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

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

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    $(1-24x+I)^{-1}=1+(24x -I)+(24x-I)^2+\dots$2011-04-15

1 Answers 1

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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 $a_0$.

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    @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.2011-04-16