Editing Theta function (section)

==Explicit values==
=== [[:Wiktionary:lemniscatic|Lemniscatic]] values ===

Proper credit for most of these results goes to Ramanujan. See [[Ramanujan's lost notebook]] and a relevant reference at [[Euler function]].  The Ramanujan results quoted at [[Euler function]] plus a few elementary operations give the results below, so they are either in Ramanujan's lost notebook or follow immediately from it. See also Yi (2004).<ref name="Yi">{{Cite journal | first = Jinhee | last = Yi | title = Theta-function identities and the explicit formulas for theta-function and their applications | journal = Journal of Mathematical Analysis and Applications  | pages = 381–400 | volume = 292 | issue = 2 | doi=10.1016/j.jmaa.2003.12.009 | year = 2004| doi-access = free }}</ref> Define,

:<math>\quad \varphi(q) =\vartheta_{00}(0;\tau) =\theta_3(0;q)=\sum_{n=-\infty}^\infty q^{n^2}</math>

with the nome <math>q =e^{\pi i \tau},</math> <math>\tau = n\sqrt{-1},</math> and [[Dedekind eta function]] <math>\eta(\tau).</math> Then for <math>n = 1,2,3,\dots</math>

:<math>\begin{align}
\varphi\left(e^{-\pi} \right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} = \sqrt2\,\eta\left(\sqrt{-1}\right)\\
\varphi\left(e^{-2\pi}\right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \frac{\sqrt{2+\sqrt2}}{2}\\
\varphi\left(e^{-3\pi}\right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \frac{\sqrt{1+\sqrt3}}{\sqrt[8]{108}}\\
\varphi\left(e^{-4\pi}\right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \frac{2+\sqrt[4]{8}}{4}\\
\varphi\left(e^{-5\pi}\right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \sqrt{\frac{2+\sqrt5}{5}}\\
\varphi\left(e^{-6\pi}\right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \frac{\sqrt{\sqrt[4]{1}+\sqrt[4]{3}+\sqrt[4]{4}+\sqrt[4]{9}}}{\sqrt[8]{12^3}}\\
\varphi\left(e^{-7\pi}\right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \frac{\sqrt{\sqrt{13+\sqrt{7}}+\sqrt{7+3\sqrt{7}}}}{\sqrt[8]{14^3}\cdot\sqrt[16]{7}}\\
\varphi\left(e^{-8\pi}\right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \frac{\sqrt{2+\sqrt{2}}+\sqrt[8]{128}}{4}\\
\varphi\left(e^{-9\pi}\right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \frac{1+\sqrt[3]{2+2\sqrt{3}}}{3}\\
\varphi\left(e^{-10\pi}\right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \frac{\sqrt{\sqrt[4]{64}+\sqrt[4]{80}+\sqrt[4]{81}+\sqrt[4]{100}}}{\sqrt[4]{200}}\\
\varphi\left(e^{-11\pi}\right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \frac{\sqrt{11+\sqrt{11}+(5+3\sqrt{3}+\sqrt{11}+\sqrt{33})\sqrt[3]{-44+33\sqrt{3}}+(-5+3\sqrt{3}-\sqrt{11}+\sqrt{33})\sqrt[3]{44+33\sqrt{3}}}}{\sqrt[8]{52180524}}\\
\varphi\left(e^{-12\pi}\right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \frac{\sqrt{\sqrt[4]{1}+\sqrt[4]{2}+\sqrt[4]{3}+\sqrt[4]{4}+\sqrt[4]{9}+\sqrt[4]{18}+\sqrt[4]{24}}}{2\sqrt[8]{108}}\\
\varphi\left(e^{-13\pi}\right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \frac{\sqrt{13+8\sqrt{13}+(11-6\sqrt{3}+\sqrt{13})\sqrt[3]{143+78\sqrt{3}}+(11+6\sqrt{3}+\sqrt{13})\sqrt[3]{143-78\sqrt{3}}}}{\sqrt[4]{19773}}\\
\varphi\left(e^{-14\pi}\right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \frac{\sqrt{\sqrt{13+\sqrt{7}}+\sqrt{7+3\sqrt{7}}+\sqrt{10+2\sqrt{7}}+\sqrt[8]{28}\sqrt{4+\sqrt{7}}}}{\sqrt[16]{28^7}}\\
\varphi\left(e^{-15\pi}\right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \frac{\sqrt{7+3\sqrt{3}+\sqrt{5}+\sqrt{15}+\sqrt[4]{60}+\sqrt[4]{1500}}}{\sqrt[8]{12^3}\cdot\sqrt{5}}\\
2\varphi\left(e^{-16\pi}\right) &= \varphi\left(e^{-4\pi}\right) + \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \frac{\sqrt[4]{1+\sqrt{2}}}{\sqrt[16]{128}}\\
\varphi\left(e^{-17\pi}\right) &= \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \frac{\sqrt{2}(1+\sqrt[4]{17})+\sqrt[8]{17}\sqrt{5+\sqrt{17}}}{\sqrt{17+17\sqrt{17}}}\\
2\varphi\left(e^{-20\pi}\right) &= \varphi\left(e^{-5\pi}\right) + \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \sqrt{\frac{3+2\sqrt[4]{5}}{5\sqrt2}}\\
6\varphi\left(e^{-36\pi}\right) &= 3\varphi\left(e^{-9\pi}\right) + 2\varphi\left(e^{-4\pi}\right) - \varphi\left(e^{-\pi}\right) + \frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34\right)} \sqrt[3]{\sqrt[4]{2}+\sqrt[4]{18}+\sqrt[4]{216}}
\end{align}</math>

