Prove that $\sum_{d|n}\mu(d)\log^m d=0$

Question

Prove that $$\sum_{d|n}\mu(d)\log^m d=0$$ if $m\ge 1$ and $n$ has more than $m$ distinct prime factors.

I tried using Induction and kind of succeeded in the sense that if we write down the case $m=2$ and $n=pqr$, $p,q,r$ being distinct primes, the general case becomes quite evident- \begin{align*} \sum_{d|pqr}\mu(d)\log^2 d &= \mu(1)\log^2 1+\mu(p)\log^2 p+\mu(q)\log^2 q+\mu(r)\log^2 r+\mu(pq)\log^2 pq+\mu(qr)\log^2 qr+\mu(pr)\log^2 pr+\mu(pqr)\log^2 pqr\\ &=0-\log^2 p-\log^2 q-\log^2 r+(\log p+\log q)^2+(\log q+\log r)^2+(\log p+\log r)^2-(\log p+\log q+\log r)^2 \end{align*} from which it is intuitively clear that the terms will cancel out to give $0$.

But, to use Induction, we need to break a $\log^{m+1}d$ into lower powers of $\log d$, and the cancellation will be very difficult to show (especially considering the fact that it would require multinomial theorem).

So, is there an easy way out? Can we somehow use Dirichlet Multiplication or Möbius Inversion Formula?

A duplicate question has been suggested, but the answer here ends with "You still need to do some work before you can apply the induction hypothesis" which is precisely where I am having my troubles.

Answer 1

There is in fact an easy way to solve this using Dirichlet convolution. I'll let you do the case $m = 1$. Suppose the result is valid for $m$, that is, if $\omega(n) > m$, then $$\sum_{d | n} \mu(d) \log^{m}(d) = 0.$$ This allows us to rewrite the $m+1$ sum in the following way: $$\sum_{d|n} \mu(d) \log^{m+1}(d)= - \sum_{d|n} \mu(d) \log^{m}(d) \log(n/d).$$ Indeed, just expand $\log$ on the right hand side and use the induction hypothesis. Now we have a Dirichlet convolution, namely $-\mu \log^m * \log$. Since $\log = \Lambda * 1$, where $\Lambda$ is the Mangoldt function and $1$ is the constant function, we can use commutativity and associativity of the operation to get $$-\mu \log^m * \log = -\mu \log^m * (\Lambda * 1) = (-\mu \log^m * 1) * \Lambda.$$ Therefore, $$\sum_{d|n} \mu(d) \log^{m+1}(d) = -\sum_{d | n} \left(\sum_{s|d} \mu(s) \log^m(s)\right)\Lambda(n/d).$$ If $\omega(d) > m$, the inner sum is zero by the induction hypothesis. If $\omega(d) \leq m$, since $\omega(n) > m+1$, $n/d$ has at least two distinct prime factors. Therefore, $\Lambda(n/d) = 0$. Thus, the entire sum is zero.

Answer 2

A way to finish the claim of the linked post is the following:

The required sum $$\sum_{d \mid s}{\mu(d)\log^m(d)}+\sum_{d\mid s}{\mu(pd)\log^m(pd)}$$ can be combined as $$\sum_{d \mid s} \left[ {\mu(d)\log^m(d)+\mu(p)\mu(d)(\log (p) + \log(d))^m} \right].$$ Using the binomial theorem and the fact that $(d,p)=1$, $$\sum_{d \mid s}{ \mu(d)\left[ \log^m(d)- \left(\log^m(d) + \sum_{0\leq k \leq m-1}{ \binom{m}{k} \log^{k}(d)} \log^{m-k}(p)\right) \right]}.$$ which simplifies to

$$-\sum_{d \mid s}{ \mu(d)\left[ \sum_{0\leq k \leq m-1}{ \binom{m}{k} \log^{k}(d)} \log^{m-k}(p) \right]}.$$ Once we apply the fundamental theorem of analytic number theory (aka switching sums), we get $$-\sum_{0\leq k \leq m-1}\binom{m}{k}\log^{m-k}(p)\sum_{d \mid s}{ \mu(d) { \log^{k}(d)} }.$$

The inner sum is $0$ by the induction hypothesis for all $k$ in the given range since the number of prime factors of $s$ is $>m-1 \geq k$.

