Dave

Just to make a point, I don't "auto-reply" to anything on these forums. However, I misinterpreted what exactly you were asking for, so I apologise. I think this idea might annoy a lot of people, and moreover I think it might be very hard to implement. I'll look around for options, but I'm not prepared to change the current system unless there's a significant amount of people that want this, or whether there's some kind of optional hack that can be enabled on a per-user basis.

At the bottom of the home-page is a "Mark all forums read" link. On each forum, you can click the "Forum Tools" pop-up menu and get an option saying "Mark this forum read".

From what I can see, Python has the standard C-style bitwise operators (i.e. <<, >> for shifting, | & ^ for OR, AND and XOR). If you're interested in 3D programming, there's the Visualization Toolkit (VTK) which is a relatively easy way of displaying 3D stuff. Either that, or you can just use a Python OpenGL wrapper. Basically there's a lot of bindings out there for Python as it seems to be one of the languages of choice over there. A lot of the cooler things like GStreamer have bindings, as well as GTK, QT and a whole host of other things. Personally, I don't particularly like the language, although I am going to learn it since I'm taking a module which is partially based off of Python in the Autumn term. But it seems to be a handy and easy to learn language, so go for it

I have to agree with Phi and the other opponents. I am certainly of the opinion that the death penalty is not something that one should have in a civilized society. My main problem is the complete and total finality of it; there's no way of bringing someone back to life if after they're dead. If you cannot be 100% sure that the person you are going to kill is guilty, then sooner or later innocent people going to die. And that is cold blooded murder, which ever way you look at it.

What did you get as the power series?

Well, assuming you don't have to give a precise answer, you could just evaluate the first few terms of the power series to get a decent approximation.

Basically, yes. More formally, if [imath]a_n[/imath] is a decreasing sequence, then [imath]a_1 \geq a_n[/imath] for all n, and if, moreover, it is convergent then [imath]\lim_{n\to\infty} a_n \leq a_n[/imath].

n! = n * (n-1) * (n-2) * ... * 3 * 2 * 1 It's called a factorial.

Sure, they can say they broke their computer - just not that they need help fixing it

Incidentally I've been talking about "the" bounds for the sequence. In fact, for a bounded sequence there are infinitely many upper and lower bounds. When I say "the" upper bound, I'm actually talking about the least upper bound - i.e. the supremum, if you've covered that at all.

Your sequence is a series of discrete points [math](a_1, a_2, a_3, \dots)[/math]. An upper bound is a number M such that [math]a_n \leq M[/math] for any N. A lower bound is a number m such that [math]a_n \geq m[/math] for any N. Notice that upper and lower bounds have no dependency on n at all. Now, yes, your sequence does converge to zero. But that doesn't mean that it's decreasing. For example, take [imath]a_n = \frac{(-1)^n}{n}[/imath]. This is a sequence which tends to zero, but is not strictly decreasing. To show a sequence is stricly decreasing, you need to show that [imath]a_{n+1} - a_n < 0[/imath] for all n - which is what I have done above. Basically, now that you know it's strictly decreasing, you know that you're not going to have places where the sequence might jump up and then start to fall down again, so you can say that the upper bound for the sequence is simply [imath]a_1[/imath]. The lower bound will trivially be zero.

I guess we're going to have to wait and see what sanctions - if any - the UN decides to impose on Iran. I'm still holding out hope for a diplomatic solution, but I have a feeling that this probably isn't going to end particularly well.

That's "the Microsoft Way" I'm not a big fan of this restictive hosting type thing. I'd much rather pay a small amount for some semi-decent hosting and a domain than be restricted to using somebody else's templates.

Unfortunately this is a relatively new/improved feature in Firefox 1.5, and as such it may be the case that there are memory leaks occurring here. Just thought I'd mention it for the sake of saying it.

Yes, just to reiterate the point: If you decide you want to do this and break your computer in the process, don't come crying to us about it. A little harsh, but I feel it needs to be said.

To be honest, I think it's a reasonably sensible idea to keep backup copies and system restore points. Working in Linux can sometimes be a pain; it's very easy to delete a fundamental library by accident or have glibc installations die on you, and all of a sudden nothing works. At least for the sacrifice of a few gigs of space, you have a certain amount of protection against this sort of thing.

I don't think they'd allow this somehow. It seems to be primarily geared to those who have little to no experience with web-authoring.

Because Firefox isn't embedded deep into the OS like IE Firefox will keep a selection of webpages in memory, since most people want to hit the back button a few times, and don't want the page to necessarily have to be parsed again. Pretty much all modern browsers do these days. I can't explain your Firefox woes, but I'm glad the IE toolbar is working out for you Let's just not have this turn into another Firefox vs. IE debate, since we've been down that road many, many times before.

Well, the ratio test should tell you that this sequence is null; i.e. the limit is zero. Since all of the terms are positive, you should be able to get a lower bound trivially. Moreover, [imath]a_{n+1} - a_n = \frac{2^{n-1}}{(n+1)!} \left( \frac{2}{n+2} - 1\right) < 0[/imath] for any n > 0. So the sequence is strictly decreasing, and it should be easy to find an upper bound from here.

Well, you should know the power series expansion for cosine and sine. Work out the power series expansion for cos(t3) and sin(t2), and then integrate term-by-term. You can work out the answer by plugging the first few terms into a calculator.

Given the type of assignment he's giving, I'm betting this is a precursor to the chain rule - which is why I explicity didn't say so

Here's a hint. If you have a function of the form [math]f(x) = (x^n+1)^m[/math], let [math]u(x) = x^n + 1[/math]. Then we get a function [math]f(u) = u^m[/math]. Construct an equation for [math]\frac{df}{dx}[/math] by considering [math]\frac{df}{du}[/math] and [math]\frac{du}{dx}[/math].

Personally, I believe that it would have been the smart thing to do. If they truly wanted uranium enriched for the sole purposes of nuclear power, then they could have perhaps considered it as a possibility. If not from Russia, then another supplier, or multiple suppliers. The point I was trying to convey, albeit poorly, is that there have been a lot of options thrown on the table which have all been rejected. They are clearly determined to perform the enrichment by themselves, and moreover they know that this isn't going to be looked kindly upon by the international community. So one has to wonder whether their intentions are honest at all.

I'm coming to the end of my third year at university, and all of these guys are right when they say that you have nothing to worry about. You are in total control of how you want to spend your time, and if that is just sitting in your room working or going out for a drink, then that's entirely fine. Everybody on my corridor in the first year was absolutely great, and it has probably been the best experience of my life so far.