the great math problems

[greghouse]

Hi, I was just wondering if there is available the proof of fermats last theorem on the web, aswell as the proof for the poincare conjecture.

I was looking over the Riemann hypothesis and I'm having some difficulties... It's claimed that the functions have trivial zeroes at the even negative integers... but when I put -2 as s in the sum equation I get 1+3+9 ... etc. Someone has explained on previous threads about RH that another formula is to be used... but can someone explain exactly why? (I understand the problem of the hypothesis pretty well but just not the equation...)

Thanks.

[cristo]

You really want to see the proof? I believe there is a link as the last reference of the Wikipedia article (http://en.wikipedia.org/wiki/Fermat's_last_theorem).

[n_bourbaki]

Wiles's work appears in The Annals, and hence is freely available via arXiv. Perelman's work has only ever appear on the arXiv and is thus also freely available.

[greghouse]

Thanx!

[n_bourbaki]

I didn't notice the second part of your post. If you're struggling with that, then I don't think you'll get much out of research papers.

The thing you're confused about is analytic continuation. The series you're using for the Riemann zeta function is only defined for a restricted set of the complex plane. But there are ways to analytically continue it away from this restricted set. I.e. there exists a (unique) meromorphic (allowing poles) function defined on C which agrees the the series expansion you know where the series expansion is defined.

A simpler example, is to consider a function we know exists for all of the complex plane (allowing for poles).

E.g. Take 1/(1-z), we can take a series expansion about zero:

1+z+z^2+z^3+z^4+.....

This series will only converge for |z|<1, because of that pole at z=1.

Now, suppose that we were just given that series without the nice 1/(1-z) interpretation for it. We can analytically continue it to a function on the entire complex plane (with a pole at 1), using some complex analysis. Ok, in this case we have the 'cheat' of being able to notice that it has a nice closed expression like 1/(1-z), but not all series give a nice elementary function like that.

Not every function can be continued beyond its radius of convergence: eg.

\sum z^{n!}

which has a pole at every complex number on the unit circle whose argument is a rational multiple of pi.

The rough notion for analytic continuation is that we patch together little overlapping discs where we extend the function piece by piece. That counter example can't be extended cos any little patch extending the unit circle will overlap with one of those poles.