@DanielFischer I'm missing your point when you say "Sequentially closed sets need not be closed"...I know that is true, I just don't see how it connects with what I stated previously?
I didn't know whether you knew that. Anyway, that was an example how the family of all sequences in $A$ differs from the family of all nets in $A$. The thing is that $\mathbb{N}$ is small, so you have a relatively little supply of sequences.
@Khallil It happens to be so for $\sin$, but generally, the negativity of the second derivative on an interval doesn't imply that it is decreasing on any subinterval.
Hi mse people! Quick question: Can we retrieve the eigenvalues of a symmetric positive-definite matrix, $A$, given its Cholesky decomposition, $A=L^TL$? Thanks a lot!
@Pedro @DanielF: Regarding the Brazilian Fields medalist, one of my math Facebook friends posted this article, with the comment that this is the first Fields medalist to pose topless :D
@nullgeppetto: I thought it was usually done $L^\top D L$ with $1$'s on the diagonal of $L$? What conditions do you have on the diagonal elements?
@TedShifrin, I am not sure I see your point... Cholesky gives $L^\top L$, where $L$ is lower triangle. I think that I would take $L^\top D L$ is I applied SVD... Probably I miss something here...
@joe For the first part you need sinv and cosu but you are only given sinu and cosv that gives a relationship between sin and cos...so sin^2(u) + cos^2(u) = 1 gives you cos(u) and you can do the same for sin(v),
$2$ is a primitive element $ \text{mod} p$ for $p=3,5,11,13,19,21,...,$ and for the remaining primes, $2^{(p-1)/2}=1$. Anyone have any idea how to prove this (if it is true)?
@Chris'ssis runs down like a smoothie .. step 1: stolz .. step 2: change of variables + writing down as a series , step 3: estimating series as reimann sum .. $$\frac{2\arcsin 1}{5}\int_0^1 \frac{x^2}{x+1}\,dx$$ :)
maybe long calculation .. but doable :P .. I need to think of a shorter/simpler one :|
@Alyosha This is the so-called "Euler's criterion" as we know it. In general, $\left( \frac{x}{p} \right) = x^{(p-1)/2} \mod p$
You can prove this by noting that $x^{p-1} - 1 = 0$ in $\Bbb Z/p\Bbb Z$ and that it factors as $(x^{(p-1)/2} - 1)(x^{(p-1)/2} + 1)$. The converse is patently obvious. So your problem follows as a mere consequence.
@skullpatrol What's pitiful is that the guys who write the articles in the newspaper in here don't know any math. They said that the Lagrange four-square theorem is actually proved by Ramanujan with no mention of Lagrange whatsoever and the higher-power generalization (the so-called Waring problem, really solved by Euler) is what Bhargava proved. What a bag of gibberish.
It's good that some Indian descended won the Fields medal, but it is absolutely not good to override the credits to the great people who proved stuffs decades ago by Bhargava either.
@JasperLoy Do you know of a good antivirus I can download from internet quick and free? My machine is not quite protected right now and I think it might be good to set up a security. The last system crashed because of a malware invasion.
@BalarkaSen If you are using Windows, the most natural antivirus is from Microsoft itself: search for Microsoft Security Essentials and install the right version.
Apparently his work with PDEs (partial differential equations) might 'end up' in the study of the climate due to it's progression in understanding randomness in a different way.
Of course, that all came from a BBC webpage, haha!
@BalarkaSen: I'm currently watching some Richard Dawkins videos, of which this (youtube.com/watch?v=5of4A7DGQvw) is my favourite!
