfrac12)^2 Γ(2n+tfrac32)^2, and H_n=_3F_2 ( n+1,n+tfrac12,n+1 2n+tfrac32,2n+tfrac32;1). Indeed, the two beta integrals in the term with index k are B(n+tfrac12+k,n+1) and B(n+1+k,n+tfrac12). Their ratios to the corresponding k=0 integrals give exactly the displayed hypergeometric parameters. The excess of the lower parameters over the upper parameters is n+1/2>0, so…
Exact mathematical passage
H_n={}_3F_2\!\left( \begin{matrix}n+1,n+\tfrac12,n+1\\ 2n+\tfrac32,2n+\tfrac32\end{matrix};1\right).
\mathcal F_nUse in source · 2 passages
Search aliases: F_n mathcal
Use in source · chapter03.tex, line 201
The two distinguished solutions can be collected into the fundamental matrix F_n= u_n · v_n u_n-1 · v_n-1, n≥1. It satisfies F_n+1=M_n F_n. Its initial determinant is det F_1= det7/4 · 13/8
Exact mathematical passage
\mathcal F_n= \begin{pmatrix}u_n&v_n\\u_{n-1}&v_{n-1}\end{pmatrix},\qquad n\geq1.
\mathcal F_{N+1}Use in source · 1 passage
Search aliases: F_N+1 mathcal
Use in source · chapter03.tex, line 211
1 · 0=-frac138. Multiplicativity of determinants now gives det F_N+1 =-frac138∏_n=1^N(-β_n)≠0. Thus adjacent values of the two solutions remain independent at every index. Each transfer reverses the determinant sign. The closed evaluation of this product is the Wronskian theorem in Chapter~[source reference], where the factors telescope and yield exact quotient increments. Here the matrix description identifies the mechanism without duplicating that full proof.
Exact mathematical passage
\det\mathcal F_{N+1} =-\frac{13}{8}\prod_{n=1}^N(-\beta_n)\ne0.
\mathcal F_{n+1}Use in source · 1 passage
Search aliases: F_n+1 mathcal
Use in source · chapter03.tex, line 205
u_n-1 · v_n-1, n≥1. It satisfies F_n+1=M_n F_n. Its initial determinant is det F_1= det7/4 · 13/8
Exact mathematical passage
\mathcal F_{n+1}=M_n\mathcal F_n
F_{n,m}Use in source · 10 passages
Search aliases: F_n,m
Use in source · appendices.tex, line 63
For n,m≥0, the integer and rational packets are S_n,m=∑_i=0^mbinom miT_n+i, H_n,m=∑_i=0^mbinom mi16^n+iv_n+i. Their raw linear form and integral filter satisfy F_n,m · =8(-1)^n(S_n,mG-H_n,m),notag widetilde J_n,m · =∑_i=0^m(-16)^ibinom miJ_n+i =16^-nF_n,m. The corrected unfiltered integral is J_n=∫_0^1 ∫_0^1 fracx^n-1/2(1-x)^ny^n(1-y)^n-1/2 (1-xy)^n+1 dd x dd y =8(-1)^n(u_nG-v_n). These are fixed normalization identities. Their factors should be checked together when translating a differently no…
Exact mathematical passage
F_{n,m}&=8(-1)^n(S_{n,m}G-H_{n,m}),\notag\\ \widetilde J_{n,m} &=\sum_{i=0}^m(-16)^i\binom miJ_{n+i} =16^{-n}F_{n,m}. \label{eq:app-filter}
g_nUse in source · 3 passages
Search aliases: g_n
Use in source · chapters08_10.tex, line 261
With the cleared gcd g_n=g_n,m(n), the primitive form is exactly Λ_n=fracK_ng_nwidetilde J_n,m(n). Suppose a future arithmetic argument establishes δ=liminf_n→∞fraclog g_nn. For finite δ, the present analytic estimate yields limsup_n→∞fraclog|Λ_n|n ≤κ_0(α)+log M(α)-δ. The same interpretation applies when δ=+∞: any sufficiently large finite lower bound for the content rate would suffice. All logarithms of nonzero forms are legitimate for the e
Exact mathematical passage
g_n=g_{n,m(n)}
g_{n,m(n)}Use in source · 3 passages
Search aliases: g_n,m(n)
Use in source · chapters08_10.tex, line 261
With the cleared gcd g_n=g_n,m(n), the primitive form is exactly Λ_n=fracK_ng_nwidetilde J_n,m(n). Suppose a future arithmetic argument establishes δ=liminf_n→∞fraclog g_nn. For finite δ, the present analytic estimate yields limsup_n→∞fraclog|Λ_n|n ≤κ_0(α)+log M(α)-δ. The same interpretation applies when δ=+∞: any sufficiently large finite lower bound for the content rate would suffice. All logarithms of nonzero forms are legitimate for the e
Exact mathematical passage
g_n=g_{n,m(n)}
H_{1,1}Use in source · 3 passages
Search aliases: H_1,1
Use in source · appendices.tex, line 169
The associated rational packet is H_1,1=16v_1+16^2v_2 =26+frac213989 =frac216329. Therefore the raw form is F_1,1=-8(2624G-frac216329) =frac1730569-20992G. A certified interval for G makes this value positive. This is the smallest counterexample in the finite audit to the proposed sign rule signF_n,m=(-1)^m.
Exact mathematical passage
H_{1,1}=16v_1+16^2v_2 =26+\frac{21398}{9} =\frac{21632}{9}.
h_jUse in source · 9 passages
Search aliases: h_j
Use in source · chapters01_03.tex, line 546
Suppose a cleared form is A_jα-B_j with A_j,B_j∈Z and (A_j,B_j)≠(0,0). Define its common content by h_j=gcd(absA_j,absB_j)>0. The primitive pair is (fracA_jh_j,fracB_jh_j), and the associated form has size fracabsA_jα-B_jh_j. Both coefficients remain integers. Thus a proved lower bound on common content may improve an irrationality argument. The word ``proved'' is essential when the improvement is used at infinitely many indices.
Exact mathematical passage
h_j=\gcd(\abs{A_j},\abs{B_j})>0. \label{eq:c02-content}
h_nUse in source · 2 passages
Search aliases: h_n
Use in source · chapters08_10.tex, line 379
There is also a formulation independent of the universal clearing choice. With D_n,h_n as in Chapter~[source reference], the primitive error is (-1)^n(D_n/h_n)(S_n,m(n)G-H_n,m(n)). An estimate directly on that expression would suffice. If denominator reduction yields a provably smaller exponential cost, the excess factor Q_n/D_n should simply be removed from the accounting. The irrationality criterion itself is unchanged.
Exact mathematical passage
D_n,h_n
H_{n,m(n)}Use in source · 1 passage
Search aliases: H_n,m(n)
Use in source · chapters08_10.tex, line 379
There is also a formulation independent of the universal clearing choice. With D_n,h_n as in Chapter~[source reference], the primitive error is (-1)^n(D_n/h_n)(S_n,m(n)G-H_n,m(n)). An estimate directly on that expression would suffice. If denominator reduction yields a provably smaller exponential cost, the excess factor Q_n/D_n should simply be removed from the accounting. The irrationality criterion itself is unchanged.
Exact mathematical passage
(-1)^n(D_n/h_n)(S_{n,m(n)}G-H_{n,m(n)})
\hboxUse in source · 1 passage
Search aliases: hbox
Use in source · chapters01_03.tex, line 622
t can recover a proved rate] Assume the setup of Corollary~[source reference], except that τ>σ is not required. Discard the finite initial set of indices before nonvanishing holds. For the remaining sufficiently large j, define h_j=gcd(absD_ju_j,absD_jv_j), which is positive because the form is nonzero at those indices. If liminf_j→∞fraclog h_jj≥κ hboxand τ+κ>σ, then α is irrational.
Exact mathematical passage
\liminf_{j\to\infty}\frac{\log h_j}{j}\geq\kappa \quad\hbox{and}\quad \tau+\kappa>\sigma, \label{eq:c02-content-rate}
JUse in source · 21 passages
Search aliases: J
Use in source · appendices.tex, line 63
For n,m≥0, the integer and rational packets are S_n,m=∑_i=0^mbinom miT_n+i, H_n,m=∑_i=0^mbinom mi16^n+iv_n+i. Their raw linear form and integral filter satisfy F_n,m · =8(-1)^n(S_n,mG-H_n,m),notag widetilde J_n,m · =∑_i=0^m(-16)^ibinom miJ_n+i =16^-nF_n,m. The corrected unfiltered integral is J_n=∫_0^1 ∫_0^1 fracx^n-1/2(1-x)^ny^n(1-y)^n-1/2 (1-xy)^n+1 dd x dd y =8(-1)^n(u_nG-v_n). These are fixed normalization identities. Their factors should be checked together when translating a differently no…
Exact mathematical passage
F_{n,m}&=8(-1)^n(S_{n,m}G-H_{n,m}),\notag\\ \widetilde J_{n,m} &=\sum_{i=0}^m(-16)^i\binom miJ_{n+i} =16^{-n}F_{n,m}. \label{eq:app-filter}
j^{-s}Use in source · 1 passage
Search aliases: j^-s
Use in source · chapters01_03.tex, line 53
For real s>1, the Dirichlet beta value is β(s)=∑_k=0^∞frac(-1)^k(2k+1)^s. Absolute convergence follows from comparison with ∑ j^-s. Thus G=β(2). The alternating series also defines β(s) for s>0, though absolute convergence is no longer available throughout that wider range. At s=1, integration of the finite geometric expansion of (1+x^2)^-1 gives β(1)=π/4. The integrated remainder is bounded by ∫_0^1x^2Ndd x=(2N+1)^-1 and tends to zero, so this passage does not require absolute convergence of th…
Exact mathematical passage
\sum j^{-s}
J_0Use in source · 8 passages
Search aliases: J_0
Use in source · appendices.tex, line 208
For comparison, (n,m)=(1,0) has S=28,H=26,D=1,h=2,Q=32,g=64. Its primitive form is 13-14G=J_1. At (0,0), the convention H=0,D=1 gives S=1,Q=8,g=8 and primitive form G=J_0/8. These boundary cases require no valuation of the integer zero.
Exact mathematical passage
G=J_0/8
J_1Use in source · 1 passage
Search aliases: J_1
Use in source · appendices.tex, line 208
For comparison, (n,m)=(1,0) has S=28,H=26,D=1,h=2,Q=32,g=64. Its primitive form is 13-14G=J_1. At (0,0), the convention H=0,D=1 gives S=1,Q=8,g=8 and primitive form G=J_0/8. These boundary cases require no valuation of the integer zero.
Exact mathematical passage
13-14G=J_1
J_kUse in source · 2 passages
Search aliases: J_k
Use in source · chapters04_07.tex, line 589
Substitute the corrected normalization J_k=8(-1)^k(u_kG-v_k) into [source reference]. The two factors (-1)^i cancel. Thus widetilde J_n,m=8(-1)^n16^-n(S_n,mG-H_n,m). Define the raw packet form F_n,m=16^nwidetilde J_n,m =8(-1)^n(S_n,mG-H_n,m). The adjective raw means that the common denominator of H_n,m has not yet been cleared. It does not mean that F_n,m is an integer linear form. The coefficient of G is an integer, but the other coefficient may be rational.
Exact mathematical passage
J_k=8(-1)^k(u_kG-v_k)
J_nUse in source · 8 passages
Search aliases: J_n
Use in source · appendices.tex, line 71
widetilde J_n,m · =∑_i=0^m(-16)^ibinom miJ_n+i =16^-nF_n,m. The corrected unfiltered integral is J_n=∫_0^1 ∫_0^1 fracx^n-1/2(1-x)^ny^n(1-y)^n-1/2 (1-xy)^n+1 dd x dd y =8(-1)^n(u_nG-v_n). These are fixed normalization identities. Their factors should be checked together when translating a differently normalized source.
Exact mathematical passage
J_n=\int_0^1\!\int_0^1 \frac{x^{n-1/2}(1-x)^ny^n(1-y)^{n-1/2}} {(1-xy)^{n+1}}\,\dd x\,\dd y =8(-1)^n(u_nG-v_n). \label{eq:app-J}
J_{n+i}Use in source · 2 passages
Search aliases: J_n+i
Use in source · appendices.tex, line 63
For n,m≥0, the integer and rational packets are S_n,m=∑_i=0^mbinom miT_n+i, H_n,m=∑_i=0^mbinom mi16^n+iv_n+i. Their raw linear form and integral filter satisfy F_n,m · =8(-1)^n(S_n,mG-H_n,m),notag widetilde J_n,m · =∑_i=0^m(-16)^ibinom miJ_n+i =16^-nF_n,m. The corrected unfiltered integral is J_n=∫_0^1 ∫_0^1 fracx^n-1/2(1-x)^ny^n(1-y)^n-1/2 (1-xy)^n+1 dd x dd y =8(-1)^n(u_nG-v_n). These are fixed normalization identities. Their factors should be checked together when translating a differently no…
Exact mathematical passage
F_{n,m}&=8(-1)^n(S_{n,m}G-H_{n,m}),\notag\\ \widetilde J_{n,m} &=\sum_{i=0}^m(-16)^i\binom miJ_{n+i} =16^{-n}F_{n,m}. \label{eq:app-filter}
KUse in source · 24 passages
Search aliases: K
Use in source · chapter03.tex, line 225
The transfer product can also be written using continuants. For integers 1≤ r≤ s, let K_r,s be the determinant of the tridiagonal matrix whose diagonal is α_r,…,α_s, lower adjacent diagonal is one, and upper adjacent diagonal is -β_r+1,…,-β_s. For instance, K_r,r=α_r, K_r,r+1=det α_r · -β_r+1
Exact mathematical passage
K_{r,s}
K_{1,n-1}Use in source · 1 passage
Search aliases: K_1,n-1
Use in source · chapter03.tex, line 246
[Initial data through continuants] For any solution w and every n≥2, w_n=K_1,n-1w_1+β_1K_2,n-1w_0.
Exact mathematical passage
\label{cb_eq_continuant_solution} w_n=K_{1,n-1}w_1+\beta_1K_{2,n-1}w_0.
K_{2,1}Use in source · 1 passage
Search aliases: K_2,1
Use in source · chapter03.tex, line 253
At n=2, the formula is w_2=α_1w_1+β_1w_0, using the empty determinant for K_2,1. At n=3, direct iteration gives w_3=(α_2α_1+β_2)w_1+α_2β_1w_0, which agrees with the formula. For later indices, both coefficients in [source reference] satisfy the same recurrence by [source reference]. Induction finishes the proof.
Exact mathematical passage
K_{2,1}
K_{2,n-1}Use in source · 1 passage
Search aliases: K_2,n-1
Use in source · chapter03.tex, line 246
[Initial data through continuants] For any solution w and every n≥2, w_n=K_1,n-1w_1+β_1K_2,n-1w_0.
Exact mathematical passage
\label{cb_eq_continuant_solution} w_n=K_{1,n-1}w_1+\beta_1K_{2,n-1}w_0.
K_nUse in source · 4 passages
Search aliases: K_n
Use in source · chapters08_10.tex, line 249
Set K_n=16^nL_2(n+m(n))^2. For m(n)/n→α, [source reference] gives fraclog K_nn⟶ κ_0(α)=log16+4(1+α). The factor 4 has a precise origin. The lcm index is 2(n+m), and the lcm is squared. Missing either factor would create an incorrect apparent decay margin. The subscript in κ_0 emphasizes that this is the clearing cost alone. Define the combined bound exponent by E(α)=κ_0(α)+log M(α).
Exact mathematical passage
K_n=16^nL_{2(n+m(n))}^2. \label{eq:K-asymptotic}
K_{r,r}Use in source · 1 passage
Search aliases: K_r,r
Use in source · chapter03.tex, line 229
The transfer product can also be written using continuants. For integers 1≤ r≤ s, let K_r,s be the determinant of the tridiagonal matrix whose diagonal is α_r,…,α_s, lower adjacent diagonal is one, and upper adjacent diagonal is -β_r+1,…,-β_s. For instance, K_r,r=α_r, K_r,r+1=det α_r · -β_r+1 1 · α_r+1 =α_rα_r+1+β_r+1. Set K_r,r-1=1 for the empty determinant. Expansion at the last row gives K_r,s=α_sK_r,s-1+β_sK_r,s-2 (s≥ r+1). The plus sign comes from the minus sign on the upper adjacent diagon…
Exact mathematical passage
K_{r,r}=\alpha_r,\qquad K_{r,r+1}=\det \begin{pmatrix}\alpha_r&-\beta_{r+1}\\1&\alpha_{r+1}\end{pmatrix} =\alpha_r\alpha_{r+1}+\beta_{r+1}.
K_{r,r+1}Use in source · 1 passage
Search aliases: K_r,r+1
Use in source · chapter03.tex, line 229
The transfer product can also be written using continuants. For integers 1≤ r≤ s, let K_r,s be the determinant of the tridiagonal matrix whose diagonal is α_r,…,α_s, lower adjacent diagonal is one, and upper adjacent diagonal is -β_r+1,…,-β_s. For instance, K_r,r=α_r, K_r,r+1=det α_r · -β_r+1 1 · α_r+1 =α_rα_r+1+β_r+1. Set K_r,r-1=1 for the empty determinant. Expansion at the last row gives K_r,s=α_sK_r,s-1+β_sK_r,s-2 (s≥ r+1). The plus sign comes from the minus sign on the upper adjacent diagon…
Exact mathematical passage
K_{r,r}=\alpha_r,\qquad K_{r,r+1}=\det \begin{pmatrix}\alpha_r&-\beta_{r+1}\\1&\alpha_{r+1}\end{pmatrix} =\alpha_r\alpha_{r+1}+\beta_{r+1}.
K_{r,r-1}Use in source · 1 passage
Search aliases: K_r,r-1
Use in source · chapter03.tex, line 235
1 · α_r+1 =α_rα_r+1+β_r+1. Set K_r,r-1=1 for the empty determinant. Expansion at the last row gives K_r,s=α_sK_r,s-1+β_sK_r,s-2 (s≥ r+1). The plus sign comes from the minus sign on the upper adjacent diagonal. This convention matches the positive trailing term in the recurrence. All these continuants are positive rationals.
Exact mathematical passage
K_{r,r-1}=1
K_{r,s}Use in source · 2 passages
Search aliases: K_r,s
Use in source · chapter03.tex, line 225
The transfer product can also be written using continuants. For integers 1≤ r≤ s, let K_r,s be the determinant of the tridiagonal matrix whose diagonal is α_r,…,α_s, lower adjacent diagonal is one, and upper adjacent diagonal is -β_r+1,…,-β_s. For instance, K_r,r=α_r, K_r,r+1=det α_r · -β_r+1
Exact mathematical passage
K_{r,s}
K_{r,s-1}Use in source · 1 passage
Search aliases: K_r,s-1
Use in source · chapter03.tex, line 236
1 · α_r+1 =α_rα_r+1+β_r+1. Set K_r,r-1=1 for the empty determinant. Expansion at the last row gives K_r,s=α_sK_r,s-1+β_sK_r,s-2 (s≥ r+1). The plus sign comes from the minus sign on the upper adjacent diagonal. This convention matches the positive trailing term in the recurrence. All these continuants are positive rationals.
Exact mathematical passage
\label{cb_eq_continuant} K_{r,s}=\alpha_sK_{r,s-1}+\beta_sK_{r,s-2} \qquad(s\geq r+1).
K_{r,s-2}Use in source · 1 passage
Search aliases: K_r,s-2
Use in source · chapter03.tex, line 236
1 · α_r+1 =α_rα_r+1+β_r+1. Set K_r,r-1=1 for the empty determinant. Expansion at the last row gives K_r,s=α_sK_r,s-1+β_sK_r,s-2 (s≥ r+1). The plus sign comes from the minus sign on the upper adjacent diagonal. This convention matches the positive trailing term in the recurrence. All these continuants are positive rationals.
