I have the generating function of $a_n$, $A(z)$, and I want to find the generating function of $(1/n)a_n$ in terms of the generating function of $a_n$. Any help is appreciated.
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Try integration of the generating function. – Tai Jul 25 '17 at 19:42
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My backgrounds is on engineering, would you please explain more or refer me to a document? – Rasoul Jul 25 '17 at 19:46
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Sure, I'll write an answer. – Tai Jul 25 '17 at 19:46
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By the way, you can reverse this. If $A(z)$ is the generating function of the sequence ${a_n}$, then $z A^\prime(z)$ is the generating function of the sequence ${ n a_n }$. – Michael Lugo Jul 25 '17 at 20:00
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You are given \begin{align*} A(z) & =\sum_{n=0}^{\infty}a_nz^{n}\\ A(z)-a_0 & =a_1z+a_2z^2+\dotsb + a_nz^n +\dotsb\\ \frac{A(z)-a_0}{z} & =a_1+a_2z^1+\dotsb + a_nz^{n-1} +\dotsb\\ \int \frac{A(z)-a_0}{z} \, dz & =a_1z+\frac{a_2}{2}z^2+\dotsb + \frac{a_n}{n}z^n +\dotsb. \end{align*}
Thus the generating function is $$\int \frac{A(z)-a_0}{z} \, dz$$
Anurag A
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We must assume $a_0=0$ in order to make sense of this problem. Or perhaps we can assume that by $a_n/n$ we mean the sequence whose zeroth term is anything (can be adjusted by the arbitrary constant for your integral) and the rest given by $a_n/n$. – spaceisdarkgreen Jul 25 '17 at 20:24