===
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). Define, :\quad \varphi(q) =\vartheta_{00}(0;\tau) =\theta_3(0;q)=\sum_{n=-\infty}^\infty q^{n^2} with the nome q =e^{\pi i \tau}, \tau = n\sqrt{-1}, and
Dedekind eta function \eta(\tau). Then for n = 1,2,3,\dots :\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} If the reciprocal of the
Gelfond constant is raised to the power of the reciprocal of an odd number, then the corresponding \vartheta_{00} values or \phi values can be represented in a simplified way by using the
hyperbolic lemniscatic sine: : \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) : \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) : \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) : \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) With the letter \varpi the
Lemniscate constant is represented. Note that the following modular identities hold: :\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} where s(q)=s\left(e^{\pi i\tau}\right)=-R\left(-e^{-\pi i/(5\tau)}\right) is the
Rogers–Ramanujan continued fraction: :\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} ===
Equianharmonic values === The mathematician
Bruce Berndt found out further values of the theta function: :\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}
Further values Many values of the theta function and especially of the shown phi function can be represented in terms of the gamma function: :\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} ==Nome power theorems==