Exact mathematical passage
\label{cb_eq_continuant} K_{r,s}=\alpha_sK_{r,s-1}+\beta_sK_{r,s-2} \qquad(s\geq r+1).
lUse in source · 2 passages
Search aliases: l
Use in source · chapters01_03.tex, line 336
Suppose l≤ G≤ h is a proven rational enclosure, and suppose a,b∈Q. Then the interval for aG-b is [al-b,ah-b], · a≥0,
Exact mathematical passage
l\leq G\leq h
L_{2(n+m(n))}^2Use in source · 1 passage
Search aliases: L_2(n+m(n))^2
Use in source · chapters08_10.tex, line 249
Set K_n=16^nL_2(n+m(n))^2. For m(n)/n→α, [source reference] gives fraclog K_nn⟶ κ_0(α)=log16+4(1+α). The factor 4 has a precise origin. The lcm index is 2(n+m), and the lcm is squared. Missing either factor would create an incorrect apparent decay margin. The subscript in κ_0 emphasizes that this is the clearing cost alone. Define the combined bound exponent by E(α)=κ_0(α)+log M(α).
Exact mathematical passage
K_n=16^nL_{2(n+m(n))}^2. \label{eq:K-asymptotic}
L_{2k}Use in source · 2 passages
Search aliases: L_2k
Use in source · chapters04_07.tex, line 722
B_n,m · =8(-1)^nC_n,mH_n,m. Both coefficients are integers. Integrality of A follows from that of S. For B, each index k=n+i satisfies k≤ N, so L_2k divides L_2N. The sharp source clearing 4·16^kL_2k^2v_k∈Z therefore proves integrality term by term after multiplication by 8L_2N^2. The binomial weights introduce no further denominators.
Exact mathematical passage
L_{2k}
L_{2k}^2Use in source · 2 passages
Search aliases: L_2k^2
Use in source · chapters04_07.tex, line 724
B_n,m · =8(-1)^nC_n,mH_n,m. Both coefficients are integers. Integrality of A follows from that of S. For B, each index k=n+i satisfies k≤ N, so L_2k divides L_2N. The sharp source clearing 4·16^kL_2k^2v_k∈Z therefore proves integrality term by term after multiplication by 8L_2N^2. The binomial weights introduce no further denominators.
Exact mathematical passage
4\cdot16^kL_{2k}^2v_k\in\Z
L_{2N}Use in source · 1 passage
Search aliases: L_2N
Use in source · chapters04_07.tex, line 723
B_n,m · =8(-1)^nC_n,mH_n,m. Both coefficients are integers. Integrality of A follows from that of S. For B, each index k=n+i satisfies k≤ N, so L_2k divides L_2N. The sharp source clearing 4·16^kL_2k^2v_k∈Z therefore proves integrality term by term after multiplication by 8L_2N^2. The binomial weights introduce no further denominators.
Exact mathematical passage
L_{2N}
L_{2N}^2Use in source · 4 passages
Search aliases: L_2N^2
Use in source · chapters04_07.tex, line 712
Let N=n+m and put C_n,m=L_2N^2. The pair produced by the historical script is A_n,m · =8(-1)^nC_n,mS_n,m,
Exact mathematical passage
\label{cb_eq_C_clear} C_{n,m}=L_{2N}^2.
L_4Use in source · 1 passage
Search aliases: L_4
Use in source · appendices.tex, line 192
The universal factor uses L_4=12, hence Q=8L_4^2=1152. Its redundant scale is Q/D=128. Consequently g=128·64=8192. The signed cleared pair is (-3 022 848,-2 768 896). Dividing by 8192 gives (-369,-338), so the primitive form is Λ_1,1=338-369G. The integral identity gives the same expression. Since widetilde J_1,1=F_1,1/16 and 16L_4^2/g=9/32, frac16L_4^2gwidetilde J_1,1 =frac9512F_1,1=338-369G. This checks the factor 8, the sign (-1)^n, and the denominator index in one exact example.
Exact mathematical passage
L_4=12
L_4^2Use in source · 3 passages
Search aliases: L_4^2
Use in source · appendices.tex, line 192
The universal factor uses L_4=12, hence Q=8L_4^2=1152. Its redundant scale is Q/D=128. Consequently g=128·64=8192. The signed cleared pair is (-3 022 848,-2 768 896). Dividing by 8192 gives (-369,-338), so the primitive form is Λ_1,1=338-369G. The integral identity gives the same expression. Since widetilde J_1,1=F_1,1/16 and 16L_4^2/g=9/32, frac16L_4^2gwidetilde J_1,1 =frac9512F_1,1=338-369G. This checks the factor 8, the sign (-1)^n, and the denominator index in one exact example.
Exact mathematical passage
Q=8L_4^2=1152
L_jUse in source · 1 passage
Search aliases: L_j
Use in source · chapters01_03.tex, line 528
In the later chapters the starting form is ℓ_j=u_jα-v_j, u_j,v_j∈Q. Choose a positive integer D_j for which D_ju_j and D_jv_j are integers. Then L_j=D_jℓ_j=(D_ju_j)α-D_jv_j has integer coefficients. Its size is D_jabsℓ_j, so the clearance factor must be included in any decay estimate.
Exact mathematical passage
L_j=D_j\ell_j=(D_ju_j)\alpha-D_jv_j \label{eq:c02-cleared-form}
L_\ellUse in source · 2 passages
Search aliases: L_ℓ ell
Use in source · chapters08_10.tex, line 15
ndequation These are fixed-pair aliases U=S_n,m and V=H_n,m for the packets defined earlier, not the integral coefficients of a different source normalization. Here T_k=16^ku_k is a positive integer. Consequently U is a positive integer. Write V=N/D in lowest terms, with D>0 and gcd(N,D)=1. In the exceptional case V=0, the same convention sets N=0,D=1. Set ℓ=2(n+m), L_ℓ=lcm(1,…,ℓ), Q=8L_ℓ^2, with L_0=1.
Exact mathematical passage
\ell=2(n+m),\qquad L_\ell=\operatorname{lcm}(1,\ldots,\ell), \qquad Q=8L_\ell^2, \label{eq:Q-clearance}
L_\ell^2Use in source · 2 passages
Search aliases: L_ℓ^2 ell
Use in source · chapters08_10.tex, line 15
ndequation These are fixed-pair aliases U=S_n,m and V=H_n,m for the packets defined earlier, not the integral coefficients of a different source normalization. Here T_k=16^ku_k is a positive integer. Consequently U is a positive integer. Write V=N/D in lowest terms, with D>0 and gcd(N,D)=1. In the exceptional case V=0, the same convention sets N=0,D=1. Set ℓ=2(n+m), L_ℓ=lcm(1,…,ℓ), Q=8L_ℓ^2, with L_0=1.
Exact mathematical passage
\ell=2(n+m),\qquad L_\ell=\operatorname{lcm}(1,\ldots,\ell), \qquad Q=8L_\ell^2, \label{eq:Q-clearance}
\lceilUse in source · 1 passage
Search aliases: lceil
Use in source · chapters04_07.tex, line 821
For a fixed row m, the allowed indices are max(1,lceil m/4rceil)≤ n≤ N. Rows are needed through m=4N. In row m, the seed array is sufficient for every n≤5N-m. At the extreme row m=4N this bound is exactly n≤ N. Thus no scanned entry needs an unconstructed seed term. The algorithm evaluates additional row entries to support later rows. Those additions do not enlarge the declared set of scanned pairs.
Exact mathematical passage
\max(1,\lceil m/4\rceil)\leq n\leq N.
\limUse in source · 4 passages
Search aliases: lim
Use in source · chapter03.tex, line 321
[External recurrence identification and rates] For the exact normalization [source reference], Zudilin's theorem and its stated asymptotic rates give lim_n→∞fracv_nu_n · =G, lim_n→∞u_n^1/n · =ϕ^5, lim_n→∞|u_nG-v_n|^1/n · =ϕ^-5. These are external mathematical inputs, recorded precisely in source contracts LC01 and LC02 [source reference].
Exact mathematical passage
\lim_{n\to\infty}\frac{v_n}{u_n}&=G,\label{cb_eq_source_limit}\\ \lim_{n\to\infty}u_n^{1/n}&=\phi^5,\label{cb_eq_source_u_rate}\\ \lim_{n\to\infty}|u_nG-v_n|^{1/n}&=\phi^{-5}. \label{cb_eq_source_error_rate}
\lim_{n\to\infty}Use in source · 4 passages
Search aliases: lim_n→∞ lim to infty
Use in source · chapter03.tex, line 321
[External recurrence identification and rates] For the exact normalization [source reference], Zudilin's theorem and its stated asymptotic rates give lim_n→∞fracv_nu_n · =G, lim_n→∞u_n^1/n · =ϕ^5, lim_n→∞|u_nG-v_n|^1/n · =ϕ^-5. These are external mathematical inputs, recorded precisely in source contracts LC01 and LC02 [source reference].
Exact mathematical passage
\lim_{n\to\infty}\frac{v_n}{u_n}&=G,\label{cb_eq_source_limit}\\ \lim_{n\to\infty}u_n^{1/n}&=\phi^5,\label{cb_eq_source_u_rate}\\ \lim_{n\to\infty}|u_nG-v_n|^{1/n}&=\phi^{-5}. \label{cb_eq_source_error_rate}
\liminfUse in source · 3 passages
Search aliases: liminf
Use in source · chapters01_03.tex, line 622
t can recover a proved rate] Assume the setup of Corollary~[source reference], except that τ>σ is not required. Discard the finite initial set of indices before nonvanishing holds. For the remaining sufficiently large j, define h_j=gcd(absD_ju_j,absD_jv_j), which is positive because the form is nonzero at those indices. If liminf_j→∞fraclog h_jj≥κ hboxand τ+κ>σ, then α is irrational.
Exact mathematical passage
\liminf_{j\to\infty}\frac{\log h_j}{j}\geq\kappa \quad\hbox{and}\quad \tau+\kappa>\sigma, \label{eq:c02-content-rate}
\liminf_{j\to\infty}Use in source · 1 passage
Search aliases: liminf_j→∞ liminf to infty
Use in source · chapters01_03.tex, line 622
t can recover a proved rate] Assume the setup of Corollary~[source reference], except that τ>σ is not required. Discard the finite initial set of indices before nonvanishing holds. For the remaining sufficiently large j, define h_j=gcd(absD_ju_j,absD_jv_j), which is positive because the form is nonzero at those indices. If liminf_j→∞fraclog h_jj≥κ hboxand τ+κ>σ, then α is irrational.
Exact mathematical passage
\liminf_{j\to\infty}\frac{\log h_j}{j}\geq\kappa \quad\hbox{and}\quad \tau+\kappa>\sigma, \label{eq:c02-content-rate}
\liminf_{n\to\infty}Use in source · 2 passages
Search aliases: liminf_n→∞ liminf to infty
Use in source · chapters08_10.tex, line 267
With the cleared gcd g_n=g_n,m(n), the primitive form is exactly Λ_n=fracK_ng_nwidetilde J_n,m(n). Suppose a future arithmetic argument establishes δ=liminf_n→∞fraclog g_nn. For finite δ, the present analytic estimate yields limsup_n→∞fraclog|Λ_n|n ≤κ_0(α)+log M(α)-δ. The same interpretation applies when δ=+∞: any sufficiently large finite lower bound for the content rate would suffice. All logarithms of nonzero forms are legitimate for the even family by [source reference]. The strict negativit…
Exact mathematical passage
\delta=\liminf_{n\to\infty}\frac{\log g_n}{n}. \label{eq:delta-content}
\limsupUse in source · 4 passages
Search aliases: limsup
Use in source · chapters01_03.tex, line 572
[Exponential clearance criterion] Let α∈R, and let u_j,v_j∈Q for j≥1. Let D_j be positive integers such that D_ju_j,D_jv_j∈Z. Assume u_jα-v_j≠0 for all sufficiently large j. Suppose limsup_j→∞absu_jα-v_j^1/j≤λ, limsup_j→∞D_j^1/j≤δ, where λ≥0, δ<∞ and δλ<1. Then α is irrational.
Exact mathematical passage
\limsup_{j\to\infty}\abs{u_j\alpha-v_j}^{1/j}\leq\lambda, \qquad \limsup_{j\to\infty}D_j^{1/j}\leq\delta, \label{eq:c02-root-bounds}
\limsup_{j\to\infty}Use in source · 2 passages
Search aliases: limsup_j→∞ limsup to infty
Use in source · chapters01_03.tex, line 572
[Exponential clearance criterion] Let α∈R, and let u_j,v_j∈Q for j≥1. Let D_j be positive integers such that D_ju_j,D_jv_j∈Z. Assume u_jα-v_j≠0 for all sufficiently large j. Suppose limsup_j→∞absu_jα-v_j^1/j≤λ, limsup_j→∞D_j^1/j≤δ, where λ≥0, δ<∞ and δλ<1. Then α is irrational.
Exact mathematical passage
\limsup_{j\to\infty}\abs{u_j\alpha-v_j}^{1/j}\leq\lambda, \qquad \limsup_{j\to\infty}D_j^{1/j}\leq\delta, \label{eq:c02-root-bounds}
\limsup_{n\to\infty}Use in source · 2 passages
Search aliases: limsup_n→∞ limsup to infty
Use in source · chapters08_10.tex, line 232
Let α_n=m(n)/n. The finite measure in [source reference] gives |widetilde J_n,m(n)| ≤ 8G M(α_n)^n. Continuity of the two branches, including their stated extension at zero, then proves limsup_n→∞ |widetilde J_n,m(n)|^1/n≤ M(α). This is an upper bound. An equality for the actual integral would require a further argument. In particular, an upper bound does not measure the effect of sign cancellation in an odd filter.
Exact mathematical passage
\limsup_{n\to\infty} |\widetilde J_{n,m(n)}|^{1/n}\leq M(\alpha). \label{eq:M-limsup}
MUse in source · 26 passages
Search aliases: M
Use in source · appendices.tex, line 116
[5pt] LC07 · Eskandari, 2026 · The bound M(α) and approximation exponent. It is a preprint input below the irrationality threshold.
Exact mathematical passage
M(\alpha)
M_1Use in source · 2 passages
Search aliases: M_1
Use in source · chapter03.tex, line 181
1 · 0. X_n+1(w)=M_nX_n(w). Hence X_N+1(w)=M_NM_N-1⋯ M_1X_1(w) (N≥1). The order of multiplication follows the temporal order: M_1 acts first. This is a product of changing matrices, not a power of one fixed matrix.
Exact mathematical passage
X_{N+1}(w)=M_NM_{N-1}\cdots M_1X_1(w)\qquad(N\geq1).
M_NUse in source · 1 passage
Search aliases: M_N
Use in source · chapter03.tex, line 181
1 · 0. X_n+1(w)=M_nX_n(w). Hence X_N+1(w)=M_NM_N-1⋯ M_1X_1(w) (N≥1). The order of multiplication follows the temporal order: M_1 acts first. This is a product of changing matrices, not a power of one fixed matrix.
Exact mathematical passage
X_{N+1}(w)=M_NM_{N-1}\cdots M_1X_1(w)\qquad(N\geq1).
M_nUse in source · 4 passages
Search aliases: M_n
Use in source · chapter03.tex, line 176
Normalize the recurrence by putting α_n=fracq(n)a_n, β_n=fracc_na_n (n≥1). Both quantities are positive rationals. For the adjacent state vector X_n(w)=(w_n,w_n-1)^ T, define M_n=α_n · β_n 1 · 0. X_n+1(w)=M_nX_n(w). Hence X_N+1(w)=M_NM_N-1⋯ M_1X_1(w) (N≥1). The order of multiplication follows the temporal order: M_1 acts first. This is a product of changing matrices, not a power of one fixed matrix.
Exact mathematical passage
\label{cb_eq_transfer} M_n=\begin{pmatrix}\alpha_n&\beta_n\\1&0\end{pmatrix}. \qquad X_{n+1}(w)=M_nX_n(w).
M_{N-1}Use in source · 1 passage
Search aliases: M_N-1
Use in source · chapter03.tex, line 181
1 · 0. X_n+1(w)=M_nX_n(w). Hence X_N+1(w)=M_NM_N-1⋯ M_1X_1(w) (N≥1). The order of multiplication follows the temporal order: M_1 acts first. This is a product of changing matrices, not a power of one fixed matrix.
Exact mathematical passage
X_{N+1}(w)=M_NM_{N-1}\cdots M_1X_1(w)\qquad(N\geq1).
M_n^{-1}Use in source · 1 passage
Search aliases: M_n^-1
Use in source · chapter03.tex, line 192
The determinant is det M_n=-β_n=-fracc_na_n<0. Every transfer is invertible, with M_n^-1=0 · 1 1/β_n · -α_n/β_n. An adjacent pair at index n+1 therefore determines the preceding pair at index n, within the stated recurrence domain. This does not extend the equation through n=0, where the trailing coefficient would vanish.
Exact mathematical passage
M_n^{-1}=\begin{pmatrix}0&1\\1/\beta_n&-\alpha_n/\beta_n\end{pmatrix}.
M_\inftyUse in source · 1 passage
Search aliases: M_∞ infty
Use in source · chapter03.tex, line 295
The leading terms of the coefficient polynomials are a_n=320n^6+O(n^5), q(n)=3520n^6+O(n^5), c_n=320n^6+O(n^5). Consequently α_n→11, β_n→1, and M_n⟶ M_∞=11 · 1 1 · 0. The characteristic polynomial of this limiting matrix is λ^2-11λ-1. Its two roots are λ_+=frac11+5sqrt52=ϕ^5, λ_-=frac11-5sqrt52=-ϕ^-5. Here ϕ=(1+sqrt5)/2. The identity ϕ^5=5ϕ+3 follows by repeatedly using ϕ^2=ϕ+1. The roots have product -1, as required by the limiting determinant, and very different magnitudes.
Exact mathematical passage
M_n\longrightarrow M_\infty=\begin{pmatrix}11&1\\1&0\end{pmatrix}.
\maxUse in source · 4 passages
Search aliases: max
Use in source · chapters04_07.tex, line 644
why the filter changes sign. Put r=sqrtxy. Since x+y≥2r, ψ(x,y)≤fracr^2(1-r)^21-r^2 =fracr^2(1-r)1+r,qquad0<r<1. Equality is achieved when x=y=r. The derivative of the last expression is frac2r(1-r-r^2)(1+r)^2. It has its interior maximum at r=ϕ^-1, where ϕ=(1+sqrt5)/2. Since 1-r=r^2 there, the maximum is maxψ=ϕ^-5≈0.09017. This is greater than 1/16. One can already see a negative filter value at x=y=1/2, where ψ=1/12 and 1-16ψ=-1/3. Near the sides of the square ψ is small and the filter is posi…
Exact mathematical passage
\label{cb_eq_psi_max} \max\psi=\phi^{-5}\approx0.09017.
\max_{(n,m)\in\mathcal D_{2000}}Use in source · 1 passage
Search aliases: max_(n,m)∈ D_2000 max in mathcal
Use in source · chapters04_07.tex, line 845
cal run takes N=2000 and K=128. The release reran the unchanged historical algorithm and compared its result with the recovered output. The finite arithmetic results agree. In the domain D_2000 there are exactly 8,006,000 pairs. Every observed residue is nonzero, so the residue lemma applies at every pair. The exact maximum valuation in this finite domain is max_(n,m)∈ D_2000v_2(S_n,m)=31. The recorded maximizing pair is (1466,2113). The record reports the first maximum encountered in its scan o…
Exact mathematical passage
\label{cb_eq_finite_max} \max_{(n,m)\in\mathcal D_{2000}}v_2(S_{n,m})=31.
\midUse in source · 5 passages
Search aliases: mid
Use in source · chapters01_03.tex, line 539
Such a D_j always exists for one finite rational pair. Write u_j=a_j/b_j and v_j=c_j/d_j in lowest terms with positive denominators. The smallest positive common clearing factor is D_j^min=lcm(b_j,d_j). Indeed, integrality of D_ja_j/b_j forces b_jmid D_j by coprimality, and integrality of D_jc_j/d_j forces d_jmid D_j. Conversely, their least common multiple clears both denominators.
Exact mathematical passage
b_j\mid D_j
\minUse in source · 3 passages
Search aliases: min
Use in source · chapters01_03.tex, line 535
Such a D_j always exists for one finite rational pair. Write u_j=a_j/b_j and v_j=c_j/d_j in lowest terms with positive denominators. The smallest positive common clearing factor is D_j^min=lcm(b_j,d_j). Indeed, integrality of D_ja_j/b_j forces b_jmid D_j by coprimality, and integrality of D_jc_j/d_j forces d_jmid D_j. Conversely, their least common multiple clears both denominators.
Exact mathematical passage
D_j^{\min}=\operatorname{lcm}(b_j,d_j). \label{eq:c02-minimal-clearance}
N^2Use in source · 1 passage
Search aliases: N^2
Use in source · chapters04_07.tex, line 806
For a positive integer N, the scanned domain is D_N={(n,m):1≤ n≤ N, 0≤ m≤4n}. The number of scanned pairs is ∑_n=1^N(4n+1)=2N^2+3N. The upper slope 4 is part of the experiment's definition. The scan supplies no information about pairs above this boundary or with n>N.
Exact mathematical passage
\label{cb_eq_scan_count} \sum_{n=1}^N(4n+1)=2N^2+3N.
n^2Use in source · 5 passages
Search aliases: n^2
Use in source · appendices.tex, line 12
The normalized rational sequences start with (u_0,u_1)=(1,7/4), (v_0,v_1)=(0,13/8). For n≥1, set p(n) · =20n^2-8n+1,notag q(n) · =3520n^6+5632n^5+2064n^4notag · -384n^3-156n^2+16n+7,notag a_n · =(2n+1)^2(2n+2)^2p(n),notag c_n · =(2n-1)^2(2n)^2p(n+1). Each sequence w=u or w=v obeys a_nw_n+1=q(n)w_n+c_nw_n-1. The forward recurrence is used only at n≥1. Its coefficients a_n are positive there. The first additional terms are u_2=frac64964, u_3=frac19471256, v_2=frac106991152, v_3=frac8025653115200.
Exact mathematical passage
p(n)&=20n^2-8n+1,\notag\\ q(n)&=3520n^6+5632n^5+2064n^4\notag\\ &\hspace{1.6em}-384n^3-156n^2+16n+7,\notag\\ a_n&=(2n+1)^2(2n+2)^2p(n),\notag\\ c_n&=(2n-1)^2(2n)^2p(n+1). \label{eq:app-recurrence-coeffs}
n^3Use in source · 4 passages
Search aliases: n^3
Use in source · appendices.tex, line 12
The normalized rational sequences start with (u_0,u_1)=(1,7/4), (v_0,v_1)=(0,13/8). For n≥1, set p(n) · =20n^2-8n+1,notag q(n) · =3520n^6+5632n^5+2064n^4notag · -384n^3-156n^2+16n+7,notag a_n · =(2n+1)^2(2n+2)^2p(n),notag c_n · =(2n-1)^2(2n)^2p(n+1). Each sequence w=u or w=v obeys a_nw_n+1=q(n)w_n+c_nw_n-1. The forward recurrence is used only at n≥1. Its coefficients a_n are positive there. The first additional terms are u_2=frac64964, u_3=frac19471256, v_2=frac106991152, v_3=frac8025653115200.
Exact mathematical passage
p(n)&=20n^2-8n+1,\notag\\ q(n)&=3520n^6+5632n^5+2064n^4\notag\\ &\hspace{1.6em}-384n^3-156n^2+16n+7,\notag\\ a_n&=(2n+1)^2(2n+2)^2p(n),\notag\\ c_n&=(2n-1)^2(2n)^2p(n+1). \label{eq:app-recurrence-coeffs}
n^4Use in source · 5 passages
Search aliases: n^4
Use in source · appendices.tex, line 12
The normalized rational sequences start with (u_0,u_1)=(1,7/4), (v_0,v_1)=(0,13/8). For n≥1, set p(n) · =20n^2-8n+1,notag q(n) · =3520n^6+5632n^5+2064n^4notag · -384n^3-156n^2+16n+7,notag a_n · =(2n+1)^2(2n+2)^2p(n),notag c_n · =(2n-1)^2(2n)^2p(n+1). Each sequence w=u or w=v obeys a_nw_n+1=q(n)w_n+c_nw_n-1. The forward recurrence is used only at n≥1. Its coefficients a_n are positive there. The first additional terms are u_2=frac64964, u_3=frac19471256, v_2=frac106991152, v_3=frac8025653115200.
Exact mathematical passage
p(n)&=20n^2-8n+1,\notag\\ q(n)&=3520n^6+5632n^5+2064n^4\notag\\ &\hspace{1.6em}-384n^3-156n^2+16n+7,\notag\\ a_n&=(2n+1)^2(2n+2)^2p(n),\notag\\ c_n&=(2n-1)^2(2n)^2p(n+1). \label{eq:app-recurrence-coeffs}
n^5Use in source · 3 passages
Search aliases: n^5
Use in source · appendices.tex, line 12
The normalized rational sequences start with (u_0,u_1)=(1,7/4), (v_0,v_1)=(0,13/8). For n≥1, set p(n) · =20n^2-8n+1,notag q(n) · =3520n^6+5632n^5+2064n^4notag · -384n^3-156n^2+16n+7,notag a_n · =(2n+1)^2(2n+2)^2p(n),notag c_n · =(2n-1)^2(2n)^2p(n+1). Each sequence w=u or w=v obeys a_nw_n+1=q(n)w_n+c_nw_n-1. The forward recurrence is used only at n≥1. Its coefficients a_n are positive there. The first additional terms are u_2=frac64964, u_3=frac19471256, v_2=frac106991152, v_3=frac8025653115200.
Exact mathematical passage
p(n)&=20n^2-8n+1,\notag\\ q(n)&=3520n^6+5632n^5+2064n^4\notag\\ &\hspace{1.6em}-384n^3-156n^2+16n+7,\notag\\ a_n&=(2n+1)^2(2n+2)^2p(n),\notag\\ c_n&=(2n-1)^2(2n)^2p(n+1). \label{eq:app-recurrence-coeffs}
n^6Use in source · 3 passages
Search aliases: n^6
Use in source · appendices.tex, line 12
The normalized rational sequences start with (u_0,u_1)=(1,7/4), (v_0,v_1)=(0,13/8). For n≥1, set p(n) · =20n^2-8n+1,notag q(n) · =3520n^6+5632n^5+2064n^4notag · -384n^3-156n^2+16n+7,notag a_n · =(2n+1)^2(2n+2)^2p(n),notag c_n · =(2n-1)^2(2n)^2p(n+1). Each sequence w=u or w=v obeys a_nw_n+1=q(n)w_n+c_nw_n-1. The forward recurrence is used only at n≥1. Its coefficients a_n are positive there. The first additional terms are u_2=frac64964, u_3=frac19471256, v_2=frac106991152, v_3=frac8025653115200.
Exact mathematical passage
p(n)&=20n^2-8n+1,\notag\\ q(n)&=3520n^6+5632n^5+2064n^4\notag\\ &\hspace{1.6em}-384n^3-156n^2+16n+7,\notag\\ a_n&=(2n+1)^2(2n+2)^2p(n),\notag\\ c_n&=(2n-1)^2(2n)^2p(n+1). \label{eq:app-recurrence-coeffs}
PUse in source · 30 passages
Search aliases: P
Use in source · chapters01_03.tex, line 361
[Rational lattice gap] Suppose α=a/b with a∈Z and b∈Z_>0. For every pair P,Q∈Z, either Qα-P=0 or absQα-P≥frac1b.
Exact mathematical passage
P,Q\in\Z
P_jUse in source · 11 passages
Search aliases: P_j
Use in source · chapters01_03.tex, line 380
[Integer form criterion] Let α∈R. Suppose there are integer pairs (P_j,Q_j) for j≥1 such that 0<absQ_jα-P_j⟶0. Then α is irrational. The same conclusion holds if these conditions are verified along any infinite subsequence.
Exact mathematical passage
(P_j,Q_j)
P_nUse in source · 6 passages
Search aliases: P_n
Use in source · chapters01_03.tex, line 479
[A convergent rational sequence with a fixed form error] Let α=1/2 and take fracP_nQ_n=fracn2n+1, n≥1. The numerator and denominator are coprime because any common divisor of n and 2n+1 divides 1. The ordinary error is frac12-fracn2n+1=frac12(2n+1)⟶0. Nevertheless, Q_nα-P_n=frac12 for every n. The approximants are distinct from the target, all coefficients are integers, and the approximations converge. The integer form does not tend to zero.
Exact mathematical passage
\frac{P_n}{Q_n}=\frac{n}{2n+1},\qquad n\geq1.
\QUse in source · 7 passages
Search aliases: Q
Use in source · chapter03.tex, line 65
[Well defined rational solutions] Every initial pair (w_0,w_1)∈Q^2 determines exactly one rational sequence satisfying [source reference]. In particular the sequences in Definition~[source reference] exist uniquely. Moreover, u_n>0 (n≥0), v_n>0 (n≥1), v_0=0.
Exact mathematical passage
(w_0,w_1)\in\Q^2
Q'Use in source · 3 passages
Search aliases: Q'
Use in source · chapters08_10.tex, line 83
[Invariance under additional clearing] If an integer pair (A,B) is replaced by (cA,cB) for an integer c>0, then its primitive pair and primitive linear form remain unchanged. More generally, any integer clearing factor Q' divisible by D gives the same unsigned primitive pair (DU/h,N/h).
Exact mathematical passage
Q'
q^{-0.62}Use in source · 1 passage
Search aliases: q^-0.62
Use in source · chapters08_10.tex, line 349
An approximation exponent below one cannot contradict rationality. If G=a/b were rational and p/q≠ a/b were reduced, then |frac ab-frac pq| =frac|aq-bp|bq≥frac1bq. A bound of order q^-0.62 is larger than this obstruction for large q. Even rational numbers have many distinct approximations of that quality. For example a/b+1/(bk) has error 1/(bk) and, after reduction, denominator at most bk. Nothing in such a bound isolates irrationality.
Exact mathematical passage
q^{-0.62}
\Q^2Use in source · 1 passage
Search aliases: Q^2 Q
Use in source · chapter03.tex, line 65
[Well defined rational solutions] Every initial pair (w_0,w_1)∈Q^2 determines exactly one rational sequence satisfying [source reference]. In particular the sequences in Definition~[source reference] exist uniquely. Moreover, u_n>0 (n≥0), v_n>0 (n≥1), v_0=0.
Exact mathematical passage
(w_0,w_1)\in\Q^2
Q_jUse in source · 15 passages
Search aliases: Q_j
Use in source · chapters01_03.tex, line 380
[Integer form criterion] Let α∈R. Suppose there are integer pairs (P_j,Q_j) for j≥1 such that 0<absQ_jα-P_j⟶0. Then α is irrational. The same conclusion holds if these conditions are verified along any infinite subsequence.
Exact mathematical passage
(P_j,Q_j)
Q_j^{-\theta}Use in source · 1 passage
Search aliases: Q_j^-θ theta
Use in source · chapters01_03.tex, line 468
This lower bound explains a familiar sufficient exponent. If distinct approximants have Q_j→∞ and absα-fracP_jQ_j≤ C Q_j^-θ for a fixed constant C>0 and an exponent θ>1, then multiplying by Q_j gives a quantity tending to zero. The criterion applies. An upper bound with θ≤1 does not supply that conclusion on its own. It may still describe useful approximation, and another argument may still prove irrationality, but this particular inequality does not cross the required threshold.
Exact mathematical passage
\abs{\alpha-\frac{P_j}{Q_j}}\leq C Q_j^{-\theta} \label{eq:c02-exponent-bound}
Q_nUse in source · 8 passages
Search aliases: Q_n
Use in source · chapter03.tex, line 117
[tbp] centeringincludegraphics[width=0.93linewidth]../visual/figures/09_recurrence_structure.pdf captionTwo preceding values determine the next rational value. The forward rule begins at n=1 and uses the stated initial pair. Here and in subsequent recurrence diagrams, A_n=a_n, Q_n=q(n) and C_n=c_n.
Exact mathematical passage
Q_n=q(n)
RUse in source · 4 passages
Search aliases: R
Use in source · chapters08_10.tex, line 170
Write ϕ=frac1+sqrt52, t=ϕ^-5, R=16ϕ^5=88+40sqrt5. The unfiltered integral has the representation J_k=∫_[0,1]^2ψ(x,y)^k ddμ(x,y), ψ=fracxy(1-x)(1-y)1-xy, ddμ=fracdd x dd ysqrtx(1-y)(1-xy). The measure has finite total mass J_0=8G. The factor 1-xy in its denominator lies outside the square root. Moving it inside changes the integral and invalidates the coefficient identification.
Exact mathematical passage
\phi=\frac{1+\sqrt5}{2},\qquad t=\phi^{-5}, \qquad R=16\phi^5=88+40\sqrt5. \label{eq:asymptotic-constants}
\RUse in source · 2 passages
Search aliases: R
Use in source · chapters01_03.tex, line 380
[Integer form criterion] Let α∈R. Suppose there are integer pairs (P_j,Q_j) for j≥1 such that 0<absQ_jα-P_j⟶0. Then α is irrational. The same conclusion holds if these conditions are verified along any infinite subsequence.
Exact mathematical passage
\alpha\in\R
r^2Use in source · 3 passages
Search aliases: r^2
Use in source · chapters04_07.tex, line 633
The kernel has a simple maximum that also explains why the filter changes sign. Put r=sqrtxy. Since x+y≥2r, ψ(x,y)≤fracr^2(1-r)^21-r^2 =fracr^2(1-r)1+r,qquad0<r<1. Equality is achieved when x=y=r. The derivative of the last expression is frac2r(1-r-r^2)(1+r)^2. It has its interior maximum at r=ϕ^-1, where ϕ=(1+sqrt5)/2. Since 1-r=r^2 there, the maximum is maxψ=ϕ^-5≈0.09017. This is greater than 1/16. One can already see a negative filter value at x=y=1/2, where ψ=1/12 and 1-16ψ=-1/3. Near the si…
Exact mathematical passage
\psi(x,y)\leq\frac{r^2(1-r)^2}{1-r^2} =\frac{r^2(1-r)}{1+r},\qquad0<r<1.
r_0Use in source · 1 passage
Search aliases: r_0
Use in source · chapter03.tex, line 161
The first ratios are r_0=0 and r_1=13/14. The subsequent values lie near the decimal for G, but proximity is only a first observation. Chapter~[source reference] proves exact nested brackets and an error sign at every index. The external identification below then makes those brackets enclosures of Catalan's constant itself.
Exact mathematical passage
r_0=0
r_1Use in source · 1 passage
Search aliases: r_1
Use in source · chapter03.tex, line 161
The first ratios are r_0=0 and r_1=13/14. The subsequent values lie near the decimal for G, but proximity is only a first observation. Chapter~[source reference] proves exact nested brackets and an error sign at every index. The external identification below then makes those brackets enclosures of Catalan's constant itself.
Exact mathematical passage
r_1=13/14
r_2Use in source · 1 passage
Search aliases: r_2
Use in source · chapter03.tex, line 139
, p(1)=13, p(2)=65, q(1)=10699, a_1=1872, c_1=260. For u, the recurrence gives u_2=frac10699(7/4)+2601872 =frac75933/41872=frac64964. For v, its zero initial term removes the trailing contribution: v_2=frac10699(13/8)+260·01872 =frac139087/81872=frac106991152. Therefore the second rational approximant is r_2=frac1069911682.
Exact mathematical passage
r_2=\frac{10699}{11682}.
r_{2k}Use in source · 2 passages
Search aliases: r_2k
Use in source · chapters04_07.tex, line 148
[Alternating brackets] There is a real number L such that r_n→ L. For every k≥0, r_2k<L<r_2k+1. The even subsequence increases strictly and the odd subsequence decreases strictly. For every n≥0, d_n-d_n+1<|L-r_n|<d_n, sign(L-r_n)=(-1)^n.
Exact mathematical passage
\label{cb_eq_bracket} r_{2k}<L<r_{2k+1}.
r_{2k+1}Use in source · 2 passages
Search aliases: r_2k+1
Use in source · chapters04_07.tex, line 148
[Alternating brackets] There is a real number L such that r_n→ L. For every k≥0, r_2k<L<r_2k+1. The even subsequence increases strictly and the odd subsequence decreases strictly. For every n≥0, d_n-d_n+1<|L-r_n|<d_n, sign(L-r_n)=(-1)^n.
Exact mathematical passage
\label{cb_eq_bracket} r_{2k}<L<r_{2k+1}.
r_{2k+2}Use in source · 1 passage
Search aliases: r_2k+2
Use in source · chapters04_07.tex, line 167
Equation [source reference] and the preceding lemma identify r_n with partial sums of a convergent alternating series: r_n=∑_j=0^n-1(-1)^j d_j, L=∑_j=0^∞(-1)^j d_j. For the even subsequence, r_2k+2-r_2k=d_2k-d_2k+1>0. For the odd subsequence, r_2k+3-r_2k+1=-d_2k+1+d_2k+2<0. Their adjacent separation d_2k tends to zero, so they have the same limit. Strictness of both subsequences places that limit strictly between every displayed pair. The alternating tail beginning with d_n gives the sign and up…
Exact mathematical passage
r_{2k+2}-r_{2k}=d_{2k}-d_{2k+1}>0
r_{2k+3}Use in source · 1 passage
Search aliases: r_2k+3
Use in source · chapters04_07.tex, line 169
Equation [source reference] and the preceding lemma identify r_n with partial sums of a convergent alternating series: r_n=∑_j=0^n-1(-1)^j d_j, L=∑_j=0^∞(-1)^j d_j. For the even subsequence, r_2k+2-r_2k=d_2k-d_2k+1>0. For the odd subsequence, r_2k+3-r_2k+1=-d_2k+1+d_2k+2<0. Their adjacent separation d_2k tends to zero, so they have the same limit. Strictness of both subsequences places that limit strictly between every displayed pair. The alternating tail beginning with d_n gives the sign and up…
Exact mathematical passage
r_{2k+3}-r_{2k+1}=-d_{2k+1}+d_{2k+2}<0
r_3Use in source · 1 passage
Search aliases: r_3
Use in source · chapter03.tex, line 157
· =frac521667445/12858500=frac8025653115200. It follows that r_3=8025653/8761950. All these values were also checked against the recovered exact rational script. A floating point calculation is unnecessary for the equality at any stage.
Exact mathematical passage
r_3=8025653/8761950
r_n^{1/n}Use in source · 1 passage
Search aliases: r_n^1/n
Use in source · chapter03.tex, line 362
diate consequence of the two source rates is lim_n→∞|G-r_n|^1/n =lim_n→∞frac|u_nG-v_n|^1/nu_n^1/n =ϕ^-10. Thus the ratio error has a smaller exponential scale than the unnormalized linear form. The distinction comes from division by the growing coefficient u_n. Also, since r_n→ G>0, one has r_n^1/n→1 and consequently v_n^1/n→ϕ^5. Both individual sequences grow, while their specified linear combination decays.
Exact mathematical passage
r_n^{1/n}\to1
\rceilUse in source · 1 passage
Search aliases: rceil
Use in source · chapters04_07.tex, line 821
For a fixed row m, the allowed indices are max(1,lceil m/4rceil)≤ n≤ N. Rows are needed through m=4N. In row m, the seed array is sufficient for every n≤5N-m. At the extreme row m=4N this bound is exactly n≤ N. Thus no scanned entry needs an unconstructed seed term. The algorithm evaluates additional row entries to support later rows. Those additions do not enlarge the declared set of scanned pairs.
Exact mathematical passage
\max(1,\lceil m/4\rceil)\leq n\leq N.
s^2Use in source · 2 passages
Search aliases: s^2
Use in source · chapters04_07.tex, line 427
The initial integral supplies a short diagnostic. Substitute x=t^2 and y=1-s^2. Then J_0 · =4∫_0^1 ∫_0^1fracdd s dd t1-t^2+t^2s^2
Exact mathematical passage
y=1-s^2
s_0Use in source · 2 passages
Search aliases: s_0
Use in source · chapters01_03.tex, line 68
Write s_N=∑_k=0^N-1frac(-1)^k(2k+1)^2, N≥0, where s_0=0 is the empty sum. Thus N counts retained terms, as it does in the visual guide. Each s_N is an exact rational number. The terms decrease strictly in magnitude to zero. Grouping successive terms shows that the even partial sums increase and the odd partial sums decrease. These two sequences approach the same limit.
