A theta function identity involving $\vartheta_2(q^3),\vartheta_3(q^3)$

Question

How can we verify the theta function identity? $$ \left ( \vartheta_2(q)^2+3\vartheta_2(q^3) ^2\right )\left ( \vartheta_3(q)^2+3\vartheta_3(q^3)^2 \right )=4\vartheta_2(q)^2\vartheta_3(q)^2. $$ Where two theta functions are defined by $$ \vartheta_2(q) =\sum_{n\in\mathbb{Z}}q^{\left ( n+\frac{1}{2} \right )^2}, \vartheta_3(q) =\sum_{n\in\mathbb{Z}}q^{n^2}$$ where $q=e^{-\pi\frac{K^\prime}{K} }$. So we have $$ \vartheta_2(q)=\sqrt{\frac{2kK}{\pi} }, \vartheta_3(q)=\sqrt{\frac{2K}{\pi} }. $$ Where $K^\prime(k)=K(k^\prime),k^\prime{}^2+k^2=1$, $K(k)$ is the complete elliptic integral of the first kind, and $k$ is an elliptic modulus.

Answer 1

You asked

How can we verify the theta function identity? $$ \left ( \vartheta_2(q)^2+3\vartheta_2(q^3) ^2\right )\left ( \vartheta_3(q)^2+3\vartheta_3(q^3)^2 \right )=4\vartheta_2(q)^2\vartheta_3(q)^2. $$

Your interesting identity is $\texttt{q12_24_108b}$ in my collection of "Dedekind eta function product identities", formulated as $$ (PQ)−3q/(PQ)=(Q/P)+q/(Q/P) \tag1 $$ where $\,P := \phi(q)/\phi(q^3),\;Q :=\psi(q^2)/\psi(q^6)\,$ and $\,\phi(q),\psi(q)\,$ are Ramanujan theta functions.

The right side of your identity is equal to $\,\vartheta_2(\sqrt{q})^4.\,$ Also, using the involution $\,\tau\to-1/(12\tau)\,$ where $\,q = e^{2\pi i\tau},\,$ your identity becomes

$$ \left ( \vartheta_4(q)^2+ \vartheta_4(q^3) ^2\right )\left ( \vartheta_3(q)^2+ \vartheta_3(q^3)^2 \right )=4\,\vartheta_4(q^3)^2\vartheta_3(q^3)^2 \tag2 $$

which is identity $\texttt{q12_24_96a}$ formulated as

$$ ((Q/P) + (P/Q)) / 2 = (- (P Q) + 3 / (P Q)) / 2 \tag3 $$ where $\, P := \phi(-q)/\phi(-q^3) ,\; Q := \phi(q)/\phi(q^3). $

Now the identity

$$\left ( \vartheta_4(q)^2+ \vartheta_4(q^3) ^2\right )/2 = \vartheta_4(q^6) \vartheta_3(q^3) \vartheta_4(q) / \vartheta_4(q^2) \tag4 $$

is equivalent to $\texttt{t12_12_60b}$ in my collection.

Also, the identity $$\left ( \vartheta_3(q)^2+ \vartheta_3(q^3) ^2\right )/2 = \vartheta_4(q^6) \vartheta_4(q^3) \vartheta_4(q^2) / \vartheta_4(q) \tag5 $$

is equivalent to $\texttt{t12_18_80b}$ in my collection which is equivalent to the previous identity when $\,q\,$ is replaced with $\,-q.$ This identity is also equivalent to the Ramanujan modular equation of degree $3$

$$ ((1-\beta)^3/(1-\alpha))^{1/8} = \frac{m+1}2 \tag6 $$ where $$ m=(\phi(q)/\phi(q^3))^2,\: \alpha=\lambda(\tau), \: \beta=\lambda(3\tau), \\ 1-\lambda(\tau) = (\phi(-q)/\phi(q))^4, \quad q = e^{2\pi i \tau} $$

and $\,\lambda(\tau)\,$ is the $\,\Gamma(2)\,$ modular lambda function. ($m$ is the "multiplier")

The product of identities $(4)$ and $(5)$ is identity $(2)$ because the right side simplifies to $\,\vartheta_4(q^6)^4.$

Answer 2

We note that the series $\sum_{n\in\mathbb{Z}}\frac{1}{\cosh(3\pi nx)},x=\frac{K^\prime}{K}$ can be represented in two ways:

1. Because $ \sum_{n\in\mathbb{Z}}\frac{1}{\cosh(\pi nx)}=\vartheta_3(q)^2 $, we have $$ \sum_{n\in\mathbb{Z}}\frac{1}{\cosh(3\pi nx)}=\vartheta_3(q^3)^2. $$
2. In this question, I provide a method to compute such hyperbolic sums. We get, $$ \sum_{n\in\mathbb{Z}}\frac{1}{\cosh(3\pi nx)} =\frac{2K}{\pi}\frac{2\alpha_{1/3}-1}{3} $$ where $\alpha_{1/3}=\operatorname{ns}\left(\frac{K(k)}{3},k\right)$.

