Mathematics - 2016-06-06

12:05 AM

@Jeff $\newcommand{\d}{\operatorname d\!}$According to Wolfram Alpha, it's not an elementary function. Specifically, it's equal to $\displaystyle\frac12x\sin(x^2)-\frac12\int\sin(x^2)\d x$, and it is known that $\displaystyle\int\sin(x^2)\d x$ isn't elementary.

Jeff

wolfram alpha gave me something about a Fresnel intgral @AkivaWeinberger

Akiva Weinberger

Yeah, $\int \sin(x^2)\d x$ is essentially the definition of the Fresnel integral (except scaled a bit)

@Jeff mathworld.wolfram.com/FresnelIntegrals.html

There's a $\frac12\pi$ in the definition for some reason, not sure why

Jeff

@akiva this is a homework problem for a calc 2 class.

Akiva Weinberger

You sure you didn't copy the problem incorrectly?

Jeff

sure.

Akiva Weinberger

12:09 AM

Like, maybe it's just $\int x\cos x^2\d x$

instead of $x^2$ twice

Jeff

that's also part of the problem

Akiva Weinberger

Huh

I dunno

Jeff

it gives several integrals to do. i'll find it again and post them

haha, the instructions say to calculate FOUR OF the following integrals: $x \cos x^2, x^2 \cos x^2 (the one i can't do), x^2 \cos x, x^2 \cos^2 x, x \cos^2 x$. @akiva

Akiva Weinberger

Why would they do that

Jeff

next time i'll read the instructions

Akiva Weinberger

12:14 AM

That just seems intentionally mean

Jeff

that's what these types of problems are. they're "workshops". supposed to be multi-step, thought-provoking questions.

so yeah... intentionally mean :D

OneRaynyDay

1:23 AM

Hey guys! Quick question about lin alg - can anyone give me some intuition on the relationship between kernel and image of a nonsquare matrix?

If we're given a 3x4 matrix, there can only be max im size of 3, right? This means the kernel cannot be nontrivial in this case? Just making sure

0celo7

1:53 AM

@MikeMiller If you don't mind helping me I'm here

Mike Miller

Well, you should start with why it's true; we can work on relevance after that.

0celo7

@MikeMiller Ok

@MikeMiller I don't see what to do with that $\mathbb R^{m-n}$

Mike Miller

what does it mean to be in $F^{-1}(\{y\} \times \Bbb R^{n-1})$

0celo7

What I can show is that $f^{-1}(y)\subset F^{-1}(\{y\}\times \mathbb R^{m-n})$

But not equality

@MikeMiller that $F(x)\in \{y\}\times\mathbb R^{m-n}$

Mike Miller

@0celo7 well, that doesn't actually make sense, because the two things you wrote there live in different spaces

0celo7

2:03 AM

@MikeMiller typo

Mike Miller

@0celo7 and decompose this further?

0celo7

@MikeMiller $(f(x),L(x))\in \{y\}\times\mathbb R^{m-n}$

$f(x)\in \{y\}$ and $L(x)\in\mathbb R^{m-n}$

Mike Miller

and what's the deal with the latter

0celo7

well this is always true

Mike Miller

so conclude

0celo7

2:06 AM

$f^{-1}(y)\subset F^{-1}(\{y\}\times \mathbb R^{m-n})$

Hmm, maybe it's $\supset$.

Mike Miller

it's both

0celo7

I don't see that

Mike Miller

you just pointed out that the latter set is the same as the set of $x$ such that $f(x) \in \{y\}$ and $L(x) \in \Bbb R^{m-n}$

agreed?

0celo7

No

Mike Miller

let me know when you agree

0celo7

2:08 AM

Because I'm not convinced that $L(f^{-1}(y))=\mathbb R^{m-n}$

Mike Miller

that's fine, since it's false

0celo7

then I don't understand what you want me to prove

@MikeMiller Ok I agree

Mike Miller

now the rest of the point is you invert $F$ to get a map $F^{-1}: V \to U$; restrict this to $\{y\} \times \Bbb R^{m-n}$ to get a chart on $f^{-1}(y) \cap V$ near $x$

0celo7

Oh of course

I think...