Exact mathematical passage
s_0=0
s_1Use in source · 1 passage
Search aliases: s_1
Use in source · chapters01_03.tex, line 101
For example, s_0=0, s_1=1, s_2=frac89, s_3=frac209225, s_4=frac1001611025. Hence frac1001611025<G<frac209225. These statements use only rational arithmetic. The strict inequality also proves 0<G<1, which will simplify later tail estimates.
Exact mathematical passage
s_0=0,\qquad s_1=1,\qquad s_2=\frac89,\qquad s_3=\frac{209}{225},\qquad s_4=\frac{10016}{11025}.
S_{1,0}Use in source · 1 passage
Search aliases: S_1,0
Use in source · appendices.tex, line 164
For (n,m)=(1,1), S_1,1=T_1+T_2=28+2596=2624=2^6·41. Its exact valuation is v_2(S_1,1)=6. Pascal evolution gives the same result from the initial row: S_1,1=S_1,0+S_2,0.
Exact mathematical passage
S_{1,1}=S_{1,0}+S_{2,0}
S_{1,1}Use in source · 9 passages
Search aliases: S_1,1
Use in source · appendices.tex, line 161
For (n,m)=(1,1), S_1,1=T_1+T_2=28+2596=2624=2^6·41. Its exact valuation is v_2(S_1,1)=6. Pascal evolution gives the same result from the initial row: S_1,1=S_1,0+S_2,0.
Exact mathematical passage
S_{1,1}=T_1+T_2=28+2596=2624=2^6\cdot41.
s_2Use in source · 1 passage
Search aliases: s_2
Use in source · chapters01_03.tex, line 101
For example, s_0=0, s_1=1, s_2=frac89, s_3=frac209225, s_4=frac1001611025. Hence frac1001611025<G<frac209225. These statements use only rational arithmetic. The strict inequality also proves 0<G<1, which will simplify later tail estimates.
Exact mathematical passage
s_0=0,\qquad s_1=1,\qquad s_2=\frac89,\qquad s_3=\frac{209}{225},\qquad s_4=\frac{10016}{11025}.
S_{2,0}Use in source · 1 passage
Search aliases: S_2,0
Use in source · appendices.tex, line 164
For (n,m)=(1,1), S_1,1=T_1+T_2=28+2596=2624=2^6·41. Its exact valuation is v_2(S_1,1)=6. Pascal evolution gives the same result from the initial row: S_1,1=S_1,0+S_2,0.
Exact mathematical passage
S_{1,1}=S_{1,0}+S_{2,0}
s_{2r}Use in source · 2 passages
Search aliases: s_2r
Use in source · chapters01_03.tex, line 73
[Alternating enclosure] For every integer r≥0, s_2r<G<s_2r+1. For every integer N≥0, 0<(-1)^N(G-s_N)<frac1(2N+1)^2.
Exact mathematical passage
s_{2r}<G<s_{2r+1}. \label{eq:c01-parity-enclosure}
s_{2r+1}Use in source · 2 passages
Search aliases: s_2r+1
Use in source · chapters01_03.tex, line 73
[Alternating enclosure] For every integer r≥0, s_2r<G<s_2r+1. For every integer N≥0, 0<(-1)^N(G-s_N)<frac1(2N+1)^2.
Exact mathematical passage
s_{2r}<G<s_{2r+1}. \label{eq:c01-parity-enclosure}
s_{2r+2}Use in source · 1 passage
Search aliases: s_2r+2
Use in source · chapters01_03.tex, line 86
The difference between two consecutive even sums is s_2r+2-s_2r =frac1(4r+1)^2-frac1(4r+3)^2>0. Similarly, s_2r+3-s_2r+1 =-frac1(4r+3)^2+frac1(4r+5)^2<0. Consecutive sums differ by a term tending to zero, so the increasing even sums and decreasing odd sums have the same limit G. Strictness follows because the next sum of the same parity lies strictly between the current endpoint and the limit on the other side.
Exact mathematical passage
s_{2r+2}-s_{2r} =\frac1{(4r+1)^2}-\frac1{(4r+3)^2}>0.
s_{2r+3}Use in source · 1 passage
Search aliases: s_2r+3
Use in source · chapters01_03.tex, line 91
The difference between two consecutive even sums is s_2r+2-s_2r =frac1(4r+1)^2-frac1(4r+3)^2>0. Similarly, s_2r+3-s_2r+1 =-frac1(4r+3)^2+frac1(4r+5)^2<0. Consecutive sums differ by a term tending to zero, so the increasing even sums and decreasing odd sums have the same limit G. Strictness follows because the next sum of the same parity lies strictly between the current endpoint and the limit on the other side.
Exact mathematical passage
s_{2r+3}-s_{2r+1} =-\frac1{(4r+3)^2}+\frac1{(4r+5)^2}<0.
s_3Use in source · 1 passage
Search aliases: s_3
Use in source · chapters01_03.tex, line 101
For example, s_0=0, s_1=1, s_2=frac89, s_3=frac209225, s_4=frac1001611025. Hence frac1001611025<G<frac209225. These statements use only rational arithmetic. The strict inequality also proves 0<G<1, which will simplify later tail estimates.
Exact mathematical passage
s_0=0,\qquad s_1=1,\qquad s_2=\frac89,\qquad s_3=\frac{209}{225},\qquad s_4=\frac{10016}{11025}.
s_4Use in source · 1 passage
Search aliases: s_4
Use in source · chapters01_03.tex, line 101
For example, s_0=0, s_1=1, s_2=frac89, s_3=frac209225, s_4=frac1001611025. Hence frac1001611025<G<frac209225. These statements use only rational arithmetic. The strict inequality also proves 0<G<1, which will simplify later tail estimates.
Exact mathematical passage
s_0=0,\qquad s_1=1,\qquad s_2=\frac89,\qquad s_3=\frac{209}{225},\qquad s_4=\frac{10016}{11025}.
s_NUse in source · 5 passages
Search aliases: s_N
Use in source · chapters01_03.tex, line 64
Write s_N=∑_k=0^N-1frac(-1)^k(2k+1)^2, N≥0, where s_0=0 is the empty sum. Thus N counts retained terms, as it does in the visual guide. Each s_N is an exact rational number. The terms decrease strictly in magnitude to zero. Grouping successive terms shows that the even partial sums increase and the odd partial sums decrease. These two sequences approach the same limit.
Exact mathematical passage
s_N=\sum_{k=0}^{N-1}\frac{(-1)^k}{(2k+1)^2},\qquad N\geq0, \label{eq:c01-partial-sum}
s_{N+1}Use in source · 2 passages
Search aliases: s_N+1
Use in source · chapters01_03.tex, line 97
To obtain the error bound, apply this enclosure to s_N and s_N+1. The limit lies strictly between them. Their distance is (2N+1)^-2, and the direction from s_N to s_N+1 has sign (-1)^N.
Exact mathematical passage
s_{N+1}
S_{n+1,m}Use in source · 2 passages
Search aliases: S_n+1,m
Use in source · chapters04_07.tex, line 538
[Pascal evolution] For n,m≥0, S_n,m+1=S_n,m+S_n+1,m. The same identity holds with H in place of S.
Exact mathematical passage
\label{cb_eq_Pascal} S_{n,m+1}=S_{n,m}+S_{n+1,m}.
S_{n,0}Use in source · 1 passage
Search aliases: S_n,0
Use in source · chapters04_07.tex, line 552
At m=0, the packet is just S_n,0=T_n. The first two nontrivial rows are S_n,1=T_n+T_n+1, S_n,2=T_n+2T_n+1+T_n+2. For example S_1,1=28+2596=2624. The Pascal identity produces exactly the same integer as direct evaluation of its defining sum. This agreement between an additive row evolution and a direct weighted sum gives two different implementations of the finite transform.
Exact mathematical passage
S_{n,0}=T_n
S_{n,1}Use in source · 1 passage
Search aliases: S_n,1
Use in source · chapters04_07.tex, line 554
At m=0, the packet is just S_n,0=T_n. The first two nontrivial rows are S_n,1=T_n+T_n+1, S_n,2=T_n+2T_n+1+T_n+2. For example S_1,1=28+2596=2624. The Pascal identity produces exactly the same integer as direct evaluation of its defining sum. This agreement between an additive row evolution and a direct weighted sum gives two different implementations of the finite transform.
Exact mathematical passage
S_{n,1}=T_n+T_{n+1},\qquad S_{n,2}=T_n+2T_{n+1}+T_{n+2}.
S_{n,2}Use in source · 1 passage
Search aliases: S_n,2
Use in source · chapters04_07.tex, line 554
At m=0, the packet is just S_n,0=T_n. The first two nontrivial rows are S_n,1=T_n+T_n+1, S_n,2=T_n+2T_n+1+T_n+2. For example S_1,1=28+2596=2624. The Pascal identity produces exactly the same integer as direct evaluation of its defining sum. This agreement between an additive row evolution and a direct weighted sum gives two different implementations of the finite transform.
Exact mathematical passage
S_{n,1}=T_n+T_{n+1},\qquad S_{n,2}=T_n+2T_{n+1}+T_{n+2}.
S_{n,m(n)}Use in source · 1 passage
Search aliases: S_n,m(n)
Use in source · chapters08_10.tex, line 379
There is also a formulation independent of the universal clearing choice. With D_n,h_n as in Chapter~[source reference], the primitive error is (-1)^n(D_n/h_n)(S_n,m(n)G-H_n,m(n)). An estimate directly on that expression would suffice. If denominator reduction yields a provably smaller exponential cost, the excess factor Q_n/D_n should simply be removed from the accounting. The irrationality criterion itself is unchanged.
Exact mathematical passage
(-1)^n(D_n/h_n)(S_{n,m(n)}G-H_{n,m(n)})
S_{n,m+1}Use in source · 1 passage
Search aliases: S_n,m+1
Use in source · chapters04_07.tex, line 538
[Pascal evolution] For n,m≥0, S_n,m+1=S_n,m+S_n+1,m. The same identity holds with H in place of S.
Exact mathematical passage
\label{cb_eq_Pascal} S_{n,m+1}=S_{n,m}+S_{n+1,m}.
\operatorname{sign}Use in source · 4 passages
Search aliases: sign operatorname
Use in source · appendices.tex, line 179
associated rational packet is H_1,1=16v_1+16^2v_2 =26+frac213989 =frac216329. Therefore the raw form is F_1,1=-8(2624G-frac216329) =frac1730569-20992G. A certified interval for G makes this value positive. This is the smallest counterexample in the finite audit to the proposed sign rule signF_n,m=(-1)^m.
Exact mathematical passage
\operatorname{sign}F_{n,m}=(-1)^m
\mathsf TUse in source · 1 passage
Search aliases: T mathsf
Use in source · chapter03.tex, line 175
Normalize the recurrence by putting α_n=fracq(n)a_n, β_n=fracc_na_n (n≥1). Both quantities are positive rationals. For the adjacent state vector X_n(w)=(w_n,w_n-1)^ T, define M_n=α_n · β_n
Exact mathematical passage
X_n(w)=(w_n,w_{n-1})^{\mathsf T}
tUse in source · 27 passages
Search aliases: t
Use in source · chapter03.tex, line 149
At n=2 the coefficients become p(2)=65, p(3)=157, q(2)=434871, a_2=58500, c_2=22608. Using the fractions just obtained, u_3 · =frac434871(649/64)+22608(7/4)58500 · =frac284763375/6458500=frac19471256, [3pt] v_3 · =frac434871(10699/1152)+22608(13/8)58500 · =frac521667445/12858500=frac8025653115200. It follows that r_3=8025653/8761950. All these values were also checked against the recovered exact rational script. A floating point calculation is unnecessary for the equality at any stage.
Exact mathematical passage
u_3 &=\frac{434871(649/64)+22608(7/4)}{58500}\\ &=\frac{284763375/64}{58500}=\frac{19471}{256},\\[3pt] v_3 &=\frac{434871(10699/1152)+22608(13/8)}{58500}\\ &=\frac{521667445/128}{58500}=\frac{8025653}{115200}.
t^2Use in source · 3 passages
Search aliases: t^2
Use in source · chapters04_07.tex, line 93
The next argument needs the positivity of u_n. It also needs a modest lower bound on it, rather than a full asymptotic theorem. Both follow from the recurrence. To see the relevant sign without relying on a plot, put t=n-1. Direct polynomial expansion gives q(n)-a_n= · 3200t^6+24000t^5+73392t^4+117072t^3 · +102768t^2+47112t+8827. Thus q(n)>a_n>0 for every n≥1. Starting from u_1>u_0>0, the recurrence implies u_n+1>fracq(n)a_nu_n>u_n. In particular u_n≥1. This elementary bound is enough for the co…
Exact mathematical passage
q(n)-a_n={}&3200t^6+24000t^5+73392t^4+117072t^3\\ &+102768t^2+47112t+8827.
t^3Use in source · 1 passage
Search aliases: t^3
Use in source · chapters04_07.tex, line 93
The next argument needs the positivity of u_n. It also needs a modest lower bound on it, rather than a full asymptotic theorem. Both follow from the recurrence. To see the relevant sign without relying on a plot, put t=n-1. Direct polynomial expansion gives q(n)-a_n= · 3200t^6+24000t^5+73392t^4+117072t^3 · +102768t^2+47112t+8827. Thus q(n)>a_n>0 for every n≥1. Starting from u_1>u_0>0, the recurrence implies u_n+1>fracq(n)a_nu_n>u_n. In particular u_n≥1. This elementary bound is enough for the co…
Exact mathematical passage
q(n)-a_n={}&3200t^6+24000t^5+73392t^4+117072t^3\\ &+102768t^2+47112t+8827.
t^4Use in source · 1 passage
Search aliases: t^4
Use in source · chapters04_07.tex, line 93
The next argument needs the positivity of u_n. It also needs a modest lower bound on it, rather than a full asymptotic theorem. Both follow from the recurrence. To see the relevant sign without relying on a plot, put t=n-1. Direct polynomial expansion gives q(n)-a_n= · 3200t^6+24000t^5+73392t^4+117072t^3 · +102768t^2+47112t+8827. Thus q(n)>a_n>0 for every n≥1. Starting from u_1>u_0>0, the recurrence implies u_n+1>fracq(n)a_nu_n>u_n. In particular u_n≥1. This elementary bound is enough for the co…
Exact mathematical passage
q(n)-a_n={}&3200t^6+24000t^5+73392t^4+117072t^3\\ &+102768t^2+47112t+8827.
t^5Use in source · 1 passage
Search aliases: t^5
Use in source · chapters04_07.tex, line 93
The next argument needs the positivity of u_n. It also needs a modest lower bound on it, rather than a full asymptotic theorem. Both follow from the recurrence. To see the relevant sign without relying on a plot, put t=n-1. Direct polynomial expansion gives q(n)-a_n= · 3200t^6+24000t^5+73392t^4+117072t^3 · +102768t^2+47112t+8827. Thus q(n)>a_n>0 for every n≥1. Starting from u_1>u_0>0, the recurrence implies u_n+1>fracq(n)a_nu_n>u_n. In particular u_n≥1. This elementary bound is enough for the co…
Exact mathematical passage
q(n)-a_n={}&3200t^6+24000t^5+73392t^4+117072t^3\\ &+102768t^2+47112t+8827.
t^6Use in source · 1 passage
Search aliases: t^6
Use in source · chapters04_07.tex, line 93
The next argument needs the positivity of u_n. It also needs a modest lower bound on it, rather than a full asymptotic theorem. Both follow from the recurrence. To see the relevant sign without relying on a plot, put t=n-1. Direct polynomial expansion gives q(n)-a_n= · 3200t^6+24000t^5+73392t^4+117072t^3 · +102768t^2+47112t+8827. Thus q(n)>a_n>0 for every n≥1. Starting from u_1>u_0>0, the recurrence implies u_n+1>fracq(n)a_nu_n>u_n. In particular u_n≥1. This elementary bound is enough for the co…
Exact mathematical passage
q(n)-a_n={}&3200t^6+24000t^5+73392t^4+117072t^3\\ &+102768t^2+47112t+8827.
T_0Use in source · 4 passages
Search aliases: T_0
Use in source · appendices.tex, line 43
Set T_n=16^nu_n. Its forward recurrence is a_nT_n+1=16q(n)T_n+256c_nT_n-1, T_0=1, T_1=28. The initial terms are 1,28,2596,311536. The universally integral expression is u_n=∑_j=0^nbinom nj binomn-tfrac12jbinomn+j-tfrac12j, with the external coefficient identity and the written half-binomial denominator argument kept distinct.
Exact mathematical passage
a_nT_{n+1}=16q(n)T_n+256c_nT_{n-1}, \qquad T_0=1,\quad T_1=28. \label{eq:app-T-recurrence}
T_1Use in source · 3 passages
Search aliases: T_1
Use in source · appendices.tex, line 43
Set T_n=16^nu_n. Its forward recurrence is a_nT_n+1=16q(n)T_n+256c_nT_n-1, T_0=1, T_1=28. The initial terms are 1,28,2596,311536. The universally integral expression is u_n=∑_j=0^nbinom nj binomn-tfrac12jbinomn+j-tfrac12j, with the external coefficient identity and the written half-binomial denominator argument kept distinct.
Exact mathematical passage
a_nT_{n+1}=16q(n)T_n+256c_nT_{n-1}, \qquad T_0=1,\quad T_1=28. \label{eq:app-T-recurrence}
T_2Use in source · 4 passages
Search aliases: T_2
Use in source · appendices.tex, line 141
v_2 · =frac10699·(13/8)1872 =frac106991152. Multiplying u_2 by 16^2 gives T_2=2596. Alternatively, the integer recurrence numerator is 16·10699·28+256·260·1 =4 859 712=1872·2596. The exact division has no remainder. This calculation illustrates the program's division check. The all-index integrality proof uses the separate half-binomial argument.
Exact mathematical passage
T_2=2596
T_3Use in source · 1 passage
Search aliases: T_3
Use in source · chapters04_07.tex, line 339
For example, the n=2 sum is u_2=1+frac152+frac10564=frac64964. Multiplication by 16^2 gives the three integers 256, 1920, and 420, whose sum is T_2=2596. This illustrates the proof at the level of individual terms. The first sequence values are T_0=1, T_1=28, T_2=2596, T_3=311536, T_4=41759524.
Exact mathematical passage
T_0=1,\quad T_1=28,\quad T_2=2596,\quad T_3=311536,\quad T_4=41759524.
T_4Use in source · 1 passage
Search aliases: T_4
Use in source · chapters04_07.tex, line 339
For example, the n=2 sum is u_2=1+frac152+frac10564=frac64964. Multiplication by 16^2 gives the three integers 256, 1920, and 420, whose sum is T_2=2596. This illustrates the proof at the level of individual terms. The first sequence values are T_0=1, T_1=28, T_2=2596, T_3=311536, T_4=41759524.
Exact mathematical passage
T_0=1,\quad T_1=28,\quad T_2=2596,\quad T_3=311536,\quad T_4=41759524.
T_{5N}Use in source · 1 passage
Search aliases: T_5N
Use in source · chapters04_07.tex, line 813
The computation first constructs T_0,…,T_5N through the exact integer recurrence. It then reduces these integers modulo 2^K. The resulting array is the packet row with m=0. At each subsequent row, Pascal evolution replaces adjacent entries by their sum modulo 2^K. Induction from [source reference] proves that every entry read by the scan is the exact residue of its packet sum.
