Recent content by Quiggy

Sorry, I don't see how any of this is supposed to help me. I just don't understand what you're getting at.

I didn't really know where the proper place was for this, but this is an intro thermodynamics class and I'm really confused over this math question (it's not strictly physics-related). Homework Statement Consider rolling N six-sided dice. Define a microstate as the number showing on...

Y'know, somehow it's reassuring that this is gaining a good amount of disagreement, since it means I'm not missing something obvious :) The whole bit with the integral split up differently and the link with the CPV discussion all help out quite a lot. I think I understand it enough now to know...

Actually, that's a pretty good explanation. Neither my textbook nor my teacher did a very good job of explaining exactly what divergence is, so I suppose that that's the reason why it didn't work out. Thanks again.

Huh. That's very strange, as I'd assume that \lim_{x\rightarrow\infty} \frac{1}{4} x^4 and \lim_{x\rightarrow-\infty} \frac{1}{4} x^4 would both approach \infty at the same rate. I don't understand why, but I'll take your word for it. Thanks!

I'm not finding area at all, I'm performing the integration \int_{-\infty}^{\infty} x^3 dx. I agree that I would have to split it up if I was finding area and would get \infty, but that's not what I'm doing. Why then did my teacher and I get different answers?

Why can't I do that? \lim_{x\rightarrow\infty} -x = -\infty, so it's not like there's a problem (that I see, anyways) with representing the lower limit -\infty as \lim_{x\rightarrow\infty} -x. I guess the other thing that's confusing me is that \int_{-1}^{1} x^3 dx = \int_{-10}^{10} x^3 dx =...

[SOLVED] Improper Integration Hi, I was working on this problem in my calculus class today and kept getting the answer of 0, while my teacher was saying that the integral diverges. I just want to know if I'm wrong, he's wrong, or something way over my head is going on here and we're both...