Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple Proof of Riemann's Hypothesis

  1. May 18, 2010 #1

    I am attempting to prove Riemann's Hypothesis and need someone to critque the proof.

    1. Does it prove anything?
    2. What more must I prove?
    3. Where can I learn more about this problem?

    See attached 51910_RH_proof.JPG

    Attached Files:

    Last edited: May 19, 2010
  2. jcsd
  3. May 18, 2010 #2
    Doesn't the series definition only correspond to the Riemann Zeta function for Re(s) > 1?
  4. May 18, 2010 #3


    User Avatar
    Science Advisor

    I think you need to keep working.One flaw is that you've assumed that if an infinite series sums to zero, all of the terms must be zero. Clearly this isn't the case. There's a really good book called "Prime Obsession" by John Derbyshire about the Riemann hypothesis.
  5. May 18, 2010 #4

    The below paragraph was supposed to be above the proof. (excuse free use of 's'):

    We will try to prove that the nontrivial or interesting Riemann zeta function zeros, i.e., the values of other than -2, -4, -6, ... such that Z(s)=0 (where Z(s) is the Riemann zeta function) all lie on the "critical line" sigma=R=1/2 (where R denotes the real part of s ).
    | _ ___ __
    It stems from Hardy’s 1914 proof that Z(!s) = !Z(s)= !0 = 0. This is the same as
    Z(1-s) = Z(s)= 0 for the domain 0 <s <1. The proof follows.

    Attached Files:

  6. May 18, 2010 #5
    So the constraint is that 0 < s < 1 . For Re = 1/2 , Re > 0 (not 1.)

    On the "flaw" I'm saying IF there exists s' s. t. (k^(1-s') - k^(s'))/k = 0 for k=1-> oo
    then k^(1-s') - k^(s')=0 and only s'=1/2 makes this true. So for s'=1/2 every term is 0 for all k shown above. So for s'=1/2 the whole series collapses to zero.
    Last edited: May 19, 2010
  7. May 19, 2010 #6
    The problem with your approach is that the infinite series that you're using for the Riemann zeta function converges only if Re(s) > 1. You can't use it for 0 < Re(s) < 1 because it (the infinite series) doesn't converge there. You need to learn about notions such as analytic continuation and the functional equation. You might start by looking at the Wikipedia article on the Riemann zeta function.


  8. May 20, 2010 #7
    Why does the series converge to 0 if s=1/2?
  9. May 20, 2010 #8

    Gib Z

    User Avatar
    Homework Helper

    It doesn't. The series only corresponds to Riemann's Zeta function if Re(s)> 1. That is what people have been trying to tell you.
  10. May 20, 2010 #9
    For zeros, s includes an imaginary component A*i besides 1/2 and I believe the Sqrt(1/4 + A^2) > 1
  11. May 20, 2010 #10
    If (A*i)^2 = -A^2 (i^2= -1) Then SQT(1/4 - A^2) could be < 1 if e.g. A^2 = 1/9, but
    I get what you are saying and now am working with XI(s) which is convergent and continuos.

    If 0 = xi(1-s) = xi(s) 0 < s < 1, Hardy 1914 and xi(s) is convergent, continuous and has the same nontrivial zeroes as z(s) why can't we say 0 = z(1-s) = z(s) ? and yes s = s + it a complex number.

    Attached Files:

    Last edited: May 20, 2010
  12. May 24, 2010 #11
    Is sum ( k^(1-s) - k^(s) ) / k ) analytic continuous for 0< Re(s) <1, pos. k ???
    in the complex plane? For s=.4 => sum ( k^.6 - k^.4)/k = (k'-k'')/k would seem close to zero for most terms. For k=10 k^(.6)= 3.98, k^(.4)=2.5 , k'-k''=3.58 => 0.358 for term 10.
    For k=100= 9.5 => 0.095 for term 100.
  13. May 28, 2010 #12
    perhaps if we consider it in the sense of analytic continuation so [tex] \sum_{n=1}^{\infty}n^{k} = \zeta(-k) [/tex] for 'k' different of k=-1
  14. May 28, 2010 #13
    Hello Rlrandallx. I've looked at this thread from time to time and feel it a bit awkward to say the least and am hesitant to say anything because I'm new here. However I do wish to sincerely give you some answers to your question ok?

    1. I doubt it proves anything but I only got to the first line before you incorrectly stated the Euler sum is the zeta function for s ne 1. That's not true. The Euler sum represents the zeta function for Re(s)>1.

    2. You must prove the analytic extension of the Euler sum, which is called the zeta function, has non-trivial zeros only on the critical line.

    3. Here is the most important thing I can suggest to you: You have got to take at least a class in Complex Variables, then get into Complex Analysis, work many problems in Complex Analysis just to become comfortable with the subject, then study the analytic properties of the zeta function rigorously for some time. The standard texts on the subject are by Edwards and Titchmarsh but are difficult to follow but if you are persistent, they will start to open up.
  15. May 28, 2010 #14
    Thanks for all the helpful remarks. I learned a lot! I was trying to build off of Hardy's (1914) results. He extended the zeta function to one where 0< Re <1 and found f(1-s) = f(s) and as I understood it, all the non-trivial zeroes were the same as that for the zeta Riemann function.

    In reading all the literature, it seems that the only thing left to really prove is that "All the non-trivials zeroes lie on the line Re=1/2 in the complex plane." So maybe someone else can do this by assuming there is a non-trivial zero at a point not on this line. Then we just show a contradiction.

    All the Best,
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook