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

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==Theorem==
 
==Theorem==
 
The following formula holds:
 
The following formula holds:
$$\dfrac{z}{1-z} {}_2\phi_1(q,q;q^2;z) = \displaystyle\sum_{k=1}^{\infty} \dfrac{z^k}{1-q^k},$$
+
$$\dfrac{z}{1-q} {}_2\phi_1(q,q;q^2;z) = \displaystyle\sum_{k=1}^{\infty} \dfrac{z^k}{1-q^k},$$
 
where ${}_2\phi_1$ denotes [[basic hypergeometric phi]].
 
where ${}_2\phi_1$ denotes [[basic hypergeometric phi]].
  
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==References==
 
==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=1Phi0(a;;z)1Phi0(b;;az)=1Phi0(ab;;z)|next=2Phi1(q,-1;-q;z)=1+2Sum z^k/(1+q^k)}}: $4.8 (6)$ (typo in text: text has sum beginning at $k=0$)
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* {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=1Phi0(a;;z)1Phi0(b;;az)=1Phi0(ab;;z)|next=2Phi1(q,-1;-q;z)=1+2Sum z^k/(1+q^k)}}: $4.8 (6)$ (typo in text: text has sum beginning at $k=0$)
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Latest revision as of 23:26, 3 March 2018

Theorem

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

Proof

References