@Mike the proof itself is not too difficult to master - the hard part really is the toplogical background of it, where (especially the Jordan Curve theorem part) it is usually presented in a 'naive' way.
Feeling quite lovely tonight! Took my girlfriend's kid rock climbing today and am sore as hell, reading about divergent series and waiting for sleep to take hold. ;)
@robjohn you mean I have this option in Acrobat, and I can edit a file using latex formulae? For example, I wanna write a proof that uses latex and put all in a pdf.
@robjohn Thanks for math.stackexchange.com/questions/171970/…, another question, in the the "Therefore," part do you identify the Real and Imaginiary part of $\int_0^\infty \frac{e^{ix}}{\sqrt x}dx$? how can you do so, since there is $\sqrt x$ in the denominator?
If matrix A is a square matrix. And A's characteristic polynomial is $p(t) = t^2 + 1$. It's not necessary true that A is non-singular right? Because, the eigenvalues are $i,-i$. and if dimA= 2 then A is a (2x2) matrix. and Thus, I can put in its non diagonal values values that will give a result that will make A a singular matrix, or even better, the zero matrix which is pure singular (Cuz its determinant is zero) @robjohn
@Studentmath ^^
Or I'm wrong and it the eigenvalues are $+i , -i$ then its determinant is $i* -i = 1$?
@r9m another out-of-the-blue integral inequality... Let $f\in C^1([0,\pi], \mathbb R) $ such that $\int_0^\pi f=0$. Prove that $\forall x\in [0,1], |f(x) |\leq\sqrt{\frac{\pi}3 \int_0^\pi {f'}^2}$
@Chris'ssis: Can you please look at this one too: http://math.stackexchange.com/questions/818494/integral-int-0-infty-frac-sqrt-sqrt-alpha2x2-alpha-exp-big-bet
I am scratching my head from days but I am unable to come up with anything. :(
@r9m The inequalities are again an art where one needs to work a lot to become good. They are fantastic too. (although I didn't work too much on this area in the last period of time)
@r9m The truth around the subject is painful ... (when I'm asked how I mananged to learn things I do now, I try to avoid the answer - I wouldn't like to discourage anyone)
Just think about: you do 50 years of math (supposing) and then you're put down by an inequality (that is not that hard) ... or any other problem. I mean you need to be powerful enough to face this reality.
There is so much stuff around!
For instance, to understand me well, if I only think of a certain type of integral family I might create hundreds of questions, and each one with its trick (of course). Everything is so vast, immense ... in this area.
@r9m Surely, too much knowledge would kill the mystery, but the mystery plays an important role in our life. Probably life would be boring without any mystery. ;)
@IlanAizelmanWS The determinant of $A$ is $1$ since the determinant of a matrix is its characteristic polynomial evaluated at $0$. If its determinant is non-zero, a matrix is non-singular.
@Chris'ssis I bet those will all be workable using the identity I mentioned. I still have not worked on them since I chose to go to sleep last night :-)
@robjohn $f:[0,\infty)\to\mathbb{R}$ is a strictly increasing function such that $f(f(x))=x^2+2$, then how can we disprove (or prove) that $f(3)$ is the same for all possible solutions of $f$?
If $ \{v_1\}^\bot = \{v_2\}^\bot $ , does it mean that $ \{v_1\} = \{v_2\} $ ? Which means that I can also write it as $Sp(v_1) = Sp(v_2)$ which I can say that $\{v1,v2\}$ are linearly dependent?
@EnjoysMath I was creating a rotation matrix to rotate a vector from one to another, this was for the case that they were opposite direction, there are probably nicer ways to do it
the lousy students who took the exam are the ones complaining. Their arguments are: 1. they weren't prepared for the topics covered in the test 2. their calculators couldn't help this time 3. the claims about complex numbers were too hard to prove
@N3buchadnezzar In my opinion, if you don't have fun at your job, you need to quit immediately. (it's about the quality of your life that is much more than the whole money in the world)
@G.T.R It's not a word of itself [well, there is a word "Wart", which means something like warden or keeper, as in Torwart, but that's something different]. "Gegenwart" is the present. Now, this day and age.
@G.T.R The "wart" in "Gegenwart" is a variant of "wärts", which is a suffix indicating direction. The details are hairy, and mostly beyond me and my etymological lexicon.
@G.T.R "Wohin gehst Du" is more common, but there's a subtle difference between the two. "Wohin gehst Du" is [usually, unless with particular emphasis] just a casual enquiry. "Wo gehst Du hin" really wants an answer.
@G.T.R It's just a reinforcing word, if something is very, really very, X, it is "unheimlich X". X may be good or bad, doesn't matter. [And of course, it has its original meaning, eerie, frightening, ...]
@r9m way back in time, it was a theatrical phenomenon where the king/god appeared out of nowhere. iirc it happens in Le Cid by Corneille. Nowadays, it's the same as out-of-thin-air
It's happening to lots of us, @DanielF, although, to be honest, one of my recent answers probably deserved a downvote or two. But if someone knows the right way to do it, I'm all ears. Any ideas?
@r9m I think I'm going to ask the inequality about $\srqt{\frac{\pi}{3}}\int_0^\pi$ on main because your solution deserves some rep and there might be other as well (i'll post the fourier series method also)
@TedShifrin Not sure. If you look at the level curves $f^{-1}(c)$ for small $c > 0$, you have two closed Jordan curves winding around the minima (plus possibly some curves elsewhere). If you let $c$ grow, at some point the level curves must touch, but then the level set isn't a submanifold, so there must be a critical point of $f$ where they meet. But making that precise, I imagine to be a nightmare.
The issue, @DanielF, as I started to realize and someone pointed out today, is that we might have $f$ not proper, even without introducing other critical points.
I encountered this integral in physics and got stuck.
$$\int_{0}^{\Large\frac{\pi}{2}} \dfrac{d\theta}{\sqrt{\cos \theta}}.$$
