@BalarkaSen I have a relatively simple proof, actually. Let $(M,g)$ be Riemannian and let $(S,h)$ be a properly embedded submanifold with $h$ the induced metric. Let $d_g,d_h$ be the distance functions wrt. the two metrics. Then $d_h(x,y)\le d_g(x,y)$ for $x,y\in S$.
What this means is that if $(x_n)\subset S$ is Cauchy wrt. $d_g$, it is also Cauchy wrt. $d_h$.
Suppose $(M,g)$ is complete. Since $S$ is properly embedded, it is closed in $M$. Then for $(x_n)\subset S$ Cauchy in $M$, it is also Cauchy in $S$, and converges in $S$ by closedness of $S$ in $M$ and completeness of $M$.
Let $M$ be any manifold. By Whitney, we may properly embed $M$ into $\Bbb R^{2n+1}$. Give $\Bbb R^{2n+1}$ the standard Euclidean metric. This is a complete Riemannian manifold. As we just showed, the pullback metric on $M$ is complete.
This also doubles as a proof that any paracompact manifold has a Riemannian metric.
It's shorter than the usual one, but Whitney is pretty powerful for it.
Oh no
Nomizu-Ozeki is that there is a conformal complete metric.
@MathWanderer Either you have Hatcher and Munkres, which are both easy to read, or Spanier and more abstract categorical books on AT which are hard to read. Take your pick ;)
@MathWanderer Eh?
Can you do all the problems?
The problems are very NOT easy. The general conscience is that they are hard.
@Agawa001 I'm pretty aware that in a class you won't be told too much about how important is to trust yourself, but I think this is a critical point if you wanna get anywhere. So, having around enough sources that suggest you to follow a certain behaviour in terms of performance, not to have greater expectations, or never dare to reach the performance of the great figures in science is the thing you might not like to happen to you.
Despite that for a long period of time I never had a positive environment for my math development, I evaded from there and followed the course that led me to some results I would have never obtained otherwise. It's not easy, but it's not impossibe at all.
Yeah, there are lots of Prof. Howard' likes, and you want to avoid them as much as you can (although sometimes it is pretty hard).
It's not that I ever felt the need to share anything from my thought related to Ramanujan, or my performance expectations, but often I did that on purpose and saw how open-minded users here are.
Actually, I did that more times also on different topics to see how people react, but I never needed a confirmation, approval from anyone for any of my decisions.
anyways, i quit my journey of "finding myself" and "self-beating for perfection" long ago, why? because working does mean simply to obey rules rather than focusing on touching the ceil of success, maybe i can retake that journey one time if the demon of inspiration reknocks my door
Find the maximum of the the function $$f(\bar x)=\sqrt{x_1^2+2x_2^2+......2016x_{2016}^2}$$ $\bar x=(x_1,x_2,....x_{2016}) \in R^{2016}$
where the domain of $f$ is $$\bar x=(x_1,x_2,....x_{2016}) \in R^{2016} :\sum_{n=1}^{2016}x_n^2=1$$.
Sol:
$f^2(\bar x)={x_1^2+2x_2^2+......2016x_{2016}^2}=...
@ManolisLyviakis Your function is $f(x) = x_1^2 + 2x_2^2 + \cdots + nx_n^2$, which you want to maximize wrt $g(x) = x_1^2 + \cdots + x_n^2 = 1$. Lagrange maximization says extremums are the solutions to $\nabla f = \lambda \nabla g$. Write it out; you get $2[x_1, 2x_2, \cdots, nx_n] = 2\lambda [x_1, \cdots, x_n]$. The solutions are given by by $\lambda = k$ and $x_i = 0$ for all $i \neq k$ and $x_k = 1$. The maximum is attained for $k = n$: that's your solution, $x_n = 1$ and everything else is $0$.
That's where $f$ is $n$, so your answer is $\sqrt{n}$.
The point is the 2016 equations (or, indeed, $n$ many equations) are very simple here.
If $\Vert x \Vert = 1$ then can I automatically say $\Vert A x \Vert = \Vert A \Vert$? Or can I only get as far as the inequality $\Vert A x \Vert \le \Vert A \Vert$?
@Kari suppose i have a symmetric matrix find its eigenvalues and say $||A||=|δ|$ the greater eigenvalue. then if im asked to find a u vector such that $||x||=1$ $\in R^2$ that satisfys the $||Ax||=||A||$ what am i suppose to do?
That's still too broad. PL-manifolds are a different category of manifolds which sit between Top and Diff. They are an interesting object of study, as in manifolds with any extra structure worth studying.
@JuanFran Because topological manifolds are way too hard to study.
A little weaker structure than a triangulation, but sure.
Every smooth manifold admits a PL structure. And PL in a sense is easier than C^0 because having a triangulation makes a lot of things easier (e.g., I think a version of the tubular neighborhood theorem holds here) but on the other hand allow "coning" a map, which makes e.g., the Poincare conjecture in high dimensions true, unlike C^\infty.
It's an interesting category which lie in the middle of them so you can use tools from both, hence worth studying.
@JuanFran They are not a tool, I don't think. They are a category worth studying in themselves.
I just gave you an example of a theorem which is false in the smooth category, but holds here. That's interesting.
I think the 4 dimensional Poincare conjecture in the PL category is open. If you can prove that, that'd be interesting, though weaker than the one in smooth category.
I mean, I can similarly say "schemes seem like a technical tool" because I am not familiar with them :P
All of this is part of the study of smoothability of topological manifolds, where I think the PL category is very very important and may indeed be used as a tool.
@robjohn As you are room owner and maintainer of the rules for this chatroom, I wanted to ask whether posting pictures should not be treated similarly as YouTube and Wikipedia links (rule 7) - for the same reason. (It probably does not matter too much, since most users are probably unaware that rules for this chatroom exists or they simply ignore them.)
@MartinSleziak The idea was to reduce the bandwidth, and increase the number of lines visible, when looking at a chatroom filled with images. That would apply to images as well as links with image content. So applying the same ideas, posting large images should be relegated to links. If the images are tiny or there is no more than one moderate sized image per 10-20 comments, I would say that the impact is minimal. More images or huge images would have a larger impact. Does that seem reasonable?
@MartinSleziak The image in this comment is pretty huge. It does take up half my screen. It also has little mathematical meaning. If an image has mathematical content, I would be more lenient.
8. Avoid posting tall images (more than $\sim\!\!150$ pixels, which is the height of a $4$ line comment), unless the image has mathematical content or is part of an ongoing mathematical discussion.
Find $x$ such that $||Ax||=||A|| $ and where $||A||$=sup{|d|:d $d$ is eigenvalue of of the symmetric matrix $A$.WHy does it suffice to find an eigenvector of the eigenvalue i found?
Find the maximum of the the function $$f(\bar x)=\sqrt{x_1^2+2x_2^2+......2016x_{2016}^2}$$ $\bar x=(x_1,x_2,....x_{2016}) \in R^{2016}$
where the domain of $f$ is $$\bar x=(x_1,x_2,....x_{2016}) \in R^{2016} :\sum_{n=1}^{2016}x_n^2=1$$.
Sol:
$f^2(\bar x)={x_1^2+2x_2^2+......2016x_{2016}^2}=...
