Dave

Yeah, I've been reading about this around the newspapers. If I remember right there was even a BBC article on it somewhere. Just another thing that Labour have done to irreversibly damage this country :/

With all due respect, I have read your posts quite clearly. I understand your desire for a rigorous proof. But metric spaces and topology in general takes such a basis in geometry that you can derive a great deal of intuition from pictures, especially at simple levels such as proving sets are open.

My purpose here is not to give you the answer, because you won't learn anything from that. It is to provide enough information to give you a good shot at proving it yourself.

The latest news is that Blair will soon testify.

Unfortunately, Blair's testimonial was completely overshadowed by the news that the head of the English football team had been a bit naughty. Which pretty much sums up the amount of political apathy currently present in this country.

While there are a number of people would like to see something similar in the US, my reaction to the UK in regard to Iraq is that the war was even less popular there and generally agitated the population more. At a time when US citizens were almost evenly polarized on the issue it seems Britons were adamantly opposed.

Oh yes. At the time, there were marches consisting of millions throughout the country. Everybody I know was, and still is, opposed to the war. Even with all of the "evidence" presented by Blair et al., there were extremely few people that wanted to get involved.

When the decision was made to go to war, I think the British people decided to pull it together for the troops. But when it became apparent that we weren't supplying them the correct equipment, and the original motivation for going to war was not in fact the case, public favour turned unanimously against it.

It is a good idea to do this inquiry. The widespread belief is that the people were being lied to. And this inquiry will restore (some of) the trust people must have in their government in order for it to function.

I don't think it will - it certainly won't restore my trust. Remember that this government set the remit for the inquiry and chose the members of the board. It feels like this is a political manoeuvre to try and clear the air before the general election to give Labour the best chance of being re-elected.

In any case, I was actually at the filming of Question Time yesterday, which had Tony's best friend on there. Just the smarmyness, the complete lack of any regrets at all, was enough to make my blood boil. And he essentially stated that even if something does come out of this inquiry, in all likelihood no criminal prosecutions would be made.

So I think the best I can hope for is that we get rid of this lot at the next general election.

Okay. So... I don't get it At this point, the problem is really trivial and you should be able to get it with everything i've given you, or at least make some kind of stab at guessing the solution. I'm going to give you a final hint: take a point in (0,1). Say, for simplicity, a half. Then take ANY ball of radius [math]\epsilon[/math]. Is there a point which is in a ball of radius [math]\epsilon[/math] and centred at the origin, but not contained within (0,1)? It will probably help if you draw a picture

Why'd you bother defining f if you don't use it?

But seriously, this problem is fairly straightforward. You should attempt a proof first, and then we'll be able to help some more.

Well that shows you know the definitions. But that's not really the point

I'll show you how to proof that (0,1) is open in [math]\mathbb{R}[/math]. Firstly, take [math]x\in (0,1)[/math]. Then, we want to show that there exists an [math]\epsilon[/math] such that [math]B(x,\epsilon) \subset (0,1)[/math].

So, draw a line and label 0 and 1, then put x somewhere on that line between the two points. To make a ball of radius [math]\epsilon[/math] fit in there, you need [math]\epsilon[/math] to be smaller than the distance from 0 to x, or x to 1; whichever is the smallest. And just to be on the safe side, we can divide that distance by two to make sure the ball still fits.

So let [math]\epsilon = \tfrac{1}{2}\min\{ x, 1-x \}[/math], and you're done. Of course you will want to formalise this argument a bit, but the key argument is there.

In terms of proving it isn't open in [math]\mathbb{R}^2[/math], since you have the extra dimension to play with, this should be easy. It's like saying; can I fit a circle onto a straight line. Of course the answer is no.

Actually the time should be based around GMT now since the server is back in the UK... I'll have to figure that one out. But yes, to receive your personalised time settings you need to remain logged in.

This is something that's been bouncing around the mod forum for some time now. I'm starting to be swayed in the direction of 'this would be a good idea'. I mean, yes, there's the repetition argument, but that can be cured by improving the search functionality, and closing duplicate threads when they pop up. And yes, whilst other sites out there do this, I don't see why we can't do both this AND just our usual science discussions.

This is a bit of an issue for me. I am supposed to travel to the US in March to see a friend of mine who is a postdoc at UNC. I was already having second thoughts - what with them taking ALL of my fingerprints and retaining that indefinitely, having to fill in a buttload of information 72 hours before the flight, and just generally being treated like a piece of crap. So this extra security is probably going to be the straw that breaks the camel's back. I think I'll just put the US on my own no-fly list.

I agree with the most part about Mokele's post about the length of time it takes to get published. In the mathematical world things aren't very different. I know people who've waited 2+ years for their papers to finally be published.

However, I have to say that I recently submitted a letter to PNAS (through the normal peer reviewed route) and was really impressed with the speed. From initial submission to receipt of peer reviews and a decision (in my case corrections) was under two months. So I guess if you can fit what you want to say into a short format, and can find an appropriate journal, you don't have to wait an age to get published.

Yeah. So Victor has previously been on here multiple times with the same 'proof', which eventually ended in a ban because he refused to listed to reason. So please report this if it gets out of hand.

I should probably point out that if the situation doesn't sort itself out in the next few days, then we'll move the server to an alternative data centre which may result in a little more downtime, but much more stable in the long run I guess.

In my mind, you are mostly right. For the standard developer, C++ probably is dead - you give a pretty good list of reasons. However, I'm not a standard developer. The kind of programs I develop are very high-performance related, generally parallelized using an interface like MPI and need to be very efficient. C++ fills a niche in that, as far as I am aware, it is a high-performance object-oriented language - perhaps the only one. So for this reason I don't think the 'true' death of C++ will happen any time soon.

+1 for Python/numpy. Awesome for doing all kinds of data processing.

The first thing to realize regarding the delta function is that it isn't a function.

 

Looking at the delta function (OK, I'm a physicist at heart) as a limit of some series of functions is a very good approach to visualize what is going on. Bignose (post #2) and Severian (post #9) talked about it terms of a normal distribution. That's a very nice way of looking at it, but of course not the way. Some of the reasons this is particularly useful: It's already normalized for you, the normal distribution is analytic everywhere, and it generalizes to multiple dimensions.

This is essentially what I wanted to say. Clearly it's not possible to define an integrable function which has a finite non-zero integral but which is only non-zero at a set of points of measure zero. If you want to know more about the technical aspects of what the delta "function" is, you should actually look up distributions. Assuming you take the correct space of test functions (look up Schwarz space of rapidly reducing functions) then you're guaranteed to do cool stuff like Fourier transforms, and many of the fundamental properties hold as well.

If you find one, then let me know. I tried a while back and found it was next to impossible to find such a thing, for some reason!

Any reason for such a question?

Additionally, if you are a member of a university, then often you will find that the campus will have a site license for many of the major publishers.

No, not really

Late to the thread but I have to say that multitouch on a computer is completely pointless. I saw a demo of it on a laptop and thought "why the hell would you want that on your laptop?" The reason it works so well on the iPhone is because of the size of the thing. I don't want to be dragging my fingers all over my considerably larger laptop screen.

Surface will supposedly have its uses in shops and soforth, but really I'm quite happy with my keyboard and mouse for my computing needs.

Please, if you're not sure of your answer, don't answer.

I wouldn't go that far - just put a label saying you're not sure about it.

Okay... or not.

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]

For this problem, I'd say that [math]\lim_{x\to 0^{+}}\; x^{\sin x} \; = \; 1[/math]

 

The domain has nothing to do with the limit laws, if that was implied. I'm guessing the problem was looking for the answer of 1 regardless, and the right-hand limit wasn't important in evaluation.

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