# 슬레이터 2

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## 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$\mathfrak{f}_2(\tau)=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n})=\sqrt{2}q^{1/24}\sum_{n\geq 0}\frac{q^{n(n+1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}$

## 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

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

## 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

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