Recent content by scholzie

That makes more sense, but it seems people leave the integration step out when explaining or using III(x). I've looked it up in numerous places and find the same conclusion. Brad Osgood of Stanford even states this in the course reader for his course "The Fourier Transform and its Applications"...

I just thought of something. If the FT of delta is 1, then the FT of comb(x) would be a train of 1s, not a train of infinities, wouldn't it? I know that the FT of comb is another comb, but is it fair to say that one is a train of infinite points while the other is a train of 1s? That doesn't...

comb(x) (or III(x), or Shah(x), whatever you want to call it) is DEFINED as the infinite sum of delta functions. III(x)=\sum_{n=-\infty}^{\infty}\delta(x-nT) for some period T. We know that \delta(x-a) is the shifted delta function where a is some constant. We also know that \delta...

Sorry, this thread should be in Calculus & Beyond. I thought I was in that forum before I posted. If a mod could move it, I'd appreciate not having to re-post it.

Homework Statement Take the Fourier Transform of f(x)=rect(x/2)*comb(x) where rect is the rectangle function and comb is the Dirac comb. Sketch the results. Homework Equations The FT of a convolution is the product of the individual FTs. The Attempt at a Solution Taking the FT is...

[C / s] = [A], so 1/[A] = [s / C] [N] / [Am] = [N / m]*[1/A], right? so [N]/[Am] = [s/C]*[N/m], or [s/m]*[N/C] I may have missed something, but I don't see a problem. Also, be careful. V = voltage, v = velocity. It was hard to read your posts with some of those things in caps. This could have...

Your amplitude is given to you (the amount pulled down). The period, 1.2s, is the amount of time it will take to go all the way up, and then back down again. Remember your unit circle? In this case, the low point corresponds to \cos{\pi}. It makes one full revolution around the unit circle in...

The statement is not 1 Farad = Ne Coulombs. It's 1 Faraday = Ne. A monovalent ion is an ion which has one single positive or negative charge. Therefore, a mole of ions with + or - charge will have N*e coulombs of charge. This unit is called a Faraday, but is better known as the Faraday Constant...

Remember, the electric force is conservative. That means that the path traveled does NOT affect the total work done. It's simply a matter of where the point starts, and where it ends. Find the potential energy at each place, and W = -\Delta PE. Note that work is only done when r changes, by the...

What have you done so far?

The cross product is the vector created by the determinant a \times b = \begin{tabular}{|c c c|} i & j & k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ \end{tabular} So, take the determinant of the above matrix, and then differentiate as normal. (Hint: the determinant will give you a vector...

Well, then maybe you should read the document linked in the very first reply instead of demanding it be spoon-fed to you. It's written clear as day: The same document said it is often better to use standard deviations or analysis of covariance to do your error propagation. A 5 minute Google...

Here's another uncertainties manual, also from RIT incidentally, with less theory than Vern's (which might not be what you're looking for, based on your original question) but with more emphasis on actually carrying through uncertainties with examples. Richmond gave this to us in in print-out...

If you imagine a "scanner," as you put it, recording the center of mass (a position!) of each slice, and you add those up, then you get an integral of positions with respect to the mass of the arbitrarily small slice, not the other way around.If you're imagining a bunch of thin slices, and...

Well, the centripital FORCE must equal the normal force of the rider, or he'll go flying off the bump, right? We care about the MAXIMUM velocity, which means we have to look at the case of highest normal force. Well that force is at a maximum when the biker is on the very top of the bump (when...