@MikeMiller you mean $f^{-1}(y)\cap U$, right?

not $V$

Mike Miller

yeah, sorry, I stuck that in the wrong place

$F^{-1}: \{y\} \times \Bbb R^{m-n} \cap V \to f^{-1}(y) \cap U$

0celo7

2:23 AM

@MikeMiller Ok, so $F^{-1}(\{y\}\times\mathbb R^{m-n}\cap V)=F^{-1}(\{y\}\times\mathbb R^{m-n})\cap F^{-1}(V)=f^{-1}(y)\cap U$?

Mike Miller

yeah

0celo7

Ok, but why isn't it true in the other direction?

Mike Miller

don't know what that means

0celo7

I'm probably being dumb, but why doesn't $F^{-1}(\{y\}\times\mathbb R^{m-n})=f^{-1}(y)$ imply $F(f^{-1}(y))=\{y\}\times\mathbb R^{m-n}$?

I mean, Milnor literally says "Note that $f^{-1}(y)$ corresponds, under $F$, to the hyperplane $\{y\}\times\mathbb R^{m-n}$."

Mike Miller

in general neither $f(f^{-1}(A))$ nor $f^{-1}(f(A)) = A$

0celo7

2:27 AM

@MikeMiller Ok I know that, but I'm curious about his wording (poor phrasing on my part)

Mike Miller

your first question is not the same as the second... I mean "why doesn't blah imply blah" is answered by "cuz it doesn't", which is what I just said

0celo7

I know I know

But Milnor literally says it!

Mike Miller

He says some words that you interpreted to mean something false

0celo7

@MikeMiller So "$A$ corresponds to $B$ under $f$" does NOT mean $f(A)=B$?

How is one supposed to know that?

Mike Miller

clearly not since as you say it's obviously false; as above it means $f^{-1}(B) = A$. now you know it, and will leave a happy and productive life

0celo7

2:31 AM

@MikeMiller but how is the reader supposed to know that

If I hadn't asked you I would have sat for another 4 hours staring at it, or more

Mike Miller

I'm sorry to hear that. I'm not really interested in talking about his choice in language.

0celo7

...thanks

SoumyoB

2:55 AM

hello guys

does anyone here know of any good books on the topic of simulating the traveling salesman problem?

Semiclassical

5:12 AM

I suppose it'd be mean to answer this question...

0

Q: How to define a function recruisively?

I'm currently trying to work out a function where f(x) = 2x + 4 The question is to define f(x) recursively I'm unsure what this means, I have an idea of how to do it but I'm not sure if it's right. The way I thought of doing it was: f(0) = 2*0 + 4 = 4 f(1) = 2*1 + 4 = 2 + 4 = 6 f(2) = 2*2 + 4...

functions

with "By defining it recursively." (as a comment)

SoumyoB

hello

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

hi

SoumyoB

@Semi

@Semiclassical I think the following should work:

Define $f(n) = f_n$.
Then $f_n = 2n+4 = 2(n-1) +4+2 = f_{n-1}+2$.
Hence $f_n = f_{n-1}+2$ is the required recurrence relation

Semiclassical

Perhaps. I was making a joke, not a judgment on the question.

SoumyoB

oh my bad

Semiclassical

5:22 AM

np

SoumyoB

so uh... anyone familiar with the traveling salesman problem?

Bernhard

6:46 AM

Hi all, I need a small hint and don't have my calculus books at hand. I need an approximation for z-->infinity. For small z I know Taylor, what do I need for large z?

My function goes like sin(sqrt(x^2+a)-x)

(I know the solution would scale with 1/x, but I want to know how to get there)

Aneesh

7:12 AM

Hi, can someone help me with a logic problem?

its the 'green eyes' or 'common knowledge' problem, I typed up a solution here: math.stackexchange.com/questions/1815471/…, but I'm not sure if its accurate

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

8:28 AM

25 million people

lattice

10:01 AM

Hey, is there someone here right now who's familiar with smooth manifolds?

Balarka Sen

I know a little bit about them.

What's your question?

lattice

I just need some verification that I got the definition of the rank of a map f:M->N right

Balarka Sen

What about them?

lattice

so to find the rank of f at a point p, I just need a chart on M near p, say (phi,U) a chart on N near f(p), say (psi,V), and then the rank of f at p is the rank of the Jacobian of psi°f°phi^-1?

*at p

Balarka Sen

That's right. You need local coordinates around $p$ and $f(p)$, and look at the rank of $Df$ at $p$.

lattice

10:05 AM

okay, thanks a lot!

Balarka Sen

Er, no problem.

Huy

@BalarkaSen: do you know what to call groups satisfying the following two properties?
1) if $g, h \in G$ are not conjugate, there exists a finite group $F$ and a homomorphism $\varphi: G \to F$ such that $\varphi(g)$ and $\varphi(h)$ are not conjugate and
2) if $\varphi: G \to G$ is an automorphism such that $\varphi(g)$ is conjugate to $g$ for all $g \in G$, then there exists some $h \in G$ such that $\varphi(g) = h g h^{-1}$ for all $g$.

Agawa001

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ that is half the population of a continent large country

all gathered in one place ? makes me wonder if the crust is unbalanced in this exact spot of the earth ?

Huy

10:23 AM

I think the first property is called conjugacy seperable

no idea for the second, something like "pointwise inner implies inner"

Balarka Sen

10:47 AM

@Huy No idea

Ask Tobias, maybe.

Evinda

11:00 AM

Hello

Lemma: Let $\psi \in C_C^{\infty}(\mathbb{R}^n), \psi \geq 0, \int \psi(x)dx=1, \psi_{\epsilon}(x)= \epsilon^{-n} \psi{\left( \frac{x}{\epsilon}\right)}, \epsilon>0$. Let $f \in L^2(\mathbb{R}^n)$ and $f_{\epsilon}(x)=f \ast \psi_{\epsilon}=\int_{\mathbb{R}^n} f(y) \psi_{\epsilon}(x-y) dy$. Then $f_{\epsilon} \in C^{\infty}(\mathbb{R}^n), f_{\epsilon} \to f$ in $L^2$ while $\epsilon \to 0$. Furthermore $||f_{\epsilon}||_{L^2(\mathbb{R}^n)} \leq ||f||_{L^2(\mathbb{R}^n)}, \epsilon>0$.
In order to show that it is differentiable:

bolbteppa

"Formally currents behave like (Schwartz) distributions on a space of differential forms. In a geometric setting, they can represent integration over a submanifold, generalizing the Dirac delta function, or more generally even directional derivatives of delta functions (multipoles) spread out along subsets of M."
lolwut, multipole is directional derivative of Delta function? You can define delta functions on more than a point, but on a submanifold???

sepideh

12:38 PM

Hi folks,

I couldn't find a room dedicated to statistics in Math.SE

The question didn't get much attention in Cross Validated, too

so I decided to move it to Math.SE

May you have a look at the following question

and tell me if there's any problem in the way of asking?

that doesn't attract much attention!!!

0

Q: Can we model this set of experiments as an stochastic process and estimate the sample size?

I have an image with the size 5575x9440 and I'm implementing a modified version of the algorithm used in this paper on it, but because the code performance is low right now, I have divided the image to 52628 submatrices of the size 25x40 (1000 pixels) and my first experiments show that some line...

statistics probability-distributions stochastic-processes random-variables sampling

Abhishek Bhatia

1:41 PM

hi guys!

I have question on how to calculate the expected value for an uniform distribution. $E(u^2)$, where u~unif(0,1).

Now this evaluates to integral of $\int_{0}^{1} u^2 f(u).du. $

In a lecture I the prof. wrote $f(u)$ as 1.

I am unsure how you take it out of the integration.

Reference: youtube.com/watch?v=Tci---bVs60 Time 42:20

SoumyoB

2:26 PM

guys I forgot whether $A^B$ means functions from $A$ to $B$ or vice versa

Akiva Weinberger

@SoumyoB If $A$ has $2$ elements and $B$ has $3$, there are $2^3$ functions from $B$ to $A$

Since we want to have $|A^B|=|A|^{|B|}$ we define $A^B$ to be the functions from $B$ to $A$

SoumyoB

ahhhh right I should have thought of it that way

Thanks @AkivaWeinberger

Akiva Weinberger

Also:

$(A^B)^C$ and $A^{B\times C}$ have a natural bijection when we define it like this

$f(c)(b)$ and $f(b,c)$

SoumyoB

yeah I guess we wanted the notation of sets as exponents to be as close to algebraic expressions as possible

by the way where are you from @AkivaWeinberger

I had a strong gut feeling you're from Russia xD

Akiva Weinberger

Nope, NYC

Where did "Russia" come from?

