Dave

Forgive me for cutting off your post at this point, but I believe this is the part that is most relevant to what I am about to say. Essentially, you are doing what statisticians and meteorologists all around the globe do every day. You're attempting to extrapolate what's going to happen in the next n days from existing data. Now, ask yourself this: how often do the weather guys get it wrong? My experience from watching weather reports around here is that it's wrong about half of the time. And that's by using huge computer simulations and all of the advanced prediction techniques at their disposal. Additionally, consider that given all of these methods at their disposal, they are only able to predict accurate weather for probably the next 24-48 hours. Hopefully this is the message that Pangloss is trying to convey. One can only observe trends in statistical data; these may be right and they may be wrong, as many stock traders out there will no doubt know. Additionally, you are trying to observe trends for a system in which you do not know all of the variables. An oil company could dig up an absolutely huge deposit tomorrow and it would nullify your 10-15 years claim immediately. There is no way that anybody can possibly predict this. I have not really taken part in this discussion so far so I shouldn't try to comment too much. But just bear in mind that data sets do not give you the complete picture. As good as the data looks, it can be completely wrong. Additionally consider that the evidence will not speak for itself - you need to provide clear, statistical reasons as to why you think your extrapolation is in any way justified. Meer observation of the data will simply not suffice.

Fixed, was a CSS bug (line-height not set correctly).

I don't have any knowledge of electrical engineering at all really. If you get ideas and are stuck on specific mathematical things then I might be able to help. I can certainly try to explain eigenvalues and eigenvectors if that might be of any use.

Yes, otherwise the other computers would not be able to access the internet. I know of software that allows you to limit the bandwidth, but none which requires any kind of password-protection.

This thread makes me feel old. Although in fairness, it was created before I even joined up.

Yes, you're correct. I can do a simple demonstration with something like the Cantor set if people are interested.

I need to go through the entire thing and sort the whole document out. But I can't at the moment, as the computer it's on is back at my halls of residence and I'm at home.

More generally we're interested in something like the Hausdorff dimension for a fractal, but I shouldn't derail the topic. It does make sense to have some notion of volume, It's not meant to be a practical idea (I assumed, as the OP posted in Maths a physical answer would have been a bit pointless). But it does form part of an extremely important theory of measures, which we can use to define integrals and a whole bunch of other stuff.

My money's on "CompletelyConfusedKid"

I just checked, it works, honest! You might need to hard-refresh the page. Or try http://www.scienceforums.net/forum/chat.php by hand. Or you might not be using the new style like you should be!

I haven't added the redirect rules for the new server yet. I'll put it on the list of things to do.

I've not plugged IRC in a long, long time. However, I've spent a couple of hours today implementing a cool new feature, one which I hope will attract some more of you on there! For those that don't know; we do have an IRC chatroom for real-time science chats (sometimes) and general miscellaneous conversation. Some of the moderators and administrators can be found on there, and so if you need instant feedback it could be useful to visit. To access the IRC chat, just click the SFN Chatroom link on the navigation bar at the top of the page, next the User CP link. Anyway, the cool new feature of the day is featured at the bottom of the forum homepage. If you scroll down and look just below the active users, you will see a list of people who are currently on IRC. Hopefully, you'll find this useful and come and chat more.. Hopefully see you there! Edit: One thing I would quickly like to say is that not everybody you see will always be around, so don't get too impatient if nobody's talking! Leave your computer on for a couple hours and idle like the rest of us

True However it won't work so well if you're trying to measure the volume of, let's say, a Sierpinski pyramid.

Essentially this is the idea. You cover your shape by rectangles, and make an approximation. However, the inf term I stated above is the greatest lower bound. So the answer is exact; it's sort of like taking a limit.

For anybody who's on an undergraduate mathematics degree and wants to learn about some of the theory behind fractal geometry, I highly recommend Fractal Geometry: Mathematical Foundations and Applications (2nd ed) by Falconer. It's a really excellent book.

