# The Möbius Function

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 $$8$$ The first 50 values ​​of the function μ (n)MöbiusMu.PNG: User:Tosderivative work: Gregors (talk) 22:04, 8 March 2011 (UTC) [Public domain]Thus the Möbius function is a multiplicative function :

$$\mu(ab) = \mu(a)\mu(b)$$,

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

$$\displaystyle \varphi(6) = \frac 61 \, – \frac 62 \, – \frac 63 + \frac 66 = 6 \, – 3 \, – 2 + 1 = 2$$

© OldTrout $$2018$$