How can I prove this functional equation?

Question

Question:- If $$F(x)=f(x)+\frac12f(x^2)+\frac13f(x^3)+\frac14f(x^4)+\ldots\,,$$ then prove that $$f(x)=F(x)-\frac12F(x^2)-\frac13F(x^3)-\frac15F(x^5)+\frac16F(x^6)-\frac17F(x^7)+\frac1{10}F(x^{10})-\ldots\,,$$ where only terms involving numbers that contain no square factor appear in the second series, and the sign is positive or negative according as the number of prime factors of the numbers is even or odd.

My friend sent me this question today, I haven't deal with such questions much so I was not able to figure out any method to start this question.

Can anybody help me to prove it!!

Answer 1

If we assume $$ f(x)=\sum_{n\geq 1}a_n x^n $$ we have $$ F(x)=\sum_{m\geq 1}\frac{f(x^m)}{m}=\sum_{m\geq 1}\sum_{n\geq 1}\frac{a_n}{m}x^{mn}=\sum_{h\geq 1}x^h \sum_{d\mid h}\frac{a_d}{h/d} $$

$$\sum_{n\geq 1}\frac{\mu(n)}{n}F(x^n)=\sum_{n\geq 1}\frac{\mu(n)}{n}\sum_{h\geq 1}x^{hn}\sum_{d\mid h}\frac{a_d}{h/d}=\sum_{h\geq 1}x^h\sum_{d\mid h}\frac{\mu(d)}{d}\sum_{e\mid h/d}\frac{ea_e}{h/d} $$

so

$$ [x^h]F(x) = \frac{1}{h}\sum_{d\mid h} d a_d $$ $$ [x^h]\sum_{n\geq 1}\frac{\mu(n)}{n}F(x^n)=\frac{1}{h}\sum_{d\mid h}\sum_{e\mid h/d}\mu(d) e a_e = \frac{1}{h}\sum_{e\mid h}\sum_{d\mid h/e}\mu(e) d a_d.$$ Given the arithmetic function $u(n)=na_n$, in terms of Dirichlet convolutions we have $$ [x^h]F(x)=\frac{1}{h}(u*1)(h) $$ $$ [x^h]\sum_{n\geq 1}\frac{\mu(n)}{n}F(x^n)=\frac{1}{h}\left(1*(\mu*u)\right)(h) $$ The claim follows from the fact that $*$ is commutative, associative and $$ (\mu*1)(n) = \sum_{d\mid n}\mu(d) $$ is a multiplicative function wich equals $1$ for $n=1$ and $0$ for any prime power.

Answer 2

I assume that all involved series converge absolutely (whence I can reorder the summation). I am not sure whether the formula is even correct if there are conditionally convergent series involved. Alternatively, if $f$ is a power series with zero constant term, then both sums make sense regardless of the base field (which has to be of characteristic $0$), just like how Jack D'Aurizio handled it.

Let $\mu$ be the usual Möbius function. From $F(x)=\sum\limits_{n\in\mathbb{Z}_{>0}}\,\dfrac{1}{n}\,f(x^n)$, we have $$\sum_{n\in\mathbb{Z}_{>0}}\,\frac{\mu(n)}{n}\,F(x^n)=\sum_{n\in\mathbb{Z}_{>0}}\,\frac{\mu(n)}{n}\,\sum_{m\in\mathbb{Z}_{>0}}\,\frac{1}{m}\,f\left(x^{mn}\right)=\sum_{m,n\in\mathbb{Z}_{>0}}\,\frac{\mu(n)}{mn}\,f\left(x^{mn}\right)\,.$$ That is, $$\sum_{n\in\mathbb{Z}_{>0}}\,\frac{\mu(n)}{n}\,F(x^n)=\sum_{n\in\mathbb{Z}_{>0}}\,\frac{1}{n}\,f(x^n)\,\sum_{d\mid n}\,\mu(d)=f(x)\,,$$ since $\sum\limits_{d\mid n}\,\mu(d)=0$ if $n\in\mathbb{Z}_{>1}$, whilst $\sum\limits_{d\mid 1}\,\mu(d)=\mu(1)=1$.