By comparing both sides, we have $$ \vartheta_3(q^3)=\sqrt{\frac{2K}{\pi}}\sqrt{\frac{2\alpha_{1/3}-1}{3} }, $$ and in similar fashion we compute $$ \vartheta_2(q^3)=\sqrt{\frac{2kK}{\pi}}\sqrt{\frac{2\alpha^{-1}_{1/3}-1}{3} }. $$

Denoting $k_n$ such that $K^\prime(k_n)/K(k_n)=\sqrt{n}$. We have concluded $$K(k_{9n})=\frac{2\alpha_{1/3}-1}{3} K(k_n),\quad k_{9n}=\frac{2\alpha^{-1}_{1/3}-1}{2\alpha_{1/3}-1} k_n. $$ where $\alpha_{s}=\operatorname{ns}\left(s\cdot K(k_n),k_n\right)$.
And a theta function identity is obtained $$ \left ( \vartheta_2(q)^2+3\vartheta_2(q^3) ^2\right )\left ( \vartheta_3(q)^2+3\vartheta_3(q^3)^2 \right )=4\vartheta_2(q)^2\vartheta_3(q)^2. $$

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Application.1: Let $n=2$, we have $$k_2=\sqrt{2}-1,K(k_2)=\frac{\left ( \sqrt{2}+1 \right )^{1/2} }{2^{13/4} \sqrt{\pi} }\Gamma\left ( \frac18 \right ) \Gamma\left ( \frac38 \right ).$$ From the addition formula of $\operatorname{ns}(z,k)$, we evaluate $$ \alpha_{1/3}=\frac{\sqrt{3}+1}{\sqrt{2}}. $$ Therefore $$k_{18}=\left ( \sqrt{2}- 1\right )^3 \left ( 2-\sqrt{3} \right )^2,\\ K(k_{18}) =\frac{\left ( \sqrt{6}+\sqrt{2} -1 \right ) \left ( \sqrt{2}+1 \right )^{1/2} }{ 3\cdot2^{13/4}\sqrt{\pi} }\Gamma\left ( \frac18 \right )\Gamma\left ( \frac38 \right ). $$ For $n=3$, we have $$k_3=\frac{\sqrt{6}-\sqrt{2}}{4},K(k_3)=\frac{3^{1/4}}{2^{7/3}\pi} \Gamma\left ( \frac{1}{3} \right )^3,\alpha_{1/3}=\frac{1+c_{27}}{2}$$ where $c_{27}=\sqrt{3+2^{4/3}(1+2^{1/3})}$. Therefore $$ k_{27}= \frac{\left ( 3-c_{27} \right ) }{c_{27}(1+c_{27})} \frac{\sqrt{6}-\sqrt{2} }{4},\\ K(k_{27}) =\frac{\sqrt{3+2^{4/3}(1+2^{1/3})}}{3^{3/4}\cdot2^{7/3}\pi} \Gamma\left ( \frac13 \right )^3. $$

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Application.2: Setting $n=1/3$, we get $$K(k_{3})=\frac{2\alpha_{1/3}-1}{3} K(k_{1/3}),\quad k_{3}=\frac{2\alpha^{-1}_{1/3}-1}{2\alpha_{1/3}-1} k_{1/3}. $$ By definition, $K(k_3)=\frac{1}{\sqrt{3 }}K(k_{1/3})$. We therefore evaluate $\alpha_{1/3}=\frac{1+\sqrt{3}}{2}$. So we find $$ k_3=(2-\sqrt{3})k_{1/3}. $$ Another relation is $k^2_3+k_{1/3}^2=1$. Follow these instructions we obtain $$ k_3=\frac{\sqrt{6} -\sqrt{2} }{4}, k_{1/3}=\frac{\sqrt{6}+\sqrt{2} }{4} $$ which confirm the known results.
A more challenging singular modulus $k_6=(2-\sqrt{3})(\sqrt{3}-\sqrt{2})$ can be calculated in this way as well. By using two quadratic transformations $$ K^\prime(k)=\frac{2}{1+k} K\left ( \frac{1-k}{1+k} \right ), K(k)=\frac{1}{1+k} K^\prime\left ( \frac{1-k}{1+k} \right ). $$ We have $$ \frac{1-k_6}{1+k_6}=k_{2/3}.\tag{v6.1} $$ Let $n=2/3$, to get $$K(k_{6})=\frac{2\alpha_{1/3}-1}{3} K(k_{2/3}),$$ $$ k_{6}=\frac{2\alpha^{-1}_{1/3}-1}{2\alpha_{1/3}-1} k_{2/3}.\tag{v6.2} $$ Also, $$ \begin{aligned} K(k_{6})&=\frac{2\alpha_{1/3}-1}{3} K(k_{2/3})\\ &=\frac{2\alpha_{1/3}-1}{3} K\left(\frac{1-k_6}{1+k_6}\right)\\ &=\frac{2\alpha_{1/3}-1}{3}\frac{1+k_6}{2}K^\prime\left(k_6\right). \end{aligned} $$ Therefore the third relation is $$ (2\alpha_{1/3}-1)(1+k_6)=\sqrt{6}\tag{v6.3}. $$ From $(\mathrm{v}6).(1)(2)(3)$, we solve $$ \begin{aligned} &k_6=\left ( 2-\sqrt{3} \right )\left ( \sqrt{3}-\sqrt{2} \right ),\\ &k_{2/3}=\left ( 2-\sqrt{3} \right )\left ( \sqrt{3}+\sqrt{2} \right ),\\ &\alpha_{1/3}=\frac{1}{\sqrt{2}}\left ( \sqrt{3}-1 \right )\left ( \sqrt{3}+\sqrt{2} \right ). \end{aligned} $$

