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Top five favourite theorems

Callan
Post Icon Posted: Submitted by Callan on 1 May 2009 - 9:18am.

Joined: 2008-09-30
Posts: 173

Taking a cue from: http://warwickmaths.org/blogs/entries/my-five-favourite-theorems , what are your five favourite theorems and why (give a proof if it isn't too long and is nice)? I'll do this later on when I've decided.

‹ Does the empty set exist? Mathematics websites ›
Callan
Post Icon Posted: 3 May 2009 - 5:06pm

Joined: 2008-09-30
Posts: 173

Here's a theorem of complex analysis which I always thought was quite nice and really useful (although not quite as baffling as Liouville's theorem for example).

Rouché's Theorem
If $ f $ and $ g $ are holomorphic and $ |g(z)|<|f(z)| $ on some closed contour $ \Gamma $ then $ (f+g) $ must have the same number of zeros as $ f. $

Proof (sketch):
Clearly the winding number of $ (f+g) $ is the same as that of $ f $ so we can use the argument principle which tells us that these have the same number of zeros in $ \Gamma. $

If you actually want to prove this use that $ |\frac{f+g}{f}-1|<1 $ and so the winding number around the origin of any closed contour is zero. This means the number of poles of $ \frac{f+g}{f} $ is equal to the number of zeros but the number of poles is the same as the number of zeros of f so f and f+g have the same number of zeros in $ \Gamma. $

Here's the really nice part though:
Corollary 1- The Fundamental Theorem of Algebra
Proof:
Take $ f(z)=z^{n} $ and $ g(z) $ to be any polynomial of degree n-1. Taking a large enough (in fact a circle with radius equal to the sum of the absolute values of the coefficients plus one) contour $ |f(z)|>|g(z)| $ meaning a general nth degree monic polynomial has n zeros.

Corollary 2- A Fixed Point Theorem
If the disc is mapped holomorphically to its interior then there must be a unique fixed point.
Proof:
Let $ f(z) $ be the function which maps the disc to its interior, so $ |f(z)|<1 $ and then look at the function which connects each point to its image point i.e. $ g(z)=f(z)-z.\; $ Now $ |f(z)|<1 $ so taking $ h(z)=-z $ this has one zero in the disc meaning $ g(z) $ has one zero, but this is a fixed point and since this is the only zero must be unique.

Corollary 3- Open Mapping Theorem
Any non-constant holomorphic map from a conneccted open subset of the plane will be an open mapping.

I don't really want to do the proof since its a little longer, but it's here:
http://en.wikipedia.org/wiki/Open_mapping_theorem_(complex_analysis)
richardhp
Post Icon Posted: 3 May 2009 - 7:11pm

Joined: 2007-10-01
Posts: 239

not really a full proof, but a little argument. i like that you can prove there are 2 irrational numbers $ a,b $ such that $ a^b $ is rational. i like the simplicity but also the fact that it's non-constructive so you have no idea which numbers satisfy that property.
Callan
Post Icon Posted: 3 May 2009 - 11:16pm

Joined: 2008-09-30
Posts: 173

Yeah thats pretty neat. Kind of related is: http://en.wikipedia.org/wiki/Gelfond–Schneider_theorem which gives the actual example $ \sqrt{2}^{\sqrt{2}} $.
Sam
Post Icon Posted: 4 May 2009 - 1:57am

Joined: 2007-10-03
Posts: 562

I really like the Uniformization Theorem: Every simply connected Riemann surface is conformally equivalent to just one of the Riemann sphere, the complex plane or the complex upper half plane.

It's very nice as it allows to classify lots of stuff, so for example any Riemann surface is a quotient of its universal covering space, which by the above must be either the sphere, the plane or the upper half plane. In particular, this tells us the universal covers of all compact Riemann surfaces: the sphere covers itself, the plane covers all polynomial tori (elliptic curves) as quotients by a given lattice, and the upper half plane covers higher genus surfaces, by action of a Fuchsian group.

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