Editing Theta function (section)

== Derivatives of theta functions ==

These are two identical definitions of the complete elliptic integral of the second kind:

:<math>E(k) = \int_{0}^{\pi/2} \sqrt{1 - k^2\sin(\varphi)^2} d\varphi</math>

:<math>E(k) = \frac{\pi}{2}\sum_{a = 0}^{\infty} \frac{[(2a)!]^2}{(1 - 2a)16^{a}(a!)^4} k^{2a}</math>

The derivatives of the Theta Nullwert functions have these MacLaurin series:

:<math>\theta_{2}'(x) = \frac{\mathrm{d}}{\mathrm{d}x}\,\theta_{2}(x) = \frac{1}{2} x^{-3/4}+\sum_{n = 1}^{\infty} \frac{1}{2}(2n + 1)^2 x^{(2n-1)(2n+3)/4}</math>

:<math>\theta_{3}'(x) = \frac{\mathrm{d}}{\mathrm{d}x}\,\theta_{3}(x) = 2+\sum_{n = 1}^{\infty} 2(n + 1)^2 x^{n(n+2)}</math>

:<math>\theta_{4}'(x) = \frac{\mathrm{d}}{\mathrm{d}x}\,\theta_{4}(x) = -2+\sum_{n = 1}^{\infty} 2(n + 1)^2 (-1)^{n+1} x^{n(n+2)}</math>

The derivatives of theta zero-value functions<ref>{{MathWorld|EllipticAlphaFunction|Elliptic Alpha Function}}</ref> are as follows:
:<math>\theta_{2}'(x) = \frac{\mathrm{d}}{\mathrm{d}x} \,\theta_{2}(x) = \frac{1}{2\pi x} \theta_{2}(x)\theta_{3}(x)^2 E\biggl[\frac{\theta_{2}(x)^2}{\theta_{3}(x)^2}\biggr]</math>
:<math>\theta_{3}'(x) = \frac{\mathrm{d}}{\mathrm{d}x} \,\theta_{3}(x) = \theta_{3}(x)\bigl[\theta_{3}(x)^2 + \theta_{4}(x)^2\bigr]\biggl\{\frac{1}{2\pi x}E\biggl[\frac{\theta_{3}(x)^2 - \theta_{4}(x)^2}{\theta_{3}(x)^2 + \theta_{4}(x)^2}\biggr] - \frac{\theta_{4}(x)^2}{4\,x}\biggr\}</math>
:<math>\theta_{4}'(x) = \frac{\mathrm{d}}{\mathrm{d}x} \,\theta_{4}(x) = \theta_{4}(x)\bigl[\theta_{3}(x)^2 + \theta_{4}(x)^2\bigr]\biggl\{\frac{1}{2\pi x}E\biggl[\frac{\theta_{3}(x)^2 - \theta _{4}(x)^2}{\theta_{3}(x)^2+\theta_{4}(x)^2}\biggr] - \frac{\theta _{3}(x)^ 2}{4\,x}\biggr\}</math>

The two last mentioned formulas are valid for all real numbers of the real definition interval: <math> -1 < x < 1 \,\cap \,x \in \R </math>

And these two last named theta derivative functions are related to each other in this way:
:<math>\vartheta _{4}(x)\biggl[\frac{\mathrm{d}}{\mathrm{d}x} \,\vartheta _{3}(x)\biggr] - \vartheta _{3}(x)\biggl[\frac{\mathrm{d}}{\mathrm{d}x} \,\theta _{4}(x)\biggr] = \frac{1}{4\,x}\,\theta_{3}(x)\,\theta_{4}(x)\bigl[\theta_{3}(x)^4 - \theta_{4}(x)^4\bigr] </math>

The derivatives of the quotients from two of the three theta functions mentioned here always have a rational relationship to those three functions:
:<math>\frac{\mathrm{d}}{\mathrm{d}x} \,\frac{\theta _{2}(x)}{\theta _{3}(x)} = \frac{\theta_{2}(x)\,\theta _{4}(x)^4}{4\,x\,\theta _{3}(x)}</math>
:<math>\frac{\mathrm{d}}{\mathrm{d}x} \,\frac{\theta _{2}(x)}{\theta _{4}(x)} = \frac{\theta_{2}(x)\,\theta _{3}(x)^4}{4\,x\,\theta _{4}(x)}</math>
:<math>\frac{\mathrm{d}}{\mathrm{d}x} \,\frac{\theta _{3}(x)}{\theta _{4}(x)} = \frac{\theta_{3}(x)^5 - \theta _{3}(x)\,\theta _{4}(x)^4}{4\,x\,\theta _{4}(x)}</math>

For the derivation of these derivation formulas see the articles [[Nome (mathematics)]] and [[Modular lambda function]]!