"Slater 92"의 두 판 사이의 차이
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| + | a=2,d_1=1,d_2=2,d_3=2,e_1=e_2=e_3=1 | ||
| 91번째 줄: | 93번째 줄: | ||
<math>x, -y, -z^{-1}</math>가 방정식 의 해 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^3%2B3x^2-1%3D0 | ||
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2010년 7월 27일 (화) 17:39 판
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
\(\sum_{n=0}^{\infty}\frac{(q^3;q^3)_{n}q^{n(n+1)}}{ (q)_{n}(q;q^{2})_n(q^2;q^2)_{n}}=\frac{(q^{9};q^{27})_{\infty}(q^{18};q^{27})_{\infty}(q^{27};q^{27})_{\infty}}{(q)_{\infty}}\)
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
\(L(x^3)-3L(x^2)-3L(x)=-\frac{7}{3}L(1)\)
books
- 2010년 books and articles
- http://gigapedia.info/1/
- http://gigapedia.info/1/
- http://www.amazon.com/s/ref=nb_ss_gw?url=search-alias%3Dstripbooks&field-keywords=
[[4909919|]]
articles