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Denote $f(z)=\theta_3(z)^2$, where $\theta_3(z)=\theta(z/2), \theta(z)=\sum_n q^{n^2}$ are Jacobi's $\theta$-functions.

And $\lambda(z)=\left( \frac{\theta_2(z)}{\theta_3(z)}\right)^4$ is the Legendre modular function, where $\theta_2(z)=\sum_n q^{({n+1/2})^2}$

There is a power series expression

$$f(z)=\sum_{n=0}^{\infty} \binom{2n}{n}\left(\frac{\lambda(z)}{16}\right)^n$$

How to get the expression?

J. M. ain't a mathematician
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BY zhang
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  • the coefficients of $f(z) = \sum_{n \ge 0} a_n \lambda(z)^n$ are given by $\int_{\lambda^{-1}(C)} \frac{f(z)}{\lambda(z)^{n+1}}\lambda'(z)dz$ where $\lambda^{-1}(C)$ is a contour enclosing $i\infty$, depending on the Fourier coefficients of $1/\lambda(z)$ and $\lambda'(z)$ – reuns Oct 26 '16 at 11:34
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    Shouldn't the binomial coefficient be squared? Then this is essentially a hypergeometric series. See here. In fact, that series gives the elliptic integral $\theta_3^2=\frac{2}{\pi}K(k)$ where $\lambda=k^2$. Cf. DLMF 19.5. – ccorn Oct 29 '16 at 11:54

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