If the reciprocal of the [[Gelfond constant]] is raised to the power of the reciprocal of an odd number, then the corresponding  <math> \vartheta_{00} </math> values or <math> \phi </math> values can be represented in a simplified way by using the [[Lemniscate elliptic functions|hyperbolic lemniscatic sine]]:

:<math> \varphi\bigl[\exp(-\tfrac{1}{5}\pi)\bigr] = \sqrt[4]{\pi}\,{\Gamma\left(\tfrac{3}{4}\right)}^{-1} \operatorname{slh}\bigl(\tfrac{1}{5}\sqrt{2}\,\varpi\bigr)\operatorname{slh}\bigl(\tfrac{2}{5}\sqrt{2}\,\varpi\bigr) </math>

:<math> \varphi\bigl[\exp(-\tfrac{1}{7}\pi)\bigr] = \sqrt[4]{\pi}\,{\Gamma\left(\tfrac{3}{4}\right)}^{-1} \operatorname{slh}\bigl(\tfrac{1}{7}\sqrt{2}\,\varpi\bigr)\operatorname{slh}\bigl(\tfrac{2}{7}\sqrt{2}\,\varpi\bigr) \operatorname{slh}\bigl(\tfrac{3}{7}\sqrt{2}\,\varpi\bigr) </math>

:<math> \varphi\bigl[\exp(-\tfrac{1}{9}\pi)\bigr] = \sqrt[4]{\pi}\,{\Gamma\left(\tfrac{3}{4}\right)}^{-1} \operatorname{slh}\bigl(\tfrac{1}{9}\sqrt{2}\,\varpi\bigr)\operatorname{slh}\bigl(\tfrac{2}{9}\sqrt{2}\,\varpi\bigr) \operatorname{slh}\bigl(\tfrac{3}{9}\sqrt{2}\,\varpi\bigr) \operatorname{slh} \bigl(\tfrac{4}{9}\sqrt{2}\,\varpi\bigr) </math>