This is quite an interesting question. Since you posted in the mathematics forum, I'm assuming you want a maths-style answer, so I'll attempt to give you one There's a number of ways that one might define an area/volume. Probably the most intuitive is an area of mathematics called measure theory, originally created by Lebesgue at the turn of the 20th century. One object that we have a clear definition of area/volume for is a rectangle (or oblong in three dimensions). It's clear that if we have a rectangle [math]R = \{ (x,y) \ | \ a \leq x \leq b, c \leq y \leq d \}[/math] with [math]b > a, d > c[/math] then its area is simply given by (b-c)*(c-d). Now, the problem of measuring the area of any set can be approached as follows. We cover for the set by a union of disjoint rectangles. Then we look at smaller and smaller covers and try to find the greatest lower bound of the area. In more mathematical notation, we define [math]\mu^* : \mathbb{P}(\mathbb{R}^2) \to \mathbb{R}[/math] by: [math]\mu^*(A) = \inf_{\cup_k R_k \supset A} \left\{ \sum_k (b_k - a_k)(d_k - c_k) \right\}[/math] where [math]R_k = \{ (x,y) \ | \ a_k \leq x \leq b_k, c_k \leq y \leq d_k \}[/math]. This can be easily extended to calculate volumes of sets in [math]\mathbb{R}^3[/math]. I'm terribly sorry if you don't get this - I'll try to provide a more intuitive post if so!

I would use expansion by minors: [math]\left| \begin{array}{cccc} 10 & -10 & 0 & 0 \\ -10 & 17 & -2 & -5 \\ 0 & -2 & 7 & -1 \\ 0 & 5 & -1 & 26 \end{array} \right| = 10 \left| \begin{array}{ccc} 17 & -2 & -5 \\ -2 & 7 & -1 \\ 5 & -1 & 26 \end{array} \right| + 10 \left| \begin{array}{ccc} -10 & -2 & -5 \\ 0 & 7 & -1 \\ 0& -1 & 26 \end{array} \right|[/math] Notice the change in sign on the second determinant.

I won't provide full solutions, but I will give you hints. The hardest step here is a single application of Green's theorem. You should get that: [math]\int_{C} F® \, dr = \int_0^3 \int_1^2\left( \frac{\partial}{\partial x} (ye^x) - \frac{\partial}{\partial y}(x \log y) \right) \, dy \, dx [/math] (I would appreciate someone double checking this because it's been a long time since I've used Green's theorem). The double integral on the right hand side is easy to evaluate. Try to look at what you've got there. You should know that the equation [math]x^2+y^2 = a^2[/math] gives you a circle of radius a. [math]y \geq 0[/math] means that we're going to be reduced to half a circle. And [math]0 \leq z \leq h[/math] simply means that we take this half-circle and extend it upwards by length h. So, S is basically just a cylinder of height h which has been cut in half. Hopefully this well help you visualise what's going on. Well, this is a bit tricky but simple once you've worked out what to do. If we have a function [math]f(x,y)[/math], then [math]\nabla f(x,y) = \left( \frac{\partial f}{\partial x}(x,y),\frac{\partial f}{\partial y}(x,y) \right)[/math] So, if we want [math]\nabla f = F[/math], then f must satisfy: [math]\frac{\partial f}{\partial x} = x \log y, \frac{\partial f}{\partial y} = y e^x[/math] Now, if we simply integrate the first equation with respect to x, the second with respect to y we obtain the desired result. (Be careful about integration with respect to a single variable; instead of a constant of integration, you get a function which is dependent upon the variable you didn't integrate with respect to. e.g. in the first equation we get f(x,y) = ... + g(y)).

Yeah, front page is a little borked at the moment. Hopefully this'll get sorted reasonably soon.

I'd appreciate any constructive feedback on the changes just made after server move. In my opinion, the most drastic change is the lack of link underlines, which some people may not like. The rest is fairly cosmetic and mostly the same as before.

The server is back up with an updated version of the new style. Please give us feedback on speed and the style in the style thread. Thanks!

So you're all aware: server will be going down in about 15-20 minutes.

Yeah, I'm not totally happy with either colour right now. The problem is trying to find a shade colour which will work with both light and dark backgrounds. I decided to cop out and I've made the bar slightly lighter (on the development server) as well as mking a few other changes to make things align properly.

I'll see what I can do about contrast with quotes. I agree they should be a bit darker; I would also like to see a bit more spacing between that and the text. Right now I'm working on the front page; the server move should happen later on today, and I'll probably encorporate any changes that I make to the style tomorrow. And it would have worked too, had it not been for those pesky...

Although I may change this