Let \[ g(z)=\sum_{n=0}^\infty g_nz^n, \qquad h(z)=\sum_{n=0}^\infty h_nz^n \] be entire. Find a formula for the coefficients \(f_n\) in the Taylor expansion at \(0\) of \[ f(z)=\frac1{2\pi i}\int_{|w|=1}g(z/w)h(w)\frac{dw}{w}. \]
Solution
1. For each fixed \(z\in\CC\) and every \(w\) with \(\abs{w}=1\),\[ g(z/w) = \sum_{n=0}^{\infty}g_nz^nw^{-n} \]and\[ h(w) = \sum_{m=0}^{\infty}h_mw^m, \]with both series absolutely and uniformly convergent on the unit circle.
Because \(g\) is entire, its power series converges absolutely at \(\abs{z}\): \[ \sum_{n=0}^{\infty}\abs{g_n}\abs{z}^n<\infty. \] For \(\abs{w}=1\), \[ \abs{g_nz^nw^{-n}} = \abs{g_n}\abs{z}^n, \] so the Weierstrass \(M\)-test gives uniform absolute convergence of the first series. Likewise, entire-ness of \(h\) gives \[ \sum_{m=0}^{\infty}\abs{h_m}<\infty, \] and the \(M\)-test gives uniform absolute convergence of the second series on \(\abs{w}=1\).
2. The product admits the uniformly absolutely convergent expansion\[ g(z/w)h(w) = \sum_{n,m\geq0} g_nh_mz^nw^{m-n} \]on \(\abs{w}=1\).
By step 1, \[ \sum_{n,m\geq0} \abs{g_nh_mz^nw^{m-n}} = \left( \sum_{n\geq0}\abs{g_n}\abs{z}^n \right) \left( \sum_{m\geq0}\abs{h_m} \right) <\infty, \] uniformly for \(\abs{w}=1\). Hence the double series converges uniformly and absolutely there and represents the product.
3. One has\[ f(z) = \sum_{n=0}^{\infty}g_nh_nz^n. \]
By step 2, termwise contour integration is valid: \[ \begin{aligned} f(z) &= \sum_{n,m\geq0} g_nh_mz^n \frac1{2\pi i} \int_{\abs{w}=1} w^{m-n-1}\,dw. \end{aligned} \] The standard contour integral gives \[ \frac1{2\pi i} \int_{\abs{w}=1} w^{m-n-1}\,dw = \begin{cases} 1,&m=n,\\ 0,&m\neq n. \end{cases} \] Therefore only the terms with \(m=n\) survive, yielding the displayed series.
4. The Taylor coefficients of \(f\) at the origin are\[ \boxed{f_n=g_nh_n}. \]
Step 3 is already the Taylor expansion of \(f\) in powers of \(z\), so its \(n\)th coefficient is \(g_nh_n\).
5. Q.E.D.
Step 4 gives the requested formula.