"Bailey pair and lemma"의 두 판 사이의 차이

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See
  
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[[q-analogue of summation formulas]]
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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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Also note that <math>(a)_{n+r}=(a)_{r}(aq^{r})_{n}</math> and <math>(a)_{\infty}=(a)_{r}(aq^{r})_{\infty}</math>
 
Also note that <math>(a)_{n+r}=(a)_{r}(aq^{r})_{n}</math> and <math>(a)_{\infty}=(a)_{r}(aq^{r})_{\infty}</math>

2011년 11월 12일 (토) 05:24 판

introduction
  •  q-Pfaff-Sallschutz sum

 

 

Bailey pair
  • the sequence \(\{\alpha_r\}, \{\beta_r\}\) satisfying the following is called a Bailey pair relative to a
    \(\beta_L=\sum_{r=0}^{L}\frac{\alpha_r}{(q)_{L-r}(aq)_{L+r}}\)
  • conjugate Bailey pair  \(\{\delta_r\}, \{\gamma_r\}\)
    \(\gamma_L=\sum_{r=L}^{\infty}\frac{\delta_r}{(q)_{r-L}(aq)_{r+L}}=\sum_{r=0}^{\infty}\frac{\delta_{r+L}}{(q)_{r}(aq)_{r+2L}}\)
  • Note that the summation for a conjugate Bailey pair is different from that of a Bailey pair

 

 

 

examples of Bailey pair

 

 

how to obtain and check Bailey pair?
  • various complicated q-series identities

 

 

why do we care about Bailey pair?
  • When we have a Bailey pair, we can produce q-series identities
    • (1) Bailey lemma gives an identity involving q-series
    • (2) using the definition of Bailey pair
      \(\beta_L=\sum_{r=0}^{L}\frac{\alpha_r}{(q)_{L-r}(aq)_{L+r}}\)

 

 

Bailey lemma
  • Bailey lemma involved a Bailey pair and a conjugate Bailey pair
  • If the sequence \(\{\alpha_r\}, \{\beta_r\}\), \(\{\delta_r\}, \{\gamma_r\}\) satisfy the following
    \(\beta_L=\sum_{r=0}^{L}{\alpha_r}{u_{L-r}v_{L+r}}\), \(\gamma_L=\sum_{r=L}^{\infty}{\delta_r}{u_{r-L}v_{r+L}}\)
    then,
    \(\sum_{n=0}^{\infty}\alpha_n\gamma_{n}=\sum_{n=0}^{\infty}\beta_n\delta_{n}\)

 

 

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
    \(\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\)
     \(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\)

 

(proof)

By the basic analogue of Gauss' theorem 

\(\sum_{n=0}^{\infty}\frac{(a,q)_{n}(b,q)_{n}}{(c ,q)_{n}(q ,q)_{n}}(\frac{c}{ab})^{n}=\frac{(c/a;q)_{\infty}(c/b;q)_{\infty}}{(c;q)_{\infty}(c/(ab);q)_{\infty}}\),

See

[[q-analogue of summation formulas|]]

 

Also note that \((a)_{n+r}=(a)_{r}(aq^{r})_{n}\) and \((a)_{\infty}=(a)_{r}(aq^{r})_{\infty}\)

Put \(a=yq^{r},b=zq^{r},c=xq^{2r}\). 

Then we get (*)

\(A=\sum_{n=0}^{\infty}\frac{(yq^{r})_{n}(zq^{r})_{n}}{(xq^{2r})_{n}(q)_{n}}(\frac{x}{yz})^{n}=\frac{(xq^{r}/y)_{\infty}(xq^{r}/z)_{\infty}}{(xq^{2r})_{\infty}(x/(yz))_{\infty}}=B\)

From the left hand side,

\(A=\sum_{n=0}^{\infty}\frac{(yq^{r})_{n}(zq^{r})_{n}}{(xq^{2r})_{n}(q)_{n}}(\frac{x}{yz})^{n}=\frac{(x)_{2r}}{(y)_{r}(z)_{r}}\sum_{n=0}^{\infty}\frac{(y)_{n+r}(z)_{n+r}}{(x)_{n+2r}(q)_{n}}(\frac{x}{yz})^{n}=\frac{(x)_{2r}y^{r}z^{r}}{(y)_{r}(z)_{r}x^{r}}\sum_{n=0}^{\infty}\frac{(y)_{n+r}(z)_{n+r}}{(x)_{n+2r}(q)_{n}}(\frac{x}{yz})^{n+r}\)

Now let

\(\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\) so that

\(A=\frac{(x)_{2r}y^{r}z^{r}}{(y)_{r}(z)_{r}x^{r}}\sum_{n=0}^{\infty}\frac{\delta_{n+r}}{(x)_{n+2r}(q)_{n}}\)

From the right hand side of (*), we get

\(B=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(x)_{2r}}{(x/y)_{r}(x/z)_{r}}\)

Therefore,

\(A=\frac{(x)_{2r}y^{r}z^{r}}{(y)_{r}(z)_{r}x^{r}}\sum_{n=0}^{\infty}\frac{\delta_{n+r}}{(x)_{n+2r}(q)_{n}}=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(x)_{2r}}{(x/y)_{r}(x/z)_{r}}=B\)

By simplifying the above equation, we obtain

\(\sum_{n=0}^{\infty}\frac{\delta_{n+r}}{(x)_{n+2r}(q)_{n}}=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{1}{(x/y)_{r}(x/z)_{r}}\frac{(y)_{r}(z)_{r}x^{r}}{y^{r}z^{r}}\)

\(\gamma_n=\frac{(x/y;q)_{\infty}(x/z;q)_{\infty}}{(x;q)_{\infty}(x/yz;q)_{\infty}}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\)  with \(\delta_n=\frac{(y)_n(z)_n x^n}{y^n z^n}\) gives a conjugate Bailey pair

(a different notation

 \(\gamma_n=\prod{{x/y,x/z;q}\choose {x,x/yz;}}\frac{(y)_n(z)_n x^n}{(x/y)_{n}(x/z)_{n}y^n z^n}\)

is also used sometimes) ■

  • 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

 

 

encyclopedia

 

 

books

[[4909919|]]

 

 

 

expositions

 

 

 

articles

 

 

question and answers(Math Overflow)

 

 

blogs

 

 

experts on the field

 

 

links
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