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QUESTION

Prove that if $f$ and $g$ are multiplicative, then so is the
function $f.g$ defined by $f.g(n)=f(n)g(n)$ for all $n\in N$.

If $g(n)\neq0$ for all $n\in Z$, prove that the function
$\frac{f}{g}$ defined by $\frac{f}{g}(n)=\frac{f(n)}{g(n)}$ for
all $n\in N$ is also multiplicative.

Deduce that the function $h(n)$ defined by
$h(n)=\sum_{d|n}\frac{\mu(d)d|(d)}{\sigma(d)}$ is multiplicative.
By finding the value of $h$ when $n$ is a prime power, find a
formula for $h(n)$ in terms of the prime factorisation of $n$.




ANSWER


Let gcd$(m.n)=1$. Then $f.g(mn)=f(mn).g(mn)=f(m)f(n).g(m)g(n)$ as
$f$ and $g$ are both multiplicative. But
$f.g(m).f.g(n)=f(m)g(m)f(n)g(n)$ by definition, so comparing this
with the equation above, $f.g(mn)=f.g(m)f.g(n)$, showing that
$f.g$ is multiplicative, as required.

Similarly, if gcd$(m,n)=1,\
\frac{f}{g}(mn)=\frac{f(mn)}{g(mn)}=\frac{f(m)f(n)}{g(m)g(n)}$ by
multiplicativity of $f$ and $g$. But
$\frac{f}{g}(m).\frac{f}{g}(n)=\frac{f(m)}{g(m)}.\frac{f(n)}{g(n)}$,
and again, comparing with the equation above,
$\frac{f}{g}(mn)=\frac{f}{g}(m).\frac{f}{g}(n)$, so $\frac{f}{g}$
is multiplicative.

Now $\mu,d$ and $\sigma$ are all multiplicative, so by the above
$\frac{\mu d}{\sigma}$ is multiplicative too, and $h(n)$ is given
by $h(n)=\sum_{d|N}\left(\frac{\mu.d}{\sigma}\right)(d)$, so
$h(n)$ is multiplicative by th.8.1.

Now $h(p^k)=\sum_{d|p^k}\left(\frac{\mu
d}{\sigma}\right)(d)=\sum_{d|p^k}\frac{\mu(d)d(d)}{\sigma(d)}$. If
$d|p^k$, then $d$ is a power of $p$, so by definition of $\mu,
\mu(d)$ will be 0 unless $d=1$ or $d=p$. Thus

\begin{eqnarray*}
h(p^k)&=&\frac{\mu(1)d(1)}{\sigma(1)}+\frac{\mu(p)d(p)}{\sigma(p)}\textrm{
(all other terms vanishing)}\\ &=&1-\frac{2}{p+1} \textrm{ (using
}\mu(1)=d(1)=\sigma(1)=1,\ \mu(p)=-1, d(p)=2,\\&& \sigma(p)=p+1)\\
&=&\frac{p-1}{p+1}
\end{eqnarray*}

Thus by multiplicativity $h(n)\prod_{p|n}\frac{(p-1)}{(p+1)}$.



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