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On page 237 of Bruce C. Berndt's book "Ramanujan's Notebooks Part IV" is Entry 72 (p. 322).

"Let $$P=\frac{f(-q)}{q^{1/2}f(-q^{13})}$$ and $$Q=\frac{f(-q^{3})}{q^{3/2}f(-q^{39})}$$

then $$PQ+\frac{13}{PQ}=\Big(\frac{Q}{P}\Big)^{2}-3\frac{Q}{P}-3-3\frac{P}{Q}+\Big(\frac{P}{Q}\Big)^{2}."$$

In my opinion, it represents a 13-degree modular equation when interpreted in the "$q_{3}$" theory.

$$m=\Big(\frac{\beta}{\alpha}\Big)^{1/3}+\Big(\frac{1-\beta}{1-\alpha}\Big)^{1/3}-\frac{13}{m}\Big(\frac{\beta(1-\beta}{\alpha(1-\alpha)}\Big)^{1/3}-3\Big(\frac{\beta(1-\beta)^{3}}{\alpha(1-\alpha)^{3}}\Big)^{1/12}-3\Big(\frac{\beta^{3}(1-\beta)}{\alpha^{3}(1-\alpha)}\Big)^{1/12}-3.$$

When $\beta=1-\alpha$

then

$m=\sqrt{13}$

and

$\beta=\frac{1}{2}-\frac{1}{54}\sqrt{9842\sqrt{13}-34757}$.

Can anyone tell me if my interpretation is correct?

To respond to Somos: it is sufficient in the formula “$$\big(\frac{1} {\alpha_{3n}}\big)^{1/3}=\frac{2}{3} \frac{(G_{9n})^{6}}{(G_{n})^{2}}+\frac{\sqrt{2}}{3} \frac{G_{n}} {(G_{9n})^{3}}$$“ of How to calculate singular moduli $\alpha_{3,n}$ of Ramanujan' s "$q_{3}$" theory? , ask $n=\frac{13}{3}$ and use class invariants $G_{39}$ e $G_{13/3}$ with the values:

$G_{39}=2^{1/4}\Big(\frac{\sqrt{13}+3}{2}\Big)^{1/6}\Big(\sqrt{\frac{5+\sqrt{13}}{8}}+\sqrt{\frac{-3+\sqrt{13}}{8}}\Big)$

$G_{\frac{13}{3}}=2^{1/4}\Big(\frac{\sqrt{13}+3}{2}\Big)^{1/6}\Big(\sqrt{\frac{5+\sqrt{13}}{8}}-\sqrt{\frac{-3+\sqrt{13}}{8}}\Big).$

Because the equation

$m=\Big(\frac{\beta}{\alpha}\Big)^{1/3} +\Big(\frac{1-\beta}{1-\alpha}\Big)^{1/3}-\frac{13}{m}\Big(\frac{\beta(1-\beta)}{\alpha(1-\alpha)}\Big)^{1/3}-3\Big(\frac{\beta(1-\beta)^{3}}{\alpha(1-\alpha)^{3}}\Big)^{1/12}-3-3\Big(\frac{\beta^{3}(1-\beta)}{\alpha^{3}(1-\alpha)}\Big)^{1/12}$

is a 13-degree modular equation in the signature 3. Using Corollary 3.2 of “RAMANUJAN’S THEORIES OF ELLIPTIC FUNCTIONS TO ALTERNATIVE BASES” BRUCE C. BERNDT, S. BHARGAVA AND FRANK G. GARVAN – TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY – Volume 347, Number 11, November 1995, p.4175:

$b(q)=(1-\alpha)^{1/3}z_{1}$

$c(q)=\alpha^{1/3}z_{1}$

$b(q^{13})=(1-\beta)z_{13}$

$c(q^{13})=\beta^{1/3}z_{13}$

$m=\frac{z_{1}}{z_{13}}$

We obtain

$1=\frac{b(q^{13})}{b(q)}+\frac{c(q^{13})}{c(q)}-13\frac{b(q^{13})}{b(q)} \frac{c(q^{13})}{c(q)}-3\Big(\frac{b(q^{13})}{b(q)}\Big)^{1/4}\Big(\frac{c(q^{13})}{c(q)}\Big)^{3/4}-3\Big(\frac{b(q^{13})}{b(q)}\Big)^{1/2}\Big(\frac{c(q^{13})}{c(q)}\Big)^{1/2}-3\Big(\frac{b(q^{13})}{b(q)}\Big)^{3/4}\Big(\frac{c(q^{13})}{c(q)}\Big)^{1/4}$

Now using Lemma 5.1 0f “RAMANUJAN’S THEORIES…” p.4181:

$b(q)=\frac{f(-q)^{3}}{f(-q^{3})}$ and $c(q)=3q^{1/3}\frac{f(-q^{3})^3}{f(-q)}$

$b(q^{13})=\frac{f(-q^{13})^{3}}{f(-q^{39})}$ and $c(q^{13})=3q^{13/3}\frac{f(-q^{39})^3}{f(-q^{13})}$

Multiplying both side of the resultant equation by

$\frac{f(-q)f(-q^{3})}{q^{2}f(-q^{13})f(-q^{39})}$

And using the definition of $P$ and $Q$ we have

$P Q=\Big(\frac{P}{Q}\Big)^{2}-3\frac{P}{Q}-3\frac{Q}{P}-\frac{13}{P Q}+\Big(\frac{Q}{P}\Big)^{2}-3$

q.d.e.

giuseppe mancò
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    Your equation $, m = ... - 3,$ is correct assuming $,\beta=1-\alpha,,$ $,m=\sqrt{13},,$ and $,\beta=1/2-\sqrt{...},$ but I don't see how you can claim this is a "13-degree modular equation when interpreted in the $q_3$ theory". – Somos Sep 12 '19 at 13:19
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    What the hell is $f$. Do you understand that the first thing is to find for what congruence subroup they are modular, of what weight, with what poles/zeros and what first few coefficients ? Do you understand that two modular forms are equal if they have the same first few coefficients ? Do you understand that two meromorphic modular forms are equal if they have the same poles and the same first few coefficients ? – reuns Sep 12 '19 at 20:22
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    f is a function of name q, i.e.:$f=f(-q)=1-q-q^{2}+q^{5}+q^{7}-…$ with $q=e^{-\pi\sqrt{n}}$ – giuseppe mancò Sep 13 '19 at 07:07

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