:<math> \varphi\bigl[\exp(-\tfrac{1}{11}\pi)\bigr] = \sqrt[4]{\pi}\,{\Gamma\left(\tfrac{3}{4}\right)}^{-1} \operatorname{slh}\bigl(\tfrac{1}{11}\sqrt{2}\,\varpi\bigr)\operatorname{slh}\bigl(\tfrac{2}{11}\sqrt{2}\,\varpi\bigr) \operatorname{slh}\bigl(\tfrac{3}{11}\sqrt{2}\,\varpi\bigr) \operatorname{slh} \bigl(\tfrac{4}{11}\sqrt{2}\,\varpi\bigr) \operatorname{slh}\bigl(\tfrac{5}{11}\sqrt{2}\,\varpi\bigr) </math>

With the letter <math> \varpi </math> the [[Lemniscate constant]] is represented.

Note that the following modular identities hold:

:<math>\begin{align}
2\varphi\left(q^4\right) &= \varphi(q)+\sqrt{2\varphi^2\left(q^2\right)-\varphi^2(q)}\\
3\varphi\left(q^9\right) &= \varphi(q)+\sqrt[3]{9\frac{\varphi^4\left(q^3\right)}{\varphi(q)}-\varphi^3(q)}\\
\sqrt{5}\varphi\left(q^{25}\right) &= \varphi\left(q^5\right)\cot\left(\frac{1}{2}\arctan\left(\frac{2}{\sqrt{5}}\frac{\varphi(q)\varphi\left(q^5\right)}{\varphi^2(q)-\varphi^2\left(q^5\right)}\frac{1+s(q)-s^2(q)}{s(q)}\right)\right)
\end{align}</math>

where <math>s(q)=s\left(e^{\pi i\tau}\right)=-R\left(-e^{-\pi i/(5\tau)}\right)</math> is the [[Rogers–Ramanujan continued fraction]]:

:<math>\begin{align}
s(q) &= \sqrt[5]{\tan\left(\frac{1}{2}\arctan\left(\frac{5}{2}\frac{\varphi^2\left(q^5\right)}{\varphi^2(q)}-\frac{1}{2}\right)\right)\cot^2\left(\frac{1}{2}\operatorname{arccot}\left(\frac{5}{2}\frac{\varphi^2\left(q^5\right)}{\varphi^2(q)}-\frac{1}{2}\right)\right)}\\
&= \cfrac{e^{-\pi i/(25\tau)}}{1-\cfrac{e^{-\pi i/(5\tau)}}{1+\cfrac{e^{-2\pi i/(5\tau)}}{1-\ddots}}}
\end{align}</math>

=== [[Equianharmonic]] values ===
The mathematician [[Bruce Berndt]] found out further values<ref>{{cite journal |last1=Berndt |first1=Bruce C |last2=Rebák |first2=Örs |title=Explicit Values for Ramanujan's Theta Function {{not a typo|ϕ}}(q) |journal=Hardy-Ramanujan Journal |date=9 January 2022 |volume=44 |pages=8923 |doi=10.46298/hrj.2022.8923 |s2cid=245851672 |doi-access=free |arxiv=2112.11882 }}</ref> of the theta function:

:<math>\begin{array}{lll}
\varphi\left(\exp( -\sqrt{3}\,\pi)\right) &=& \pi^{-1}{\Gamma\left(\tfrac{4}{3}\right)}^{3/2}2^{-2/3}3^{13/8} \\
\varphi\left(\exp(-2\sqrt{3}\,\pi)\right) &=& \pi^{-1}{\Gamma\left(\tfrac{4}{3}\right)}^{3/2}2^{-2/3}3^{13/8}\cos(\tfrac{1}{24}\pi) \\
\varphi\left(\exp(-3\sqrt{3}\,\pi)\right) &=& \pi^{-1}{\Gamma\left(\tfrac{4}{3}\right)}^{3/2}2^{-2/3}3^{7/8}(\sqrt[3]{2}+1) \\
\varphi\left(\exp(-4\sqrt{3}\,\pi)\right) &=& \pi^{-1}{\Gamma\left(\tfrac{4}{3}\right)}^{3/2}2^{-5/3}3^{13/8}\Bigl(1+\sqrt{\cos(\tfrac{1}{12}\pi)}\Bigr) \\
\varphi\left(\exp(-5\sqrt{3}\,\pi)\right) &=& \pi^{-1}{\Gamma\left(\tfrac{4}{3}\right)}^{3/2}2^{-2/3}3^{5/8}\sin(\tfrac{1}{5}\pi)(\tfrac{2}{5}\sqrt[3]{100}+\tfrac{2}{5}\sqrt[3]{10}+\tfrac{3}{5}\sqrt{5}+1)
\end{array}</math>

