1Phi0(a;;z)1Phi0(b;;az)=1Phi0(ab;;z)

From specialfunctionswiki
Revision as of 23:26, 3 March 2018 by Tom (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search

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