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A possible proof for Riemann Hypothesis

3 points by nutanc 10y ago

I work with a professor,Dr. Kumar Eswaran, on several computer science problems especially in the neural network field. He is the quintessential prof who is always lost in thought and interested in solving unsolvable problems. So I was pleasantly surprised last week when he told me that he had found a proof for the Riemann Hypothesis. The Professor was able to crack the Riemann Hypothesis because he used the properties of primes and analytic continuation and had a new way of handling slowly converging series and was able to use (at crucial point) concepts borrowed from Donald Knuth regarding random numbers and random sequences. Knuth had said that for any sequence to be truly random it has to be non-cyclic. The proof required to show that a sequence of +1's and -1's , obtained from the prime factorization of the infinite sequence of integers, had to be shown to be random and to asymptotically behave like the tosses of a coin. Though I am good at computers and math, I am not an authority in math. So I thought I will put up the proof here and invite comments as Dr.Kumar does not know too much about Hackernews. Links to proof: https://arxiv.org/pdf/1609.06971v4.pdf https://www.researchgate.net/publication/309205618_The_Dirichlet_Series_for_the_Liouville_Function_and_the_Riemann_Hypothesis

mathematician23 10y ago
The function he is trying to prove is analytic is a ratio of the zeta function at two places, so can be thought of as a ratio of two different complex functions. What if they both have a zero for some complex number s, i.e. zeta(2s)/zeta(s) = 0/0, and l'hospitals rule ends up showing its a removable singularity and so the ratio is still analytic, like he says. In that case, he hasn't shown there are no zeros in the critical strip outside the critical line.
- sainik 10y ago
  This does not happen. The zeros on the critical line appear as poles in F(s). There are no other poles,in F(s), because The trivial zeros of the zeta function which occur at -2,-4,-6 etc actually cancel out from the numerator and the denominator (i.e. from zeta(2s) and from zeta(s)). The paper then shows that the only poles in F(s) are those on the critical line, thus proving R.H.
  - mathematician23 10y ago
    A pole for F(s) means zeta(s) = 0 but zeta(2s) not=0 for some complex s. So yes, the zeros on the critical line appear as poles in F(s). Looking outside the critical line, at negative even integers, you will get 0/0 and using L-Hospitals rule, they may end up cancelling out, meaning F(s) remains analytic at these locations. But my point is how do you know this cancellation isn't occurring at other locations, outside the critical line? I am not saying there are other poles outside the critical line, I am saying the author hasn't ruled out 0/0 happening in the critical strip. This is really just high school calculus we are talking about, 0/0 is an indeterminate form and more work must be done to see what type of singularity it is. The author saying F(s) is analytic outside the critical line but within the critical strip means there could still be removable singularities, i.e. 0/0, that end up cancelling out. These are like holes in our graphs back in Cal I. Riemann's Hypothesis is 157 years old now, it is not going to be proven by a 20 page paper.
  - mathematician23 10y ago
    There are many equivalent statements to the Riemann Hypothesis (RH) and statements that, if true, would imply the RH holds. But F(s) = zeta(2s)/zeta(s) being analytic in the critical strip, outside the critical line, DOES NOT imply the RH is true. That's my point, it doesn't matter if the author really showed F(s) is analytic in that region or not. Showing something is equivalent to the RH is a major area of research and so is finding statements that directly imply the RH. The function F(s) is very well known and a simple function, so if there were any simple statements involving it that directly implies the RH, we would know by now. I hope this helps.
    - vskp2016 10y ago
      The paper clearly demonstrates that if F(s) has poles on the critical line, RH is proved.This is because the poles in F(s) correspond to zeros of zeta(s) and it is well known that there are no other zeros outside the critical strip (except for the trivial zeros). The paper has proved that the only poles of F(s),occur on this critical line. Ok bye!
  - mathematician23 10y ago
deepnotderp 10y ago
Wow, this is super interesting. I'm gonna wait on results. Btw,love that you published it in arxiv.
- nutanc 10y ago
  Yes, arxiv for the win :)
noobermin 10y ago
Not a mathematician, but I'd wait to see what refree report/editors say(s). If this is validated, this is huge.
cokernel 10y ago
It's not a good sign that it's classified as math.GM. One would normally expect a real proof to be classified more finely. To be fair, the lack of proper classification does not strictly imply that it's not a real proof.

3 points by nutanc 10y ago

mathematician23 10y ago

The function he is trying to prove is analytic is a ratio of the zeta function at two places, so can be thought of as a ratio of two different complex functions. What if they both have a zero for some complex number s, i.e. zeta(2s)/zeta(s) = 0/0, and l'hospitals rule ends up showing its a removable singularity and so the ratio is still analytic, like he says. In that case, he hasn't shown there are no zeros in the critical strip outside the critical line.

sainik 10y ago
This does not happen. The zeros on the critical line appear as poles in F(s). There are no other poles,in F(s), because The trivial zeros of the zeta function which occur at -2,-4,-6 etc actually cancel out from the numerator and the denominator (i.e. from zeta(2s) and from zeta(s)). The paper then shows that the only poles in F(s) are those on the critical line, thus proving R.H.
- mathematician23 10y ago
  A pole for F(s) means zeta(s) = 0 but zeta(2s) not=0 for some complex s. So yes, the zeros on the critical line appear as poles in F(s). Looking outside the critical line, at negative even integers, you will get 0/0 and using L-Hospitals rule, they may end up cancelling out, meaning F(s) remains analytic at these locations. But my point is how do you know this cancellation isn't occurring at other locations, outside the critical line? I am not saying there are other poles outside the critical line, I am saying the author hasn't ruled out 0/0 happening in the critical strip. This is really just high school calculus we are talking about, 0/0 is an indeterminate form and more work must be done to see what type of singularity it is. The author saying F(s) is analytic outside the critical line but within the critical strip means there could still be removable singularities, i.e. 0/0, that end up cancelling out. These are like holes in our graphs back in Cal I. Riemann's Hypothesis is 157 years old now, it is not going to be proven by a 20 page paper.
- mathematician23 10y ago
  There are many equivalent statements to the Riemann Hypothesis (RH) and statements that, if true, would imply the RH holds. But F(s) = zeta(2s)/zeta(s) being analytic in the critical strip, outside the critical line, DOES NOT imply the RH is true. That's my point, it doesn't matter if the author really showed F(s) is analytic in that region or not. Showing something is equivalent to the RH is a major area of research and so is finding statements that directly imply the RH. The function F(s) is very well known and a simple function, so if there were any simple statements involving it that directly implies the RH, we would know by now. I hope this helps.
  - vskp2016 10y ago
    The paper clearly demonstrates that if F(s) has poles on the critical line, RH is proved.This is because the poles in F(s) correspond to zeros of zeta(s) and it is well known that there are no other zeros outside the critical strip (except for the trivial zeros). The paper has proved that the only poles of F(s),occur on this critical line. Ok bye!
- mathematician23 10y ago

deepnotderp 10y ago

Wow, this is super interesting. I'm gonna wait on results. Btw,love that you published it in arxiv.

nutanc 10y ago
Yes, arxiv for the win :)

noobermin 10y ago

Not a mathematician, but I'd wait to see what refree report/editors say(s). If this is validated, this is huge.

cokernel 10y ago

It's not a good sign that it's classified as math.GM. One would normally expect a real proof to be classified more finely. To be fair, the lack of proper classification does not strictly imply that it's not a real proof.