[Guest]

prove: 1/2π∮arctan(f(u,v)/g(u,v) )dl = n where f and g are continuous functions in the u ^_x v plane, n is an integer, the path of integration is a closed curve including no points (u₀,v₀) such that f(u₀,v₀) = 0 = g(u₀,v₀)

(as a convention arctan(f/g) = +infinity if f>0,g=0 and arctan(f/g)= -infinity if f<0, g=0)

Sorry if all those conditions are hard to digest.

I don't know how easy hard this is because I didn't really use calculus (per se) or abstract algebra to derive this. If it can be proven with either of those then it might be easy, but it takes a slightly more outside the box path to get to this the same way I did. Good luck!

[Guest]

wow no attempts yet. maybe its because I just noticed a typo which makes it make no sense: where i wrote infinity I meant to write pi/2...I was clearly sleepy when I wrote this. should I tell the answer or is it still too soon?

[superprismatic]

Well, there have been about 200 views of the problem. The interesting part of this problem seems to be how to get there from a

"back door". I guess nobody has any good idea about how to do that. It's been a week, so it's probably not to soon to give the

answer. But please use a spoiler in case someone out there is still actively working on it. I, for one, am interested in how you

got it. But, I'm in no hurry -- I'll just check this thread from time to time. Thanks for the problem!

[bonanova]

I'd hold off a bit with the answer.

I just saw the problem and I think there's a theorem from my school days that may prove fruitful

[Guest]

[bonanova]

I expect that it can be shown using Green's theorem

with a judiciously chosen rectangular curve. But I

don't see the general result, or the back door.

Interesting problem.

[Guest]

So I’ve created a monster. I failed pretty hard at solving this myself so I’ll explain the answer and why I somehow doubt the answer.

I don’t know how this is possible with simple calculus. A lot of it seems like a red herring in fact. I noted that you can simply replace h(x) with tan(w(x)) and sent g(x)=1 and you should be able to get any general function that way with the proper substitution for w. But this requires pretending to not realize the w(x) would really have to have a limited range (in order for tan(w(x)) to not be discontinuous) and I still see no way to use calculus to arrive at the result.

I then decided to try to disprove it by, among other ways, simply defining a function who’s derivative is an arbitrary constant all the way along the curve as a counterexample (if this sounds like cheating just look up the definition of electric field in terms of voltage and displacement and what happens when a magnetic field inside a closed conducting loop changes at a constant rate). But I had no way of showing that this function can fit the description of f.

This is just a formal definition of the conley index.

Consider a dynamical system u’=g(u,v), v’=h(u,v). This is just a fancy way of saying a vector field where the x-component is g and the y component is h. if you define the index of a closed path as the signed number of times the vector <g,h> rotates along the path. Since the path is closed the index has to be an integer (otherwise this would mean that <g,h> takes on multiple values at one point.

Defining this with an integral you get that the index is 1/2pi times the integral of the change of theta along the path. Theta simply equals arctan(h/g) and the original monster of a problem is created

The topological sense I referred to is actually not dependant on the area inside the curve as is often the case in vector calculus but on the fixed points (points where u’=v’=0). The index of a function is equal to a-b where a+b is the total number of fixed points within the region.

This explains how the integral of a change along a path doesn’t simply always equal 0.

But this still bothers me. It seems like this only works when you have a basis such that one parameter has an equivalent to zero (in this case theta is equivalent to zero whenever it is a multiple of 2pi). My intuition says that this should carry over to any other basis because even if you choose a basis with a random equivalent of zero, to integrate it you’d need a jacobian matrix multiplier which would scale it back to 2pi. Alas I am way too lazy to actually work out such an ugly abstract integral as that.

Thusly this problem is still open to a non-backdoor solution (or counterexample).

I apologize if this was really hard to follow, it’s hard to explain without pictures and I didn’t feel like making some yet. If you really want to see pictures though, just let me know and I’ll work on some.