Editing Theta function (section)

== Derivation of the theta values ==

=== Identity of the Euler beta function ===
In the following, three important theta function values are to be derived as examples:

This is how the [[Euler beta function]] is defined in its reduced form:

:<math>\beta(x) = \frac{\Gamma(x)^2}{\Gamma(2x)}</math>

In general, for all natural numbers <math>n \isin \mathbb{N}</math> this formula of the Euler beta function is valid:

:<math>\frac{4^{-1/(n + 2)}}{n + 2}\csc\bigl(\frac{\pi}{n + 2}\bigr)\beta\biggl[\frac{n}{2(n + 2)}\biggr] = \int_{0}^{\infty} \frac{1}{\sqrt{x^{n+2} + 1}} \,\mathrm {d}x</math>

=== Exemplary elliptic integrals ===

In the following some ''Elliptic Integral Singular Values''<ref>{{cite web|access-date=2023-04-07|title=Elliptic Integral Singular Value
|url=https://archive.lib.msu.edu/crcmath/math/math/e/e103.htm|website=msu.edu}}<!-- auto-translated by Module:CS1 translator --></ref> are derived:

{| class="wikitable"
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The ensuing function has the following lemniscatically elliptic antiderivative:

:<math>\frac{1}{\sqrt{x^4 + 1}} = \frac{\mathrm{d}}{\mathrm{d}x}\,\frac{1}{2} F\biggl[2\arctan(x); \frac{1}{2}\sqrt{2}\,\biggr]</math>

For the value <math>n = 2</math> this identity appears:

:<math>\frac{1}{4\sqrt{2}}\csc\bigl(\frac{\pi}{4}\bigr)\beta\bigl(\frac{1}{4}\bigr) = \int_{0}^{\infty} \frac{1}{\sqrt{x^4 + 1}} \,\mathrm{d}x = \biggl\{{\color{blue}\frac{1}{2} F\biggl[2\arctan(x); \frac{1}{2}\sqrt{2}\,\biggr]}\biggr\}_{x = 0}^{x = \infty} =</math>

:<math>= \frac{1}{2} F\bigl(\pi; \frac{1}{2}\sqrt{2}\bigr) = K\bigl(\frac{1}{2}\sqrt{2}\bigr) </math>

This result follows from that equation chain:

:<math>{\color{ForestGreen}K\bigl(\frac{1}{2}\sqrt{2}\bigr) = \frac{1}{4}\beta\bigl(\frac{1}{ 4}\bigr)}</math>
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The following function has the following equianharmonic elliptic antiderivative:

:<math>\frac{1}{\sqrt{x^6 + 1}} = \frac{\mathrm{d}}{\mathrm{d}x}\,\frac{1}{6}\sqrt [4]{27}F\biggl[2\arctan\biggl(\frac{\sqrt[4]{3}\,x}{\sqrt{x^2 + 1}}\biggr);\frac{1 }{4}(\sqrt{6} + \sqrt{2})\biggr]</math>

For the value <math>n = 4</math> that identity appears:

:<math>\frac{1}{6\sqrt[3]{2}}\csc\bigl(\frac{\pi}{6}\bigr)\beta\bigl(\frac{1}{3} \bigr) = \int_{0}^{\infty} \frac{1}{\sqrt{x^6 + 1}} \,\mathrm{d}x = \biggl\{{\color{blue}\frac{1}{6}\sqrt[4]{27}F\biggl[2\arctan\biggl(\frac{\sqrt[4]{3}\,x}{\sqrt{x^2 + 1}}\biggr);\frac{1}{4}(\sqrt{6} + \sqrt{2})\biggr]}\biggr\}_{x = 0}^{x = \infty} =</math>

:<math>= \frac{1}{6}\sqrt[4]{27} F\bigl[2\arctan(\sqrt[4]{3});\frac{1}{4}(\sqrt {6} + \sqrt{2})\bigr] = \frac{2}{9}\sqrt[4]{27} K\bigl[\frac{1}{4}(\sqrt{6} + \sqrt{2})\bigr] = \frac{2}{3}\sqrt[4]{3} K\bigl[\frac{1}{4}(\sqrt{6} - \sqrt{2}) \bigr] </math>

This result follows from that equation chain:

:<math>{\color{ForestGreen}K\bigl[\frac{1}{4}(\sqrt{6} - \sqrt{2})\bigr] = \frac{1}{2\sqrt[3 ]{2}\sqrt[4]{3}}\beta\bigl(\frac{1}{3}\bigr)}</math>
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And the following function has the following elliptic antiderivative:
:<math>\frac{1}{\sqrt{x^8 + 1}} =</math>

:<math>= \frac{\mathrm{d}}{\mathrm{d}x}\,\frac{1}{4}\sec \bigl(\frac{\pi}{8}\bigr)F\biggl\{2\arctan\biggl[\frac{2\cos(\pi/8)\,x}{\sqrt{x^4 + \sqrt{2}\,x^2 + 1} - x^2 + 1}\biggr];2\sqrt[4]{2}\sin\bigl(\frac{\pi}{8}\bigr) \biggr\} + \frac{1}{4}\sec\bigl(\frac{\pi}{8}\bigr)F\biggl\{\arcsin\biggl[\frac{2\cos(\pi/ 8)\,x}{x^2 + 1}\biggr];\tan\bigl(\frac{\pi}{8}\bigr)\biggr\}</math>

