"Slater 92"의 두 판 사이의 차이

2020년 12월 28일 (월) 04:02 기준 최신판

Note

an explanation for dilogarithm ladder twisted Chebyshev polynomials and dilogarithm identities
Loxton & Lewin\(x, -y, -z^{-1}\)가 방정식 \(x^3+3x^2-1=0\)의 해라고 하자.\(3L(x^3)-9L(x^2)-9L(x)+7L(1)=0\)\(3L(y^6)-6L(y^3)-27L(y^2)+18L(y)+2L(1)=0\)\(3L(z^6)-6L(z^3)-27L(z^2)+18L(z)-2L(1)=0\)

type of identity

Slater's list
B(3)

Bailey pair 1

Use the folloing\(\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\), \(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\)
Specialize\(x=q^2, y=-q, z\to\infty\).
Bailey pair\(\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}\)\(\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}\)

Bailey pair 2

Use the following \(\sum_{r=0}^{n}\frac{(1-aq^{2r})(-1)^{r}q^{\frac{1}{2}(r^2+r)}(a)_{r}(c)_{r}(d)_{r}a^{r}}{(a)_{n+r+1}(q)_{n-r}(q)_{r}(aq/c)_{r}(aq/d)_{r}c^{r}d^{r}}=\frac{(aq/cd)_{n}}{(q)_{n}(aq/c)_{n}(aq/d)_{n}}\)
Specialize\(a=q,c=-q,d=\infty\)
Bailey pair\(\alpha_{0}=1\), \(\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)\)\(\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{2})_{n-r}(q)_{n+r}}=\frac{1}{(q)_{n}(-q)_{n}}\)

Bailey pair

Bailey pairs\(\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}\)\(\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}\)\(\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)\)\(\beta_n=\frac{1}{(q)_{n}(-q)_{n}}\)

q-series identity

\(\sum_{n=0}^{\infty}\frac{(q^3;q^3)_{n}q^{n(n+1)}}{ (q)_{n}(q;q^{2})_n(q^2;q^2)_{n}}=\frac{(q^{9};q^{27})_{\infty}(q^{18};q^{27})_{\infty}(q^{27};q^{27})_{\infty}}{(q)_{\infty}}\)

The On-Line Encyclopedia of Integer Sequences
- http://www.research.att.com/~njas/sequences/?q=

Bethe type equation (cyclotomic equation)

Let \(\sum_{n=0}^{\infty}\frac{q^{n(an+b)/2}}{ \prod_{j=1}^{r}(q^{c_j};q^{d_j})_n^{e_j}}=\sum_{N=0}^{\infty} a_N q^{N}\).

Then \(\prod_{j=1}^{r}(1-x^{d_j})^{e_j}=x^a\) has a unique root \(0<\mu<1\). We get

\(\log^2 a_N \sim 4N\sum_{j=1}^{r}\frac{e_j}{d_j}L(1-\mu^{d_j})\)

a=2,d_1=1,d_2=2,d_3=2,e_1=e_2=e_3=1

\(\frac{(1-x)(1-x^2)^2}{(1-x^3)}=x^2\)

\(x^3+3x^2-1=0\)

\(x, -y, -z^{-1}\)가 방정식 의 해 http://www.wolframalpha.com/input/?i=x^3%2B3x^2-1%3D0

dilogarithm identity

\(L(x^3)-3L(x^2)-3L(x)=-\frac{7}{3}L(1)\)

