Recent content by resolvent1

Out of curiosity, what was the signals book you used? Thanks.

Homework Statement I'm trying to show that if \sum_{n=1}^\infty x_n converges, then so does \sum_{n=1}^\infty \frac{x_n}{1-x_n} Homework Equations Unknown. Possibly using the limit of the sum of x_n to bound the partial sums of x_n/(1-x_n)? The Attempt at a...

Basic math skills are more important - if you want to study physics at the university level, you'll eventually need to take SATs, and good scores on that can help out subtantially.

Well, if you're interested in doing them analytically as mich as possible, then the natural way to solve it is to use the Taylor series expansions (preferably centered at 0) for the trig functions. (naturally because these functions are analytic).

I think you can do something like the following. Assuming the integrals of f and g are equal for every set A (and f and g are obviously measurable): Consider the function f(x) - g(x) . Let E be the set of x's in A where this function is positive. Define E_{1/n} = \left{ x \in E : f(x) -...

The definition I'm using for smooth is: A function f : \mathbb{R} \rightarrow \mathbb{R} is said to be smooth at a point x \in \mathbb{R} iff it is defined in a neighborhood of x and \lim_{h \rightarrow 0} \frac{ f(x+h) + f(x-h) - 2f(x) }{ h } = 0 .

I'm trying to prove that if a function is continuous on [a,b] and smooth on (a,b) then there's a point x in (a,b) where f'(x) exists. The definition of smoothness is: f is smooth at x iff \lim_{h \rightarrow 0} \frac{ f(x+h) + f(x-h) - 2f(x) }{ h } = 0 . I'm starting with the simpler case...

Yeah, I knew that several hours ago, just not the dominating function. THank you for your help.

Thanks, I think that works perfectly. (I can't believe I didn't see that before - sorry - and that problem was really starting to piss me off.)

Yes, you can use polar coordinates. What your professor told you to use is the change of variables formula, by setting x = r sin(theta), etc. This defines a transformation from xyz space to r-theta-z space - thus the integral over, say, a cylinder in xyz space is equal to the integral over a...

Sorry, that should be an n, not a k. It's true that x^{2n} \leq x^2 on [-1,1], so \frac{1}{\sqrt[3]{1-x^{2n}}} \leq \frac{1}{1-x^2} on [-1,1], but \frac{1}{1-x^2} isn't in L^1(-1,1) So I'm not sure what to do with it.

Homework Statement Determine whether the limit \lim_{n \rightarrow \infty} \int_{-1}^1 \frac{1}{\sqrt[3]{1-x^{2n}}} dx exists and evaluate the integral if it doesHomework Equations Dominated convergence theorem (I think) and a power series representation.The Attempt at a Solution I've...

No, you have to use the change of variables formula.

THanks, I've got it.

I've been trying to use Taylor's theorem with h = (y-x)/2 to show that a twice differentiable function for which the second derivative is positive is midpoint convex (ie, f( (1/2)*(x+y) ) \leq (1/2) * (f(x)+f(y)) ). (It's not a homework problem.) The problem I end up with this is that I'm not...