"베버(Weber) 모듈라 함수"의 두 판 사이의 차이

2009년 10월 24일 (토) 18:26 판

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

\(\mathfrak{f}_1(\tau)=\frac{\eta(\frac{\tau}{2})}{\eta(\tau)}=q^{-1/48} \prod_{n=1}^{\infty} (1-q^{n-\frac{1}{2}})\)

\(\mathfrak{f}_2(\tau)=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n})\)

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

\(\mathfrak{f}_1(2\tau)\mathfrak{f}_2(\tau)=\sqrt2\)

\(f(\tau)f_1(\tau)f_2(\tau)=\sqrt2\)

\(f(\tau)^8=f_1(\tau)^8+f_2(\tau)^8\)

\(f(\tau+1)=\zeta_{48}^{-1}f_1(\tau)\)

\(f_1(\tau+1)=\zeta_{48}^{-1}f(\tau)\)

\(f_2(\tau+1)=\zeta_{24}f_2(\tau)\)

\(f(-\frac{1}{\tau})=f(\tau)\)

\(f_1(-\frac{1}{\tau})=f_2(\tau)\)

\(f_2(-\frac{1}{\tau})=f_1(\tau)\)

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
\(f(2\tau)=q^{-1/24}\prod_{n=1}^{\infty} (1+q^{2n-1})=q^{-1/24}\sum_{n\geq 0}\frac{q^{n^2}}{(1-q^2)(1-q^4)\cdots(1-q^{2n})}\)
\(f_2(\tau)=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n})=\sqrt{2}q^{1/24}\sum_{n\geq 0}\frac{q^{n(n+1)/2}}{(1-q)(1-q^2)\cdots(1-q^n)}\)

@@ 13번째 줄: / 13번째 줄: @@
 *  class field theory에서 중요한 역할을 함<br>
-<math>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}})</math>
+<math>\mathfrak{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}})</math>
-<math>f_1(\tau)=\frac{\eta(\frac{\tau}{2})}{\eta(\tau)}=q^{-1/48} \prod_{n=1}^{\infty} (1-q^{n-\frac{1}{2}})</math>
+<math>\mathfrak{f}_1(\tau)=\frac{\eta(\frac{\tau}{2})}{\eta(\tau)}=q^{-1/48} \prod_{n=1}^{\infty} (1-q^{n-\frac{1}{2}})</math>
-<math>f_2(\tau)=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n})</math>
+<math>\mathfrak{f}_2(\tau)=\sqrt{2}\frac{\eta(2\tau)}{\eta(\tau)}=\sqrt{2}q^{1/24} \prod_{n=1}^{\infty} (1+q^{n})</math>
 여기서  <math>\eta(\tau) = q^{1/24} \prod_{n=1}^{\infty} (1-q^{n})</math> 는 [[데데킨트 에타함수]]
@@ 27번째 줄: / 27번째 줄: @@
 <h5 style="margin: 0px; line-height: 2em;">항등식</h5>
-<math>f_1(2\tau)f_2(\tau)=\sqrt2</math>
+<math>\mathfrak{f}_1(2\tau)\mathfrak{f}_2(\tau)=\sqrt2</math>
 <math>f(\tau)f_1(\tau)f_2(\tau)=\sqrt2</math>