It is a fundamental fact, often quoted these days in its connection with Monstrous Moonshine, that the q-expansion (i.e., the Laurent expansion in a neighborhood of $\tau = i\infty$) of the j-invariant is given by
$$j(\tau) = \frac{1}{q} + 744 + 196884q + 21493760q^2+\dots, \quad q = e^{2 \pi i \tau}.$$
I do not recall, however, ever seeing a modern treatment, nor even a hint, of how one might go about obtaining this expansion. Does anybody know a nice way to compute these coefficients? (I mean a way which does not invoke Moonshine, not that I'd expect that to make the computation more pleasant.) Is there a standard way to do it?
I did find one approach published by H.S. Zuckerman in the late 1930s*, which makes use of a "fifth order multiplicator equation" for $j(\tau)$ — distilled from Fricke and Klein's Vorlesungen uber die Theorie der elliptischen Modulfunktionen — and an identity of Ramanujan for the generating function of partition numbers of the form $p(25n + 24)$. Is this typical?
*Zuckerman, Herbert S., The computation of the smaller coefficients of $J(\tau)$. Bull. Amer. Math. Soc. 45, (1939). 917–919.}
Best Answer
The trick is to write the $j$-invariant function in terms of Eisenstein series, whose $q$-expansions have a simple expression. See Silverman's "Advanced Topics in the Arithmetic of Elliptic Curves", Chapter 1, Section 7, and in particular, Proposition 7.4 and Remark 7.4.1.
In particular $$ j(\tau) = 1728\frac{g_2(\tau)^3}{\Delta(\tau)}=1728 \frac{g_2(\tau)^3}{g_2(\tau)^3-27g_3(\tau)^2} = \frac{\left(1+240\sum_{n\geq 1}\sigma_3(n)q^n\right)^3}{q\prod_{n\geq 1}(1-q^n)^{24}}, $$ or, if you prefer, $$ j(\tau) = 1728 \frac{\left(1+240\sum_{n\geq 1}\sigma_3(n)q^n\right)^3}{\left(1+240\sum_{n\geq 1} \sigma_3(n)q^n\right)^3 - \left(1-504\sum_{n\geq 1}\sigma_5(n)q^n \right)^2}. $$