Dave

Welcome back, J'Dona Yes, this thread has rather degraded; I don't know whether there's actually much discussion left in it anymore. That's why I stopped posting in it a while back.

That's a pretty conclusive list of things Unfortunately the latex isn't working (for some reason) so I haven't had time to check it through properly. Thanks

Seems to be okay to me. You don't necessarily need the cos(stuff) = cos(bleh) line.

You don't need the \[ and \]. All of the stuff encapsulated by math tags is already put into the displaymath environment.

You don't need to bother with the enclosing \[ and \]. As far as I'm aware, there's no quick way, but if you'd like me to define a new command, I can put that in and then add it to the documentation.

I can start them up again soon if people would like, since I'm on holiday next week

The proof is quite nice; you have to use the fact that: [math]\sum_{k=1}^{n} \left[(k+1)^3-k^3\right] = (n+1)^3-1[/math] which is fairly obvious; just start writing it out and you'll see that all the terms cancel apart from the ones given. Now also note that: [math]\sum_{k=1}^{n} \left[(k+1)^3-k^3\right] = \sum_{k=1}^{n} (3k^2+3k+1)[/math] By swapping things around and equating bits and pieces, you get: [math](n+1)^3-1-n-\tfrac{3}{2}n(n+1) = 3\sum_{k=1}^{n} k^2[/math]. So by simplifying everything down: [math]\sum_{k=1}^{n} k^2 = \frac{n(2n-1)(n+1)}{6}[/math].

I have to say that from a strictly mathematical point of view, we don't really care all that much about what the variables in the equation actually represent. There's nothing in any definition of the derivative of a function to say something completely different for time-based functions.

Seeing that x = 1 is a fairly trivial solution of the equation. Knowing it's the only solution is a bit of a beast though; the way I thought of doing it was to look at the function f(x) = 3x + log(x) - 3, then plot f(x) on a graph and see where it intersects the x-axis (it only does it at one place). It's not really a "proof" though.

He shouldn't have said that, because it doesn't really make sense. A better way (and what I believe he probably meant) is "one, two, three, and so on".

Bleh. I don't know about the pinging out YT, it only seems to affect you for some reason. You could try connecting to a different port, or maybe check for problems your end. As for the applet, I really don't know when the last time it actually worked. Maybe blike can fix it?

Unfortunately, we had a problem yesterday with the IRC server; it died unexpectedly and took the services down with it. We've got the server back up now, but services haven't been brought back up because of some other problems. Hopefully, they should be back up soon. Sorry for any inconvenience.

It's a rather clever idea, because once you get a few people interested, they're going to e-mail their friends to sign up, and then they'll e-mail their friends, etc. They've gotta have a lot of users by now if (seemingly) most of the people I know both online and irl have got these e-mails. Plus it seems to be quite a useful service. However, I already have my address book online

You might want to try Avid FreeDV (http://www.avid.com), I've heard it's pretty good for just bolting video together. The interface takes a bit of getting used to, but (imo) it's the best of any video editor out there.

It's hard enough trying to project a three-dimensional image onto a two-dimensional surface, let alone something that's four-dimensional

Damn, forgot I was that ugly

I missed the "in Java" bit, sorry. In most of the programming languages I know, a char is only 1 byte, but fair enough

It means that the number is always going to be positive. There's no bit to signify whether the number is positive or negative. A char is just a 1-byte data type (holds integers from 0-255), and the associated integers are matched up with the ASCII table.

Personally I would say that nothing could beat a good book, some paper and a few hours worth of work Online stuff doesn't really cut it for me.

Why even bother? Just go and grab yourself a cheap old pentium 3 for 50 quid.

I think you're missing a factor of 2 there somewhere. Btw, I got my volume of integration formula wrong at the top, changed it now though.

There's no need for that in this forum. Moving this to the stats forum.

exp is the exponential function - (e^log(x)) = x

The equation of your ellipse is (in this case) [math]\frac{x^2}{16} + \frac{y^2}{a^2} = 1[/math] where a is the semi-major length. Consider the rotation about the x-axis of the ellipse: [math]V = \pi\int_{-2}^{2} y^2 \, dx = a \int_{-2}^{2} 1-\frac{x^2}{16}\,dx[/math]. If you know that [math]V = \frac{800\pi}{27}[/math] then by evaluating the integral you can calculate a precisely.

w00t! Cheers blike