Difference between revisions of "1Phi0(a;;z)1Phi0(b;;az)=1Phi0(ab;;z)"

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(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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Revision as of 21:47, 17 June 2017

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

References