"Q-analogue of summation formulas"의 두 판 사이의 차이

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* '''[GR2004]''' (1.5.1) Heine's q-analogue of Gauss' summation formula<br><math>_2\phi_1(a,b;c,q,c/ab)=\frac{(c/a;q)_{\infty}(c/b;q)_{\infty}}{(c;q)_{\infty}(c/(ab);q)_{\infty}}</math> or <br><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><br>
 
* '''[GR2004]''' (1.5.1) Heine's q-analogue of Gauss' summation formula<br><math>_2\phi_1(a,b;c,q,c/ab)=\frac{(c/a;q)_{\infty}(c/b;q)_{\infty}}{(c;q)_{\infty}(c/(ab);q)_{\infty}}</math> or <br><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><br>
* '''[GR2004]''' (1.7.2) q-analogue of Saalschutz's summation formula<br><math>_3\phi_2(a,b,q^{-n};c,abc^{-1}q^{1-n};q,q)=\frac{(c/a,c/b;q)_{n}}{(c,c/ab;q)_{n}}</math> or<br><math>_3\phi_2(a,b,q^{-n};c,abc^{-1}q^{1-n};q,q)=\frac{(c/a,c/b;q)_{n}}{(c,c/ab;q)_{n}}</math><br>  <br>
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* '''[GR2004]''' (1.7.2) q-analogue of Saalschutz's summation formula<br><math>_3\phi_2(a,b,q^{-n};c,abc^{-1}q^{1-n};q,q)=\frac{(c/a,c/b;q)_{n}}{(c,c/ab;q)_{n}}</math> or<br><math>\sum_{n=0}^{\infty}\frac{(a,q)_{n}(b,q)_{n}(q^{-n},q)_{n}}{(c)_{n}(abc^{-1}q^{1-n} ,q)_{n}(q ,q)_{n}}q^{n}=\frac{(c/a,c/b;q)_{n}}{(c,c/ab;q)_{n}}</math><br>  <br>
 
 
 
 
  
 
 
 
 

2010년 6월 23일 (수) 20:57 판

introduction
  • [GR2004] (1.5.1) Heine's q-analogue of Gauss' summation formula
    \(_2\phi_1(a,b;c,q,c/ab)=\frac{(c/a;q)_{\infty}(c/b;q)_{\infty}}{(c;q)_{\infty}(c/(ab);q)_{\infty}}\) or 
    \(\sum_{n=0}^{\infty}\frac{(a,q)_{n}(b,q)_{n}}{(c ,q)_{n}(q ,q)_{n}}(\frac{c}{ab})^{n}=\frac{(c/a;q)_{\infty}(c/b;q)_{\infty}}{(c;q)_{\infty}(c/(ab);q)_{\infty}}\)
  • [GR2004] (1.7.2) q-analogue of Saalschutz's summation formula
    \(_3\phi_2(a,b,q^{-n};c,abc^{-1}q^{1-n};q,q)=\frac{(c/a,c/b;q)_{n}}{(c,c/ab;q)_{n}}\) or
    \(\sum_{n=0}^{\infty}\frac{(a,q)_{n}(b,q)_{n}(q^{-n},q)_{n}}{(c)_{n}(abc^{-1}q^{1-n} ,q)_{n}(q ,q)_{n}}q^{n}=\frac{(c/a,c/b;q)_{n}}{(c,c/ab;q)_{n}}\)
     

 

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