Answer 3

Let the elliptic modulus $l$ and elliptic integral $L$ correspond to nome $q^3$ so that $l$ is of degree $3$ over $k$ and let $m=K/L$ be the multiplier.

Then the desired identity can be written as $$\left(\frac{2kK}{\pi}+\frac{6lL}{\pi}\right) \left(\frac{2K}{\pi}+\frac{6L}{\pi}\right) =4\cdot\frac{2kK}{\pi}\cdot\frac{2K}{\pi}$$ or $$\left(1+\frac{3l}{mk}\right) \left(1+\frac{3}{m}\right) =4\tag{1}$$ Using well known modular equations of degree $3$ $$k^2 = \frac{(3 + m)^{3}(m - 1)}{16m^{3}}, \\ l^2 = \frac{(m - 1)^{3}(3 + m)}{16m}$$ we get $$\frac{l} {k} =\frac{m(m-1)} {m+3} $$ and then the desired identity $(1)$ holds.

Answer 4

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Higher powers: We have $$ K(k_{25n})=\frac{2\alpha_{1/5}-2\alpha_{3/5}+1}{5}K(k_n),\\ k_{25n}=\frac{2\alpha_{1/5}^{-1}-2\alpha_{3/5}^{-1}+1}{2\alpha_{1/5}-2\alpha_{3/5}+1}k_n. $$ In general, for any positive odd $m$, $$ K(k_{nm^2})=\frac{2\sum_{0\le j\le(m-1)/2 }(-1)^j \alpha_{(2j+1)/m} +(-1)^{(m+1)/2}}{m}K(k_{n}),$$ $$ k_{nm^2}=\frac{2\sum_{0\le j\le(m-1)/2 }(-1)^j \alpha_{(2j+1)/m}^{-1} +(-1)^{(m+1)/2}}{2\sum_{0\le j\le(m-1)/2 }(-1)^j \alpha_{(2j+1)/m} +(-1)^{(m+1)/2}}k_n. $$ Let $n=1$. It can be calculated that $\alpha_{1/5}-\alpha_{3/5}=\frac{1+\sqrt{5}}{2},\alpha_{1/5}^{-1}-\alpha_{3/5}^{-1}=1-5^{1/4}$. And $$ k_{25}=\frac{\left ( 3-2\cdot5^{1/4} \right ) \left ( \sqrt{5}-2 \right ) }{ \sqrt{2} }, $$ $$ K(k_{25}) =\frac{\left(2+\sqrt{5} \right)}{20\sqrt{\pi} }\Gamma\left ( \frac14 \right)^2. $$ Now we focus on the computation of $k_5=\sqrt{\frac{1}{2}-\sqrt{\sqrt{5}-2 } }$. The idea is the same as the previous proof of $k_3$. The equation system should be $$\begin{aligned} &\alpha_{1/5}-\alpha_{3/5}=\frac{\sqrt{5}-1}{2},\\ &k_5=\frac{2\alpha_{1/5}^{-1}-2\alpha_{3/5}^{-1} +1}{\sqrt{5}}k_{1/5},\\ &k_5^2+k_{1/5}^2=1,\\ &\alpha_{3/5}=\frac{k_{1/5}^2- \frac{k_{1/5}^2-\alpha_{1/5}^2}{1-\alpha_{1/5}^2} \alpha_{1/5}^2 }{\frac{k_{1/5}^2-\alpha_{1/5}^2}{1-\alpha_{1/5}^2}-\alpha_{1/5}^2}. \end{aligned}$$ Which actually gives the value of $k_5$.
An additional value: $$ K(k_{49}) =\frac{\sqrt{7+4\sqrt{7}+2\sqrt{35+16\sqrt{7} } } }{28\sqrt{\pi}} \Gamma\left ( \frac14 \right )^2, $$ $$ k_{49}=\frac{2c_{49}-1}{\sqrt{7+4\sqrt{7}+2\sqrt{35+16\sqrt{7} }}} $$ where $c_{49}$ is the root of $P(x)=x^8-4x^7+56x^5-140x^4+168x^3+56x^2-32x-12$ that is close to $0.5$.