Mathematics - 2022-04-21

Ted Shifrin

00:03

I think it’s just undefined ….

I did say something very stooopid earlier. But the question — do the three impose constraints on one another — is a good one.

Jakobian

08:32

If $X$ is a compact topological space, do we have $\dim X^n = n \dim X$ ?

Alessandro Codenotti

09:32

@Jakobian Sounds extremely unlikely

There's surely something about this in the book by Pears, there is a whole section dedicated to product theorems

Even though they tend to be conditions under which $\dim X\times Y\leq \dim X+\dim Y$

Jakobian

12:01

@AlessandroCodenotti I didn't find anything helpful there

Koro

12:14

I think I finally figured out the error I was making while using the third isomorphism theorem in identifying a quotient ring. I have posted an answer to my own question here math.stackexchange.com/a/4432741/266435

JansthcirlU

12:44

isn't it strange that all number bases can be used to uniquely sum to any integer, but that there is only one set of prime numbers that can be used to uniquely multiply to any integer?

PyGamer0

14:08

i have a question

or am i being stupid and just not realising an obvious thing

hopper

14:23

Let $\mu,\nu:\mathbb{R}^d\to \mathbb{R}$ be probability densities. Let $T:\mathbb{R}^d\to \mathbb{R}^d$ be some vector field.

I define the push forward of a density $\mu$ by $T$ as $T_{\#}\mu(x)=\mu(T^{-1}(x))|JT^{-1}(x)|$ for $x\in\mathbb{R}^d$. Where $J$ is the $J$ is the jacobian, and the $|\cdot|$ is the determinant of that matrix. Now for some vector field $w$ I have a PDE which is solved by some function $u:[0,1]\times \mathbb{R}^d\to \mathbb{R}$

$$ \partial_t u+\text{div}(wu)=0,~~~~u(0)=X_{\#}\mu,~u(1)=\nu. $$

hopper

14:34

I thinking to introduce the mapping $(1-t)X^{-1}$

or better that definining $\tilde{u}$=(1-t)X^{-1}_{\#}u+tu

Ted Shifrin

15:58

@PyGamer0 no, you’re just wrong. Try the simplest example, like $f(x)=x^2-2x+1$.

Prithu biswas leftmse

@TedShifrin In the red line, Isn't α[h] being treated like a function?

dsillman2000

16:11

Can anybody testify to the truth of this statement? "If $R$ is a commutative ring, then $R+r := \{r + x \mid x\in R\}$ is isomorphic to $R$." It feels true to me when I think about it, but IDK if it's proven

robjohn

@Prithubiswasleftmse $\alpha_h$ is not a function of $h$, as there may be more than one point that satisfies $\frac{f(a+h)-f(a)}h=f'(\alpha_h)$. We are guaranteed that there is at least one.

Prithu biswas leftmse

@robjohn Then how can we say "Now $\alpha_h$ approaches $a$ as $h$ approaches $0$" when $\alpha_h$ is not a function of $h$?

robjohn

@Prithubiswasleftmse because $a\lt\alpha_h\lt a+h$, so by the squeeze theorem, we know that any $\alpha_h$ we choose will be close to $a$

dsillman2000

Nvm, I figured out my question. disregard above

Prithu biswas leftmse

@robjohn wait, which squeeze theorem are you referring to? Just asking because I am not sure.

this one?

robjohn

16:22

sure. Look, the definition of convergence says that if we can say for any $|h|\gt0$ that $a\lt\alpha_h\lt a+h$ then $\alpha_h\to a$. If you can't get your mind around this concept of convergence, then you can choose a particular $\alpha_h$ for each $h$ and make it a function.

However, the mean value theorem does not define $\alpha_h$ as a function.

