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Old 2008-02-21, 16:20   #1
Damian
 
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Default Why is RH so difficult to prove?

I know that it is related to the primes numbers distribution, and everything related to primes numbers is hard to prove.

But there is a 1.000.000$ prize for demonstrating it, so I think a lot of people must be interested on finding a proof.

And, after all, it only says that the non-trivial zeroes of the zeta function lies on the Re=1/2 line. If you consider the fuction SymetricZeta(z) = Zeta(z)*Zeta(1-z) you get a function with the sames non-trivial zeroes, and completly simetrical about the Re=1/2 line. Also the real part is zero in the Im=0 and the Re=1/2 lines, working as some kind of "coordinate lines" where the trivial and non trivial zeroes can stand. You can look at my avatar for a plot of the SymetricZeta function to look what I mean.

It seems almost obvious to me that the hipothesys has to be true. Other way it would break this simetry.
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Old 2008-02-21, 16:43   #2
R.D. Silverman
 
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The question itself is ridiculous. The same question may be asked of
any conjecture that has remain unproved for any moderate amount of
time. Noone knows why a theorem is hard until it is proved. Take,
for example, the Bieberbach conjecture. For a half century it was
thought to be out-of-reach until DeBrange proved it. Now it is
understandable by undergrads.
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Old 2008-02-21, 17:34   #3
ewmayer
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Quote:
Originally Posted by Damian View Post
If you consider the fuction SymetricZeta(z) = Zeta(z)*Zeta(1-z) you get a function with the sames non-trivial zeroes, and completly simetrical about the Re=1/2 line.
Quote:
It seems almost obvious to me that the hipothesys has to be true. Other way it would break this simetry.
A non sequitur ... your conclusion of "obviousness" makes the unjustified assumption that "symmetric about Re = 1/2" plus some other meaningless handwaving implies "must lie on Re = 1/2".

As Bob notes, it may turn out to indeed be "obvious" once someone has a crucial insight or the right mathematical tools are developed ... but we won't know until it is proven, one way or the other.

Actually, even if it is proven, we may not know ... the initial proof might be very non-obvious, but that would not rule out that a much-simpler proof exists.
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Old 2008-02-21, 17:38   #4
Damian
 
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Thanks for your reply.
I know what you mean, and I'm not asking for a proof of RH.
But I think each person can have it's own opinion as to why RH hasn't been proved.
For example, I think the reason may be related to the fact that it isn't easy to calculate the exact value of the Zeta function at any point. For example, wich is the value of Zeta(1+1i) ? Not an aproximation but an exact value. Surely it is an irrational or even trascendental number. But we know for example that Zeta(2) = Pi^2/6 (Euler proof of the Basel problem), and for negative integers (the bernoulli numbers) but what about numbers that are not on the real line?
Not knowing the exact values of Zeta mean we can only aproximate them. And I think it is difficult to get a proof of something that can only be approximated.
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Old 2008-02-21, 19:51   #5
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Damian,

The symmetry about the line with real part =1/2 with regards to zeta(z)*zeta(1-z) comes from the symmetrization of the function, and not any special properties of zeta. For example, let f(z)=x^2+1. The function f(z)f(1-z) is symmetric around the line with real part =1/2, but the zeros of f(z) are not related to 1/2 in any meaningful way.

Second, the proof is difficult because nobody has found a successful way to bound the zeros of the function. Qualitatively, you might express the problem by looking at the Mobius function, and trying to explain how much \sum_{x<n}\mu(x) varies as n increases. Nobody has found a good way to smooth out the randomness of this sum.

Edited to add: A better function to look at than zeta(z)*zeta(1-z) is the function obtained by taking one side of the functional equation and then putting the line with real part=1/2 on the real line. The resulting function is symmetric, hypothetically all zeros are real and simple, and the function is real on the real line. The Taylor series for the resulting function is pretty nice.

Last fiddled with by Zeta-Flux on 2008-02-21 at 19:54
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Old 2008-02-22, 06:48   #6
devarajkandadai
 
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Default Why is RH so difficult to prove?

Just out of curiosity I am interested in knowing

a) Who is giving the prize?

b) Which committee scrutinises the proof?

c) Procedure & related matters

A.K.Devaraj
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Old 2008-02-22, 08:09   #7
jinydu
 
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It is a Millennium Problem:

http://www.claymath.org/millennium/
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Old 2008-02-22, 11:51   #8
davieddy
 
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Quote:
Originally Posted by jinydu View Post
It is a Millennium Problem:

http://www.claymath.org/millennium/
Not to mention one of Hilbert's problems from 1900.
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Old 2008-02-22, 12:26   #9
victor
 
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"If I were to awaken after having slept for a thousand years, my first question would be: Has the Riemann hypothesis been proven?" - Hilbert
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Old 2008-02-22, 16:48   #10
Damian
 
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Thanks all for your replys.
This may be a bit of topic but I was thinking which is the simplest known proof that \pi is irrational.
Based on the assumption that \zeta(2) = \fra{\pi^2}{6} is irrational, wouldn't it imply that \pi is also irrational?
But I don't know if the demonstration that \zeta(2) is irrational uses the fact that \pi is irrational, in that case it wouldn't help.
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Old 2008-02-22, 16:57   #11
R.D. Silverman
 
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[QUOTE=Damian;126372]Thanks for your reply.
I know what you mean, and I'm not asking for a proof of RH.
But I think each person can have it's own opinion as to why RH hasn't been proved.
QUOTE]


No. Every person is NOT entitled to such an opinion.
In particular, you are not so entitled.

Every *informed* person is entitled to an opinion. Your knowledge
of mathematics has not even reached the point where you have
mastered high school level math.
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