Exact mathematical passage
T_0,\ldots,T_{5N}
T_kUse in source · 1 passage
Search aliases: T_k
Use in source · chapters08_10.tex, line 14
Fix integers n,m≥0, and write U=∑_i=0^mbinom miT_n+i, V=∑_i=0^mbinom mi16^n+iv_n+i. These are fixed-pair aliases U=S_n,m and V=H_n,m for the packets defined earlier, not the integral coefficients of a different source normalization. Here T_k=16^ku_k is a positive integer. Consequently U is a positive integer. Write V=N/D in lowest terms, with D>0 and gcd(N,D)=1. In the exceptional case V=0, the same convention sets N=0,D=1. Set ℓ=2(n+m), L_ℓ=lcm(1,…,ℓ), Q=8L_ℓ^2, with L_0=1.
Exact mathematical passage
T_k=16^ku_k
T_{n+1}Use in source · 3 passages
Search aliases: T_n+1
Use in source · appendices.tex, line 43
Set T_n=16^nu_n. Its forward recurrence is a_nT_n+1=16q(n)T_n+256c_nT_n-1, T_0=1, T_1=28. The initial terms are 1,28,2596,311536. The universally integral expression is u_n=∑_j=0^nbinom nj binomn-tfrac12jbinomn+j-tfrac12j, with the external coefficient identity and the written half-binomial denominator argument kept distinct.
Exact mathematical passage
a_nT_{n+1}=16q(n)T_n+256c_nT_{n-1}, \qquad T_0=1,\quad T_1=28. \label{eq:app-T-recurrence}
T_{n+2}Use in source · 1 passage
Search aliases: T_n+2
Use in source · chapters04_07.tex, line 554
At m=0, the packet is just S_n,0=T_n. The first two nontrivial rows are S_n,1=T_n+T_n+1, S_n,2=T_n+2T_n+1+T_n+2. For example S_1,1=28+2596=2624. The Pascal identity produces exactly the same integer as direct evaluation of its defining sum. This agreement between an additive row evolution and a direct weighted sum gives two different implementations of the finite transform.
Exact mathematical passage
S_{n,1}=T_n+T_{n+1},\qquad S_{n,2}=T_n+2T_{n+1}+T_{n+2}.
T_{n-1}Use in source · 2 passages
Search aliases: T_n-1
Use in source · appendices.tex, line 43
Set T_n=16^nu_n. Its forward recurrence is a_nT_n+1=16q(n)T_n+256c_nT_n-1, T_0=1, T_1=28. The initial terms are 1,28,2596,311536. The universally integral expression is u_n=∑_j=0^nbinom nj binomn-tfrac12jbinomn+j-tfrac12j, with the external coefficient identity and the written half-binomial denominator argument kept distinct.
Exact mathematical passage
a_nT_{n+1}=16q(n)T_n+256c_nT_{n-1}, \qquad T_0=1,\quad T_1=28. \label{eq:app-T-recurrence}
\tanUse in source · 5 passages
Search aliases: tan
Use in source · chapters01_03.tex, line 173
[Arctangent and tangent forms] The following equalities hold: G=∫_0^1fracarctan xx dd x =-∫_0^π/4log(tanθ) ddθ. The integrand arctan x/x is assigned its continuous value 1 at x=0.
Exact mathematical passage
G=\int_0^1\frac{\arctan x}{x}\,\dd x =-\int_0^{\pi/4}\log(\tan\theta)\,\dd\theta. \label{eq:c01-tangent-forms}
u_jUse in source · 15 passages
Search aliases: u_j
Use in source · chapters01_03.tex, line 523
In the later chapters the starting form is ℓ_j=u_jα-v_j, u_j,v_j∈Q. Choose a positive integer D_j for which D_ju_j and D_jv_j are integers. Then L_j=D_jℓ_j=(D_ju_j)α-D_jv_j has integer coefficients. Its size is D_jabsℓ_j, so the clearance factor must be included in any decay estimate.
Exact mathematical passage
\ell_j=u_j\alpha-v_j,\qquad u_j,v_j\in\Q. \label{eq:c02-rational-form}
u_kUse in source · 2 passages
Search aliases: u_k
Use in source · chapters04_07.tex, line 589
Substitute the corrected normalization J_k=8(-1)^k(u_kG-v_k) into [source reference]. The two factors (-1)^i cancel. Thus widetilde J_n,m=8(-1)^n16^-n(S_n,mG-H_n,m). Define the raw packet form F_n,m=16^nwidetilde J_n,m =8(-1)^n(S_n,mG-H_n,m). The adjective raw means that the common denominator of H_n,m has not yet been cleared. It does not mean that F_n,m is an integer linear form. The coefficient of G is an integer, but the other coefficient may be rational.
Exact mathematical passage
J_k=8(-1)^k(u_kG-v_k)
u_{n-1}Use in source · 4 passages
Search aliases: u_n-1
Use in source · chapter03.tex, line 201
The two distinguished solutions can be collected into the fundamental matrix F_n= u_n · v_n u_n-1 · v_n-1, n≥1. It satisfies F_n+1=M_n F_n. Its initial determinant is det F_1= det7/4 · 13/8
Exact mathematical passage
\mathcal F_n= \begin{pmatrix}u_n&v_n\\u_{n-1}&v_{n-1}\end{pmatrix},\qquad n\geq1.
u_n^{1/n}Use in source · 2 passages
Search aliases: u_n^1/n
Use in source · chapter03.tex, line 321
[External recurrence identification and rates] For the exact normalization [source reference], Zudilin's theorem and its stated asymptotic rates give lim_n→∞fracv_nu_n · =G, lim_n→∞u_n^1/n · =ϕ^5, lim_n→∞|u_nG-v_n|^1/n · =ϕ^-5. These are external mathematical inputs, recorded precisely in source contracts LC01 and LC02 [source reference].
Exact mathematical passage
\lim_{n\to\infty}\frac{v_n}{u_n}&=G,\label{cb_eq_source_limit}\\ \lim_{n\to\infty}u_n^{1/n}&=\phi^5,\label{cb_eq_source_u_rate}\\ \lim_{n\to\infty}|u_nG-v_n|^{1/n}&=\phi^{-5}. \label{cb_eq_source_error_rate}
v_{20+i}Use in source · 1 passage
Search aliases: v_20+i
Use in source · chapters08_10.tex, line 128
The first row illustrates the exact check without printing its very large numerator and denominator. The independently reduced rational packet has denominator D=den(∑_i=0^66binom66i16^20+iv_20+i). Its recorded quotient Q/D is 1 703 936, and the minimal pair has gcd 8. Multiplication gives 1 703 936·8=13 631 488, exactly the historical value. The complete decimal integers U,N,D,Q are retained in the diagnostic JSON. They are inputs to an exact equality check, rather than values inferred from roun…
Exact mathematical passage
D=\operatorname{den}\left(\sum_{i=0}^{66}\binom{66}{i}16^{20+i}v_{20+i}\right).
v_3Use in source · 2 passages
Search aliases: v_3
Use in source · appendices.tex, line 26
c_n · =(2n-1)^2(2n)^2p(n+1). Each sequence w=u or w=v obeys a_nw_n+1=q(n)w_n+c_nw_n-1. The forward recurrence is used only at n≥1. Its coefficients a_n are positive there. The first additional terms are u_2=frac64964, u_3=frac19471256, v_2=frac106991152, v_3=frac8025653115200.
Exact mathematical passage
u_2=\frac{649}{64},\quad u_3=\frac{19471}{256},\qquad v_2=\frac{10699}{1152},\quad v_3=\frac{8025653}{115200}. \label{eq:app-small-sequences}
v_jUse in source · 15 passages
Search aliases: v_j
Use in source · chapters01_03.tex, line 523
In the later chapters the starting form is ℓ_j=u_jα-v_j, u_j,v_j∈Q. Choose a positive integer D_j for which D_ju_j and D_jv_j are integers. Then L_j=D_jℓ_j=(D_ju_j)α-D_jv_j has integer coefficients. Its size is D_jabsℓ_j, so the clearance factor must be included in any decay estimate.
Exact mathematical passage
\ell_j=u_j\alpha-v_j,\qquad u_j,v_j\in\Q. \label{eq:c02-rational-form}
v_kUse in source · 3 passages
Search aliases: v_k
Use in source · chapters04_07.tex, line 589
Substitute the corrected normalization J_k=8(-1)^k(u_kG-v_k) into [source reference]. The two factors (-1)^i cancel. Thus widetilde J_n,m=8(-1)^n16^-n(S_n,mG-H_n,m). Define the raw packet form F_n,m=16^nwidetilde J_n,m =8(-1)^n(S_n,mG-H_n,m). The adjective raw means that the common denominator of H_n,m has not yet been cleared. It does not mean that F_n,m is an integer linear form. The coefficient of G is an integer, but the other coefficient may be rational.
Exact mathematical passage
J_k=8(-1)^k(u_kG-v_k)
v_{n+1}Use in source · 2 passages
Search aliases: v_n+1
Use in source · appendices.tex, line 33
The Wronskian in this edition uses the orientation W_n=u_n+1v_n-u_nv_n+1 =frac(-1)^n+1p(n+1)2(2n+1)^2(2n+2)^2. Reversing its two terms reverses the sign. The formula applies to n≥0, with the base case supplied by the initial pair and the induction step supplied by [source reference].
Exact mathematical passage
W_n=u_{n+1}v_n-u_nv_{n+1} =\frac{(-1)^{n+1}p(n+1)}{2(2n+1)^2(2n+2)^2}. \label{eq:app-Wronskian}
v_{n-1}Use in source · 2 passages
Search aliases: v_n-1
Use in source · chapter03.tex, line 201
The two distinguished solutions can be collected into the fundamental matrix F_n= u_n · v_n u_n-1 · v_n-1, n≥1. It satisfies F_n+1=M_n F_n. Its initial determinant is det F_1= det7/4 · 13/8
Exact mathematical passage
\mathcal F_n= \begin{pmatrix}u_n&v_n\\u_{n-1}&v_{n-1}\end{pmatrix},\qquad n\geq1.
v_n^{1/n}Use in source · 1 passage
Search aliases: v_n^1/n
Use in source · chapter03.tex, line 362
ce rates is lim_n→∞|G-r_n|^1/n =lim_n→∞frac|u_nG-v_n|^1/nu_n^1/n =ϕ^-10. Thus the ratio error has a smaller exponential scale than the unnormalized linear form. The distinction comes from division by the growing coefficient u_n. Also, since r_n→ G>0, one has r_n^1/n→1 and consequently v_n^1/n→ϕ^5. Both individual sequences grow, while their specified linear combination decays.
Exact mathematical passage
v_n^{1/n}\to\phi^5
v_pUse in source · 1 passage
Search aliases: v_p
Use in source · chapters08_10.tex, line 58
If N=0, the reduced convention gives D=1. Since U>0, one has h=U, g=QU, and the unsigned primitive pair is (1,0). All the stated identities follow directly in this case. Assume now N≠0. Since gcd(D,N)=1, multiplication of U by D adds no prime factor that can divide N. Prime by prime, min{v_p(DU),v_p(|N|)}=min{v_p(U),v_p(|N|)}. This proves the first equality. Next write Q=cD, where c is a positive integer. The unsigned cleared pair is (cDU,cN), so its gcd is cgcd(DU,N)=ch. Dividing both coefficie…
Exact mathematical passage
\min\{v_p(DU),v_p(|N|)\}=\min\{v_p(U),v_p(|N|)\}
WUse in source · 9 passages
Search aliases: W
Use in source · appendices.tex, line 33
The Wronskian in this edition uses the orientation W_n=u_n+1v_n-u_nv_n+1 =frac(-1)^n+1p(n+1)2(2n+1)^2(2n+2)^2. Reversing its two terms reverses the sign. The formula applies to n≥0, with the base case supplied by the initial pair and the induction step supplied by [source reference].
Exact mathematical passage
W_n=u_{n+1}v_n-u_nv_{n+1} =\frac{(-1)^{n+1}p(n+1)}{2(2n+1)^2(2n+2)^2}. \label{eq:app-Wronskian}
W_0Use in source · 3 passages
Search aliases: W_0
Use in source · appendices.tex, line 149
The same small case checks the Wronskian orientation: W_0 · =frac74·0-1·frac138 =-frac138, W_1 · =frac64964·frac138 -frac74·frac106991152 =frac65288. Both agree with [source reference]. They show why the sign is (-1)^n+1 for this orientation.
Exact mathematical passage
W_0&=\frac74\cdot0-1\cdot\frac{13}{8} =-\frac{13}{8},\\ W_1&=\frac{649}{64}\cdot\frac{13}{8} -\frac74\cdot\frac{10699}{1152} =\frac{65}{288}.
w_0Use in source · 7 passages
Search aliases: w_0
Use in source · chapter03.tex, line 56
The forward equation starts at n=1. Substituting n=0 into its polynomial expression would remove the trailing term and impose 4w_1=7w_0. Although the initial values of u happen to satisfy that extra equation, those of v do not. It is therefore not a valid additional recurrence condition for the pair. At n=1, both w_0 and w_1 are already specified and w_2 is the first value to be computed.
Exact mathematical passage
4w_1=7w_0
W_1Use in source · 1 passage
Search aliases: W_1
Use in source · appendices.tex, line 149
The same small case checks the Wronskian orientation: W_0 · =frac74·0-1·frac138 =-frac138, W_1 · =frac64964·frac138 -frac74·frac106991152 =frac65288. Both agree with [source reference]. They show why the sign is (-1)^n+1 for this orientation.
Exact mathematical passage
W_0&=\frac74\cdot0-1\cdot\frac{13}{8} =-\frac{13}{8},\\ W_1&=\frac{649}{64}\cdot\frac{13}{8} -\frac74\cdot\frac{10699}{1152} =\frac{65}{288}.
w_1Use in source · 7 passages
Search aliases: w_1
Use in source · chapter03.tex, line 56
The forward equation starts at n=1. Substituting n=0 into its polynomial expression would remove the trailing term and impose 4w_1=7w_0. Although the initial values of u happen to satisfy that extra equation, those of v do not. It is therefore not a valid additional recurrence condition for the pair. At n=1, both w_0 and w_1 are already specified and w_2 is the first value to be computed.
Exact mathematical passage
4w_1=7w_0
w_2Use in source · 2 passages
Search aliases: w_2
Use in source · chapter03.tex, line 60
The forward equation starts at n=1. Substituting n=0 into its polynomial expression would remove the trailing term and impose 4w_1=7w_0. Although the initial values of u happen to satisfy that extra equation, those of v do not. It is therefore not a valid additional recurrence condition for the pair. At n=1, both w_0 and w_1 are already specified and w_2 is the first value to be computed.
Exact mathematical passage
w_2
w_3Use in source · 1 passage
Search aliases: w_3
Use in source · chapter03.tex, line 255
At n=2, the formula is w_2=α_1w_1+β_1w_0, using the empty determinant for K_2,1. At n=3, direct iteration gives w_3=(α_2α_1+β_2)w_1+α_2β_1w_0, which agrees with the formula. For later indices, both coefficients in [source reference] satisfy the same recurrence by [source reference]. Induction finishes the proof.
Exact mathematical passage
w_3=(\alpha_2\alpha_1+\beta_2)w_1+\alpha_2\beta_1w_0,
W_nUse in source · 7 passages
Search aliases: W_n
Use in source · appendices.tex, line 33
The Wronskian in this edition uses the orientation W_n=u_n+1v_n-u_nv_n+1 =frac(-1)^n+1p(n+1)2(2n+1)^2(2n+2)^2. Reversing its two terms reverses the sign. The formula applies to n≥0, with the base case supplied by the initial pair and the induction step supplied by [source reference].
Exact mathematical passage
W_n=u_{n+1}v_n-u_nv_{n+1} =\frac{(-1)^{n+1}p(n+1)}{2(2n+1)^2(2n+2)^2}. \label{eq:app-Wronskian}
w_{n+1}Use in source · 4 passages
Search aliases: w_n+1
Use in source · appendices.tex, line 21
c_n · =(2n-1)^2(2n)^2p(n+1). Each sequence w=u or w=v obeys a_nw_n+1=q(n)w_n+c_nw_n-1. The forward recurrence is used only at n≥1. Its coefficients a_n are positive there. The first additional terms are u_2=frac64964, u_3=frac19471256, v_2=frac106991152, v_3=frac8025653115200.
Exact mathematical passage
a_nw_{n+1}=q(n)w_n+c_nw_{n-1}. \label{eq:app-recurrence}
W_{n-1}Use in source · 1 passage
Search aliases: W_n-1
Use in source · chapters04_07.tex, line 49
At n=0 the determinant is -13/8, which agrees with the displayed formula because p(1)=13. For n≥1, insert [source reference] into both terms of [source reference]. The terms containing q(n)u_nv_n cancel, leaving W_n=fracc_na_n(u_n-1v_n-u_nv_n-1) =-fracc_na_nW_n-1. Iterating this equality gives W_n=(-1)^nW_0 ∏_j=1^n frac(2j-1)^2(2j)^2p(j+1)(2j+1)^2(2j+2)^2p(j). There are two independent telescopes. The polynomial factors leave p(n+1)/p(1). The integer factors leave 4/((2n+1)^2(2n+2)^2). Substitut…
Exact mathematical passage
W_n=\frac{c_n}{a_n}(u_{n-1}v_n-u_nv_{n-1}) =-\frac{c_n}{a_n}W_{n-1}.
\widetilde J_{1,1}Use in source · 2 passages
Search aliases: widetilde J_1,1 widetilde
Use in source · appendices.tex, line 201
_4=12, hence Q=8L_4^2=1152. Its redundant scale is Q/D=128. Consequently g=128·64=8192. The signed cleared pair is (-3 022 848,-2 768 896). Dividing by 8192 gives (-369,-338), so the primitive form is Λ_1,1=338-369G. The integral identity gives the same expression. Since widetilde J_1,1=F_1,1/16 and 16L_4^2/g=9/32, frac16L_4^2gwidetilde J_1,1 =frac9512F_1,1=338-369G. This checks the factor 8, the sign (-1)^n, and the denominator index in one exact example.
Exact mathematical passage
\widetilde J_{1,1}=F_{1,1}/16
\widetilde J_{n,m(n)}Use in source · 5 passages
Search aliases: widetilde J_n,m(n) widetilde
Use in source · chapters08_10.tex, line 226
Let α_n=m(n)/n. The finite measure in [source reference] gives |widetilde J_n,m(n)| ≤ 8G M(α_n)^n. Continuity of the two branches, including their stated extension at zero, then proves limsup_n→∞ |widetilde J_n,m(n)|^1/n≤ M(α). This is an upper bound. An equality for the actual integral would require a further argument. In particular, an upper bound does not measure the effect of sign cancellation in an odd filter.
Exact mathematical passage
|\widetilde J_{n,m(n)}| \leq 8G\,M(\alpha_n)^n. \label{eq:actual-M-bound}
\widetilde U_{n,m(n)}Use in source · 1 passage
Search aliases: widetilde U_n,m(n) widetilde
Use in source · chapters08_10.tex, line 315
Eskandari's coefficient growth statement is lim_n→∞fraclog|widetilde U_n,m(n)|n =γ(α) =-log16+log R+αlog(1+R), where widetilde U=8(-1)^n16^-nU. This is an external asymptotic result in the preprint, not a conclusion of the finite recurrence audit. Together with the filter bound it controls the approximation error |G-frac VU| =frac|widetilde J||widetilde U|. Before any additional quantified gcd saving, the reduced denominator q is at most the absolute cleared coefficient |A|. Its available expone…
Exact mathematical passage
\lim_{n\to\infty}\frac{\log|\widetilde U_{n,m(n)}|}{n} =\gamma(\alpha) =-\log16+\log R+\alpha\log(1+R), \label{eq:gamma-alpha}
XUse in source · 8 passages
Search aliases: X
Use in source · chapter03.tex, line 175
Normalize the recurrence by putting α_n=fracq(n)a_n, β_n=fracc_na_n (n≥1). Both quantities are positive rationals. For the adjacent state vector X_n(w)=(w_n,w_n-1)^ T, define M_n=α_n · β_n
Exact mathematical passage
X_n(w)=(w_n,w_{n-1})^{\mathsf T}
x^2Use in source · 16 passages
Search aliases: x^2
Use in source · appendices.tex, line 236
The Euler interval uses a_k=(2k+1)^-2 and the positive difference convention Δ a_k=a_k-a_k+1. The identity Δ^ja_0=∫_0^1(1-x^2)^j(-log x) dd x gives 0≤Δ^ja_0≤1. The positive Euler series for G is ∑_j≥02^-j-1Δ^ja_0. Keeping the first 1001 terms therefore gives a lower endpoint and an upper endpoint obtained by adding 2^-1001. All retained terms are rational. The analytic tail bound is a written argument. The computed interval is its finite arithmetic application.
