Dave

Well, there's lots of things one can mean by a 3d graph, but I'll assume that you're going to talk about a function like z = x+y which defines some surface. Basically, they're exactly the same as 2d graphs. Imagine you have some surface, and you take a cross-section through that surface (let's say this is parallel to the x-axis). Then, you effectively have some 2d graph for every value of x for one value of y. So you can build up the graph of the function by plotting lots of these curves for every value of y. That's all it effectively does.

We'll post the IP address in the notice we put up when the boards on this server are shut down. Yeah, I'm sorry for all of that other downtime. The new setup seems a lot more reliable, and I've certainly tried to make it fairly robust.

I don't think this is the approach you should be taking. The very last integral you state is quite non-obvious to integrate at first sight. However, you can solve your original equation by simply letting [math]u = a\sinh t[/math]. Then it transforms into: [math]a^2 \int \cosh^2 t \, dt[/math] by using [math]\cosh^2 t - \sinh^2 t = 1[/math]. Evaluating this integral is easy by using the double angle formulae for cosh2 in terms of cosh(2x). The result follows from taking the inverse of your substitution.

I see where you're going with this, but I'm afraid that this is ill-defined. What you currently have is a function [math]f_{x_0} : \mathbb{N} \to \mathbb{R}[/math], defined by: [math]f_{x_0}(n) = \underbrace{\tan \circ \cdots \circ \tan}_{n \text{ times}}(x_0)[/math] And what you wish to do is extend the domain of this function from the natural numbers to all real numbers. Unfortunately, without some kind of special form for [math]f_{x_0}[/math], you can effectively pick any extension you like and it won't necessarily be the 'correct' extension. Please get back to me if you think I'm wrong. For the second half of your question, you might wish to consider using something like Matlab or its open-source alternative, GNU Octave. They're pretty powerful and are probably capable of doing this sort of thing.

That's a lot better, and far more readable.

I like Phi's suggestion, I think they do need to be swapped. I'll also try to get the alternating colours sorted. However, feel free to use the default vBulletin style until we get things sorted out.

The function you've described could be quite interesting. Generally, these iterated functions can be investigated using cobweb diagrams. Take a look at the Logistic map, which is somewhat simpler than the function you're looking at and introduces some of the simpler ideas behind this sort of theory.

Downtime and Upgrades Hi guys. Firstly I have to apologise for the large amounts of downtime that we've been suffering, and the beta style which we've had to implement for the time being. With a bit of luck, in the next week or so we will be moving away from our current servers, so hopefully this will be the last significant period of downtime you'll have to endure. The reason for this latest downtime is twofold. Firstly, the forum software needed a security upgrade. Secondly, we've tried to re-organise the forum index in order to make it easier for you all to post. Some forums have disappeared because they were particularly unpopular. However, for the most part, they have just been moved around. Don't fear - General Discussions is still here, but at the bottom of the page. Amateur Science A new section is the amateur science section. We're hoping that it will prove to be popular; it's a place for everybody to discuss any experiment ideas they have, or maybe try to get feedback on stuff they've already done. It's not constrained to just one discipline, so if you're interested, don't be afraid to post! Anyway, I hope you enjoy the changes, and please post your comments/feedback in this thread.

I'm somewhat reticent to take their data at face value without some sort of analysis of the algorithm. But even at a quick glance, it doesn't look like a significant improvement; much less than 1% for the larger range of random numbers, which is all we really care about.

It's quite beta right now obviously We're working on it, hopefully it should be ready within the next few days/end of next week.

If you have a single equation and two unknowns, it's going to be impossible to get two equations defining x and y out of it. You can get x in terms of y, or y in terms of x; that's about it.

Just a really quite, pedantic note. You should be careful when using multiple factorial symbols, as 4!! is generally regarded as the double factorial: [math]n!! = \begin{cases} n(n-2) \cdots 5 \cdot 3 \cdot 1, & n \text{ odd} \\ n(n-2) \cdots 4 \cdot 2, & n \text{ even} \end{cases}[/math] Indeed, multiple factorial symbols are generally regarded as multifactorials. Just so you know

That statement doesn't really make sense and needs some kind of clarification. Do you mean the vectors forming the columns/rows of the matrix?

To be honest, it's not that much of an issue. I'm buying a Mac in the summer, and I intend to have a moderately small XP partition for a few games, but everything else will be done under OS X. There's an excellent bundle of applications, plus my other main stuff (Photoshop, photo editing stuff etc) all runs under OS X natively. I'm hoping that WINE (or Crossover Office) will progress enough in the next few months to run Windows apps without needing XP to be installed at all. But that may be asking too much for the time being

If one takes two vectors in [math]v_1, v_2 \in \mathbb{R}^3[/math], then indeed the subspace generated by the spanning set V of v1 and v2 has dimension of at most 2. If the dimension is indeed 2 (i.e. v1 and v2 are linearly independent), then there exists a bijection between [math]\mathbb{R}^2[/math] and V. So V is "equal" to [math]\mathbb{R}^2[/math] up to a bijection. This doesn't mean the spanning space actually is [math]\mathbb{R}^2[/math], so be careful as to how you phrase your wording in this regard. Secondly, is it possible for this subspace to be equal to [math]\mathbb{R}^3[/math]? No. Clearly [math]\mathbb{R}^3[/math] has dimension 3. As discussed previously, the maximal dimension of V is 2, and so the complement of V in [math]\mathbb{R}^3[/math] must be non-empty. I don't quite have the time to answer your other questions right now, but I will hopefully be able to do so a little later. Does this help at all?

Not quite. [math]i^{2n+3} = (i^2)^{n}i^3 = (-1)^n (-i) = (-1)^{n+1}i[/math].

I'm afraid I have to disagree. Discrete mathematics has an important part to play in computer science, but that certainly does not imply that the rest of mathematics is completely pointless. For instance, the study of partial differential equations and modelling of fluid flow relies totally on the Navier-Stokes equations. Whilst the only way of solving these (in general) is to indeed discretise the problem, without the study of continuous functions one could say very little about the solutions obtained, nor about any behaviour observed in a quantitative fashion. Likewise, whilst quantum mechanics relies on the fact that energy is discretized, it is seldom employed when engineers need to construct bridges. In fact, most of the mathematics studied during an engineering degree will likely consist of calculus and real analysis. Both of these fields are extremely important, not only in the mathematical community, but in the scientific community in general. From my (albeit limited) perspective, this is the way things are, and they are not likely to change any time in the near future.

Thanks for the proof! My group theory isn't so good and I don't generally have the time to answer these questions in detail, but I hate to leave a thread with no answer at all.

Okay, first split the sum up: [math]\sum_{l=0}^{n-1} (2l+1) = 2\sum_{l=0}^{n-1} l + \sum_{l=0}^{n-1} 1[/math] Now, the right hand summation is easy: [math]\sum_{l=0}^{n-1} 1 = n[/math]. For the left-hand summation; it's a well known fact that [math]1 + 2 + \cdots + n = \frac{n(n+1)}{2}[/math] (I can prove this if you want, but there's millions of proofs out there). So, we get [math]2\sum_{l=0}^{n-1} l = n(n-1) = n^2 - n[/math]. Adding the two sums together we get the final answer.

If we let [math]u = \cos x[/math] then [math] du = -\sin x \, dx[/math]; hence our integral is transformed into: [math]-\int \frac{1}{u^2} \, du = \sec u + c[/math]

Actually, since the summation starts from n = 0, the first term will be 100 = 1. Try to work out an expression for the summation, then square it.

To avoid me reposting everything about 70 million times, I wrote my views on my blog. Either way, it's an amazing phone - going to shake up the industry, that's for sure.

I'm glad it worked out okay I have no idea whether that formula works or not, but if you want to make doubly sure I would try to prove it by induction.

You'll be wanting to use Sylow's theorem for this. It's such an easy application that I don't really want to do it for you.

The easiest way is probably to show that the group is generated by a single element.