Exercise 2.10

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 a multiplicative function, show that the function $g(n) = \sum_{d \mid n} f(d)$ 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}
If $n \wedge m =1$,
\begin{align*}
g(nm)&= \sum_{\delta \mid nm} f(\delta)\
&= \sum_{d\mid n, d'\mid m }f(dd')\
\end{align*}
Actually, 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 $\delta = dd'$.

If $d \mid n, d' \mid m$, with $n\wedge m = 1$, then $d \wedge d' = 1$, so
\begin{align*}
g(nm) &=\sum_{d\mid n } \sum_{d'\mid m} f(d) f(d')\
&= \sum_{d\mid n}f(d) \sum_{d'\mid m} f(d')\
&=g(n) g(m)
\end{align*}
$g$ is a multiplicative function.
\end{proof}

Exercise 2.10

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Exercise 2.10

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