Dave

That's not actually correct. You can see from the equation [imath]e^{\pi i} = -1[/imath] that ln(-1) can actually be any number [imath](2n+1)\pi i[/imath]. I think the more likely answer is that you've been given or wrongly written down an incorrect set of initial conditions.

I don't see why you're using scanf and printf when you have cin and cout? For example, you can clean pretty much everything up and dispense with stdio.h: #include <iostream> using namespace std; // you need this so that you don't have to do std::cout int main() { float xo,xt,yo,yt; float grad; cout << "Enter the first co-ordinate:" << endl; cin >> xo >> yo; cout << "And the second one:" << endl; cin >> xt >> yt; grad = (yo-yt)/(xo-xt); cout << "Chord of (" << xo << "," << yo << ") and (" << xt << "," << yt << ")" << endl << "Gradient: " << grad << endl; return 0; } Note that the current standard says you should drop the '.h' at the end of standard include files. For instance, iostream.h -> iostream, string.h -> string etc. I've changed your co-ordinates to be floats instead of ints, otherwise you won't get a proper value for the gradient. Also, you need the "using namespace std" since all of the standard streams like cout, cin are all kept in std (read up on namespaces). If you don't include that, you'll get undefined variable errors. Hope this helps.

That's very strange; I just had precisely the same question as this in my Measure Theory assignment last week. It's fairly easy. It's fairly easy to show that [imath]\mu^*(A) = \mu^*(A_{+a})[/imath]. Just consider the set of rectangles covering A and translate them by a, then prove that they provide a cover for A+a. Now if A is measurable, then for every epsilon > 0, there exists an elementary set B such that [imath]\mu(A \triangle B) < \epsilon[/imath] (triangle = symmetric difference of A and B). Consider B+a and by using the definition you have there, show you get the same answer (i.e. [imath]\mu(A_{+a} \triangle B_{+a}) < \epsilon[/imath] - you need to use some set theoretic identities for this. Hope this helps.

I don't think you need to look at the fourth derivative; the third will suffice. This article at Wikipedia might help to clarify matters.

I've got a possible answer to why the question may be wrong. If you use the summation technique posted by builgate you must have the answer to be: [math]\sum_{n=0}^{50} (2n+1) = 51^2 = 2601[/math]. It's quite probable that someone made a numerical mistake, perhaps summing (2n-1) to give an answer of 502 = 2500 by mistake. I've looked at both the methods and can't see a problem, so... I don't know

Frankly you've done a lot of hard work for very little reason. You have already calculated the value of [math]\cos \frac{\pi}{12}[/math]. So now, use a series of easy arithmetic relations to get your answer: [math]\sec \frac{-\pi}{12} = \frac{1}{\cos \frac{-\pi}{12}} = \frac{1}{\cos \frac{\pi}{12}}[/math] Substitute in your answer and you're done.

Frankly, this thread has gone on long enough. Unless people decide to post some sensible, reasoned arguments then I will simply close the thread.

I think it would be wise for Iran to step away from this situation and apologise as much as possible in a very short amount of time. Unfortunately, I don't believe that this is actually going to happen; their leader seems far too arrogant, and will probably not back down easily. The thing that concerns me most at the moment is that Israel isn't going to take these comments lightly. I'm just wondering whether this single comment is going to snowball out of control, and whether we're going to have a very serious issue on our hands. Edit: there's some comments from the Israeli press at BBC News if anyone's interested in reading them.

I'm not too sure about that "and". Personally I think it should be a "whenever", because then that statement is equivalent to: [math]\forall \epsilon > 0 \, \exists \, \delta > 0 \text{ such that } 0 < |x-a| < \delta \Rightarrow |f(x) - L| < \epsilon[/math]. However, this is pretty much completely beyond the scope of this thread, so if you don't understand it, don't worry about it

Well, [math]\sec \frac{-\pi}{12} = \frac{1}{\cos \frac{-\pi}{12}}[/math]. Now since [math]\cos \frac{-\pi}{12} = \cos \frac{\pi}{12}[/math] and you've already evaluated [math]\cos\frac{\pi}{12}[/math] you can work out the answer

Yes, it's another thing on the "to do" list I've been a bit busy recently with uni, and I really haven't had as much time as I had hoped to play around with SFN. I'll sort it in a couple of days.

It has a vertical translation of 3 (the +3 at the end), a phase shift of [imath]2\pi[/imath] and oscillates twice as much as a normal cos graph (the 2x).

Yes, since you have 2 expressions for cos(2a), hence they must be equal.

Well, because [imath]\cos(A+B) = \cos A \cos B - \sin A \sin B[/imath]. It's the standard compound angle formula.

Because [imath]\cos(2x) = \cos(x+x) = \cos^2 x - \sin^2 x = (1-\sin^2 x) - \sin^2 x = 1 - 2\sin^2 x[/imath]. i.e. because sin2 x + cos2x = 1.

You can only edit a post up to 4 hours after you've posted it. It's a rule we have on here to stop people from deliberately removing all of their forum posts, which breaks the continuity of the thread.

I can't see an error in your reasoning. I'll have a look at it tomorrow when I'm not so tired

I'm assuming that this means [imath]\mathcal{T} = \left\{ X, \phi, A \right\}[/imath] for some [imath]A \subset X[/imath]? Or am I being silly?

There's some very nice features in Firefox 1.5 that are coming up. Amongst numerous speed improvements and bug fixes, rendering time for cached pages is a lot faster. There's the automated upgrade system which is much better than having a new installer for each copy of Firefox; it simply patches your application and relevant files instead. SVG support is coming on a treat, and it really does look great. I've been running 1.5b1 and 1.5b2 for a while now with next to no problems. The only major downside is that most of the extensions and themes aren't compatible with the changes they've been making in 1.5. But there is a noticable improvement in both rendering time and just general application performance. I haven't measured the footprint yet, but in terms of RAM it seems to take a little less, at least. By all accounts so far, IE7 is nothing but IE6 with tabbed browsing and PNG transparency. Most of the reviews I've read on b1 indicate that things like the 3px jog and numerous other CSS annoyances are yet to be fixed. In their defence, they say that there will be more CSS fixed in b2, but I'm not convinced. Frankly, if it's taken 4 years for PNG and some tabs, then god only knows how long it's going to take for them to implement a proper CSS model.

Yes, but projective 3-space is a lot different to the set of reals in which we're talking about. In that space, parallel lines are allowed to meet at points labelled infinity, and basically makes everything a lot nicer to work with. Or so I've heard at least - I've never really done all that much with hyperbolic geometry.

I'm not quite sure on my interpretation of your question. Are you asking for all functions f that have a fixed point at R, or the fixed points of the function f, or what?

I'm obviously not saying that at all.

Except for the small point that there is no such thing as n "being" infinity, so neither function can ever attain that value.

Using epsilon-delta? A much easier way is to prove the quotient rule