Derivation of the termwise Hadamard product of two generating functions.

Question

Let $f$ and $g$ be sequences of functions and $F$ and $G$ their corresponding generating functions, \begin{eqnarray*} & F(z)=\sum_{n\in\mathbb{N}} f_n z^n \\ & G(z)=\sum_{n\in\mathbb{N}} g_n z^n \\ \end{eqnarray*} The termwise multiplication of these sequences is an operation called Hadamar product, $$F \cdot G (z) := \sum_{n\in\mathbb{N}} f_ng_n z^n $$ As it can be found in Wikipedia, Hadamard products and diagonal generating functions, it states that $$ \boxed{F \cdot G (z) = \frac{1}{2\pi} \int_{0}^{2\pi} F\big(\sqrt{z} e^{it}\big) G\big(\sqrt{z} e^{-it}\big) dt}.$$

How do you get this formula? I have thoroughly looked for it without finding any derivation, in Wikipedia article there is a cite to a book, but this formula does not seem to appear in any form there.

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This question is related to:

Algorithm for computing Hadamard product of two rational generating functions,

Does having closed forms of two generating functions guarantee that one can find the closed form of their term-by-term product? and

Is there an easier formulation for the Hadamard product of certain pair of series? as well.

Answer 1

By Cauchy product, $$F(\sqrt{z} x)\, G(\sqrt{z}/x) = \sum_{m=0}^\infty f_m (\sqrt{z})^m \sum_{k=0}^\infty g_k (\sqrt{z})^k \, x^{m-k} $$ Use the Cauchy integral representation of the Kronecker delta (1 if m=k, 0 otherwise) $$\frac{1}{2\pi\,i}\oint \frac{dx}{x} x^{m-k} = \delta_{m,k} $$ Thus $$\frac{1}{2\pi\,i}\oint \frac{dx}{x} F(\sqrt{z} x)\, G(\sqrt{z}/x) = \sum_{m=0}^\infty f_m \, g_m z^m $$ Now make the substitution $x=e^{i\,t}.$ Thus we get the OP's answer, corrected by the leading factor of $1/(2 \pi)$,

$$ \frac{1}{2 \pi} \int_{0}^{2 \pi} F(\sqrt{z} e^{it})\, G(\sqrt{z} e^{-i\,t}) dt = \sum_{m=0}^\infty f_m \, g_m z^m $$

Answer 2

Two derivations from operstional calculus:

For

$$A(x) = \sum_{n \geq 0} a_n x^n$$

and

$$\widetilde{A}(x) = \sum_{n \geq 0} a_n \frac{x^n}{n!} = e^{a.x}$$

with $(a.)^n = a_n$, the Hadamard product is given by

$$\sum_{n \geq 0} a_n x^n \frac{D_{x=0}^n}{n!} G(x)= \sum_{n\geq 0} a_ng_n x^n $$

with $d/dx= D_x$,

or more concisely,

$$\widetilde{A}(:xD_{x=0}:)G(x)= \exp(a.:xD_{x=0}:)G(x)=G(a.x)= (A*G)(x)$$

with $:xD_x:^n = x^nD_x^n$, by definition, a notational convenience.

The derivatives may be coded as the Cauchy contour integrals

$$g_n = \frac{D^n_{z=0}}{n!}G(z) = \frac{1}{2\pi i} \oint_{|z|<\epsilon} \frac{G(z)}{z^{n+1}}dz$$

where $\epsilon$ is less than the radii of the circles of convergence of the two series.

So, with appropriate changes of variables,

$$H(x)= (F*G)(x)$$

$$ = \frac{1}{2\pi i} \sum_{n \geq 0} f_n x^n \oint_{|z|<\epsilon} \frac{G(z)}{z^{n+1}}dz.$$

$$= \frac{1}{2\pi i} \oint_{|z|<\epsilon} \frac{F(\frac{x}{z})G(z)}{z}dz$$

$$= \frac{1}{2\pi i} \oint_{|z|<\alpha} \frac{F(\frac{\sqrt{x}}{z})G(z\sqrt{x})}{z}dz$$

$$= \frac{1}{2\pi} \int_0^{2\pi} F(\sqrt{x}\alpha^{-1}e^{-it})G(\sqrt{x}\alpha e^{it})dt$$

$$= \frac{1}{2\pi} \int_{0}^{2\pi} F(\sqrt{x}e^{-it})G(\sqrt{x}e^{it})dt,$$

assuming both series reps are convergent for $\alpha=1$. The last real integral is convergent for all functions bounded in the segment of integration.

For some discussion of the validity of these formulas, see "Hadamard grade of power series" by Allouche and France.

Alternatively, note (cf. this MSE answer)

$$\exp(txD_x)f(x)=f(e^t x).$$

Then

$$\frac{1}{2\pi} \int_{-\pi}^{\pi} F(ue^{-it})G(ve^{it})dt$$

$$= \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{-ituD_u}e^{itvD_v} dt F(u)G(v)$$

$$= \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{-it(uD_u-vD_v)} dt F(u)G(v)$$

$$=\frac{sin[\pi(uD_u-vD_v)]}{\pi(uD_u-vD_v)}F(u)G(v)$$

$$= \sum_{j,k \geq 0} sinc(\pi(j-k)) f_j g_k u^jv^k$$

$$= \sum_{k \geq 0} f_k g_k (uv)^k.$$