슬레이터 8

이 항목의 수학노트 원문주소

• 슬레이터 8

개요

• 항등식\[\sum_{n=0}^{\infty}\frac{(q^2;q^2)_{n}q^{n(n+1)/2}}{ (q)_{n}^2}=\frac{(-q)_{\infty}}{(q^2;q^4)_{\infty}}\]

항등식의 분류

• 슬레이터 목록 (Slater's list)
• B(3)

켤레 베일리 쌍의 유도

• q-가우스 합 에서 얻어진 다음 결과를 이용\[\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\], \(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\)\[\gamma_{n}=\sum_{r=0}^{\infty}\frac{\delta_{n+r}}{(x)_{r+2n}(q)_{r}}\]
• 다음 특별한 경우\[x=q^2, y=-q, z\to\infty\].
• 얻어진 켤레 베일리 쌍 (슬레이터 2 와 같음)\[\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}=\frac{(q^2;q^2)_{n}}{(q)_{n}}q^{\frac{n(n+1)}{2}}\]\[\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}\]

베일리 쌍의 유도

• 다음을 이용\[\sum_{r=0}^{n}\frac{(1-aq^{2r})(-1)^{r}q^{\frac{1}{2}(r^2+r)}(a)_{r}(c)_{r}(d)_{r}a^{r}}{(a)_{n+r+1}(q)_{n-r}(q)_{r}(aq/c)_{r}(aq/d)_{r}c^{r}d^{r}}=\frac{(aq/cd)_{n}}{(q)_{n}(aq/c)_{n}(aq/d)_{n}}\]
• 다음의 특수한 경우\[a=q, c\to\infty, d\to\infty\]
• 얻어진 베일리 쌍\[\alpha_{n}=\frac{(-1)^{n}q^{\frac{3}{2}n^2+\frac{1}{2}n}(1-q^{2n+1})}{1-q}\]\[\beta_n=\frac{1}{(q)_{n}}\]\[\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{2})_{n-r}(q)_{n+r}}\]

베일리 쌍

• 베일리 쌍과 켤레 베일리 쌍\[\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}=\frac{(q^2;q^2)_{n}}{(q)_{n}}q^{\frac{n(n+1)}{2}}\]\[\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}\]\[\alpha_{n}=\frac{(-1)^{n}q^{\frac{3}{2}n^2+\frac{1}{2}n}(1-q^{2n+1})}{1-q}\]\[\beta_n=\frac{1}{(q)_{n}}\]

q-series 항등식

• 항등식\[\sum_{n=0}^{\infty}\frac{(q^2;q^2)_{n}q^{n(n+1)/2}}{ (q)_{n}^2}=\frac{(-q)_{\infty}}{(q^2;q^4)_{\infty}}\]

• 베일리 쌍(Bailey pair)과 베일리 보조정리\[\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}\]\[\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{(q^2;q^2)_{n}q^{n(n+1)/2}}{ (q)_{n}^2}\]\[\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\frac{(-1)^{n}q^{\frac{3}{2}n^2+\frac{1}{2}n}(1-q^{2n+1})}{1-q}\cdot \frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{2n^2+n}-q^{2n^2+3n+1})=\sum_{n=-\infty}^{\infty}(-1)^{n}q^{2n^2+n} =\frac{(-q)_{\infty}}{(q^2;q^4)_{\infty}}\]

• [product[%281-q%5E%284n-1%29%29%281-q%5E%284n-3%29%29%281-q%5E%284n%29%29,%7Bn,1,10%7D] http://www.wolframalpha.com/input/?i=expand[product[(1-q^(4n-1))(1-q^(4n-3))(1-q^(4n)),{n,1,10}]]]
• The On-Line Encyclopedia of Integer Sequences
  • http://www.research.att.com/~njas/sequences/?q=

베테 타입 방정식 (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

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

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

다이로그 항등식

\(4L(x)-L(x^2)=\frac{\pi^2}{4}\)

\(x=\sqrt{2}-1\)