Leaky Nun

16:43

@dsillman2000 it isn't isomorphic, they're the same set

robjohn

17:07

indeed

DanielSank

17:28

Send halp:

0

Q: Integral expression for covariance matrix in diffusion process

Consider the Fokker-Planck equation $$\frac{\partial \rho}{\partial t} = \sum_{i,j=1}^2 \mathbf{\Gamma}_{ij}\frac{\partial}{\partial x_i}(x_j \rho) + \mathbf{D}_{ij}\frac{\partial^2 \rho}{\partial x_i \partial x_j}$$ where $$\rho = \begin{pmatrix} x \\ v \end{pmatrix} \quad \mathbf{\Gamma} = \beg...

Prithu biswas leftmse

17:43

@robjohn Hmm. It seems like I am not familiar with this concept of convergence.

jay

is my use of chain rule right? $f:\mathbb{R}^n\to \mathbb{R}$ and $g:\mathbb{R}^n\to \mathbb{R}^n$ so $\nabla (f(g(x))= Jg \nabla f(y)|_{y=g(x)}$?

monoidaltransform

if $f:X\rightarrow Y$ is continuous and $f_{}:H_1(X)\rightarrow H_1(Y)$ is an isomorphism, must $f^{}:H^1(Y)\rightarrow H^1(X)$ also be an isomorphism?

Ted Shifrin

@Prithubiswasleftmse Yes, the $\alpha$ coming from the MVT depends on $h$. There may not be a unique $\alpha$, but for any choice of $\alpha_h$ the claim still holds because $a<\alpha_h<a+h$, so $0<\alpha_h-a<h$ and so $\alpha_h-h\to 0$ as $h\to 0^+$.

@robjohn Figure anything out re gmail? I did the suggested for one of my accounts, and now I have 17,746 emails to sort through and of which to erase almost all. If I do my main account, that number will probably be over 100K.

robjohn

I thought I tried to say the same thing, but I guess I failed.

Ted Shifrin

Oh, sorry, I didn't even check to see if there had been a response. Duh.

robjohn

17:58

@TedShifrin I have not found anything that will allow me to turn off the "allow untrusted applications" setting in GMail.

I am still working on it.

Ted Shifrin

@jay By $Jg$ you mean the full jacobian matrix, not determinant. Evaluated at what point? Why not write $\nabla f(g(x))$?

jay

$Jg$ is full jacobian matrix evaluated at $x$

Ted Shifrin

@robjohn: It does seem that setting up the account anew incorporates an explicit approval process. I wish google just gave us the option to be naughty and pay for the consequences.

@jay No, not evaluated at $x$. You need to say precisely where it's evaluated (correctly).

jay

I was just trying to use the formula

right so its not $\nabla (f(g(x))= Jg(x) \nabla f(y)|_{y=g(x)}$

Ted Shifrin

Oh sorry. It is $x$. You're right. Just write $Jg(x)$. No, this formula is not correct.

You have to write it with derivatives, not gradient, and then you can transpose it if you wish.

So it should be $Df(g(x))Dg(x)$. If you want this as a vector, rather than as a linear map, you transpose it, getting $Dg(x)^\top Df(g(x))^\top = Jg(x)^\top\nabla f(g(x))$.

And you don't write $\nabla f(g(x))$. You write $\nabla (f\circ g)(x)$.

You take the gradient of $f\circ g$ and evaluate that gradient at $x$.

monoidaltransform

18:03

The answer is Yes by UCT

jay

what about $\nabla f(y)|_{y=g(x)}$ notation is ok ?

@TedShifrin cheers so I missed the transpose

Ted Shifrin

It's OK, but unnecessarily ponderous.

Do you write $(f\circ g)'(x) = f'(y)|_{y=g(x)}g'(x)$? I sure don't.

But if you don't understand the difference between $f'(g(x))$ and $(f\circ g)'(x)$, maybe it's important to write it that way.

jay

okok :) super thanks

what does the 100k under your name in chat mean?

