Mathematics - 2017-01-09 (page 7 of 7)

That is indeed a misconception, but my additional hypothesis makes it right. But I want you to be precise. What do I add to $s_{2n}$ to get $s_{2n+1}$?

Null

10:25 PM

$a_{2n+1}$

Ted Shifrin

No, $a_{2n+1}$.

There we go.

And what did I assume about the terms $a_k$?

Null

that they approach 0

Ted Shifrin

So if $s_{2n+1}\to\ell$, what about $s_{2n}$?

Null

$s_{2n}=s_{2n+1}-a_{2n+1}$

Ted Shifrin

Good. Proceed.

Not quite.

Null

10:27 PM

mmh

Ted Shifrin

It was right a moment ago. Put that back.

OK. Now what happens when $n\to\infty$?

Akiva Weinberger

@Null Yup

Null

$s_{2n}\to l$?

@AkivaWeinberger timetraveler confirmed :D

Ted Shifrin

The limit of the sum is the sum of the limits, right, @Null?

Null

ah, and then we keep adding a limit

Ted Shifrin

10:30 PM

Huh?

Null

mmh

Akiva Weinberger

@TedShifrin Are you saying we can always interchange $\lim$ and $\sum$?? (jk)

Ted Shifrin

DogAteMy: I have multivariable exercises to give you if you're being a troublemaker.

Akiva Weinberger

Sure, why not

Other than the ones in the book?

Ted Shifrin

I gave my students additional ones not in the book, yes.

Have you gotten to the definition of the derivative?

Null

10:32 PM

if the limit is 0, we have at the end 0-0+0-0...? @TedShifrin

Ted Shifrin

No, @Null.

Akiva Weinberger

Yeah.

Ted Shifrin

You just have small number plus/minus small number.

Null

@TedShifrin ok

Ted Shifrin

OK, DogAteMy. Using only the definition of the derivative, show me that $f(\mathbf x) = \|\mathbf x\|$ is differentiable at any $\mathbf a\ne \mathbf 0$ and find the derivative. (I.e., you will need to show that the appropriate error term goes to 0.) As you'll find out, this function is one of my favorites.

So, @Null, now you are armed to consider your original question and a quest for a counterexample. You know that you can't have the terms going to 0. What's the easiest series you can think of where the terms don't go to 0 but the odd partial sums converge?

Akiva Weinberger

10:35 PM

Hrm. OK. Well, the Jacobian is $\frac1{x^2+y^2}\begin{bmatrix}x&y\end{bmatrix}$, right?

Ted Shifrin

Not quite.

Akiva Weinberger

$\frac1{\sqrt{x^2+y^2}}\begin{bmatrix}x&y\end{bmatrix}$, sorry

Ted Shifrin

OK. So at $\mathbf a$ that's $\mathbf a/\|\mathbf a\|$.

Akiva Weinberger

Ah. Yes.

Ted Shifrin

You'll need paper and pen/pencil to do this.

Akiva Weinberger

10:37 PM

So I need to show that $\bf(\|x+h\|-\|x\|-h\cdot x/\|h\|)/\|h\|\to0$?

SteamyRoot

Someone editing a post with "punctuation" by putting spaces within mathmode.

Akiva Weinberger

…???

Ted Shifrin

Well done, DogAteMy. Proceed.

Akiva Weinberger

Well $\|\bf x+h\|\le\|x\|+\|h\|$

Ted Shifrin

I think that'll get you in trouble.

Akiva Weinberger

10:41 PM

Ah, yeah, probably.

Ted Shifrin

Oh, and I just remembered another good question, which one of my students asked a few years ago.

@Null: What's the simplest sequence you can think of that does not go to 0?

Akiva Weinberger

@TedShifrin "Have you ever thought about how all the dogs that were alive when you were born are now dead?"

Probably wasn't the question, but could have been

Ted Shifrin

DogAteMy: What about all my relatives and friends, for that matter?

Are you killing time at school now?

