Difference between revisions of "Mobius function"

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One unique fact about the Mobius function, which leads to the Mobius inversion formula, is that
 
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>
 
<cmath>\sum_{d|n} \mu(d) = \begin{cases} 1 & n = 1, \\ 0 & \text{otherwise}. \end{cases}</cmath>
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Property 1: The function <math>\mu(n)</math> is multiplicative .
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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>.
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The Mobius function is also closely related to the [[Riemann zeta function]], as
 
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>
 
<cmath>\frac{1}{\zeta(s)} = \sum \frac{\mu(k)}{n^s}.</cmath>

Revision as of 02:24, 12 March 2012

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 $\mu(m)\mu(n)=(-1)^{k} (-1)^{h}=(-1)^{k+h}=\mu(mn)$.


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