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Could any one explain to me what is the Mobius inversion formula and what is it connection with the principle of inclusion and exclusion?

My understanding is that Mobuis inversion formula can find the inverse of a function using algebraic subtraction/addition in a similar way as the inclusion-exclusion formula doing adding/subtraction.

if this is correct my next question is how do people find the Mobius inverse function?

Quant
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    It just so happens that in an answer to this question someone just pointed me to the book A Course in Enumeration by Martin Aigner, in which Section $5.2$ explains the connection between Möbius inversion and inclusion-exclusion. (Note that the letters 'o' and 'ö', though graphically similar and historically related, stand for completely different phonemes, so replacing an 'ö' by an 'o' is about as bad a misspelling as, say, replacing an 'a' by an 'i'.) – joriki Jun 01 '16 at 14:17
  • Many thanks for the reference. However, I do not have access to any library. Is it possible for you to summarize it? Furthermore, please comment on my point view, i.e. "Mobuis inversion formula can find the inverse of a function using algebraic subtraction/addition in a similar way as the inclusion-exclusion formula doing adding/subtraction." Is that correct? I understand Mobius is a Germany. However, my computer cannot type umlaub...should I type Moebius? – Quant Jun 03 '16 at 03:34

1 Answers1

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The definition of the Möbius function should resemble the adding and subtracting of inclusion-exclusion:

$$ \mu(n) = (-1)^{\omega(n)} [ n \text{ squarefree}] $$

$\omega(n)$ counts the number of distinct prime factors of $n$. The inclusion-exclusion is useful for counting and sums. Here's the simplest example I can think of:

$\varphi(n)$ counts the positive integers up to $n$ that are coprime to $n$.

To count the totient, we count how many are coprime to $n$ using inclusion-exclusion:
Start with $n$. Then for each prime divisor $p$, $n / p$ are divisible by $p$, so exclude those. Then $n / pq $ for distinct $p, q$ have been over-excluded, so add those back in, etc.

$$\varphi(n) = n - \sum_{p | n} \frac{n}{p} + \sum_{\substack{p < q \\ p, q | n }} \frac n {p q} - \cdots $$

The trick is we can combine all these sums into one sum over the divisors of $n$. Every divisor is either square-free, the product of $\omega(n)$ distinct prime factors, or has a square factor. This gives us the nice identity

$$\varphi(n) = \sum_{d | n} \mu(d) \frac n d$$

where $\mu(d)$ encodes the sign from inclusion-exclusion, including ignoring divisors $d$ with square factors by virtue of $\mu(d) = 0$. In the notation of Dirichlet convolution, $\varphi = \mu * \operatorname{Id}$. We can do similar identities with divisor-related functions like $G(n) = \sum_{a < b \le n} [\gcd(a,b) = 1] $, by summing over every possible $d \le N$. Then we can even calculate sums involving $\lfloor n / d \rfloor$ in sub-linear time with the Dirichlet hyperbola method.

qwr
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