convolution method

Let $f$, $g$, and $h$ be multiplicative functions such that $f=g*h$, where $*$ denotes the convolution (http://planetmath.org/MultiplicativeFunction) of $g$ and $h$. The convolution method is a way to ${\displaystyle \sum _{n\le x}}f(n)$ by using the fact that $f=g*h$:

This method for calculating ${\displaystyle \sum _{n\le x}}f(n)$ is advantageous when the sums in of $g$ and $h$ are easier to handle.

As an example, the sum ${\displaystyle \sum _{n\le x}}{\mu }^2(n)$ will be calculated using the convolution method.

Since ${\mu }^2={\mu }^2*\epsilon ={\mu }^2*(\mu *1)=({\mu }^2*\mu )*1$, the functions ${\mu }^2*\mu$ and $1$ can be used.

To use the convolution method, a nice way to $({\mu }^2*\mu )(n)$ needs to be found. Note that ${\mu }^2*\mu$ is multiplicative (http://planetmath.org/MultiplicativeFunction), so it only needs to be evaluated at prime powers.

Let $k\in \mathbb{N}$. Then

Since ${\mu }^2*\mu$ is multiplicative (http://planetmath.org/MultiplicativeFunction), then

The convolution method yields: