# j-invariant Fourier expansion

I'm reading about Fourier expansion of modular functions, but I have trouble understanding the following equation:

Is it some inherent property of the denominator, as it is?

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$I$ stands for a power series which is $O(x^2)$ and which has integer coefficients. Different $I$'s stand for different power series. E.g. $(1-24x+\text{someseries})^{-1}=(1+24x+\text{someotherseries})$. – user8268 Apr 15 '11 at 13:58
That's exactly what I don't understand - how it is done? Seems not at all that trivial, that the coefficients remain integer in the new series too. – Pavel Apr 15 '11 at 14:09
$(1-24x+I)^{-1}=1+(24x -I)+(24x-I)^2+\dots$ – user8268 Apr 15 '11 at 14:17
Right... Thanks – Pavel Apr 15 '11 at 14:32

As user8268 explained in the comments, any power series $1+a_1x+a_2x^2+\dots$ with integral coefficients and constant term 1 has an inverse with integral coefficients. You can use the nice expansion for $(1+a_1x+a_2x^2+\dots)^{-1}$ given by $$\frac{1}{1-(-a_1x-a_2x^2-\ldots)} = 1 + (-a_1x-a_2x^2-\ldots) + (-a_1x-a_2x^2-\ldots)^2 + \ldots$$
We will not get an integral power series for $(a_0+a_1x+a_2x^2 + \ldots)^{-1}$ if the leading term $a_0$ is not invertible; we will have $$\frac{1}{a_0} + \frac{(-a_1x-a_2x^2-\ldots)}{a_0{}^2} + \frac{(-a_1x-a_2x^2-\ldots)^2}{a_0{}^3} + \ldots$$ But you can see that the only denominators we get are powers of $a_0$.
Got to say - it's still looks kind of strange to me. I know that $\frac{1}{1-x} = 1 + x + x^2 ...$ iff $\|x\| < 1$. Shouldn't the following hold also for the polynomial above? – Pavel Apr 15 '11 at 20:51
@Pavel: Sure these power series have some radius of convergence. And it is possible that in the expression $J=g_2{}^3/\Delta$, each power series has a different radius of convergence. (Though I believe in each case it is 1.) But since they all converge in some neighborhood, the power series are equal. – Jonas Kibelbek Apr 16 '11 at 23:33