# Riemann hypothesis

1. Sep 29, 2008

### epkid08

I have two questions:

Why hasn't the hypothesis been proved yet? Is it because we don't know why re(s) has to be 1/2 and thus can't prove it, or is it because we know why re(s) has to be 1/2 but we just don't know how to prove it.

Why exactly does re(s) have to be 1/2?

$$\zeta (s)=1/(1-2^(1-s)) \sum_{n=0}^{\infty}1/(2^(n+1)) \sum_{k=0}^{n} (-1)^k (n ; k) (k+1)^{-s}$$

If re(s) is greater than 0, and if im(s) is any real number, doesn't the function always converge to zero? In the above equation, as k approaches infinity, the denominator of the term, $$(k+1)^{1-s}$$, always approaches infinity, for any value of re(s)>0, which in turn, has the series approaching zero, always. Please help me understand where I am wrong.

Last edited: Sep 29, 2008
2. Sep 29, 2008

### CRGreathouse

That the terms decrease in magnitude means that the sum (possibly) converges, not that its value is zero. $\zeta(1/4)=-0.81\ldots\neq0,\zeta(1/4 + i)=0.0438\ldots-0.5600\ldots i\neq0.$

3. Sep 30, 2008

### mhill

Riemann Zeta satisfies a 'reflection' equation $$\zeta (1-s) = \xi (s) \zeta(s)$$

so if s is a zero then 1-s is also another zero (exceptions are -2,-4,-6,..............)

if all the zeros have real part 1/2 many math applications become simpler.

Selberg proved that a zeta function running over the lenghts of Geodesics had the zeros

$$\lamda = s(s-1)$$ with 'lambda' an eigenvalue of a Laplacian, then since the eigenvalues are real s=1/2+ir r=real ---> RH is true.

it would be enough finding a certain surface whose geodesic lenght equal primes.

i do not know if Selberg Zeta has a functional equation, Selberg Trace is a generalization of Poisson summation formula and Riemann Zeta functional equation can be proved using Poisson summation formula.

4. Sep 30, 2008

### Tzar

We know that the non trivial zeros must have 0<Re(s)<1 but that's about it. We have many examples of zeros with Re(s)=1/2 and in fact is has been proven that there infinitely many zeros with that property. However we still haven't proved that you can't have a zero with 0<Re(s)<1 but Re(s) not equal to 1/2.

5. Sep 30, 2008

### CRGreathouse

I think we even know that a positive proportion (40%) of zeros are on the line.

6. Oct 1, 2008

### mhill

I always felt curious about that.. how can you know that the 40 % 30 % or even larger number of zeros are on the line Re(1/2) is there a general method to say if X is true then an Y % of zeros have real part 1/2 .

Perhaps is related to the density of Lehmer Zeros ??

7. Oct 1, 2008

### Tzar

You probably mean at least 40% then :) I think I heard of that result as well. I would assume that the proof would need to be heavily combinatorial, but if someone knows more, please tell us!

CRGreathouse you might also find the following results interesting:

http://mathworld.wolfram.com/VisiblePoint.html" [Broken]

Last edited by a moderator: May 3, 2017
8. Oct 1, 2008

### CRGreathouse

I said that 40% of the nontrivial zeros were on the critical line. I didn't make any claim about the 60%.

9. Oct 1, 2008

### Tzar

oh ok sorry

10. Oct 4, 2008

### muppet

I'm just reading Du Sautoy's The music of the Primes now and there's something I'm curious about. Are the corrections to Riemann's function for counting the number of primes in a given interval- which can apparently be used to predict exactly how many primes one will find up to a certain number- conditional on the truth of the Riemann hypothesis or not? It's not always clear in the book what Riemann proved in that paper and what he didn't.

11. Oct 4, 2008

### CRGreathouse

Riemann's explicit formula (the "correction" you mention) is not dependent on the RH. But we also can't use it for proven calculation, since it's only true for an infinite number of zeros. Heuristically, just using a few zeros seems to help a lot, but there's no proof (that I know of) that this helps at all in an asymptotic sense.

12. Oct 4, 2008

### muppet

Sorry... we can't use it for proven calculation of what?

13. Oct 4, 2008

### CRGreathouse

Values of pi(x) or theta(x), of course.

14. Oct 4, 2008

### soandos

what is theta(x)?

15. Oct 4, 2008

### CRGreathouse

16. Oct 5, 2008

### muppet

That's what I thought you meant, but it didn't make sense to me that that could be the case at the time! Having re-read the relevant pages and trying to get past the extended metaphor, is it the case that
• Each zero makes a contribution to the number of primes yielded by pi(n)
• It can be proven that pi(n) will yield the correct number of primes, given the correct input of zeroes of the Riemann zeta function
• It presently appears that the right answers are yielded by an input of zeros consistent with the Riemann hypothesis?

17. Oct 5, 2008

### CRGreathouse

pi(n) *is* the answer. Using a finite number of zeta zeros the answer can be approximated. Using an infinite number yields the true value. For small numbers (< 10^23, say) pi can be computed directly, of course.

Here, read this page and see if it helps. It uses Psi instead of pi or theta, but the point is the same.
http://www.math.ucsb.edu/~stopple/explicit.html

18. Oct 5, 2008

### muppet

Yes, that clears things up. Thanks!