Dave

My own personal opinion is that these people have severe mental health problems and are extremely misguided. I do believe they just need to be segregated from society - Bettina's opinion of them is rather a good example of why. Other than that... what else is there to say? (Well, rather a lot I would imagine, but I don't think I'm in a position to say anything of great value).

At the end of the day, if you want decent webhosting, you're going to have to pay for it or host it yourself. Often the latter really isn't an option though

That's a much better explanation It's what I was trying to say, only in more detail

BBC News for me. el Reg is also checked throughout the day

Indeed.

I will say this now as to avoid further confusion. We will not tolerate posts made in regards to production of hazardous/deadly materials. It is quite obvious from reading this thread that these instructions could, in the wrong hands, give quite appauling consequences. In fact, this is a matter that we take very seriously, and are currently discussing in the moderator forum. Please remember that not everyone reading these forums are responsible adults. You may inadvertantly cause accidents, or worse, by posting material ascertaining to the production of hazardous chemicals. akcapr: Consider this a formal warning. We won't tolerate this kind of behaviour. This thread is closed. Anyone who thinks this thread should be re-opened should either PM or e-mail me.

The router is "smart" as someone said earlier; it keeps track of the outgoing connections and decides where incoming data should be sent. It's what NAT is all about.

Indeed Only just saw this thread, but glad that things are looking up.

I don't think I've got the adequate knowledge to answer that properly. The way I visualized it was to imagine a particle on a line, but that's probably wrong

I should probably point out that the Schrodinger equation is often introduced for a wavefunction of only two variables; x and t. i.e. we only consider what happens in one-dimensional space. In this case, the equation simplifies down to: [math] -\frac{ \hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + U \psi = i \hbar \frac{\partial \psi}{\partial t} [/math]

I'd think the problem is a bit more widespread than that. A lot of people buy copies of XP from your local market thinking that the thing they're picking up at a "discount" price is going to be a legal version; obviously it won't be. Plus, you have to take into account the businesses that use illegal site copies. This is where they're losing most of their money. I think the idea of having to pay for updates is just outrageous. It's more or less equivalent to blackmailing people - "if you don't pay for the updates then your machine will get h4x0rd!" What they neglect to mention is that most of these updates are security fixes for problems that have arisen during the development of their code; effectively they're charging you for fixing their mistakes. Edit: didn't notice that nobody mentioned paying for updates, only that they might require product authentication. Whoops

I'd choose NTFS over FAT32 any day of the week to be honest.

5614: There are keygens for XP. It hasn't affected anyone that I know of at all.

You only need to forward ports if you're running servers behind the router. I have a seperate webserver running on my other machine, and I portforward a port to that.

snore2walk: Please quote the URL of the place you're referencing from in the future. We don't like plagarism here Thanks.

My condolences to him personally. I thought he was a great guy.

Well, the TCP/IP protocol is smart, not necessarily the routers themselves

I think I'll wait before I post the next few threads. There's no point going on for the time being.

Well, they're obviously not equivalent except for one special case. For a function [math]f:\mathbb{R} \to \mathbb{R}[/math] we define the derivative by: [math]\frac{df}{dx} = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}[/math] For a function [math]f:\mathbb{R}^n \to \mathbb{R}[/math] we can define a whole host of partial derivatives for each component xi: [math]\frac{\partial f}{\partial x_1} = \lim_{h \to 0} \frac{f(x_1+h,x_2,\dots, x_n) - f(x_1,\dots,x_n)}{h}[/math] [math]\frac{\partial f}{\partial x_2} = \lim_{h \to 0} \frac{f(x_1, x_2+h, x_3,\dots, x_n) - f(x_1,\dots,x_n)}{h}[/math] And so on. The easy answer is that if n = 1, then the two definitions are equivalent.

...or not.

The tuition fees thing is just a vote-grabber. I can't see it happening myself. Yes, I found that to be rather amusing.

Not a problem It's useful to know integration by parts, but things like this can save you a lot of time in the exam room. It's a pretty common question to be asked.

Your typical NAT setup on a bog-standard router assumes that anyone can connect to anywhere they like and that all incoming connections are blocked. You've set up port 2302 to be port forwarded to your computer, so all incoming connections are sent there. It's like saying that because you're forwarded port 80 to your computer, nobody else can use the web Edit: I just realised that this might answer your question. When you "connect" to someone you effectively set up a stream between them and you so that data can be transferred. I don't know how to explain this better; sorry

No. The former is called a partial derivatives and deals with function of more than one variable. I'd rather not get into that though.

Whoa there tiger. A much easier way is to realise that: [math]\sin(mx)\sin(nx) = \tfrac{1}{2}\left(\cos(m-n)x - \cos(m+n)x \right)[/math] The proof will drop out fairly trivially then.