The following is definition of $S^n = \{(x_1,....,x_{n + 1} \in R^{n + 1} : x_1^2 + .... + x_{n + 1}^2 = 1 \}$ right ?

just need to be super precise in my lecture

???

@L33ter ya .. except you missed a parenthesis $(x_1,....,x_{n + 1})$ :P

and $B^n = \{ (x_1,....,x_n) \in R^n : x_1^2 + .... x_n^2 \leq 1\}$

right @r9m

yeah

indeed I did

@Mahnax $\lim_{(x,y)\rightarrow(0,0)} \frac{x^4+y^4}{x^2+y^2} = \lim_{(x,y)\rightarrow(0,0)} x^2+y^2 - \frac{2x^2y^2}{x^2+y^2}$ and you have $0< \frac{x^2y^2}{x^2+y^2} < x^2$ finish by squeeze.

Hi all

@r9m That's really nifty! Thank you so much.

Who likes dynamical systems and periodic functions ?

$$0< \frac{x^4+y^4}{x^2+y^2} \le\frac{(x^2+y^2)^2}{x^2+y^2}=x^2+y^2. $$

@OFFSHARING :-) good to see you back! :)

@mick I have an interest in dynamical systems .. :) I plan on taking a course eventually :)

@r9m Hey ;). I'm not really back. I realize it's hard to suddenly stop any activity in a math community, but this is my aim.

Hope to come here more and more rare.

@OFFSHARING why you need to stop? :o

@r9m I might come but not to talk about math, and this would not be good.

@OFFSHARING ?!

@r9m hehe, at least for a good while I wanna share nothing about my work, and here I'm tempted to do it.

Oh @OFFSHARING I feel you.

@OFFSHARING aha! :) so that's the reason! 'kay!! I understand :D

I'm out

@r9m ;)

@Agawa001 Sorry, but no... Could you resend it to me?

Could I also ask you something else about the upwind method? Or aren't you familiar with it? @Agawa001

Hi, a had a question... $\aleph_0 \in \Bbb N$?

@evinda i dont promise uto answer,im at cyberspace righnow, but pose ur problem may i have something to contribute in

@Masacroso heeeyyyyy buddie

hi @Agawa001

@Masacroso how is it doing, it was long time

Reading math papers is haaaaard. xD

Yes, sometimes @PerplexedGuest, sometimes... it depends how they are written. Change book if you encounter something extremely hard to understand.

My anterior question is because I cant understand why in this it is "obvious" that $\Bbb N$ is closed. It is not obvious to me in any way... maybe Im overlooking something very simple, Idk.

Not books, papers.

Reading up for research. xD

@r9m check out my last question then ! 😊

@mick ! Nice question! :) but I don't know much about the stuff even to make any helpful comments ..

Thanks ... I assume you upvoted. 😊

@mick ya! Not sure what that downvote was for .. :|

For the originally poor written OP , i had to run ...

I guess the idea of An irrational period for integer indexed iterations is controversial too. Could probably be stated more formal INSTEAD of intuitive ...

If i wrote sin(n) + cos(n) i i get dynamics on a circle ! Probably more popular , but not necc ( mathwise )

Consider putting on MO too ..

How can I show that $$\sum_{n=1}^{\infty} \frac{\sqrt{1+2^n x}}{n!}$$ uniform converge on $\mathbb{R}$?

@Cortizol What is the definition of uniform convergence you have?

@AliCaglayan Standard one en.wikipedia.org/wiki/Uniform_convergence#To_series

@Cortizol Did you consider using Abels test?

@AliCaglayan I don't see how can we get bound for $\sqrt{1+2^n x}$

@MickLH Correct me if I am wrong, but that only means that for every $x \in \mathbb{R}$ series is convergent

@Cortizol The series doesn't converge uniformly on all of $\mathbb{R}$. (Due to the $\sqrt{1+ 2^n x}$, you have a problem for $x < 0$ anyway, since the radicand is then negative for large enough $n$, but the failure of uniform convergence already happens for positive $x$, since $\sqrt{1 + 2^n x}$ is unbounded.)

@Cortizol I don't know, I just saw $\mathbb{R}$ and added $x > -2^{-n}$ and did a ratio test

@DanielF !!!

Hi @MickLH

hi @TedShifrin

Hi @Ted.

It's a rare privilege to see you, @DanielF.

@TedShifrin Rare it is, but privilege?

good evening @Ted

@DanielF ... I miss the old days in here with you and Pedro around, yes.

hi @Huy

@DanielFischer Sorry, my bad, $x \in [0, \infty)$, not $\mathbb{R}$.

I literally can not continue being here while that bracket is closed with a parenthesis, bbl.

@Cortizol That gets rid of the problem of negative radicands, but the series still isn't uniformly convergent. It's uniformly convergent on $[0,K]$ for every $K \in (0,+\infty)$, however. That is probably sufficient for whatever you want to do with the series.

Are you sure, @DanielF. Morally, you can just pull out $\sqrt x$ and you have a convergent series. So I would think it is uniform.

@TedShifrin If you have a uniformly convergent series, all but a finite number of terms must be bounded. $\sqrt{x}$ isn't bounded on $[0,+\infty)$.

@DanielFischer It is problem from some old book in my language. It doesn't have solution, but is says that this series uniform convergent on $[0,\infty)$. Maybe they are wrong...

But it's comparable to $\sqrt x\sum 2\cdot 2^{n/2}/n!$, is it not? What am I missing?

@Cortizol Possible. Wouldn't be the first mistake in a book, won't be the last.

@TedShifrin $\sqrt{x}\cdot \varepsilon_n$ is unbounded for $\varepsilon_n > 0$.

Ah, duh, yeah, of course. Thanks. I think I should go back to retirement.

But perhaps that helps @Cortizol understand it :)

Hi @Mike.

Goodnight @MikeM

Morning.

Two weeks left, @MikeM?

@MickLH Say what?!

This week is the only stressfulnweeknleft. Next week is just grafing

Wow, can't believe everything ends 3 weeks before Xmas.

Oh, right, and you told me Ko is teaching your section this week. So not so stressful.

No, we're trading. I have section Fri.

Ohhhh ...

Have a meeting wed I need another week to prep for and then I need to grade stuff I'm late on for afri.

meeting with Cyprian, you mean?

@TedShifrin

Hi Karim

I am typing my notes for my presentation next week

it will be on the black board not over latex

I hope I don't run over time

Most people who aren't experienced lecturers find that time goes 3x faster than you expect.

Practice giving the talk, and allow for interruptions with questions. Usually you will run way, way over.

yeah I am already running over now as I was practicing

Well, it's already tomorrow here, so I'm now off to bed. 'twas nice seeing you, @Ted, even if only short. Also night @Mike and everybody else.

Practicing is important, so you don't stumble over stuff.

Night, @DanielF, and sorry about my mental lapse ... Schlaf gut!

@TedShifrin someone opened a range with [ and closed it with )

It was seriously drilling into my OCD-type tendencies

But it was correct, @MickLH :)

@Ted: i, not y. And da.

Oh, sorry about the misspelling.

Sorry, I'm lost, I'm not trying to assert my result at all

He'll expect a letter of apology in the mail next week.

I just saw he asked "how to show" not "is it" so I added the assumptions he forgot to type

@MikeM: He would say "who the hell is this?" if he got such a letter.

anyhow, @MickLH, sorry to have derailed you. Get back to your work :)

@TedShifrin If only it were that easy...

Well, I still feel guilty for disrupting you.

$\text{Don}'(t!)$

Okay, so I have a very trivial question, but I'm dogtired so I just don't see it. Let's say we have $\pi: G \to \mathbb{Z}/4\mathbb{Z}$ a surjective group homomorphism with kernel $\mathbb{Z}/3\mathbb{Z}$. Then $G \cong \mathbb{Z}/12\mathbb{Z}$. What is a generator for $G$?

It is not obvious that $G\cong \Bbb Z/12\Bbb Z$.

No, not in this case, but I am doing this for elliptic curves, and I should have mentioned that

Are you using some stuff to prove this?

This part I have proved, and I have a generator for the kernel, $(3:1:0)$ where $G$ is the group of points of the elliptic curve $E: Y^2Z = X^3 + XZ^2$ over $\mathbb{Z}/9\mathbb{Z}$

And $\pi$ is just mapping points $\mod{3}$

Idk seems obvious to me.

By first isomorphism theorem, I'd say

@Krijn: From the algebra standpoint, pick a generator of the kernel and pick an element that maps to a generator of the image.

@TedShifrin and then add them together?

Well, try it out. Yes, you're definitely hypothesizing that the group is abelian, so you need that.

@TedShifrin I now feel guilty for interrupting you, but I'll be back in a few minutes and I might miss you before I get back... Before you go, if you have a few words on the topic I would love fresh avenues of research. I am working on very high dimensional optimization, and I'm stuck in the quasi-Newton tunnel vision and can't see anything else!

Oh yes, but that is very obvious in this case

I'm not sure I know anything on that stuff, @MickLH.

As if computer science isn't already obsessed with optimization... I predict the obsession getting an order of magnitude stronger...

Oh great.

Before, research in optimization was sortof just to speed things up

But now that we can do enough arithmetic ops per second to make it feasible, machine learning can usually be formulated as extremely high dimensional optimization

Well, with bigger and bigger data, such things are still at the forefront.

I'm off to sleep, I need it so much

Good day/night everyone

Goodnight

Night, @Krijn.

Recently while reading a paper that claims to have a cutting edge result on a topic I'm researching, I read that they are approximating the Hessian with a Fisher information matrix...

Quick question (sorry to interrupt your discussion MikeLH): It is stated in some notes that "A line through the origin and the whole plane are never orthogonal subspaces."

ocw.mit.edu/courses/mathematics/…

It sounds like they're talking about subspaces of just the plane, @FreshAir.

In $\Bbb R^3$, yes, your example works. Not in $\Bbb R^4$, though.

lol @TedShifrin our top prof has a spelling prob

he always mis-spell stuff

I complain when my students misspell, Karim :P

no worries @FreshAir, I wasn't really starting a topic, just being playful because they describe the approximation as "well known"... but I have to go learn about Fisher information matrices now as I think the "well known" part came from another field ;)

so as I was writing my notes for my lecture I misspelled homomorphism so I was laughing my ass off

haha

lol

Most people want to abbreviate it homo, but then that has other issues ... :D

hahahaha

lol

I guess group homo. is out too then?

Yup, at least put an m on the end of it.

hom. ?

no

lol hom is actually used in maps