Recent content by pylauzier

Ok I found out how to solve the problem. Using the superposition principle, substract 2 step functions to generate a pulse only for the first 0.0005 seconds. Do the same thing on the output.

Oh god I'm an idiot, thanks everyone. I guess I shouldn't be doing maths at 1:30am when sleep deprived :P

I didn't forget it, I simply started at the step right after it in my initial post. Notice the squared 8, I squared the equation to remove the square roots. If you simplify the (x+1)2 and (x-1)2 terms, you end up with 2x2 + 2, so the equation becomes 2x2 + 2y2 = 82 - 2 x2 + y2 = 7

Well when I did the problem I didn't intuitively notice that the equation would give me an ellipse, that's why I developped it the way I did. I'm trying to figure out what I did wrong, since I got the equation of a circle. Why isn't |z+1| + |z-1| = 8 equal to sqrt [(x+1)2 + y2] +...

|z-1| = sqrt ((x-1)2 + y2 ), no?

Homework Statement I'm asked to describe geometrically the set of points in the complex plane describing some equations. I got them all right except this one: |z+1| + |z-1| = 8 Homework Equations |z| = sqrt( x2 + y2 ) The Attempt at a Solution Well, I know that an equation of...

Oh right, that sure is simpler. Thanks again!

Your mistake is in the 3rd to last step, where you have one term with a 2x-3y^2 denominator. In the following step you simplified the equation by multiplying (2x-3y^2) with the (2x-3y^2)^2 term. However the two 4y terms on the numerator were not being divided by 2x-3y^2.

Sorry if this is the wrong section, it wouldn't let me post in the Science Learning Material area. Basically I started self-studying Mary L Boas book Mathematical Methods in the Physical Sciences. I've seen a lot of great things about the book and I know it's very popular in Mathematical...

Homework Statement A problem in my book asks the reader to evaluate lim x→0 xln(2x) using L'Hôpital. Homework Equations None The Attempt at a Solution I think I got it right but I simply wanted to make sure I did no illegal manipulations (I'm self-studying this stuff). I...

Thanks a lot for your help!

Ok, I think I understand now. The nth term will be smaller than (1/2^n), since all the terms are smaller than one and because there is n 1/2 terms. So lim n -> ∞ (n!)2 / (2n)! =< lim n -> ∞ (1/2^n) Since lim n -> ∞ (1/2^n) = 0, it follows that lim n -> ∞ (n!)2 / (2n)! =< 0

I'm not sure I understand what I'm supposed to compare. I thought the comparison test was for series (sums of terms), not simple limits. If I'm not asking for too much, would you care pointing me towards the right direction? I even searched in my old calc II book and there is no mention of a...

I know how to compare sums, but I've never learned to compare sequences. What am I looking for? For example if I simplify the product given by n=3, I can see that the denominator in my sequence is bigger than the denominator of 1/n. Since lim n-> infinity of 1/n tends to zero, does that prove...

edit: Wait, I messed up.