Generating functions "inner product"

Question

Given a function $$G(z) = \sum_{i=0}^{\infty} a_{i}z^{i}$$

What is the closed form of: $$ \sum_{i=0}^{\infty}\left(a_{i}\sum_{j=0}^{i}a_{j}\right)z^{i} $$

in terms of $G(z)$ ?

Thank you!

Answer 1

Since a closed form seems to be out of reach, here is an expression via generating functions which might be useful.

We use the coefficient of operator to denote the coefficient of $z^n$ in a series. We also use the following identity for a generating function $G(z)=\sum_{j=0}^\infty a_j z^j$

\begin{align*} [z^n]\frac{1}{1-z}G(z)&=[z^n]\sum_{j=0}^\infty\left(\sum_{i=0}^j a_i\right) z^j =\sum_{i=0}^n a_i\tag{1} \end{align*}

We obtain \begin{align*} \sum_{n=0}^\infty \left(a_n\sum_{j=0}^n a_j\right)z^n &=\sum_{n=0}^\infty a_n\left([t^n]\frac{G(t)}{1-t}\right)z^n\tag{2}\\ &=[t^0]\frac{G(t)}{1-t}\sum_{n=0}^\infty a_n\left(\frac{z}{t}\right)^n\tag{3}\\ &=[t^0]\frac{1}{1-t}G(t)G\left(\frac{z}{t}\right) \end{align*}

Comment:

• In (2) we use the identity (1).

• In (3) we use the linearity of the coefficient of operator and use the formula \begin{align*} [t^{p+q}]G(t)=[t^p]t^{-q}G(t) \end{align*}