Conjectured identity for the ratio of Ramanujan theta functions

Question

Following Ramanujan, we define theta functions as follows $$\chi(q):=\prod_{n = 1}^{\infty}\left(1+q^{2n-1}\right),\\\phi(q)=\sum_{n=-\infty}^{\infty}q^{n^2},\\\displaystyle \psi(q)=\sum_{n = 0}^{\infty}q^{\frac{n(n+1)}{2}}\\\text{and}\\ f(-q)=\sum_{n =-\infty}^{\infty}(-1)^n q^{\frac{n(3n-1)}{2}}$$ where $\displaystyle |q|\lt1$, based on his general theta function triple product $\displaystyle f(a,b)=\sum_{n =-\infty}^{\infty} a^{\frac{n(n-1)}{2}}b^{\frac{n(n+1)}{2}}=(-a;ab)_\infty(-b;ab)_\infty(ab;ab)_\infty$

The q-pochhammer symbol is customarily defined as $\displaystyle (a;q)_\infty=\prod_{n = 1}^{\infty}\left(1-aq^{n-1}\right)$

How do we prove the conjectured identity

$\begin{aligned}\dfrac{\chi^9(q)}{\chi^3(q^3)}+8=9\dfrac{\phi^3(q^3)}{\phi(q)}\cdot\dfrac{\psi(-q^3)}{\psi^3(-q)}\end{aligned}$

Whereby the expansion of the right hand side divided by $9$ can readily be found in this oeis article

Answer 1

Written in terms of Ramanujan theta functions , the identity to prove is

$$ \frac{\chi^9(q)}{\chi^3(q^3)} + 8 = 9 \frac{\phi^3(q^3)}{\phi(q)} \cdot \frac{\psi(-q^3)}{\psi^3(-q)}. \tag1 $$

As a first step replace $q$ with $-q$ in equation $(1)$ to get

$$ \frac{\chi^9(-q)}{\chi^3(-q^3)} + 8 = 9 \frac{\phi^3(-q^3)}{\phi(-q)} \cdot \frac{\psi(q^3)}{\psi^3(q)}. \tag2 $$

Referring to my list of Commonly used functions in modular equations written in terms of Ramanujan's $\,f(-q)\,$ to find that

\begin{align} \phi(q) &=& \frac{f^5(-q^2)}{f^2(-q)f^2(-q^4)}, \quad \phi(-q) &=& \frac{f^2(-q)}{f(-q^2)}, \\ \psi(q) &=& \frac{f^2(-q^2)}{f(-q)}, \quad \psi(-q) &=& \frac{f(-q)f(-q^4)}{f(-q^2)}, \\ \chi(q) &=& \frac{f^2(-q^2)}{f(-q)f(-q^4)}, \quad \chi(-q) &=& \frac{f(-q)}{f(-q^2)}. \tag3 \end{align}

Use these equivalences and switch to using the related Dedekind eta function to get

$$ \frac{\eta(q)^9\eta(q^6)^3}{\eta(q^2)^9\eta(q^3)^3} + 8 = 9 \frac{\eta(q)\eta(q^3)^5}{\eta(q^2)^5\eta(q^6)}. \tag4 $$

Multiply both sides by a common denominator to get

$$ \eta(q)^9\eta(q^6)^4 + 8\,\eta(q^2)^9\eta(q^3)^3\eta(q^6) = 9\,\eta(q)\eta(q^2)^4\eta(q^3)^8 \tag5 $$

which is in my "Dedkind eta product identities" as $\texttt{t6_13_33}$. Its first formulation is

equivalent to Fine, p. 87, Eq. (33.31) : $8 [1]^3[2]^3/([3][6]) - 9 [1]^4[3]^4/([2]^2[6]^2) + [1]^{12}[6]^2/([2]^6[3]^4) = 0$

where Fine refers to Nathan J. Fine, Basic Hypergeometric Series and Applications, AMS 1988.

---

Note that one of my formulations of $\texttt{t6_13_33}$ is

$$ \frac{\phi^3(-q)}{\phi(-q^3)} = 9\frac{\phi^3(-q^3)}{\phi(-q)} - 8\frac{\psi^3(q)}{\psi(q^3)}. $$

Multiply both sides by the same factor to leave $8$ constant and move it to other side to get

$$ \frac{\phi^3(-q)}{\phi(-q^3)}\cdot\frac{\psi(q^3)}{\psi^3(q)} + 8 = 9\frac{\phi^3(-q^3)}{\phi(-q)}\cdot\frac{\psi(q^3)}{\psi^3(q)}. $$

Now apply the identity $\,\chi^3(-q) = \phi(-q)/\psi(q)\,$ to get equation $(2)$ and hence to the original equation $(1)$.

