Mathematics - 2015-02-04 (page 5 of 6)

@robjohn The expansion of $\cosh(x)$ at $x=0$ is $$1+\frac{x^2}{2}+\frac{x^4}{24}+ \dots $$ right?? So is the expansion of $\cosh(xy+z)$ at $(0, 0, 0)$ the following??

$$1+\frac{(xy+z)^2}{2}+\frac{(xy+z)^4}{24}+ \dots $$

@MaryStar Yes, and you will get some terms of higher order than $4$ out of that... throw those out.

@robjohn Do you mean that we take only $$1+\frac{(xy+z)^2}{2}+\frac{(xy+z)^4}{24}$$ and throw the tems $\frac{(xy+z)^5}{5!}+\dots $ out ??

@MaryStar No, $$1+\frac{z^2}2+xyz+\frac{x^2y^2}2+\frac{z^4}{24}+\dots$$

Those are all the terms with degree less than or equal to $4$

Hi guys

5:08 PM

I have a doubt in topology, a very basic one. Any one up for it?

@iwriteonbananas Do you have any conditions that need to be satisfied for that to be a submanifold?

@SwapnilTripathi Just ask the question, if someone can answer it, they will. If it is on main, just post a short link to it.

@robjohn Hey there! I am still alive, lol.

@ABeautifulMind was there some doubt?

Yes u r jasper

@SayanChattopadhyay Yes, he is Jasper.

Huy

5:12 PM

How's it going, @MikeMiller?

I m saying he is alive@robjohn

@SayanChattopadhyay He is making a joke.

@SayanChattopadhyay I know; I was just playing with the indefiniteness of the statement.

By int_Y(A) do we mean int(Y\cap A)? Where int denotes interior of a set.

Ok!! Thank you !!! :-) @robjohn

5:13 PM

We need context to answer that, @Swapnil. Many authors use many notations to mean many things.

There are even different definitions for the boundary.

@ABeautifulMind Its all about the boundary

@robjohn: note that non singularity of $f$ is not sufficient for $f(M)$ to be a submanifold; in particular consider a map $\Bbb R \to \Bbb R^2$ that draws a figure 8.

I like big boundaries and I can not lie

@N3buchadnezzar Have you found a girl yet?

5:15 PM

Do u have a wife Jasper

@SayanChattopadhyay No, and I might never have one.

She should take care of u right

Why....never leave hope

@MikeMiller I was proving: If (Y,T_Y) is the subspace topology of (X,T), then for any set A\subseteq X, int_X(A)\subseteq int_Y(A)

@MikeMiller Ah, okay... so we need to have injectiveness... That is what I was trying to get from iwriteonbananas.

@ABeautifulMind Have you found a pet/banana yet?

5:16 PM

@N3buchadnezzar I do not keep pets.

I think jasper u should have a friend who can help u

I've gotten to the point where I get that int_X(A)\subseteq int(A\cap Y) @MikeMiller

@ABeautifulMind girls pets, whats the difference

@SayanChattopadhyay Yes, I have friends in real life and online that I talk to. I can also see psychiatrists and therapists if I want to, thanks.

@robjohn his thing descends to an injective immersion (=embedding) of the torus. I'm not sure if that's the way he wants to phrase it, as I don't know if his course considers abstract manifolds instead of ones that live in euclideannspace.

5:18 PM

@robjohn can you please elaborate why rank jacobian = 2 implies that the local diffeo is non singular in a small enough patch?

@Swapnil My phone is at 1% battery so I will not be able to help. Sorry.

sorry was away

@MikeMiller Wow... mine starts shutting functions down at around 20%

No problem. Thank you @MikeMiller

Jasper always remember u r not the only one........keep on fighting u will always win

2

5:21 PM

@SayanChattopadhyay Have you watched the movie A Beautiful Mind?

@iwriteonbananas The Taylor series in $u$ and $v$ for the diffeomorphism is locally invertible if the Jacobian is non-singular. As you get closer to the point of expansion, the series approximates an invertible linear function.

Nope...

Hi all.

@SayanChattopadhyay You should watch it. Also watch Good Will Hunting. These are my two favourite movies.

Hi@mikeonly

Oh I have seen good will hunting

U should watch the imitation game

5:24 PM

@iwriteonbananas: Did you not figure out my hint?

