# 정팔면체와 모듈라 연분수

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## 개요

• [Duke2005] (9.1) $u(\tau)={\sqrt{2}q^{1/8} \over 1+ } {q \over 1+q+} {q^2 \over 1+q^2+} {q^3 \over 1+q^3} \cdots=\sqrt{2}q^{1/8}\prod_{n=1}^{\infty}(1+q^{n})^{(-1)^{n}}=\sqrt{2}q^{1/8}\frac{(-q^{2};q^{2})_{\infty}} {(-q;q^{2})_{\infty}}$

$v(\tau)={q^{1/2} \over 1+q + } {q \over 1+q^2+} {q^2 \over 1+q^3} \cdots=q^{1/2}\prod_{n=1}^{\infty}(1-q^{n})^{(\frac{8}{n})}=q^{1/2}\frac{(q^{1};q^{8})_{\infty}(q^{7};q^{8})_{\infty}}{(q^{3};q^{8})_{\infty}(q^{5};q^{8})_{\infty}}$

$$u(\tau)={\sqrt{2}q^{1/8} \over 1+ } {q \over 1+q+} {q^2 \over 1+q^2} {q^3 \over 1+q^3} \cdots=\sqrt{2}q^{1/8}\prod_{n=1}^{\infty}(1+q^{n})^{(-1)^{n}}=\sqrt{2}q^{1/8}\frac{(-q^{2};q^{2})_{\infty}} {(-q;q^{2})_{\infty}}=\sqrt{2}q^{1/8}\frac{(q^{1};q^{4})_{\infty}(q^{3};q^{4})_{\infty}}{(q^{2};q^{4})_{\infty}(q^{2};q^{4})_{\infty}}$$

(증명)

$$(-q^2;q^{2})_{n}=\frac{(q^4;q^4)_{n}}{(q^2;q^2)_{n}}=\frac{1}{(q^2;q^4)_{n}}$$

$$(-q;q^{2})_{n}=\frac{(-q;q)_{n}}{(-q^{2};q^{2})_{n}}=\frac{(q^{2};q^{2})_{n}(q^{2};q^{2})_{n}}{(q^{4};q^{4})_{n}(q;q)_{n}}=\frac{(q^{2};q^{4})_{n}}{(q^{1};q^{4})_{n}(q^{3};q^{4})_{n}}$$

$$(-q^2;q^{2})_{\infty}=\frac{(q^4;q^4)_{\infty}}{(q^2;q^2)_{\infty}}=\frac{1}{(q^2;q^4)_{\infty}}$$

$$(-q;q^{2})_{\infty}=\frac{(q^{2};q^{2})_{\infty}(q^{2};q^{2})_{\infty}}{(q^{4};q^{4})_{\infty}(q;q)_{\infty}}=\frac{(q^{2};q^{4})_{\infty}}{(q^{1};q^{4})_{\infty}(q^{3};q^{4})_{\infty}}$$ ■

## Ramanujan-Göllnitz-Gordon continued fraction

• Ramanujan-Göllnitz-Gordon 연분수$1/v(\tau) \sim 1+q+{q^{2} \over 1+q^{3} + } {q^{4} \over 1+q^{5}+} {q^{6} \over \cdots}=\frac{(q^{3};q^{8})_{\infty}(q^{4};q^{8})_{\infty}(q^{5};q^{8})_{\infty}}{(q^{1};q^{8})_{\infty}(q^{4};q^{8})_{\infty}(q^{7};q^{8})_{\infty}}=\frac{(q^{3};q^{8})_{\infty}(q^{5};q^{8})_{\infty}}{(q^{1};q^{8})_{\infty}(q^{7};q^{8})_{\infty}}$

## eta quotient

$$u(\tau)=\sqrt{2}q^{1/8}\frac{(-q^{2};q^{2})_{\infty}} {(-q;q^{2})_{\infty}}=\sqrt{2}\frac{\eta(\tau)\eta^{2}(4\tau)}{\eta^{3}(2\tau)}$$

(proof)

$$\frac{(-q^{2};q^{2})_{\infty}} {(-q;q^{2})_{\infty}}=\frac{(q^4;q^4)_{\infty}}{(q^2;q^2)_{\infty}}\frac{(q^{4};q^{4})_{\infty}(q;q)_{\infty}}{(q^{2};q^{2})_{\infty}(q^{2};q^{2})_{\infty}}$$■