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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!

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  • $\begingroup$ The generating function of $\sum{i=0}^\infty\left(\sum_{j=0}^ia_j\right)z^i$ is $G(z)/(1-z)$. However, the introduction of the extra factor of $a_i$ gives a Hadamard product (also called Schur or entrywise product). In general, given two generating functions, there's not an easy formula for their Hadamard product. So, you'll probably have to handle your calculations on a case-by-case basis. $\endgroup$
    – Rus May
    Nov 18 '16 at 13:51
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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*}

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