The Riemann hypothesis has been proved wrong!

[Thing]

the guy who was working on a math/physics approach was Fields medalist Alain Connes.

There's also mean some interesting work done along these lines by Michael Berry and Jon Keating at Bristol. In fact the roots (so to speak) of this idea go back to the 1950s. The deal is that the imaginary part of the non-trivial zeros of the zeta function have a lot of the properties that the eigenvalues of dynamical systems do. If you could construct a dynamical system whose eigenvalues were the non-trivial zeros of zeta then you'd pretty much have the thing licked. Unfortunately it's also been proven that such a dynamical system would be far from ordinary so it's very unlikely that someone will just stumble across it.

 

[jsfisher]

perhaps to redeem this thread it would be interesting to discuss Riemann....

Now, according to the sites, the trivial solutions to this problem are given by choosing s= -2, -4, -6.....but that gives you;

[latex]$$ \zeta (-2) = 1+ \frac{1}{2^{-2}} + \frac{1}{3^{-2}} + \frac{1}{4^{-2}}...[/latex]

which tends to infinity....and not to zero. So what's with that?

I believe the difficulty is that the summation form for Riemann's Zeta function (which is really Euler's Zeta function) holds only for s real and greater than 1.

Riemann's version is expressed as an integral, which reduces to Euler's where appropriate, but not when s < 0. I don't "do" latex, but I can point you here for something that looks like an integral.

 

[jsfisher]

I don't "do" latex, but I'm willing to play:

[latex]
$$ \zeta(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x - 1}{dx}
[/latex]

 

[andyandy]

I don't "do" latex, but I'm willing to play:

[latex]
$$ \zeta(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^x - 1}{dx}
[/latex]

cool - much appreciated...now....with s=-2

[latex]
$$ \zeta(-3) = \frac{1}{\Gamma(-3)} \int_0^\infty \frac{x^{-3}}{e^x - 1}{dx}
[/latex]

let's see how that can be solved...

hmm. Not very easily. And neither by Wolfram's integrator....

 

[jsfisher]

cool - much appreciated...now....with s=-2

[latex]
$$ \zeta(-3) = \frac{1}{\Gamma(-3)} \int_0^\infty \frac{x^{-3}}{e^x - 1}{dx}
[/latex]

let's see how that can be solved...

hmm. Not very easily. And neither by Wolfram's integrator....

Well, actually it should be

[latex]
$$ \zeta(-2) = \frac{1}{\Gamma(-2)} \int_0^\infty \frac{x^{-3}}{e^x - 1}{dx}
[/latex]

 

[Yllanes]

cool - much appreciated...now....with s=-2

[latex]
$$ \zeta(-3) = \frac{1}{\Gamma(-3)} \int_0^\infty \frac{x^{-3}}{e^x - 1}{dx}
[/latex]

let's see how that can be solved...

That formula doesn't work for Re s <= 0. The Gamma function diverges for negative integers.

As for the proof, it is quite simple. Start with

[latex]
\[
\int_0^\infty\mathrm{d}x\ \frac{x^{s-1}}{\mathrm{e}^x-1} =
\int_0^\infty\mathrm{d}x\ x^{s-1}\mathrm{e}^{-x} \frac{1}{1-\mathrm{e}^{-x}}
\]
[/latex]

Now you use the formula for the sum of the geometric series and take the sum out of the integral.

[latex]
\[
\int_0^\infty\mathrm{d}x\ x^{s-1}\mathrm{e}^{-x} \sum_{n=0}^\infty \mathrm{e}^{-nx}=
\sum_{n=1}^\infty\int_0^\infty\mathrm{d}x\ x^{s-1} \mathrm{e}^{-nx}
\]
[/latex]

All you have to do is integrate x^{s-1} e^{-nx} (which is just the definition of the Gamma function anyway). Doing this you get:

[latex]
\[
\int_0^\infty\mathrm{d}x\ \frac{x^{s-1}}{\mathrm{e}^x-1} = \Gamma(s)\sum_{n=1}^\infty \frac{1}{n^s}
\]
[/latex]

Do you see where the requirement Re(s) > 0 comes from?

 

[jsfisher]

Do you see where the requirement Re(s) > 0 comes from?

