The Möbius function is denoted by \(\mu(n)\), where \(\mu\) is the Greek letter mu.
Depending on the factorisation of \(n\), the function takes the values \(\{-1, 0, 1\}\).
\(\mu(n) =
\begin{cases}
1, & \text{if}\, n = 1\\
0, & \text{if}\, p^2 | n\\
(-1)^k, & \text{if}\, n \,\text{is the product of}\, k \,\text{distinct primes}
\end{cases} \)
The last says that if \(k\) is even, then \(\mu(n) = 1\) and if \(k\) is odd, then \(\mu(n) = -1\).
Consider the numbers \(6, 7\) and \(8\).
\(6 = 2\cdot 3 \) so \(\mu(6) = 1\)
\(7 = \) prime so \(\mu(7) = -1\)
\(8 = 2^3\) so \(\mu(8) = 0\) since \(2^2\) divides \(8\)
MöbiusMu.PNG: User:Tosderivative work: Gregors (talk) 22:04, 8 March 2011 (UTC) [Public domain]
\(\mu(ab) = \mu(a)\mu(b)\),
where gcd\((a, b) = 1\).
The sum of the Möbius function over all the positive divisors of \(n\) (including \(1\) and itself) is zero except when \(n = 1\).
\(\displaystyle \sum_{d|n} \mu (d) =
\begin{cases}
1, & \text{if} \, n = 1\\
0, & \text{if} \, n \gt 1
\end{cases} \)
For example, the divisors of \(6\) are \(1, 2, 3\) and \(6\) so \(1 -1 -1 + 1 = 0\).
The Möbius function is connected to Euler’s totient function by Möbius inversion :
Let \(f\) and \(g\) be arithmetic (multiplicative) functions such that
\(\displaystyle f(n) = \sum_{d|n} g(d)\) for all \(n\), then
\(\displaystyle g(n) = \sum_{d|n} \mu(d) \, \, f\left(\frac nd\right) \)
From the Gaussian identity of Euler’s totient function we have :
\(\displaystyle f(n) = \sum_{d|n}\varphi(d)\)
When \(n = 6, \, \, f(6) = 1 + 1 + 2 + 2 = 6\).
So, by the inversion theorem :
\(\displaystyle g(n) = \varphi(n) = \sum_{d|n} \mu(d) \frac nd\)
When \(n = 6\) then
\(\displaystyle \varphi(6) = \sum_{d|6} \mu(d) \frac 6d \)
Thus
\(\displaystyle \varphi(6) = \frac 61 \, – \frac 62 \, – \frac 63 + \frac 66 = 6 \, – 3 \, – 2 + 1 = 2\)
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