5

Schläfli's form of elliptic modular equation of weight $5$ is $$\frac{u^3}{v^3}+\frac{v^3}{u^3}=2\left(u^2v^2-\frac{1}{u^2v^2}\right)$$ Where $$\begin{cases}u=2^{-1/4}q^{-1/24}(1+q)(1+q^3)(1+q^5)\cdots \\ v=2^{-1/4}q^{-5/24}(1+q^5)(1+q^{15})(1+q^{25})\cdots \end{cases}$$

How can I verify this? It seems like that I can change the product into an infinite sum via Jacobi triple product but I couldn't proceed further

Roland F
  • 5,122
  • I found a simpler proof available in my blog post https://paramanands.blogspot.com/2012/03/ramanujans-class-invariants.html?m=0 (the modular equation in question is listed as equation $(9)$ there) – Paramanand Singh Apr 06 '25 at 04:33
  • 1
    There is a counter part as well which is obtained by replacing $q$ with $-q$ as $u^3/v^3-v^3/u^3=2(u^2v^2+u^{-2}v^{-2})$ where $u=2^{-1/4}q^{-1/24}\prod_{n\geq 1}(1-q^{2n-1})=f(q)\text{(say)}$ and $v=f(q^5)$. – Paramanand Singh Apr 06 '25 at 04:44
  • I know we can deduce the Schläfli form from the degree 5 modular equation. But in order to deduce the degree 5 eq., the only method I know is using Jacobi's method (Multiplier). I was finding an algebraic way using only the product expansions of $u$ and $v$, but now I think about it this method may not be always feasible; and may be asking for too much. I'll leave this question here for reference for others –  Apr 06 '25 at 04:53

1 Answers1

1

I don't know much about that topic and googl Schläfli yields extremely complex Ramanujan results; but with some help of Mathematica

$$ f(\text{q$\_$})= \left\{\frac{u^3}{v^3}+\frac{v^3}{u^3},2 \left(u^2 v^2-\frac{1}{u^2 v^2}\right)\right\} \\ \text{/.}\, \left\{u:\to \frac{\prod _{k=0}^{\infty } \left(q^{2 k+1}+1\right)}{\sqrt[4]{2} \sqrt[24]{q}},v:\to \frac{\prod _{k=0}^{\infty } \left(p^{2 k+1}+1\right)}{\sqrt[4]{2} q^{5/24}}\right\} \text{/.}\, \left\{p\to q^5\right\} $$

it yields in Q-Pochhammer symbols

$$\left\{ \ \frac{q \left(-q;q^2\right)_{\infty }^6+\left(-q^5;q^{10}\right)_{\infty }^6}{\sqrt{q} \left(-q;q^2\right)_{\infty }^3 \left(-q^5;q^{10}\right)_{\infty }^3},\quad \frac{\left(-q;q^2\right)_{\infty }^4 \left(-q^5;q^{10}\right)_{\infty }^4-4 q}{\sqrt{q} \left(-q;q^2\right)_{\infty }^2 \left(-q^5;q^{10}\right)_{\infty }^2} \ \right\}$$

The plot is showing equality up to $q\lt 1$

 g[q_] = Assuming[0 < q < 1/2, 
   FunctionExpand@FullSimplify[Numerator@ 
      Together[Subtract @@ f[q]]]]

$$q \left(-q;q^2\right)_{\infty }^6+\left(-q^5;q^{10}\right)_{\infty }^6+4 q \left(-q^5;q^{10}\right){}_{\infty } \left(-q;q^2\right){}_{\infty }-\left(-q^5;q^{10}\right)_{\infty }^5 \left(-q;q^2\right)_{\infty }^5$$

  g[q] /. {QPochhammer[-q, q^2] -> a, QPochhammer[-q^5, q^10] -> b}

-a^5 b^5 + b^6 + a^6 q + 4 a b q

The plot indicates the numerator $g(q)$ to be zero in $(0\lt q\lt 1)$ (for some random rational points), but Mathematica cannot solve that problem algebraically, no wonder if one has a look at the google results.

Schlafli's form of elliptic modular equation of weight 5

Roland F
  • 5,122