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00:11
So consider $\mathbb{S}^1\times \mathbb{R}$. Since there exists a smooth covering map $\pi: \mathbb{R}^2\rightarrow \mathbb{S}^1\times \mathbb{R}$, $\mathbb{S}^1\times \mathbb{R}$ admits a Riemanian metric making it into a complete Riemannian manifold under $\pi^{*}g$, where $g$ is euclidean metric. here, $\pi$ is also a local isometry, and so this means under this metric $\mathbb{S}^1\times \mathbb{R}$ has constant sectional curvature 0, right?
00:24
Your sentences are all garbled. You should proofread. The product metric on the product of any two $1$-dimensional manifolds has curvature $0$.
Yes, but i'm not looking at product metric @TedShifrin
What you’ve written makes no sense.
What metric are you looking at?
00:40
My notes say the following: There exists a smooth covering map $\pi: \mathbb{R}^2\rightarrow \mathbb{S}^1\times \mathbb{R}$. Therefore, $\mathbb{S}^1\times \mathbb{R}$ has a metric for which $\pi$ is a local isometry and under such a metric, $\mathbb{S}^1\times \mathbb{R}$ is complete
01:06
Do you see why what you wrote the first time is garbage? Your notes are still sloppy. What metric on the plane?
nobody minces words like ted.
Euclidean metric, ofcourse @TedShifrin
When does a covering map induce a metric on the quotient space?
At any rate, if it is the Euclidean metric of course, this is a product metric on the cylinder.
We haven't discussed quotient spaces and manifold structures on quotient spaces
That makes no sense.
You have a covering space, so interpret my question in your context and think. You don’t need any of this covering space talk here, but your prof did it for a reason, I suppose.
How does $\Bbb R\times \Bbb R$ naturally cover $S^1\times\Bbb R$?
01:18
Let $p(t)=e^{2\pi it}$ then the covering map is $p\times id_{\mathbb{R}}$ I assume
Right.
We assume.
No need to talk about all this fancy stuff.
Product metric pulls back to Euclidean metric on the plane.
01:47
Can I modify the product metric on the cylinder to give me a riemannian metric which preserves completeness but now the cylinder has sectional curvature atleast 1?
02:24
ARG! We are leaving for Tucson tomorrow morning, and won't be back for more than a week, yet the small child decided that no, leftovers aren't good enough (despite the fact that they are all going to have gone off by the time I get home), so she pulled out a pint of beans from the freezer and refried them.
I don't know if I should be upset that she is wasting food, or impressed that she refried her own damn beans.
Ugh... I hate teaching evening classes.
Right, back to teaching.
 
5 hours later…
07:25
If $u(x,t) $ is a C^2 function satisfying the following conditions : $u_{tt} - c^2 u_{xx} + ku = 0$, $u(x,0) = 0$ for all $-\infty<x<\infty$, $\lim_{|x|\to\infty} u_t(x,0) = 0$ and $\lim_{|x|\to\infty}u(x,t) = 0$ for all $t>0$ then what is the sign of $E(t)- E(0)$ for $t>0$ where $E(t) = \frac{1}{2}\int_{-\infty}^\infty u^2_t(x,t)+c^2u_x^2(x,t)dx$ ?
$E(t) - E(0) = \frac{1}{2}\int_{-\infty}^\infty u_t^2(x,t) - u_t^2(x,0) + c^2 (u_x^2(x,t) - u_x^2(x,0)) dx$. I think the second condition implies $u_x(x,0) = 0$ and last one implies $\lim_{|x|\infty} u_x(x,t) = 0$. But I don't know how to use the given pde.
It's deciding the sign of $E(t) - E(0)$ so I think differentiating w.r.t. $t$ is not a great idea. It only decides if it's increasing or decreasing.
 
1 hour later…
08:36
@Jakobian the Vietoris topology is is the topology induced by the Hausdorff metric (when the latter makes sense)
(And 99% of the times you encounter a Vietoris topology, it comes from a metric space. More often than not it is compact as well)
@Koro $\frac{c^{p-1}}{(1+c^p)^2}<\frac{c^{p}}{(1+c^p)}\frac{1}{c}<1$ but i need $<\frac12$
08:51
I guess $E(t) - E(0) =0$. I'm pretty sure differentiating w.r.t. $t$ should be used so that the given pde can be applied somehow.
But it seems nobody care about the question.
09:05
@Vrouvrou you agree that $c^p>1$?
 
2 hours later…
11:25
@Vrouvrou what is the maximum of $\frac{c^p}{\left(1+c^p\right)^2}$?
 
1 hour later…
12:42
bounty started on this question
13:29
Does $\lim_{x\to\infty} u(x,t) = 0$ imply $\lim_{x\to\infty} u_x(x,t) = 0$ and $\lim_{x\to\infty} u_t(x,t) = 0$?
I'm quite sure former is true but not sure the latter one.
13:54
There is a famous trick for the first one (ignoring t).
14:33
But the second one may not be true I think.
15:31
how are my chat algebros doin?
@Koro
:>
@Prithubiswasleftmse but didn't leave MSE completely :D
MSE is the google of math, you can't leave it lol
@PurpleHaze But google doesn't close my queries.
True true
I want to write some learning software for math (Django website) in my spare time
Right now got a construction job
Kind of like "learn proof of X" content, but users can add in more proofs, expand certain points of proof and so on... The bottleneck is not getting the computer to understand math, it's helping the user learn math quicker
For example, I want to show that the standard complex satisfies $d^2 = 0$, but am iffy about expanding that by hand. The proof page will show the different steps in expansion and allow different ways of viewing the formulae
Support for commutative diagrams I have experience with on a Django site using Quiver CD editor as a frontend. But I need to write support for parsing & rearranging formulas
The user should be able to enter in regular mathjax and have the system parse it and auto-create some parts of the content
16:02
Hello @PurpleHaze! :)
Hey you
:D
I'm going to hook up a desktop that has webdev stuff, but it doesn't have web connectivity yet, but I'll leave this laptop open
I'll probably be sealing a tile floor most of the day and helping parents to unpack stuff
@PurpleHaze you can check out any time you like, but you can never leave
4
16:34
Does $\mathbb{R}^2$ admit a complete metric with sectional curvate $k\geq 1$?
16:44
Does it admit any metric with curvature $\ge 1$?
I would say no, but i'm not sure of any theorem that would allow me to deduce such a conclusion
Try to find one.
So all Riemannian metrics on $\mathbb{R}^2$ (compatible with the standard topology) give sectional curvature 0?
17:08
Is the theorem Killing-Hopf theorem @TedShifrin
that would mean that $\mathbb{R}^2$ is isometric to $\mathbb{R}^2$ with standard metric or the upper half plane
You need hands-on experience with playing with calculations and concrete examples. You also need to understand the hypotheses of the theorems you find.
I meant it when I said to try to find a metric of curvature $\ge 1$ on the plane.
Oh, I thought you meant for me to try to find a theorem that would allow me to deduce such a conclusion
17:48
Exercise: If $F$ is a continuous distribution function on $(\mathbb R, \mathscr B, \mu_{\mathcal L})$ with distribution $\mu_F$, use Fubini's theorem to show that $\int_{\mathbb R} F(x) \, d\mu_F(x) = \frac{1}{2}$, and show that if $X_1, X_2$ are i.i.d random variables with common distribution $F$, then $P(\{X_1 \leq X_2 \}) = 1/2$ and $\text E(F(X_1)) = 1/2$.
I am unsure of what to do but I will try to type some thoughts.
First of all, this seems vaguely connected with the probability integral transform, although that observation may not be permitted to help solve this problem.
Second, the integral and the expectation look very similar - I am not sure if they are the same thing, though.
@maths How about metrics of everywhere positive curvature on the plane?
Third thought: Fubini's theorem says (roughly) that an integral over a product space can be computed as an iterated sequence of integrals over the spaces that make up the product. I don't see a product space here.
Might have more thoughts coming; not sure.
A couple definitions might be worth writing down.
A distribution function is a montone increasing, right continuous function $F \colon \mathbb R \to \mathbb R$.
If $F \colon \mathbb R \to \mathbb R$ is a distribution function, there is a unique Borel measure $\mu_F$ on $\mathscr B$ such that $\mu_F((a, b]) = \hat F(b) - \hat F(a)$ for all $a, b \in \hat{\mathbb R}$, $a < b$.
($\hat F$ is the extension of $F$, to the extended reals I think.)
(I hope this doesn't come across as spamming this chat room, sorry.)
I probably need to think more about what that integral means, in terms of defintions.
Maybe the product space is the product of $(\mathbb R, \mathscr B, \mu_{\mathcal L})$ and $(\mathbb R, \mathscr B, \mu_F)$. Not sure; need to think about it.
This link seems relevant, but Graham Kemp's answer uses Riemann integration, and seems to assume that $F$ is differentiable, which seems unwarranted in my case.
Maybe I need to write an integral over a product space, use Fubini's Theorem to turn that into two individual integrals, with an integral with respect to Lebesgue measure on the inside, so I can evaluate that somehow, and then get the result Dunno. Just riffing on another exercise I completed that went sort of like that.
I feel bad about posting so many messages here. Maybe I'll go write up a question and post it on the main site.
18:17
Question posted, if anyone cares. Thanks for reading.
18:40
@TedShifrin Did Gauss have a bonnet?
if $\sum a_n, a_n\gt 0$ converges, then can we say that $a_n\lt 1/n$ for all large n?
@Koro: You figured that out about $a_n\lt\frac1n$?
7
Q: Showing that if $\sum a_n,a_n\ge 0$ converges, then $\sum_{n=1}^\infty\frac {1}{n^2a_n}$ diverges.

KoroShowing that $\sum_{n=1}^\infty\frac {1}{n^2a_n}$ diverges if $\sum a_n,a_n\ge 0$ converges. Since $\sum a_n$ converges, there exists $N$ such that for any $m\gt N$, the following holds: $$a_{m+1}+a_{m+2}+\cdots+a_{2m}<\frac 12$$ By AM$\ge$ HM, $$\frac{\sum_{n=m+1}^{2m}a_n}{m}\ge \frac m{\sum_{n...

@robjohn I have been thinking about that since someone posted an answer (now deleted) using this idea. And then someone posted that query in the comments to my above linked post.
@Koro Ah, I thought that you deleted it because you had the answer.
@robjohn Only when he was compact?
@Koro That certainly is false.
18:50
if $a_n$'s are allowed to be negative also then I have a counterexample $a_n=(-1)^n n^{-1/2}$
Then it's extremely false.
But I've been thinking of counterexample when $a_n$'s are positive.
Hint/fun fact: the harmonic series diverges, but if you only sum the reciprocal of numbers that don't contain a 9, it converges
@Alessandro: I've posted that here several times.
And hi :P
Ah I see
It's a cute fact
Also hi!
18:56
Yes. I first encountered that as an exercise in Simmons's wonderful calculus book, I believe.
I don't know why I know this. Could very well be from a time when you posted it here
If you don't know why you know, I would wager yes.
i never noted that before.
But that wasn't even what I had in mind when I commented to Koro.
probably this was posted here before I started using Chat.
18:58
Yeah there are way easier examples, I just wanted to share this fun fact :P
And I don't know why and how that solves my question.
@AlessandroCodenotti: are you referring to the linked post or my question asked in comments above?
It does solve your question, put $a_n=1/n$ if $n$ contains no $9$ and $a_n=0$ otherwise
Re your question: First of all, you are not allowing constants (let alone more complicated things). But make $a_n = 1/n^2$ unless $n=k^3$ for some $k$. And then what?
@Koro To the question you posted in chat here
@AlessandroCodenotti :-)
@TedShifrin then $a_n:=\frac 1{n^{1/3}}$
yep, that should work.
19:03
We want something convergent, don't we? I'm confuzled.
But, anyhow, you're on the intended track now.
Unrelated @Ted but I got today's wordle in 3. I think I'm improving
@robjohn :) I guess you can call his professorial cap a bonnet
⬜⬜⬜⬜🟨
🟨🟨⬜⬜🟩
🟩🟩🟩🟩🟩
Despite the terrible start
@AlessandroCodenotti I try exhausting as many vowels as possible in the first attempt.
That's easier on the italian version since "aiole" is a real word
19:13
@AlessandroCodenotti $a_n=0$ is not allowed :(.
Put $a_n=1/n$ to a billion or anything else wich is not zero but small enough for the series to converge
putting $a_n=1/n$ to a higher power satisfies $a_n\le 1/n$.
I want to do that only for those that contain a 9
And a_n=1/n for those not containing a 9
yeah, but none of these cases violates $a_n\le1/n$ and yet $\sum a_n$ converges.
:(
Oh you don't have a strict inequality
Well put 1/n+very small epsilon that depends on n
19:19
Okay. I want to keep strict inequality only. :)
then it's fine :).
Or put $a_{2^n}=2^{n-1}$ and for indices that are not power of two put $a_n=0$ (use the same trick as above to not use 0)
19:38
@AlessandroCodenotti I got today's in 3, too.
@PM2Ring That doesn't look like the bonnet of a car.
@Koro Consider $a_n=\frac1{n^2}$ when $n$ is not a perfect fourth power and $a_n=\frac1{n^{1/2}}$ when $k$ is a perfect fourth power.
20:05
@TedShifrin Agreed. But Judy's Easter bonnet is almost large enough to cover a car engine. :)
> In your Easter bonnet, with all the frills upon it,
You'll be the grandest lady in the Easter Parade.
I'll be all in clover and when they look you over,
I'll be the proudest fellow in the Easter Parade.…
It's a sweet song, but it's also pretty corny. I don't mind some Irving Berlin stuff, but for music from that era I prefer Jerome Kern and Dorothy Fields.
@AlessandroCodenotti my browser doesn't render any of those unicode characters.
I've got a suggestion to put some emojis in the chat description?
🧮🧮♾️
20:26
@robjohn They're just coloured squares. 0x2b1c WHITE LARGE SQUARE, 0x1f7e8 LARGE YELLOW SQUARE, 0x1f7e9 LARGE GREEN SQUARE

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