@MikeMiller: In my lecture notes on GR, after defining tensors, for some reason (it doesn't seem to be relevant to me) it is stated that $(T_p^*)^* \cong T_p$. I don't remember a lot about dual spaces but I recall from functional analysis that a space is called reflexive is it's isometrically isomorphic to its bidual. There's always a isometrical embedding iirc, but reflexivity requires surjectivity. Is that correct? If so, does that imply that all tangent spaces are reflexive spaces?
EDITED. The major obstacle when calculating the integral
$$ I
= \int_{1}^{\infty} (u^{\alpha} - (u-1)^{\alpha})^{2} \, du
= \int_{0}^{1} x^{-2\alpha-2}(1 - (1-x)^{\alpha})^{2} \, dx $$
is that, within the range $|\alpha| < \frac{1}{2}$, the term $x^{-2\alpha-2}$ has singularity at $x = 0$ which...
@r9m I wonder if there are other awesome ways of computing the integral ...
@Chris'ssis I have no idea ! I couldn't do it myself :) (otherwise I would have posted a solution .. I had a lot of doubts .. lemme read sos's solution :) ...)
@Chris'ssis NICE ! Its an interesting class for sure ! :D
@r9m In case you have doubts $$\frac{\pi \csc (\pi (s-1)) (-\, _2\tilde{F}_1(1,-s;s+2;1) \Gamma (-s)+2 i (\sin (2 \pi (s-1))+i) \Gamma (-2 (s-1)-3))}{\Gamma (-s)^2}$$
@r9m after that I did it magically nice
@r9m I's also pose the following question $$\int_0^{\infty} ((x+1)^{s+1}-x^{s+1}) ((x+1)^s-x^s) \ dx$$
@KhallilBenyattou gcd(a, b), where a and b are not both zero, may be defined as the smallest positive integer d which can be written in the form d = a·p + b·q, where p and q are integers.
@AlexanderGruber I wanna publish by Springer, but it will be also available through amazon. :-)
@AlexanderGruber I'm going to tame beasts in my book. You'll see that is my point ... :-)
@AlexanderGruber In a way I'm the American dream (related to the area I attend) I think since I show that in every single person lies a superman, you only need to discover it through extremely hard work and passion.
@AlexanderGruber My aim in the book is not only to teach you how to attend beasts, but I will also teach you how to think when you're in the area of integrals, series and limits.
@AlexanderGruber The first thing I'm interested in is the quality of the book. I have both the ebook and the hard copy of Ovidiu Furdui's book and I can assure you they are amazing (referring to the quality).
@AlexanderGruber You have to use their template for writing the book. I also learn these details (at the moment I'm more focused on finishing my modules that will be put in my book).
@Chris'ssis I've been writing a book about finite groups for a few years. Basically a collection of interesting examples... kind of like a more detailed version of what I did here.
I may just make it available to the public for free, though. Publishers charge a lot of overhead, and I'm not really trying to make any money off of it.
The thing that is very intriguing to me is that some are not prepared to accept that some might be able to produce research papers of high quality without having any math degree and without belonging to any institution. Some people I use to talk to (especially those that have some math degrees) seem shocked when discussing about advanced integrals, series and limits.
Maybe this situation might seem somewhat cool at the beginning, but after a while you realize that it is terribly annoying. OK, you did it being self-educated only, say, that is nice, OK, but what's next? I'm annoyed seeing people shocked.
I was about to publish my proof of the Basel problem, but then I realized that I didn't want to go through the hassle of gaining permission from StackExchange to use my own damn intellectual property...
@mixedmath I don't know. Ramanujan was self-educated, and I don't wanna compare to him, of course, but this possiblity exists, he proved one can reaches very high peaks by self-education only.
@MikeMiller Oh, you're right.... I must have misread the legal info when I read it a couple of weeks ago...
Skimming through it now, don't see anything to justify my previous conclusion
@Chris'ssis It's not a "new" proof. I encountered Apostol's proof from the 1980's that used multivariable calculus, and I decided to reverse the order of integration. His was a simple two-page proof, I don't know how long mine would end up after being published
hmmm, I need to rewrite my proof to the first problem submitted by Ramanujan (I mean I wanna rewrite it to make it clearer - it uses high school tools only)
Can anyone help me with question 1 here? math.stackexchange.com/questions/1092514/…. I used the hint that David gave me in the answer to get cos(3x)+cos(x). I know i have to find 2cos(4x) in the LHS, so for the RHS, I am using 2cos(2x+2x), but then I will get a polynominal after expanding/cleaning things up after that. What do I do?
I am not sure if this question technically belongs here, but I think this is the best place for it on the Stack Exchange network. If there is somewhere else it should go, please let me know and I will remove it.
So, I am using the Processing programming language to create an animation where a b...
@Lord_Farin That's true. But it's also the best word to use when your state flag has a bad-ass topless lady holding a sword over a vanquished foe, and when your motto translates to "Death always to tyrants."
@Lord_Farin I wanted to show that $n \subset m \rightarrow n \in m \lor n=m$. The induction hypothesis is that $\forall n(n \subset m \rightarrow n \in m \lor n=m)$ We want to show that $n \subset m' \rightarrow n \in m \lor n=m'$ We assume that $n \subset m \cup \{m\}$ and we distinguish two cases for $n$: $n \subset m$ and $n \not\subset m$. I have understood the case $n \subset m$. For the other case: We know that $n \subset m \cup \{m\}$ and we want to prove that $n=m'$. So doesn't it suffice to show that $m \cup \{m\} \subset n$ ?
@DavideCervone In this post, the construct $\overline{T}^i$ no longer renders. It seems that it worked previously. $\overline{T}_i$ still works. — Fundamental1 min ago
OK, thanks. I'll look into it. There were changes to \overline and \underline for other reasons, and that will need to be fixed. In the meantime, you can use {\overline{T}}^2 if you want. — Davide Cervone7 mins ago
EDITED. The major obstacle when calculating the integral
$$ I
= \int_{1}^{\infty} (u^{\alpha} - (u-1)^{\alpha})^{2} \, du
= \int_{0}^{1} x^{-2\alpha-2}(1 - (1-x)^{\alpha})^{2} \, dx $$
is that, within the range $|\alpha| < \frac{1}{2}$, the term $x^{-2\alpha-2}$ has singularity at $x = 0$ which...
I am not sure if this question technically belongs here, but I think this is the best place for it on the Stack Exchange network. If there is somewhere else it should go, please let me know and I will remove it.
So, I am using the Processing programming language to create an animation where a b...