Exact mathematical passage
\Delta^ja_0=\int_0^1(1-x^2)^j(-\log x)\,\dd x
x^{2k}Use in source · 5 passages
Search aliases: x^2k
Use in source · chapters01_03.tex, line 155
For 0≤ x<1, the geometric expansion is frac11+x^2=∑_k=0^∞(-1)^kx^2k. The integrated absolute values of the proposed terms have finite sum: ∑_k=0^∞∫_0^1x^2k(-log x) dd x =∑_k=0^∞frac1(2k+1)^2<∞. Therefore the series of integrable functions can be summed under the integral. The moment identity gives the defining series for G. The point x=1 has measure zero and does not affect the integral.
Exact mathematical passage
\frac1{1+x^2}=\sum_{k=0}^{\infty}(-1)^kx^{2k}.
x^{2N}Use in source · 1 passage
Search aliases: x^2N
Use in source · chapters01_03.tex, line 53
^s. Absolute convergence follows from comparison with ∑ j^-s. Thus G=β(2). The alternating series also defines β(s) for s>0, though absolute convergence is no longer available throughout that wider range. At s=1, integration of the finite geometric expansion of (1+x^2)^-1 gives β(1)=π/4. The integrated remainder is bounded by ∫_0^1x^2Ndd x=(2N+1)^-1 and tends to zero, so this passage does not require absolute convergence of the beta series at 1.
Exact mathematical passage
\int_0^1x^{2N}\dd x=(2N+1)^{-1}
x^aUse in source · 1 passage
Search aliases: x^a
Use in source · chapters01_03.tex, line 129
[A basic moment] For every real a>-1, ∫_0^1x^a(-log x) dd x=frac1(a+1)^2. The integral is understood as an improper integral at 0.
Exact mathematical passage
\int_0^1x^a(-\log x)\,\dd x=\frac1{(a+1)^2}. \label{eq:c01-moment}
x^{n-1/2}Use in source · 2 passages
Search aliases: x^n-1/2
Use in source · appendices.tex, line 71
widetilde J_n,m · =∑_i=0^m(-16)^ibinom miJ_n+i =16^-nF_n,m. The corrected unfiltered integral is J_n=∫_0^1 ∫_0^1 fracx^n-1/2(1-x)^ny^n(1-y)^n-1/2 (1-xy)^n+1 dd x dd y =8(-1)^n(u_nG-v_n). These are fixed normalization identities. Their factors should be checked together when translating a differently normalized source.
Exact mathematical passage
J_n=\int_0^1\!\int_0^1 \frac{x^{n-1/2}(1-x)^ny^n(1-y)^{n-1/2}} {(1-xy)^{n+1}}\,\dd x\,\dd y =8(-1)^n(u_nG-v_n). \label{eq:app-J}
X_1Use in source · 1 passage
Search aliases: X_1
Use in source · chapter03.tex, line 181
1 · 0. X_n+1(w)=M_nX_n(w). Hence X_N+1(w)=M_NM_N-1⋯ M_1X_1(w) (N≥1). The order of multiplication follows the temporal order: M_1 acts first. This is a product of changing matrices, not a power of one fixed matrix.
Exact mathematical passage
X_{N+1}(w)=M_NM_{N-1}\cdots M_1X_1(w)\qquad(N\geq1).
x_1Use in source · 1 passage
Search aliases: x_1
Use in source · chapters08_10.tex, line 389
There is a simple logical reason to resist extrapolation. Given any finite list of positive values x_1,…,x_N, it has a positive extension that tends to zero and another positive extension that tends to infinity. Keep the list fixed and set the tail to e^-n in the first case or e^n in the second. Similarly, any finite list of positive integer gcds can be extended by 1 forever or by integers with an arbitrarily prescribed positive exponential rate.
Exact mathematical passage
x_1,\ldots,x_N
X_nUse in source · 2 passages
Search aliases: X_n
Use in source · chapter03.tex, line 175
Normalize the recurrence by putting α_n=fracq(n)a_n, β_n=fracc_na_n (n≥1). Both quantities are positive rationals. For the adjacent state vector X_n(w)=(w_n,w_n-1)^ T, define M_n=α_n · β_n
Exact mathematical passage
X_n(w)=(w_n,w_{n-1})^{\mathsf T}
x_NUse in source · 1 passage
Search aliases: x_N
Use in source · chapters08_10.tex, line 389
There is a simple logical reason to resist extrapolation. Given any finite list of positive values x_1,…,x_N, it has a positive extension that tends to zero and another positive extension that tends to infinity. Keep the list fixed and set the tail to e^-n in the first case or e^n in the second. Similarly, any finite list of positive integer gcds can be extended by 1 forever or by integers with an arbitrarily prescribed positive exponential rate.
Exact mathematical passage
x_1,\ldots,x_N
X_{N+1}Use in source · 1 passage
Search aliases: X_N+1
Use in source · chapter03.tex, line 181
1 · 0. X_n+1(w)=M_nX_n(w). Hence X_N+1(w)=M_NM_N-1⋯ M_1X_1(w) (N≥1). The order of multiplication follows the temporal order: M_1 acts first. This is a product of changing matrices, not a power of one fixed matrix.
Exact mathematical passage
X_{N+1}(w)=M_NM_{N-1}\cdots M_1X_1(w)\qquad(N\geq1).
X_{n+1}Use in source · 1 passage
Search aliases: X_n+1
Use in source · chapter03.tex, line 176
Normalize the recurrence by putting α_n=fracq(n)a_n, β_n=fracc_na_n (n≥1). Both quantities are positive rationals. For the adjacent state vector X_n(w)=(w_n,w_n-1)^ T, define M_n=α_n · β_n 1 · 0. X_n+1(w)=M_nX_n(w). Hence X_N+1(w)=M_NM_N-1⋯ M_1X_1(w) (N≥1). The order of multiplication follows the temporal order: M_1 acts first. This is a product of changing matrices, not a power of one fixed matrix.
Exact mathematical passage
\label{cb_eq_transfer} M_n=\begin{pmatrix}\alpha_n&\beta_n\\1&0\end{pmatrix}. \qquad X_{n+1}(w)=M_nX_n(w).
yUse in source · 23 passages
Search aliases: y
Use in source · appendices.tex, line 71
widetilde J_n,m · =∑_i=0^m(-16)^ibinom miJ_n+i =16^-nF_n,m. The corrected unfiltered integral is J_n=∫_0^1 ∫_0^1 fracx^n-1/2(1-x)^ny^n(1-y)^n-1/2 (1-xy)^n+1 dd x dd y =8(-1)^n(u_nG-v_n). These are fixed normalization identities. Their factors should be checked together when translating a differently normalized source.
Exact mathematical passage
J_n=\int_0^1\!\int_0^1 \frac{x^{n-1/2}(1-x)^ny^n(1-y)^{n-1/2}} {(1-xy)^{n+1}}\,\dd x\,\dd y =8(-1)^n(u_nG-v_n). \label{eq:app-J}
y^2Use in source · 3 passages
Search aliases: y^2
Use in source · chapters01_03.tex, line 204
[Elementary double integral] Catalan's constant also satisfies G=∫_0^1∫_0^1fracdd x dd y1+x^2y^2.
Exact mathematical passage
G=\int_0^1\int_0^1\frac{\dd x\,\dd y}{1+x^2y^2}. \label{eq:c01-double}
y^{2k}Use in source · 1 passage
Search aliases: y^2k
Use in source · chapters01_03.tex, line 217
There is also a direct series justification. Expanding (1+x^2y^2)^-1 on the square except at the corner (1,1) gives geometric terms (-1)^kx^2ky^2k. Their integrated absolute values sum to ∑_k=0^∞(∫_0^1x^2k dd x)^2 =∑_k=0^∞frac1(2k+1)^2<∞. Thus summation and integration may be interchanged, giving the defining series again.
Exact mathematical passage
(-1)^kx^{2k}y^{2k}
y^nUse in source · 2 passages
Search aliases: y^n
Use in source · appendices.tex, line 71
widetilde J_n,m · =∑_i=0^m(-16)^ibinom miJ_n+i =16^-nF_n,m. The corrected unfiltered integral is J_n=∫_0^1 ∫_0^1 fracx^n-1/2(1-x)^ny^n(1-y)^n-1/2 (1-xy)^n+1 dd x dd y =8(-1)^n(u_nG-v_n). These are fixed normalization identities. Their factors should be checked together when translating a differently normalized source.
Exact mathematical passage
J_n=\int_0^1\!\int_0^1 \frac{x^{n-1/2}(1-x)^ny^n(1-y)^{n-1/2}} {(1-xy)^{n+1}}\,\dd x\,\dd y =8(-1)^n(u_nG-v_n). \label{eq:app-J}
zUse in source · 13 passages
Search aliases: z
Use in source · chapters04_07.tex, line 298
Work with formal power series, so no convergence issue arises. The central binomial expansion gives (1+4z)^-1/2=∑_k≥0(-1)^kbinom2kkz^k. Every coefficient is an integer. Multiplying by the integer polynomial (1+4z)^h preserves this property. The coefficient of z^j in the product is 4^jbinomh-tfrac12j. More explicitly it is ∑_ℓ=0^min(h,j) binom hℓ 4^ℓ(-1)^j-ℓ binom2(j-ℓ)j-ℓ, an integer sum.
Exact mathematical passage
(1+4z)^{-1/2}=\sum_{k\geq0}(-1)^k\binom{2k}{k}z^k.
z^jUse in source · 1 passage
Search aliases: z^j
Use in source · chapters04_07.tex, line 302
Work with formal power series, so no convergence issue arises. The central binomial expansion gives (1+4z)^-1/2=∑_k≥0(-1)^kbinom2kkz^k. Every coefficient is an integer. Multiplying by the integer polynomial (1+4z)^h preserves this property. The coefficient of z^j in the product is 4^jbinomh-tfrac12j. More explicitly it is ∑_ℓ=0^min(h,j) binom hℓ 4^ℓ(-1)^j-ℓ binom2(j-ℓ)j-ℓ, an integer sum.
Exact mathematical passage
z^j
z^kUse in source · 1 passage
Search aliases: z^k
Use in source · chapters04_07.tex, line 298
Work with formal power series, so no convergence issue arises. The central binomial expansion gives (1+4z)^-1/2=∑_k≥0(-1)^kbinom2kkz^k. Every coefficient is an integer. Multiplying by the integer polynomial (1+4z)^h preserves this property. The coefficient of z^j in the product is 4^jbinomh-tfrac12j. More explicitly it is ∑_ℓ=0^min(h,j) binom hℓ 4^ℓ(-1)^j-ℓ binom2(j-ℓ)j-ℓ, an integer sum.
Exact mathematical passage
(1+4z)^{-1/2}=\sum_{k\geq0}(-1)^k\binom{2k}{k}z^k.
|^{1/n}Use in source · 3 passages
Search aliases: |^1/n
Use in source · chapter03.tex, line 321
[External recurrence identification and rates] For the exact normalization [source reference], Zudilin's theorem and its stated asymptotic rates give lim_n→∞fracv_nu_n · =G, lim_n→∞u_n^1/n · =ϕ^5, lim_n→∞|u_nG-v_n|^1/n · =ϕ^-5. These are external mathematical inputs, recorded precisely in source contracts LC01 and LC02 [source reference].
Exact mathematical passage
\lim_{n\to\infty}\frac{v_n}{u_n}&=G,\label{cb_eq_source_limit}\\ \lim_{n\to\infty}u_n^{1/n}&=\phi^5,\label{cb_eq_source_u_rate}\\ \lim_{n\to\infty}|u_nG-v_n|^{1/n}&=\phi^{-5}. \label{cb_eq_source_error_rate}
\alpha_rUse in source · 2 passages
Search aliases: α_r alpha
Use in source · chapter03.tex, line 226
The transfer product can also be written using continuants. For integers 1≤ r≤ s, let K_r,s be the determinant of the tridiagonal matrix whose diagonal is α_r,…,α_s, lower adjacent diagonal is one, and upper adjacent diagonal is -β_r+1,…,-β_s. For instance, K_r,r=α_r, K_r,r+1=det α_r · -β_r+1
Exact mathematical passage
\alpha_r,\ldots,\alpha_s
\alpha_{r+1}Use in source · 1 passage
Search aliases: α_r+1 alpha
Use in source · chapter03.tex, line 229
The transfer product can also be written using continuants. For integers 1≤ r≤ s, let K_r,s be the determinant of the tridiagonal matrix whose diagonal is α_r,…,α_s, lower adjacent diagonal is one, and upper adjacent diagonal is -β_r+1,…,-β_s. For instance, K_r,r=α_r, K_r,r+1=det α_r · -β_r+1 1 · α_r+1 =α_rα_r+1+β_r+1. Set K_r,r-1=1 for the empty determinant. Expansion at the last row gives K_r,s=α_sK_r,s-1+β_sK_r,s-2 (s≥ r+1). The plus sign comes from the minus sign on the upper adjacent diagon…
Exact mathematical passage
K_{r,r}=\alpha_r,\qquad K_{r,r+1}=\det \begin{pmatrix}\alpha_r&-\beta_{r+1}\\1&\alpha_{r+1}\end{pmatrix} =\alpha_r\alpha_{r+1}+\beta_{r+1}.
\alpha_sUse in source · 2 passages
Search aliases: α_s alpha
Use in source · chapter03.tex, line 226
The transfer product can also be written using continuants. For integers 1≤ r≤ s, let K_r,s be the determinant of the tridiagonal matrix whose diagonal is α_r,…,α_s, lower adjacent diagonal is one, and upper adjacent diagonal is -β_r+1,…,-β_s. For instance, K_r,r=α_r, K_r,r+1=det α_r · -β_r+1
Exact mathematical passage
\alpha_r,\ldots,\alpha_s
\beta_{r+1}Use in source · 2 passages
Search aliases: β_r+1 beta
Use in source · chapter03.tex, line 228
The transfer product can also be written using continuants. For integers 1≤ r≤ s, let K_r,s be the determinant of the tridiagonal matrix whose diagonal is α_r,…,α_s, lower adjacent diagonal is one, and upper adjacent diagonal is -β_r+1,…,-β_s. For instance, K_r,r=α_r, K_r,r+1=det α_r · -β_r+1
Exact mathematical passage
-\beta_{r+1},\ldots,-\beta_s
\beta_sUse in source · 2 passages
Search aliases: β_s beta
Use in source · chapter03.tex, line 228
The transfer product can also be written using continuants. For integers 1≤ r≤ s, let K_r,s be the determinant of the tridiagonal matrix whose diagonal is α_r,…,α_s, lower adjacent diagonal is one, and upper adjacent diagonal is -β_r+1,…,-β_s. For instance, K_r,r=α_r, K_r,r+1=det α_r · -β_r+1
Exact mathematical passage
-\beta_{r+1},\ldots,-\beta_s
\gammaUse in source · 2 passages
Search aliases: γ gamma
Use in source · chapters08_10.tex, line 315
Eskandari's coefficient growth statement is lim_n→∞fraclog|widetilde U_n,m(n)|n =γ(α) =-log16+log R+αlog(1+R), where widetilde U=8(-1)^n16^-nU. This is an external asymptotic result in the preprint, not a conclusion of the finite recurrence audit. Together with the filter bound it controls the approximation error |G-frac VU| =frac|widetilde J||widetilde U|. Before any additional quantified gcd saving, the reduced denominator q is at most the absolute cleared coefficient |A|. Its available expone…
Exact mathematical passage
\lim_{n\to\infty}\frac{\log|\widetilde U_{n,m(n)}|}{n} =\gamma(\alpha) =-\log16+\log R+\alpha\log(1+R), \label{eq:gamma-alpha}
\DeltaUse in source · 11 passages
Search aliases: Δ Delta
Use in source · appendices.tex, line 235
The Euler interval uses a_k=(2k+1)^-2 and the positive difference convention Δ a_k=a_k-a_k+1. The identity Δ^ja_0=∫_0^1(1-x^2)^j(-log x) dd x gives 0≤Δ^ja_0≤1. The positive Euler series for G is ∑_j≥02^-j-1Δ^ja_0. Keeping the first 1001 terms therefore gives a lower endpoint and an upper endpoint obtained by adding 2^-1001. All retained terms are rational. The analytic tail bound is a written argument. The computed interval is its finite arithmetic application.
Exact mathematical passage
\Delta a_k=a_k-a_{k+1}
\deltaUse in source · 25 passages
Search aliases: δ delta
Use in source · chapters01_03.tex, line 182
For 0<δ<1, integration by parts gives ∫_δ^1fracarctan xx dd x =-arctan(δ)logδ -∫_δ^1fraclog x1+x^2 dd x. The boundary expression tends to zero as δdownarrow0, since 0≤arctanδ≤δ and δabslogδ→0. The first equality follows from Proposition~[source reference].
Exact mathematical passage
0<\delta<1
\delta'Use in source · 4 passages
Search aliases: δ' delta
Use in source · chapters01_03.tex, line 582
Since D_j≥1, the second hypothesis implies δ≥1. Choose positive numbers λ'>λ and δ'>δ with λ'δ'<1. Such a choice is possible by the strict product inequality, including when λ=0.
Exact mathematical passage
\delta'>\delta
\Delta^0Use in source · 1 passage
Search aliases: Δ^0 Delta
Use in source · chapters01_03.tex, line 316
[Small Euler intervals] The first difference values are Δ^0a_0=1, Δ a_0=frac89, Δ^2a_0=frac184225. They give E_0=frac12, E_1=frac1318, E_2=frac371450. Thus, for example, frac371450<G<frac17091800. These short transformed sums are useful for checking conventions. Their principal advantage appears as N grows, because the certified width decreases geometrically.
Exact mathematical passage
\Delta^0a_0=1,\qquad \Delta a_0=\frac89,\qquad \Delta^2a_0=\frac{184}{225}.
\Delta^2Use in source · 1 passage
Search aliases: Δ^2 Delta
Use in source · chapters01_03.tex, line 316
[Small Euler intervals] The first difference values are Δ^0a_0=1, Δ a_0=frac89, Δ^2a_0=frac184225. They give E_0=frac12, E_1=frac1318, E_2=frac371450. Thus, for example, frac371450<G<frac17091800. These short transformed sums are useful for checking conventions. Their principal advantage appears as N grows, because the certified width decreases geometrically.
Exact mathematical passage
\Delta^0a_0=1,\qquad \Delta a_0=\frac89,\qquad \Delta^2a_0=\frac{184}{225}.
\Delta^jUse in source · 8 passages
Search aliases: Δ^j Delta
Use in source · appendices.tex, line 236
The Euler interval uses a_k=(2k+1)^-2 and the positive difference convention Δ a_k=a_k-a_k+1. The identity Δ^ja_0=∫_0^1(1-x^2)^j(-log x) dd x gives 0≤Δ^ja_0≤1. The positive Euler series for G is ∑_j≥02^-j-1Δ^ja_0. Keeping the first 1001 terms therefore gives a lower endpoint and an upper endpoint obtained by adding 2^-1001. All retained terms are rational. The analytic tail bound is a written argument. The computed interval is its finite arithmetic application.
Exact mathematical passage
\Delta^ja_0=\int_0^1(1-x^2)^j(-\log x)\,\dd x
\delta_0Use in source · 5 passages
Search aliases: δ_0 delta
Use in source · chapters08_10.tex, line 357
[Sufficient content margin] Choose even integers m(n)≥0 such that m(n)/n→α≥0. Use the coefficient clearing theorem, the corrected filter identity, and the prime number theorem as specified above. If there exists a real number δ_0 with liminf_n→∞fraclog g_n,m(n)n≥δ_0 and δ_0>κ_0(α)+log M(α), then G is irrational.