Answer 3

For $m=1$ one can argue directly counting how may times each $\log(p_i)$ appears and using the binomial Theorem:

$$\sum_{d|n}\mu(d)\log(d)=\sum_i \log(p_i)-\sum_{i_1<i_2}\log(p_{i_1}p_{i_2})+\sum_{i_1<i_2<i_3}\log(p_{i_1}p_{i_2}p_{i_3})...=$$ $$\sum_i \log(p_i)\left({k-1\choose0}-{k-1\choose1}+{k-1\choose2}-...\right)=0$$

For the general case, there is a nice trick using the formal derivative which will allow us to finish the problem. We recall: $f'(n):=f(n)\log(n)$ has very nice properties such as:

$$(f*g)'=f'*g+f*g'$$

With this we are able to deduce that if $n$ has at least $m+1$ prime factors:

$$\mu \log^m* u(n)=(\mu \log^{m-1})'*u(n)=-\mu\log^{m-1}*u'(n)$$

Where the boundary term $(-\mu \log^{m-1}*u)'(n)$ has vanished due to the induction hypothesis: $$(-\mu \log^{m-1}*u)'(n)=-\log(n)\sum_{d|n}\mu(d)\log^{m-1}(d)$$ It also holds that $u'=u*\Lambda$ from which it follows that:

$$\mu \log^m *u(n)=-\mu \log^{m-1}*\Lambda*u(n)$$ Proceeding like this it is easy to see: $$\mu \log^m *u(n)=(-1)^{m}\mu*\Lambda^{m} *u=(-1)^m \mathbb{1}*\Lambda^m(n)=(-1)^m\Lambda^m(n)$$

Thus we see one only needs to verify that:

$\Lambda^m(n)=0$ whenever $n$ has at least $m+1$ prime divisors.

When $m=1$ this follows from the definition of the Mangoldt function. For larger $m$ we may use induction:

$$\Lambda^m(n)=\sum_{d|n}\Lambda^{m-1}(n/d)\Lambda(d)=\sum_{p_i}\sum_{j=1}^{a_i}\Lambda^{m-1}(n/p_i^{j})\log(p_i) $$

However $n/p_i^{j}$ has at least $m-1$ prime factors from which it follows that:

$$\Lambda^m(n)=\sum_{p_i}\sum_{j=1}^{a_i}0\log(p_i)=0 $$

Answer 4

A little late to the show, but take my 5 cents on this. Arkady has, in my opinion, the best argument, even though I find it too neat. Yes, I think too neat is a thing. I just wouldn't bother to investigate such a case because I wouldn't think that it'd yield the result. Nonetheless, here's a brief proof in the language of Dirichlet multiplication, and I'm skipping the base case because it's trivial.

We assume that for $\omega(n)>m\geq1 \Rightarrow (\mu u'^{m}*u)(n)=0$, and that $n=p_{1}p_{2}\dots p_{l}$ which doesn't harm the general case, because $\mu(k)=0$ for $k$ that is not a square-free integer.

Now, for an integer $r$ s.t. $\omega(r)>m+1$, the inductive hypothesis holds because $\omega(r)>m+1>m \Leftrightarrow \omega(r)>m$, and upon differentiating it we arrive at

$(\mu u'^{m+1}*u)(r) + (\mu u'^{m}*u')(r)=0$, which can be equivalently rewritten as $(\mu u'^{m+1}*u)(r)=-(\mu u'^{m}*u')(r)$.

Remembering that $\Lambda*u=u'$ we can further rewrite the RHS of the above identity as $-(\mu u'^{m}*(\Lambda * u))(r)$, and as $-((\mu u'^{m}*u)*\Lambda)(r)$.

Thus, we have that

$(\mu u'^{m+1}*u)(r)=$

$=-\sum_{d|r}(\mu u'^{m}*u)(d)\Lambda(\frac{r}{d})=$

$=-\sum_{d|r}\sum_{s|d}\mu(s)\log^{m}(s)\Lambda(\frac{r}{d})=$

$=\sum_{d|r}\sum_{s|d}\mu(s)\log^{m}(s)\sum_{s'|\frac{r}{d}}\mu(s')\log(s')=$

$=\sum_{d|r}\sum_{s|\frac{r}{d}}\mu(s)\log^{m}(s)\sum_{s'|d}\mu(s')\log(s')$,

from which we see that the sum $\sum_{s'|d}\mu(s')\log(s')=0$ everywhere where $d\neq p_{j}$ because of our inductive hypothesis$(\omega(d)>1$ except for $d=p_{j} \Leftrightarrow \omega(d)=\omega(p_{j})=1)$.

This means that

$(\mu u'^{m+1}*u)(r)=$

$=\sum_{d|r}\sum_{s|\frac{r}{d}}\mu(s)\log^{m}(s)\sum_{s'|d}\mu(s')\log(s')=$

$=\sum_{j=1}^{l}\sum_{s|\frac{r}{p_{j}}}\mu(s)\log^{m}(s)\sum_{s'|p_{j}}\mu(s')\log(s')$

and by the inductive hypothesis the sum $\sum_{s|\frac{r}{p_{j}}}\mu(s)\log^{m}(s)=0$ because $\omega(\frac{r}{p_{j}})\geq m+1 > m \Leftrightarrow \omega(\frac{r}{p_{j}}) > m$.

Therefore $\omega(r)>m+1 \rightarrow(\mu u'^{m+1} * u)(r) = 0$, which completes the proof.