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Let $f$ $g$ and $h$ be multiplicative functions such that $f=g*h$ where $*$ denotes the convolution of $g$ and $h$ The convolution method is a way to calculate $\displaystyle \sum_{n \le x} f(n)$ by using the fact that $f=g*h$
$\begin{array}{ll} \displaystyle \sum_{n \le x} f(n) & \displaystyle = \sum_{n \le x} \sum_{d|n} g(d) h \left( \frac{n}{d} \right) \\ & \displaystyle = \sum_{d \le x} \sum_{m \le \frac{x}{d}} g(d) h(m) \\ & \displaystyle = \sum_{d \le x} g(d) \sum_{m \le \frac{x}{d}} h(m) \end{array}$
This method for calculating $\displaystyle \sum_{n \le x} f(n)$ is advantageous when the sums in terms 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 $\mu^2=\mu^2*\varepsilon=\mu^2*(\mu*1)=(\mu^2*\mu)*1$ the functions $\mu^2*\mu$ and $1$ can be used.
To use the convolution method, a nice way to calculate $(\mu^2*\mu)(n)$ needs to be found. Note that $\mu^2*\mu$ is multiplicative, so it only needs to be evaluated at prime powers.
Let $k \in \mathbb{N}$ Then
$$ (\mu^2*\mu)(p^k) = \sum_{j=0}^k \mu^2(p^j) \mu(p^{k-j}) = \mu(p^k)+\mu(p^{k-1}) = \begin{cases} 0 & \text{if } k \neq 2 \\ -1 & \text{if } k=2. \end{cases}$$
Since $\mu^2*\mu$ is multiplicative, then $(\mu^2*\mu)(n)= \begin{cases} 0 & {if } n { is not a square} \\ \mu(\sqrt{n}) & {if } n { is a square.} \end{cases}$ The convolution method yields:
$\begin{array}{ll} \displaystyle \sum_{n \le x} \mu^2(n) & \displaystyle = \sum_{d \le x} (\mu^2*\mu)(d) \sum_{m \le \frac{x}{d}} 1 \\ & \\ & \displaystyle = \sum_{d \le x} (\mu^2*\mu)(d) \left( \frac{x}{d} + O(1) \right) \\ & \\ & \displaystyle = \sum_{a \le \sqrt{x}} \mu(a) \left( \frac{x}{a^2} + O(1) \right) \\ & \\ & \displaystyle = x \sum_{a \le \sqrt{x}} \frac{\mu(a)}{a^2} + O \left( \sum_{a \le \sqrt{x}} |\mu(a)| \right) \\ & \\ & \displaystyle = x \left( \sum_{a \in \mathbb{N}} \frac{\mu(a)}{a^2} - \sum_{a>\sqrt{x}} \frac{\mu(a)}{a^2} \right) + O \left( \sum_{a \le \sqrt{x}} 1 \right) \\ & \\ & \displaystyle = x \sum_{a \in \mathbb{N}} \frac{\mu(a)}{a^2} + O \left( x \sum_{a>\sqrt{x}} \left| \frac{\mu(a)}{a^2} \right| \right) + O(\sqrt{x}) \\ & \\ & \displaystyle = \frac{6}{\pi^2}x + O \left( x \sum_{a>\sqrt{x}} \frac{1}{a^2} \right) + O(\sqrt{x}) \\ & \\ & \displaystyle = \frac{6}{\pi^2}x +
O \left( x \left( \frac{1}{\sqrt{x}} \right) \right) + O(\sqrt{x}) \\ & \\ & \displaystyle = \frac{6}{\pi^2}x + O(\sqrt{x}) \end{array}$
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