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.
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] +...
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...
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...
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...