Difference between revisions of "Mobius function"

Revision as of 00:05, 22 September 2020

The Mobius function is a multiplicative number theoretic function defined as follows: $\mu(n) = \begin{cases} 0 & d^2 | n, \\ (-1)^k & n = p_1p_2\cdots{p_k} .\end{cases}$ In addition, $\mu(1) = 1$ .

The Mobius function is useful for a variety of reasons.

First, it conveniently encodes Principle of Inclusion-Exclusion. For example, to count the number of positive integers less than or equal to $n$ and relatively prime to $n$ , we have

$\phi(n)$	$=n$
	$- \frac{n}{p_1} - \frac{n}{p_2} - \cdots - \frac{n}{p_k}$
	$+ \frac{n}{p_1p_2} + \frac{n}{p_1p_3} + \cdots + \frac{n}{p_{k-1}p_k}$
	$\vdots$
	$+ (-1)^k \frac{n}{p_1 p_2 \cdots p_k},$

more succinctly expressed as $\phi(n) = \sum_{d|n} \mu(d) \frac{n}{d}.$

One unique fact about the Mobius function, which leads to the Mobius inversion formula, is that $\sum_{d|n} \mu(d) = \begin{cases} 1 & n = 1, \\ 0 & \text{otherwise}. \end{cases}$

Property 1: The function $\mu(n)$ is multiplicative .

Proof:If $n=1$ or $p^2|n$ for a prime $p$ , we are done.Else let $m=\prod_{j=1}^{k} {p_j}$ and $n=\prod_{i=1}^{h} {q_i}$ where $gcd(m,n)=1$ ,then $\mu(m)\mu(n)=(-1)^{k} (-1)^{h}=(-1)^{k+h}=\mu(mn)$ .

Property 2:If $F(n)=\sum_{d|n} f(d)$ for every positive integer $n$ , then $f(n)=\sum_{d|n}\mu(d)F(\frac{n}{d})$ . Proof:We have $\sum_{d|n}\mu(d)F(\frac{n}{d})=\sum_{d|n}\mu(d)\sum_{k|n/d}f(k)=\sum_{k|n}\sum_{d|n/k}\mu(d)f(k) =\sum_{k|n}f(k)\sum_{d|n/d}\mu(d)= f(n)$ .

The Mobius function is also closely related to the Riemann zeta function, as $\frac{1}{\zeta(s)} = \sum \frac{\mu(k)}{n^s}.$

@@ Line 24: / Line 24: @@
 One unique fact about the Mobius function, which leads to the Mobius inversion formula, is that
 <cmath>\sum_{d|n} \mu(d) = \begin{cases} 1 & n = 1, \\ 0 & \text{otherwise}. \end{cases}</cmath>
+Property 1: The function <math>\mu(n)</math> is multiplicative .
+Proof:If <math>n=1</math> or <math>p^2|n</math> for a prime <math>p</math>, we are done.Else let <math>m=\prod_{j=1}^{k} {p_j}</math> and <math>n=\prod_{i=1}^{h} {q_i}</math> where <math>gcd(m,n)=1</math>,then <math>\mu(m)\mu(n)=(-1)^{k} (-1)^{h}=(-1)^{k+h}=\mu(mn)</math>.
+Property 2:If <math>F(n)=\sum_{d|n} f(d)</math> for every positive integer <math>n</math>, then <cmath>f(n)=\sum_{d|n}\mu(d)F(\frac{n}{d})</cmath>.
+Proof:We have <cmath>\sum_{d|n}\mu(d)F(\frac{n}{d})=\sum_{d|n}\mu(d)\sum_{k|n/d}f(k)=\sum_{k|n}\sum_{d|n/k}\mu(d)f(k) =\sum_{k|n}f(k)\sum_{d|n/d}\mu(d)= f(n)</cmath> .
 The Mobius function is also closely related to the [[Riemann zeta function]], as
 <cmath>\frac{1}{\zeta(s)} = \sum \frac{\mu(k)}{n^s}.</cmath>

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Difference between revisions of "Mobius function"

Revision as of 00:05, 22 September 2020