"슬레이터 3"의 두 판 사이의 차이

2011년 11월 15일 (화) 04:59 판

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

다음을 이용 [Slater51] (4.2)
\(\sum_{r=-[n/2]}^{r=[n/2]}\frac{(1-aq^{4r})(q^{-n})_{2r}a^{2r}q^{2nr+r}(d)_{q^2,r}(e)_{q^2,r}}{(1-a)(aq^{n+1})_{2r}d^re^r(aq^2/d)_{q^2,r}(aq^2/e)_{q^2,r}}=\frac{(q^2/a,aq/d,aq/e,aq^2/de;q^2)_{\infty}}{(q,q^2/d,q^2/e,a^2q/de;q^2)_{\infty}}\frac{(q)_{n}(aq)_{n}(a^2/de)_{q^2,n}}{(aq)_{q^2,n}(aq/d)_{n}(aq/e)_{n}}\)
다음의 특수한 경우
\(a=q,d\to\infty,e\to\infty\)
얻어진 베일리 쌍
\(\alpha_{2r}=2\), \(\alpha_{2r+1}=-2\)
\(\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{\alpha_r}{(q^{2})_{n-r}(q)_{n+r}}=\frac{(-1)^n}{(q^2;q^2)_{n}}\)

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

베일리 쌍(Bailey pair)과 베일리 보조정리
\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}\)
\(\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}\)
\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{(-q)_{\infty}}{(q)_{\infty}}\sum_{n=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}\)

@@ 38번째 줄: / 38번째 줄: @@
 *  다음의 특수한 경우<br><math>a=q,d\to\infty,e\to\infty</math><br>
 *  얻어진 베일리 쌍<br><math>\alpha_{2r}=2</math>, <math>\alpha_{2r+1}=-2</math><br><math>\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)^n}{(q^2;q^2)_{n}}</math><br>
+<h5 style="line-height: 3.428em; margin: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">q-series 항등식</h5>
+*  항등식<br><math>\prod_{n=1}^{\infty}(1+q^n)=\sum_{n=1}^{\infty}\frac{q^{n(n+1)/2}}{(q)_n}\sim \frac{1}{\sqrt{2}}\exp(\frac{\pi^2}{12t}+\frac{t}{24})</math><br>
+* [[베일리 쌍(Bailey pair)과 베일리 보조정리]]<br><math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}</math><br><math>\sum_{n=0}^{\infty}\beta_n\delta_{n}=\sum_{n=0}^{\infty}\frac{q^{\frac{n(n+1)}{2}}}{(q)_{n}}</math><br><math>\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}</math><br>
+* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]<br>
+** http://www.research.att.com/~njas/sequences/?q=