# Doubt in theory of Logarithmic differentiation of generating functions from Apostol

While self studying analytic number theory from Tom M Apostol introduction to analytic number theory I have following doubt in section 14.10 .

I am adding it's image highlighting the part in which I have doubt.

My doubt is - how $$\sum_{m=1}^{\infty} G'_A (x^m) x^m$$ = $$\sum_{m=1}^{\infty} \sum_{n\epsilon A} f(n) x^{mn}$$ .

I think there should be a additional term $$x^{mn+m}$$ on RHS as x × $$G'_A(x) = \sum_{n \epsilon A} f(n) x^n$$ and then I have to put x = $$x^m$$ and then multiply by $$x^m$$ .

• "how is $G'_A (x^m) x^m = \sum_{m=1}^{\infty} \sum_{n\epsilon A} f(n) x^{mn}$". It isn't. It's $\sum_{m=1}^\infty G'_A (x^m) x^m = \sum_{m=1}^{\infty} \sum_{n\epsilon A} f(n) x^{mn}$. – Angina Seng Jan 7 '20 at 4:57
• @Lord Shark the Unknown yes that was a mistake, I have edited it. – Ben Jan 7 '20 at 5:12

Using your expression, i.e., where you multiplied both sides of the expression for $$G'_A(x)$$ by $$x$$, you could replace all $$x$$ with $$x^m$$ to get
\begin{aligned} x\left(G'_A(x)\right) & = \sum_{n \epsilon A} f(n) x^n \\ x^m\left(G'_A(x^m)\right) & = \sum_{n \epsilon A} f(n) \left(x^m\right)^n \\ G'_A(x^m)x^m & = \sum_{n \epsilon A} f(n) x^{mn} \end{aligned}\tag{1}\label{eq1A}
\begin{aligned} G_A(x) & = \sum_{n\epsilon A} \frac{f(n)}{n} x^{n} \\ G'_A(x) & = \sum_{n\epsilon A} f(n) x^{n-1} \\ G'_A(x^m) & = \sum_{n\epsilon A} f(n) \left(x^m\right)^{n-1} \\ G'_A(x^m) & = \sum_{n\epsilon A} f(n) x^{mn - m} \\ G'_A(x^m)x^m & = \sum_{n\epsilon A} f(n) x^{mn - m}x^m \\ G'_A(x^m)x^m & = \sum_{n\epsilon A} f(n) x^{mn} \end{aligned}\tag{2}\label{eq2A}
In either case, taking the sum from $$m = 1$$ to $$\infty$$ of both sides gives the part you're asking about, i.e,.
$$\sum_{m=1}^{\infty} G'_A(x^m)x^m = \sum_{m=1}^{\infty} \sum_{n\epsilon A} f(n) x^{mn} \tag{3}\label{eq3A}$$