Difference between revisions of "1Phi0(a;;z)1Phi0(b;;az)=1Phi0(ab;;z)"
From specialfunctionswiki
(Created page with "==Theorem== The following formula holds: $${}_1\phi_0(a;;z){}_1\phi_0(b;;z)={}_1\phi_0(ab;;z),$$ where ${}_1\phi_0$ denotes basic hypergeometric phi. ==Proof== ==Referen...") |
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==Theorem== | ==Theorem== | ||
The following formula holds: | The following formula holds: | ||
| − | $${}_1\phi_0(a;;z){}_1\phi_0(b;; | + | $${}_1\phi_0(a;;z){}_1\phi_0(b;;az)={}_1\phi_0(ab;;z),$$ |
where ${}_1\phi_0$ denotes [[basic hypergeometric phi]]. | where ${}_1\phi_0$ denotes [[basic hypergeometric phi]]. | ||
| Line 7: | Line 7: | ||
==References== | ==References== | ||
| − | * {{BookReference|Higher Transcendental Functions Volume I|1953| | + | * {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=1Phi0(a;;z) as infinite product |
|next=(z/(1-q))2Phi1(q,q;q^2;z)=Sum z^k/(1-q^k)}}: $4.8 (5)$ | |next=(z/(1-q))2Phi1(q,q;q^2;z)=Sum z^k/(1-q^k)}}: $4.8 (5)$ | ||
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Unproven]] | [[Category:Unproven]] | ||
Latest revision as of 23:26, 3 March 2018
Theorem
The following formula holds: $${}_1\phi_0(a;;z){}_1\phi_0(b;;az)={}_1\phi_0(ab;;z),$$ where ${}_1\phi_0$ denotes basic hypergeometric phi.
Proof
References
- 1953: Arthur Erdélyi, Wilhelm Magnus, Fritz Oberhettinger and Francesco G. Tricomi: Higher Transcendental Functions Volume I ... (previous) ... (next): $4.8 (5)$
