Dave

I've found Wikipedia is a good reference for many general equations of the Fourier transform (although you have to be a little careful of the normalization that you're using). Anyway, yes, the convolution identity you've got up there will work in reverse and I believe the proof will work in exactly the same way.

Certainly it's true in my experience. I've found that mental arithmetic is of very little consequence in a mathematics degree, since the emphasis is on proof. So as long as you know how to do addition, you're fine; whether or not you can necessarily do it is not really the point.

Indeed. My mental arithmetic has declined immensely since completing my degree

Could you elaborate on precisely what you mean by "cone"?

It may be more helpful to omit any kind of dot or cross notation. When I learned the product rule back at A-level we used: [math]\frac{d}{dx} (uv) = v \frac{du}{dx} + u \frac{dv}{dx}[/math]

I believe Capn was upgrading vB to the latest version. This may or may not cure the bug, I guess. This is one of the known side-effects, yes Others include a sudden obsession to count the floral patterns on random plates to make sure they match

You use \ln. You can click on the latex image to display the code, just so you know.

Yes please. This thread should have been closed a looong time ago.

Unfortunately I don't have time to do the whole thing, but I've done a few of the questions and that should be something, at least. Hopefully somebody else can come and fill in the blanks. I don't think this is correct, as the area you're actually integrating over isn't correct. However, I've been known to be wrong myself with these things. To be sure, split the triangle up into two parts, and then the answer should be: [math]\int_{0}^{\frac{1}{2}} \int_{0}^{2x} x \, dy dx + \int_{\frac{1}{2}}^1 \int_{0}^{2-2x} x \, dy dx[/math] [math]\nabla f(\vec{x}) = (y^2-4x, 2xy)[/math], and so [math]\nabla_{(3,-4)} f(2, -1) = (3, -4) \cdot \nabla f(2, -1) = (3, -4) \cdot (-7, -4) = -21 + 16 = -5[/math] (a) is correct since [imath]\sum a_n[/imath] will converge by the alternating series test, but [imath]\sum |a_n|[/imath] is similar to the harmonic series and so will diverge. (b) Not quite. It's the same behaviour as the series above, only in another guise. [math]\sqrt{n}[/math] and [math]\log(n)[/math] are both strictly increasing functions, so [math]a_n = 1/\sqrt{n}\log(n)[/math] will be a strictly decreasing sequence. Hence, by the alternating series test, it will converge. However, the absolute series is not convergent, since [math]\log n \leq \sqrt{n} \Rightarrow \frac{1}{\log n} \geq \frac{1}{\sqrt{n}} \Rightarrow \frac{1}{\sqrt{n} \log n} \geq \frac{1}{n}[/math] So, by the comparison test, [imath]\sum |a_n| \geq \sum \frac{1}{n}[/imath] and the series diverges. © looks correct but I haven't checked it. I get the same answer. Yeah, it looks like you've tried to change into cylindrical polar co-ordinates when in fact spherical will be easier to work with. All you have to do is calculate the integrals over the two spheres and then subtract to get the region in-between. So, for the smaller sphere, I apply the transformation: [math]x = r \sin\theta \cos\varphi, y = r\sin\theta \sin\varphi, z = r\cos\theta[/math] and then evaluate: [math]\int_{0}^{2\pi} \int_{0}^{\pi} \int_0^1 r\cos\theta \cdot r^2 \sin\theta dr d\varphi d\theta[/math] Didn't check the stability but fixed points are correct. I think the answer is actually: [math]\int^{3}_{0}\int^{3y}_{y^2}2x dxdy[/math] Have a look at the plot and orient your head so that you can see what's going on. If you put x = 0, then your limit becomes [math]\lim_{x,y \to 0,0}\frac{y^2}{y^2} = 1[/math] Similarly if I put y = 0, then we get [math]\lim_{x,y \to 0,0}\frac{x^2}{x^2} = 1[/math] So these agree. However, if I put x = y, we get: [math]\lim_{x,y \to 0,0}\frac{3x^2}{2x^2} = \frac{3}{2}[/math] So you're correct, but your explanation isn't quite there. Looks good.

I quite agree.

I don't get this. Just look up the method on Wikipedia or Google it, there's about a million examples of Gauss-Jordan elimination on there. Two seconds brought up this: http://ceee.rice.edu/Books/CS/chapter2/linear44.html.

Yeah, I'm classifying this as junk until it gets cleaned up. Please re-post only if you fix it to make sense.

As you know, I'm based in the UK so probably not the best judge of this, but I do watch TDS a lot as it's broadcast on More4 here the following day. We also get Fox News which I tune into a few times a month before I'm forced to turn it off to avoid being severely ill from the propaganda. You can't possibly argue that TDS claims to be objective - as swansont said, it's even posted on their website. However, in my mind, it does serve a very useful purpose, and that is to show up the news networks for not covering the important issues of the day. Seriously, when a news network gives Anna Nicole Smith's death the most continuous coverage since the September 11th bombings, you need to be concerned about what they're choosing to broadcast. Frankly, some of the stuff coming out of the networks is shockingly bad, and they need to be called out on it. That this job (from my albeit limited perspective) seems to be left down to a relatively small cable show on Comedy Central is completely out of order, but at least it does a decent job of isolating important stories that either fall through the cracks or aren't given decent coverage by the networks.

The actual font itself is called Computer Modern.

Quite right. Thanks for that.

My server's dead (power failure knocked out the motherboard and now is non-bootable). I have all the logs, but for the time being I have nothing to put a bot onto

Just apply the quotient rule: [math]f(x) = \frac{g(x)}{h(x)} \Rightarrow f'(x) = \frac{h'(x)g(x) - g'(x)h(x)}{(h(x))^2}[/math] So set [imath]g(x) = x[/imath], [imath]h(x) = 1 - \ln(x-1)[/imath], work out their derivatives and go from there.

Could you give the definition of semi-positive? I've not seen that term.

Happy holidays from myself and the rest of the SFN staff - I hope you all have a good one, and a great new year!

Also, you should note that the Gamma function is not defined for any negative integer (I believe there are poles at each one of those points).

Would appreciate it if you shared some of these ideas on the forum instead of linking to ad-riddled pages!

Short answer: there isn't. Unfortunately when we look at the graphs of functions, it sometimes leads us to infer statements which are not true. That's why it's essential to impose logic and rigour to ensure that the results we prove are correct. The truth of the matter is that [imath]\sqrt{x}[/imath] does tend to infinity, but it does it extremely slowly. Here's the proof. We want to show that [imath]f(x) = \sqrt{x}[/imath] is unbounded. What does that mean? Simply, that for any possible value [imath]M \in \mathbb{R}[/imath], there exists another value [imath]m \in \mathbb{R}[/imath] such that [imath]\sqrt{m} > M[/imath]. So, given [imath]M[/imath], we pick any [imath]m > M^2[/imath] - this is obviously possible. Then by taking square roots of the inequality, we get [imath]\sqrt{m} = f(x) > |M|[/imath]. Hence f is unbounded, and since f is strictly increasing ([imath]a < b \Rightarrow f(a) < f(b)[/imath]), we can conclude that [imath]\lim_{x\to\infty} f(x) = \infty[/imath]. That is a rock solid argument - sometimes you should not believe what your eyes are saying!

Missed most of this conversation now, but I thought I'd make a few comments. Generally, when talking about polynomials, we are referring to elements of the space [math]\mathbb{R}[X] = \left\{ \left. \sum_{k=0}^n a_k X^k \ \right| n \in \mathbb{N}, \ a_k \in \mathbb{R}\right\}[/math] Notice that this is a far more general definition of a polynomial: here, X could be a real number, or it could be a function, matrix, etc. So in a sense, the summation notation is incorrect - elements of this space are just finite sequences of real numbers. Indeed, to vaguely refer back to a point that the tree made above, if one assumes [imath]X[/imath] is some non-zero real number [imath]1/x[/imath], this ring gives you finite series of the form [math]\sum_{k=0}^n \frac{a_k}{x^k} \in \mathbb{R}[x^{-1}][/math] (In fact, if you define the space of formal power series [imath]\mathbb{R}[[x]][/imath] by removing the restriction on [imath]n[/imath] being finite and ignore the fact that these series do not always converge, you can prove some cool things in terms of asymptotic relations and Laplace transforms). Defining the degree of a polynomial is then rather trivial. You can define the function [imath]\text{deg} : \mathbb{R}[x] \to \mathbb{N}[/imath] by [imath]\text{deg}(\sum_{k=0}^n a_k X^k) = n[/imath]. In regards to Taylor/Maclaurin series, Klaynos is quite correct in saying that a smooth function cannot be written as a polynomial (which are deemed to be of finite order). However, you can write them as the limiting sequence of polynomials. Essentially, Taylor's theorem says that for any n+1 times differentiable function [imath]f : [a,x] \to \mathbb{R}[/imath], the following expansion is possible: [math]f(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{n!}(x-a)^k + R_n(x)[/math] where the 'remainder' [imath]R_n(x) = O(x^{n+1})[/imath]. For smooth (infinitely differentiable) functions, one can prove that [imath]R_n(x) \to 0[/imath] as [imath]n \to \infty[/imath].

Don't know what this is, but it's certainly not mathematics.