The Dirichlet Convolution μ * 1 = ꙇ

# The Dirichlet Convolution μ * 1 = ꙇ

Recall from The Möbius Function page that the Möbius function is an arithmetic function defined by:

(1)\begin{align} \quad \mu(n) = \left\{\begin{matrix}0 & \mathrm{if} \: n \: \mathrm{is \: NOT \: square free.}\\ (-1)^{\omega (n)} & \mathrm{if} \: n \: \mathrm{is \: square \: free.} \end{matrix}\right. \end{align}

We proved that $\mu$ is a multiplicative function. We will now prove a very important Dirichlet convolution, which says the Dirichlet convolution with $\mu$ and $1$ is the $\iota$ function.

Proposition 1: $\mu * 1 = \iota$. |

**Proof:**Since $\mu$ and $1$ are both multiplicative functions, their Dirichlet convolution $\mu * 1$ is also multiplicative. So we only need to show that $\mu * 1$ and $\iota$ agree on prime powers. Let $p$ be a prime and let $k \in \mathbb{N}$. Then:

\begin{align} \quad \mu(1) = 1 \end{align}

- And:

\begin{align} \quad (\mu * 1)(p^k) &= \sum_{d|p^k} \mu (d) \\ &= \mu(1) + \mu(p) + \mu(p^2) + ... + \mu(p^k) \\ &= 1 + (-1) + 0 + ... + 0 \\ &= 0 \end{align}

- Therefore we see that $(\mu * 1)(n) = 1$ if $n = 1$ and $(\mu *1)(n) = 0$ if $n \geq 1$. Hence:

\begin{align} \quad \mu * 1 = \iota \quad \blacksquare \end{align}