@SoumyoB

SoumyoB

2:41 PM

well it was just a gut feeling...

may be because of your name?

also I have a stereotype in mind that Russians are expert as hell in math

Balarka Sen

3:11 PM

Weinberger is a sort of German surname, not Russian.

user 1618033

Jewish name

Balarka Sen

"Akiva" is Jewish, yes.

@AkivaWeinberger What's up?

Akiva Weinberger

Just had my last math class

Balarka Sen

Last?

Akiva Weinberger

We just did some exercises (Rudin Chapter 3, sequences and series), kinda boring honestly

Balarka Sen

3:19 PM

I think analysis is boring if not put in a larger context, yes. But then I do not know too much analysis.

Akiva Weinberger

Last this year, yeah. Tomorrow's my last day of school but my teacher has a funeral to go to that day apparently

Balarka Sen

Oh, are you going to a university, then?

Akiva Weinberger

The subset as a whole isn't boring, the exercises kind of were

What?

No, last day of school this year, I mean

Balarka Sen

Gotcha.

Akiva Weinberger

Going into 11th grade

Balarka Sen

3:22 PM

@AkivaWeinberger The exercises may seem ad-hoc, if that's what you are saying.

But I found sometimes putting them in a larger context helps them motivate.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

3:46 PM

:O

Akiva Weinberger

4:00 PM

@AkivaWeinberger *Subject, not subset

There was one exercise about how, assuming $a_n\ge0$, if $\sum a_n$ converges then so does $\sum\sqrt{a_n}/n$

I don't know if I solved it the intended way

Uria Mor

@AkivaWeinberger In what schools do they teach Rudin? (just curious)

SoumyoB

4:19 PM

if $X$ is a stochastic process, then how do we prove that it's a Martingale if its expectation remains constant throughout the entire filtration?

Balarka Sen

@AkivaWeinberger Cauchy-Schwarz?

Evinda

Hey @MikeMiller
Could you maybe take a look at my question?

0

Q: Show property of convolution

Proposition: Let $ u, v \in L^2(\mathbb{R}^n) $ then $ \widehat{u \ast v}=\widehat{u} \cdot \widehat{v}$. Proof: We want to show that $\mathcal{F}^{-1} (\widehat{u} \cdot \widehat{v})=u \ast v $. We have that $ |\widehat{u} \cdot \widehat{v}| \leq \frac{1}{2}|\widehat{u}|^2+\frac{1}{2} |\wideha...

lp-spaces distribution-theory

Akiva Weinberger

4:47 PM

@BalarkaSen Ohh. This?:$$\sum\frac{\sqrt{a_n}}n\le\sqrt{\sum a_n\sum\frac1{n^2}}$$

That works a lot better than what I had.

I had $\sum\sqrt{a_n}/n\le\sum(a_n+1/n^2)$, which is proven by considering separately the cases $a_n>1/n^2$ and $a_n\le1/n^2$.

Evinda

If $|u(-x)| \to 0$ while $x \to +\infty$ it doesn't also hold that $|u(x)| \to 0$ as $x \to +\infty$, right?

Akiva Weinberger

Is $u$ any function? What about $u(x)=e^x$? @Evinda

Balarka Sen

@AkivaWeinberger That's right.

Akiva Weinberger

So, right, it doesn't always hold

Evinda

@AkivaWeinberger A function $u \in H_s$, where $H_s=\{ u \in S'(\mathbb{R}^n): (1+ |\xi|^2)^{\frac{s}{2}} \widehat{u}(\xi) \in L^2(\mathbb{R}^n)\}$

Mike Miller

5:19 PM

@BalarkaSen Did you have thoughts about either of the things I asked?

Balarka Sen

@MikeMiller Nope, sorry, I have been busy today. Probably will get back to you tomorrow.

DeNiSkA

5:34 PM

anyone knows how do we put this kind of thing on wolfram or desmos $$
f(x) =
\begin{cases}
1+x,\,\,\,\,\,\,\,\,\,\,\ 0\leq x\leq2 \\
3-x,\,\,\,\,\,\,\,\,\,\,\, 2<x\leq3
\end{cases}
$$

Semiclassical

5:55 PM

This problem hurts my eyes to look at: math.stackexchange.com/q/1816176/137524

Agawa001

6:14 PM

@user1618033 hey, writing details to ur book now ?

user 1618033

hehe, precisely! These hours were painful, I rewrote a couple of parts from a proof again and again and I wasn't content with what I had till now when I've found a brillinant way of proving things.

@Agawa001 Some hours here may be painful (until you find that satisfactory way of writing things).

user147690

6:39 PM

I can't see a latex symbol for some reason in the lists, and I can't get it on dextify, it's like a smaller version of $\vee$ in the superscript, like $L^\vee$ (for the dual of an invertible sheaf is the context)

user147690

Smaller and flatter

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

@Semiclassical Professor Hofstadter's lecture on YouTube hurt my ears :P

Mike Miller

$\lor$

$\lor$ vs $\vee$

huh, guess they're the same

Hiroto Takahashi

Hi everyone

somone know what does 'plane conic' mean?

Mike Miller

in any case, \vee is a common choice

Hiroto Takahashi

6:46 PM

I was searching and I don't find any exact definition

Only about section conics

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

a conic cut by a plane

Hiroto Takahashi

I have the following question: Show that any smooth projective curve of genus zero over a field K is isomoprhic to a plane conic over K

But what does plane conic mean?

Mike Miller

A conic in $\Bbb P^2$.

user147690

Ahh sure, I just found the one I had in mind, although it uses a package

user147690

$\sqsmile$ and $\sqfrown$

Mike Miller

6:51 PM

at the very least I've seen something about the size of vee in print, tho of course I don't know what they actually used

user147690

Seems fair, I'll try out the symbol they 'made' for me, or I'll just use the \vee. Since apparently the package is bad

user147690

in TeX, LaTeX and Friends, 3 mins ago, by yo'

@AlexClark Don't load mnsymbol, it's insane

user147690

in TeX, LaTeX and Friends, 2 mins ago, by yo'

@AlexClark it's intended for use with Minion fonts, and it changes a lot of things. It's visually incompatible with any other font, IMHO

Mike Miller

like the little yellow cartoon ddues?

user147690

I certainly hope not haha

user147690

6:56 PM

Mike's answer was more relevant^

Balarka Sen

7:47 PM

ok, I think I am free tonight. time to do math.

Akiva Weinberger

Why is there complex analysis but no quaternion analysis

Does the lack of commutativity make everything worse?

arctic tern

yes. I wrote about some quaternion calculus here.

Balarka Sen

@AkivaWeinberger.

but it should be very much more complicated than complex analysis. the whole reason complex analysis is different than real analysis is because of the automorphism on C coming from multiplication by i. Here you have a whole lot more "natural" automorphisms.

arctic tern

8:02 PM

most of the cool stuff from quaternions comes from the geometry and lie theory, not generalization of complex analysis

Akiva Weinberger

I'm curious what happens if you try to integrate $1/z$ on a 3-sphere around the origin.

arctic tern

what do you mean by dz when z is a quaternion?

Balarka Sen

There should be some analogue for differential forms, but everything would be noncommutative :S

Akiva Weinberger

Why would that be harder than in $\Bbb C$ or $\Bbb R$?

Oh, sorry, I don't actually know about forms

arctic tern

what do you mean by integrating 1/z on a 3-sphere? if you just mean integrate 1/z with respect to |dz|, then even that would be 0 over C by symmetry.

Akiva Weinberger

8:09 PM

Hm. Forget integrating on 3-spheres — just integrating on a circle would be interesting. If the path is a subset of $\Bbb C\subseteq\Bbb H$, it depends on if the origin is inside it, but you can homotopy one to the other, so what happens there?

arctic tern

you mean you can homotopy a circle in H to a circle in C?

you'd have to be careful about homotopying in H changing the value of the integral probably.

for instance you can homotopy a (loop not around 0 in C) through H to a (loop around 0 in C)

Akiva Weinberger

Right, that's essentially what I just said

Balarka Sen

@AkivaWeinberger I mean, over C, you declare dz = dx + idy. But here you'd need to write it as dx1 + idx2 + jdx3 + kdx4, and everything would be noncommutative.

Akiva Weinberger

Ah.

You can still define integration over paths and loops, though, right?

arctic tern

you get two versions depending on if you put dz in front of or behind the integrand, but yes

Akiva Weinberger

8:14 PM

Oh, I see.

Balarka Sen

you have to store the information about the algebra of H somehow. i don't know how to do that easily, is all I am saying.

arctic tern

the differential equation x'(t)=x(t)a(t) would be interesting over the quaternions. essentially if you draw a path on a 2-sphere, there is a path in SO(3) which effectively rotates the sphere so a particle moves along the path, and to find the "best" such path you'd solve that differential equation in H.

Akiva Weinberger

Huh, apparently Mike knows a little about this

If there is you should be able to find it in here. Note that the correct definition of a quaternion holomorphic function is that it satisfies a certain version of the Cauchy-Riemann equations, not that the limit $\lim_{h \to 0} h^{-1}(f(x+h)-f(x)$ exists. That paper proves that such functions are actually linear, IIRC. — Mike Miller Oct 14 '15 at 21:21

Mike Miller

I read that paper once is all. I'm sure anon knows more than I do about this.

Akiva Weinberger

That qualifies as "a little"

Another link: zipcon.net/~swhite/docs/math/quaternions/analysis.html

Mike Miller

8:21 PM

I was mostly just amused at what happens when you try to write down the naive notion of quaternion differentiable. It's probably reasonabłe to assume that the CR equariona give a much better theory.

There are also good notions of quaternionic geometry, but in defining those you tend not to use quaternionic charts etc.

TheGreatDuck

I was told my Q&A post pair is unclear

math.stackexchange.com/questions/1814961/… I think I fixed it

is it clear now?

Huy

@MikeMiller: any idea?

TheGreatDuck

You guys talking about quaternions?

Mike Miller

the first condition is "conjugacy-separable group"

Akiva Weinberger

Mike, I can't say I understand that arxiv paper

Mike Miller

8:33 PM

I suspect there's no name for condition (2) - see here

Akiva Weinberger

@TheGreatDuck I asked about "quaternion analysis," which apparently is a thing (though not well-known, and the noncommutativity messes everything up)

Mike Miller

I can't say I remember anything about it, other than the one theorem

Balarka Sen

suggestion: learn some complex analysis (or more fundamentally, real analysis, which you seem to be studying of late) before learning quaternionic analysis.

i am sure a time will come, then, when you could understand that paper. :)

Akiva Weinberger

I know complex analysis

Mostly

Balarka Sen

good for you.

Mike Miller

8:40 PM

@BalarkaSen To be honest, that sounded a bit condescending

Balarka Sen

my suggestion? I truly didn't mean to sound condescending.

Mike Miller

Ah, the thing I was thinking of wasn't in the afore-linked paper; see rather theorem 1 here

as expected it's just sort of bad luck that the naive notion of derivative doesn't work, and you want to use a different definition, which gives something flavored rather like complex analysis

Tyler Gaona

I must be making a simple mistake, can someone help me out? I'm trying to verify that if $s_k = \sum_{k=0}^\infty \frac{1}{2^k}$ then $s_k = \frac{2^k - 1}{2^{k-1}}$. It's clear that $s_1 = \frac{2^1 - 1}{2^0} = 1$. Suppose it holds for some $k \geq 1$. Now $$s_{k+1} = s_k + \frac{1}{2^{k+1}} = \frac{2^k - 1}{2^{k-1}} + \frac{1}{2^{k+1}} = \frac{2^{k+2} - 2^2 + 1}{2^{k+1}} = \frac{2^{k+1} - \frac{3}{2}}{2^k}.$$ Of course I'm expecting that $\frac{3}{2}$ to instead be 1.

Balarka Sen

I just naturally assumed Akiva doesn't know a lot of complex analysis as he's working on real analysis, so suggested that he should do that before learning the quaternionic version. I of course do not know much real or complex analysis, let alone quaternionic analysis.

(sorry for late follow-up; got disconnected)

Mike Miller

I don't think people really much do quaternionic analysis per se.

Balarka Sen

8:51 PM

I can believe that.

arctic tern

@TylerGaona note that $s_k=\sum_{k=0}^\infty 1/2^k$ does not actually make any sense

you presumably mean $s_k=\sum_{j=0}^k 1/2^j$, which is the $k$th partial sum of $\sum_{j=0}^\infty 1/2^j$.

Tyler Gaona

@arctictern @arctictern yeah sorry i meant the finite sum $s_k = \sum_{j=0}^k \frac{1}{2j}$

that is, $2^j$

Akiva Weinberger

('Cause on the right, $k$ is a dummy variable, and on the left it isn't)

arctic tern

in which case $s_k=(2^{k+1}-1)/2^k$ is the correct formula. (note $s_1=1+1/2=3/2$.) your stuff has indices off by 1.

Tyler Gaona

Oh I see it now. Thanks a lot!

Balarka Sen

9:03 PM

Interesting.

Mike Miller

You can do alright at answering the question for $r=1$. For $r=2$ it's already too hard for me.

Balarka Sen

We can cover $M_{m, n, 1}$ in general with open sets parametrized by $(\Bbb R^n - \{0\}) \times \cdots \times (\Bbb R^n - \{0\}) \times \Bbb R^{m-1}$ similarly, I think.

So I suppose the same argument generalizes. The normal flips sign while moving from chart-to-chart.

Yeah, weird. I don't know anything much about the topology of $M_{m, n, r}$ ("how does it look like?") either.

Mike Miller

There's some combinatorics involved which you haven't done... in particular $M_{m,1,1}$ is obviously always orientable

Balarka Sen

9:18 PM

Yikes, right.

Eh, I guess I can only certainly say $M_{2, n, 1}$ is not orientable then.

Hmm, not immediately clear how to generalize this (the question mentions $M_{3, 3, 1}$ is orientable, so I suppose not all nontrivial M_{m, n, 1}$'s are nonorientable).

Forget the normal thing I said, that's nonsense.

l' - ' - ' - ' - 'l

9:55 PM

Hello.

Eric Stucky

hey

how goes?

l' - ' - ' - ' - 'l

Pretty good, hopefully will get some math done today. Yourself?

Eric Stucky

Similar situation, except more tired I think :)

I'm not yet hungr, but if I wait long enough I will be and the stores will be closed

I just got back from a long weekend and a couple weeks at my grandma's house, so the fridge is pretty empty.

Hoping that MSE chat would help me pass some time, but it's the slow time of day

l' - ' - ' - ' - 'l

Ahhh, yeah I'm not quite tired but only because I woke up at an embarrassingly late time.

Eric Stucky

What are you doing nowadays with the maths?

l' - ' - ' - ' - 'l

10:08 PM

Learning some topology and working on, more or less, a combinatorial problem. You?

Eric Stucky

This is my kind of guy :P

I'm blogging a lot right now, which mostly consists of writing up combinatorics talks and topology prelim problems.

At some point I need to get off my ass and study complex analysis.

But that day isn't today.

Oh, there's going to be a combo reading group starting soon. Not sure what topic we're going to be on yet, but I'm crossing my fingers for polytopes.

l' - ' - ' - ' - 'l

An online one, or in real life? The problem I'm working on has to do with polytopes, so I'd love to learn some things about them.

And I've just checked your blog, looks really nice.

Eric Stucky

IRL, unf. Minneapolis, if you're nearby :P

thanks :D

It's been autoposting while I was on the weekend; I should check what's gone up recently XD

Oh my gods that horrendous hyperboloid post got 4 notes wat

l' - ' - ' - ' - 'l

:C, do you know any good resources for polytopes?

Eric Stucky

Besides Polytopes, presumably? :P

Let me see what the recommended book was for the group.

l' - ' - ' - ' - 'l

10:18 PM

Thanks!

Eric Stucky

(Oh, the book I was thinking about is actually called Convex Polytopes. It's by Grünbaum.)

The suggested book for the group is Lectures on Polytopes by Ziegler.

l' - ' - ' - ' - 'l

Cool, thanks.

Eric Stucky

10:43 PM

I wouldn't laugh if it weren't so damn funny.

I mean, I'm open to serious criticism, from minor corrections to wholesale takedowns. If anyone's interested, have at it.

But yeah, as much fun as I'm having with the blog, I am really looking forward to abandoning all contact with the vagueblogosphere.

Mike Miller

why so?

Balarka Sen

To me, in a proof of a theorem or a solution to a problem, I find "insight" (or, from a more biased point of view, the geometric content!) and "trick" (again, from a biased point of view, the symbol-pushing) to be very distinct mathematical methods.

I like to think any proof is made up of those two bits. Some amount of trick is needed as much as insight.

Eric Stucky

Hmm, balarka, it's interesting to me that you describe trick as e.g. symbol pushing. I agree that symbol pushing tricks exist (multiplying by 1 &c), but my experience hasn't given me too many of them.

Mike, I'm not cut out for it. I want to be disagreed with privately or publicly. The half-public nature of vagueblogging sends me off the rails.

Maybe in another year and a half I'll have thicker skin.

