# Riemann Hypothesis

• Nov 11th 2005, 01:26 PM
Jameson
Riemann Hypothesis
What do you guys think about the RH? Does anyone want to give a brief explanation of it? I have a basic understanding but would love for someone with a solid math background to go into detail.

The real part of the Zeta Function is at a high school level, but the analytic continuation starts to get above me.
• Nov 14th 2005, 07:22 PM
ThePerfectHacker
The Riemann Hypothesis appeared in 1859. It is based on the Zeta function, which was actually first used by Leonard Euler (1707-1783) in 1777. The Zeta function is defined for all real numbers greater than 1. (I would write the function here but I do not have an equation editor). Its signifigance as Euler showed is that this Zeta function can be expressed diffrently involving ONLY PRIME NUMBERS. This result is called the "Euler-Product Formula" (you can look it up on www.wikipedia.com). Thus there is a relationship between the Zeta function (all the numbers) and prime numbers in a single equation!
Later on Riemann extended the definition for the Zeta function (meaning it was defined for complex numbers besides for the reals), thus this generalization of the Zeta function came to be know as Riemann's Zeta Function. Riemann was interested in finding it zeros (the values for which the function is zero, i.e. solving the zeta equation). There are trivial solution (the obvious non-interesing ones) and the non-trivial. Riemann realized that all the non-trivial solutions had the form 1/2+ai.
Thus, "The non-trivial zeros of the Zeta function have real part equal to 1/2."
Hence to prove, the statement before this one is to prove the Riemann Hypothesis. The Hypothesis is based on zeros, the zeros are based on the zeta function, the zeta function is based on the primes. Thus, this problem concerns prime numbers.
• Nov 16th 2005, 03:14 PM
Jameson
Thank you for that background on the Zeta Function. Do you know a proof of the analytic continuation for the zeta function?
• Nov 16th 2005, 07:02 PM
ThePerfectHacker
You mean to prove that the zeta function is countinous for x>1?
• Nov 19th 2005, 09:23 AM
Jameson
No, I mean show the continuation of the domain of s, from s>1 to s<1.
• Nov 20th 2005, 05:50 AM
hpe
Quote:

Originally Posted by Jameson
No, I mean show the continuation of the domain of s, from s>1 to s<1.

See the mathworld page, in particular formulae (13) - (15).
• Nov 20th 2005, 01:18 PM
ThePerfectHacker
Okay If s>1. Then to proof the zeta function has continouation we have to show:
It is defined at s (which it is)
limZ(x) exits and is equal to Z(s) (which it is).
x->c

For the second condition I employed a theorem about infinite series called Riemann's Rearrangement Theorem which states the if a an infinite series is absolutely convergent than it can be rearranged in way to produce the same result.

It is interesting to note that the zeta function is decreasing and continous for s>1 thus it must be surjective for s>1 thus the Zeta function is a bijective map for s>1. Thus, we can speak about an inverse function for the zeta function.
• Nov 21st 2005, 06:37 AM
hemza
it is sufficient to prove that (see picture) for all n>5040. Where sum(d|n)(d) is the sum of all divisors of n. and gamma is Euler constant.

I tried it for weeks but I always hit one problem in the solving process that is unknown to me.

Louis de Branges de Bourcia, is a professor at Purdue University. He gave a proof of the Riemann H. that was not yet contradicted by anyone so if it stays 2 years in total he will win.
• Nov 22nd 2005, 02:53 PM
ThePerfectHacker
Quote:

Originally Posted by hemza

Louis de Branges de Bourcia, is a professor at Purdue University. He gave a proof of the Riemann H. that was not yet contradicted by anyone so if it stays 2 years in total he will win.

You seriously make no sense! This is MATH not SCIENCE! If it a proof IT IS A PROOF! It can never be contradicted! In science you would propose a solution and hope nobody finds a contradiction because that is what science is all about. You would make sense if you said that people are checking his proof and searching for mistakes in the proof.
• Nov 22nd 2005, 04:56 PM
hpe
Quote:

Originally Posted by ThePerfectHacker
You would make sense if you said that people are checking his proof and searching for mistakes in the proof.

That is evidently what hemza had in mind. According to experts, the paper by De Branges, available here, does not really contain a proof of the RH. The difficulties with DeBranges' approach are well-known and may be found here. So at this time the RH remains unproven.
• Nov 22nd 2005, 06:55 PM
ThePerfectHacker
Thank you hpe; question if it is not complete i.e. he makes conjectures then how can it be a proof after 2 years?
• Nov 23rd 2005, 04:02 AM
hemza
The hypotesis in itself is not important but the one who proves it has to come up with a whole new theory for this misterious and not well known field of mathematics. A simple prove of the RH is not the objective (as said by a number theory professor at Ottawa University).

Louis has tried many times to prove RH but he did not succeed (I think the link given above by my college is an old one (1998) i.e. : an old try). Louis has finally invented a theory which seems good in june or july 2004. Because it is a new theory and it is so difficult, the organization who gives the price of 1M\$ gives 2 years to the mathematicians to read (and try to understand this new theory) and find an error in the prove if there is one. If not, it is accepted and Louis wins.

So what Louis has submitted IS A PROVE but it is so high level and so complicated math that it takes a lot of time for mathematicians (those who can actually read this stuff) to debug it.

The point is : it is not because at the end there is "and that is what we had to prove" that makes a prove correct. It must be clear of all mistakes and nothing has to be taken as granted if it is not proven before. This kind of thing has already happened before. Someone submits what seems like a good prove but he assumes for example commutativity in a set where there is no proven commutativity and that nullifies his prove.

That is it. It is very difficult math.
• Nov 23rd 2005, 06:05 PM
hpe
Quote:

Originally Posted by hemza
That is it. It is very difficult math.

And De Branges is a difficult character. But that is another matter.
• Nov 23rd 2005, 06:43 PM
ThePerfectHacker
What type of concepts of math does it use? Topology?
• Nov 24th 2005, 04:52 AM
hpe
Quote:

Originally Posted by ThePerfectHacker
What type of concepts of math does it use? Topology?

Hilbert spaces of entire functions. He used this theory successfully to prove the Bieberbach conjecture about 20 years ago.