Sigh, sure enough: I replaced one partial version of the Riemann Zeta function with another partial version. See this link.

 

[andyandy]

Well, actually it should be

[latex]
$$ \zeta(-2) = \frac{1}{\Gamma(-2)} \int_0^\infty \frac{x^{-3}}{e^x - 1}{dx}
[/latex]

oops.....

That formula doesn't work for Re s <= 0. The Gamma function diverges for negative integers.

As for the proof, it is quite simple. Start with

cheers - very interesting.....

 

[illogical]

The Riemann Zeta Function (Edwards)

i haven't read it, but it's listed in many bibliographies.

Prime Obsession (Derbyshire)

high school mathematics.

The Riemann Hypothesis (Sabbagh)

equation free!

Stalking the Riemann Hypothesis

Peter Woit says it's good, but i haven't finished it.

Dover Publications

they print a variety of number theory textbooks, from high school to graduate level.

http://www.dartmouth.edu/~chance/chance_news/recent_news/primes.html

 

[illogical]

did Sabbagh make an error? he states that 100% of the (nontrivial) zeros are on the critical line. seems like a misunderstanding of limits, Hardy's proof, or something of that nature.

 

[andyandy]

did Sabbagh make an error? he states that 100% of the (nontrivial) zeros are on the critical line. seems like a misunderstanding of limits, Hardy's proof, or something of that nature.

not that i know much on this, but isn't that just what the riemann hypothesis actually states? The trivial solutions are as i understand the negative integers, -2, -4 etc. but the non trivial solutions are all hypothesised to lie on a line with Re(s) = 1/2....though this is not proved

 

[becomingagodo]

Riemann hypothesis has been solved

http://arxiv.org/abs/math.GM/0307136

Beginning from the formal resolution of Riemann Zeta function, by using the formula of inner product between two infinite-dimensional vectors in the complex space, the author proved the world's baffling problem -- Riemann hypothesis raised by German mathematician B. Riemann in 1859.

Well, I was looking around arxiv and searching for Riemann hypothesis and it has been solved. I can't believe it has been solved. Man, I really wanted the million dollar prize money.

 

[Stir]

Skeptical response:

I've seen no mention of this in newspapers or magazines (and I read a lor of 'em). What's the status of confirmation of the 'proof'?

 

[becomingagodo]

I've seen no mention of this in newspapers or magazines (and I read a lor of 'em). What's the status of confirmation of the 'proof'?

I don't know.

Doesn't it take like two years before the proof to be consider a proof. Perelmann proof took some time to be confirmed, so it would proberly take alot of time to confirm this.

 

[Jekyll]

Skeptical response:

I've seen no mention of this in newspapers or magazines (and I read a lor of 'em). What's the status of confirmation of the 'proof'?

There's another paper on arvix that claims proof of P=/=NP. I wouldn't worry about it to much as arvix allows people to 'publish' without peer review.

Having said that I haven't looked at the paper, and good people do put legitimate work out through avix.

Wait for peer review, and ignore the paper if it doesn't come would be my advice.

 

[Yllanes]

http://arxiv.org/abs/math.GM/0307136

Well, I was looking around arxiv and searching for Riemann hypothesis and it has been solved. I can't believe it has been solved. Man, I really wanted the million dollar prize money.

That paper is more than 4 years old and has 0 citations. It obviously is not a valid proof, or we would have heard about it by now. This is not the first time a 'proof' of the Riemann hypothesis appears on the arXiv:

http://uk.arxiv.org/abs/hep-th/0208221

This eprint server is an invaluable tool, but one must beware of papers that have not been published in any journal.

 

[illogical]

if you read carefully, you'll find it's in General Mathematics. most of the crackpot and amateur papers do not go in the other sections.

 

[illogical]

I really wanted the million dollar prize money.

i'm fairly confident the work involved in proving *any* of the millenium problems would be worth more than US $1 million. in America, there are definitely easier ways to make money.

 

[Complexity]

becomingagodo - Stop announcing things, especially mathematical news, especially anything about the Riemann Hypothesis.

 

[illogical]

Andyandy,

apparently Sabbagh is referring to the work of Granville and others. even if the "limit" goes to 100%, then there can still be zeros off the line. i think that tidbit should have been left out of the book or edited by a mathematician who can explain it.