Exercise 2.14

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

If $f(n)$ is multiplicative, show that $h(n) = \sum_{d \mid n} \mu(n/d)f(d)$ is also multiplicative.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
We show first that the Dirichlet product $f \circ g$ of two multiplicative functions $f,g$ is multiplicative.
Suppose that $n\wedge m = 1$. If $d\mid n, d'\mid m$, so $\delta = dd' \mid nm$, and conversely, if $\delta \mid nm$, as $n\wedge m = 1$, there exist $d,d'$ such that $d\mid n,d'\mid m$, and $\delta = dd'$. Thus
\begin{align*}
(f\circ g)(nm)&= \sum_{\delta \mid nm} f(\delta) g\left(\frac{nm}{\delta}ight)\
&=\sum_{d\mid n,d'\mid m} f(dd')g\left(\frac{nm}{dd'}ight)\
&=\sum_{d\mid n} \sum_{d'\mid m} f(d)f(d') g\left(\frac{n}{d}ight)g\left(\frac{m}{d'}ight)\
&=\sum_{d\mid n} f(d)g\left(\frac{n}{d}ight)\sum_{d'\mid m} f(d') g\left(\frac{m}{d'}ight)\
&=(f\circ g)(n)(f\circ g)(m)
\end{align*}
Applying this result with $g = \mu$, we obtain that $n\mapsto h(n) = \sum_{d \mid n} \mu(n/d)f(d)$ is multiplicative, if $f$ is multiplicative.
\end{proof}

Exercise 2.14

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Exercise 2.14

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