"Q- Pfaff-Saalschütz 항등식"의 두 판 사이의 차이

2011년 11월 12일 (토) 10:49 판

[GR2004] (1.7.2) q-analogue of Pfaff-Saalschutz's summation formula
\(\, _3\phi _2\left(a,b,q^{-k};c,\frac{a b q^{1-k}}{c};q,q\right)=\frac{\left(\frac{c}{a};q\right)_k \left(\frac{c}{b};q\right)_k}{(c;q)_k \left(\frac{c}{a b};q\right)_k}\) or
\(\sum_{n=0}^{\infty}\frac{q^n (a;q)_n (b;q)_n \left(q^{-k};q\right)_n}{(q;q)_n (c;q)_n \left(\frac{a b q^{1-k}}{c};q\right)_n}=\frac{\left(\frac{c}{a};q\right)_k \left(\frac{c}{b};q\right)_k}{(c;q)_k \left(\frac{c}{a b};q\right)_k}\)

\(_3\phi_2(a,b,q^{-n};c,abc^{-1}q^{1-n};q,q)=\frac{(c/a,c/b;q)_{n}}{(c,c/ab;q)_{n}}\) or
\(\sum_{n=0}^{\infty}\frac{(a,q)_{n}(b,q)_{n}(q^{-n},q)_{n}}{(c)_{n}(abc^{-1}q^{1-n} ,q)_{n}(q ,q)_{n}}q^{n}=\frac{(c/a,c/b;q)_{n}}{(c,c/ab;q)_{n}}\)

@@ 7번째 줄: / 7번째 줄: @@
 <h5>개요</h5>
-<math>\, _3\phi _2\left(a,b,q^{-k};c,\frac{a b q^{1-k}}{c};q,q\right)=\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>
+* '''[GR2004]''' (1.7.2) q-analogue of Pfaff-Saalschutz's summation formula<br><math>\, _3\phi _2\left(a,b,q^{-k};c,\frac{a b q^{1-k}}{c};q,q\right)=\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> 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>
-<math>\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}</math>
+*   <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>