"Bailey pair and lemma"의 두 판 사이의 차이

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<h5 style="background-position: 0px 100%; font-size: 1.16em; margin: 0px; color: rgb(34, 61, 103); line-height: 3.42em; font-family: 'malgun gothic',dotum,gulim,sans-serif;">introduction</h5>
 
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<h5 style="margin: 0px; line-height: 2em;">examples of Bailey pair</h5>
 
<h5 style="margin: 0px; line-height: 2em;">examples of Bailey pair</h5>
  
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<h5 style="background-position: 0px 100%; font-size: 1.16em; margin: 0px; color: rgb(34, 61, 103); line-height: 2em; font-family: 'malgun gothic',dotum,gulim,sans-serif;">Bailey lemma</h5>
 
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*  Bailey lemma involved a Bailey pair and a conjugate Bailey pair<br>
 
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*  If the sequence <math>\{\alpha_r\}, \{\beta_r\}</math>, <math>\{\delta_r\}, \{\gamma_r\}</math> satisfy the following<br><math>\beta_L=\sum_{r=0}^{L}{\alpha_r}{u_{L-r}v_{L+r}}</math>, <math>\gamma_L=\sum_{r=L}^{\infty}{\delta_r}{u_{r-L}v_{r+L}}</math><br> then,<br><math>\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}</math><br>
 
  
 
 
 
 
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(proof)
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By the basic analogue of Gauss' theorem 
 
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<math>\sum_{n=0}^{\infty}\frac{(a,q)_{n}(b,q)_{n}}{(c ,q)_{n}(q ,q)_{n}}(\frac{c}{ab})^{n}=\frac{(c/a;q)_{\infty}(c/b;q)_{\infty}}{(c;q)_{\infty}(c/(ab);q)_{\infty}}</math>,
 
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[[q-analogue of summation formulas|]]
 
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* [http://pythagoras0.springnote.com/pages/9408516 합공식의 q-analogue]
 
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* [http://pythagoras0.springnote.com/pages/9394260 q-가우스 합]
 
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Also note that <math>(a)_{n+r}=(a)_{r}(aq^{r})_{n}</math> and <math>(a)_{\infty}=(a)_{r}(aq^{r})_{\infty}</math>
 
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Put <math>a=yq^{r},b=zq^{r},c=xq^{2r}</math>. 
 
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Then we get (*)
 
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<math>A=\sum_{n=0}^{\infty}\frac{(yq^{r})_{n}(zq^{r})_{n}}{(xq^{2r})_{n}(q)_{n}}(\frac{x}{yz})^{n}=\frac{(xq^{r}/y)_{\infty}(xq^{r}/z)_{\infty}}{(xq^{2r})_{\infty}(x/(yz))_{\infty}}=B</math>
 
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From the left hand side,
 
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<math>A=\sum_{n=0}^{\infty}\frac{(yq^{r})_{n}(zq^{r})_{n}}{(xq^{2r})_{n}(q)_{n}}(\frac{x}{yz})^{n}=\frac{(x)_{2r}}{(y)_{r}(z)_{r}}\sum_{n=0}^{\infty}\frac{(y)_{n+r}(z)_{n+r}}{(x)_{n+2r}(q)_{n}}(\frac{x}{yz})^{n}=\frac{(x)_{2r}y^{r}z^{r}}{(y)_{r}(z)_{r}x^{r}}\sum_{n=0}^{\infty}\frac{(y)_{n+r}(z)_{n+r}}{(x)_{n+2r}(q)_{n}}(\frac{x}{yz})^{n+r}</math>
 
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Now let
 
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<math>\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}</math> so that
 
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<math>A=\frac{(x)_{2r}y^{r}z^{r}}{(y)_{r}(z)_{r}x^{r}}\sum_{n=0}^{\infty}\frac{\delta_{n+r}}{(x)_{n+2r}(q)_{n}}</math>
 
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From the right hand side of (*), we get
 
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<math>B=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(x)_{2r}}{(x/y)_{r}(x/z)_{r}}</math>
 
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Therefore,
 
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<math>A=\frac{(x)_{2r}y^{r}z^{r}}{(y)_{r}(z)_{r}x^{r}}\sum_{n=0}^{\infty}\frac{\delta_{n+r}}{(x)_{n+2r}(q)_{n}}=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(x)_{2r}}{(x/y)_{r}(x/z)_{r}}=B</math>
 
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By simplifying the above equation, we obtain
 
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<math>\sum_{n=0}^{\infty}\frac{\delta_{n+r}}{(x)_{n+2r}(q)_{n}}=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{1}{(x/y)_{r}(x/z)_{r}}\frac{(y)_{r}(z)_{r}x^{r}}{y^{r}z^{r}}</math>
 
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<math>\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}</math>  with <math>\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}</math> gives a conjugate Bailey pair
 
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(a different notation
 
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 <math>\gamma_n=\prod{{x/y,x/z;q}\choose {x,x/yz;}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}</math>
 
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is also used sometimes) ■
 
  
 
*  If we apply Bailey lemma to the above conjugate pair, we get<br><math>\sum_{n=0}^{\infty}\frac{(y)_n(z)_n x^n}{y^n z^n}\beta_{n}=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\sum_{n=0}^{\infty}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\alpha_{n}</math><br>
 
*  If we apply Bailey lemma to the above conjugate pair, we get<br><math>\sum_{n=0}^{\infty}\frac{(y)_n(z)_n x^n}{y^n z^n}\beta_{n}=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\sum_{n=0}^{\infty}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\alpha_{n}</math><br>

2011년 11월 12일 (토) 05:38 판

examples of Bailey pair

 

 

 

specialization
  • Choose the following (in the following, x=aq to get a Bailey pair relative to a)
    \(u_{n}=\frac{1}{(q)_n}\) ,\(v_{n}=\frac{1}{(x)_n}\),
  • There is a conjugate Bailey pair
    \(\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\)
     \(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\)

 

 

  • If we apply Bailey lemma to the above conjugate pair, we get
    \(\sum_{n=0}^{\infty}\frac{(y)_n(z)_n x^n}{y^n z^n}\beta_{n}=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\sum_{n=0}^{\infty}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\alpha_{n}\)

 

 

examples
  • Conjugate Bailey pair (\(x=q,y\to\infty, z\to\infty\))
    \(\delta_n=q^{n^2}\)
    \(\gamma_n=\frac{q^{n^2}}{(q)_{\infty}}\)
  • Bailey pair
    \(\alpha_{n}=(-1)^{n}q^{\frac{3}{2}n^2}(q^{\frac{1}{2}n}+q^{-\frac{1}{2}n})\)
    \(\beta_n=\frac{1}{(q)_{n}}\)
  • we get the Rogers-Ramanujan identity(Slater 18)
    \(\sum_{n=0}^{\infty}\frac{q^{n^2}}{ (q)_{n}}=\frac{(q^{3};q^{5})_{\infty}(q^{2};q^{5})_{\infty}(q^{5};q^{5})_{\infty}}{(q)_{\infty}}=\frac{1}{(q^{1};q^{5})_{\infty}(q^{4};q^{5})_{\infty}}\)

 

 

Bailey chain

 

 

history

 

 

related items

 

 

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