@TedShifrin!!

hi @Balarka

Hi @BalarkaSen

Hi @TedShifrin

hi Sayan

@TedShifrin I want to compute homology of the orientable genus $g$ surface. If we take a piece of the torus-like shape at the end of the surface (homeo to torus minus a disk) to be A then you get a short exact sequence from the long one (as homologies higher than 2 vanishes). But I can't visualize the bonding map $H_n(\Sigma_{g-1}) = H_n(X, A) \stackrel{\partial}{\to} H_{n-1}(X) = H_{n-1}(\Sigma_g)$

5:28 PM

Balarka u have any idea for a book in calculus

Spivak, @Sayan. Standard reference.

I remember reading on some stack exchange question that proving of FTA is like a test for each newly created algebra theory. Could it be true, how do you think?

Full name

@Balarka: So you're trying to do this with relative homology, rather than Mayer-Vietoris?

@mikeonly Sounds like a catchphrase without much real meaning.

5:30 PM

Spivak's Calculus.

@TedShifrin Haven't studied Mayer Vietoris yet, but I do have another approach.

Note that if you chuck a point outta $\Sigma_g$, it def rets onto $\vee^{2g} S^1$

So you're thinking of $T_g = T_{g-1}#T_1$, @Balarka, and using the punctured $T_1$ as your $A$?

Yes, one can prove the FTA in many different ways. That's about all it says to me. I don't think homological algebra is weaker because it doesn't prove FTA.

$A$ should be, homotopically anyhow, a closed subset, right?

So there might be hope with excision on $(X, A, Z) = (\Sigma_g, D^2, D_{+}^2)$

Wait. Are we doing excision or relative?

5:32 PM

Well, no.

$A$ needn't be closed

@TedShifrin Both. No idea which one is easier.

:P

@MikeMiller I think the idea was that merely every part (theory maybe?) of algebra has methods of proving it. By saying theory I mean topology, group theory, complex analysis, etc.

You can certainly do it cellularly, too, generalizing what works for $T_1$, sure.

Haven't studied cellular homology. The usual annihilating cells one-by-one doesn't work, as $\Sigma_g$ is a 2-cell attached to the $2g$-boquet.

And $H_2(S^2) \cong \Bbb Z$. :/

Why are you saying $H_2(S^2)$?

OK, let's focus on one approach. There are too many balls in the air here.

Greetings

5:38 PM

@TedShifrin im not sure...say we pick a point $p=f(u,v) \in S$ such that the upper 2x2 block of the jacobian of $f$ in $(u,v)$ is invertible. then the inverse function thm gives us a nbhd $U\subset \mathbb{R}^2$ such that $f$ restricted to $U$ is a diffeo, right?

@iwriteonbananas The absolute value of the cross-product of the partials is $2+\cos(u)$, and that is always at least $1$. This means the inverse has a $2\times2$ Jacobian with determinant at most $1$.

yes, bananas, where the $\Bbb R^2$ is the $x_1x_2$-plane.

so then $U$ is diffeomorphic to $f(U) \subset \mathbb{R}^3$ which contains $p$

@robjohn: We're trying to use the inverse function theorem to construct the local charts.

@Chris'ssis Hi.

5:40 PM

@TedShifrin I was just giving some details on the inverse

then we're done right?

If you put it all together correctly, sure, bananas :)

ok but the upper 2x2 block of the jacobian is not invertible for every $(u,v)$

only when $\sin(u) \neq 0$

@TedShifrin The absolute value of the cross-product of the partials is the determinant of the $2\times2$ Jacobian between the tangent spaces

Sure, so you move on to a different projection at other points, bananas.

Of course, @robjohn.

But what tangent spaces, precisely:

5:43 PM

@TedShifrin so we could look at the lower 2x2 block right?

Unless a lot happened while I was gone, the point is to see how to get a local chart from the parametrization.

Sure, bananas, or the 1st+3rd ...

@TedShifrin The map between $\mathbb{R}^2$ and the surface parametrized

oh i see now what u meant when u wrote to consider $\pi \circ f$

Yup bananas.

thanks

5:44 PM

@TedShifrin Perhaps I am using too much, sorry

@TedShifrin What I showed you yesterday was an improper double integral, of course, but if needed, I can use this notation $$\lim_{s\to 0^{+}}\int_{s}^{\infty} \int_{s}^{\infty} \frac{\sin(\alpha x)\sin( \beta y)}{x y (x+y)} \ dx \ dy$$ and then make use of the splitting (because we can perform the splitting), and then using Tonelli's theorem for non-negative functions I pass from the double intregal to the iterated integral and reach my initial objective.

