Recent content by Kelvie

Well.. whatever the particular solution is, it makes c1 != y(0), because it will always be off by Yp. But the answer(the amplitude) is clearly just the initial displacement (y(0)). Or am I missing something ? Yp can't be 0, can it ?

Please learn to check your spelling, and I think this belongs in the homework help section. Either way, you should attempt a problem before just posting it (poorly) on a forum, and expecting us to do it for you.

To start, I already know the answer.. but I can't seem to get it the "hard" way, i.e. through solving the 2nd order ODE. Redundant, but it's a block on water, and it's oscillating after a mass is removed from it. There's an initial displacement, and no \begin{align*} y(0) &= -0.025m \\...

Hmm.. I was afraid it would come to this. I was trying to solve this for a delta-epsilon proof of a limit at infinity (finding what N of epsilon could be that is < |n|. I got my two quadratic equations, so technically, the smallest one could be N? or the largest? Or do all of them work...

A bit of a newbie question, but I was wondering how does one go about solving these? For example: (I was working on a problem posted on another thread on Homework Help) |3n-4| < 9\epsilon n^2 + 3 \epsilon Epsilon is a small positive number of course :P The tricky part is when I split...

Hmm.. I got a different quadratic equation(above, probably wrong?). Time to check over my work ! By the way.. when you get values for n.. does it mean that n > that value? Wait.. after a quick google, in your last example, wouldn't n be bounded between the two values from the q. equation?

You may want to do a forum search for delta-epsilon proofs. The way to do it for a limit at infinity is as follows (correct me if I'm wrong). For: \lim_{x \to \infty} f(x) = L Assuming L is a finite number, you want to find a N such that |x| > N(\epsilon) \implies |f(x) - L| <...

If by the second statement you meant 0 \leq L < a then yeah, I guess they are equivalent statements. Thank you for your help.

Alright, this also makes sense, doesn't it? We ignore the assumption L \neq M[/itex] and \epsilon = \frac{|L-M|}{3}[/itex] [tex] \begin{align*} &|L-M| = |L -M + f(x) - f(x)| = |(L-f(x)) + (f(x) - M)| \\ &|L-M| \leq |-1||f(x) - L| + |f(x) - M| = |f(x) - L| + |f(x) - M| \\ &|L-M| \leq...

So this is "hand-in"able? :P Thanks for the timely reply, by the way.

Good evening, I am a first year engineer here and a first time poster also. I had a problem that has been bugging me for the last few days; after much head-scratching and tree-killing, I may have solved it. I am, however, not sure at all if all my assumptions along the way are correct. So...