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

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==References==
 
==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=1Phi0(a;;z) as infinite product
+
* {{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