StarDestroyer.Net BBS

Kuroneko wrote:

Durandal wrote:Oh so that's what you're doing. I'd long since forgotten about the Hell that was integration by parts.

Personally, I dislike that u,v,du,dv formulation. Yes, it is prima facie more general, but only at first glance--but if I think of integration by parts as simply the integral of the product rule (fg)' = f'g + fg', then everything comes out easier.

Durandal wrote:It's not so much the notation that infuriates me; it's that if you choose the wrong substitutions, you can wind up spending a half hour to find out that you did so.

SirNitram wrote:If this keeps up, I'm posting the reason why the population of the universe is zero.

Andrew J. wrote:

SirNitram wrote:If this keeps up, I'm posting the reason why the population of the universe is zero.

I'll save you the trouble.

The universe has infinite space, but the space occupied by people is finite. Since any finite number divided by infinity is as close to zero as makes no odds, the average poplation of the universe is zero.

Andrew J. wrote:The universe has infinite space, but the space occupied by people is finite. Since any finite number divided by infinity is as close to zero as makes no odds, the average poplation of the universe is zero.

Specialist wrote:

Andrew J. wrote:The universe has infinite space, but the space occupied by people is finite. Since any finite number divided by infinity is as close to zero as makes no odds, the average poplation of the universe is zero.

I thought it was already proven the universe is expanding, therefore it's not infinite in size.

Andrew J. wrote:The universe has infinite space, but the space occupied by people is finite. Since any finite number divided by infinity is as close to zero as makes no odds, the average poplation of the universe is zero.

Kuroneko wrote:

Durandal wrote:It's not so much the notation that infuriates me; it's that if you choose the wrong substitutions, you can wind up spending a half hour to find out that you did so.

That's not bad. What's bad is when you do half an hour of substitutions with no result, because integration by parts does not apply at all in any manner whatsoever.

Try this: I = \int_0^{\pi/2}\log\sin(x) dx.

muse wrote: student: why do we have to learn all this stupid integration and identities and stuff?

teacher: because one day some guy on a message board will ask you to prove that 1=1, and if you don't know how you'll look stupid.

muse wrote: It's scary how fast one can forget this stuff, a year ago I could do the proofs and stuff for these equations, now I can't even remember what the hell it all means. This is why I'm not in engineering or sciences.

phongn wrote:Can someone post the LaTeX code to that expression? I'd like to print it out

\documentclass[11pt]{article}
\begin{document}

\begin{eqnarray*}
\left|\left| \mathbf{\nabla} \times \left( \frac{-k}{r^{2}} \right) \mathbf{\hat{e}}_{r} \right|\right| &=& \frac{d}{dx} \left[ \frac{1}{2} \left(x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots \right) \left( \frac{x}{\sin x} - \cos x \right) \right] - \int_{0}^{\frac{\pi}{2}} \cos x \ dx \\
0 &=& 1 - 1 \\
&=& 0 
\end{eqnarray*}

\end{document}

kojikun wrote:someone should code something into phpBB that allows you to display mathematical notation. I'm sure I could figure out some way, but then, I am not the person at phpBB.