For the value <math>n = 6</math> the following identity appears:

:<math>\frac{1}{8\sqrt[4]{2}}\csc\bigl(\frac{\pi}{8}\bigr)\beta\bigl(\frac{3}{8} \bigr) = \int_{0}^{\infty} \frac{1}{\sqrt{x^8 + 1}} \,\mathrm{d}x =</math>

:<math>= \biggl\langle{\color{blue}\frac{1}{4}\sec\bigl(\frac{\pi}{8}\bigr)F\biggl\{2\arctan\biggl[\frac{2\cos(\pi/8)\,x}{\sqrt{x^4 + \sqrt{2}\,x^2 + 1} - x^2 + 1}\biggr];2\sqrt[4]{ 2}\sin\bigl(\frac{\pi}{8}\bigr)\biggr\} + \frac{1}{4}\sec\bigl(\frac{\pi}{8}\bigr)F \biggl\{\arcsin\biggl[\frac{2\cos(\pi/8)\,x}{x^2 + 1}\biggr];\tan\bigl(\frac{\pi}{8} \bigr)\biggr\}}\biggr\rangle_{x = 0}^{x = \infty} =</math>

:<math>= \frac{1}{4}\sec\bigl(\frac{\pi}{8}\bigr)F\bigl[\pi;2\sqrt[4]{2}\sin\bigl (\frac{\pi}{8}\bigr)\bigr] = \frac{1}{2}\sec\bigl(\frac{\pi}{8}\bigr)K(\sqrt{2\sqrt {2} - 2}\bigr) = 2\sin\bigl(\frac{\pi}{8}\bigr)K(\sqrt{2} - 1)</math>

This result follows from that equation chain:
:<math>{\color{ForestGreen}K(\sqrt{2} - 1) = \frac{1}{8}\sqrt[4]{2}\,(\sqrt{2} + 1)\, \beta\bigl(\frac{3}{8}\bigr)}</math>
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=== Combination of the integral identities with the nome ===

The elliptic nome function has these important values:
:<math>q(\tfrac{1}{2}\sqrt{2}) = \exp(-\pi)</math>

:<math>q[\tfrac{1}{4}(\sqrt{6} - \sqrt{2})] = \exp(-\sqrt{3}\,\pi)</math>

:<math>q(\sqrt{2} - 1) = \exp(-\sqrt{2}\,\pi)</math>

For the proof of the correctness of these nome values, see the article [[Nome (mathematics)]]!

On the basis of these integral identities and the above-mentioned '''Definition and identities to the theta functions''' in the same section of this article, exemplary theta zero values shall be determined now:
:{| class="wikitable"
|<math>\theta_{3}[q(k)] = \sqrt{2\pi^{-1} K(k)}</math>

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:<math>\theta_{3}[\exp(-\pi)] = \theta_{3}[q(\tfrac{1}{2}\sqrt{2})] = \sqrt{2\pi^ {-1}K(\tfrac{1}{2}\sqrt{2})} = 2^{-1/2}\pi^{-1/2}\beta(\tfrac{1}{4} )^{1/2} = 2^{-1/4}\sqrt[4]{\pi}\,{\Gamma\bigl(\tfrac{3}{4}\bigr)}^{-1} </math>

:<math>\theta _{3}[\exp(-\sqrt{3}\,\pi )] = \theta _{3}\bigl\{q\bigl[\tfrac{1}{4}(\sqrt {6} - \sqrt{2})\bigr]\bigr\} = \sqrt{2\pi^{-1}K\bigl[\tfrac{1}{4}(\sqrt{6} - \sqrt {2})\bigr]} = 2^{-1/6}3^{-1/8}\pi^{-1/2}\beta(\tfrac{1}{3})^{1/ 2}</math>

:<math>\theta _{3}[\exp(-\sqrt{2}\,\pi )] = \theta _{3}[q(\sqrt{2} - 1)] = \sqrt{2\pi ^{-1}K(\sqrt{2} - 1)} = 2^{-1/8}\cos(\tfrac{1}{8}\pi)\,\pi^{-1/2} \beta(\tfrac{3}{8})^{1/2}</math>

:{| class="wikitable"
|<math>\theta_{4}[q(k)] = \sqrt[4]{1 - k^2}\,\sqrt{2\pi^{-1} K(k)}</math>
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:<math>\theta_{4}[\exp(-\sqrt{2}\,\pi)] = \theta_{4}[q(\sqrt{2} - 1)] = \sqrt[4]{ 2\sqrt{2} - 2}\,\sqrt{2\pi^{-1}K(\sqrt{2} - 1)} = 2^{-1/4}\cos(\tfrac{1} {8}\pi)^{1/2}\,\pi^{-1/2}\beta(\tfrac{3}{8})^{1/2}</math>