which is absolutely convergent for each x ∈ D. Substituting e2πiτ for x we see that the theorem is proved [1]. QED ALaurent expansion for a function g can be defined as g(z) = ∞ X n=0 an(z −z0) n + ∞ X n=1 bn (z −z0) n [2]. Since we are dealing with functions with poles of finite order, there cannot be more terms with a negative degree than the order of the poles. So most of the bn terms are zero in this expansion. SoJ(τ) has a Fourier expansion, but we must now show that it has integer coefficients. It can be shown that for τ ∈Hwe have the Fourier expansions g2(τ) = 4π4 3 " 1 + 240 ∞ X k=1 σ3(k)e 2πikτ# and ∆(τ) = (2π)12 ∞ X n=1 τ(n)e2πinτ, where σ3(k) = Pd |k d3 and τ(n) is Ramanujan’s tau function, where the coefficients are integers and τ(1) = 1 and τ(2) =−24. In this definition of ∆ it is slightly more clear that ∆ = (2π)12(η)24. Using these formulas we can find the Fourier expansion for J(τ). Note that both of these formulas have integer coefficients if we ignore the factors with powers of π. Again, we will follow a proof from [1]. Theorem: If τ ∈H, we have the Fourier expansion 123J(τ) =e−2πiτ + 744 + ∞ X n=1 c(n)e2πinτ, where the c(n) are integers. Proof: In this proof we will write I for any power series inxwith integer coefficients. Hence, letting x=e2πiτ as before, we have g3 2(τ) = 64π12 27 (1 + 240x+I)3 = 64π12 27 (1 + 720x+I) (where the coefficient on x comes from the multinomial theorem), and ∆(τ) = 64π12 27 [123x(1−24x+I)]. Page 44 Riley • An Overview of Monstrous Moonshine
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