베버(Weber) 모듈라 함수

\(f(\tau)=\frac{e^{-\frac{\pi i}{24}}\eta(\frac{\tau+1}{2})}{\eta(\tau)}=q^{-1/48} \prod_{n=1}^{\infty} (1+q^{n-\frac{1}{2}})\)

\(f_1(\tau)=\frac{\eta(\frac{\tau}{2})}{\eta(\tau)}=q^{-1/48} \prod_{n=1}^{\infty} (1-q^{n-\frac{1}{2}})\)

\(f_2(\tau)=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n})\)

여기서 \(\eta(\tau) = q^{1/24} \prod_{n=1}^{\infty} (1-q^{n})\) 는 데데킨트 에타함수

q-초기하급수(q-hypergeometric series) 의 공식
\(\prod_{n=0}^{\infty}(1+zq^n)=\sum_{n\geq 0}\frac{q^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} z^n\)
\(z=q^{1/2}\) 인 경우
\(\prod_{n=1}^{\infty} (1+q^{n-\frac{1}{2}})=\sum_{n\geq 0}\frac{q^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)} (q^{1/2})^n=\sum_{n\geq 0}\frac{q^{n^2/2}}{(1-q)(1-q^2)\cdots(1-q^n)} \)
\(\prod_{n=1}^{\infty} (1+q^{2n-1})=\sum_{n\geq 0}\frac{q^{n^2}}{(1-q^2)(1-q^4)\cdots(1-q^{2n})} \)
\(z=q\) 인 경우
\(\prod_{n=1}^{\infty} (1+q^{n})=\sum_{n\geq 0}\frac{q^{n(n-1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}q^n=\sum_{n\geq 0}\frac{q^{n(n+1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}\)
위의 결과로부터 다음을 얻을 수 있다
\(f(2\tau)=q^{-1/24}\prod_{n=1}^{\infty} (1+q^{2n-1})=q^{-1/24}\sum_{n\geq 0}\frac{q^{n^2}}{(1-q^2)(1-q^4)\cdots(1-q^{2n})}\)
\(f_2(\tau)=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n})=\sqrt{2}q^{1/24}\sum_{n\geq 0}\frac{q^{n(n+1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}\)

목차