Ted Shifrin

It means I just went over 100k rep points. :P

jay

cool :)

nice

Ted Shifrin

18:06

Your rep should show up under your avatar, too, if what you write is long enough :)

@robjohn Amusingly, many dozens of the messages I have to delete are google security alerts because it thought I wasn't me signing into my accounts from my iPad in various places not home.

Govind75

Can I do a taylor expansion for a function which is only $n$-times differentiable?

Ted Shifrin

You can do a Taylor expansion of degree $n$. But you have no control over the error.

If you want an error estimate, then do an expansion of degree $(n-1)$.

Govind75

18:22

Maybe I am approaching this problem incorrectly then

I have a $1$-D differential eqn. given by $\dot{x} = f(x)$ and $f$ is $n$-times differentiable with equilibrium $x^*$. But I have $f^{(k)}(x^*) = 0$ for all $k \le n - 1$, if $f^{(n)} (x^*) \neq 0$ what is the stability for different cases of $n$ and signs of the $n$th derivative

I figured I'd look at $n$ odd first and analyse in a small $\delta$ neighbourhood of the equilibrium.

Surely the easiest way to do this is through a taylor expansion but idk now

Ted Shifrin

It would be nice if the $n$th derivative were continuous at $x^*$. Without that, we can probably make up standard counterexamples.

Yes, a Taylor polynomial is the only approach, but the hypothesis is not quite enough.

Govind75

What do I need further?

Ted Shifrin

18:40

I already told you.

Probably your course is sloppy about what “differentiable” means.

robjohn

@TedShifrin I get a lot of those. Google should clean that atuff up before requiring people to change clients, if any client change will work. I have my Apple Mail client set up as sercurely as it can be, but Google still doesn't work without "allow insecure apps" set.

Ted Shifrin

@robjohn If you draft a petition, I will sign dozens of times.

robjohn

They want the authentication set properly. I would think that if you verify your identity at a given IP address, that should be good enough with username and password encrypted. Certainly, when you change IP addresses, I can see needing to reauthenticate.

Ted Shifrin

That’s typically when I used to run into trouble … phone and iPad whilst traveling. But these days I get occasional issues at home.

robjohn

I would even go for a very lightweight app running to keep my authentication current on a given machine. I just don't want to use a browser to read my email.

Ted Shifrin

18:52

Me either. I use the gmail app on the iPad and iphone just to check, but typically like to run apple mail on both when traveling. Right now the home desktop is the main issue.

robjohn

That is my issue. I want to use Apple Mail on my laptop. I hate keeping track of mail and the thread presentation on browsers is not to my taste.

Ted Shifrin

Years ago the math dept server at UGA stopped doing mail and the UGA mail server (outlook based) forbade apple mail, so I forwarded to a proxy gmail address, set up so that answers appear with the uga return address. That was already a giant hassle.

But I’m sure you and I are the only ones disgruntled by gmail.

robjohn

I object to the idea of browser mail in the first place. You have to be connected to Google just to access ANY email, whether old or new. I like to have my old email on my computer and readable whether connected or not. I also don't like that my email is stored on their server as more than a transport mechanism. It should be stored on my computer, not theirs.

Ted Shifrin

I agree 1000%. I used to POP for that reason, but now I give up on that.

robjohn

19:55

@Ted: Supposedly Apple Mail has supported OAuth since El Capitan and OAuth2 since Mojave. I don't see it in my mail app, but maybe I need to delete and recreate the email account to get OAuth (or OAuth2). This possibly being an arduous process, I will keep investigating.

Ted Shifrin

20:15

@robjohn the arduous process is refiltering years of emails. Do you know a way to check if a particular email connection already has OAuth “approval”?

leslie townes

all of this is reminding me to update my local backups of my email. i lost track of that when i got a new desktop just before the pandemic.

Ted Shifrin

All I can say is …. Good luck! Put munchkin in charge.

robjohn

20:33

@TedShifrin I know that mine doesn't use OAuth by looking at the mail logs. I see that OAuth is offered by Google's server, but my client does not use it. My mail client does not have OAuth as an option, but supposedly OAuth has been in MacOS since El Capitan. My mail accounts were set up before El Capitan and simply copied. There should be a way to simply turn on OAuth authentication, but I don't see it.

