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

2010년 7월 14일 (수) 23:33 판

초기하 급수의 합공식
[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 Pfaff-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
\(\sum_{n=0}^{\infty}\frac{(a,q)_{n}(b,q)_{n}(q^{-n},q)_{n}}{(c)_{n}(abc^{-1}q^{1-n} ,q)_{n}(q ,q)_{n}}q^{n}=\frac{(c/a,c/b;q)_{n}}{(c,c/ab;q)_{n}}\)
q-analogue of Whipple's theorem
Jackson's q-analogue of Dougall's theorem

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 * [http://pythagoras0.springnote.com/pages/5980283 초기하 급수의 합공식]<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 Pfaff-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>
+*  q-analogue of Whipple's theorem<br>
+*  Jackson's q-analogue of Dougall's theorem<br>