Mathematics - 2018-03-19 (page 3 of 3)

user193319

5:04 PM

@Dattier Okay. So, is what you're suggesting is based upon the following two facts: (1) $A \subseteq B$ implies $\int_A f \le \int_B f$, and something like (2) $\inf_{x \in A} f(x) \cdot m(A) \le \int_A f$?

Dattier

(2) yes, (1) I don't think

for the conclusion I think is more easy to use, a propriety of the measures

user193319

Which property? The continuity property which states $m(E) = \lim_{n \to \infty} m(E_n)$?

0celo7

@user193319 yes.

user193319

Okay. I still don't see how to get $m(E_n)=0$ unless both (1) and (2) are true (or some appropriate variation).

0celo7

you just need (2)

Dattier

5:10 PM

yes, because in this case, $E_{n}\subset E_{n+1}$

user193319

@0celo7 Okay. I'll try proving that first.

0celo7

@user193319 do you know the corresponding thing for the sup

user193319

No. I don't think either had been proven in my book yet, so I suppose that means I have to.

0celo7

what book

user193319

Royden and Fitzpatrick's Real Analysis

0celo7

5:13 PM

I don't think you should have to prove that...it's important enough to be a theorem

Do you not have any monotonicty results for intetgration?

user193319

The only monotonicity result I have is that if $f \le g$ on $E$, then $\int_E f \le \int_E g$.

Dattier

Measure (mathematics)

In mathematical analysis, a measure on a set is a systematic way to assign a number to each suitable subset of that set, intuitively interpreted as its size. In this sense, a measure is a generalization of the concepts of length, area, and volume. A particularly important example is the Lebesgue measure on a Euclidean space, which assigns the conventional length, area, and volume of Euclidean geometry to suitable subsets of the n-dimensional Euclidean space Rn. For instance, the Lebesgue measure of the interval [0, 1] in the real numbers is its length in the everyday sense of the word – specifically...

Continuity from below[edit]
If E1, E2, E3, ... are measurable sets and En is a subset of En + 1 for all n, then the union of the sets En is measurable, and
{\displaystyle \mu \left(\bigcup _{i=1}^{\infty }E_{i}\right)=\lim _{i\to \infty }\mu (E_{i}).} μ(⋃i=1∞Ei)=limi→∞μ(Ei).

well is not a problem works with $F_n=\{x\in E, \frac{1}{n+1}\leq f(x) <\frac{1}{n}\}$ , it's the same think, and here you can use the axiome of the measures

because $F_n$ is a partition of $E$, @user193319 : ok ?

user193319

@Dattier Sorry for not responding. Yes. I think I understand now. Thank your help!

Dattier

@user193319 with pleasure

Lozansky

"Let $p_0(x) = 1, p_1(x) = x$ and define $p_n$ for $n\geq 2$ by the recursion formula $(n+1)p_{n+1}(x) = (2n+1)xp_n(x)-np_{n-1}(x)$, for $n=1,2,...$. Prove that $p_n = P_n$ (Legendre polynomials)"

Dattier

5:29 PM

@Lozansky what's the definition you have of Legendre polynomials ?

Lozansky

@Dattier Orthogonal polynomials on a $L^2$ space with $P_n(1) = 1$ norm

But of course I want to use Rodrigues' formula

Proof by induction yada yada

Dattier

I think, It's more easy to show p_j and p_k is orthogonal and verify p_n(1)=1, ok ?

Lozansky

Anyway, I get $(n+1)p_{n+1} = (n+1)\dfrac{1}{2^{n+1}(n+1)!}D^{n+1}((x^2-1)^{n+1}) = (n+1)[xD^n((x^2-1)^n)+nD^{n-1}((x^2-1)^n)]$

I think I am pretty close, @Dattier

Just that last term that needs to be fixed

Dattier

you want to show p_0(x)=1 and p_1(x)=x ?

Lozansky

No?

That's given

Dattier

5:37 PM

what's "the last term that needs to be fixed"

Lozansky

Oh

$nD^{n-1}((x^2-1)^n)$

I want the order of the derivative to match the exponent

overexchange

@micsthepick yes

you are right

Dattier

@Lozansky : have show your formula verify the recursion formula ?

Lozansky

@Dattier That's backwards. I want to show that, given the recursion formula, it satisfies Rodrigues' formula

Dattier

@Lozansky No, if you want take this path, I think you must show, the Rodrigues' formula, satisfies the recursion formula, ok ?

Lozansky

5:44 PM

@Dattier Yeah I think we mean the same thing? I plug in Rodrigues' formula into the recursion expression

Dattier

well just verify the formula recursion, ok ?

Lozansky

@Dattier That's what I'm trying to do

Dattier

don't forgot the derivate is linear

use this propriety

@Lozansky

0celo7

@BalarkaSen Do you believe in elliptic regularity? My plan there didn't work out, but I still have a really cool proof of the existence of a solution.

Dattier

@Lozansky now you see ?

Lozansky

5:55 PM

@Dattier See what?

Dattier

what you can do for prouve the formula

Lozansky

Not really

@Dattier What I want to show is that $(n+1)P_{n+1} = (2n+1)xP_n-nP_{n-1}$ using Rodrigues' formula

Dattier

don't forgot the derivate is linear

use this propriety

Lozansky

@Dattier Lol, how?

Dattier

An example $D^nx^n+2 \times D^{n-1} x^{n-1}=D^{n-1}(Dx^n+2x)=D^{n-1}(nx^{n-1}+2x\times x)$ ok ?

@Lozansky now you see ?

If when I will back, you have not find the solution, I will give it.

AustereTiger

6:09 PM

Hey all. I have posted this question on the main forum and no one could answer it for me there so I've come on here instead.

Determine whether the following series is divergent, conditionally convergent or absolutely convergent.

The infinite sum starting at k=0 of $\frac{ksin(1+k^3)}{k+\ln(k)}$.

Any help would be vastly appreciated.

Shobhit

6:49 PM

let $X \times Y$ be the cartesian product of two topological spaces, then is the following definition correct? : $W \subseteq X \times Y$, then call $W$ open if given $(x,y) \in W$, there exists open sets $U$ in $(X, \delta)$ and $V$ in $(Y,\sigma)$ such that $(x,y) \in U \times V \subseteq W$. If not please tell me what should be the correct definition.

Perturbative

@Shobhit The product topology of two topological spaces has as its basis products of open sets of $X$ and $Y$

Balarka Sen

@0celo7 I do

Daminark

@Balarka lol sounds like preaching almost

Perturbative

So an arbitrary open set of $X \times Y$ would be a union of the products of open sets in $X$ and products of open sets in $Y$

Shobhit

yes, keeping that in mind, i formulated the above definition, i just wanted to confirm that its alright. @Perturbative

Daminark

6:55 PM

Also hey @Tobias and @Perturbative!

@Daminark Hi

Hey @Daminark

How's it going?

@Perturbative oh ok. So what i wrote is wrong :(

0celo7

@BalarkaSen Ok, then the plan is as follows: First, prove a spectral theorem for $\Delta$, i.e. $L^2=$sum of eigenspaces, each eigenspace is finite-dimensional

Perturbative

6:56 PM

@Daminark Prepping for my first proper real analysis test tomorrow, how's things going with you

0celo7

This is more or less easy

Then you want to define a solution operator for $\Delta$

Daminark

I'm on spring break, and spring quarter is about to start

Perturbative

@Shobhit Actually what you wrote looks good to me

0celo7

you know what it should be on eigenfunctions, namely just dividing by the eigenvalue

this can then be extended to L^2 by density (you have to be careful here b/c $\Delta$ isn't continuous)

and that's basically it

it's really simple given Rellich's theorem and elliptic regularity

Corellian

A missile is en route to fall on a city with population 500 thousand. You have the means to alter the path of the weapon but in an unpredictable way. If you divert the missile, it will either fall safely in deep ocean or divert to a city of population 1 million. Either of these is equally likely should you choose to distort its path. Do you let the missile stay its course or divert it from the initial trajectory?

0celo7

6:59 PM

but constructing the solution operator using the eigenfunctions is pretty cool and I haven't seen it anywhere before

Perturbative

@Shobhit Sorry I haven't dealt with product topologies in a while, I didn't mean to discourage you, what you wrote looks right (you've basically shown that $Int(W) = W$ which shows that $W$ is open in $X \times Y$)

But please double check on main, I haven't done that stuff in a while

@Daminark Cool cool

Semiclassical

Linear algebra question :

Suppose I have two rectangular matrices $A,B$ of the same shape. Then they have the same rank iff there are square matrices P,Q such that $ PA=BQ$.

That’s a known result.

Perturbative

@Daminark We had strikes today at campus actually

Shobhit

@Perturbative its cool. I think i should change "for given $(x,y)$" to "for every $(x,y)$", then all such $U \times V$ will be member of basis of $X \times Y$. Will this be ok?

Daminark

Huh, were any classes cancelled?

Semiclassical

7:02 PM

What if one requires that $P$ is the identity matrix?

Perturbative

@Shobhit Well you've picked $(x, y)$ arbitrarily (I assume) so it would hold for every (X, y) \in W$

Semiclassical

Presumably that imposes a stronger constraint than just equal rank

Shobhit

yes, thats true. Thank you for your help. @Perturbative

Perturbative

@Dami Yeah, whole academic programme got shut down

@Shobhit No problem :)

Tobias Kildetoft

@Perturbative We will soon be going on strike. Or rather, some places some people will go on strike, and as a response the university will lock out the rest of us

Semiclassical

7:03 PM

Nm, I found it

Perturbative

What maths are you up to these days? @Dami Still doing the dank algebraic topology stuff?

@TobiasKildetoft And here I thought campus strikes didn't happen that often in other places around the world

Tobias Kildetoft

@Perturbative This is not really a campus thing. The entire public sector will basically be shut down in about 3 weeks

Perturbative

Damn

I guess the people on sabbatical won't be too happy, they won't be getting a three week refund on the time they could've got at home for free if they weren't on sabbatical

Anyway my test prep calls, night y'all

Daminark

I think most people on sabbatical are tenured profs, in which case I don't think they're paid based on time, right?

So next quarter I will be doing AT, it'll be more of the standard business though

Dattier

7:34 PM

@Lozansky : I forgot this x outside, it takes another trick to unlock the situation if I find I put it here.

Xander Henderson

@Daminark Sabbatical is not just for tenured professors.

Lozansky

@Dattier Sure thing, just ping me. I'm doing some other stuff right now

Xander Henderson

It is possible for non-tenured profs to go on sabbatical.

Typically, they are funded by other sources while on sabbatical, as the "idea" of a sabbatical is that you are off and doing research.

Dattier

@Lozansky ok

Tobias Kildetoft

Still not quite sure how I am supposed to not be working while locked out. Will it require hitting my head with something first?

ÍgjøgnumMeg

7:45 PM

Just made the mistake of thinking that $\sigma(B) = B$ implies $\sigma(b) = b$ for all $b \in B$ and now I want to die

Tobias Kildetoft

@ÍgjøgnumMeg I actually think that is a fairly common mistake to make

ÍgjøgnumMeg

@TobiasKildetoft I know but I've been doing algebra solidly for 2 or 3 years and it's such a silly mistake

lol

0celo7

what is $\sigma$

ÍgjøgnumMeg

it's just a map, the point was just that I mistook a set equality for an elementwise equality

Tobias Kildetoft

We all make silly mistakes at times. I had spent a ton of calculation on some basis changes until MatheinBoulomenos pointed out that what I was trying to figure out followed trivially by CRT

And a few days ago I commented on a question essentially making it seem very obvious why something should hold. This turned out to take a ton more argument than I had thought, and some of the objects I work the most with gave examples of why my simple reasoning was not enough.

ÍgjøgnumMeg

7:57 PM

@TobiasKildetoft Lol, it's funny when it's something you work with all the time

Complacency I guess

overexchange

8:27 PM

@micsthepick Did you get my question?

Tuki

hello chat

Mikhail

8:48 PM

I'm feeling bad for not being able to take a fourier transform, or rather maybe I took it correctly but have no clue

Xander Henderson

9:06 PM

Fourier transform of what?

Daminark

A function, likely

Mikhail

0

Q: Fourier Tranform of a Parabolic Dish

I'm trying to draw the world's best dish by drawing it in the frequency domain and taking the inverse 3D Fourier transform. I've used this approach to draw other shapes at high precision. So my dish is a surface that lives in $R^3$ defined as: $\text{Real}\left[\sqrt{1-x^2-y^2}\right]$ This is...

fourier-transform

Basically I'm messing up the dz dimension integral, I think I need to stick a $\delta$ in there somewhere

Always confused about how to take transform of $R^2$ embedded in $R^3$

micsthepick

9:28 PM

@overexchange so according to your first chat post, you wanted to come up with some expression that is equivalent to 2018?

Semiclassical

9:56 PM

@Mikhail I’m a big confused by the setup. Isn’t $z=\sqrt(1-x^2-y^2}$ a hemisphere ?

A paraboloid would actually be easier since there’d be no square root

Akiva Weinberger

10:11 PM

Cello hat

Mikhail

@Semiclassical Yeah, its a hemisphere I'm thinking maybe take the transform of the signal including the imaginary numbers (aka remove the real in coordinate space) and when I draw my fancy shape take the inverse, FT on that end. Right now I'm handling the "real" requirement by limiting the integral to only real values.

Akiva Weinberger

If it's a hemisphere then it's not a parabolic dish

Mikhail

Indeed, anyways I fixed the question.

Akiva Weinberger

Relevant: physics.stackexchange.com/q/25275

Xander Henderson

10:43 PM

@Daminark I prefer to take the Fourier transform of a measure.

It is more fun that way

Joe Shmo

has anyone seen Ted lately?

0celo7

@JoeShmo wow no love for me :'(

Joe Shmo

@0celo7 one love mun

cmon its not like you have multivariable analysis lectures on youtube that can explain the material even to an idiot like me

0celo7

11:00 PM

ted doesn't like that kind of talk

Joe Shmo

Jamaican?

0celo7

what?

where does Jamaica fit in here

Joe Shmo

one love mun

0celo7

oh

no

Joe Shmo

youre butchering my jokes here ocelo. im trying to be hysterical.

0celo7

11:02 PM

"an idiot like me"

Joe Shmo

oh there he is

Ted Shifrin

Oh oh ... I picked a horrrrrrid time to show up.

?

howdy ted

Howdy, Shmo.

11:03 PM

hi

Ted Shifrin

hi 0celo

Joe Shmo

the comp sci kids are talking about the peter pan syndrome next to me. cute.

millennials

(full disclosure -- i am a millennial)

Ted Shifrin

I don't even know what I am. Ancient.

0celo7

you're a millennial if you're <39 or something

Joe Shmo

baby boomer?

Ted Shifrin

11:05 PM

Did you actually have a math query for me?

Joe Shmo

nope. just saying hi

0celo7

I do, Ted.

Joe Shmo

watching your implicit value theorem proof

end of lecture 21, beginning of 22

Ted Shifrin

So you have manifolds all figured out?

Joe Shmo

um, no.

0celo7

11:06 PM

@TedShifrin To what extent to holomorphic maps $\Bbb C^k\to\Bbb C^k$ preserve angles?

Ted Shifrin

The meat of the proof is inverse function theorem, the way I present stuff.

Joe Shmo

an embedding of R^n into R^(n+k) ?

Ted Shifrin

@0celo: There's no reason to expect them to when $k>1$. You'd need the derivative to be a scalar multiple of the identity.

Joe Shmo

^^yuh

Ted Shifrin

What is that supposed to mean, @JoeShmo? That's an example of a submanifold, but far from the general case. Consider spheres, tori, etc.

0celo7

11:08 PM

@TedShifrin Hmm, $k\ge 2$ is of course what I'm curious about. Is it neccessary that the derivative is propto the identity?

Joe Shmo

do you have lectures about manifolds @TedShifrin?

Ted Shifrin

Well, I lied. It has to be a scalar multiple of a unitary matrix.

0celo7

Right.

Ted Shifrin

But that's still relatively rare.

0celo7

A different matrix at each point or is it so rigid that it's the same one everywhere?

Ted Shifrin

11:10 PM

@JoeShmo: There are a few (they're introduced in the first semester in a couple of lectures) and then I talked about the three equivalent formulations for a submanifold of $\Bbb R^n$. Did partitions of unity with the proof of Stokes, etc.

Joe Shmo

That's 3500, not 3510, right?

Ted Shifrin

You mean the unitary part, 0celo? I haven't thought about this, but I don't see why it shouldn't vary.

Yeah, @JoeShmo, but the substantive examples are in 3510. I guess Patty didn't label the lecture with manifolds.

0celo7

@TedShifrin Ok. Would you believe me if I told you that a map $\Bbb C^k\to\Bbb C^k$, conformal in the Riemannian sense, is injective? Maybe this is elementary.

Ted Shifrin

It should be after the implicit function theorem, just before differential forms.

0celo7

Note: not assumed to be a diffeomorphism.

Ted Shifrin

11:12 PM

0celo, holomorphic map?

0celo7

Well, holomorphic does not imply R-conformal, as you've just confirmed.

But suppose it does preserve angles.

Ted Shifrin

Are you assuming holo? Otherwise it's just a Riemannian statement in dim 2k.

0celo7

Yes, it is a Riemannian statement in dimension 2k.

Ted Shifrin

Oh.

0celo7

I was wondering if such a thing was known to complex people in that setting.

Ted Shifrin

11:14 PM

I'm hardly the arbiter of what's known, but it's not known to me.

But this isn't the sort of stuff I ever thought about.

So it's obviously a very global result, or else $f(z)=z^2$ gives a counterexample.

0celo7

Ah, durr.

This is just Liouville's theorem.

Ted Shifrin

So the derivative can never drop rank.

This almost sounds like the Jacobian conjecture.

0celo7

Or does that assume bijectivity. Hmm...

This should be somewhere in Spivak, because it should hold for $\Bbb R^3\to\Bbb R^3$.

Ted Shifrin

No, it's just about open subsets of $\Bbb R^n$.

It is. I just looked. Beginning of volume 4.

But nothing special about parity of dimensions for Liouville, obviously.

0celo7

"Conformal map." I'm assuming he doesn't need injectivity there?

overexchange

11:19 PM

@micsthepick That's correct

Ted Shifrin

Not that I see. Nor does the proof use injectivity.

But they all end up injective if they're compositions of inversions and similarities.

0celo7

Right.

Joe Shmo

aced the second assignment @TedShifrin, sounds like some of the other people in the class are struggling. I owe it all to you...

Ted Shifrin

So in the holomorphic category it would seem we can only have similarities.

I doubt that's true, @JoeShmo, but I'm very glad you're doing well, despite the frustration.

0celo7

@TedShifrin Well this dashes any hope of me finding a counterexample for something. Hmm.

Ted Shifrin

11:23 PM

Sorry I didn't help, 0celo.

0celo7

No problem. The only people who can help are probably Yau and Rick.

Ted Shifrin

Well, I'm friendly with Rick. Not with Yau. :P

Daminark

Hello

Ted Shifrin

Hi Demonark

0celo7

My advisor asked me not to ask Rick until we've thought about this more.

@Daminark hi

Ted Shifrin

11:25 PM

My go-to person for many years has always been Robert Bryant. I haven't ever asked him something he can't nail in moments.

0celo7

Is he a GR person?

Ted Shifrin

Nope. Although I'd be shocked if he doesn't know some of that too.

0celo7

We tried to get Bray up here but he doesn't like to travel, apparently.

I had lots of questions for that guy.

Ted Shifrin

well, he has multiplicity 2 ... but I assume you mean a particular 1 of them.

0celo7

We're getting his student who was in the Bulletin a couple months ago, hopefully he can answer those questions.

@TedShifrin The one at Duke.

Ted Shifrin

11:30 PM

I figured.

overexchange

@micsthepick cs61a.org/assets/slides/02-Names_full.pdf#page=14

Daminark

Is Bryant the forms guy?

Ted Shifrin

LOL, among other things, yes.

Fargle

Hello chat.

Ted Shifrin

Fargle!!!

I was just thinking about you earlier as I commented a few times on Ken Knox's Facebook post.

Fargle

11:43 PM

lol

0celo7

@TedShifrin What is he up to now? I never met him when he was here.

@Fargle Huh, we're at the same school. Who are you?

Fargle

I should communicate with him a bit more. I really enjoyed his class.

Ted Shifrin

He's teaching at a small school in western MA.

Fargle

@0celo7 I used to be there--not anymore.

0celo7

Deja vu. Have we had this conversation before?

Ted Shifrin

11:44 PM

i have to keep annoying you, Fargle :)

0celo7

Like, years ago.

Fargle

We may have, lol

I have a terrible memory for specific social interactions, web or otherwise

Ted Shifrin

Fargle doesn't even remember who I am.

Fargle

@Ted By coincidence, I was just looking at your diff geo notes again today.

Ted Shifrin

Oh, well, maybe he does.

Were you trying to suffer, Fargle?

Daminark

11:46 PM

Some folk are masochists

Ted Shifrin

And some folk actually like to learn mathematics well :)

0celo7

@Fargle Where are you now?

Fargle

Haha, not at all. I just wanted to take another crack at it. I feel like I'm somehow getting more out of the first bits, even thought that may be something like squeezing juice from a stone.

Ted Shifrin

There's plenty of challenging exercises even in the first chapter.

Personally, I find surfaces a richer subject.

Fargle

@0celo7 Other side of Nashville--Austin Peay.

0celo7

11:48 PM

@Fargle You should apply for funds for our mean curvature flow conference in May.

Fargle

@0celo7 That sounds like it could be fun. I'll see what I can do.

Ted Shifrin

Then you'd better read Chapter 2, Fargle :P

0celo7

@Fargle math.utk.edu/barrett

Ted Shifrin

Did either of you ever take a class from Jim Conant?

0celo7

No, but I have some of his books that he left behind.

Ted Shifrin

11:49 PM

He's now my boss at AoPS ... and we know zillions of the same people.

0celo7

I heard it was a strange situation that made him leave.

Ted Shifrin

nah ... not strange. He just left academia to be the head of this AoPS school so that he could live in the same city as his wife. Not an uncommon desire for people of that persuasion.

He and I exchange occasional geometer/topologist barbs :)

0celo7

Well his wife seemed to have randomly moved while he was at a conference or something. That's the scuttlebutt, anyway.

Ted Shifrin

I think she's been in San Diego for quite a while, but anyhow this is not the place for this ....

Fargle

I don't think I had a class with him, but I did know the name.

Ted Shifrin

11:51 PM

We're both finding the teaching experience rather different from what we were used to at universities.

0celo7

I have his Hungerford and his Hempel.

And his Lang.

Ted Shifrin

LOL, OK. I'll tell him they're in the hands of a geometric analyst.

Fargle

Is Dr. Thistlethwaite still there?

Ted Shifrin

Now that I have to move again I wish I'd kept even fewer books.

0celo7

Morwen is alive and well.

Did you take his 400 level topology?

Fargle

11:54 PM

Awesome. I only interacted with him for a bit (he was my honors mentor, but during the Year Which Was The Reason I'm Not There Anymore) but I got quite a lot out of it. Plus, he just seemed pretty cool.

No, would have liked to though.

Ted Shifrin

Your year subtitle reminds me a lot of A.A. Milne and Winnie the Pooh :)

0celo7

It seems a lot more interesting than the 500 sequence that I took from him. More knots and less separation axioms.

Ted Shifrin

wait — it's past @Alessandro's bedtime. This must be his phantom.