# Specific values of Jacobi theta functions

See Jacobi theta functions for other properties of these functions.

## Values at infinity

$\lim_{\tau \to i \infty} \theta_{1}\!\left(z , \tau\right) = 0$
$\lim_{\tau \to i \infty} \theta_{2}\!\left(z , \tau\right) = 0$
$\lim_{\tau \to i \infty} \theta_{3}\!\left(z , \tau\right) = 1$
$\lim_{\tau \to i \infty} \theta_{4}\!\left(z , \tau\right) = 1$

## Trivial values

$\theta_{1}\!\left(0 , \tau\right) = 0$
$\theta^{(2 r)}_{1}\!\left(0 , \tau\right) = 0$
$\theta^{(2 r + 1)}_{2}\!\left(0 , \tau\right) = 0$
$\theta^{(2 r + 1)}_{3}\!\left(0 , \tau\right) = 0$
$\theta^{(2 r + 1)}_{4}\!\left(0 , \tau\right) = 0$

## Square lattice

### Theta constants for the square lattice

$\theta_{3}\!\left(0 , i\right) = \frac{{\pi}^{1 / 4}}{\Gamma\!\left(\frac{3}{4}\right)}$
$\theta_{3}\!\left(0 , i\right) = \frac{\Gamma\!\left(\frac{1}{4}\right)}{\sqrt{2} {\pi}^{3 / 4}}$
$\theta_{3}\!\left(0 , i\right) \notin \overline{\mathbb{Q}}$
$\theta_{2}\!\left(0 , i\right) = \theta_{4}\!\left(0 , i\right) = \left[{2}^{-1 / 4}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , i\right) \in \left[1.0864348112133080145753161215102234570702057072452 \pm 1.89 \cdot 10^{-50}\right]$
$\theta_{4}\!\left(0 , i\right) \in \left[0.91357913815611682140724259340122208970196391639347 \pm 9.67 \cdot 10^{-52}\right]$

### Values for simple rational arguments

$\theta_{1}\!\left(\frac{n}{4} , i\right) = \begin{cases} 0, & n \equiv 0 \pmod {4}\\{\left(-1\right)}^{\left\lfloor n / 4 \right\rfloor} \theta_{4}\!\left(0 , i\right), & n \equiv 2 \pmod {4}\\{\left(-1\right)}^{\left\lfloor n / 4 \right\rfloor} \left[{2}^{-7 / 16} \sqrt{\sqrt{2} - 1} {\left(\sqrt{2} + 1\right)}^{1 / 4}\right] \theta_{3}\!\left(0 , i\right), & \text{otherwise}\\ \end{cases}$
$\theta_{2}\!\left(\frac{n}{4} , i\right) = \begin{cases} {\left(-1\right)}^{\left\lfloor \left( n + 1 \right) / 4 \right\rfloor} \theta_{4}\!\left(0 , i\right), & n \equiv 0 \pmod {4}\\0, & n \equiv 2 \pmod {4}\\{\left(-1\right)}^{\left\lfloor \left( n + 1 \right) / 4 \right\rfloor} \left[{2}^{-7 / 16} \sqrt{\sqrt{2} - 1} {\left(\sqrt{2} + 1\right)}^{1 / 4}\right] \theta_{3}\!\left(0 , i\right), & \text{otherwise}\\ \end{cases}$
$\theta_{3}\!\left(\frac{n}{4} , i\right) = \begin{cases} \theta_{3}\!\left(0 , i\right), & n \equiv 0 \pmod {4}\\\theta_{4}\!\left(0 , i\right), & n \equiv 2 \pmod {4}\\\left[{2}^{-7 / 16} {\left(\sqrt{2} + 1\right)}^{1 / 4}\right] \theta_{3}\!\left(0 , i\right), & \text{otherwise}\\ \end{cases}$
$\theta_{4}\!\left(\frac{n}{4} , i\right) = \begin{cases} \theta_{4}\!\left(0 , i\right), & n \equiv 0 \pmod {4}\\\theta_{3}\!\left(0 , i\right), & n \equiv 2 \pmod {4}\\\left[{2}^{-7 / 16} {\left(\sqrt{2} + 1\right)}^{1 / 4}\right] \theta_{3}\!\left(0 , i\right), & \text{otherwise}\\ \end{cases}$

## Theta constants for non-square lattices

### Conversion from index 2 and 4 to index 3

$\theta_{2}\!\left(0 , y i\right) = \frac{1}{\sqrt{y}} \theta_{3}\!\left(0 , 1 + \frac{i}{y}\right)$
$\theta_{2}\!\left(0 , 1 + y i\right) = \frac{1 + i}{\sqrt{2 y}} \theta_{3}\!\left(0 , 1 + \frac{i}{y}\right)$
$\theta_{4}\!\left(0 , y i\right) = \theta_{3}\!\left(0 , 1 + y i\right)$
$\theta_{4}\!\left(0 , 1 + y i\right) = \theta_{3}\!\left(0 , y i\right)$

