"슬레이터 1"의 두 판 사이의 차이

수학노트
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*  Specialize<br><math>a=1,c=0,d=\infty</math><br>
 
*  Specialize<br><math>a=1,c=0,d=\infty</math><br>
 
*  Bailey pair<br><math>\alpha_{0}=1</math>, <math>\alpha_{r}=(-1)^{r}(1+q^r)q^{\frac{1}{2}r(r-1)}</math><br><math>\beta_{0}=1</math>, <math>\beta_{r}=0</math><br><math>\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</math><br>
 
*  Bailey pair<br><math>\alpha_{0}=1</math>, <math>\alpha_{r}=(-1)^{r}(1+q^r)q^{\frac{1}{2}r(r-1)}</math><br><math>\beta_{0}=1</math>, <math>\beta_{r}=0</math><br><math>\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</math><br>
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<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%;">베일리 쌍</h5>
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*   <br> Bailey pairs<br><math>\delta_n=q^{n^2}</math><br><math>\gamma_n=\frac{1}{(q)_{\infty}}q^{n^2}</math><br>  <br><math>\alpha_{0}=1</math>, <math>\alpha_{r}=(-1)^{r}(1+q^r)q^{\frac{1}{2}r(r-1)}</math><br><math>\beta_{0}=1</math>, <math>\beta_{r}=0</math><br>
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<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>
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<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>
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* [[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=0}^{\infty}(-1)^{n}(q^{\frac{3n^2+n}{2}}-q^{\frac{3n^2+5n+2}{2}})=(-q)_{\infty}</math><br>
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* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]<br>
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** http://www.research.att.com/~njas/sequences/?q=

2011년 11월 14일 (월) 10:50 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

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

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