Talk:Equivalence of Definitions of Riemann Zeta Function

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The proof that is currently up, the equivalency of the sum of reciprocal powers and product of reciprocal 1 - prime reciprocal powers definitions of the zeta function, is not just applicable to the function $f(p) = p^{-z} \ $, but to any completely multiplicative function. Zelmerszoetrop 04:15, 9 April 2009 (UTC)

Interesting! A new one on me. Might be worth putting up a separate page for that. --Matt Westwood 06:36, 9 April 2009 (UTC)
Agreed - the proof is precisely the same as it appears here, the only reason I'm not doing it tonight is that I'd also have to define completely multiplicative, and would probably end up putting up all those basic number-theoretic terms as well, and it's late. But the proof of that fact is exactly the same for all f(n) that are completely multiplicative, just like what's up now. Zelmerszoetrop 07:22, 9 April 2009 (UTC)

Okay done completely multiplicative, just so happens I've been doing a fair bit of number theoretic stuff this last couple of days (really fundamental do-in-your-sleep sort of stuff) so check it out, a lot of it may already be there by now. I'll get on with it a bit more over the last few days, they've given me a 4-day weekend in celebration of a particularly gruesome murder that happened a couple of millennia ago which apparently they're still arguing about ...--Matt Westwood 22:38, 9 April 2009 (UTC)

I'm not sure the proof you give here is correct - there seems to be a 1 missing from all your factors.

Now we have the proof up for Product Form of Sum on Completely Multiplicative Function we can probably lose the proof out of this page anyway, and just have a link to that instead. --Matt Westwood 09:37, 10 April 2009 (UTC)

- in the first proof how do you argue about the multiplication between 2 infinite sum? smb can explain it ? thk -- gamma

Good point, it does need to be shown that this is valid. --Matt Westwood 05:36, 16 April 2009 (UTC)

... We also need to add the conditions under which the sum is absolutely convergent. That limits what $z$ is. --Matt Westwood 05:32, 17 April 2009 (UTC)

I'm interested as to why you removed the "Alternative Proof" section, as that last bit at the bottom stands alone and is separate from the bit above it, surely? --Matt Westwood 19:41, 24 April 2009 (UTC)

I didn't remove the material, just the title. The theorem has been expanded, and now a second part is needed for the proof. If we would like to keep the alternative proof of part 1 together with part 1, and keep part 2 of the proof flowing immediately after part 1, it is necessary to remove the section heading for the alternate proof. Zelmerszoetrop 19:57, 24 April 2009 (UTC)

I don't understand, the "alternative proof" was the application of the general result that you alluded to, and in that sense can replace the "part 1", never mind, feel free to get on with it, I'll stick to the stuff I'm doing. --Matt Westwood 20:00, 24 April 2009 (UTC)

I'm Struggling enough thinking how to prove just the functional equation on a single page, so moved + transcluded. Not very tidy as it is I know. --Linus44 11:30, 31 March 2011 (CDT)

No worries, it works, it's in. We can tidy it up later, may do it myself when I'm in the mood. --prime mover 14:49, 31 March 2011 (CDT)