Dave

That's cool; I think to a certain extent you're right: too many options can lead to it detracting from the posts. But 1 or 2 isn't going to kill anyone.

At the end of the day, they're only custom titles. Just because a certain type of forum has them doesn't mean that this forum will automatically turn into that type.

Howso?

Don't think it has a specific mod; most of our mods are supermods anyway.

Yes, did sound a bit like that in retrospect. Sorry about that

I'm not 100% sure, but the proof goes something along the lines of using that subtraction formula in the definition of continuity.

Hmm. I did it, but I'm not sure it's entirely true. We know that [math]f(2x) = xf(x) + 1[/math]. Now say [math]g(x) = xf(x) + 1 - f(2x)[/math]. Then we have that [math]g'(x) = f(x) + xf'(x) - 2f(2x)[/math] by the chain rule (effectively this is implicit differentiation). By differentiating more and more times, it's pretty obvious that you get [math]g^{(n)}(x) = nf^{(n-1)}(x) + xf^{(n)}(x) - 2^n f^{(n)}(x)[/math] as required.

Download probably went wrong, it's there for me. I know it's certainly not part of the mainstream browser (although it's good enough to be there).

You're hired

An IRC network basically comprises of various servers all linked together to provide a method for people to talk real-time over the internet. Short-hand: it provides chatrooms.

Nope. Basically, this is the definition: A function [math]f(x):\mathbb{R}\to\mathbb{R}[/math] is sequentially continuous at a point c if and only if for any sequence [math](x_n) \to c, f(x_n) \to f©[/math] for some [math]x\in\mathbb{R}[/math]. A consequence of this is that every continuous function is sequentially continuous and vice versa. So for f(x) = x, take some sequence [math]a_n \to c[/math]. Then [math]f(a_n) = a_n[/math], so [math]\lim_{n\to\infty} f(a_n) = \lim_{n\to\infty}a_n = c = f©[/math] as required. This effectively proves that f(x) is continous for all of the reals - similar proof for f(x) = -x.

Must've posted loads of this already - but what the hell. Why don't you guys get on IRC? It's fun for all the family Server: irc.sorcery.net Channel: #physics Be there or be square

Read literally the upside-down A means "for all" - the entire thing means that that condition (the f(x) bit) is true for all x in the set of real numbers. btw, to generate a proper real numbers symbol, use \mathbb: [math]\forall x\in\mathbb{R}[/math]

We should break out the cattle prods and get him back

(sorry about the rubbish ampersand, it's php applying specialchars, try and fix that asap)

I think I told you in your PM, but just for clarity's sake: [math]|x| = \begin{cases}x,& x\geq 0\\ -x,& x \leq 0 \\ \end{cases}[/math]. x and -x are very easy to prove continuous (sequential continuity is probably the easiest way) and if you show the limit exists at 0 (i.e. [math]\lim_{x\to 0} (x) = \lim_{x\to 0} (-x)[/math]) then you've got your proof.

Anyone seen any of the other mods around here recently? greg's not been around for aaaaaaages (as far as I'm aware)

No idea. Has more than enough posts to qualify for supermod though anyway tbh.

Nicely done, I'd not really thought about it like that. Stumped on the first limit though.

Done (1) - pretty obvious. Working on 2, haven't quite got there yet, but I think I know how to do it

Haven't found a solution yet, but I'll post what I have. If you construct your chord and then run a radius to one of the edges of the chord and through the centre of it, you get: [math]\left( \frac{L}{2} \right)^2 +(R-H)^2 = R^2[/math]. Just by looking at the diagram, it's pretty obvious that as L->0 so does H, but I get the limit coming out as [math]\frac{1}{8R-1}[/math]? I dunno. Can't be bothered beating the solution out atm.

And certainly not new.

Lots, mainly pure though. Doing a couple physics modules as well to keep my hand in.

Yeah, I've been looking at that. It looks like a good idea.

Compared to disabling right click with JS (something that can be overcome anyway), it's not that much of a problem.