Exercises on asymptotic estimate in analytic numbere theory

Question

I have a problem doing this (apparently) simle exercise: show that uniformly in $k$, for $x\to\infty$ one has$$\sum_{n\leq x,\ (n,k)=1} 1=x \frac{\varphi(k)}{k}+O(2^{\omega(k)}),$$where $\omega(k)$ is the number of distinct primes of $k$ (without the molteplicity). I started in this way $$\sum_{n\leq x,\ (n,k)=1} 1=\sum_{n\leq x} \sum_{d|n,\ d|k} \mu(d),$$with $\mu(d)$ Moebius function.$$=\sum_{d\leq x,\ d|k}\sum_{n\leq x,\ d|n}\mu(d)=\sum_{d\leq x,\ d|k}\mu(d)(\frac{x}{d}+O(1)). $$I don't know if it could be a right way to do this exercise. Does someone have a better idea? Thanks!

Answer 1

You may note that from the Möbius inversion we have $$\phi\left(n\right)=n\sum_{d\mid n}\frac{\mu\left(d\right)}{d}$$ so $$x\sum_{d\leq x,\,d\mid k}\frac{\mu\left(d\right)}{d}+O\left(\sum_{d\leq x,\,d\mid k}\left|\mu\left(d\right)\right|\right)$$ $$=x\frac{\phi\left(k\right)}{k}+O\left(2^{\omega\left(k\right)}\right).$$ The identity $$\sum_{d\leq x,\,d\mid k}\left|\mu\left(d\right)\right|=2^{\omega\left(k\right)}$$ follows from the fact that the Möbius function is multiplicative so $$\sum_{d\leq x,\,d\mid k}\left|\mu\left(d\right)\right|=1+\sum_{i}\left|\mu\left(p_{i}\right)\right|+\sum_{i,j}\left|\mu\left(p_{i}p_{j}\right)\right|+\dots$$ $$=1+\dbinom{\omega\left(k\right)}{1}+\dbinom{\omega\left(k\right)}{2}+\dots=\sum_{m=0}^{\omega\left(k\right)}\dbinom{\omega\left(k\right)}{m}=2^{\omega\left(k\right)}.$$