3
$\begingroup$

for positive integer $n$, how can we show

$$ \sum_{d | n} \mu(d) d(d) = (-1)^{\omega(n)} $$

where $d(n)$ is number of positive divisors of $n$ and $mu(n)$ is $(-1)^{\omega(n)} $ if $n$ is square free, and $0$ otherwise. Also, what is

$$ \sum_{d | n} \mu(d) \sigma (d) $$ where $\sigma(n)$ is the sum of positive divisors of $n$

$\endgroup$
  • 3
    $\begingroup$ Bad form to use the symbol $d$ with two different meanings in one formula. $\endgroup$ – Gerry Myerson Apr 4 '13 at 12:27
  • 1
    $\begingroup$ $d(n)$ is usually written $\tau(n)$. $\endgroup$ – lhf Apr 4 '13 at 12:38
2
$\begingroup$

$$\sum_{d\mid n}\mu(d)f(d)$$ $$=\mu(1)d(1)+\sum_{p_i|n}\mu(p_i)f(p_i)+\sum_{p_ip_j|n}\mu(p_ip_j)f(p_ip_j)+\sum_{p_ip_jp_k|n}\mu(p_ip_jp_k)f(p_ip_jp_k)+\cdots+\sum_{p_ip_j\cdots|n}\mu(p_i^{r_1}p_j^{r_j}p_k^{r_k}\cdots)f(p_i^{r_1}p_j^{r_j}p_k^{r_k}\cdots)$$ where $p_i,p_j,p_k$ are distinct primes and at least one of the integers $r_i,r_j,r_k>1$

As $\mu$ is multiplicative, $\mu(p_i^{r_1}p_j^{r_j}p_k^{r_k}\cdots)=\mu(p_i^{r_1})\mu(p_j^{r_j})\mu(p_k^{r_k})\cdots=0$ as at least one of the integers $r_i,r_j,r_k>1$

$$\sum_{d\mid n}\mu(d)f(d)=\mu(1)f(1)+\sum_{p_i|n}\mu(p_i)f(p_i)+\sum_{p_ip_j|n}\mu(p_ip_j)f(p_ip_j)+\sum_{p_ip_jp_k|n}\mu(p_ip_jp_k)f(p_ip_jp_k)+\cdots$$

If $f(n)$ is also multiplicative with $f(1)=1$,

$$\sum_{d\mid n}\mu(d)f(d)$$ $$=\mu(1)f(1)+\sum_{p_i|n}\mu(p_i)f(p_i)+\sum_{p_ip_j|n}\mu(p_i)\mu(p_j)f(p_i)f(p_j)+\sum_{p_ip_jp_k|n}\mu(p_i)\mu(p_j)\mu(p_k)f(p_i)f(p_j)(f)p_k)+\cdots$$

$$=1+\sum_{p_i|n}\mu(p_i)f(p_i)+\sum_{p_ip_j|n}\mu(p_i)\mu(p_j)f(p_i)f(p_j)+\sum_{p_ip_jp_k|n}\mu(p_i)\mu(p_j)\mu(p_k)f(p_i)f(p_j)(f)p_k)+\cdots$$

$$=\prod(1+\mu(p_i)f(p_i))=\prod_{p_i\mid n}(1-f(p_i))$$

As $d$ or $\tau$ is multiplicative with $\tau(1)=1,\tau(p)=1+1=2$ where $p$ is prime

$$\sum_{d\mid n}\mu(d)\tau(d)=\prod_{p_i\mid n}(1-2)=(-1)^{\omega(n)}$$ where $\omega(n)$ is the number of prime factors of $n$

As $\sigma$ is multiplicative with $\sigma(1)=1$ and $\sigma(p)=p+1$ where $p$ is prime

$$\sum_{d\mid n}\mu(d)\sigma(d)=\prod_{p_i\mid n}\{1-(p_i+1)\} = \prod_{p_i\mid n}(-p_i)= (-1)^{\omega(n)}\prod_{p_i\mid n} p_i$$ where $\omega(n)$ is the number of prime factors of $n$

$\endgroup$
  • $\begingroup$ It would appear that you have left out terms involving $d=p_1p_2$ and $d=p_1p_2p_3$ and the like from your equations. $\endgroup$ – Gerry Myerson Apr 4 '13 at 12:29
  • $\begingroup$ @GerryMyerson, rectified the horrible mistake. $\endgroup$ – lab bhattacharjee Apr 4 '13 at 15:17
2
$\begingroup$

The sums you ask about will be multiplicative functions, so you need only evaluate them in the case where $n$ is a prime power, then multiply over the primes dividing $n$. Perhaps this is what @lab was getting at in the other answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.