(whoops, thinko on that one...)

Balarka Sen

11:01 PM

Eric: I have faced such situations in algebra and analysis, where a point comes when you "just keep doing the math" without making the geometry clear (background: usually I approach problems by thinking about it geometrically - once I know what I have to do, geometrically, for getting a proof, I try to translate the vague geometric intuition to mathematics). You need previous experience on this bit, though.

To me, that is what a trick is. You just do something which is not clear why one would do, but you do it because your experience says you to, and everything works out.

Eric Stucky

Hmm, okay, I think I misunderstood. That makes sense.

Somehow this doesn't register as the word 'trick' to me, but I think it's a useful category.

That would be a pretty serious issue with the post, frankly. In this language, to call a 'trick' a theorem somewhat misses the point.

(I mean, yes, it is a theorem that $\varepsilon = \sum \varepsilon 2^{-n}$, for instance, but that classification does miss the point, I think.)

Hi Eric :)

Heya

I have a question

@EricStucky Fair enough, I don't know what you would call a trick.

Evinda

11:08 PM

How can we prove the case $p=\infty$ directly?

Eric Stucky

Is W^1,infty still the set of Lipschitz functions in higher dimension?

Bounded with bounded derivative, seems reasonable.

So, it's no longer differentiable but it now integrable

Hm, maybe? I don't know what C^a,b means.

Evinda

$u \in C^{0, \gamma}$ means that u is Hoelder continuous

Eric Stucky

oh XD

So the question 'is W^1,infty...' is actually quite relevant :P

Mike Miller

@EricStucky on a compact domain, yes, according to a colleague in the room

Evinda

When $p=\infty$ then $||u||_{W^{k,p}(\Omega)}=\sum_{|\alpha| \leq k}$ ess \sup_{\Omega} |D^{\alpha} u| $

Eric Stucky

11:17 PM

Mike: Well, that seems like enough, since we know U is bounded.

But we gotta prove it ;P

Evinda

How? :/

Eric Stucky

Hmm, Evinda, I think I believe it now: the idea seems to be something like: it's everywhere bounded (in U), and everywhere the derivative is bounded, so you should have some guarantee about the existence of the limit points.

The fact that the boundary is required to be 'nice' makes this seem plausible.

Evinda

But how can we show the inequality?

Eric Stucky

That should be a technicality: I think the inequality says that the Lipschitz constant is (basically) bounded by the bound on the first derivative.

The classical case uses the mean value theorem, so I guess I would try that?

Evinda

Yes, I think it is used at an other theorem, but I haven't really understood how

Could you explain to me how we can apply it?

Eric Stucky

11:32 PM

Perhaps you should consider the one-dimensional case carefully, then.

You're probably not going to get a proof out of me; I don't know the formalisms. This is what I've got.

@BalarkaSen: I tried to explain it in the post. When I see the word, I think of a technique that only applies in a limit selection of problems you're interested in, or a technique that doesn't scale well to large problems.

(At least, that is what I think in an elementary context, which it seemed pretty clear was the OP's intention, since they mentioned multiplication.)

Evinda

Since $\overline{u} \in W^{1, \infty}(\mathbb{R}^n)$ we have that $\overline{u}'$ exists and $\overline{u}, \overline{u}'$ are essentially bounded.
So now we apply the mean value theorem at $\overline{u}'$ ?

Eric Stucky

Worth a shot. What happens?

Evinda

You are reffering to me? :D

Eric Stucky

yea

Evinda

I have been studying all these days the theory because I have an exam tomorrow and I want to understand the proof :D

Balarka Sen

11:38 PM

Eric: I see.

Mike Miller

a trick is something clever, not something tedious

Evinda

Do we maybe look for a relation between the sets $C^{0, \gamma}(\Omega)$ and $W^{1, \infty}(\mathbb{R}^n)$ ?

Eric Stucky

Well, yes. You do need to produce a function in C^0,1(closure(U)), which you don't have right now. But if you do this in any reasonable way, it shouldn't do anything to help or hurt the proof of the inequality.

mkay, I am going to eat. Then maybe I will try to write some more posts if I'm not too bitter.

tyty Balarka & Mike :)

Balarka Sen

bon apetit

Agawa001

and it is upvoted to 2

Mike Miller

11:51 PM

why are you bitter

Balarka Sen

such is life

Agawa001