슬레이터 2

개요

• 항등식\[\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=0}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})\]
• 베버(Weber) 모듈라 함수 의 하나\[\mathfrak{f}_2(\tau)=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n})=\sqrt{2}q^{1/24}\sum_{n\geq 0}\frac{q^{n(n+1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}\]

 

 

항등식의 분류

• 슬레이터 목록 (Slater's list)
• E(3)

 

 

켤레 베일리 쌍의 유도

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

 

 

베일리 쌍의 유도

• Use the following [Slater51] (4.1)\[\sum_{r=0}^{n}\frac{(1-aq^{2r})(-1)^{r}q^{\frac{1}{2}(r^2+r)}(a)_{r}(c)_{r}(d)_{r}a^{r}}{(a)_{n+r+1}(q)_{n-r}(q)_{r}(aq/c)_{r}(aq/d)_{r}c^{r}d^{r}}=\frac{(aq/cd)_{n}}{(q)_{n}(aq/c)_{n}(aq/d)_{n}}\]
• Specialize\[a=q,c=-q,d=\infty\]
• Bailey pair\[\alpha_{0}=1\], \(\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)\)\[\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{2})_{n-r}(q)_{n+r}}=\frac{1}{(q)_{n}(-q)_{n}}\]

 

 

베일리 쌍

\(\delta_n=(-q)_{n}q^{\frac{n(n+1)}{2}}\)

\(\gamma_n=\frac{(-q)_{\infty}}{(q^2)_{\infty}}q^{\frac{n(n+1)}{2}}\)

\(\alpha_{n}=(-1)^{n}q^{n^2}(1-q^{2n+1})/(1-q)\)

\(\beta_n=\frac{1}{(q)_{n}(-q)_{n}}\)

 

 

q-series 항등식

\(\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=0}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})\)

• 베일리 쌍(Bailey pair)과 베일리 보조정리\[\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}\]\[\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}\]\[\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}\] (오일러의 오각수정리(pentagonal number theorem) 가 사용됨)

• 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

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

a=1,d=1,e=1

The equation  becomes \(1-x=x\).

\(4L(\frac{1}{2})=\frac{1}{2}(\frac{2}{3}\pi^2)=\frac{1}{3}\pi^2\)

 

 

다이로그 항등식

\(L(\frac{1}{2})=\frac{1}{12}\pi^2\)