Mathematics - 2023-02-01

12:01 AM

@XanderHenderson my uncle, who's currently making more barbecue

You know what? I'll do it myself. I'm going to eat a frozen hotdog, drink lots of water, and then light myself up. Perfectly cooked hotdog in my tummy

Xander Henderson

@ペガサスSeiya So did it come off of the grill?

ペガサス Seiya

@XanderHenderson yep

Xander Henderson

There you go. It was cooked on the grill.

So, as long as it is not just a solid tube of char, it is not overcooked. And if it were overcooked, it would be, essentially a tube of carbonized meat, and quite unpleasant to eat (almost certainly not too easy to chew).

ペガサス Seiya

@XanderHenderson it would look funny though

Even funnier if you got someone to eat that

Jakobian

12:52 AM

I have too little reasons to learn, too much expectations about myself, and too much doubt about my future

Sine of the Time

@Jakobian you're not alone

i am

hi @copper.hat

Hi @SineoftheTime!

it's 2 a.m. :'(

12:58 AM

its only 17:00 in california

Sine of the Time

a lot of users from US

copper.hat

probably just time zone dependent.

I just look for convex questions and bug Leslie & Ted.

Ted Shifrin

Hard to dispute that claim!

Obliv

1:20 AM

what is an example of a function with no known derivative and integral

i know $e^{x^2}$ has no known antiderivative but i think u can easily take the derivative

leslie townes

it has many known antiderivatives. you are implicitly requiring antiderivatives to be of some special form.

Obliv

Oh really? In my diff eq class my professor made it seem like it didn't have one.

maybe with the math we know

leslie townes

there are different ways of formulating that, but in the usual way, the set of things of that form is closed under differentiation, so you will not have a 'known' function with an ugly derivative. but, possible to have a 'known' function whose only antiderivatives are ugly.

Obliv

not sure I understand what you're saying. from what I gather, some diff eq have no "simple" solutions but you're saying by definition they must have some solution that may be ugly?

leslie townes

i would say something like, there's no 'precalculus formula' for an antiderivative of e^(x^2), although e^(x^2) itself might count as a 'precalculus formula.' this is not a precise definition but captures the world of functions someone has in your toolkit around the time that they begin to take calculus in the USA.

this is somewhat different from a function not existing, or not being 'known' in any other sense. there are functions with precalculus formulas that are not really 'known' so much as 'things people are used to.'

e.g. people think of sin(x) as known, but what is sin(7) and how do you know?

a^b is known, but what is 2^pi? 2 times itself pi many times?

Obliv

1:27 AM

wait one at a time lol

isn't $\sin(x)$ defined specifically as opposite/hypotenuse of a right angle triangle

leslie townes

maybe it is, so what is sin(7) to three decimal places?

Obliv

is that in degrees or radians

leslie townes

you pick.

Ted Shifrin

Radians, always.

Is Munchkin ready to be president of her class?

Obliv

the issue is $\pi$ is transcendental

leslie townes

1:32 AM

i sure hope not.

now pi isn't known!

Ted Shifrin

@Obliv Huh?

Obliv

I am desperately trying to not say the wrong thing

Ted Shifrin

Why is that the issue?

leslie townes

obliv it is kind of a numbers version of the thing with functions up above. you can adopt a view of number in which, say, rational numbers and roots and such are 'known' and pi is not, and even formalize that distinction.

but the distinction is not really whether something is known or knowable or accessible to any kind of analysis. it's just, a choice of where to draw the line in your universe.

Obliv

to calculate $\sin(7)$ I was constructing a triangle but then I realized the input angle is defined by $\pi$

leslie townes

1:36 AM

if only power functions are 'known' then you can differentiate all of them, and find antiderivatives for most of them, except for x^(-1). does that mean that there is no known antiderivative of 1/x? maybe, but only if "known" means something special.

Ted Shifrin

You can do just fine with $\pi/4$ and $\pi/6$.

Obliv

@leslietownes I think if you define any function within the boundaries of an axiomatic system there will have to exist a representation of its domain, range, derivatives, antiderivatives, etc. because everything you used to define the function was confined to a system.

what would be interesting is if you needed an infinite amount of axiomatic set theory symbols to represent a functions derivative/antiderivative

but I don't think that'd ever be necessary unless you needed an infinite amount of those symbols to define it in the first place.

Jakobian

@SineoftheTime I'm not looking for company

there's this question that bothers me

at what point is the weakness your own fault

Obliv

When you deem it to be

Jakobian

rational answer isn't very insightful

Ted Shifrin

1:49 AM

Is it my fault I’m not Gauss? Maybe my parents’ fault. Why is this fault relevant?

