"슬레이터 1"의 두 판 사이의 차이
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5번째 줄: | 5번째 줄: | ||
− | + | <h5>노트</h5> | |
− | + | * [[오일러의 오각수정리(pentagonal number theorem)]] | |
54번째 줄: | 54번째 줄: | ||
<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> | <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> | ||
− | <math>\prod_{n=1}^{\infty}(1 | + | <math>\prod_{n=1}^{\infty}(1-q^n)=1+\sum_{n=1}^{\infty}(-1)^{n}(q^{\frac{3 n^2-n}{2}}+q^{\frac{3 n^2+n}{2}})</math> |
− | * [[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}=1</math><br><math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{1}{(q)_{\infty}}\sum_{n= | + | * [[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}=1</math><br><math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{1}{(q)_{\infty}}(1+\sum_{n=1}^{\infty}(-1)^{n}(q^{\frac{3 n^2-n}{2}}+q^{\frac{3 n^2+n}{2}}))</math><br> |
* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]<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= | ** http://www.research.att.com/~njas/sequences/?q= |
2011년 11월 14일 (월) 10:55 판
이 항목의 수학노트 원문주소
노트
항등식의 종류
켤레 베일리 쌍의 유도
- Use the following
\(\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\), \(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{\delta_{n}}{(x/y)_{n}(x/z)_{n}}\) - Specialize
\(x=q, y\to\infty, z\to\infty\). - Bailey pair
\(\delta_n=q^{n^2}\)
\(\gamma_n=\frac{1}{(q)_{\infty}}q^{n^2}\)
베일리 쌍의 유도
- Use the following
\(\sum_{r=0}^{n}\frac{(1-aq^{2r})(-1)^{r}q^{\frac{1}{2}(r^2+r)}(a)_{r}(c)_{r}(d)_{r}a^{r}}{(a)_{n+r+1}(q)_{n-r}(q)_{r}(aq/c)_{r}(aq/d)_{r}c^{r}d^{r}}=\frac{(aq/cd)_{n}}{(q)_{n}(aq/c)_{n}(aq/d)_{n}}\) - Specialize
\(a=1,c=0,d=\infty\) - Bailey pair
\(\alpha_{0}=1\), \(\alpha_{r}=(-1)^{r}(1+q^r)q^{\frac{1}{2}r(r-1)}\)
\(\beta_{0}=1\), \(\beta_{r}=0\)
\(\beta_n=\sum_{r=0}^{n}\frac{\alpha_r}{(x)_{n-r}(q)_{n+r}}=\sum_{r=0}^{n}\frac{(-1)^{r}(1+q^r)q^{\frac{1}{2}r(r-1)}}{(q)_{n-r}(q)_{n+r}}=0\)
베일리 쌍
-
Bailey pairs
\(\delta_n=q^{n^2}\)
\(\gamma_n=\frac{1}{(q)_{\infty}}q^{n^2}\)
\(\alpha_{0}=1\), \(\alpha_{r}=(-1)^{r}(1+q^r)q^{\frac{1}{2}r(r-1)}\)
\(\beta_{0}=1\), \(\beta_{r}=0\)
q-series 항등식
\(\prod_{n=1}^{\infty}(1-q^n)=1+\sum_{n=1}^{\infty}(-1)^{n}(q^{\frac{3 n^2-n}{2}}+q^{\frac{3 n^2+n}{2}})\)
- 베일리 쌍(Bailey pair)과 베일리 보조정리
\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}\)
\(\sum_{n=0}^{\infty}\beta_n\delta_{n}=1\)
\(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\frac{1}{(q)_{\infty}}(1+\sum_{n=1}^{\infty}(-1)^{n}(q^{\frac{3 n^2-n}{2}}+q^{\frac{3 n^2+n}{2}}))\)