Dave

Yes, that's the additive identity. Generally, I write [imath]0_L[/imath] to avoid confusion between that and the multiplicative identity (which lies in the field over the vector space). In pretty much all the linear algebra I've done, it's called the zero vector, zero element, etc. Simply put, the identity function is [imath]0_L(v) = 0_W \ \forall v \in V[/imath]. This is easily checked. Fix [imath]v \in V, T \in L[/imath]. Then, clearly, [imath]T(v) \in W[/imath]. Hence, [imath]T(v) + 0_L(v) = T(v) + 0_W = T(v)[/imath] since W is a vector space. Let me try and clarify my previous post a bit. Here's what you posted earlier: What I was trying to say, in a rather convoluted fashion, is that there's absolutely no guarantee that you can do this since you know nothing about V and W. If the two vector spaces are disjoint (i.e. [imath]V \cap W = \phi[/imath]) then there's no possible way that the function [imath]0_L[/imath], which maps elements of V into W, could map a vector [imath]v \in V[/imath] to itself. Does this clear it up a bit?

Not too sure about that: [imath]0_L \in L[/imath] necessarily, which means [imath]0_L : V \to W[/imath]. It is quite possible that [imath]V \neq W[/imath] - indeed it is entirely possible that [imath]V \cap W = \phi[/imath] - so [imath]0_L[/imath] can't be the identity function. I'm a little confused as to the original poster's question. If you're taking it as given that L is a vector space, then this is a trivial question because it's one of the axioms (additive inverse). However, if you're actually trying to prove that L is a vector space, then do as HallsofIvy suggests.

Yes, it is. Let me clarify: I'm saying that it is a requirement that [imath]f \circ g = \text{id}_B[/imath] and [imath]g \circ f = \text{id}_A[/imath]. My desire to use the least amount of notation possible unfortunately caused some misunderstanding.

Not quite; it is necessary that [math]f(g(x)) = g(f(x)) = x[/math]. imath tags do display math a little differently (namely they typeset inline maths). For example, [math]\lim_{x\to\infty}[/math] vs. [imath]\lim_{x\to\infty}[/imath]

No. But group theory provides us with some very useful existing results (isomorphism theorems, Sylow's theorems, to name just a few). So if you want to do something useful with your new objects, you're going to have to start from the ground upwards. [This rest of the post is not necessarily related to the tree in particular, or the quote above]. The amount of sarcasm and use of rhetorical questions in this thread is way over the top and completely unnecessary. If you must banter over such issues then try not to offend everybody when you press the post button. Acting high and mighty simply makes you look unnecessarily arrogant.

There's very little to learn past this point Essentially, we just define [math]a=\log_b c[/math] as the number which satisfies the equation [imath]b^a = c[/imath]. That's all it is - we can derive everything else about logarithms from this point. Perhaps if you identified precisely what's troubling you I could help more?

Presuming you have two bijections, [imath]f : A \to B[/imath] and [imath]g : B \to A[/imath], then one takes the composite of the functions. If [imath]f \circ g = \text{id}_B[/imath] and [imath]g \circ f = \text{id}_A[/imath] (where [imath]\text{id}_X[/imath] is the identify function [imath]\text{id}_X (x) = x[/imath]) then [math]g = f^{-1}[/math].

Man, I don't know what I was thinking when I wrote that stuff about Hausdorff dimension. It's not like I took a course on it or anything Hopefully the stuff I wrote about the Cantor set won't be a load of twaddle either, otherwise I should just shoot myself now. Edit: Annoyed at myself for being such an idiot, so thought I'd explain why I got so confused. Aside from the fact I haven't touched it in a while, the Hausdorff dimension relies on the s-dimensional Hausdorff measure [imath]\mathcal{H}[/imath] to actually determine the dimension (namely, [math]s = \text{dim}_{\text{H}}(G) \Leftrightarrow s = \inf \{ t > 0 \ | \ \mathcal{H}^t (G) = \infty \}[/math]). I plucked infinity out of the air from this, rather stupidly.

Well that may or may not be right. My idea was to explicitly solve the ODE, then the resulting function y(x) should resemble an exponential function. Then find the Taylor expansion of y to find the co-efficients [math]a_k[/math].

You're virtually there. When you substitute y(x) into the equation, one has that: [math]y'(x) + 2y(x) = 0[/math] This is a separable ODE which can easily be solved. You just need to figure out what to do with the constant that comes along with it (some kind of initial condition would be handy, but you haven't stated there is one).

I'd just like to point out that, since this was posted on the mathematics fora, I would ask that any discussion of the topic remain solely on mathematical interpretations of the word (i.e. not physically related).

You should probably note that this is a triviality, since the first summation has a factor of k involved: clearly it is the case that: [math]\sum_{k=1}^{\infty} ka_kx^{k-1} = \sum_{k=0}^{\infty} ka_kx^{k-1}[/math] I don't think the question you've asked is quite worded correctly: here is my corrected version, you need to tell me whether it's right or not: If this is what you actually meant, then write [math]f(x) = \sum_{k=0}^{\infty} a_k x^k[/math], and try to re-write the equation in terms of f. From there it is simple.

