Dave

I think you're thinking of a method used to calculate the deteminant of a matrix. Inverses of matrices are done fairly easily by row reduction. Determinants are just as easy with the method you describe - I can go into it if you want.

Basically LaTeX is a program that allows you to properly typeset mathematics. On top of this, there are packages that you can use to extend this to typesetting chemistry equations, and we use something called mhchem. It's very easy to use, and hopefully it'll make sure that people definately know what you're going on about when you post something here. I know it helps a lot in the math and physics forums - typesetting can sometimes be a problem there. I hope it gets used here as well

You can try factoring it, but you're probably not going to remove the sin2. I'm afraid I don't have time to look at this problem further atm; I shall look at it in a few hours time.

The method I gave will give it in terms of sin(a).

Well, yes. But Zeno was saying that you can only take a finite number of steps, and that summation involves a limit which is never actually attained

Indeed. [math]2\sin(a)\cos(a) = \sin(2a)[/math]. Substitute this directly into your formula and use the fact that [math]\cos^2(x) = 1 - \sin^2(x)[/math].

Once you've decided on the features you want (CPU type, RAM type, etc) then it's pretty easy to go around and find the appropriate components. Just check the specs of the individual components and you should be fine and dandy. Also, there's plenty of online resources out there for this kind of thing. Remember: google is your friend

I was thinking about using subfora at some point. I don't know though, have to think about that one.

I'm pretty sure there's a couple of threads on it.

Yeah, there's no problems there. It's just 2000 and XP that cause all the problems

It's not in the PC bit. PC3200 = DDR400 = 400MHz.

Not a problem That one didn't seem to have a solid answer, and it's always good for these kind of threads to have one.

That might be (I certainly didn't know that), but the majority of posts that we get around here containing infinity aren't used in the proper context. Maybe I'm just being picky, I don't know.

I just have a nack for remembering theorems. I don't know how to explain it.

Let [math]f:[a,b] \to \mathbb{R}[/math]. If f is differentiable on (a,b) and continuous on [a,b] then there exists [math]c \in (a,b)[/math] such that: [math]f'© = \frac{f(b) - f(a)}{b-a}[/math]

My problem with that definition is that it doesn't really imply how f acts on x. I don't quite understand the notation of B[(x,y) in f] either. At some point, you're going to have to start defining things in terms of other things. In my opinion, relations are rather a nice, neat and concise way of doing this and follow pretty much from set theory. I don't feel that I'm qualified enough to say much else, I'm afraid. I haven't worked with set theory very much at all over the past few years and my first order logic is rather sketchy at best. FYI, if a student was comfortable with spacial co-ordinates and not much else, I wouldn't start throwing abstract maths at them

There is that. However, plugging the integral into mathematica tells me the answer is 87.162, so you may have evaluated it wrongly; the working looks okay though. Indeed.

I was trying to show it in easier terms that everyone would understand. If I wanted to prove it formally I'd use the MVT and that fact above.

I never really implied that did I? If you want to define it a bit more thoroughly you might think about it in terms of an equivalence relation or some other binary relation. Have a look at: http://www.cs.odu.edu/~toida/nerzic/content/function/definitions.html

If you're not prepared to put up with the boring stuff don't expect to understand the interesting stuff as well as you should A function takes an element in some set A, and relates it to a function in a set B. That's about all there is to it, as far as I'm aware. Why start a thread on "linear algebra review" if what you want to learn has nothing to do with linear algebra?

It's a closed integral. Basically, you can integrate functions along paths in the plane. Think of a wire in three dimensions; you might have some function that tells you the density of the wire at some particular point. To find the total density of the wire, you'd have to do a line integral along the length of the wire. When the two ends of the wire are connected to each other, you're integrating around a closed loop and you can use that symbol.

Nope, [math]70 \leq F(t) \leq 90[/math] (80 -10 doesn't equal 80 ). Remember that sin is periodic. arcsin will give you one answer because to be a well-defined function it can only give you one answer. Once you find one value, it's easy to find the next value by looking at the symmetry of the graph. Also, did you have your calculator on degrees mode? That sounds about right to me. Think about this one for a second: the derivative will give you the gradient of the curve. If you have a horizontal tangent, what does this imply?

Okay, this is fine. Just as long as you understand that there's not really a lot to them; they're just a way of corresponding elements in one set with elements in another set. I don't get what you mean by "exactly one". I don't really see how this fits into the picture either. From personal experience - this won't get you anywhere fast. Concentrate on learning a single topic first, then move onto other things. Functions and set theory is a bit of a mountain by itself. If you want to learn about linear transformations, then you need to get your head around the concept of the vector space. You then need to concentrate on learning about linear independence, spanning and basis vectors. After this, you might want to detour and read up on subspaces, which are rather interesting. Then, move onto linear transformations and from there you can get a solid grounding.