Jakobian

because you want to keep yourself in check

Ted Shifrin

It is my fault if I am lazy and don’t make my best efforts.

But maybe we’re not all supposed to be PhD Math superstars.

Obliv

Why would other people be able to offer you insight on how you should keep yourself in check?

Jakobian

how do you tell if you're lazy or just in an emotional state that makes it harder for you to take effort in the things you should be doing

Obliv

Experience

Jakobian

1:53 AM

@Obliv because they can think just as much?

Ted Shifrin

If you need counseling, get counseling. But I will tell you that it’s hard to be successful in math with emotional baggage.

But we’re not here to be psychologists and deal with neuroses.

Obliv

I know this isn't related to math, but in a "grand unified theory" for physics, doesn't that require all of mathematics since mathematics is in the world that physics is trying to describe?

Or is that too philosophical

Ted Shifrin

Why should physics not use all the math available as needed?

Obliv

I think it's too philosophical because then you have russell's paradox with the set of all sets

yeah I just don't know what they mean by that claim of a grand unified theory. It seems so naive to think you can have a single equation to describe everything observable.

just means you don't have enough observations

Ted Shifrin

There’s a good reason most mathematicians don’t bog down in advanced set theory.

Jakobian

2:02 AM

I think, a lot of the things physicist do don't have to be described precisely as we do it in mathematics

Ted Shifrin

Well, I think they think they’re rigorous without being mathematically pedantic.

Jakobian

it's a bit that the "local picture" doesn't need to be dependent that much on what axiomatic system we currently are in, the argument works for the purpose its intended to work

Obliv

The generalizations of mathematics serves as a useful extension for research on subjects where a single neat equation describes a phenomenon

right, and that local picture can seem like everything in the sense of a grand unified theory but it's still local

because we live locally :P and that's why it gets philosophical because you can adopt a solipsistic look, or any other view of what counts as real or observable etc

ペガサス Seiya

2:57 AM

Calc 3 test results are out

94% for me. Highest again.

Didn't think I'd be top this time (I was sleepy during the test) but hey, here we are

Cotton Headed Ninnymuggins

4:52 AM

@ペガサスSeiya congrats Snorlax

A point in space

5:15 AM

@MISC {611966,
TITLE = {Is the set of decreasing functions from $\Bbb N$ to $\Bbb N$ countable?},
AUTHOR = {Spenser (https://math.stackexchange.com/users/39285/spenser)},
HOWPUBLISHED = {Mathematics Stack Exchange},
NOTE = {URL:https://math.stackexchange.com/q/611966 (version: 2013-12-18)},
EPRINT = {https://math.stackexchange.com/q/611966},
URL = {https://math.stackexchange.com/q/611966}
}

In this answer how is the bijection between $S_{N+1}$ and $S_{N}\times \mathbb{N}$ established?

Ted Shifrin

5:37 AM

He explained that explicitly. What precisely is your question?

Actually, he should have said something slightly more explicit. If the function is not identically $N+1$, then there is a first point $n_0$ at which it drops down. Then $f(n_0) = m\le N$. Now proceed. :62889339

Cotton Headed Ninnymuggins

5:54 AM

I'm in a combinatorics class that asks how many 6-digit numbers are divisible by 3 but not by 2. I don't know if there's a simpler approach to what I did, but I simply said 100002 + 3(299999) = 999999 so there's 300000 numbers divisible by 3.

Next, 100002 + 6(149999) = 999996 so 150000 are divisible by 6. So 300000-150000=150000 numbers divisible by 3 not 2 (since divisible by 6 means both divisible by 2 AND 3)

A. P.

Hi, I am answering the following question

4

A: Double integral over general regions between 2 curves.

You do not need a double integral. $$\int\limits_{x=0}^1 \sqrt{2 x} - x^2/2\ dx = \frac{1}{6} \left(4 \sqrt{2}-1\right)$$

leslie townes

it's not entirely clear how you're using the arithmetic on the left hand sides to perform those counts, but those counts and the reasoning (i.e. "divisible by 3 but not by 2" being the same thing as "divisible by 3 but not by 6") are OK.

A. P.

The two answers seem to agree on the limits of integration? However, I think they are incorrect as I explain in the comments. I'm wrong?

The intersection give the points (0,0) and (2,2) over the real plane.

Cotton Headed Ninnymuggins

Yeah my arithmetic is just showing that to get from the lowest number divisible by $x$ to the highest one divisible by $x$, we're adding $n*x$ amount to get there

There's probably a nicer way to show that lol

leslie townes

yeah. you're implicitly identifying the set of numbers in that range that are divisible by 3 as the set of numbers of the form 100002 + 3k. for k between 0 and 299999, and there are as many of those as there are k between 0 and 299999. that's solid.

and half of those are even (corresponding to taking even k in that range), and half are not.

Cotton Headed Ninnymuggins

6:07 AM

Snazzy

Cotton Headed Ninnymuggins

6:35 AM

In words, why are there the same number of even subsets as odd subsets of [n]?

Ted Shifrin

@A.P. They look fine to me. What do you think is wrong?

David Stork is clearly wrong.

leslie townes

6:51 AM

cotton: 'even subset' means a subset with an even number of elements? lots of ways to see that. you could find some clever way of counting them (as functions of n) and showing that they are equal. or write down a bijection between those two sets.

Cotton Headed Ninnymuggins

7:14 AM

Yes, even-sized subsets

A. P.

7:26 AM

@TedShifrin I saw that two users agreed with the limits of integration [0,1], which using the same graph from one of them does not holds. I was a little hesitant though because there were several upvotes for the answer and no comments on it. In my answer, I show that the questioner's path is correct and the answer is 4/3.

I think it can be a typo, though a rather strange one given that the questioner gives the proper bounds and on the graph as well.

Gwyn

8:41 AM

In the proof of "if the two subsequences $a_{2k}$ and $a_{2k+1}$ converges to the same limit $L$ then the sequence ${a_n}$ converges to $L$", I obtained two natural indexes $N_1$ and $N_2$ from the convergences of $a_{2k}$ and $a_{2k+1}$ and I considered $\max\{2N_1,2N_2+1\}$. This works, but I am not completely sure if I understood this.

My reasoning is that I get $|a_{2k}-L|<\epsilon$ for $n \ge N_1$ and $|a_{2k+1}-L|<\epsilon$ for $n \ge N_2$. Since any positive integer is either even or odd, I distinguished two cases. If $n$ is even, then $n=2k$ and if $n \ge \max\{2N_1,2N_2+1\}$ in particular $2k=n \ge 2N_1 \implies k \ge N_1$ and so $|a_n-L|=|a_{2k}-L|<\epsilon$ because of the hypothesis $a_{2k+1} \to L$. Similarly, if even is odd $2k+1=n \ge 2N_2+1 \implies k\ge N_2$ and so $|a_n-L|=|a_{2k+1}-L|<\epsilon$.

Since each integer is either even or odd, this works for any $n\in\mathbb{N}$ such that $n\ge \max\{2N_1,2N_2+1\}$ and so the limit definition is satisfied for ${a_n}$. Is this the correct reasoning why we must consider $\max\{2N_1,2N_2+1\}$? To satisfy universal quantifier "for any $n\in\mathbb{N}$" right before the implication "$n \ge \max\{2N_1,2N_2+1\} \implies |a_n-L|<\epsilon$"?

leslie townes

in the summary above, you're missing the overall goal of the proof, which is to let e > 0 be given, and to exhibit a positive integer N for which n >= N implies |a_n - L| < e. you get there in the end, but that's the entire point of the proof. given e > 0, do that.

so given your e > 0, you feed this e into the limit statements you know, getting N_1 and N_2, and as you sketch out, if N happens to be larger than that max, then you can guarantee |a_n - L| < e, with slightly different justifications depending on whether n is odd or even.

the thing that makes the case analysis possible is that when you take N to be that max, then n >= N allows you to deduce both of the statements you need for that case analysis.

Gwyn

Oh yes, I didn't write it because of "space issue" in chat, but I was aware of that) I must prove this for an arbitrary $\epsilon>0$ and so I must introduce that it is given at the beginning. Yes, I agree with the part of the both statement, thanks for the clarifications)

leslie townes

8:57 AM

n >= max (A,B) implies both n >= A and n >= B. that's almost a definition of "max." you see it used a lot in sequence proofs because N >= max(A,B) is a way of ensuring that both n >= A and n >= B. so if you had a case analysis where n >= A makes some nice things happen, and n >= B makes some other nice things happen, n >= max(A,B) will make all of the nice things happen.

Thorgott

12:08 PM

I love it when nice things happen

Xander Henderson

12:51 PM

@Thorgott Nice things happen here?

Thorgott

never did I imply my statement was non-empty

one potato two potato

0

Q: Hatcher exercise $\bf 2.2.30\left(e\right)$

Compute the homology of the mapping torus $T_f$ of the map $f:S^1\times S^1\to S^1\times S^1$ that interchanges the two factors and then reflects one of the factors. From the l.e.s of the mapping torus, $$0\to H_3(T_f)\to H_2(S^1\times S^1)\xrightarrow{1-f_*} H_2(S^1\times S^1)\to H_2(T_f).$$ F...

Nasser

1:18 PM

isn't strange that to find an integrating factor for a second order ode (which is not exact) in order to make it exact, one has to solve another second order ode? math.stackexchange.com/questions/1725717/… so would this not end up in an infinite loop? If this other ode is not exact, how to solve it? Find an integrating factor? Which means solving a third ode, and then a 4th and so on. There got to be a better way.

It is the second answer in the above linked to post. The answer is correct. but notice the adjoint ode is also second order and varying coefficient. Just like the one we are trying to solve in the first place.

Not Tfue

2:02 PM

worst thing ever

Xander Henderson

2:16 PM

@Nasser Sometimes, things just don't get easier.

Other times, they do.

Semiclassical

Trying to remember if there’s a name for the following; given a convex 2D region A and a positive length r, let A(r) be the set of all points which are within r of A.

Thorgott

$\varepsilon$-neighborhood, but $\varepsilon=r$

Xander Henderson

@Semiclassical $A+B(0,r)$

(in the sense of a Minkowski sum)

Semiclassical

Ah, nice

Xander Henderson

or an $r$-neighborhood of $A$, as @Thorgott suggests.

Note that $A$ need not be convex, nor a subset of $\mathbb{R}^2$.

The definition works in any metric space.

Semiclassical

2:29 PM

True. I was thinking about the case of an ellipse specifically

Which seems painful

Xander Henderson

@Semiclassical Life is pain.

Semiclassical

Yes

It’s motivated by a (bad) question on the main site

MathWorld labels the boundary as Ellipse Parallel Curves

Which makes sense

ペガサス Seiya

3:01 PM

$19 to anyone who knows Obama's last name

Koro

Suppose that X is a topological vector space over R, A is a convex open subset of X. Let $b\notin A$, then there is a continuous linear functional $L\in$ X* such that $L(b)=1$ and $L(x)<1$ for all x in A.

I don't understand why this L should be continuous.

Alessandro Codenotti

What do you mean? Is there a step in particular in the proof that's unclear? From the little I remember this is an application of Hahn-Banach

Koro

3:17 PM

yes, so I understand how existence of L satisfying L(b)=1, L(x)<1 for all x in A is shown.

But I'm having difficulty understanding why L should be continuous.

The outline of proof is as follows: 1) A is shown to be absorbing. 2) Using a result, it follows from 1) that A={x in X: p_A(x)<1}, where p_A is the Minkowski functional of A.
3) Since b is not in A, $p_A(b)\ge 1$. Consider the subspace span (b), and define $L(\lambda b)= \lambda,\lambda \in \mathbb R$. It is observed that $L(\lambda b)\le p_A(\lambda b)$

4) By Hahn Banach theorem, L is extended to a linear functional $\tilde L$ on X, $\tilde L(x)\le p_A(x)$ for all x in X.

Why is $\tilde L$ continuous?

Alessandro Codenotti

Because continuity of the extension is the main point of Hahn-Banach

Koro

yes, if we are in normed linear spaces.

but here we have a topological vector space.

parz

hi

Alessandro Codenotti

Hahn-Banach should work everywhere, I'm not sure where to look to find it stated in full generality though

Koro

The proof further says that L is continuous because if x is in $(-A)\cap A$, then $-1<L(x)<1$. I don't understand how it implies continuity.

I'm trying to instead see if $L^{-1}(-1/n,1/n)$ is open in X or not.

that will atleast give me the continuity at 0.

Oh I understood it now.

In topological vector spaces, if a linear functional is bounded in an open neighborhood of 0, then the functional is continuous.

Thorgott

4:34 PM

isn't the whole point in Hahn-Banach to get a continuous extension that remains bounded above by a sublinear functional (in this case, the Minkowski functional)?

Jakobian

5:48 PM

well... Hahn-Banach doesn't even say anything about continuity (explicitly)

it's the "being bounded by a functional" part that gives it continuity

Koro

6:04 PM

@Jakobian yeah

@Thorgott not sure, I thought it was for getting an extension that is still dominated by the sublinear functional.

Suppose that I have a continuous linear functional $\phi$ defined on a topological vector space X. Suppose that A is an open set of X. There is a real no. r such that for all a in A, $\phi(a)\le r$. Then, can I say that $\phi(a)<r$ as A is open?

I think no because $\{x \in X: \phi^{-1}(-\infty, r)\}$ may not be the same as $A$.

perfect, pinch zoom on macbook is also not working now.

robjohn

6:22 PM

Did you change any settings? Sometimes an update changes some settings.

ペガサス Seiya

^Only if he has automatic updates enabled

Koro

nope it seems to be a global issue, that feature is unreliable (sometimes it works, sometimes it doesn't). At the moment of writing this comment, it is not working.

It's working now.

This happens very frequently after an update (months ago).

robjohn

Some apps recognize pinch an spread, and some don’t. When using an app that recognizes those gestures, I haven’t noticed them being unreliable.

Koro

but recognising sometimes, and not recognising other times is also an issue.

leslie townes

koro: no need to write the {x in X : __ } to describe phi^{-1}(-infty, r) there. when you say "can i say that phi(a) < r" do you mean, that it holds for an arbitrary a in A?

robjohn

6:37 PM

@Koro never seen that on my macbook

leslie townes

koro: you can't say that if phi is the zero functional and r is 0. if phi is nonzero, you can say that. (if there's a in A with phi(a) = r, there's an open ball of some positive radius centered at a that is contained in A, and this open ball contains vectors of the form ta for t slightly larger than 1.) this exercise might be an introduction to the 'open mapping theorem'

Koro

@robjohn my OS version is ventura 13.0.1 and I'm using opera browser.

I don't recall from which update it started happening.

Nov 15, 2022 at 20:56, by Koro

$\mathcal{L}$ renders a box in a macbook.

robjohn

I’m using Mojave and Firefox. I’ve never used Opera.

leslie townes

koro: you may need to adjust the above argument when r is zero even if phi is nonzero, but that's the basic idea.

robjohn

@Koro that depends on the fonts installed

Koro

6:42 PM

the linked issue is fixed by: right click-->show math as etc.

@leslietownes yes, for all a in A.

@leslietownes I was thinking along these lines. I'll get back to this soon.

Thorgott

7:12 PM

@Koro continuous extension

Sine of the Time

curiosity: how hard is to get to MIT or Princeton to do the PhD?

copper.hat

7:29 PM

is identity theft allowed?

mick

i voted to reopen

1

Q: $\sum_{n>0}\frac{f(n)}{g(n)\ln(2n)^2} = C$?

Im looking for closed form integer functions [$\mathbb{N}\to\mathbb{Z}$] $f(n),g(n)$ such that $\sum_{n>0}\dfrac{f(n)}{g(n)\ln(2n)^2} = C $ Where $C$ is a closed form number. Something like $$\sum_{n>0} \dfrac{1+n^2}{(1+n^4)\ln(2n)^2} = \pi^e $$ But then a true statement ofcourse. Is that even po...

calculus closed-form

closing without second thought is silly

please vote to reopen , thanks

Sine of the Time

@mick It's more appropriate to discuss about this in the chat "CURED"

@copper.hat I don't think so, why?

Koro

@leslietownes thanks, I got it :-).

the ball may contain elements of the form ta, t<1 too.

Mats Granvik

7:45 PM

$$\lim\limits_{n \rightarrow \infty}\int_{-n}^{n} \binom{n-1}{k-1} f\left(k+s\right) \, dk = \lim\limits_{n \rightarrow \infty}\sum _{k=-n}^{k=n} \binom{n-1}{k-1} f \left(k+s\right)$$

leslie townes

yeah. if r is 0 you can consider a + tv where phi(v) is nonzero.

Koro

i think that we only need phi non zero for the result to be true.

For every open set U of X, phi(U) can be shown to be open.

mick

@SineoftheTime thanks

i wish math could be used to cure diseases :(

KZ-Spectra

Hi everyone, I am looking for some discussion on my question here: math.stackexchange.com/q/4630067/965485 if anyone has insights, just let me know.

Koro

Taking any f(x) in f(U). By openness of U, there is a d>0, B(x, d) sits inside U. Let f(x_0)=c>0. Hence f(x_0/c)=1. Set x'=x_0/c. Since scalar multiplication is continuous, $\lim_{t\to 0}(x+tx')=x$. There exists $\delta'>0, |t|<\delta'\implies x+tx'\in B(x,d)$

Sine of the Time

7:53 PM

@mick spain without s

Koro

$f(x+tx')=y+t\in f(B(x,d))\subset f(U)$, hence f(U) is open.

So if $f(a)<=r$ for all a in open set A, then f(a)<r.

Sine of the Time

hi @ペガサスSeiya

mick

@SineoftheTime ?

Sine of the Time

@mick spain without s is spain -s = pain

mick

oh you joke

i was serious

@ペガサスSeiya WHO IS STRONGER ? Me or you ?

Koro

7:59 PM

like someone with fever getting cured by just applying Gram Schmidt orthogonalization or something.

Sine of the Time

@mick clearly me

mick

@SineoftheTime clearly ?

Sine of the Time

yep

the proof is left to the reader

mick

your not even a saiyan

i look like chuck norris

the burden of proof is on your side

Sine of the Time

@mick you look like chuck norris on steroids

mick

8:01 PM

i do

Xander Henderson

@SineoftheTime "Hard"?

Sine of the Time

@XanderHenderson is the word "hard" inappropriate?

mick

NP hard is ok

all the rest sounds sexual

Xander Henderson

@SineoftheTime What do you mean?

Sine of the Time

@XanderHenderson just wondering if it's possible to do the PhD at MIT or you have to be a genius

Xander Henderson

8:04 PM

@SineoftheTime According to MIT itself, their acceptance rate for graduate programs is 6.7% (crimsoneducation.org/in/blog/campus-life-more/…).

So only one of every 20 or so applicants gets in.

But I think that chalking that up to some kind of innate ability (e.g. "genius") is not the right way to think about it.

It probably helps, but I doubt that it is either necessary nor sufficient.

Sine of the Time

Yeah, I think so. Btw I live in Europe, so I don't know precisely how is the university in the US

I know you can rely on recomendation letters

Xander Henderson

@SineoftheTime Well, that is likely to make things even harder.

@SineoftheTime Do you mean you can't rely on recommendations?

Because that would be more accurate...

mick

if you solve the Riemann hypothesis you get in

Sine of the Time

@XanderHenderson I know a guy who had a recommenation letter from a professor and was admitted to MIT, here from Europe

Xander Henderson

@SineoftheTime Yes, that can happen. But, again, it is neither necessary nor sufficient.

I would imagine that the vast majority of applicants cannot rely on recommendations alone.

monoidaltransform

8:08 PM

Hi, sorry, as a sanity check if $A$ and $B$ are subspaces such that $A\subseteq B$ of $X$ and $C_{A}$ is a component of $A$, does it follows that $C_{A}=C_{B}\cap A$, where $C_{B}$ is a component of $B$?

Sine of the Time

@XanderHenderson Maybe they're also rich ;)

Xander Henderson

@SineoftheTime That should make no difference. Indeed, it would likely be illegal in the US for the admissions process to consider wealth.

Koro

Rejected student sends school a rejection letter of her own

►

Xander Henderson

At most US institutions, an admissions committee is going to be looking at a portfolio. This might include recommendations, past research experience or publications, recommendations, undergraduate grades, GRE (or similar) scores, whether or not the applicant's research interests align with anyone on the faculty, and so on.

You can't expect any one thing to be THE one thing that gets one through the door.

Sine of the Time

@XanderHenderson yes, I was just wondering. It'd be a dream to study at Princeton or MIT,

Xander Henderson

8:12 PM

@SineoftheTime Why?

Sine of the Time

@XanderHenderson It's a prestigious university and you're stimulated

Xander Henderson

@SineoftheTime Well, if you are interested in the prestige of the place, then you are applying for the wrong reasons, and will rightly be turned away.

And, with respect to "stimulation", nearly any graduate program is going to provide that.

Sine of the Time

@XanderHenderson yes, obviously I'll not go to brag about it. It's a good place to study

Xander Henderson

@SineoftheTime For whom?

Sine of the Time

@XanderHenderson Seems that you're not agreeing about that

Xander Henderson

8:16 PM

@SineoftheTime I didn't say that. I just think that your impression of the place is very shallow.

robjohn

@Koro Finally back on my MacBook, and $\mathcal{L}$ renders fine.

Xander Henderson

Remember that graduate programs are a kind of apprenticeship program. You are learning to be an expert in some very small niche of academia. You should be going to a place where there is some faculty or research group that is doing something you are interested in pursuing further.

Sine of the Time

@XanderHenderson It's not paradise, obviously, and it's clearly a place of competitiveness and "envy". But I think it's a good idea to do the PhD abroad

Xander Henderson

@SineoftheTime Sure. But, again, you should consider the particular faculty or research group you are going to be joining.

Sine of the Time

@XanderHenderson I guess so, anyway it's still soon to think about the PhD

Ted Shifrin

8:20 PM

For graduate work, you really have to pay attention to what field(s) you are thinking of working in for your doctorate.

Xander already said that. Never mind.

Sine of the Time

@TedShifrin I like calculus, differential equation and linear algebra

Ted Shifrin

You're two or three years too early to be thinking about this.

You're just naming standard first- and maybe second-year mathematics.

