How to prove $\sum_{ d \mid n} \mu(d)f(d)=\prod_{i=1}^r (1-f(p_i))$?

Question

I have to prove for $n \in \mathbb{N}>1$ with $n=\prod \limits_{i=1}^r p_i^{e_i}$. $f$ is a multiplicative function with $f(1)=1$:

$$\sum_{ d \mid n} \mu(d)f(d)=\prod_{i=1}^r (1-f(p_i))$$

How I have to start? Are there different cases or can I prove it in general?

Any help would be fine :)

Hint 1: If you can show that both sides define a multiplicative function of $n$, then it suffices to verify this for prime powers $n=p^a$. Hint 2 (different approach): Notice that only squarefree $d$'s have non-zero contribution to the LHS. Can you get something similar from the RHS by expanding it? — Martin Sleziak, Nov 15 '11 at 14:35
It is recommended not to use titles consisting only of formula (with no text). http://meta.math.stackexchange.com/questions/3135/why-no-use-displaystyle-in-titles/3136#3136 — Martin Sleziak, Nov 15 '11 at 14:50
In case the above hints are not sufficient, it might help you further if you have a look at Theorem 1 in Chapter 2 of W.W.L.Chen's Elementary number theory notes. — Martin Sleziak, Nov 15 '11 at 15:23
You're right, I meant Theorem 2A (the first one in that chapter). Thanks for noticing my mistake @Ali. — Martin Sleziak, Nov 15 '11 at 15:37
https://math.stackexchange.com/questions/24362/prove-that-sum-dn-mud-sigmad-1k-prod-i-1k-p-i — Intuition, May 29 '19 at 21:38

score 8 · Accepted Answer · edited Jan 23 '21 at 14:02

8

Please see Theorem 2.18 on page $37$ in Tom Apostol's Introduction to analytic number theory book.

The proof goes as follows:

Define $$ g(n) = \sum\limits_{d \mid n} \mu(d) \cdot f(d)$$

Then $g$ is multiplicative, so to determine $g(n)$ it suffices to compute $g(p^a)$. But note that $$g(p^a) = \sum\limits_{d \mid p^{a}} \mu(d) \cdot f(d) = \mu(1)\cdot f(1) + \mu(p)\cdot f(p) = 1-f(p)$$

edited Jan 23 '21 at 14:02

VIVID

11,667

answered Nov 15 '11 at 16:09

@Martin: I can't view the page in google books. – Nov 15 '11 at 16:36

score 0 · Answer 2 · answered Feb 04 '24 at 18:54

I feel silly that I didn't see user9413's solution. Instead I proved $$\forall\ell\in\mathbb{N}\,\forall X\in\mathcal{P}(P)[|X|=\ell\implies\sum_{d\,|\,i,\,d>0}\mu(d)f(d)=\prod_{p\in X}(1-f(p))],$$ where $P$ is the set of all primes and $i=\prod_{p\in X}p$, by strong induction on $\ell$. This works because for any $n\in\mathbb{N}\setminus\{1\}$ with prime factorization $n=p_1^{a_1}...p_m^{a_m}$, we have $\sum_{d\,|\,n,\,d>0}\mu(d)f(d)=\sum_{d\,|\,j,\,d>0}\mu(d)f(d)$, where $j=\prod_{p\in\{x\in P\,:\,x\,|\,n\}}p$. Anyways here it is - if there are any errors hopefully the powers that be in number theory here can elucidate. Enjoy!

The base case is when $\ell=1$, we have for any $X=\{p_1\}\in\mathcal{P}(P),$ $$\sum_{d\,|\,i,\,d>0}\mu(d)f(d)=\sum_{d\,|\,p_1,\,d>0}\mu(d)f(d)=\mu(1)f(1)+\mu(p_1)f(p_1)=1-f(p_1)=\prod_{p\in X}(1-f(p))$$as desired. Note we assume $f$ is not the zero function, and so $f(1)=1$.

The induction step is to take $k>1$ and assume for all $k>\ell>0$, $$\text{for any }X\in\mathcal{P}(P),\quad\text{ if $|X|=\ell$ }\quad\text{ then }\quad\sum_{d\,|\,i,\,d>0}\mu(d)f(d)=\prod_{p\in X}(1-f(p)).$$Then for $k$ in particular, and any $X\in\mathcal{P}(P)$ with $|X|=k,$ say $X=\{p_1,...,p_{k-1},p_k\}$, we have \begin{aligned}\sum_{d\,|\,p_1...p_{k-1}p_k,\,d>0}\mu(d)f(d)&=\sum_{d\,|\,p_1...p_{k-1},\,d>0}\mu(d)f(d)+\sum_{\text{$d\,|\,p_1...p_{k-1}p_k,\,d>0,$}\atop\text{ $p_k\,|\,d$}}\mu(d)f(d)\\&=\prod_{p\in X\setminus\{p_k\}}(1-f(p))-f(p_k)\cdot\sum_{d\,|\,p_1...p_{k-1},\,d>0}\mu(d)f(d)\\\\&=\prod_{p\in X\setminus\{p_k\}}(1-f(p))-f(p_k)\cdot\prod_{p\in X\setminus\{p_k\}}(1-f(p))\\&=(1-f(p_k))\cdot\prod_{p\in X\setminus\{p_k\}}(1-f(p))\\&=\prod_{p\in X}(1-f(p)),\end{aligned}where the second equality uses the multiplicative property of both $\mu$ and $f$ as well as the induction hypothesis. This completes the proof.

How to prove $\sum_{ d \mid n} \mu(d)f(d)=\prod_{i=1}^r (1-f(p_i))$?

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