Recent content by linearfish

I follow that, thanks. Does the 'a' not really matter since u -> inf? I think that was the problem I was running into.

Homework Statement Suppose that f is differentiable on the interval (a, inf) and f'(x) -> 0 as x -> inf. Show that f(x)/x -> 0 as x -> inf. The Attempt at a Solution Conceptually I understand this problem. The derivative gets very small so as x gets large, f(x) gets very close to being a...

That makes sense. It's not the answer in my notes but maybe I wrote it down wrong.

When one sleep deprived person is trying to converse with another, chaos is bound to ensue. I appreciate the help from both of you. I'll admit though that I'm still at a loss. I know what the normal should be so that the problem works out but I don't know how to get that. (I will also admit...

Okay, I get that. So is my calculation off somewhere else?

Yes, thanks. I'm still confused, though. I thought that if a plane cut through the center of a sphere then it was a great circle, hence its area was a quarter of the surface area of the sphere.

Homework Statement Use Stokes Theorem to compute \int_{L}^{} y dx + z dy + x dx where L is the circle x2 + y2 + z2 = a2, x + y + z = 0 The Attempt at a Solution I feel like this problem shouldn't be that hard but I can't get the right answer: (pi)a2/3. I calculated the curl of F as...

Oh, duh. I guess it didn't occur to me to hold x constant, so the points (0,pi) and (0,3*pi) both map to (-1,0), hence f is not one-to-one. Thanks. Does the rest look okay?

Not to bump this thread but I think I have a better solution now: \left| (1-x) \left(\frac{1 - (-x)^{N+1}}{1+x}-\frac{1}{1+x}\right)\right| = \frac{1 - x}{1 + x} \left| -(-x)^{N+1} \right| = \frac{1 - x}{1 + x}(x)^{N+1} \leq (1-x)x^{N+1} Keeping in mind that we are on the interval [0,1], we...

Homework Statement Let f = (f1,f2) be the mapping of R2 into R2 given by f1=excos(y), f2=exsin(y). (1) What is the range of f? (2) Show that every point of R2 has a neighborhood in which f is one-to-one. (3) Show that f is not one-to-one on R2. The Attempt at a Solution (1) Since ex is...

Thanks, that does help.

Right, I meant that as a contradiction. Thanks again for the help.

Let me try this: I claim: \sum_{n=0}^N (-1)^n x^n (1-x) = (1-x) \frac{1 - (-x)^{N+1}}{1+x} \sum_{n=0}^{\infty} (-1)^n x^n (1-x) = \frac{1-x}{1+x} If the first converges uniformly to the second, then as n goes to infinity: \left| (1-x) \left(\frac{1 -...

As far as I know, Green's Theorem is normally stated for positively oriented curves (counterclockwise). If a curve is oriented clockwise, is it just the negative version? \oint Pdx + Qdy = - \int\int \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \int\int \frac{\partial...

Yes, that should be an M, thank you. I can see why M > 0. If M < 0 we would have 0 \leq \left| f'(x) \right| \leq M \left| f(x) \right| \leq 0 and we are done. Now if M0 > 0, then we must have M (x_0 - a) \geq 1 for all x in (a, x0), but this fails if we choose x to be a +...