Talk:Dirichlet convolution

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Explain why its a Dirichlet ring, and not a field_(mathematics). Naively, it seems to have additive and multiplicative units and inverses. linas 05:00, 6 Jan 2005 (UTC)

Because only arithmetic functions for which f(1) is not 0 have inverses. I have clarified this in the article. Gandalf61 10:25, May 21, 2005 (UTC)

Weak multiplicative functions

Where do these come from? Are they the same as multiplicative (in the number-theoretic sense)? Richard Pinch (talk) 21:45, 3 September 2008 (UTC)[reply]

Yeah it looks like it just means multiplicative. The results proved in that section is that (f*g)(mn)=(f*g)(m)·(f*g)(n) for relatively prime integers m,n, which is just the definition of multiplicative. A google search for "weak multiplicative functions" returned essentially 2 results (one of which was this wiki article), so I'm removing the term weak. RobHar (talk) 23:56, 3 September 2008 (UTC)[reply]

Actually, I decided to just remove the whole section, as it was poorly written, it reiterates a statement made in the previous section, and the proof is already in the article Multiplicative functions. RobHar (talk) 00:00, 4 September 2008 (UTC)[reply]

Thanks for sorting that out. Richard Pinch (talk) 07:00, 4 September 2008 (UTC)[reply]

Special case of general convolution concept

The article says, correctly, that the Dirichlet convolution is analogous to the ordinary convolution of functions on the reals. But it seems to me that one can say more than that, namely, both are special cases of the same concept, convolution on a monoid. (Going a bit farther than the usual case of groups, which is discussed at convolution.) Is that worth putting in the article? Is there a good reference for it? -- Spireguy (talk) 16:28, 23 July 2010 (UTC)[reply]

What does this mean?

The article states "Given a completely multiplicative function f then f (g*h) = (f g)*(f h). "

Is (f g) pointwise multipllication?

Virginia-American (talk) 19:22, 22 December 2011 (UTC)[reply]

Yes it is. It's defined in the article on completely multiplicative functions.

Virginia-American (talk) 15:45, 23 December 2011 (UTC)[reply]