=== Further values ===
Many values of the theta function<ref>{{cite journal |last1=Yi |first1=Jinhee |title=Theta-function identities and the explicit formulas for theta-function and their applications |journal=Journal of Mathematical Analysis and Applications |date=15 April 2004 |volume=292 |issue=2 |pages=381–400 |doi=10.1016/j.jmaa.2003.12.009 |doi-access=free }}</ref> and especially of the shown phi function can be represented in terms of the gamma function:

:<math>\begin{array}{lll}
\varphi\left(\exp( -\sqrt{2}\,\pi)\right) &=& \pi^{-1/2}\Gamma\left(\tfrac{9}{8}\right){\Gamma\left(\tfrac{5}{4}\right)}^{-1/2}2^{7/8} \\
\varphi\left(\exp(-2\sqrt{2}\,\pi)\right) &=& \pi^{-1/2}\Gamma\left(\tfrac{9}{8}\right){\Gamma\left(\tfrac{5}{4}\right)}^{-1/2}2^{1/8}\Bigl(1+\sqrt{\sqrt{2}-1}\Bigr) \\
\varphi\left(\exp(-3\sqrt{2}\,\pi)\right) &=& \pi^{-1/2}\Gamma\left(\tfrac{9}{8}\right){\Gamma\left(\tfrac{5}{4}\right)}^{-1/2}2^{3/8}3^{-1/2}(\sqrt{3}+1)\sqrt{\tan(\tfrac{5}{24}\pi)} \\
\varphi\left(\exp(-4\sqrt{2}\,\pi)\right) &=& \pi^{-1/2}\Gamma\left(\tfrac{9}{8}\right){\Gamma\left(\tfrac{5}{4}\right)}^{-1/2}2^{-1/8}\Bigl(1+\sqrt[4]{2\sqrt{2}-2}\Bigr) \\
\varphi\left(\exp(-5\sqrt{2}\,\pi)\right) &=& \pi^{-1/2}\Gamma\left(\tfrac{9}{8}\right){\Gamma\left(\tfrac{5}{4}\right)}^{-1/2} \frac{1}{15}\,2^{3/8} \times \\
 && \times \biggl[\sqrt[3]{5}\,\sqrt{10+2\sqrt{5}}\biggl(\sqrt[3]{5+\sqrt{2}+3\sqrt{3}}+\sqrt[3]{5+\sqrt{2}-3\sqrt{3}}\,\biggr)-\bigl(2-\sqrt{2}\,\bigr)\sqrt{25-10\sqrt{5}}\,\biggr] \\
\varphi\left(\exp( -\sqrt{6}\,\pi)\right) &=& \pi^{-1/2}\Gamma\left(\tfrac{5}{24}\right){\Gamma\left(\tfrac{5}{12}\right)}^{-1/2}2^{-13/24}3^{-1/8}\sqrt{\sin(\tfrac{5}{12}\pi)} \\
\varphi\left(\exp(-\tfrac{1}{2}\sqrt{6}\,\pi)\right) &=& \pi^{-1/2}\Gamma\left(\tfrac{5}{24}\right){\Gamma\left(\tfrac{5}{12}\right)}^{-1/2}2^{5/24}3^{-1/8}\sin(\tfrac{5}{24}\pi)
\end{array}</math>