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Given the Euler function $\phi(q)=\prod_{n = 1}^{\infty}(1-q^{n})$ which is a modular form where $q=\exp(2\pi i \tau)$, $|q|\lt1$

Then what is the limit $\lim_{q\rightarrow 1}\frac{\phi^5(q)_{\infty}}{\phi(q^5)_{\infty}}$ ?

Nicco
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    You can use the expression for Dedekind's eta function in terms of elliptic integral $K$ and modulus $k$. Thus we have $$q^{1/24}\phi(q)=2^{-1/6}\sqrt{\frac{2K}{\pi}}k^{1/12}k'^{1/3}$$ – Paramanand Singh Aug 16 '18 at 06:55
  • @Paramanand Singh: that's actually a good idea – Nicco Aug 16 '18 at 07:09

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When $q\to 1^-$, $\tau \to 0$, hence $-1/\tau \to i\infty$, $$\begin{aligned}\frac{{{\phi ^5}{{(q)}_\infty }}}{{\phi {{({q^5})}_\infty }}} = \frac{{{\eta ^5}(\tau )}}{{\eta (5\tau )}} &= \frac{{{{\left( {\frac{{ - 1}}{{\tau i}}} \right)}^{5/2}}{\eta ^5}(\frac{{ - 1}}{\tau })}}{{{{\left( {\frac{{ - 1}}{{5\tau i}}} \right)}^{1/2}}\eta (\frac{{ - 1}}{{5\tau }})}} \\ &= -\frac{{\sqrt 5 }}{{{\tau ^2}}}\frac{{{e^{ - \frac{{5\pi i}}{{12\tau }}}}\prod\limits_{n = 1}^\infty {{{\left( {1 - {e^{ - \frac{{2\pi ni}}{\tau }}}} \right)}^5}} }}{{{e^{ - \frac{{\pi i}}{{60\tau }}}}\prod\limits_{n = 1}^\infty {\left( {1 - {e^{ - \frac{{2\pi ni}}{{5\tau }}}}} \right)} }}\\ &\sim -\frac{{\sqrt 5 }}{{{\tau ^2}}}{e^{ - \frac{{2\pi i}}{{5\tau }}}} \to 0\end{aligned}$$ Note that the two infinite products both $\to 1$.

pisco
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Given the following identity $\frac{(q)^5_{\infty}}{(q^5)_{\infty}}=1+5\sum_{n=1}^{\infty}\frac{(-1)^n nq^{\frac{n(n+1)}{2}}}{\prod_{k=1}^{n-1}\Big(R_{k}+q^k\Big)}$ from my old post where $\frac{1}{R_{n}}=R(q^n,q)$ is the Generalized Rogers Ramanujan continued fraction

If we consider the fact that $\lim_{x\rightarrow 0+}\frac{1}{R(e^{-nx},e^{-x})}=\Phi$ where $\Phi=\frac{1+\sqrt5}{2}$ is the golden ratio

Then we are led to $$\lim_{q\rightarrow 1}\frac{\phi^5(q)_{\infty}}{\phi(q^5)_{\infty}}=1+5\sum_{n=1}^{\infty}\frac{(-1)^n n}{(1+\Phi)^n}$$

Whereby the infinite series on the RHS converges to zero.

Nicco
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