Exact mathematical passage
\delta_0
\varepsilonUse in source · 6 passages
Search aliases: ε varepsilon
Use in source · chapters08_10.tex, line 368
The coefficient clearing theorem and division by the exact gcd make Λ_n=a_nG-b_n with a_n,b_n∈Z. Evenness of m(n) makes widetilde J_n,m(n)>0, so [source reference] gives Λ_n>0. Choose ε>0 so that 3ε<δ_0-κ_0(α)-log M(α). For all sufficiently large n, the clearing asymptotic gives K_n≤exp((κ_0+ε)n), the filter bound gives |widetilde J_n,m(n)|≤exp((log M+ε)n), and the liminf assumption gives g_n,m(n)≥exp((δ_0-ε)n). Therefore 0<Λ_n≤ exp((κ_0+log M-δ_0+3ε)n) ⟶0. If G=a/b with integers a,b and b>0, th…
Exact mathematical passage
\varepsilon>0
\etaUse in source · 3 passages
Search aliases: η eta
Use in source · chapters08_10.tex, line 333
lde J||widetilde U|. Before any additional quantified gcd saving, the reduced denominator q is at most the absolute cleared coefficient |A|. Its available exponent is c_A(α)=4(1+α)+log R+αlog(1+R). The source's resulting guaranteed approximation exponent is η(α)= fraclog R+αlog(1+R)-log16-log M(α) 4(1+α)+log R+αlog(1+R). The source reports a maximum near 0.62335 at α≈3.31765 and states its theorem in the rounded form η≥0.62~[source reference]. The present edition records that as a preprint resul…
Exact mathematical passage
\eta(\alpha)= \frac{\log R+\alpha\log(1+R)-\log16-\log M(\alpha)} {4(1+\alpha)+\log R+\alpha\log(1+R)}. \label{eq:eta-alpha}
\thetaUse in source · 12 passages
Search aliases: θ theta
Use in source · chapters01_03.tex, line 173
[Arctangent and tangent forms] The following equalities hold: G=∫_0^1fracarctan xx dd x =-∫_0^π/4log(tanθ) ddθ. The integrand arctan x/x is assigned its continuous value 1 at x=0.
Exact mathematical passage
G=\int_0^1\frac{\arctan x}{x}\,\dd x =-\int_0^{\pi/4}\log(\tan\theta)\,\dd\theta. \label{eq:c01-tangent-forms}
\kappaUse in source · 16 passages
Search aliases: κ kappa
Use in source · chapters01_03.tex, line 622
t can recover a proved rate] Assume the setup of Corollary~[source reference], except that τ>σ is not required. Discard the finite initial set of indices before nonvanishing holds. For the remaining sufficiently large j, define h_j=gcd(absD_ju_j,absD_jv_j), which is positive because the form is nonzero at those indices. If liminf_j→∞fraclog h_jj≥κ hboxand τ+κ>σ, then α is irrational.
Exact mathematical passage
\liminf_{j\to\infty}\frac{\log h_j}{j}\geq\kappa \quad\hbox{and}\quad \tau+\kappa>\sigma, \label{eq:c02-content-rate}
\kappa_0Use in source · 13 passages
Search aliases: κ_0 kappa
Use in source · chapters08_10.tex, line 254
Set K_n=16^nL_2(n+m(n))^2. For m(n)/n→α, [source reference] gives fraclog K_nn⟶ κ_0(α)=log16+4(1+α). The factor 4 has a precise origin. The lcm index is 2(n+m), and the lcm is squared. Missing either factor would create an incorrect apparent decay margin. The subscript in κ_0 emphasizes that this is the clearing cost alone. Define the combined bound exponent by E(α)=κ_0(α)+log M(α).
Exact mathematical passage
\frac{\log K_n}{n}\longrightarrow \kappa_0(\alpha)=\log16+4(1+\alpha). \label{eq:kappa-alpha}
\lambdaUse in source · 13 passages
Search aliases: λ lambda
Use in source · chapter03.tex, line 299
1 · 0. The characteristic polynomial of this limiting matrix is λ^2-11λ-1. Its two roots are λ_+=frac11+5sqrt52=ϕ^5, λ_-=frac11-5sqrt52=-ϕ^-5. Here ϕ=(1+sqrt5)/2. The identity ϕ^5=5ϕ+3 follows by repeatedly using ϕ^2=ϕ+1. The roots have product -1, as required by the limiting determinant, and very different magnitudes.
Exact mathematical passage
\label{cb_eq_characteristic} \lambda^2-11\lambda-1.
\lambda'Use in source · 4 passages
Search aliases: λ' lambda
Use in source · chapters01_03.tex, line 582
Since D_j≥1, the second hypothesis implies δ≥1. Choose positive numbers λ'>λ and δ'>δ with λ'δ'<1. Such a choice is possible by the strict product inequality, including when λ=0.
Exact mathematical passage
\lambda'>\lambda
\lambda^2Use in source · 1 passage
Search aliases: λ^2 lambda
Use in source · chapter03.tex, line 299
1 · 0. The characteristic polynomial of this limiting matrix is λ^2-11λ-1. Its two roots are λ_+=frac11+5sqrt52=ϕ^5, λ_-=frac11-5sqrt52=-ϕ^-5. Here ϕ=(1+sqrt5)/2. The identity ϕ^5=5ϕ+3 follows by repeatedly using ϕ^2=ϕ+1. The roots have product -1, as required by the limiting determinant, and very different magnitudes.
Exact mathematical passage
\label{cb_eq_characteristic} \lambda^2-11\lambda-1.
\lambda_+Use in source · 1 passage
Search aliases: λ_+ lambda
Use in source · chapter03.tex, line 303
1 · 0. The characteristic polynomial of this limiting matrix is λ^2-11λ-1. Its two roots are λ_+=frac11+5sqrt52=ϕ^5, λ_-=frac11-5sqrt52=-ϕ^-5. Here ϕ=(1+sqrt5)/2. The identity ϕ^5=5ϕ+3 follows by repeatedly using ϕ^2=ϕ+1. The roots have product -1, as required by the limiting determinant, and very different magnitudes.
Exact mathematical passage
\label{cb_eq_characteristic_roots} \lambda_+=\frac{11+5\sqrt5}{2}=\phi^5,\qquad \lambda_-=\frac{11-5\sqrt5}{2}=-\phi^{-5}.
\lambda_-Use in source · 1 passage
Search aliases: λ_- lambda
Use in source · chapter03.tex, line 303
1 · 0. The characteristic polynomial of this limiting matrix is λ^2-11λ-1. Its two roots are λ_+=frac11+5sqrt52=ϕ^5, λ_-=frac11-5sqrt52=-ϕ^-5. Here ϕ=(1+sqrt5)/2. The identity ϕ^5=5ϕ+3 follows by repeatedly using ϕ^2=ϕ+1. The roots have product -1, as required by the limiting determinant, and very different magnitudes.
Exact mathematical passage
\label{cb_eq_characteristic_roots} \lambda_+=\frac{11+5\sqrt5}{2}=\phi^5,\qquad \lambda_-=\frac{11-5\sqrt5}{2}=-\phi^{-5}.
\Lambda_{1,1}Use in source · 1 passage
Search aliases: Λ_1,1 Lambda
Use in source · appendices.tex, line 197
The universal factor uses L_4=12, hence Q=8L_4^2=1152. Its redundant scale is Q/D=128. Consequently g=128·64=8192. The signed cleared pair is (-3 022 848,-2 768 896). Dividing by 8192 gives (-369,-338), so the primitive form is Λ_1,1=338-369G. The integral identity gives the same expression. Since widetilde J_1,1=F_1,1/16 and 16L_4^2/g=9/32, frac16L_4^2gwidetilde J_1,1 =frac9512F_1,1=338-369G. This checks the factor 8, the sign (-1)^n, and the denominator index in one exact example.
Exact mathematical passage
\Lambda_{1,1}=338-369G. \label{eq:app-small-primitive}
\Lambda_nUse in source · 8 passages
Search aliases: Λ_n Lambda
Use in source · chapters08_10.tex, line 262
With the cleared gcd g_n=g_n,m(n), the primitive form is exactly Λ_n=fracK_ng_nwidetilde J_n,m(n). Suppose a future arithmetic argument establishes δ=liminf_n→∞fraclog g_nn. For finite δ, the present analytic estimate yields limsup_n→∞fraclog|Λ_n|n ≤κ_0(α)+log M(α)-δ. The same interpretation applies when δ=+∞: any sufficiently large finite lower bound for the content rate would suffice. All logarithms of nonzero forms are legitimate for the even family by [source reference]. The strict negativit…
Exact mathematical passage
\Lambda_n=\frac{K_n}{g_n}\widetilde J_{n,m(n)}. \label{eq:Lambda-bottleneck}
\muUse in source · 6 passages
Search aliases: μ mu
Use in source · chapters04_07.tex, line 614
On the open unit square, set ψ(x,y)=fracxy(1-x)(1-y)1-xy, ddμ=fracdd x dd ysqrtx(1-y)(1-xy). The factor 1-xy is outside the square root. Moving that factor inside, as in fracdd x dd ysqrtx(1-y)(1-xy), would represent a different integral. With the measure in [source reference], J_n=∫_(0,1)^2ψ^n ddμ, widetilde J_n,m=∫_(0,1)^2ψ^n(1-16ψ)^m ddμ. The second identity follows by the ordinary binomial theorem inside a finite sum. The total measure is J_0=8G<∞.
Exact mathematical passage
\label{cb_eq_psi_mu} \psi(x,y)=\frac{xy(1-x)(1-y)}{1-xy},\qquad \dd\mu=\frac{\dd x\,\dd y}{\sqrt{x(1-y)}(1-xy)}.
\piUse in source · 6 passages
Search aliases: π pi
Use in source · chapters01_03.tex, line 53
quation β(s)=∑_k=0^∞frac(-1)^k(2k+1)^s. Absolute convergence follows from comparison with ∑ j^-s. Thus G=β(2). The alternating series also defines β(s) for s>0, though absolute convergence is no longer available throughout that wider range. At s=1, integration of the finite geometric expansion of (1+x^2)^-1 gives β(1)=π/4. The integrated remainder is bounded by ∫_0^1x^2Ndd x=(2N+1)^-1 and tends to zero, so this passage does not require absolute convergence of the beta series at 1.
Exact mathematical passage
\beta(1)=\pi/4
\rhoUse in source · 7 passages
Search aliases: ρ rho
Use in source · chapters04_07.tex, line 766
[Exactness below the precision ceiling] Let K≥1, let z≠0 be an integer, and let ρ be its least nonnegative residue modulo 2^K. If ρ≠0, then v_2(z)=v_2(ρ)<K. If ρ=0, the residue gives only v_2(z)≥ K.
Exact mathematical passage
\rho
\sigmaUse in source · 9 passages
Search aliases: σ sigma
Use in source · chapters01_03.tex, line 602
[Logarithmic rate accounting] Under the integrality and eventual nonvanishing hypotheses above, suppose limsup_j→∞fraclog D_jj≤σ, limsup_j→∞fraclogabsu_jα-v_jj≤-τ, with finite σ and τ>σ. Then α is irrational.
Exact mathematical passage
\limsup_{j\to\infty}\frac{\log D_j}{j}\leq\sigma, \qquad \limsup_{j\to\infty}\frac{\log\abs{u_j\alpha-v_j}}{j}\leq-\tau, \label{eq:c02-log-bounds}
\tauUse in source · 8 passages
Search aliases: τ tau
Use in source · chapters01_03.tex, line 602
[Logarithmic rate accounting] Under the integrality and eventual nonvanishing hypotheses above, suppose limsup_j→∞fraclog D_jj≤σ, limsup_j→∞fraclogabsu_jα-v_jj≤-τ, with finite σ and τ>σ. Then α is irrational.
Exact mathematical passage
\limsup_{j\to\infty}\frac{\log D_j}{j}\leq\sigma, \qquad \limsup_{j\to\infty}\frac{\log\abs{u_j\alpha-v_j}}{j}\leq-\tau, \label{eq:c02-log-bounds}
\phiUse in source · 12 passages
Search aliases: ϕ phi
Use in source · chapter03.tex, line 303
1 · 0. The characteristic polynomial of this limiting matrix is λ^2-11λ-1. Its two roots are λ_+=frac11+5sqrt52=ϕ^5, λ_-=frac11-5sqrt52=-ϕ^-5. Here ϕ=(1+sqrt5)/2. The identity ϕ^5=5ϕ+3 follows by repeatedly using ϕ^2=ϕ+1. The roots have product -1, as required by the limiting determinant, and very different magnitudes.
Exact mathematical passage
\label{cb_eq_characteristic_roots} \lambda_+=\frac{11+5\sqrt5}{2}=\phi^5,\qquad \lambda_-=\frac{11-5\sqrt5}{2}=-\phi^{-5}.
\phi^{-1}Use in source · 1 passage
Search aliases: ϕ^-1 phi
Use in source · chapters04_07.tex, line 642
The kernel has a simple maximum that also explains why the filter changes sign. Put r=sqrtxy. Since x+y≥2r, ψ(x,y)≤fracr^2(1-r)^21-r^2 =fracr^2(1-r)1+r,qquad0<r<1. Equality is achieved when x=y=r. The derivative of the last expression is frac2r(1-r-r^2)(1+r)^2. It has its interior maximum at r=ϕ^-1, where ϕ=(1+sqrt5)/2. Since 1-r=r^2 there, the maximum is maxψ=ϕ^-5≈0.09017. This is greater than 1/16. One can already see a negative filter value at x=y=1/2, where ψ=1/12 and 1-16ψ=-1/3. Near the si…
Exact mathematical passage
r=\phi^{-1}
\phi^{-10}Use in source · 1 passage
Search aliases: ϕ^-10 phi
Use in source · chapter03.tex, line 354
An immediate consequence of the two source rates is lim_n→∞|G-r_n|^1/n =lim_n→∞frac|u_nG-v_n|^1/nu_n^1/n =ϕ^-10. Thus the ratio error has a smaller exponential scale than the unnormalized linear form. The distinction comes from division by the growing coefficient u_n. Also, since r_n→ G>0, one has r_n^1/n→1 and consequently v_n^1/n→ϕ^5. Both individual sequences grow, while their specified linear combination decays.
Exact mathematical passage
\label{cb_eq_ratio_error_rate} \lim_{n\to\infty}|G-r_n|^{1/n} =\lim_{n\to\infty}\frac{|u_nG-v_n|^{1/n}}{u_n^{1/n}} =\phi^{-10}.
\phi^{-5}Use in source · 4 passages
Search aliases: ϕ^-5 phi
Use in source · chapter03.tex, line 303
1 · 0. The characteristic polynomial of this limiting matrix is λ^2-11λ-1. Its two roots are λ_+=frac11+5sqrt52=ϕ^5, λ_-=frac11-5sqrt52=-ϕ^-5. Here ϕ=(1+sqrt5)/2. The identity ϕ^5=5ϕ+3 follows by repeatedly using ϕ^2=ϕ+1. The roots have product -1, as required by the limiting determinant, and very different magnitudes.
Exact mathematical passage
\label{cb_eq_characteristic_roots} \lambda_+=\frac{11+5\sqrt5}{2}=\phi^5,\qquad \lambda_-=\frac{11-5\sqrt5}{2}=-\phi^{-5}.
\phi^2Use in source · 1 passage
Search aliases: ϕ^2 phi
Use in source · chapter03.tex, line 308
The characteristic polynomial of this limiting matrix is λ^2-11λ-1. Its two roots are λ_+=frac11+5sqrt52=ϕ^5, λ_-=frac11-5sqrt52=-ϕ^-5. Here ϕ=(1+sqrt5)/2. The identity ϕ^5=5ϕ+3 follows by repeatedly using ϕ^2=ϕ+1. The roots have product -1, as required by the limiting determinant, and very different magnitudes.
Exact mathematical passage
\phi^2=\phi+1
\phi^5Use in source · 6 passages
Search aliases: ϕ^5 phi
Use in source · chapter03.tex, line 303
1 · 0. The characteristic polynomial of this limiting matrix is λ^2-11λ-1. Its two roots are λ_+=frac11+5sqrt52=ϕ^5, λ_-=frac11-5sqrt52=-ϕ^-5. Here ϕ=(1+sqrt5)/2. The identity ϕ^5=5ϕ+3 follows by repeatedly using ϕ^2=ϕ+1. The roots have product -1, as required by the limiting determinant, and very different magnitudes.
Exact mathematical passage
\label{cb_eq_characteristic_roots} \lambda_+=\frac{11+5\sqrt5}{2}=\phi^5,\qquad \lambda_-=\frac{11-5\sqrt5}{2}=-\phi^{-5}.
\psi^nUse in source · 3 passages
Search aliases: ψ^n psi
Use in source · chapters04_07.tex, line 623
On the open unit square, set ψ(x,y)=fracxy(1-x)(1-y)1-xy, ddμ=fracdd x dd ysqrtx(1-y)(1-xy). The factor 1-xy is outside the square root. Moving that factor inside, as in fracdd x dd ysqrtx(1-y)(1-xy), would represent a different integral. With the measure in [source reference], J_n=∫_(0,1)^2ψ^n ddμ, widetilde J_n,m=∫_(0,1)^2ψ^n(1-16ψ)^m ddμ. The second identity follows by the ordinary binomial theorem inside a finite sum. The total measure is J_0=8G<∞.
Exact mathematical passage
J_n=\int_{(0,1)^2}\psi^n\,\dd\mu, \qquad \widetilde J_{n,m}=\int_{(0,1)^2}\psi^n(1-16\psi)^m\,\dd\mu.
\ellUse in source · 8 passages
Search aliases: ℓ ell
Use in source · chapters01_03.tex, line 513
The second example can also be written as a failure of denominator clearing. The rational coefficient form ℓ_n=α-(frac12+2^-n)=-2^-n is nonzero and tends to zero. Its second coefficient is not an integer. Multiplication by 2^n makes both coefficients integral and gives the constant form -1. The analytic gain has been exactly consumed by the arithmetic scale.
Exact mathematical passage
\ell_n=\alpha-\left(\frac12+2^{-n}\right)=-2^{-n}
\ell_jUse in source · 3 passages
Search aliases: ℓ_j ell
Use in source · chapters01_03.tex, line 523
In the later chapters the starting form is ℓ_j=u_jα-v_j, u_j,v_j∈Q. Choose a positive integer D_j for which D_ju_j and D_jv_j are integers. Then L_j=D_jℓ_j=(D_ju_j)α-D_jv_j has integer coefficients. Its size is D_jabsℓ_j, so the clearance factor must be included in any decay estimate.
Exact mathematical passage
\ell_j=u_j\alpha-v_j,\qquad u_j,v_j\in\Q. \label{eq:c02-rational-form}
\ell_nUse in source · 1 passage
Search aliases: ℓ_n ell
Use in source · chapters01_03.tex, line 513
The second example can also be written as a failure of denominator clearing. The rational coefficient form ℓ_n=α-(frac12+2^-n)=-2^-n is nonzero and tends to zero. Its second coefficient is not an integer. Multiplication by 2^n makes both coefficients integral and gives the constant form -1. The analytic gain has been exactly consumed by the arithmetic scale.
Exact mathematical passage
\ell_n=\alpha-\left(\frac12+2^{-n}\right)=-2^{-n}
\toUse in source · 24 passages
Search aliases: → to
Use in source · chapter03.tex, line 294
The leading terms of the coefficient polynomials are a_n=320n^6+O(n^5), q(n)=3520n^6+O(n^5), c_n=320n^6+O(n^5). Consequently α_n→11, β_n→1, and M_n⟶ M_∞=11 · 1
Exact mathematical passage
\alpha_n\to11
\prodUse in source · 2 passages
Search aliases: ∏ prod
Use in source · chapter03.tex, line 211
1 · 0=-frac138. Multiplicativity of determinants now gives det F_N+1 =-frac138∏_n=1^N(-β_n)≠0. Thus adjacent values of the two solutions remain independent at every index. Each transfer reverses the determinant sign. The closed evaluation of this product is the Wronskian theorem in Chapter~[source reference], where the factors telescope and yield exact quotient increments. Here the matrix description identifies the mechanism without duplicating that full proof.
