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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

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

$$ \mu(n) = \begin{cases} (-1)^k & \text{if $n$ is square-free with $k$ prime factors} \\ 0 & \text{if $n$ has a square factor} \end{cases} $$

Here's an example of inclusion-exclusion. $\varphi(n)$ counts the positive integers up to $n$ that are coprime to $n$. The cototient $n - \varphi(n)$ counts the positive integers up to $n$ that share at least one prime factor with $n$.

To count the cototient, we count numbers using inclusion-exclusion based on how many distinct prime factors are shared with $n$:

$$n - \varphi(n) = \sum_p \frac n p - \sum_{p, p'} \frac n {p p'} + \cdots $$

Now, every divisor $d$ of $n$ is either square-free with $k$ distinct prime factors, or has a square factor. Rearranging we get

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

where $\mu(d)$ encodes the sign from inclusion-exclusion, including ignoring divisors $d$ with square factors by virtue of $\mu(d) = 0$. This proves $\mu * \operatorname{Id} = \varphi$ and by Möbius inversion, $\varphi * 1 = \operatorname{Id}$.

qwr
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