Dave

Essentially, yes

Dak, my suggestion is to think of something that you want/need to code that is reasonably simple, then do a google search. Chances are that you'll find an open-source project that already does what you want, and you can start hacking away. At least, that's what I did. Hmm. Don't think that's quite what he's after.

Don't worry about not getting the 3D stuff, it's not essential for you at this point. You need a bit more solid grounding in vector calculus to really get going with higher dimensional integrals.

Since I probably won't be able to explain it any better, there is a general rule for this operation - more generally known as differentiating under the integral. See the Wikipedia page for a bit more information.

The answer is yes, for far easier reasons than the one described above (although they are correct). Essentially, a graph is one particular way of describing a function. In fact, the definition is explicitly dependent upon a function - the graph of a function [math]f : A \to B[/math] is defined by: [math]G(f) = \{ (x, f(x)) \ | \ x \in A \}[/math] So in other words, without a function the term 'graph' is meaningless. On the other hand, if you're saying "if I draw a line on a piece of paper, can I find a function which represents it", then the answer is 'not necessarily'. If that line is self-intersecting, then there is no possible way that it could represent the graph of the function, since the function would not be well-defined.

Congratulations to all of the winners and thanks to all those who took the time to vote. With a bit of luck I'll actually get around to creating some graphics for you all or something

If you are seriously interested in learning mathematics at a university level, you should consider reading Foundations of Mathematics by Stuart and Tall. It was the set text for the foundations course at Warwick and does a good job of introducing some of the elementary building blocks of mathematics (simple logic, set theory, proof by contradiction/induction, equivalence relations etc).

i.e. in the intersection of the two sets

Having been able to go into the common room of the maths department here at Warwick this year, I'd have to go for the usually option. Mathematics is a very social subject, there's a lot of sharing of knowledge and certainly colloboration is encouraged more than some other disciplines. Unlike physics, where there are patents and such to be had, there is very little to gain from stealing ideas from others and so there is much more social interaction.

It all depends on what setting you pick. If you take the real number line, then you can project this onto [math]S^1 \backslash \{a\}[/math], and then a is called 'the point at infinity', giving the projective line. So in this sense, "[math]\infty = -\infty[/math]". Conversely, since [math]\mathbb{R}[/math] is a totally ordered set it might be more convenient to define the extended reals: [math]\overline{\mathbb{R}} = \mathbb{R}\cup\{\infty\}\cup\{-\infty\}[/math]. The Riemann sphere is exactly the same thing as identifying a point at infinity by using stereographic projection onto the sphere.

You can always just install ubuntu-minimal and configure flux from there.

Okay, so if I'm understanding you correctly, you're unsure about this step: If [math]A(x) = \int_a^x f(x) \, dx[/math] then [math]\frac{d}{dx} A(x) = f(x)[/math] I think you may be getting a little confused with the notation that's being used. All this is saying is that if we evaluate an indefinite integral, then its derivative is the function being integrated. Let's do a simple example, using [math]f(x) = x^2[/math]. [math]A(x) = \int_a^x f(t) \, dt = \int_a^x t^2 \, dt = \frac{x^3}{3} - \frac{a^3}{3}[/math] Notice that the value of a is some constant that we are free to pick (i.e. it defines a constant of integration). Hopefully this term is familiar. Now, when we differentiate A, we get: [math]\frac{d}{dx} A(x) = \frac{d}{dx} \left( \frac{x^3}{3} - \frac{a^3}{3} \right) = \frac{d}{dx} \left( \frac{x^3}{3} \right) = x^2 = f(x)[/math] which is precisely the result that we are guaranteed by the FTC. I hope this helps you a bit!

Looks like Atheist's proof was slightly more convoluted than I first thought. You don't need to use prime decomposition, only that odd and even numbers are adjacent.

I don't see how it doesn't cover the 0/1 combination at all; 0 is divisible by any non-zero integer just by definition. Even if you don't believe this, you have just agreed that 0 is an even number, and hence it must be divisible by 2 by definition anyway. For a slightly more formal proof, try the following. Let [imath]a \in \mathbb{Z}[/imath] be any integer. Let us assume that a is even. So there exists [imath]k \in \mathbb{Z}[/imath] such that [imath]a=2k[/imath]. Hence, [imath]a(a+1) = 2k(2k+1) = 4k^2 + 2k[/imath]. Clearly 2 divides the right hand side, and so [imath]a(a+1)[/imath] is divisible by 2. A similar argument follows for a odd.

Sorry for the downtime yesterday; unfortunately a bug in the VPS software used by our provider managed to knock out our primary IP address so the website was inaccessible for quite a few hours yesterday. Hopefully it's been fixed, and shouldn't theoretically happen again.

That might be the case, but I've always seen SFN as a place where people can learn new things without fear of being shot down. Being 'stupid' doesn't mean that you shouldn't be able to post - being ignorant and ignoring what everyone's telling you is a completely different point. Clearly matt grime didn't like this and so he left. That's his choice, but I certainly won't be petitioning to get him back. I'll take this into consideration - please PM me with ideas for people you'd like to see as experts. I'd love to see some more posting in the maths forums, it would certainly motivate me to answer more often.

I'm pretty sure the second integral is wrong there. (It should be [math]u'v[/math] and you have [math]u'v'[/math]).

It doesn't quite work like that: [math]\int_a^b uv' \, dx = [uv]_a^b - \int_a^b u'v\, dx[/math]

I would guess from your description that it's probably just another type of saddle point. I'm honestly not sure - I haven't done enough on tensor theory to be able to comment. Any physics guys care to answer? The Hessian of a function [math]f:\mathbb{R}^n \to \mathbb{R}[/math] is just the nxn matrix of all second order partial derivatives.

Classification of stationary points is not particularly simple, but at least for smooth functions I'm reasonably sure that there are only three types: those being local minima, local maxima and saddle point. To answer the first part of your question: we know that if the Hessian of a function at a stationary point is positive definite then the point is a minimum, but the converse is not necessarily true. Essentially, if our Hessian is degenerate (i.e. at least one zero eigenvalue) then we'll have to resort to more rudimentary methods. Generally this involves considering the partial derivatives in each direction in a neighbourhood around the stationary point. Finally, I'm pretty sure that a stationary point that is additionally a point of inflection is actually a saddle point. Check out the Wikipedia entry.

Bignose is quite correct; vector division isn't defined in any form so you can't use your stated "identity".

I'm not keen on the newer style and positioning of the username, but I guess we can just keep it as a subtheme that people can choose if they want. Personally I'm quite liking the new SFN theme but I'm slightly biased

Moved this to the Education section, hopefully you'll get a better response there

Hi guys, I know that this won't concern a lot of you (unless you happen to run your own vBulletin forums), but I'm releasing the plugin that displays the lovely LaTeX scripting into the wild. I know some people are interested in this, hence the announcement. You can grab it here, but be warned it's still quite beta and liable to break!

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.