Exact mathematical passage
\det\mathcal F_{N+1} =-\frac{13}{8}\prod_{n=1}^N(-\beta_n)\ne0.
\prod_{j=1}^nUse in source · 1 passage
Search aliases: ∏_j=1^n prod
Use in source · chapters04_07.tex, line 54
At n=0 the determinant is -13/8, which agrees with the displayed formula because p(1)=13. For n≥1, insert [source reference] into both terms of [source reference]. The terms containing q(n)u_nv_n cancel, leaving W_n=fracc_na_n(u_n-1v_n-u_nv_n-1) =-fracc_na_nW_n-1. Iterating this equality gives W_n=(-1)^nW_0 ∏_j=1^n frac(2j-1)^2(2j)^2p(j+1)(2j+1)^2(2j+2)^2p(j). There are two independent telescopes. The polynomial factors leave p(n+1)/p(1). The integer factors leave 4/((2n+1)^2(2n+2)^2). Substitut…
Exact mathematical passage
W_n=(-1)^nW_0 \prod_{j=1}^n \frac{(2j-1)^2(2j)^2p(j+1)}{(2j+1)^2(2j+2)^2p(j)}.
\prod_{n=1}^NUse in source · 1 passage
Search aliases: ∏_n=1^N prod
Use in source · chapter03.tex, line 211
1 · 0=-frac138. Multiplicativity of determinants now gives det F_N+1 =-frac138∏_n=1^N(-β_n)≠0. Thus adjacent values of the two solutions remain independent at every index. Each transfer reverses the determinant sign. The closed evaluation of this product is the Wronskian theorem in Chapter~[source reference], where the factors telescope and yield exact quotient increments. Here the matrix description identifies the mechanism without duplicating that full proof.
Exact mathematical passage
\det\mathcal F_{N+1} =-\frac{13}{8}\prod_{n=1}^N(-\beta_n)\ne0.
\sum_{i=0}^{66}Use in source · 1 passage
Search aliases: ∑_i=0^66 sum
Use in source · chapters08_10.tex, line 128
The first row illustrates the exact check without printing its very large numerator and denominator. The independently reduced rational packet has denominator D=den(∑_i=0^66binom66i16^20+iv_20+i). Its recorded quotient Q/D is 1 703 936, and the minimal pair has gcd 8. Multiplication gives 1 703 936·8=13 631 488, exactly the historical value. The complete decimal integers U,N,D,Q are retained in the diagnostic JSON. They are inputs to an exact equality check, rather than values inferred from roun…
Exact mathematical passage
D=\operatorname{den}\left(\sum_{i=0}^{66}\binom{66}{i}16^{20+i}v_{20+i}\right).
\sum_{i=0}^{m}Use in source · 1 passage
Search aliases: ∑_i=0^m sum
Use in source · chapters08_10.tex, line 9
Fix integers n,m≥0, and write U=∑_i=0^mbinom miT_n+i, V=∑_i=0^mbinom mi16^n+iv_n+i. These are fixed-pair aliases U=S_n,m and V=H_n,m for the packets defined earlier, not the integral coefficients of a different source normalization. Here T_k=16^ku_k is a positive integer. Consequently U is a positive integer. Write V=N/D in lowest terms, with D>0 and gcd(N,D)=1. In the exceptional case V=0, the same convention sets N=0,D=1. Set ℓ=2(n+m), L_ℓ=lcm(1,…,ℓ), Q=8L_ℓ^2, with L_0=1.
Exact mathematical passage
U=\sum_{i=0}^{m}\binom miT_{n+i},\qquad V=\sum_{i=0}^{m}\binom mi16^{n+i}v_{n+i}. \label{eq:primitive-UV}
\sum_{j=0}^nUse in source · 2 passages
Search aliases: ∑_j=0^n sum
Use in source · appendices.tex, line 49
Set T_n=16^nu_n. Its forward recurrence is a_nT_n+1=16q(n)T_n+256c_nT_n-1, T_0=1, T_1=28. The initial terms are 1,28,2596,311536. The universally integral expression is u_n=∑_j=0^nbinom nj binomn-tfrac12jbinomn+j-tfrac12j, with the external coefficient identity and the written half-binomial denominator argument kept distinct.
Exact mathematical passage
u_n=\sum_{j=0}^n\binom nj \binom{n-\tfrac12}{j}\binom{n+j-\tfrac12}{j}, \label{eq:app-half-binomial}
\sum_{j=0}^{N}Use in source · 2 passages
Search aliases: ∑_j=0^N sum
Use in source · chapters01_03.tex, line 267
[Certified Euler enclosure] Define E_N=∑_j=0^NfracΔ^j a_02^j+1, N≥0. Then G=∑_j=0^∞fracΔ^j a_02^j+1, E_N<G<E_N+2^-N-1. More precisely, G-E_N =∫_0^1frac-log x1+x^2 (frac1-x^22)^N+1 dd x.
Exact mathematical passage
E_N=\sum_{j=0}^{N}\frac{\Delta^j a_0}{2^{j+1}},\qquad N\geq0. \label{eq:c01-E}
\sum_{j=0}^{n-1}Use in source · 1 passage
Search aliases: ∑_j=0^n-1 sum
Use in source · chapters04_07.tex, line 162
Equation [source reference] and the preceding lemma identify r_n with partial sums of a convergent alternating series: r_n=∑_j=0^n-1(-1)^j d_j, L=∑_j=0^∞(-1)^j d_j. For the even subsequence, r_2k+2-r_2k=d_2k-d_2k+1>0. For the odd subsequence, r_2k+3-r_2k+1=-d_2k+1+d_2k+2<0. Their adjacent separation d_2k tends to zero, so they have the same limit. Strictness of both subsequences places that limit strictly between every displayed pair. The alternating tail beginning with d_n gives the sign and up…
Exact mathematical passage
r_n=\sum_{j=0}^{n-1}(-1)^j d_j, \qquad L=\sum_{j=0}^{\infty}(-1)^j d_j.
\sum_{j=0}^{\infty}Use in source · 3 passages
Search aliases: ∑_j=0^∞ sum infty
Use in source · chapters01_03.tex, line 272
[Certified Euler enclosure] Define E_N=∑_j=0^NfracΔ^j a_02^j+1, N≥0. Then G=∑_j=0^∞fracΔ^j a_02^j+1, E_N<G<E_N+2^-N-1. More precisely, G-E_N =∫_0^1frac-log x1+x^2 (frac1-x^22)^N+1 dd x.
Exact mathematical passage
G=\sum_{j=0}^{\infty}\frac{\Delta^j a_0}{2^{j+1}}, \qquad E_N<G<E_N+2^{-N-1}. \label{eq:c01-euler-enclosure}
\sum_{j\geq0}Use in source · 1 passage
Search aliases: ∑_j≥0 sum geq
Use in source · appendices.tex, line 239
The Euler interval uses a_k=(2k+1)^-2 and the positive difference convention Δ a_k=a_k-a_k+1. The identity Δ^ja_0=∫_0^1(1-x^2)^j(-log x) dd x gives 0≤Δ^ja_0≤1. The positive Euler series for G is ∑_j≥02^-j-1Δ^ja_0. Keeping the first 1001 terms therefore gives a lower endpoint and an upper endpoint obtained by adding 2^-1001. All retained terms are rational. The analytic tail bound is a written argument. The computed interval is its finite arithmetic application.
Exact mathematical passage
\sum_{j\geq0}2^{-j-1}\Delta^ja_0
\sum_{k=0}^{j}Use in source · 2 passages
Search aliases: ∑_k=0^j sum
Use in source · chapters01_03.tex, line 239
Let a_k=frac1(2k+1)^2, k≥0, and define the positive difference convention Δ a_k=a_k-a_k+1. The sign convention is fixed here because the other common convention would change every odd difference. Iteration gives Δ^j a_0=∑_k=0^j(-1)^kbinom jkfrac1(2k+1)^2. This is an exact rational number for every j.
Exact mathematical passage
\Delta^j a_0=\sum_{k=0}^{j}(-1)^k\binom jk\frac1{(2k+1)^2}. \label{eq:c01-finite-difference}
\sum_{k=0}^{N-1}Use in source · 1 passage
Search aliases: ∑_k=0^N-1 sum
Use in source · chapters01_03.tex, line 64
Write s_N=∑_k=0^N-1frac(-1)^k(2k+1)^2, N≥0, where s_0=0 is the empty sum. Thus N counts retained terms, as it does in the visual guide. Each s_N is an exact rational number. The terms decrease strictly in magnitude to zero. Grouping successive terms shows that the even partial sums increase and the odd partial sums decrease. These two sequences approach the same limit.
Exact mathematical passage
s_N=\sum_{k=0}^{N-1}\frac{(-1)^k}{(2k+1)^2},\qquad N\geq0, \label{eq:c01-partial-sum}
\sum_{k\geq0}Use in source · 1 passage
Search aliases: ∑_k≥0 sum geq
Use in source · chapters04_07.tex, line 298
Work with formal power series, so no convergence issue arises. The central binomial expansion gives (1+4z)^-1/2=∑_k≥0(-1)^kbinom2kkz^k. Every coefficient is an integer. Multiplying by the integer polynomial (1+4z)^h preserves this property. The coefficient of z^j in the product is 4^jbinomh-tfrac12j. More explicitly it is ∑_ℓ=0^min(h,j) binom hℓ 4^ℓ(-1)^j-ℓ binom2(j-ℓ)j-ℓ, an integer sum.
Exact mathematical passage
(1+4z)^{-1/2}=\sum_{k\geq0}(-1)^k\binom{2k}{k}z^k.
\sum_{n=1}^NUse in source · 1 passage
Search aliases: ∑_n=1^N sum
Use in source · chapters04_07.tex, line 806
For a positive integer N, the scanned domain is D_N={(n,m):1≤ n≤ N, 0≤ m≤4n}. The number of scanned pairs is ∑_n=1^N(4n+1)=2N^2+3N. The upper slope 4 is part of the experiment's definition. The scan supplies no information about pairs above this boundary or with n>N.
Exact mathematical passage
\label{cb_eq_scan_count} \sum_{n=1}^N(4n+1)=2N^2+3N.
\sum_{\ell=0}^{\min(h,j)}Use in source · 1 passage
Search aliases: ∑_ℓ=0^min(h,j) sum ell min
Use in source · chapters04_07.tex, line 304
Work with formal power series, so no convergence issue arises. The central binomial expansion gives (1+4z)^-1/2=∑_k≥0(-1)^kbinom2kkz^k. Every coefficient is an integer. Multiplying by the integer polynomial (1+4z)^h preserves this property. The coefficient of z^j in the product is 4^jbinomh-tfrac12j. More explicitly it is ∑_ℓ=0^min(h,j) binom hℓ 4^ℓ(-1)^j-ℓ binom2(j-ℓ)j-ℓ, an integer sum.
Exact mathematical passage
\sum_{\ell=0}^{\min(h,j)} \binom h\ell 4^\ell(-1)^{j-\ell} \binom{2(j-\ell)}{j-\ell},
\intUse in source · 21 passages
Search aliases: ∫ int
Use in source · appendices.tex, line 71
widetilde J_n,m · =∑_i=0^m(-16)^ibinom miJ_n+i =16^-nF_n,m. The corrected unfiltered integral is J_n=∫_0^1 ∫_0^1 fracx^n-1/2(1-x)^ny^n(1-y)^n-1/2 (1-xy)^n+1 dd x dd y =8(-1)^n(u_nG-v_n). These are fixed normalization identities. Their factors should be checked together when translating a differently normalized source.
Exact mathematical passage
J_n=\int_0^1\!\int_0^1 \frac{x^{n-1/2}(1-x)^ny^n(1-y)^{n-1/2}} {(1-xy)^{n+1}}\,\dd x\,\dd y =8(-1)^n(u_nG-v_n). \label{eq:app-J}
\int_{(0,1)^2}Use in source · 1 passage
Search aliases: ∫_(0,1)^2 int
Use in source · chapters04_07.tex, line 623
On the open unit square, set ψ(x,y)=fracxy(1-x)(1-y)1-xy, ddμ=fracdd x dd ysqrtx(1-y)(1-xy). The factor 1-xy is outside the square root. Moving that factor inside, as in fracdd x dd ysqrtx(1-y)(1-xy), would represent a different integral. With the measure in [source reference], J_n=∫_(0,1)^2ψ^n ddμ, widetilde J_n,m=∫_(0,1)^2ψ^n(1-16ψ)^m ddμ. The second identity follows by the ordinary binomial theorem inside a finite sum. The total measure is J_0=8G<∞.
Exact mathematical passage
J_n=\int_{(0,1)^2}\psi^n\,\dd\mu, \qquad \widetilde J_{n,m}=\int_{(0,1)^2}\psi^n(1-16\psi)^m\,\dd\mu.
\int_0^1Use in source · 15 passages
Search aliases: ∫_0^1 int
Use in source · appendices.tex, line 71
widetilde J_n,m · =∑_i=0^m(-16)^ibinom miJ_n+i =16^-nF_n,m. The corrected unfiltered integral is J_n=∫_0^1 ∫_0^1 fracx^n-1/2(1-x)^ny^n(1-y)^n-1/2 (1-xy)^n+1 dd x dd y =8(-1)^n(u_nG-v_n). These are fixed normalization identities. Their factors should be checked together when translating a differently normalized source.
Exact mathematical passage
J_n=\int_0^1\!\int_0^1 \frac{x^{n-1/2}(1-x)^ny^n(1-y)^{n-1/2}} {(1-xy)^{n+1}}\,\dd x\,\dd y =8(-1)^n(u_nG-v_n). \label{eq:app-J}
\int_0^{\pi/2}Use in source · 2 passages
Search aliases: ∫_0^π/2 int pi
Use in source · chapters04_07.tex, line 428
The initial integral supplies a short diagnostic. Substitute x=t^2 and y=1-s^2. Then J_0 · =4∫_0^1 ∫_0^1fracdd s dd t1-t^2+t^2s^2 · =4∫_0^1fracarcsin ttsqrt1-t^2 dd t =4∫_0^π/2θcscθ ddθ. Since an antiderivative of cscθ is logtan(θ/2), integration by parts yields J_0=-4∫_0^π/2logtan(θ/2) ddθ =-8∫_0^π/4logtan z dd z=8G. Both boundary terms vanish. At zero this uses θlogtan(θ/2)→0, and at π/2 the logarithm is zero.
Exact mathematical passage
J_0 &=4\int_0^1\!\int_0^1\frac{\dd s\,\dd t}{1-t^2+t^2s^2}\\ &=4\int_0^1\frac{\arcsin t}{t\sqrt{1-t^2}}\,\dd t =4\int_0^{\pi/2}\theta\csc\theta\,\dd\theta.
\int_0^{\pi/4}Use in source · 2 passages
Search aliases: ∫_0^π/4 int pi
Use in source · chapters01_03.tex, line 173
[Arctangent and tangent forms] The following equalities hold: G=∫_0^1fracarctan xx dd x =-∫_0^π/4log(tanθ) ddθ. The integrand arctan x/x is assigned its continuous value 1 at x=0.
Exact mathematical passage
G=\int_0^1\frac{\arctan x}{x}\,\dd x =-\int_0^{\pi/4}\log(\tan\theta)\,\dd\theta. \label{eq:c01-tangent-forms}
\int_0^{\infty}Use in source · 1 passage
Search aliases: ∫_0^∞ int infty
Use in source · chapters01_03.tex, line 138
Set x=e^-t. Then -log x=t, dd x=-e^-tdd t, and the integral becomes ∫_0^∞t e^-(a+1)t dd t. Integration by parts gives (a+1)^-2. The boundary term at infinity is zero because an exponential with positive decay rate dominates t. The boundary term at zero is also zero.
Exact mathematical passage
\int_0^{\infty}t e^{-(a+1)t}\,\dd t.
\int_{[0,1]^2}Use in source · 2 passages
Search aliases: ∫_[0,1]^2 int
Use in source · chapters08_10.tex, line 176
Write ϕ=frac1+sqrt52, t=ϕ^-5, R=16ϕ^5=88+40sqrt5. The unfiltered integral has the representation J_k=∫_[0,1]^2ψ(x,y)^k ddμ(x,y), ψ=fracxy(1-x)(1-y)1-xy, ddμ=fracdd x dd ysqrtx(1-y)(1-xy). The measure has finite total mass J_0=8G. The factor 1-xy in its denominator lies outside the square root. Moving it inside changes the integral and invalidates the coefficient identification.
Exact mathematical passage
J_k=\int_{[0,1]^2}\psi(x,y)^k\,\dd\mu(x,y), \quad \psi=\frac{xy(1-x)(1-y)}{1-xy}, \quad \dd\mu=\frac{\dd x\,\dd y}{\sqrt{x(1-y)}(1-xy)}. \label{eq:bottleneck-measure}
\int_\delta^1Use in source · 1 passage
Search aliases: ∫_δ^1 int delta
Use in source · chapters01_03.tex, line 183
For 0<δ<1, integration by parts gives ∫_δ^1fracarctan xx dd x =-arctan(δ)logδ -∫_δ^1fraclog x1+x^2 dd x. The boundary expression tends to zero as δdownarrow0, since 0≤arctanδ≤δ and δabslogδ→0. The first equality follows from Proposition~[source reference].
Exact mathematical passage
\int_\delta^1\frac{\arctan x}{x}\,\dd x =-\arctan(\delta)\log\delta -\int_\delta^1\frac{\log x}{1+x^2}\,\dd x.
\approxUse in source · 3 passages
Search aliases: ≈ approx
Use in source · chapters01_03.tex, line 25
The familiar decimal display begins G≈0.91596559417721901505460351493238411077. The approximation sign matters. A decimal is a coordinate for seeing the number, while a rational enclosure is evidence for a particular numerical assertion. A proof of irrationality requires a different assertion again.
Exact mathematical passage
G\approx0.91596559417721901505460351493238411077.
\neUse in source · 14 passages
Search aliases: ≠ ne
Use in source · chapter03.tex, line 211
1 · 0=-frac138. Multiplicativity of determinants now gives det F_N+1 =-frac138∏_n=1^N(-β_n)≠0. Thus adjacent values of the two solutions remain independent at every index. Each transfer reverses the determinant sign. The closed evaluation of this product is the Wronskian theorem in Chapter~[source reference], where the factors telescope and yield exact quotient increments. Here the matrix description identifies the mechanism without duplicating that full proof.
Exact mathematical passage
\det\mathcal F_{N+1} =-\frac{13}{8}\prod_{n=1}^N(-\beta_n)\ne0.
\cdotsUse in source · 2 passages
Search aliases: ⋯ cdots
Use in source · chapter03.tex, line 181
1 · 0. X_n+1(w)=M_nX_n(w). Hence X_N+1(w)=M_NM_N-1⋯ M_1X_1(w) (N≥1). The order of multiplication follows the temporal order: M_1 acts first. This is a product of changing matrices, not a power of one fixed matrix.
Exact mathematical passage
X_{N+1}(w)=M_NM_{N-1}\cdots M_1X_1(w)\qquad(N\geq1).
\longrightarrowUse in source · 8 passages
Search aliases: ⟶ longrightarrow
Use in source · chapter03.tex, line 295
The leading terms of the coefficient polynomials are a_n=320n^6+O(n^5), q(n)=3520n^6+O(n^5), c_n=320n^6+O(n^5). Consequently α_n→11, β_n→1, and M_n⟶ M_∞=11 · 1 1 · 0. The characteristic polynomial of this limiting matrix is λ^2-11λ-1. Its two roots are λ_+=frac11+5sqrt52=ϕ^5, λ_-=frac11-5sqrt52=-ϕ^-5. Here ϕ=(1+sqrt5)/2. The identity ϕ^5=5ϕ+3 follows by repeatedly using ϕ^2=ϕ+1. The roots have product -1, as required by the limiting determinant, and very different magnitudes.
Exact mathematical passage
M_n\longrightarrow M_\infty=\begin{pmatrix}11&1\\1&0\end{pmatrix}.
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