Dave

I thought about this a bit, but it's clear that the column-space and null-space must be intersection-free. So I presume that you mean rank(A) = nullity(A), in which case the answer is pretty straight-forward; just choose a matrix like [math]\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}[/math]

I'm a bit late to the discussion, but I'd like to point out that this is entirely correct. When taking a limit, you're considering regions centered around the point of interest, so your function at the very least must be defined there. Now, in your proof, you take [imath]\lim_{x\to 0} \log x[/imath] which is not defined, since log is only defined on the half-line [imath](0,\infty)[/imath]. So the evaluation of that limit is not possible.

Matrices are really quite useful, but probably not in most of the things you're doing at the moment. Essentially, they are representations of linear maps under a particular basis, and multiplication of matrices corresponds to the composition of two linear maps (that's why it's such a funny operation). Anyway. The only thing I really wanted to say is that you shouldn't use the word 'matrixs' - the plural of matrix is matrices

Wow, that really is quite bad. It's just completely insane that people would go out there and actually say that Obama is a communist. I mean, that just doesn't even make any sense at all. Unfortunately, the Republican party appears to be extremely good at subtly encouraging this kind of behaviour by motivating irrational fear by essentially telling a load of porkies. I hate politics sometimes.

It might be a bit hard to visualise, but I suggest an analytic approach. Take the power series of [imath]e^x[/imath], and substitute [imath]x=i\theta[/imath]. You'll soon see power series for both sine and cosine appear as if by magic

Sigh. Please, use the search button. This topic has been covered so extensively on these forums that I am really sick and tired of closing them.

Essentially the Laplace transform is just a function itself; it takes one function, then integrates it to get another: [math]\mathcal{L}[f](s) = \int_0^{\infty} e^{-st} f(t) \, dt[/math] So, to calculate the Laplace transform of a function f, one simply calculates the integral on the right hand side to get another function in terms of s. For example, for the function [math]f(t) = e^{-t}[/math], we have that the Laplace transform is: [math]\mathcal{L}[f](s) = \int_0^{\infty} e^{-st} e^{-t} \, dt = \int_0^\infty e^{-(s+1)t} \, dt[/math] so [math]\mathcal{L}[f](s) = \left[ \frac{e^{-(s+1)t}}{s+1} \right]_\infty^0 = \frac{1}{s+1}[/math]. The Laplace transform is generally useful for finding the solutions to various ODEs, the reason being that: [math]\mathcal{L}[f'](s) = s\mathcal{L}[f](s) - f(0)[/math] You can prove this by writing out the left hand side, and then using integration by parts. So, for the equation that you have to solve, we can apply the Laplace transform to both sides: [math]\mathcal{L}[f'(t) + f(t)](s) = \mathcal{L}[e^{-t}](s)[/math] Now the Laplace transform is linear (easy to prove) so you can split up the left hand side. Also we use the transform I proved earlier for [imath]e^{-t}[/imath]: [math]\mathcal{L}[f'](s) + \mathcal[f](t)(s) = \frac{1}{s+1}[/math] Use the property of the Laplace transform of a derivative to get: [math]s\mathcal{L}[f](s) - f(0) + \mathcal{L}[f](s) = \frac{1}{s+1}[/math] and finally we simplify everything down, using the fact that [imath]f(0) = 0[/imath]: [math](s+1)\mathcal{L}[f](s) = \frac{1}{s+1} \Rightarrow \mathcal{L}[f](s) = \frac{1}{(s+1)^2}[/math] So the problem boils down to: can we find an f whose Laplace transform is [imath](s+1)^{-2}[/imath]? Or, in other words, what is the inverse Laplace transform of [imath](s+1)^{-2}[/imath]? Unfortunately the inverse Laplace transform is difficult to calculate, so basically you have to look it up in a table. In our case, we're lucky and this is a common inverse: [math]f(t) = te^{-t}[/math] You can find the list on Wikipedia - just search for Laplace Transform. If you put this into the original equation, you can see that this is indeed a solution to the ODE problem. Hope this helps.

I'm really not sure just how useful this would be - what about when you start trying to find primes with only a few hundred digits? The factorial at this point will be absolutely enormous. Finding primes with a few thousand digits would likely prove computationally inefficient.

Also, a lot of our local council money (~£820m) is tied up in Iceland at the moment. So 'comical' really isn't the word I'd use to describe it, really.

Almost certainly the easiest way is to use Dak's method (defining a list of strings). I should think that, in Python, a program like this would be able to be coded in a few lines.

Thanks to you all for the feedback! Currently my thoughts are to get rid of homework help, and replace it with the tagging scheme that Klaynos has suggested. I think, after some consideration, that it would probably provide the kind of system that we're all fairly happy with. Since there's been no real opposition to the change in forum layout, I may go and implement that tonight. Thanks for all your help!

Unfortunately part of the problem is that today, there is apparently an auction to deal with the credit default swaps outstanding from Lehman Brothers, and apparently there is a good chance of that being somewhere in the $400bn mark. If that is the case, then that's going to have a majorly negative impact, and the $700bn bailout is suddenly looking quite small

Thanks for the link. I expected that it was only a matter of time really.

To be honest, I think Iceland is in far greater economic trouble than Pakistan for the time being. Their top three major banks have gone under in the last week, all being heavily dependent upon debt to keep themselves afloat, and a good proportion of that in foreign currency apparently. I've read that the banks have about $300k of debt for every person in Iceland, and their loans totalled something like 5-6 times as much as they have in deposits. Their currency tanked and the government has had to bail them all out. From reading some of the US news sites, it's not been widely reported which I guess I can understand, with there only being about 320,000 people in the country, but a lot of UK savers (~300,000) have got money over there which has all been frozen by (I guess) the Icelandic government. The banks are essentially entirely reliant in that money remaining in place because it's their main deposit base. Unfortunately the UK government hasn't taken kindly to all of this, and whilst they've guaranteed all savers getting all of their money back, they're not too impressed with Iceland, and have used anti-terrorism (!) laws to freeze as many Icelandic assets that they can get their hands on in order to get some of the money back. Very sorry state of affairs over there at the moment.

Tsadi, no bother. It's just we've had bad experiences with people deleting all of their posts and ruining the thread continuity, hence the policy

Thanks for the feedback so far guys, much appreciated. Don't worry, that will still be there! This image is of a little development forum which obviously doesn't have that many posts. I hacked around a little bit to get it to show post/thread counts, downside is there's no thread link with the hack. Generally the finished product would look a little different to that as we'll be using the styles we have here. Since most of the comments so far have been about the HH forum, I should probably point out some of my own reasoning for wanting to get rid of it. I don't like the entire "this is a homework problem, we're not going to help you so do it yourself" attitude that the no homework policy brings about. It doesn't really encourage people to come here and ask questions, and at the end of the day people need help with their problems in order to advance. If they're not getting the proper support from their teachers or supervisors, then they're going to rely on places like this to get some assistance. Even if they don't stay here, Google certainly has a lot of answers out there. Lots of these questions in HH are going unanswered. I think that part of this is that we have gone to great pains to separate out all of the science-related topics and create new forums for them, but a homework problem of any subject type is just bunged in one generic forum. It reduces the number of replies that any poster might get, because (I assume) often people tend to stick to one area on here and don't want to trawl through trying to find relevant threads.

The idea is to just allow people to answer the questions how they want: full answers or not. I don't really see a problem with that.

The SFN staff have been contemplating a number of changes to take the forums forward and we want your input on a couple of things. Forum front page It's generally felt that the layout might be a bit long and cluttered for people to get through. So we're proposing to condense the front page: take a look at the attachment at the bottom of this post, and tell us what you think of the potential layout. Homework Help Our proposal is to remove Homework Help entirely, and let members ask homework questions anywhere. My own personal feeling is that this will help members to post more, and attract more people into the forums. What do you think? Please use the thread in the Suggestions and Comments forum to make your opinions known!

This is the thread for discussing the changes outlined in the announcements forum. Let us know what you think about both the forum front page and homework help.

It just follows from the definition of nCr. For example, for nC3: [math]\binom{n}{k} = \frac{n!}{k!(n-k)!} \Rightarrow \binom{n}{3} = \frac{n!}{3!(n-3)!}[/math] Now if we write this out, simply notice that most of the terms on the top and bottom cancel each other out: [math]\binom{n}{3} = \frac{1}{3!} \cdot \frac{n \cdot(n-1) \cdot (n-2) \cdot (n-3) \cdot (n-4) \cdots 2 \cdot 1}{(n-3) \cdot (n-4) \cdots 2 \cdot 1} = \frac{n(n-1)(n-2)}{3!}[/math]

There really isn't anything to prove here - this is just looking at the definitions and matching things up. For example, for proof of linear combination implies subset: [math]w \in \text{span}\{x_1, x_2\} \Rightarrow w = \alpha_1 x_1 + \alpha_2 x_2; \alpha_i \in K \Rightarrow w = \alpha_1 (\beta_1 u + \beta_2 v) + \alpha_2 (\gamma_1 u + \gamma_2 v ); \beta_i, \gamma_i \in K[/math] Hence by letting [math]a_i = \alpha_1\beta_i + \alpha_2\gamma_i \in K[/math], [math]w = a_1 u + a_2 v \Rightarrow w \in \text{span}\{ u, v \} \Rightarrow \text{span}\{x_1,x_2\} \subset \text{span}\{u,v\}[/math]

The only thing that springs to mind (and purely because you said Fermat) was the Taniyama-Shimura conjecture. But this is not really an equation, more of a conjecture about the relationship between elliptic curves and modular forms.

I meant to close this thread last night and forgot. PM me with objections.

Guys, I would like to remind you that this is a mathematics, not relativity, forum. Bearing this in mind, the answer to your question is: no, [imath]\pi[/imath] is irrational because it is a mathematical definition. You can debate 'relativistic circles' or whatever until the cows come home, but that is not mathematics. Unless someone can convince me to keep this thread open, I will close it later on this evening.

Sure, if only it were that simple. But it is now very clear from these hacked e-mails that the idea is make subpoenas and audits of that account hard/impossible to get. Even if the court granted the subpoena request and managed to get them to hand over data (which I imagine would be a lengthy process in the first place), it would be much easier for them to delete the sensitive documents and then claim they were 'accidentally' erased, with no backups kept. There is no defense for these actions.