Talk:Möbius inversion formula

Definition secdtion

In the definition section, the article says

The formula is also correct if f and g are functions from the positive integers into some abelian group.

How is one to make sense of g(d) μ(n/d) if these values only have a group operation (addition) defined on them?

It seems that it should say "... into some commutative ring."

132.206.150.179 03:18, 10 December 2006 (UTC)

No. The function g takes values in the group, but the values of μ are plain integers, and every Abelian group is a Z-module. In other words, the meaning is

$g(d)\mu (n/d)={\begin{cases}g(d),&{\text{if }}\mu (n/d)=1,\\-g(d),&{\text{if }}\mu (n/d)=-1,\\0,&{\text{if }}\mu (n/d)=0.\end{cases}}$

I reordered the formula to make it look like left module multiplication, which is hopefully more clear. -- EJ 12:40, 19 December 2006 (UTC)

I think the Moebius inversion formula was first introduced by Dirichlet and some other mathematician (perhaps Liouville). We should check the historical information ! —Preceding unsigned comment added by 132.206.33.23 (talk) 18:31, 28 February 2008 (UTC)

Left versus right convolution

There seems to be a notational inconsistency. The first section says that if g = f * 1, then f = \mu * g. But if we substitute the second equation into the first, we're getting g = \mu * g * 1, which isn't what's supposed to happen. The point of Mobius inversion is that \mu and 1 are inverses in the incidence algebra of the natural numbers, i.e. under Dirichlet convolution. So the theorem should read: if g = f * 1, then (by applying \mu on the right), g * \mu = f * 1 * \mu = f (since \mu and 1 are inverses). I tried to make this notational change but I can't seem to get the wiki syntax correct - when I changed it from \mu * g to g * \mu, I got an "unknown parse error". Maybe someone else can clean this up? Thanks. (edit: also, I forgot to sign. The above edits by 169.229.3.78 (talk))

If $d|n$, then $(n/d)|n$ so convolution is commutative. --Jerome Baum (talk) 22:32, 5 July 2014 (UTC)