Most difficult maths

Income tax return.

Ahh - I married an accountant, so Income Tax is easy! Well easier than disguising an XPS2....

Writing 5-colour trignometric raytracing optical design software for telescopic apochromatic triplets in BBC Basic.

Oh alright then, it's not your idea of maths. But great fun nonetheless.

Nime

Bam's ladder problem...

Paul

It would have to be the parametrisation of the magnetic fields of a system of coils in terms of its multipole moments (described here ). We made pretty heavy use of Mathcad's symbolic algebra to get the result out and I probably couldn't do it by hand.
John
Ps. Good Will Hunting was complete bollocks as anyone who has ever done any academic research would know...

We had to use 4th Order Morovsky Transforms to work out exactly how much of a cock Robin Williams is. In retrospect it would have been much easier just to watch Good Will Hunting.

Matthew

Calculating the whereabout's of 'S' when Maths traversed the Atlantic ?

Calculating out what goes on inside Fritz' head......................

Answer is : <0

B4 i√U (RU/16)

I remember doing stuff on Tensors in theoretical mechanics in my junior honours year. Totally mind boggling and confirmed that I wasn't the mathematician I thought I was going to be. Most folk who study maths at university should really be doing physics or engineering because unless things have changed drastically since the 70s you don't really get exposed to pure maths in school.

Question: According to the Gauss-Bonnet theorem, if X is a compact, even-
dim hypersurface in R^(k+1), then integral of K over X = Vol(S^k^)*Chi(X)/2.
However, all of the proofs I have seen of this theorem assert that X must be a
k-manifold, i.e. without vertices, edges, etc. Now, I have seen a variant
where if X is a polygonal surface, and V is the set of all vertices in X, then
sum of K over V = 2*pi*Chi(X) where K is now defined to be 2*pi-sum of angles
at vertex. It seems to me that this new definition of curvature allows
vertices to concentrate curvature on sunon-regular surfaces. This, then,
brings me to the crux of the problem. Is it possible to generalize the Gauss
Bonnet theorem to non-regular hypersurfaces? If X is a compact, non-regular
surface, then (integral of K over X)+(sum of K over V)=2*pi = 2*pi*Chi(X). I
have tried to compute this on a 2-sphere with one pole pulled to a vertex of a
cone, and the formula holds. Is it possible for you to provide a proof for
this conjecture (or disprove it)?

If only this quick reply box were big enough...

Paul

quote:

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Originally posted by Martin D:
Is it possible for you to provide a proof for
this conjecture (or disprove it)?

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It's possible, yes. But extremely unlikely.

-- Ian

"Basic Topology" with problems such as: show that the Euler characteristic of a compact, connected, triangulable topological group is zero. Wow, I can't remember all the stuff I've forgotten.

-Dan

The most difficult maths will be once again where the Yanks lost the $ over the Atlantic and named it math ?

Fritz Von Parisclubwotalarf³

Ya gotta hand it to these guys aintyer Fritz:

Maths willy waving! Wottever next!

I suddenly lost interest in topology when it became compact. Know what I mean like?

Nime >= Pi!