Plough it into your brain. I remember being sat in an exam, being faced with a similar problem and completely blanking on how to do it. It was very fustrating.
As an aside, if f^{2} was being used to mean f \circ f, then the function given wouldn't work because we would have f^{2}(x)=f(0)=1 for x>0.
ETA: As well as f^{2}(x)=f(1)=0 for x \leq 0.
Assuming that robertjordan hasn't completely misunderstood the question, they definitely aren't looking for an identity function.
Normally f^{2}=f \circ f, so I think Mark44 thought that that's what the question writer meant by f^{2}. Certainly if the question writer did mean f \circ f then a...
Homework Statement
Homework Equations
L[c]:=\int_{a}^{b}(\sum_{i,j=1}^{2}g_{ij}(c(t))c_{i}'(t)c_{j}'(t))^{1/2}dt
The Attempt at a Solution
So g_{ij}(x,y)=0 for i{\neq}j, c_{1}'(t)=-Rsin(t), c_{2}'(t)=Rcos(t)
so...
I can't imagine that you would lose points for this, but for the sake of pedantry you might want to point out that P and Q would have to both be positive integers. Just because 2^0=3^0 and 2^P, 3^Q aren't even and odd respectively when P and Q are negative.
I still say the question was wrong, as the question says that you have x in [0,pi] which clearly doesn't work, x in [0,pi) would have been fine though. Basically that series does not distinguish between x=0 and x=pi, so that's why you either can't have x=pi.
Perhaps the question itself was wrong, it seems like it wouldn't matter would you get for a_{n} as you'd always get \pi=0. I suspect it should have been x{\in}[0,\frac{\pi}{2}]. Since \sin(n0)=\sin(n\pi)
If the lecturer doesn't know your name, perhaps ask and run away if your name is asked for.
You seem to have just made a simple mistake, you should get x on top, not 1. Also that 2 shouldn't be there.
http://www.wolframalpha.com/input/?i=%28arctan%28x%5E2%29%29%27%2F%28%28sinx%2Btanx%29%5E2%29%27