---

An alternate proof of equation $(5)$ uses the space of modular forms of weight $1$ for $\Gamma_1(6)$. This space has dimension $2$ with basis

$$ M_1 = 1 + 6q^2 + 6q^6 + 6q^8 + 12q^{14} + 6q^{18} + O(q^{24}),\\ M_2 = q - q^2 + q^3 + q^4 - q^6 + 2q^7 - q^8 + q^9 + O(q^{12}) \tag6 $$

using the Magma calculator command Basis( ModularForms( Gamma1(6), 2), 30). The $\,M_1 = g(q^2)\,$ where $\,g()\,$ is the generating function (g.f.) of OEIS sequence A004016 while $\,M_2\,$ is the g.f. of OEIS sequence A093829. This modular form space contains Ramanujan theta quotients which appear in the OEIS. They are:

\begin{align} A122859:\; U_1 &= \phi(-q)^3/\phi(-q^3) &=& \;\;1 - 6q + 12q^2 + O(q^3), \\ A123330:\; U_2 &= \phi(-q^3)^3/\phi(-q) &=& \;\;1 + 2q + 4q^2 + O(q^3), \\ A107760:\; U_3 &= \psi(q)^3/\psi(q^3) &=& \;\;1 + 3q + 3q^2 + O(q^3), \\ A093829:\; U_4 &= q\,\psi(q^3)^3/\psi(q) &=& \;\;0 + q - q^2 + O(q^3). \tag7 \end{align}

Note that $\,M_2 = U_4\,$ while $\,M_1\,$ is a linear combination of the four quotients in several ways. For example, $\,M_1 = U_2 - 2U_4 = U_3 - 3U_4 = 3U_2 - 2U_3.$

The dimension of the modular space being $2$ implies that there are linear relations between the four quotients. These are expressed as $\texttt{t6_13_30},\,$ $\texttt{t6_13_33},\,$ $\texttt{t6_13_34},\,$ and $\texttt{t6_13_35}$ in my Dedekind eta product identities list.

Note the key idea is that, since the modular space is spanned by $\,M_1,M_2,\,$ then any of its elements which is $\,O(q^2)\,$ must be identically equal to $0$. This condition is easily verified by finding the first few terms of its $q$ power series expansion.

Answer 2

One standard trick to deal with these identities is to translate them into elliptic integrals and moduli.

Let $k, K $ be elliptic modulus and integral corresponding to $q$ and $l, L$ correspond to $q^3$. Also let $m=K/L$ be the multiplier so that $L=K/m$. Then we have $$\phi(q) =\sqrt{2K/\pi},\chi(q) =2^{1/4}q^{1/24}(2kk')^{-1/12},\psi(-q)=q^{-1/8}(kk') ^{1/4}\sqrt{K/\pi}$$ and $$\phi(q^3)=\sqrt{2K/\pi m}, \chi(q^3)=2^{1/4}q^{1/8}(2ll')^{-1/12},\psi(-q^3)=q^{-3/8}(ll')^{1/4}\sqrt{K/\pi m} $$ The identity in question translates to $$\frac {2(ll')^{1/4}}{(kk')^{3/4}}+8=9\cdot\frac{2(ll')^{1/4}}{m^2(kk')^{3/4}} $$ which is equivalent to $$\frac{9-m^2}{m^2}\cdot\frac {(ll')^{1/4}}{(kk')^{3/4}}=4$$ We have the following modular equations of degree $3$: $$k^2 = \frac{(3 + m)^{3}(m - 1)}{16m^{3}}, l^2 = \frac{(m - 1)^{3}(3 + m)}{16m}\\ k'^2=\frac{(3-m)^3(m+1)}{16m^3},l'^2=\frac{(m+1)^3(3-m)}{16m}$$ and hence $$(ll') ^{1/4}=\frac{1}{2}\left(\frac{(m^2-1)^3(9-m^2)}{m^2}\right)^{1/8},(kk')^{3/4}=\frac{1}{8}\left(\frac{(9-m^2)^3(m^2-1)}{m^6}\right)^{3/8}$$ We thus have $$\frac{(ll') ^{1/4}}{(kk')^{3/4}}=4\cdot\frac{m^2}{9-m^2}$$ which is exactly what was to be proved.