Ted Shifrin

Other than setting up a new account. Is there an easy way for me to check if my mac mail account in my desktop is OAuth-approved?

I think I set up the account not too long ago.

robjohn

21:04

@TedShifrin what version of the OS are you using?

Ted Shifrin

@robjohn Latest. 12.3.1

Ted Shifrin

21:22

In the mail log I see the occasional "Return to authenticated state."

Now I'll be clever and turn off the account I know is authenticated.

monoidaltransform

if $M_{2}$ is an orientable surface of genus $2$

what does the generator of $H_2(M_2)$ look like?

Ted Shifrin

Look like? As in the case of any compact orientable manifold, it's the fundamental class.

monoidaltransform

I meant geometrically

Ted Shifrin

Triangulate, orient cells compatibly, and take the sum of the top-dimensional faces.

monoidaltransform

Like for $H_2(\mathbb{S}^2)$, it consists of the upper and lower hemisphere. Its the difference with a suitable orientation

Ted Shifrin

21:29

No, that's nonsense.

What I said before is a lot clearer.

robjohn

21:59

@TedShifrin Much newer than mine. I am still using 10.14.6 (Mojave)

Ted Shifrin

Damn, you antique old man!

The log seems to suggest I’m OK on the desktop. Time to wrestle with ipad and phone.

robjohn

22:24

@TedShifrin To see if you are actually okay, you need to see if your GMail account has "Less secure app access" turned on. To find that out, go to your GMail account on the web, click on the gear icon (hover=Settings, near the upper right of the page), click on "See all settings", click on "Accounts and Import", then click on "Other Google Account settings", then click on "Security" on the left sidebar, scroll down to "Less secure app access"

If it says "off" and you are getting mail, you are okay.

Ted Shifrin

Ah, I don’t think I’ve turned it off. Great suggestion. Thanks!

Ted Shifrin

22:41

Oh yeah, it's off now. Great. I think redoing that one account yesterday may have been unnecessary. Oh well. Only 17,717 messages to go through. Thanks, @robjohn. I owe you one.

robjohn

22:58

@TedShifrin I just got mine working on my dinosaur OS using an app password for my Mail.app

I can stop worrying about this :)

Ted Shifrin

Interesting. So when do you input the pw? P.S. For security reasons, I’m shocked you’re still a dinosaur. Although you can probably run Illustrator and Excel.

robjohn

23:14

@TedShifrin I run neither of those. I wish I could run Photoshop, but if I start doing more astrophotography, I will have to deal with that. I had to generate the password for my app and computer on the same page as you verified the "Less secure app access". Then I enter it in my email program for that account.

Ted Shifrin

Oh, that’s not so bad! They should have made this clearer years ago! Sigh.

robjohn

I had to turn on 2SV (two stage verification), but that was no problem. I may need to generate a new password one in a while. I should see what happens when I use a VPN.

Ted Shifrin

Good question.

robjohn

@TedShifrin I also have TLS running on Mail (under advanced POP access in Mail) which makes a plain password secure.

it's like adding the "s" to http

@TedShifrin I agree. I have been fretting over this for weeks.

Ted Shifrin

And you’re very computer-skilled. I’m more average, but then there are all the dummies.

robjohn

23:21

Yes. Google is being cryptic. Probably because they want the average person to use Google's access technology.

Upgrade everything and use Webmail

I think most people use webmail these days.

robjohn

23:48

Just checked UCLA VPN and my iPhone's hotspot. Both work with my mail. This is good.

I was dreading the deadline, but things seem to be okay.

Ted Shifrin

I bought some new gin today. Martinis to celebrate!

robjohn

I will have some orange bubbly when I get home from walking the dog.

BBL

Transcript for

$Mathematics$ Mathematics