### Algebraic ratios for real part 0

$\theta_{3}\!\left(0 , \frac{i}{2}\right) = \left[\sqrt{\frac{\sqrt{2} + 1}{2}} \cdot {2}^{1 / 4}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , \frac{i}{3}\right) = \left[{\left(2 \sqrt{3} + 3\right)}^{1 / 4}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , \frac{i}{4}\right) = \left[\frac{1 + {2}^{-1 / 4}}{\sqrt{1 + \sqrt{2}}} \sqrt{\frac{\sqrt{2} + 1}{2}} \cdot {2}^{1 / 2}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , 2 i\right) = \left[\frac{\sqrt{\sqrt{2} + 2}}{2}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , 3 i\right) = \left[\frac{\sqrt{\sqrt{3} + 1}}{{2}^{1 / 4} \cdot {3}^{3 / 8}}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , 4 i\right) = \left[\frac{1 + {2}^{-1 / 4}}{2}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , 5 i\right) = \left[\frac{1}{\sqrt{5 \sqrt{5} - 10}}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , 5 i\right) = \left[\frac{\sqrt{5 + 2 \sqrt{5}}}{{5}^{3 / 4}}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , 6 i\right) = \left[\frac{{\left(-4 + 3 \sqrt{2} + {3}^{5 / 4} + 2 \sqrt{3} - {3}^{3 / 4} + 2 \sqrt{2} \left({3}^{3 / 4}\right)\right)}^{1 / 3}}{2 \left({3}^{3 / 8}\right) {\left(\left(\sqrt{2} - 1\right) \left(\sqrt{3} - 1\right)\right)}^{1 / 6}}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , 7 i\right) = \left[\sqrt{\frac{\sqrt{13 + \sqrt{7}} + \sqrt{7 + 3 \sqrt{7}}}{14} {\left(28\right)}^{1 / 8}}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , 9 i\right) = \left[\frac{1 + {\left(2 \left(\sqrt{3} + 1\right)\right)}^{1 / 3}}{3}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , 45 i\right) = \left[\frac{3 + \sqrt{5} + \left(\sqrt{3} + \sqrt{5} + {60}^{1 / 4}\right) {\left(2 + \sqrt{3}\right)}^{1 / 3}}{3 \sqrt{10 + 10 \sqrt{5}}}\right] \theta_{3}\!\left(0 , i\right)$

### Algebraic ratios for real part 1

$\theta_{3}\!\left(0 , 1 + i\right) = \left[{2}^{-1 / 4}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , 1 + \frac{i}{2}\right) = \left[\frac{{\left(\sqrt{2} - 1\right)}^{2 / 3} {\left(4 + 3 \sqrt{2}\right)}^{1 / 12}}{{2}^{7 / 24}}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , 1 + 2 i\right) = \left[{2}^{-1 / 8}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , 1 + 4 i\right) = \left[{2}^{-7 / 16} {\left(\sqrt{2} + 1\right)}^{1 / 4}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , 1 + 6 i\right) = \left[\frac{{\left(1 + \sqrt{3} + \sqrt{2} {\left(27\right)}^{1 / 4}\right)}^{1 / 3}}{{2}^{11 / 24} \cdot {3}^{3 / 8} {\left(\sqrt{3} - 1\right)}^{1 / 6}}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , 1 + 8 i\right) = \left[{2}^{-7 / 8} {\left(16 + 15 \cdot {2}^{1 / 4} + 12 \sqrt{2} + 9 \cdot {8}^{1 / 4}\right)}^{1 / 8}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , 1 + 10 i\right) = \left[\frac{{2}^{7 / 8}}{\left({5}^{1 / 4} - 1\right) \sqrt{5 \sqrt{5} + 5}}\right] \theta_{3}\!\left(0 , i\right)$
$\theta_{3}\!\left(0 , 1 + 12 i\right) = \left[\frac{{2}^{-19 / 48} \cdot {3}^{-3 / 8} {\left(2 - 3 \sqrt{2} + {3}^{5 / 4} + {3}^{3 / 4}\right)}^{1 / 3}}{{\left(\sqrt{2} - 1\right)}^{1 / 12} {\left(\sqrt{3} + 1\right)}^{1 / 6} {\left(-1 - \sqrt{3} + \sqrt{2} \left({3}^{3 / 4}\right)\right)}^{1 / 3}}\right] \theta_{3}\!\left(0 , i\right)$

### Other values

$\theta_{3}\!\left(0 , \sqrt{6} i\right) = {\left(\frac{\sqrt{6}}{96 {\pi}^{3}} \frac{\Gamma\!\left(\frac{1}{24}\right) \Gamma\!\left(\frac{5}{24}\right) \Gamma\!\left(\frac{7}{24}\right) \Gamma\!\left(\frac{11}{24}\right)}{18 + 12 \sqrt{2} - 10 \sqrt{3} - 7 \sqrt{6}}\right)}^{1 / 4}$
$\theta_{3}\!\left(0 , \sqrt{6} i\right) = \sqrt{\frac{2}{\pi} K\!\left({\left(2 - \sqrt{3}\right)}^{2} {\left(\sqrt{2} - \sqrt{3}\right)}^{2}\right)}$

Copyright (C) Fredrik Johansson and contributors. Fungrim is provided under the MIT license. The source code is on GitHub.

2021-03-15 19:12:00.328586 UTC