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},$$ |
where ${}_2\phi_1$ denotes [[basic hypergeometric phi]]. | where ${}_2\phi_1$ denotes [[basic hypergeometric phi]]. | ||
Revision as of 21:55, 17 June 2017
Theorem
The following formula holds: $$\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: Harry Bateman: Higher Transcendental Functions Volume I ... (previous) ... (next): $4.8 (6)$ (typo in text: text has sum beginning at $k=0$)