@TedShifrin Let $f:[a, b] \rightarrow \mathbb{R}$ continuous, differentiable on $(a, b)$ and the limit $\lim_{x \rightarrow a} f'(x)$ exists. Show that $f$ is differentiable in $a$ and that $f'(a)=\lim_{x \rightarrow a} f'(x)$.

To show that $f$ is differentiable in $a$ do we use the fact that $f$ is continuous at $a$?? So that $\lim_{x \rightarrow a} f(x)=f(a)$ ??

No, @robjohn, the point was to give the map from an open set of the submanifold to the plane, and we don't know what a diffeomorphism of abstract manifolds is :P

@MaryStar: I've already told you what to focus on. I'm working on Balarka's question now.

@Chris'ssis: Tonelli's theorem does not apply on the square. The function is NOT integrable.

goes into a quiet corner to work and ignores everyone

2

hahaha

Iota

tdoaps

5:50 PM

Basic topology question: what do I get from identifying the two bases of a cylinder?

Iota

Follow on twitter, instagram and facebook

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@Exterior torus

Should be a cylinder, no?

ah.. I'm lost. Can you point me to some pretty pictures :D?

Iota

@TedShifrin well, I didn't simply apply Tonelli, but before doing that I made the splittings and a series of inequalities. In the integrals obtained I can apply Tonelli ...

5:58 PM

@mikeonly I would hardly call topology or complex analysis a theory of algebra. Anyway, I think the quote's silly, that's all.

@MikeMiller It may be mostly problems of my English. So what are they in relation to algebra?

@mikeonly They're just fields of mathematics. Algebra refers to things like group theory, ring theory, category theory, etc.

Well, let me modify it a bit (this is another version)

@MikeMiller Oh, fields. Ok, that's the word I should have better used. Thanks.

@Exterior Depends on how you glue the top to the bottom. The way you're thinking, by the identity map $S^1 \to S^1$, results in the torus.

6:02 PM

@MikeMiller sorry, sorry, I got confused with the pushout of $X\times [0,1]\leftarrow X \rightarrow X\times [0,1]$ where the arrows are $e_i :x\mapsto(x,i)$ for $i=0,1$

That pushout is a cylinder right?

I don't think so. It should just be $X_0 \times [0,1] \amalg X_1 \times [0,1]/~$, with $(x,0)_0 \sim (x,1)_1$ (if the notation makes sense).

Which should just be $X \times [0,1]$ again.

I guess $X \times I$ is what you mean by cylinder.

Dunno why I said "I don't think so" above, cuz I do.

Right, seems we're saying the same thing

so yes?

yes

I mean pushouts in $\mathsf{Top}$ are exactly disjoint unions modulo identifying images

okay, thanks!

@TedShifrin It's like in this example where all works perfectly \begin{aligned}
\displaystyle \left|\int_s^{\epsilon}\int_s^{\epsilon} \frac{\sin(\alpha x)\sin( \beta y)}{x y (x+y)} \ dx \ dy\right|&\le \int_s^{\epsilon}\int_s^{\epsilon} \frac{|\sin(\alpha x)\sin( \beta y)|}{x y (x+y)} \ dx \ dy \\
& \le\alpha \beta\int_s^{\epsilon}\left(\int_s^{\epsilon} \frac{1}{ x+y} \ dx \right) \ dy \\
&=2\alpha \beta\left( (e+s) \log \left(\frac{2}{\epsilon+s}\right)+\epsilon \log (\epsilon)+s \log (s)\right)

6:10 PM

(warning that when you say cylinder, most people think you mean $S^1 \times I$ without further clarification...)

Oh, I was talking about cylinders in homotopical context

Thanks for the tip

6:22 PM

@ABeautifulMind are you here?

@user153330 Yes.

@ABeautifulMind i found a holy book, just like the ones you like, here it is : amazon.com/Handbook-Analysis-Foundations-Eric-Schechter/dp/…

it's called The Bible. :3

@Hippalectryon i'm talking about this committingtoachallenge.wordpress.com/about-2

@TedShifrin what do i need to search to find your lectuers on this stuff?

6:25 PM

@iwriteonbananas in this channel youtube.com/channel/UCp9W-et2Zbx7u5_VMiXGtPQ

shocking videos

@ABeautifulMind did you saw it? what do you think of it?

@user153330 Well, it's just a handbook with many results. Not really something to study from.

@ABeautifulMind i find it very accessible

@user153330 Use whatever you like. There are many good books.

