Let's define the function which gives the numerator associated with denominator \(n\) in the products above.
Let \(N=2\cdot 3\cdot 5\cdots q\) be the product of the first few primes, up to \(q\). Then we define \(\mu(n)\) as follows: \begin{equation*}\prod_{p\mid N}\left(1-\frac{1}{p}\right)=\sum_{d\mid N}\frac{\mu(d)}{d}\end{equation*} The product is over prime factors of \(N\) but the sum is over all factors of \(N\).
Yes, this is the same Moebius (or Möbius) as the Moebius strip.
Using the example in the chapter introduction, \begin{equation*}D(3)=(1-1/2)(1-1/3) = \left(1-\frac{1}{2}-\frac{1}{3}+\frac{1}{6}\right)\end{equation*} implies that \(\mu(2)=-1\) while \(\mu(6)=1\).
There is no product of \((1-1/p)\) that will yield a four in the denominator, since \((1-1/2)\) only occurs once in such a product. So \(\mu(4)=0\), as the example above already implies.
Before describing this function further, let's think more about the product \(\prod_{p<N}\left(1-\frac{1}{p}\right)\).
First, as the comment at the end of the last subsection points out, 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 \(-1\). For a moment, think about why 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.
This essentially proves the following proposition.
If \(n=p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}}\) then a nice formula for \(\mu(n)\) is \begin{equation*}\mu(n)=\begin{cases}0 & \text{ any }e_{i}>1 \\(-1)^{k} & \text{ otherwise}\end{cases}\end{equation*}
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 \begin{equation*}\mu(1)=1,\text{ and }\sum_{d\mid n}\mu(d)=0\; .\end{equation*} 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.
\begin{equation*}\sum_{d\mid n}\mu(d)=0\end{equation*}
We can always check things like this by computing.
The function \(\mu\) is multiplicative.
Let's check it:
