Mathematics - 2018-08-19

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Nick

00:50

Good morning Room 36 :D

Akiva Weinberger

01:26

user image

Application of cycloid track: moving heavy stuff

the forgotten technology

A vid explaining who he is: https://www.youtube.com/watch?v=uYQBDhkBfr0

Some boring details https://www.dailymotion.com/video/x17g5i9

2 hours later…

Faust

03:29

RAWR

2 hours later…

Adeek

05:03

@loch @TedShifrin I passed my defence

I still have to revise my thesis but I passed :D

@Daminark as well

I will talk to you guys soon when I see ya'all

user131753

05:30

Let $X$ be an infinite topological space and consider any point $a\in X$. Then consider the space $X\times X$ with the product topology and consider the subspace $X\times \{a\}$. We know that there exists an embedding of $X\times \{a\}$ to $X\times X$. But does anyone know under which condition(s) this embedding is a homeomorphism?

Mike Miller

Huh? It's not surjextive unless X = {a}

user131753

@MikeMiller I won't be so sure. Recall that the additive groups $\mathbb{R}\times \{0\}$ and $\mathbb{R}\times \mathbb{R}$ are isomorphic if you assume AC.

Mike Miller

But the inclusion is not an isomorphism.

An isomorphism is, in particular, a bijection

user131753

@MikeMiller I see what your point is.

user131753

In that case read the last line as "But does anyone know under which condition(s) these two spaces are homeomorphic"?

user131753

05:40

Is that ok @MikeMiller?

Mike Miller

That question is harder for sure.

For instance the Cantor set is homeomorphic to its square.

No finite dimensional manifold can be (it has a well-defined dimension) but Hilbert space is isometric to its square.

user131753

@MikeMiller In fact to answer one of my question I needed to show that two spaces of the form I mentioned above are not homeomorphic. But I couldn't find any topological property to do this. So I was wondering whether there exists any criteria for the "completmentary" situation.

user131753

In particular in my case the topological space $X$ was non-haudorff.

Mike Miller

06:04

@user170039 I would try to find some 'invariant' of X that doubles or becomes infinity ot something when passing to X x X.

2 hours later…

Mary Star

08:08

Hello!!

We have the density function $f_x(x)=\frac{2c^2}{x^3}, x\geq 0, c\geq 0$.

I want to calculate the maximum Likelihood estimator for $c$.

We have the Likelihood Function $$L(c)=\prod_{i=1}^nf_{X_i}(x_i;c)=\prod_{i=1}^n\frac{2c^2}{x_i^3}$$

The logarithm of the Likelihood function is $$\ell (c)=\ln L(c)=\ln \left (\prod_{i=1}^n\frac{2c^2}{x_i^3}\right )=\sum_{i=1}^n\ln \frac{2c^2}{x_i^3}=\sum_{i=1}^n \left [\ln (2c^2)-\ln (x_i^3)\right ]\\ =n\ln (2c^2)-\sum_{i=1}^n \ln (x_i^3) =n\left (\ln 2+\ln c^2\right )-3\sum_{i=1}^n \ln (x_i)\\ =n\left (\ln 2+2\ln c\right )-3\sum_{i=1}^n \ln (x…

2 hours later…

Secret

10:04

Random question:

Given the ordinal interval $[0,\omega]$ how many steps do we need to go from $\omega$ to $4$?

Secret

10:59

put it in another way, what is the closure of $\{\omega\}$

1 hour later…

Krijn

11:59

Hey all

Hey @Mike! Long time no see

Leaky Nun

12:23

@Secret re steps: it's not well-defined because the reverse order of a well order isn't a well order

re closure: the topology is Hausdorff, and in particular singletons are closed

Secret

I see

Leaky Nun

although $[0,\omega]$ is compact

that makes it a compact infinite hausdorff space

(that should be the correct order)

Secret

so that means pretty much if we start at $\omega$, without any tricks, we can walk nowhere

I might explore the necessary and sufficient condition for compactness later, though I knew they are quite common in hausedoff topologies

I recall there's a theorem, except I forgot the name

Secret

12:43

mathonline.wikidot.com/…

ah this one

but I am guessing that existence of compact set is really a property of the topology, thus it is not due to something more fundemental

Mr. Xcoder

Hi! I have this function: $$f(x)=\begin{cases}3-2x&\text{if }x\le 1\\\ln(x)+x&\text{if }x>1\end{cases}$$I'm told to show that $\sum_{x=-2013}^\infty f'(x)=-4028$, but I don't think that is the case. I have: $$f'(x)=\begin{cases}-2&\text{if }x\le 1\\\dfrac{1}{x}+1&\text{if }x>1\end{cases}$$ So the first sum (from $-2013$ to $1$) is $-4030$... But isn't the second one (from $2$ to $\infty$) divergent?? 1) we're adding $1$'s indefinitely and 2) the sum of reciprocials diverges anyway... Am I right?

I think the problem statement is likely wrong.

Alessandro Codenotti

13:12

@Mathei @ÍgjøgnumMeg Do you have an ANT book to suggest for a gentle introduction to p-adics? (before you suggest it Neukirch's chapter 2 does not count as a gentle introduction)

Leaky Nun

13:33

@AlessandroCodenotti Neukirch chapter 2

I mean, $\Bbb Z_p$ is just $\varprojlim \Bbb Z/p^n\Bbb Z$ :P

what do you mean by "p-adics"? do you have a more specific topic?

Alessandro Codenotti

I want to learn about local fields in characteristic 0, so apparently about finite extensions of the p-adic numbers, but I suppose I should learn what the p-adic numbers are and their most important properties to begin with

Leaky Nun

13:48

@AlessandroCodenotti I think the exponential map is quite important

and the Teichmuller things

and Hensel lemma

ÍgjøgnumMeg

14:24

@Alessandro Milne's notes are good

@Alessandro jmilne.org/math/CourseNotes/ANT.pdf

@Alessandro Chapter 7 deals with local fields in general

Alessandro Codenotti

14:38

Thanks, they look good indeed!

ÍgjøgnumMeg

14:50

@Alessandro I used this a lot for my undergrad dissertation :)

user131753

Consider the additive abelian group $\mathbb{R}^2$. Consider its diagonal $\Delta$. Then is it true that, $$\Delta=\displaystyle\bigcap_{\alpha\in I}\left(\bigcup_{i=1}^{n_\alpha} (G_i^\alpha\times H_i^\alpha)\right)$$ where $G_i^\alpha,H_i^\alpha$ are subgroups of the additive abelian group $\mathbb{R}$ for all $i\in\{1,\ldots,n_\alpha\}$ and for all $\alpha\in I$?

Secret

15:37

Abstract algebraic logic

In mathematical logic, abstract algebraic logic is the study of the algebraization of deductive systems arising as an abstraction of the well-known Lindenbaum–Tarski algebra, and how the resulting algebras are related to logical systems. == History == The archetypal association of this kind, one fundamental to the historical origins of algebraic logic and lying at the heart of all subsequently developed subtheories, is the association between the class of Boolean algebras and classical propositional calculus. This association was discovered by George Boole in the 1850s, and refined by oth...

Oooooooooooooooooooooooooooo...

Tanuj

16:11

Hello guys

I have to find the number of terms to be retained in the Taylor series expansion of $ln(1-x^2)$.

While evaluating the remainder term, I need to find out the $(n+1)th$ derivative of $ln(1-x^2)$

How can I possibly figure out what it is?

Adeek

Hey @loch @Daminark @TedShifrin I have passed my defence

Holo

16:27

@Tanuj You use the fact that $$\frac{d}{dx}\ln(1-x^2)=-\frac{ 2 x}{1 - x^2}=-2x\left[\frac1{1-x^2}\right]$$

Oskar Tegby

Why is a function that is monotone, non-negative, and pointwise bounded integrable?

The pointwise bound is $x^{-\alpha}$, and we're integrating with from zero to one. That bound isn't of much help there, so I don't know why it would be of any value. Do these properties together imply some stronger property?

Secret

pointwise bounded means none of the values are allowed to shoot to infinity?

Oskar Tegby

Well, $x^{-\alpha}$ goes to infinity as $x$ goes to zero.

N. Maneesh

chat.stackexchange.com/transcript/message/46293529#46293529 @Secret can you give some hint for the proof?

Secret

hmm... so that means somehow the rate of blowing up to infinity is slower than the function f, thus avoiding non integrable singularities at zero

Oskar Tegby

16:37

Yeah

Oskar Tegby

16:47

0

Q: Showing that a non-negative, monotone, and pointwise bounded function is integrable

I'm trying to solve the following problem from on old exam on real analysis. Thus, only such methods may be used. The problem is presented below. Let $\{f_n\}$ be a sequence of non-negative monotone functions which are pointwise bounded from above by $x^{-\alpha}$ where $\alpha>1$. Prove that...

real-analysis riemann-integration

Secret

16:58

(looking at that answer) Ah right, I almost forgot that integrals are ultimately limits, thus having an epsilon sticking around is ok as long it limits away

Oskar Tegby

No? We're dividing by epsilon. Read my comment.

Secret

17:16

@N.Maneesh The largest topology makes sense to me, but I don;t know enough about smallest topologies to deduce why the intersections have to be finite

N. Maneesh

@Secret intersection onw is simple. I was talking BOT UNION.

Secret

Since W' is a basis, by definition of the basis of a topology, the union of all open sets in the basis gives the topology

user131753

17:34

Is it only me or clicking on @Adeek's profile page shows "page not found"?

Oskar Tegby

@Secret I think that there was a type in that exam. I'm going to skip the exercise completely. There are nicer things to do than trying to do.

that*

On what set does $f_n=\frac{x}{1+nx^2}$ converge uniformly?

Secret

yeah, cause $\int_0^1 \frac{1}{x^{\alpha}}dx, \alpha > 1$ cannot converge by p-test

Oskar Tegby

@user170039 I don't see his StackExchange profile either.

Okay, but does that mean that it can't be integrable?

Secret

Yeah, it just blow up to minus infinity

user131753

@OskarTegby That's weird

Oskar Tegby

17:40

It must be that it should say $x^\alpha$, not $x^{-\alpha}$.

Then it's trivial.

What's the use of it being monotone and non-negative then?

@user170039 Maybe it's deleted by some reason.

Secret

Well, nonnegative is certainly useful such that any convergence cannot be due to cancellations of some of the $f_n$

I am not very sure about how monotone helps, other than it removes the need to consider functions $f_n$ which there is some wiggles midway before it get asymptote to zero

Oskar Tegby

Okay. Do you have any insights about my new question? I can't figure out on which set it should be uniformly convergent on. I just get that $x=0$ is bad when computing the limit.

Secret

I will be able to help on pointwise convergence, I cannot say I have done pretty well on uniform convergence so I will hesitate to try to answer that

My epsilo delta gymnastics is still pretty weak

Oskar Tegby

You seem to prefer algebra. Am I right?

I think that I'm going to compute the Cauchy sequence.

Secret

Indeed I am, things goes step by step in algebra, my head is still pretty bad in dealing with many things happening at the same time, which is common in analysis proofs

While I managed to find a way to make typical epislon delta proofs less art and more skill (by trying to minimsie the epsilons when writing the proofs in the backward direction to calculate the required epsilon) this method still fails for some limits and integrals thus I have not really algorithmise all epsilon delta proofs yet, and thus make the procedure more step by step

Another thing that is common in analysis proofs is inequalities. My brain tend to shutdown when I see more than 3 inequalities in one line

Oskar Tegby

17:56

I feel.

I like analysis, but it can feel unstructured and ugly at times.

Secret

General topology distract me from this BSOD by keeping me busy with weird things like infinite sets, and open sets are still kinda "chunky" thus things happen step by step. Also I am much better handling subset chains than inequalities with absolute value signs everywhere

this is why I read munkres without any analysis background (other than my 1st year calculus), as I knew if I can get used to weirdness like non T0 topologies, then analysis should become more straightforward because analyis deals with things much nicer

Oskar Tegby

That book is so thick, though.

Secret

it is, and it took some time to go through. I also learn how to deal with functions abstractly from that book

I think I was in chapter 3 before my chemistry PhD starts

loch

@Adeek hey congrats!

Oskar Tegby

Are you a PhD in chemistry?

Secret

18:02

Yup, and right now in my mac, I am coding the last part of my 2D scan code and submit some jobs before going to sleep

Oskar Tegby

Which university?

Secret

The University of Sydney

Adeek

@loch thanks :D

I still have to adjust and do revisions to my thesis this month and next

but I am done

Oskar Tegby

Oh! Cool! I was almost going to the University of Melbourne this year, but decided to stay here instead.

loch

18:37

@Adeek Yeah at least the pressure’s off!

Adeek

Yeah

I am not good also under pressure sometime I tend to procastinate

because I have a perfectionist mind set

haha

Ted Shifrin

19:19

@Adeek Hearty congratulations, Karim. :)

And hi, @Oskar @loch

Alessandro Codenotti

Hi @Ted

Ted Shifrin

Heya demonic @Alessandro

Oskar Tegby

Hi, @TedShifrin! How are you? Do you mind looking at a problem that I'm stuck with? I've been trying to find what function that $f_n(x)=\frac{x}{1+nx^2}$ converges to uniformly, and on what subset of $\mathbb{R}$, but I get stuck. I've come to the following.

$$\left|f_n(x)-f_m(x)\right|=\left|\frac{x}{1+nx^2}-\frac{x}{1+mx^2}\right|=\left|\frac{x(1+mx^2)-x(1+nx^2)}{(1+nx^2)(1+mx^2)}\right|=\left|\frac{x+mx^3-x+nx^3}{(1+nx^2)(1+mx^2)}\right|=\left|\frac{x^3(m+n)}{(1+nx^2)(1+mx^2)}\right|=\left|\frac{x^3(m+n)}{1+mx^2+nmx^4}\right|.$$

Ted Shifrin

There's a standard technique for dealing with uniform convergence, @Oskar. What do you have for the limit function $f$?

Oskar Tegby

What do you mean?

Ted Shifrin

19:28

What's the pointwise limit?

Oskar Tegby

Isn't it just zero?

Ted Shifrin

Yes.

Oskar Tegby

Okay, but what about the subset of uniform convergence?

Ted Shifrin

So now you look at the function ($n$ fixed) $|f_n-f| = |f_n|$, in this case, and ask what its maximum value is.

Oskar Tegby

Okay.

Ted Shifrin

19:31

Necessary and sufficient (on whatever domain) for uniform convergence is that that sequence of maximum values converge to $0$. You understand why?

Oskar Tegby

Hm... I'm not sure. Do you mean for which $x$ it has its maximum?

Not really.

Ted Shifrin

Well, yes, and what the maximum value is.

Oskar Tegby

Okay.

Ted Shifrin

If you're going to have $|f_n(x)-f(x)|<\epsilon$ for all $n\ge N$, then you'll need $M_n = \max|f_n-f|<\epsilon$ for all $n\ge N$.

Oskar Tegby

What do we do with $M_n$?

Ted Shifrin

19:34

What do you mean? You start by finding it :) ... AND by seeing for what $x_n$ you have $f_n(x_n) = M_n$. [That's relevant if you're going to decide on what domain convergence is or is not uniform.]

So you find $x_n$ and $M_n$ for me.

Oskar Tegby

Oh! I misread your comment. Sorry.

Adeek

Thanks @TedShifrin you know I am in a weird state where the way I view humans now is kinda of like data

Oskar Tegby

Okay. I'll have a look at it. I'm going to bed soon. It's pretty late here in Sweden. I'll get back to you with $x_n$ and $M_n$. :)

Adeek

I should probably go out more out of my apartment so I don't completely like become some kinda of weird academic robot

haha

I have seen people like that I don't want to be one of them

Ted Shifrin

It's not that late, is it? 10:30 or something? But fine.

The world is full of people I don't want to recognize as human, Karim.

Oskar Tegby

19:39

It's just 9:30, but I'm just really tired. I don't know why.

Adeek

Yeah that is also how I think as well @TedShifrin

Ted Shifrin

We're tiresome, @Oskar :P

Adeek

For example people who sent one guy to prison because of his way of viewing the world

I don't think such people are human

and if you kinda say something about that then your some kinda of traitor

Ted Shifrin

In some places they get shot, Karim.

Adeek

humanity is weird

yeah and that as well !

Oskar Tegby

19:41

Haha! No, @Ted. I love this chat. :)

Ted Shifrin

Good night @Oskar :P

Oskar Tegby

Good night! :)

Adeek

It makes sense to me because I guess they have been living in a certain world view since growing up and if someone kinda of shakes this world view then their life is a lie. Whence people protect this world view with everything that they have.

I am not sure though if humanity would be better off if we are pure logical beings.

Ted Shifrin

You cannot reduce life globally to simple logic.

7

That's Ted's philosophical statement for the year.

Adeek

but I think when you accept certain world view according to philosophy you stop critical thinking. Having lived in that part of the world I can confirm it 100 %.

I think the more you use critical thinking the more you develop that part of the brain.

Ted Shifrin

19:46

Well, 35-40% of the US "citizens" worship at the god of Fox News. Anything else is lies and "unfair" to their racist leader.

Adeek

I am not sure if like if you look at humanity is it only small percentage of humans are smart or is it the way they got brought up ? I was thinking maybe it is only small percentage of humans are smart/philosophical and rest are kinda of filler, so it is easy to kinda of control them using whatever mechanism.

I think the situation in the west probably the reason it is well developed is because their is probably like constitution which protect people, so talented people who really benefit humanity don't get persecuted like in the dark ages.

But I think probably given a population only small percentage of people think outside the box.

Ted Shifrin

Most of us, even ones who are taught to think critically in certain situations, tend to stay comfortable in our own little boxes. ... Anyhow, back to math(s) :)

Adeek

yeah I see

yeah

I will be going to UK next year for workshop for 6 month @TedShifrin

Ted Shifrin

6 months! wow! that's a long workshop.

Adeek

I really want to meet Michael Atiyah

yeah

It will cover many aspects of geometry + physics

Ted Shifrin

19:57

I knew Atiyah back in my grad school days ... I'm afraid he's past his prime now. But so am I.

Alessandro Codenotti

What's the geometric intuition behind the local homology groups $H_n(X,X\setminus\{x\})$? We're only considering cycles whose border is the single point $x$ so it feels like they should be somewhat close to $\pi_n(X,x)$ but I'm not sure

Adeek

How was his personality @TedShifrin ?

Ted Shifrin

Super friendly back then. I have no idea now.

Adeek

I want to ask him personal questions since he also lived in my country as well.

Ted Shifrin

@Alessandro, no, no. The boundary is in $X-\{x\}$.

So this, by excision, localizes to $(D^n, D^n-\{0\})$.

Oh, I can shaddup now.

hi, Mike.

Alessandro Codenotti

20:02

Oh, of course, what I was thinking about is $H_n(X,\{x\})$, which is actually the same as reduced homology of $X$, so it agrees with $H_n(X)$ for all $n>0$ and then there are some theorems I've only heard of on the relationship between $H_n$ and $\pi_n$

user31415

@AlessandroCodenotti "We're only considering cycles whose border is the single point $x$" - you are considering cycles whose boundary misses the single point x - so it's really cycle that goes around x though

Alessandro Codenotti

Hi @user31415, sorry for disappearing yesterday, I fell asleep! In the case with $k$ preimages I convinced myself the local degree is $1$ in each of them because the map I was considering restricts to a homeomorphism on a small enough neighbourhood of those points

user31415

Hey, no worries - I think homeomorphism itself is not sufficient, because that doesn't take care of orientation (so it can be 1 or -1)

regardless, AkivaWeinberger's justification was correct

Alessandro Codenotti

@user31415 Hm, right, that's a good point

I don't quite see how to pass to the boundary as suggested by Akiva

user31415

using excision, $H_n(X, X-{x}) \cong H_n(D^n, D^n - 0)$

then use short exact sequence, $H_n(D^n, D^n - 0) \cong H_{n-1}(D^n - 0) \cong H_{n-1} (S^{n-1}$

i think the only point is then to check that the induced map on $H_{n-1} (S^{n-1})$ still comes from the $z -> z^k$ map

Alessandro Codenotti

20:13

When you say short exact sequence you mean the one for chain complexes to be unrolled in the homology long exact sequence?

user31415

i should say long exact sequence i guess, mb

the one where $H_n(D^n) \to H_n(D^n, D^n - 0) \to H_{n-1} (D^n - 0) \to H_{n-1}(D^n) \to ...$

Alessandro Codenotti

Oh, ok, I see how that works now

@user31415 Right and since the two outermost groups here are trivial the two middle ones are isomorphic

And then $H_{n-1}(D^n-0)\simeq H_{n-1}(S^{n-1})$ because the former space deformation retracts on the latter

user31415

sure

Alessandro Codenotti

I see, thanks

user31415

:)

zed111

20:26

How do you define/calculate d(gradient of f)/df where f=f(x,y,z) is real?

Semiclassical

20:40

you typically don't. the only context I know of where you'd do something like that would be functional differentiation

Leaky Nun

21:06

@zed111 you calculate ∇(f+δf) - ∇f and try to extract a factor of δf, and this is in the context of functional differentiation

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