Akiva Weinberger

…oh, yeah, perhaps

Also, it's 5:43pm, school's been out for a while.

Zime tones.

Ted Shifrin

I figured.

I know that. But sometimes you're there late and I can't tell.

When you talked about buying a snack, it threw me off.

Akiva Weinberger

10:44 PM

Fair.

SteamyRoot

@GFauxPas Joyeux anniversaire!

Akiva Weinberger

Oh! Do I use $\|\bf x+h\|=(x+h)\cdot(x+h)$?

Ted Shifrin

I hope not, as you typed it.

Akiva Weinberger

No, I don't, since it's the square root.

Ted Shifrin

LOL

Akiva Weinberger

10:45 PM

Derp.

Null

why are most excercises about proofing and not "find a counterexample" :/ i find those nice

^unrelated

$a_n:=1$ for all $n$ is simple

Alessandro Codenotti

@TedShifrin what was that? I'm curious now

Ted Shifrin

My students used to hate it when I gave them "Prove or give a counterexample." But I did it a lot.

OK, @Null. Now play with that just a smidgeon.

Alessandro Codenotti

@null sometimes (real analysis I'm looking at you) the counterexample is very contrived and ugly though

Ted Shifrin

@Alessandro: If $f\colon\Bbb R^n$ is continuous along all curves coming into $0$, is it continuous at $0$? (We know it's false if it's known to be continuous only on lines or parabolas or .... ) Along those lines, can you give me a function that's continuous along all curves $|y| = |x|^n$ ($n\in\Bbb N$) but not continuous at $0$?

Alessandro: You keep blaming me for wasting your time when you have work to do ... and yet ... you ask for it!

Alessandro Codenotti

10:50 PM

I'm blaming my curiosity, not you :P

Ted Shifrin

I'm quite impressed by where you are mathematically now compared to when we started chatting. You're learning a lot.

Akiva Weinberger

@TedShifrin Oh god I hope so

Alessandro Codenotti

Thanks! For the second part can't I use a function which is 0 on those curves and 1 everywhere else?

Ted Shifrin

DogAteMy: Proof, then?

Oh, what if I ask you to make it continuous everywhere but at $0$, @Alessandro? Can you do that?

Alessandro Codenotti

And I think part 1 is false, but that's just a wild guess and I'm going to think about it. A curve here is the continuous image of $(0,1)$?

Ted Shifrin

10:54 PM

Sure ...

Or can you make it continuous everywhere except along one fixed curve? ... :D

@Null: Any ideas?

Akiva Weinberger

Ah, I see. Find an infinite sequence of points which go to $0$ but whose values don't go to $f(0)$. Make the curve visit these points one at a time, finishing at the point $0$.

Ted Shifrin

Can you guarantee you can do that continuously, DogAteMy?

Alessandro Codenotti

@TedShifrin uhm a function which is constant on $\Bbb R^n\setminus\{0\}$ and has a different value there?

Akiva Weinberger

Thinking of the curve as the image of $\phi:[0,1]\to\Bbb R^n$, we let $\phi(1/n)$ be $a_n$, $\phi(0)$ be $0$, and at all other points interpolate linearly.

@TedShifrin Sure. It maps limit points to limit points, no?

Mike Miller

h

Ted Shifrin

10:59 PM

Good, DogAteMy. Very snappily done.

@Alessandro: What you just suggested will not be continuous at $0$ along the given families of curves.

I'm getting too confuzled here.

Alessandro Codenotti

Ah, ok, I see what you meant now, a function which is continuous everywhere except at 0, but which is continuous along every curve going to 0

Ted Shifrin

Not along every curve, no.

On these families.

Oh, I messed up. The original question should have had families. $|y|=c|x|^n$, as $c>0$ varies.

Warm up by considering continuous along all lines through the origin. Then all lines and parabolas. Then ...

It's hard keeping up with you and DogAteMy.

Null

@TedShifrin so we are looking for a sequence which sum converges for all odd number of steps, but spazzes for even steps?

Ted Shifrin

Right @Null.

Akiva Weinberger

I'm still trying to think of why $\bf\|x+h\|-\|x\|\approx_{\|h\|}(h/\|h\|)\cdot x$

Ted Shifrin

11:05 PM

And I suggested you weren't far away with your constant sequence (and a tiny bit of previous ideas).

Null

what is the "whom" for neutral things?

Ted Shifrin

Hmm, something's missing there, DogAteMy.

Akiva Weinberger

@Null Example?

@TedShifrin By $\approx_{\bf\|h\|}$ I mean the difference, divided by $\|h\|$, goes to $0$

Ted Shifrin

Double check, DogAteMy. I think you had it right earlier.

GFauxPas

Null, example?

Null

11:06 PM

to the person whom belongs the red car: drive it away please. i dont even know if this is english

GFauxPas

to whom

although "to who" is used colloquially

Akiva Weinberger

29 mins ago, by Akiva Weinberger

So I need to show that $\bf(\|x+h\|-\|x\|-h\cdot x/\|h\|)/\|h\|\to0$?

Ted Shifrin

Oh, you had it wrong before.

You got your letters muddled.

GFauxPas

better, "to whom the red car belongs"

Ted Shifrin

Whoever owns the red car, please come get it before I steal it.

GFauxPas

11:08 PM

yes that's even better, but doesn't illustrate how to use "whom" :P

Ted Shifrin

Would the owner of the red car please move it before it is towed away?

Akiva Weinberger

To the person who owns the car

Ted Shifrin

Oh @GFauxPas.

Akiva Weinberger

To the person to whom the red car belongs

Null

well basicly, the sequence "whose" sum converges??

i dont know which word should there be

Akiva Weinberger

11:09 PM

Yeah.

Ted Shifrin

He's trying to use whom with the object, not the person.

GFauxPas

oh

Ted Shifrin

Yes, @Null. The sequence whose associated series converges.

Akiva Weinberger

Whose is also for inanimate things.

Ted Shifrin

Yuppers. I wish I'd known what we were doing :D

GFauxPas

11:09 PM

just make sure you pronounce it correctly or else you'll sound dumb, it's pronounce "wuh-hose"

Ted Shifrin

Anyhow, DogAteMy, you have a letter wrong in that.

Say what? @GFauxPas

GFauxPas

just kidding

Akiva Weinberger

$/\|x\|$ in the last term in the numerator?

Ted Shifrin

right

Akiva Weinberger

So I need to show that $\bf(\|x+h\|-\|x\|-h\cdot x/\|x\|)/\|h\|\to0$.

Ted Shifrin

11:11 PM

LOL, NO.

OK, YES.

:D

Kaj Hansen

http://math.stackexchange.com/questions/2090955/integration-using-substitution-int-frac1-sqrt1-x2-arcsin3x-dx/2090961#2090961.
It seems like my least-effort posts always get the most upvotes.

Ted Shifrin

I was complaining earlier about a random downvote, @Kaj. I don't think it was my arch-enemy back visiting.

SteamyRoot

Join the club.

Jessy Cat

Hey guys

4

Q: Basic proof that $GL(2, \mathbb{Z})$ is not nilpotent

I need to show that $GL(2,\mathbb{Z})=\left\{ \begin{pmatrix} a&b \\ c&d \end{pmatrix} \mid a,b,c,d \in \mathbb{Z}, ad-bc = \pm 1\right\}$ is not nilpotent. I have seen this question but the answer given there is too advanced for where I am currently in my studies. In order to show something is ...

Ted Shifrin

But it's not surprising that the elementary ones get more upvotes, @Kaj. There's a much bigger audience than for graduate algebra/topology questions.

Alessandro Codenotti

11:12 PM

I think I can construct one that works up to $|y|=c|x|^n$ for a fixed $n$, let's work with functions $\Bbb R^2\to\Bbb R$ because they're easy to visualize. If I want a function continuous in $0$ along every line I take the $2$ parabolas $y=x^2$ and $y=-x^2$ and define $f$ as $1$ if $(x,y)$ lies on or above the graph of the first or on or below the graph of the second and $0$ everywhere else

SteamyRoot

It's normal, though, @KajHansen. Usually, the least effort posts are those of a level or branch of mathematics more people can understand and value.

Ted Shifrin

Right, @Alessandro.

Jessy Cat

What is meant by the "second center" of a group?

Alessandro Codenotti

Ah, I also set it equal to $1$ on the $x$-axis because of annoying corner cases

Ted Shifrin

@Steamy: You and I just said the same thing. We concur.

SteamyRoot

11:13 PM

Indeed we do.

Ted Shifrin

That's a line through the origin, too, @Alessandro :D

Jessy Cat

I found he center of $GL(2,\mathbb{Z})$ to be $\pm$ the identity matrix

Alessandro Codenotti

yep, but it's not above $x^2$ for any $x$ so I don't like it :P

Ted Shifrin

Oh, gotcha.

SteamyRoot

@Jessy that's a good start.

Alessandro Codenotti

11:16 PM

in the same way if I want that to work for the families up to $c|x|^n$ I use the graphs of $y=|x|^{n+1}$ and $y=-|x|^{n+1}$ instead of the $2$ parabolas

SteamyRoot

Now, the second center is $Z_2 = \{ x \in \operatorname{GL}(2,\mathbb{Z}) \mid \forall y \in \operatorname{GL}(2,\mathbb{Z}) : [x,y] \in Z(\operatorname{GL}(2,\mathbb{Z}) )\}$.

Ted Shifrin

Cool, @Alessandro. So what do we use when $n=\infty$? :D

Jessy Cat

Can you type that on the question page? I don't think there's chatjax for iPads.

SteamyRoot

You should ask @Ted for getting chatjax to work on an iPad, I think

Ted Shifrin

I eventually got it working, but it was tricky to get the bookmark to load on the iPad and iPhone. I think robjohn put instructions on that page?

It's at the bottom "Troublesome Mobile Browsers"

Jessy Cat

11:20 PM

I'll try to figure out how to install it myself.

Alessandro Codenotti

eh, that's a problem. Because I can't use a function which is smaller than $x^n$ for every $n$ in a neighbourhood of $0$ since that would have to be constant

SteamyRoot

@JessyCat Anyway, I'll type it out as an answer

Ted Shifrin

The instructions are there. It's totally sneaky.

Null

$x^{n+1}<x^n$ for 0<x<1

Ted Shifrin

Are you sure, @Alessandro?

@Null: Smaller than $|x|^n$ for all $n$.

Null

11:22 PM

@TedShifrin well, i'm working on mine, so I better butt out hehe

Ted Shifrin

Yup. You shouldn't be taking so long on yours!

Jessy Cat

@SteamyRoot thanks.

Alessandro Codenotti

ah, wait, $e^{\frac1x}$

Null

I can't see how that is possible!

Ted Shifrin

Almost, @Alessandro.

Alessandro Codenotti

11:23 PM

no, not that one

change of variables are hard

Ted Shifrin

you're missing the same idea that Null is missing now :D

Mike Miller

@Ted Ko's syllabus this quarter is really ambitious.

Ted Shifrin

Did he not do winter last year?

Mike Miller

We changed the plan for Fall/Winter. Now it's manifold generalities in the fall instead of differential topology.

math.ucla.edu/%7Ehonda/…

Null

found a drooling thing: $a_1=1$, $a_2=2$, $a_{n+2}=\frac{a_n}{a_{n+1}}$

Alessandro Codenotti

11:28 PM

$\frac{1}{e^{(1/x)}}$ should be smaller than $x^n$ for every $n$ and $x$ small enough

Null

but this is obv not it!

Ted Shifrin

A lot of analysis thrown in there, @MikeM. Good luck if they don't know a pile of real analysis.

@Alessandro: Yes, $e^{-1/|x|}$.

And DogAteMy already answered my initial question ...

So you can go back to what you're supposed to be doing :D

Jessy Cat

It worked!

thanks @robjohn

Null

what means "final exam is take home"?

SteamyRoot

@JessyCat Posted an answer, let me know if it's unclear or needs more details.

Alessandro Codenotti

11:30 PM

it took me embarassingly long to get that change of variables right from $\lim\limits_{x\to+\infty} e^x-x^n=+\infty$ for every $n\in\Bbb N$...

what was the initial question again?

Ted Shifrin

If $f$ is continuous at $0$ along every curve through $0$, is $f$ continuous at $0$?

Jessy Cat

@SteamyRoot now, in general, what is meant by the "second center"? I thought it was $C(G/C(G))$ where I'm using $C$ to denote the center.

Mike Miller

@Ted I'm there for them.

SteamyRoot

Not quite, actually.

Jessy Cat

hmm.

SteamyRoot

11:33 PM

The second center is the group $Z_2$ in my answer; although there's an equivalent definition close to yours

Jessy Cat

lemme go see what you wrote

Ted Shifrin

yeah, but you're mean, @MikeM :D

SteamyRoot

$Z_2$ is the group such that $Z_2/Z(G) = Z(G/Z(G))$

(I'm very used to denoting the center as $Z$, if that wasn't clear yet >.<)

Akiva Weinberger

Did @AlessandroCodenotti solve his problem?

Ted Shifrin

It's quite customary, @Steamy.

Yes, but he's thinking about the part you got now.

Jessy Cat

11:34 PM

@SteamyRoot the center is not trivial. It's plus and minus the identity

Akiva Weinberger

What was your answer @AlessandroCodenotti

SteamyRoot

Ohhh

I misread, then. Hmm.

That invalidates my proof, I'm afraid

Jessy Cat

Otherwise it would be Z2 and we'd all live happily ever after (it's very nice when it's Z2).

Alessandro Codenotti

that a function can be continuous in $0$ along all curves $|y|=c|x|^n$ without actually being continuous in $0$ @Akiva

Jessy Cat

Better delete it then

SteamyRoot

11:35 PM

Yeah, already did.

Jessy Cat

ok

I'm r

SteamyRoot

Should've thought it through.

Wonder if it's salvageable, though

Jessy Cat

trying to show that it's not nilpotent

Akiva Weinberger

@AlessandroCodenotti …Right, but what was the function

Jessy Cat

but without all the fancy crap.

i just want to do something with the upper central series b/c at this point, that's all I really have at my disposal.

the second guy's answer I would need to prove a preliminary result, and I'd rather not have to do that if at all possible.

SteamyRoot

11:37 PM

Hmmm, well, I think I can fix it rather easily

Jessy Cat

Ooh, you upvoted me too b/c you're awesome.

SteamyRoot

No, I upvoted the question because it's a well-written question

Alessandro Codenotti

I did it for a function $\Bbb R^2\to\Bbb R$ because it's nicer to visualize. You draw the graph of $g(x)=e^{-1/|x|}$ and $h(x)=-e^{-1/|x|}$ and define $f(x,y)$ as $1$ if $(x,y)$ lies on or above the graph of $g$, on or below the graph of $h$ or on the $x$-axis. $f(x,y)=0$ in all other cases @Akiva

Null

@TedShifrin can you give me a hint?

Jessy Cat

Well, still. It warmed my cockles.

SteamyRoot

11:39 PM

Ha, nice to hear :)

Ted Shifrin

@Null: Take your original constant sequence and consider changing some signs.

Null

@TedShifrin but then the sum doesnt converge, even for odd upper values of n

Ted Shifrin

Try it ...

Alessandro Codenotti

I started with doing it for lines and $c|x|^n$ up to a bounded $n$ here @Akiva

Null

you mean $\sum(-1)^k$?

Ted Shifrin

11:41 PM

Almost, @Null.

Jessy Cat

I gotta get dinner on the table. Brb

Mike Miller

@Ted Maybe so, but I try to be a good help.

Ted Shifrin

I know, @MikeM :) If they come to your recitations ... Are this year's kids better?

heya @heather

heather

hello @TedShifrin

it's been a bit

Ted Shifrin

Happy New Year to you, heather.

heather

11:49 PM

and to you as well =)

i come with not some calculus/topology/abstract algebra question but with a geometry question =/

Ted Shifrin

Are you channeling Caesar?

heather

er...no?

Ted Shifrin

I actually answered a high school geometry question on main the other day.

heather

i just found the situation ironic

how do you find the centroid of a triangle when given the coordinates of the vertices of the triangle?

Ted Shifrin

What's your definition of centroid?

heather

11:51 PM

::checks notes::

"The medians of a triangle are concurrent and intersect each other in a ratio of 2:1."

where medians basically are midpoints, as far as I can tell

Ted Shifrin

Medians are line segments from a vertex to the midpoint of the opposite side.

Do you know how to work with vectors?

heather

oh, that makes sense

@TedShifrin yes...?

Ted Shifrin

So go 2/3 down a median and figure out where you are :P

When you're done, I'll tell you a cool secret.

SteamyRoot

Hrmf. Cba to boot to Linux; I'll just download GAP for Windows.

heather

so...wait, you basically find a perpendicular line to one of the coordinates, and go 2/3rds of the way down?

i'm just confused as to the exact procedure here

Ted Shifrin

11:54 PM

no perpendicular!

take a vertex, take the midpoint of the opposite side ... go 2/3 the way along the median.

heather

erm...i know i need to find the midpoint of at least one of the lines. oh, wait, duh...

Ted Shifrin

you know how to find midpoints!

heather

yeah

i feel like an idiot now, lol

Ted Shifrin

Nah.

Mike Miller

@Ted What was wrong with last year's kids?

Ted Shifrin

11:55 PM

When you finish your problem, tell me and I'm gonna tell you a surprise.

heather

this makes sense now, i think ::goes to give it a shot::

Akiva Weinberger

Hey, I think I figured it out, but it'll take a while to type up @TedShifrin

heather

okay

Ted Shifrin

@MikeM: Didn't you complain (or was it this fall?) that very few of 'em came to your problem sessions?

Mike Miller

Other than @AlexW, of course. He's fundamentally broken.

Akiva Weinberger

11:55 PM

But the idea is $\|x+h\|-\|x\|=2x\cdot h/(\|x+h\|+\|x\|)$

Ted Shifrin

Good, DogAteMy :)

Typo, though.

You meant $-\|x\|$.

You seem to have a recurring issue with that.

Akiva Weinberger

Indeed I did. / Indeed I do.

Equation was still right, though.

Ted Shifrin

LOL

No, it wasn't.

pouts

Akiva Weinberger

No? Interchange $x$ and $h$ in the formula

Ted Shifrin

Yeah, but look at the denominator on the RHS.

Mike Miller

11:57 PM

I don't think so... numbers dwindled a bit but whatever. Also, I don't do probl k sessions.

Ted Shifrin

No, DogAteMy, dammit :D

Akiva Weinberger

They were both switched

Ted Shifrin

OK, @MikeM, I remembered your complaining at some point. I don't know what to call them — I said recitations earlier.

Oh, they were both switched? I missed the second one. sulks for being wrong again

Akiva Weinberger

…

pats on head

Null

@TedShifrin do you mean, that at the $2n+1$th term the sum is zero, but one after that it is -1? (or 1, and 0) but that wouldnt be convergent. thatswhy i ask

Ted Shifrin

11:59 PM

Huh? @Null ... We're trying to give an example where the odd partial sums converge but the whole series does not.

You wrote down $-1+1-1+1-1+1-\dots$. So the even partial sums are all $0$.

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$Mathematics$ Mathematics