"Slater 92"의 두 판 사이의 차이
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Pythagoras0 (토론 | 기여) |
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| (사용자 3명의 중간 판 14개는 보이지 않습니다) | |||
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| − | + | ==Note== | |
| − | * [[twisted Chebyshev polynomials and dilogarithm identities|an explanation for dilogarithm ladder]] | + | * [[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 | + | * 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> |
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| − | + | ==type of identity== | |
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* [[Slater list|Slater's list]] | * [[Slater list|Slater's list]] | ||
| − | * B(3) | + | * B(3) |
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| − | + | ==Bailey pair 1== | |
| − | * Use the folloing | + | * 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 | + | * Specialize<math>x=q^2, y=-q, z\to\infty</math>. |
| − | * Bailey pair | + | * 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> |
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| − | + | ==Bailey pair 2== | |
| − | + | * 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<math>a=q,c=-q,d=\infty</math> | ||
| + | * 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> | ||
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| − | + | ==Bailey pair == | |
| − | + | * 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> | |
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| − | + | ==q-series identity== | |
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<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> | <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> | ||
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| − | * [http://www.research.att.com/ | + | * [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences] |
** http://www.research.att.com/~njas/sequences/?q= | ** http://www.research.att.com/~njas/sequences/?q= | ||
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| − | + | ==Bethe type equation (cyclotomic equation)== | |
| − | + | 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>. | \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 | |
<math>\log^2 a_N \sim 4N\sum_{j=1}^{r}\frac{e_j}{d_j}L(1-\mu^{d_j})</math> | <math>\log^2 a_N \sim 4N\sum_{j=1}^{r}\frac{e_j}{d_j}L(1-\mu^{d_j})</math> | ||
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a=2,d_1=1,d_2=2,d_3=2,e_1=e_2=e_3=1 | a=2,d_1=1,d_2=2,d_3=2,e_1=e_2=e_3=1 | ||
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<math>\frac{(1-x)(1-x^2)^2}{(1-x^3)}=x^2</math> | <math>\frac{(1-x)(1-x^2)^2}{(1-x^3)}=x^2</math> | ||
| 89번째 줄: | 79번째 줄: | ||
<math>x^3+3x^2-1=0</math> | <math>x^3+3x^2-1=0</math> | ||
| − | <math>x, -y, -z^{-1}</math>가 | + | <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] |
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| − | + | ==dilogarithm identity== | |
<math>L(x^3)-3L(x^2)-3L(x)=-\frac{7}{3}L(1)</math> | <math>L(x^3)-3L(x^2)-3L(x)=-\frac{7}{3}L(1)</math> | ||
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| − | + | ==related items== | |
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| − | + | [[분류:개인노트]] | |
| − | + | [[분류:math and physics]] | |
| − | + | [[분류:migrate]] | |
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2020년 12월 28일 (월) 04:02 기준 최신판
Note
- an explanation for dilogarithm laddertwisted 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
\(\sum_{n=0}^{\infty}\frac{(q^3;q^3)_{n}q^{n(n+1)}}{ (q)_{n}(q;q^{2})_n(q^2;q^2)_{n}}=\frac{(q^{9};q^{27})_{\infty}(q^{18};q^{27})_{\infty}(q^{27};q^{27})_{\infty}}{(q)_{\infty}}\)
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
\(L(x^3)-3L(x^2)-3L(x)=-\frac{7}{3}L(1)\)