1Phi0(a;;z)1Phi0(b;;az)=1Phi0(ab;;z)
From specialfunctionswiki
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): $4.8 (5)$