6:30 PM

Nice theory lessons ... I'd like to see a rigorous solution to $$\lim_{s\to 0^{+}}\int_{s}^{\infty} \int_{s}^{\infty} \frac{\sin(\alpha x)\sin( \beta y)}{x y (x+y)} \ dx \ dy, \quad \alpha, \beta>0$$ What's the use of the idea of knowing a lot of stuff if we cannot put it in practice? I know how to do it but not rigorously (I mean, I'm also going to do that, no problem).

@user153330 why dont those vids have 1 mil views yet

News bulletin Fundamental changed his name to Famous Blue Raincoat

6

@iwriteonbananas it's all because of @Hippalectryon and baby hippa

@Chris'ssis what do u mean u did it not rigorously?

@iwriteonbananas Not rigorously means you don't know how to do it, lol.

lol

6:41 PM

@ABeautifulMind In the style people post solutions on this site, I put the solution in 5 min, say.

(that means I don't explain everything)

what does $FTC$ mean ? @TedShifrin

Fundamental Theorem of Calculus, probably

tks @Exterior

AlexR

7:30 PM

Does anyone have an Idea if there is an ongoing contest with this question?
http://math.stackexchange.com/questions/1133471/find-kth-term-of-this-sequence?noredirect=1#comment2314311_1133471
It was asked three times already and none provided context.

8:52 PM

@hippa

@Ramanewbie

if

A[1,2]

[3,4]

B[2,3]

and

, then if C=AB,

[4,2]

C[3,5]

No, because your formatting is awful.

of course... wait, I'm searching how to post matrices in $LaTeX$ @hippa

"good morning" @MikeMiller

@Ramanewbie \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} -> $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$

9:01 PM

@daniel thanks !

@Chris'ssis are both $s$ and $\epsilon$ tending to $0$?

9:14 PM

@robjohn Yes.

@DanielF When studying spaces of distributions, the reason we consider "functionals" $C^\infty_0 \to \Bbb C$ instead of codomain $\Bbb R$ is all for the sake of the Fourier transform, correct?

9:26 PM

Having in mind the meaning of the double integral as presented above, I used the fact that \begin{aligned}
\int_0^{\infty} \int_0^{\infty} \frac{\sin(\alpha x)\sin( \beta y)}{x y (x+y)} \ dx \ dy &=\int_0^{\epsilon}\int_0^{\epsilon} \frac{\sin(\alpha x)\sin( \beta y)}{x y (x+y)} \ dx \ dy+\int_0^{\epsilon}\int_{\epsilon}^{\infty} \frac{\sin(\alpha x)\sin( \beta y)}{x y (x+y)} \ dx \ dy \\
&+\int_{\epsilon}^{\infty} \int_0^{\epsilon}\frac{\sin(\alpha x)\sin( \beta y)}{x y (x+y)} \ dx \ dy +\int_{\epsilon}^{\infty}\int_{\epsilon}^{\infty} \frac{\sin(\alpha x)\sin( \beta y)}{x y (x+y)} \ dx \ dy

@robjohn that splitting was necessary for the next steps in my proof.

Using that $\displaystyle \frac{1}{x+y}=\int_0^{\infty} e^{-(x+y)s} \ ds$, we obtain that
$$\int_{\epsilon}^{\infty}\int_{\epsilon}^{\infty} \frac{\sin(\alpha x)\sin( \beta y)}{x y (x+y)} \ dx \ dy=\int_{\epsilon}^{\infty}\int_{\epsilon}^{\infty} \int_0^{\infty}\frac{\sin(\alpha x)\sin(\beta y)}{x y} e^{-(x+y)s} \ ds \ dx \ dy. $$

9:49 PM

@robjohn above is $\epsilon \to 0$ (where I put $\epsilon \to \infty$)

@MikeMiller You can also consider real-valued distributions. But if you take complex-valued test functions, then $\mathbb{C}$ is the natural thing. And if you look at the Fourier transform (which if you consider tempered distributions is almost inevitable), then of course you need $\mathbb{C}$. I wouldn't say "all for the sake", but the Fourier transform is a pretty convincing reason to take complex scalars.

Pedro Tamaroff

@DanielFischer Daniel.

@PedroTamaroff Hola. How's things?

Pedro Tamaroff

@DanielFischer Not bad at all. I'm a bit bored with vacations, but this weekend I'll see some friends.

How are you?

@PedroTamaroff Not bad. I've been to the cinema yesterday, and thus I don't need to go again this year ;)

Studentmath

9:55 PM

Which movie?

Pedro Tamaroff

@DanielFischer Hehe. What did you watch? I'll probably go with my brother on Sunday.

Studentmath

It's a once-in-a-year movie, better be worth it