Inner product with generating functions

Question

Let there be two real series $F=(f_0, f_1, f_2, \dots)$ and $G=(g_0, g_1, g_2, \dots)$ which satisfy $f(x)=\sum_n f_n x^n$ and $g(x)=\sum_n g_n x^n$, where $f(x)$ and $g(x)$ are given analytic functions. Is there a way to calculate the inner product $I=\sum_{i=1}^\infty f_i g_i$ using $f(x)$ and $g(x)$ without having to evaluate infinite sums?

Answer 1

Let $R_1$ and $R_2$ be the radii of convergence of the the two power series $\sum_n f_n z^n$ and $\sum_n g_n z^n$. I have a positive answer when $R_1R_2>1$: choose $0<r_1<R_1$ and $0<r_2<R_2$ such that $r_1r_2=1$. Then $$\frac{1}{2\pi}\int_0^{2\pi} f(r_1e^{i\theta})g(r_2e^{-i\theta}) \mathrm{d}\theta = \sum_{k=0}^{+\infty} f_k g_k.$$

Indeed, for each real number $\theta$ the series defining $f(r_1e^{i\theta})$ and $g(r_2e^{-i\theta})$ are absolutely convergent, so their Cauchy product, is absolutely convergent. But their Cauchy product is is $\sum_n w_n(\theta)$, where $$w_n(\theta):= \sum_{k=0}^n f_k \times (r_1e^{i\theta})^k \times g_{n-k} \times (r_2e^{-i\theta})^{n-k} = \sum_{k=0}^n f_k g_{n-k} r_1^k r_2^{n-k} e^{i(2k-n)\theta}.$$

For every $n \ge 0$ and every real number $\theta$, $$|w_n(\theta)| \le \sum_{k=0}^n |f_k| |g_{n-k}| r_1^k r_2^{n-k}$$ Since $$\sum_{n=0}^{+\infty} \Big(\sum_{k=0}^n |f_k| |g_{n-k}| r_1^k r_2^{n-k}\Big) = \Big(\sum_{k=0}^{+\infty} |f_k| r_1^k\Big) \times \Big(\sum_{\ell=0}^{+\infty} |g_\ell| r_2^\ell\Big) < +\infty,$$ the series $\sum w_n(\theta)$ is normally convergent with regard to $\theta$, so $$\int_0^{2\pi} \Big(\sum_{n=0}^{+\infty} w_n(\theta) \Big) \mathrm{d}\theta = \sum_{n=0}^{+\infty} \int_0^{2\pi} w_n(\theta) \mathrm{d}\theta.$$ A direct computation shows that $\int_0^{2\pi} w_n(\theta) \mathrm{d}\theta$ equals $2\pi f_{n/2}g_{n/2}$ if $n$ is even, and $0$ otherwise. The conclusion follows.