Sine of the Time

@TedShifrin yep, second-year

Xander Henderson

For example, if you were interested in analysis on fractals, then you might consider Kyoto, as Jun Kigami is there. Or University of Conneticut, to work with Alexander Teplayev. Or wherever it is that Uta Freiberg is currently working.

Five years ago, I would have added UC Riverside to that list, as Michel Lapidus works in the field, but I suspect that he is very close to retirement.

And is unlikely to take on new students.

Sine of the Time

I see

Ted Shifrin

8:23 PM

That wouldn't surprise me. Then again, some people stay on well into their 70s and still take students. I'm shocked by how many of the old folks at MIT and Penn are still there, for example.

Sine of the Time

what about Fourier Analysis, do you know where would be good to study?

Ted Shifrin

There are so many aspects. Generally, that is the field of harmonic analysis.

Lots and lots of places.

Xander Henderson

@TedShifrin Yeah, but folk in his family have had some difficult health problems in the last couple of years, and I understand that remote teaching due to COVID has pretty much wiped him out. My guess is that he retires in 2 or 3 years, after getting his last student out the door.

He'll be 70 by then, at any rate.

Ted Shifrin

Yeah, I turn 70 in a few days.

Xander Henderson

@TedShifrin Happy birthday.

Ted Shifrin

8:27 PM

Not yet. Not yet. :)

Xander Henderson

Fine. "Happy birthday" rescinded!

Have a terrible day!

>:(

Sine of the Time

Do you know any specialist about Riemann Hypothesis?

Xander Henderson

@SineoftheTime Depends on what you mean by "expert", "know" and "Riemann hypothesis".

Ted Shifrin

Sine, this is not a productive discussion. You're just picking things out at random at this point.

Sine of the Time

@TedShifrin sorry :(

Ted Shifrin

8:29 PM

Just worry about learning mathematics. You can do your own googling to see who writes papers on various subjects.

Xander Henderson

@SineoftheTime If you are in your second year, then you still have a couple of years of work to do before applying to graduate programs. I would suggest that, in that time, you get to be friendly with a few faculty at your current institution, and learn what kind of things they are working on. Build those personal relationships.

Then, when it comes time to apply to graduate schools, talk to them about where they thing you would be a good fit.

Academia is built on personal ties. Almost no one works in isolation.

Sine of the Time

@XanderHenderson yes that's a good idea, but it's difficult to build a personal relation, expecially because there's indifference and separation

Koro

Suppose that G is a topological group, H is a subgroup of G, then closure (H) is also a subgroup.

It is clear that the identity e is in cl (H).

Xander Henderson

BOOO! Nothing is clear! Why not just say "The identity, e, is in cl(H)."?

Koro

For every x in cl(H), the map $m: \{x\}\times $ cl(H)-->G: m(x,y)=xy is continuous, and therefore $m^{-1}(e)$ is non empty.

so there exists y in cl(H), xy=e. That is, cl(H) is closed under inverses.

I'm not sure how to show that xy is in cl(H) for every x,y in cl(H).

ペガサス Seiya

8:37 PM

@mick I've killed Hades, I'm stronger

leslie townes

i would say more about why m^{-1}(e) is non empty

Thorgott

this is very obvious if you know about nets

but all this really rests on is that $f(\overline{A})\subseteq\overline{f(A)}$ for continuous $f$

Koro

Leslie: now that you say it, I think I concluded that by mistake.

leslie townes

koro: i suspected as much, because the thing you say immediately after is a rephrased version of "m^{-1}(e) is nonempty"

the result is true, i would just include more detail about why

Thorgott

I think you can get away with less work

ペガサス Seiya

8:43 PM

@XanderHenderson you're the MiG-29 jet that ambushes aircrafts passing by for fun

Xander Henderson

@ペガサスSeiya I have already told you that I find these conversation-as-war metaphors to be very uncomfortable and inappropriate. Please stop.

Koro

8:55 PM

@leslietownes I think I'll have to define some funky map to prove it.

Alessandro Codenotti

You can also argue that if $h_1,h_2\in\bar{H}$, then so is $h_1h_2$ by showing that all of its nbhds meet $\bar{H}$, since every nbhd of $h_i$ contains $h_i'$ by picking small enough nbhds and using continuity of the multiplication (this is just a rephrasing of the nets argument to avoid nets, making it more confusing, but if you don't know about nets it might be a good approach)

(have you already proved that closed subgroups are open? This is unrelated but a nice surprising exercise)

Koro

no, I was trying to prove closure under the product.