related items

@@ 1번째 줄: / 1번째 줄: @@
 ==Note==
-* [[twisted Chebyshev polynomials and dilogarithm identities|an explanation for dilogarithm ladder]]<br>[[twisted Chebyshev polynomials and dilogarithm identities|twisted Chebyshev polynomials and dilogarithm identities]]<br>
+* [[twisted Chebyshev polynomials and dilogarithm identities|an explanation for dilogarithm ladder]][[twisted Chebyshev polynomials and dilogarithm identities|twisted Chebyshev polynomials and dilogarithm identities]]
-*  Loxton & Lewin<br><math>x, -y, -z^{-1}</math>가 방정식 <math>x^3+3x^2-1=0</math>의 해라고 하자.<br><math>3L(x^3)-9L(x^2)-9L(x)+7L(1)=0</math><br><math>3L(y^6)-6L(y^3)-27L(y^2)+18L(y)+2L(1)=0</math><br><math>3L(z^6)-6L(z^3)-27L(z^2)+18L(z)-2L(1)=0</math><br>
+*  Loxton & Lewin<math>x, -y, -z^{-1}</math>가 방정식 <math>x^3+3x^2-1=0</math>의 해라고 하자.<math>3L(x^3)-9L(x^2)-9L(x)+7L(1)=0</math><math>3L(y^6)-6L(y^3)-27L(y^2)+18L(y)+2L(1)=0</math><math>3L(z^6)-6L(z^3)-27L(z^2)+18L(z)-2L(1)=0</math>
 ==type of identity==
 * [[Slater list|Slater's list]]
-*  B(3)<br>
+*  B(3)
 ==Bailey pair 1==
-*  Use the folloing<br><math>\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}</math>,  <math>\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}</math><br>
+*  Use the folloing<math>\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}</math>,  <math>\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}</math>
-*  Specialize<br><math>x=q^2, y=-q, z\to\infty</math>.<br>
+*  Specialize<math>x=q^2, y=-q, z\to\infty</math>.
-*  Bailey pair<br><math>\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}</math><br><math>\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}</math><br>
+*  Bailey pair<math>\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}</math><math>\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}</math>
 ==Bailey pair 2==
-*  Use the following <br><math>\sum_{r=0}^{n}\frac{(1-aq^{2r})(-1)^{r}q^{\frac{1}{2}(r^2+r)}(a)_{r}(c)_{r}(d)_{r}a^{r}}{(a)_{n+r+1}(q)_{n-r}(q)_{r}(aq/c)_{r}(aq/d)_{r}c^{r}d^{r}}=\frac{(aq/cd)_{n}}{(q)_{n}(aq/c)_{n}(aq/d)_{n}}</math><br>
+*  Use the following <math>\sum_{r=0}^{n}\frac{(1-aq^{2r})(-1)^{r}q^{\frac{1}{2}(r^2+r)}(a)_{r}(c)_{r}(d)_{r}a^{r}}{(a)_{n+r+1}(q)_{n-r}(q)_{r}(aq/c)_{r}(aq/d)_{r}c^{r}d^{r}}=\frac{(aq/cd)_{n}}{(q)_{n}(aq/c)_{n}(aq/d)_{n}}</math>
-*  Specialize<br><math>a=q,c=-q,d=\infty</math><br>
+*  Specialize<math>a=q,c=-q,d=\infty</math>
-*  Bailey pair<br><math>\alpha_{0}=1</math>, <math>\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)</math><br><math>\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{2})_{n-r}(q)_{n+r}}=\frac{1}{(q)_{n}(-q)_{n}}</math><br>
+*  Bailey pair<math>\alpha_{0}=1</math>, <math>\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)</math><math>\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{2})_{n-r}(q)_{n+r}}=\frac{1}{(q)_{n}(-q)_{n}}</math>
-==Bailey pair ==
+==Bailey pair ==
-*  Bailey pairs<br><math>\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}</math><br><math>\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}</math><br><math>\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)</math><br><math>\beta_n=\frac{1}{(q)_{n}(-q)_{n}}</math><br>
+*  Bailey pairs<math>\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}</math><math>\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}</math><math>\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)</math><math>\beta_n=\frac{1}{(q)_{n}(-q)_{n}}</math>
 ==q-series identity==
@@ 49번째 줄: / 49번째 줄: @@
 <math>\sum_{n=0}^{\infty}\frac{(q^3;q^3)_{n}q^{n(n+1)}}{ (q)_{n}(q;q^{2})_n(q^2;q^2)_{n}}=\frac{(q^{9};q^{27})_{\infty}(q^{18};q^{27})_{\infty}(q^{27};q^{27})_{\infty}}{(q)_{\infty}}</math>
-* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]<br>
+* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]
 ** http://www.research.att.com/~njas/sequences/?q=
 ==Bethe type equation (cyclotomic equation)==
-Let <math>\sum_{n=0}^{\infty}\frac{q^{n(an+b)/2}}{
+Let <math>\sum_{n=0}^{\infty}\frac{q^{n(an+b)/2}}{
   \prod_{j=1}^{r}(q^{c_j};q^{d_j})_n^{e_j}}=\sum_{N=0}^{\infty} a_N q^{N}</math>.
-Then <math>\prod_{j=1}^{r}(1-x^{d_j})^{e_j}=x^a</math>  has a unique root <math>0<\mu<1</math>. We get
+Then <math>\prod_{j=1}^{r}(1-x^{d_j})^{e_j}=x^a</math>  has a unique root <math>0<\mu<1</math>. We get
 <math>\log^2 a_N \sim 4N\sum_{j=1}^{r}\frac{e_j}{d_j}L(1-\mu^{d_j})</math>
 a=2,d_1=1,d_2=2,d_3=2,e_1=e_2=e_3=1
 <math>\frac{(1-x)(1-x^2)^2}{(1-x^3)}=x^2</math>
@@ 79번째 줄: / 79번째 줄: @@
 <math>x^3+3x^2-1=0</math>
-<math>x, -y, -z^{-1}</math>가 방정식 의 해 [http://www.wolframalpha.com/input/?i=x%5E3%2B3x%5E2-1%3D0 http://www.wolframalpha.com/input/?i=x^3%2B3x^2-1%3D0]
+<math>x, -y, -z^{-1}</math>가 방정식 의 해 [http://www.wolframalpha.com/input/?i=x%5E3%2B3x%5E2-1%3D0 http://www.wolframalpha.com/input/?i=x^3%2B3x^2-1%3D0]
 ==dilogarithm identity==
@@ 89번째 줄: / 89번째 줄: @@
 <math>L(x^3)-3L(x^2)-3L(x)=-\frac{7}{3}L(1)</math>
 ==related items==
-==books==
-* [[2010년 books and articles]]<br>
-* http://gigapedia.info/1/
-* http://gigapedia.info/1/
-* http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=
-==articles==
-* http://www.ams.org/mathscinet
-* [http://www.zentralblatt-math.org/zmath/en/ ]http://www.zentralblatt-math.org/zmath/en/
-* [http://arxiv.org/ ]http://arxiv.org/
-* http://pythagoras0.springnote.com/
-* [http://math.berkeley.edu/%7Ereb/papers/index.html http://math.berkeley.edu/~reb/papers/index.html]
-* http://dx.doi.org/
-*
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