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: | ||
| − | $$\dfrac{z}{1- | + | $$\dfrac{z}{1-q} {}_2\phi_1(q,q;q^2;z) = \displaystyle\sum_{k=1}^{\infty} \dfrac{z^k}{1-q^k},$$ |
where ${}_2\phi_1$ denotes [[basic hypergeometric phi]]. | where ${}_2\phi_1$ denotes [[basic hypergeometric phi]]. | ||
| Line 7: | Line 7: | ||
==References== | ==References== | ||
| − | * {{BookReference|Higher Transcendental Functions Volume I|1953| | + | * {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=1Phi0(a;;z)1Phi0(b;;az)=1Phi0(ab;;z)|next=2Phi1(q,-1;-q;z)=1+2Sum z^k/(1+q^k)}}: $4.8 (6)$ (typo in text: text has sum beginning at $k=0$) |
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Unproven]] | [[Category:Unproven]] | ||
Latest revision as of 23:26, 3 March 2018
Theorem
The following formula holds: $$\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
- 1953: Arthur Erdélyi, Wilhelm Magnus, Fritz Oberhettinger and Francesco G. Tricomi: Higher Transcendental Functions Volume I ... (previous) ... (next): $4.8 (6)$ (typo in text: text has sum beginning at $k=0$)
