Section23.1The Moebius function
Subsection23.1.1Möbius mu
Let's define \(\mu(n)\) as the numerator associated with denominator \(n\) in the products above.
Yes, this is the same Moebius as the Moebius strip.
Example23.1.2
Using the example in the chapter introduction, \[D(3)=(1-1/2)(1-1/3) = \left(1-\frac{1}{2}-\frac{1}{3}+\frac{1}{6}\right)\] implies that \(\mu(2)=-1\) while \(\mu(6)=1\).
However, there is no product of \((1-1/p)\) that will yield a four in the denominator, since \((1-1/2)\) only occurs once in these products. So \(\mu(4)=0\).
Subsection23.1.2A formula
Before looking at more description of this function, let's think more about this product.
- It seems to create denominators with each prime factor to just the first power; we couldn't get a square or cube of any given \(p\) in the denominator.
- Similarly, the numerators really can only be products of \(1\) and \(-1\); there are no other numerators available.
- Finally, the number of prime factors in the denominator should be the same as the number of times \(-1\) is part of the product in the numerator.
Proposition23.1.3
A nice formula for \(\mu(n)\) is given thus. If \(n=p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}}\) then \[\begin{cases}\mu(n)=0 & \text{ any }e_{i}>1 \\\mu(n)=(-1)^{k} & \text{ otherwise}\end{cases}.\]Subsection23.1.3Another definition
The \(\mu\) function is so important that we will want several more approaches as well. It is the mark of an important concept that there are ways to define it from many directions.
One important way that \(\mu\) is often defined is via a recurrence relation. That is, one defines \[\mu(1)=1,\text{ and }\sum_{d\mid n}\mu(d)=0\; .\] Now, we haven't proved this identity yet, and probably the reader hasn't even noticed it. But if we can prove the identity works for \(\mu\), then since \(\mu(1)=1\) is true, this would give an alternate definition.
Proposition23.1.4Recursive definition of \(\mu\)
\[\sum_{d\mid n}\mu(d)=0\]Proof
Remark23.1.5
Sage note:
We can always check things like this.
Fact23.1.6
The function \(\mu\) is multiplicative.Proof
Let's check it: