Editing Theta function (section)

=== Modulus dependent theorems===

In combination with the elliptic modulus, the following formulas can be displayed:

These are the formulas for the square of the elliptic nome:

:<math>\theta_{4}[q(k)] = \theta_{4}[q(k)^2]\sqrt[8]{1 - k^2}</math>
:<math>\theta_{4}[q(k)^2] = \theta_{3}[q(k)]\sqrt[8]{1 - k^2}</math>
:<math>\theta_{3}[q(k)^2] = \theta_{3}[q(k)]\cos[\tfrac{1}{2}\arcsin(k)]</math>

And this is an efficient formula for the cube of the nome:

:<math> \theta_{4}\biggl\langle q\bigl\{\tan\bigl[\tfrac{1}{2}\arctan(t^3)\bigr]\bigr\}^3 \biggr\rangle = 
\theta_{4}\biggl\langle q\bigl\{\tan\bigl[\tfrac{1}{2}\arctan(t^3)\bigr]\bigr\} \biggr\rangle \,3^{-1/2} \bigl(\sqrt{2\sqrt{t^4 - t^2 + 1} - t^2 + 2} + \sqrt{t^2 + 1}\,\bigr)^{1/2} </math>

For all real values <math> t \in \R </math> the now mentioned formula is valid.

And for this formula two examples shall be given:

First calculation example with the value <math> t = 1 </math> inserted:

:{| class="wikitable"
|
<math> \theta_{4}\biggl\langle q\bigl\{\tan\bigl[\tfrac{1}{2}\arctan(1)\bigr]\bigr\}^3 \biggr\rangle = 
\theta_{4}\biggl\langle q\bigl\{\tan\bigl[\tfrac{1}{2}\arctan(1)\bigr]\bigr\} \biggr\rangle \,3^{-1/2} \bigl(\sqrt{3} + \sqrt{2}\,\bigr)^{1/2} </math>
|-
|
<math> \theta_{4}\bigl[\exp(-3\sqrt{2}\,\pi)\bigr] = 
\theta_{4}\bigl[\exp(-\sqrt{2}\,\pi)\bigr] \,3^{-1/2} \bigl(\sqrt{3} + \sqrt{2}\,\bigr)^{1/2} </math>
|}

Second calculation example with the value <math> t = \Phi^{-2} </math> inserted:

:{| class="wikitable"
|
<math> \theta_{4}\biggl\langle q\bigl\{\tan\bigl[\tfrac{1}{2}\arctan(\Phi^{-6})\bigr]\bigr\}^3 \biggr\rangle = 
\theta_{4}\biggl\langle q\bigl\{\tan\bigl[\tfrac{1}{2}\arctan(\Phi^{-6})\bigr]\bigr\} \biggr\rangle \,3^{-1/2} \bigl(\sqrt{2\sqrt{\Phi^{-8} - \Phi^{-4} + 1} - \Phi^{-4} + 2} + \sqrt{\Phi^{-4} + 1}\,\bigr)^{1/2} </math>
|-
|
<math> \theta_{4}\bigl[\exp(-3\sqrt{10}\,\pi)\bigr] = 
\theta_{4}\bigl[\exp(-\sqrt{10}\,\pi)\bigr] \,3^{-1/2} \bigl(\sqrt{2\sqrt{\Phi^{-8} - \Phi^{-4} + 1} - \Phi^{-4} + 2} + \sqrt{\Phi^{-4} + 1}\,\bigr)^{1/2} </math>
|}

The constant <math> \Phi </math> represents the [[Golden ratio]] number <math> \Phi = \tfrac{1}{2}(\sqrt{5} + 1)</math> exactly.