Recent content by Stimpon

im so dumb

What answer should you be getting?

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...

Try showing that it's not a cauchy sequence instead and then just say "therefore it is not convergent."

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

I'm not going to tell you which one but using a trigonometric identity is probably the simplest way.

I'm pretty sure that's wrong. http://www.wolframalpha.com/input/?i=1%2F2%28sinh%5E%28-1%29%28x%5E2%29-ln%281%2Bx%5E4%29%5E.5%29%27

He's shown that the curve must be discontinuous but not that it must be discontinuous at infinitely many points.

Ah I think I didn't edit my post in time then, the upper limit should have been \frac{\pi}{4}, sorry. Your method is fine. And just draw the lines y=0 and y=x and you should be able to see where y satisfies 0{\leq}y{\leq}x. Or think about it this way, where does (x,y) satisfy 0{\leq}y and...