Dave

It's all to do with exchange particles. Each type of force has its own exchange particle. I can't think of a decent way to describe it, but the nuclear force has the gluon exchange particle - quarks exchange this particle and because of it they're stuck together with the nuclear force. For the electromagnetic force (which is what keeps electrons in their associated energy levels) the proton is an exchange particle. As for the Bohr radius - the particles will only exchange within a certain distance of each other. That was a pretty lame description, but I've only had the experience of A-level two years ago There's a much better explanation here.

Basically, [math]R[/math] and [math]\phi[/math] are just constants which satisfy: [math]R^2 = a^2 + b^2[/math], [math]\tan\phi = \tfrac{b}{a}[/math]. If you expand the RHS you'll see that it's equal to the LHS.

Indeed; I forgot to put that in Answers are still right though; edited the above posts to make sure that it makes sense now.

Just a couple of revisions now, and a bit of a changelog: v2.1: - New feature: LaTeX images are only displayed when you're logged in. This is to stop spiders/crawlers from re-generating the LaTeX images unintentionally, otherwise our cache gets filled up rather quickly. - New feature: Implemented mhchem package to make typesetting chemistry easier. Example: [math]\ce{H2O ->[\text{process}] N2O}[/math] - Fixed silly alt/javascript bug with multi-line latex. - Fixed various errors with vB. - Downsized the images a bit to fit better with inline text. v2.0 - Complete re-write of the backend module. - Re-wrote latexrender to include MySQL support for better image caching. - Fixed array problem (at last).

Update: I now have it down to: [math]\int \frac{1}{(a\sin x + b\cos x)^2} \, dx = \frac{b\sin x- a\cos x}{(a^2 + b^2)(b\cos x + a\sin x)}[/math] I'll be damned if I can simplify it down though Hint: Use the fact that [math]a\sin x + b\cos x \equiv R\cos(x-\phi)[/math].

I'm having some problems with this solution. I'm usually fairly good with the integrals, but it's been a while. Mathematica gives this answer: [math]\int \frac{1}{(a\sin x + b\cos x)^2} \, dx = \frac{\sin x}{b(b\cos x + a\sin x)}[/math] Excluding the constant of integration, of course. However, I can't get this answer by any means I've tried, which is annoying. If you differentiate it and simplify it down, it definately works. Hmm.

Do you mean indefinite integral?

I got pwnd. Except we were already talking about sine/cosine in terms of the reals, so fnar

It means they're topics that have a lot of posts and/or are frequently read (otherwise known as "hot" topics).

It's interesting... but I don't think I can quite take it at face value. One thing's for sure - he must be pretty convinced about it, otherwise he's going to be throwing his career down the toilet rather rapidly.

Hmm. Get more people reading them and I'll post more

From BBC News: All I have to say is: bahahahahahaha.

What? More like a tenth of that. I point you to: http://www.wordiq.com/definition/Light_bulb#Efficiency (From the other thread).

? You can't draw a graph for it for negative x; there's just too many discontinuities.

First google hit for "matrix trace". For your information, the trace of a matrix is the sum of the diagonal elements. Please also be aware that people don't necessarily want their threads invaded to cover all aspects of GR To Meir Achuz: to typeset LaTeX, you encapsulate with [ math] [/math] (obviously omitting the space).

The transformer gets quite hot inside. That's constantly giving out heat all of the time, plus all the larger capacitors and things like that. If you take one apart, you'll see why

(If, by always, you mean never ) You could also try booting up from the XP install disc, go to the recovery console and type fixmbr - this might do the trick. I don't know really.

In terms of Linux, there's a couple of options. For the hardcore performance people, Gentoo is quite a nice option and has one of the best package-management systems around (imo) - it's certainly my choice of OS. Debian is pretty good if you can't be bothered compiling everything. Fedora Core I've heard good things about, but I'm not overly keen myself. There's quite a few others (read: lots), but they're the ones I prefer to talk about.

A full install would probably fit inside a couple of gig. Doesn't allow much leeway for programs to be installed though (you could argue that programs can be installed elsewhere I suppose).

It would also help very much if you used LaTeX, although that's completely optional i suppose.

Indeed. Always go for the way that you understand best. Comes out trumps every time (unless it's wrong ).

From what I've seen, the methods are pretty much all the same. Dapthar's is probably the best because it doesn't involve any substitution.

Indeed. The limit is about 500 pixels to stop the page breaking up and whatnot.

Okay, here's my kind of "sketch" proof: [math]x^2 - 4x + 1 = (x-2)^2 - 3[/math] Now let [math]y = x-2[/math], so the limit becomes: [math]\lim_{y\to -\infty} \left( 2 + y + \sqrt{y^2 -3} \right)[/math] Now, [math]2 + y + \sqrt{y^2 -3} = 2 + \frac{3}{y - \sqrt{y^2 - 3}}[/math] The fraction on the right tends to zero, so the limit is 2. I did it Dapthar's way at first, but got unstuck at the end also. What do you guys think? Personally, I think it's a little bit dodgy and I haven't actually proved that the fraction on the right tends to zero, but I think that it's pretty easy to do so (sandwich rule?)

Huh? If you divide the limit by x, you're also going to need to multiply it by x to keep equality on both sides. That method works well for fractions, but in this case, I think you need something else.