Alessandro Codenotti

9:11 PM

I am aware, that's what my first message is referring to

Koro

yeah, I tried to it using contradiction: Fix a,b in cl (H). Suppose on the contrary that ab is not in cl(H). It follows that there is an open set U containing ab, such that $U\cap $ cl(H)= $\emptyset$. By continuity of the group multiplication, there exist nbds of a and b respectively (let's call them $U_1$ and $U_2$ respectively) such that $\{u_1 u_2: u_1 \in U_1, u_2 \in U_2\}$ is in U.

Since $a\in $ cl(H), $U_1\cap H\ne \emptyset$, and similarly for $U_2\cap H$.

but not sure how to get contradiction from here.

Koro

9:27 PM

hmm, choose $a_i\in U_i\cap H$. Then $a_1a_2\in U\cap H=\emptyset$, which is a contradiction.

amWhy

I see that the "The mean heart" is among us!! :-) Happy February 1st! Isn't tomorrow Groundhog Day?

goerie.com/story/news/2023/01/31/…

robjohn

@amWhy that's one old groundhog.

amWhy

@robjohn Indeed! :D

Ted Shifrin

10:14 PM

Will the groundhog see his shadow? Probably not back east.

@robjohn Did your doctor checkup go well?

robjohn

@TedShifrin The one a week ago? The catheter needs to stay in until next Tuesday and they'll check again.

Ted Shifrin

Ah :( Wow. Has this been a week already?

robjohn

It's been three weeks since the surgery, but a week since the cystogram.

Ted Shifrin

Well, I hope this saga will soon be entirely in the rear-view mirror.

robjohn

Like Lubbock, TX.

ペガサス Seiya

10:28 PM

Ground beef vs air beef

Who wins?

ZaWarudo

10:51 PM

do you have suggestions if I feel I got a toxic habit to try to prove anything by myself (independently of its difficulty, so even theorems), mostly because I feel insecure and because I genuinely think that this is the only way to learn? Do I risk to "waste" (I prefer "exert") my time and risk to remain fossilized too much hard results that someone must show me how to do? Consider that I am mostly self taught. I feel very bad when I watch the solution of a problem, but I have no comparisons.

robjohn

@ペガサスSeiya I would not want to be below the air beef

Ted Shifrin

@ZaWarudo It depends on your long-term goals, and some theorems no one is expected to figure out on his/her own.

leslie townes

zawarudo it might help to work linearly through a single book. the problem with being self-taught from the internet is that there is no single fixed point of reference across all of the websites and videos and other materials out there. so there's no way to say "OK, we've already proved X, Y, Z so now we can use X, Y, Z in our approach to more complicated problems."

self taught folks seem more likely to fall into the trap of always feeling like they need to go all the way back to definitions, because you might not be able to do otherwise in a coherent way, with an incoherent set of materials that don't refer to one another

and so sometimes its harder for them to develop the sense of relying on theorems to prove other theorems.

Ted Shifrin

Oh gosh ... self-taught from the internet? Scary thought.

Sometimes good old textbooks (particularly good ones) are the way to go.

2

ZaWarudo

11:09 PM

I am self taught in the sense that I used to study physics but then I felt in love with math and switched towards pure math, however in physics we always made computations and so I was not able to do proofs or too much theoretical exercises; so I bought some real analysis, abstract algebra and geometry books and I'm studying from them. After that, I bought some problem books and I try to solve as much as I can. However, I have to do this by myself because I have no time to go back to lectures

Mainly because I am 30 now, I would love to do research but I am scared about the age and the competition

Anyway, sorry if this is not the right place to ask, but I saw that some questions like these are not well always received on MSE

Ted Shifrin

11:21 PM

To me the most important part of math teaching has always been giving students feedback on their homework/exams. It's very hard to learn good proof skills without effort and feedback. I don't quite get how that works with self-learning. Answer manuals are not always the right answer. But I don't recommend trying to "discover" all the proofs of all the theorems, no.

ZaWarudo

thank you very much Ted, the feedback of a professor like you is very precious.. (I saw some of your lectures on youtube!) listening this from someone who has become a college professor this surely will help to get rid of this bad habit.

Ted Shifrin

Anyhow, I’ve answered plenty of “self-learning” questions (and plenty not so) based on my books both in here and on main. Just ask @D.C.theIII ;)

ペガサス Seiya

ZaWarudo

I cannot read that name in a normal tone and voice...

ZaWarudo

that was the purpose of this nickname :D

ペガサス Seiya

11:53 PM

@ZaWarudo Yare Yare

Ted Shifrin

What’s wrong with normal tone?

Transcript for

$Mathematics$ Mathematics