Talk:Möbius inversion formula
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Definition secdtion
[edit]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)
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 - 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
[edit]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 , then so convolution is commutative. --Jerome Baum (talk) 22:32, 5 July 2014 (UTC)