Möbius Inversion
Suppose for some (not necessarily multiplicative) number-theoretic function \(f\)
Can we make \(f(n)\) the subject of this equation?
We’ll see that we can find a function \(\mu\) such that
and we call this process Möbius inversion.
We have
If we want this equal to \(f(n)\) we need \(\mu\) to satisfy
A little thought leads to this unique solution, known as the Möbius function:
Notice \(\mu\) is multiplicative, which implies \(f(n)\) is multiplicative if \(F(n)\) is. In summary,
Theorem:
if and only if
and \(f(n)\) is multiplicative if and only if \(F(n)\) is multiplicative.
Example: From before \(n = \sum_{d|n} \phi(n) \). Write \(n = p_1^{k_1}...p_m^{k_m}\). Then
which is another way to derive the formula for \(\phi\).
Gauss encountered the Möbius function over 30 years before Möbius when he showed that the sum of the generators of \(\mathbb{Z}_p^*\) is \(\mu(p-1)\). More generally, if \(\mathbb{Z}_n^*\) has a generator, then the sum of all the generators of \(\mathbb{Z}_n^*\) is \(\mu(\phi(n))\). This can be seen by considering the sums of the roots of polynomials of the form \(x^d - 1\) where \(d | \phi(n)\).