Difference between revisions of "Logarithmic derivative of Jacobi theta 2 equals negative tangent + a sum of sines"
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(Created page with "==Theorem== The following formula holds: $$\dfrac{\vartheta_2'(u)}{\vartheta_2(u)} = -\tan(u)+4\displaystyle\sum_{k=1}^{\infty} (-1)^k \dfrac{q^{2k}}{1-q^{2k}} \sin(2ku),$$ wh...") |
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==Theorem== | ==Theorem== | ||
The following formula holds: | The following formula holds: | ||
| − | $$\dfrac{\vartheta_2'(u)}{\vartheta_2(u)} = -\tan(u)+4\displaystyle\sum_{k=1}^{\infty} (-1)^k \dfrac{q^{2k}}{1-q^{2k}} \sin(2ku),$$ | + | $$\dfrac{\vartheta_2'(u,q)}{\vartheta_2(u,q)} = -\tan(u)+4\displaystyle\sum_{k=1}^{\infty} (-1)^k \dfrac{q^{2k}}{1-q^{2k}} \sin(2ku),$$ |
where $\vartheta_2$ denotes the [[Jacobi theta 2]], $\tan$ denotes the [[tangent]], and $\sin$ denotes [[sine]]. | where $\vartheta_2$ denotes the [[Jacobi theta 2]], $\tan$ denotes the [[tangent]], and $\sin$ denotes [[sine]]. | ||
Revision as of 07:14, 27 June 2016
Theorem
The following formula holds: $$\dfrac{\vartheta_2'(u,q)}{\vartheta_2(u,q)} = -\tan(u)+4\displaystyle\sum_{k=1}^{\infty} (-1)^k \dfrac{q^{2k}}{1-q^{2k}} \sin(2ku),$$ where $\vartheta_2$ denotes the Jacobi theta 2, $\tan$ denotes the tangent, and $\sin$ denotes sine.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous): 16.29.2