When first reading this post, I'd originally intended to write quite a long reply as it's a very interesting question. But the more I thought about it, the more I've realised that I actually have very little understanding of how to discuss this kind of thing. But I'll try to sum up my thoughts here. Let's consider the vector space [math]\mathbb{R}^n[/math] (under the operation of addition). Then, since we can span the space by precisely [math]n[/math] vectors, we say that it has dimension n. Intuitively, this makes sense: one can imagine n different degrees of freedom. Edit: This is rubbish, ignore Now, I'm just going to assume you know what both the box dimension and Hausdorff dimension are. If you don't, you'll need to look then up. But rest assured, they are very sensibly defined, and work quite well. For instance, on the space [math][0,1] = \{ x \in \mathbb{R} \ | \ 0 \leq x \leq 1 \}[/math] we have that [math]\text{dim}_{\text{B}} [0,1] = \text{dim}_{\text{H}} [0,1] = 1[/math], which again, makes sense - one degree of freedom. But when we apply this to [math]\mathbb{R}^n[/math], we find [math]\text{dim}_{\text{B}} (\mathbb{R}^n) = \text{dim}_{\text{H}} (\mathbb{R}^n) = \infty[/math]. This doesn't really tally with what we did before - quite a glaring inconsistency, in fact. This is, as far as I know, correct. Not only this, but indeed the box/Hausdorff dimensions do not agree for particular sets. For instance, the set [imath]G = \{ 0, 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots \}[/imath] has [imath] \text{dim}_{\text{H}} \, G = 0[/imath] (because it is countable) and [imath]\text{dim}_{\text{B}} \, G = \frac{1}{2}[/imath]. Don't really know what else to say, so I'll end my response here. I'll try to answer questions you have though That's a pretty good intuitive interpretation of dimension. (Essentially what I outlined in the post above). However, consider something like the Cantor set. At first glance, it's a set which has Lebesgue measure 0; this should indicate that it's essentially a set of 'points' (something like [math]\{0, 1/2, 3/4, \dots\}[/math]), so should have dimension 0, the same as a point. Indeed, the Cantor set is totally disconnected. But then you take a closer look: every point in the set is an accumulation point. That is, if we're at a point in the Cantor set and look in any (small) neighbourhood, we'll find at least one other point there. In fact, the space is uncountable. So it essentially boils down to this: is it sensible to define a notion of dimension where an uncountable set is allocated dimension 0? In my mind, probably not. Certainly the dimension should be < 1, but in my mind it should also be > 0. Under both the Hausdorff and box dimension we actually have the dimension being [math]\log 2/\log 3[/math], which seems reasonably sensible. Anyway, this isn't a criticism, just trying to highlight how difficult identifying a sensible notion for dimension actually is.

Just finished a Mathematics MMath degree, going on to do a PhD.

Working here...

My advice is to remember that when you buy a DSLR, you're buying into the system whether it's Canon, Nikon, Olympus etc. Once you have the body, then you'll want a telephoto/wide-angle/fast prime/etc lens as well as flashguns, battery grips, remotes, memory cards and a whole load of other junk. So ensure you get the camera maker that you're most confident in, otherwise you'll have to sell up and start all over again since the brands aren't really compatible. I did my research and decided that with Canon making some of the best gear on the market today, and the fact that the Digital Rebel XT/350D was the best introductory DSLR at the time, I'd go with them. The camera is great, and I also got two additional (albeit expensive) lenses. The XTi is a very good SLR for a beginner, although you might want to consider the XT. It's still in production (or at least there's a lot of stock left) and about $100-$150 cheaper thant the XTi. It all depends on whether you want the extra features or whether you want to put the extra money towards new lenses. Regarding lenses... stick with the 18-55 until you know what type of photography you're into, then make your decision as to what lenses you want to purchase. Buying good quality lenses is far more important than buying a more expensive camera body. Don't rush into it, or you'll end up regretting not buying a better lens, trust me

Got mine four years ago Good luck for anybody that has GCSE/AS/A2 results out soon.

It may have been/probably was moved from another section into this forum.

I was going to point out the very same fact, but you appear to have gotten here first Additionally - since we are talking about topology here - two spaces are generally topologically 'equivalent' if you can construct a homeomorphism between them. Or maybe you could define equivalence by constructing a homology, or perhaps doing some clever tricks with identification spaces. There's lots of ways in which you might wish to define equivalence.

Not sure whether Snail's post will clarify so I'm adding my own comment here. The simplest way of thinking about the dimensionality of [math]S^2[/math] is to consider a 'small' square on the surface of the sphere. Because the section is small, to all intents and purposes it would look like a square in the unit plane (which is of two dimensions). It is this property which makes [math]S^2[/math] a 2-dimensional (smooth) manifold. Indeed, we can construct an explicit homeomorphism [math]f:[0,2\pi) \times [0,\pi] \to S^2[/math] by [math]f(\theta, \varphi) = (\cos\theta \sin\varphi, \sin\theta\sin\varphi, \cos\varphi)[/math]. If one considers the domain as a vector space, then it is clearly of dimension two and, f being an isomorphism, so is [math]S^2[/math].

There's nothing wrong with the complex numbers, but finding different ways to transform them can help to simplify problems. Whilst I'm not exactly sure what the OP intended, you could almost certainly connect the two spaces through a homeomorphism making them effectively two representations of the same thing.

MathWorld and Wikipedia are not exhaustive, and should be considered tools at best. I have recently completed a course covering Ecalle's alien derivative which most certainly is a topic in mathematics, but extremely unlikely to appear on Wikipedia.

Following up to this, and given recent staff discussion on the issue, it is highly unlikely that TFN will be reinstated. We're considering options at the moment.

You may wish to consider using the LaTeX feature to typeset the SDE properly: [math]\sigma[/math].

You need to make sure you're calculating it properly. To calculate [math]y= x^{x^{x^x}}[/math], you calculate [math]x_1 = x^x[/math], then [math]x_2 = x^{x_1}[/math], then [math]y = x^{x_2}[/math] (i.e. you start from the top and work down).