Difference between revisions of "2Phi1(q,-1;-q;z)=1+2Sum z^k/(1+q^k)"

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(Created page with "==Theorem== The following formula holds: $${}_2\phi_1(q,-1;-q;z)=1+2\displaystyle\sum_{k=1}^{\infty} \dfrac{z^k}{1+q^k},$$ where ${}_2\phi_1$ denotes basic hypergeometric ph...")
 
 
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==References==
 
==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=(z/(1-q))2Phi1(q,q;q^2;z)=Sum z^k/(1-q^k)|next=z/(1-sqrt(q))2Phi1(q,sqrt(q);sqrt(q^3);z)=Sum z^k/(1-q^(k-1/2))}}: $4.8 (7)$
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* {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=(z/(1-q))2Phi1(q,q;q^2;z)=Sum z^k/(1-q^k)|next=z/(1-sqrt(q))2Phi1(q,sqrt(q);sqrt(q^3);z)=Sum z^k/(1-q^(k-1/2))}}: $4.8 (7)$
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Latest revision as of 23:27, 3 March 2018

Theorem

The following formula holds: $${}_2\phi_1(q,-1;-q;z)=1+2\displaystyle\sum_{k=1}^{\infty} \dfrac{z^k}{1+q^k},$$ where ${}_2\phi_1$ denotes basic hypergeometric phi.

Proof

References