Dave

I don't quite understand what you're trying to accomplish with this. I'm not even sure you've got the idea of a linear transformation. From what I gather of your post (with the one and only one, you seem to think that a linear transformation has to be bijective, which isn't the case. So, in short, what are you getting at? As far as I'm aware, given two vector spaces U and V and a function [math]L : U \to V[/math], L is linear if: [math]L(\alpha \mathbf{u} + \beta \mathbf{w}) = \alpha L(\mathbf{u}) + \beta L(\mathbf{v})[/math] I just don't see the appeal of this method.

I don't really want to step in and ruin this thread, but there's no need for things like this around here. As we clearly state in the rules, ad hominem attacks are just not on. In short: if you have valid arguments, then go right ahead and post them. However, if you're not willing to take constructive criticism and don't want to post in a respectable, mature fashion then please don't bother posting at all. We can do without personal insults on here, thanks.

Indeed. We've had a thread on this before: http://www.scienceforums.net/forums/showthread.php?t=5219

I don't think there are any more myself for the first one. As for the second, I have no idea

The reason we don't allow deletions is for those people that insist on posting things then going back and deleting everything after people don't agree with them (we've had a couple of instances). Also, I don't think vB supports e-mails upon deletion - we'd also be sending out a load of e-mails to people who post junk/advertise around here as well.

I would assume so. It's a model that uses Evolutionary Game Theory to predict ESS's and whatnot.

Very sorry, I got my limit the wrong way around. It should be: [math]\lim_{h \to 0} \frac{\sqrt{2(x+h) - 1} - \sqrt{2x-1}}{h}[/math] This gives the correct answer of [math]\frac{1}{\sqrt{2x-1}}[/math].

\wedge: [math]\wedge[/math]. I'm fairly certain that if you google for "LaTeX symbols", you'll be able to find a whole host of them.

In regards to your other question; the proper way of doing it, I suppose, is to say: "[math] \vec A \bullet \vec B \equiv |\vec A| |\vec B| cos(\theta) [/math], where [math]\theta[/math] is the angle between vectors A and B." However, your shorthand appears to be okay to me.

\neg: [math]\neg[/math].

Er, right. That above post just didn't make sense. Also, please note that infinity is not a number. It's a concept. I have half a mind to go around and delete posts that make use of such "facts" that 1/infinity = 0. It's not right and it doesn't really have any meaning at all.

The question says that the electon is in its ground state.

*sigh* Doesn't anyone read stickies anymore?

I think the idea was to see what was wrong with his "proof" of sorts.

Fair enough. I just don't have time to go through every single integration (I have my own exams to worry about). I've also never seen that particular method of integration so I stand corrected. However, I do stand by the fact that the integral is not trivial

Yeah, the addition would take quite a long time. An infinitely long time in fact - it's a non-integer power.

What? I can assure you that the answer is most certainly not trivial. Just a cursory glance at what Mathematica gives will reveal this much. Another 10 minutes working on the problem will confirm it

My friend, who likes to play with fans, did the following: 1) Located his computer near a window. 2) Fed a tube out of the window. 3) Attached a large, 120cm fan (which was modded to give more speed) to the computer case. 4) Attach the hose to the fan. 5) Profit. Ice-cold air = best cooling ever. However, you want to be careful that someone doesn't pour a load of water down it for a laugh and/or chuck stones down it

Tsk, all these decimal things The answer I got, however, is 91.3223cm2, so I don't know who's right now

Yes, the view of "EllipticF" signifies that this isn't exactly an easy integral

My solution uses integration, and I'm sure it's not the easiest way of doing it. Look at the image below for what I'm trying to accomplish. Basically, we get the following integral for the shaded area: [math]A = \int_0^5 \sqrt{100-x^2} \, dx = \tfrac{25}{6} (3\sqrt{3} + 2\pi)[/math] Then the area of one unshaded section is [math]50 - A = \tfrac{25}{6}(12 - 3\sqrt{3} - 2\pi)[/math]. Multiplying this by 4 and then subtracting everything from 100 will give you the answer, which I get to be: [math]\tfrac{50}{3}(3\sqrt{3} +2\pi -6)[/math]

This is fairly straightforward. You should get, as your limit: [math]\lim_{h \to 0} \frac{\sqrt{2x-1}-\sqrt{2(x+h) - 1}}{h}[/math] Now, by multiplying by the conjugate of the top, we get: [math]\lim_{h \to 0} \frac{-2h}{h(\sqrt{2x-1}+\sqrt{2(x+h) - 1})}[/math] This should give you your answer.

A barrel with a large rocket engine on the bottom of it would be better than the shuttles atm. Hope this doesn't screw things up again though. They really need some new machines - I mean, these ones have been in service since I was born, and that was 20 years ago.

rofl. Funniest thing I've read all week

It was some kind of racial abuse at a meeting they had in December. I can't remember now, but they both stood up and said something silly.