"합공식의 q-analogue"의 두 판 사이의 차이

2012년 8월 26일 (일) 06:20 판

초기하 급수의 합공식
q-Chu-Vandermonde
\(_2\phi_1(q^{-n},b;c;q,q)=\frac{(c/b;q)_n}{(c;q)_n}b^n\)
[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
\(\sum_{n=0}^{\infty}\frac{q^n (a;q)_n (b;q)_n \left(q^{-k};q\right)_n}{(q;q)_n (c;q)_n \left(\frac{a b q^{1-k}}{c};q\right)_n}=\frac{\left(\frac{c}{a};q\right)_k \left(\frac{c}{b};q\right)_k}{(c;q)_k \left(\frac{c}{a b};q\right)_k}\)
q-analogue of Whipple's theorem
Jackson's q-analogue of Dougall's theorem

@@ 14번째 줄: / 14번째 줄: @@
 * '''[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{q^n (a;q)_n (b;q)_n \left(q^{-k};q\right)_n}{(q;q)_n (c;q)_n \left(\frac{a b q^{1-k}}{c};q\right)_n}=\frac{\left(\frac{c}{a};q\right)_k \left(\frac{c}{b};q\right)_k}{(c;q)_k \left(\frac{c}{a b};q\right)_k}</math><br>
 *  q-analogue of Whipple's theorem<br>
-*  Jackson's q-analogue of Dougall's theorem<br>
+* [[Jackson's q-analogue of Dougall's theorem]]<br>