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Talk:Möbius inversion formula

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Definition secdtion

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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)[reply]

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)[reply]

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)[reply]

Left versus right convolution

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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)[reply]