I was encouraged to post this here instead of Facebook, which is probably an excellent idea.

Anyways, I was doing some math and I came up with a sum that involved prime factors and such, expressed in terms of the Riemann Zeta function. It got me curious to look at if ζ(s)=ζ(2s)=0 for any s. If yes, then then the Riemann Hypothesis is false (so I am inclined to believe that no such complex s exists). If RH is true, then that would say that there isn't an s, but that might be the more difficult way of proving no such s exists.

So, does anybody have any ideas or approaches to showing an s does or does not exist?

EDIT:
For the interested, I was very originally looking back at a solution to a problem I found in a journal and on my way to the solution, I gave a generalized form for ∏(1+p^-s+p^(-2s)+...+p^(-ks)) as ζ(s)/ζ(ks+s) or something (I don't have the notebook here and I am rushed), and I was looking at ∏(1+p^-s)ζ(s), expanding the zeta function as a sum, which gives a sum similar to the zeta function, but with the n-th coefficient as a function of the number of prime factors of n. From the formula above:
∏(1+p^-s)ζ(s)=ζ(s)^2/ζ(2s)

I just wanted to make sure that whenever the numerator was 0, the denominator wouldn't be (or if it is, hopefully it is something like a simple pole that can be factored out).

Okay, gotta go to work, have fun!
  
Register to Join the Conversation
Have your own thoughts to add to this or any other topic? Want to ask a question, offer a suggestion, share your own programs and projects, upload a file to the file archives, get help with calculator and computer programming, or simply chat with like-minded coders and tech and calculator enthusiasts via the site-wide AJAX SAX widget? Registration for a free Cemetech account only takes a minute.

» Go to Registration page
Page 1 of 1
» All times are UTC - 5 Hours
 
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum

 

Advertisement