Mathematics - 2025-02-07

Xander Henderson

00:19

@CroCo I don't think that this is a mathematical topic. Though, if anything, it is either statistics (which is pretty far removed from mathematics anymore), or (maybe) operations research.

But the excerpt you provide really doesn't give much context.

Based on the abstract, I think that there is also a bit of psychology involved...

(Again, not math.)

Thorgott

I agree with Xander

Xander Henderson

@Thorgott Ouch. That has to hurt. :P

CroCo

00:46

@XanderHenderson so it has nothing to do with measure theory? So basically this is subjective assessment used to measure the human operator workload in numerous fields like aviation, robotics teleoperation and of course psychology

Xander Henderson

@CroCo Measure theory, in general, has nothing to do with measuring things. Measure theory is, in one word, about integration.

Thorgott

@XanderHenderson well, it's fine as long as it's not on the topic of category theory

Xander Henderson

@Thorgott Hey, you can be wrong about that. :P

CroCo

I see. Thanks a lot.

Jakobian

01:22

@XanderHenderson well that is a sort of measurement

coffeemath

musk-e gee instute for the criminally insane. sorry, i only know math at the phd level [suny stony brook 1980 daniel m cass thesis advisor anthony v phillips] and also do little more than make bad jokes.

Jakobian

I think that's where the name comes from after all. It has nothing to do with measuring things as would a physicist or engineer, say, understand it

My friend constantly is convincing me to go back to university since math is what I do anyway. I don't know. Maybe I should

although its not like I have the capacity to write those long theses about a subject that others do

I'm a bad mathematician with (practically) no other alternatives

Thorgott

01:50

@Jakobian idk, I feel like you regularly display a lot of tenacity and/or thoroughness in terms of investigating the topics you care about here

though of course it's not exactly comparable to writing a thesis

Jakobian

02:26

@Thorgott I just have time to do so

hbghlyj

13:20

I have a question about the proof of this example (the cubic nature the Heisenberg group)

I don't understand why (1) the integral is independent of the bounding disk and (2) by the area inequality, it gives a lower bound on the area of such a bounding disk.

Thorgott

13:40

@Jakobian well, one also gets time to write a thesis

it's a complex decision and I won't claim to know what the right one is for you

but my impression is not that your mathematical ability would not be up to par

Shean

14:21

https://math.stackexchange.com/q/5031776/987127
Help plz

pie

17:03

Is there a name for polynomial that only differ in the sign of the coff of $x$ like $ax^2+bx+c$ and $ax^2-bx +c$

Xander Henderson

That would be a set of polynomials, not just one.

But I know of no name for such a collection.

pie

@XanderHenderson Well I need some name for an mse question, can you recommend me some name?

Xander Henderson

Why do you need a name?

This makes no sense...

This feels like an XY problem to me. :/

pie

if $p$, $\bar{p}$ are the polynomial with change the coff of $x$

For $n \in \mathbb{N}$ let $p_1, p_2, p_3, \dot, p_n$ be a second degree polynomial with $a_k$ the the coff of $x$ in $p_k$ let $q_1, q_2, q_3, \dot, q_n$ be a first degree polynomial with $b_k$ the the coff of $x$ in $q_k$

then $$\int \frac{1}{1+x^{4n}}= \sum_{k=1}^n \left(a_k\frac{\ln(p_k)-\ln(\bar{p_k})}{8n}\right)+\sum_{k=1}^n \left(ab_k\frac{\ln(q_k)-\ln(\bar{q_k})}{4n}\right)$$

I was gonna ask how to find such polynomials, I tried till n=10 and this pattern holds, idk if this is interesting enough tbh

Xander Henderson

I don't understand. Is $x$ a polynomial?

pie

17:13

@XanderHenderson no a variable, $p_k , q_k$ are polynomials of $x$

Xander Henderson

Your integral has nothing to do with those polynomials...

Are you asking if there exist polynomials of this form such that $\int (1+x^{4n})^{-1}$ have the form on the right?

pie

@XanderHenderson hmmm is there a closed form for this integral? I tried to find a general solution for $\in\frac{1}{1+x^n}$ and noticed that the when n is multiple of $4$ say $n=4m$ then there are n terms m of them are in form $\frac{a_k}{2n}\ln(p_k)$ the othe m is the form of $\frac{-a_k}{2n}\ln(\bar{p_k})$ and the other m in form of $\frac{b_k\arctan{q_k}}{n}$ and the last m terms have the form $-\frac{b_k\arctan{\bar{q_k}}}{n}$

Xander Henderson

I have no idea if there is a closed form. My intuition is that there should be (in terms of some inverse trig functions, probably), but that it is going to be a bit of a mess.

pie

There is some mistakes in the above expression

For $n \in \mathbb{N}$ there exist second degree polynomial $p_{1,n}, p_{2,n}, p_{3,n}, \dots, p_{n,n}$ $a_{k,n}$ the the coff of $x$ in $p_{k,n}$ and there exist a first degree polynomial $q_{1,n}, q_{2,n}, q_{3,n}, \dots, q_{n,n}$ be with $b_{k,n}$ the the coff of $x$ in $q_{k,n}$

then $$\int \frac{dx}{1+x^{4n}}= \sum_{k=1}^n \left(a_{k,n}\frac{\ln(p_{k,n}(x))-\ln(\bar{p}_{k,n}(x))}{8n}\right)+\sum_{k=1}^n \left(b_k\frac{\arctan(q_{k,n}(x))-\arctan(\bar{q}_{k,n}(x))}{4n}\right)$$

my questions are how to prove this form holds for all such $n$ also how to find $p_{k,n}$ and $q_{k,n}$

I need to name $p$ and $\bar{p}$ and more importantly i need some acceptable symbol other than using the bar, any suggestions

Akiva Weinberger

18:29

I asked a question on Math Stack Exchange

2

Q: Showing that the number of spanning directed trees of a directed graph does not depend on root or direction

A directed tree is called an in-tree if all of the edges points towards a root, and an out-tree if all edges point away from a root. (Equivalently, an in-tree is any connected directed graph for which every vertex has out-degree $1$ except for one, called the root, which has out-degree $0$; an ou...

leslie townes

interesting question, akiva. i am not in combinatorics but this strikes me as the kind of thing where maybe a bijective proof could be hard (or where the ones you can get are kind of artificial, e.g. backing out some formal thing from less artificial bijections on much larger sets)

as i guess has already appeared in the comments

ILikeMathematics

19:36

$f(x) = |x|^{-x^2} = e^{-x^2 \ln(|x|)}$ for $x \neq 0$ is continuous on $\mathbb R$ if we define $f(0) = 1$. Is it also uniformly continuous?

@XanderHenderson I guess this is a bit similar to the one we did last time, where we used the definition. But I have the feeling this won't be uniformly continuous here

Do you have any idea for a counterexample?

It would be very messy with the definition here

Jakobian

19:59

@AlessandroCodenotti hi. Do you know what an extremally disconnected space is?

Alessandro Codenotti

Yes, why?

Jakobian

You recall that a dense subspace of extremally disconnected space is extremally disconnected, right?

It's a pretty standard fact that both open, and dense subspaces of extremally disconnected are extremally disconnected, but closed subspaces are not necessarily so

Alessandro Codenotti

Sure

Jakobian

Now there's a weakening of extremally disconnected propety, where instead of demanding that $\overline{U}$ is open for every open $U$, we demand its open for a cozero set $U$. Such spaces are called basically disconnected

Now one might wonder if dense subspace of basically disconnected space is basically disconnected or not

0

A: About subspaces of $F$-spaces

For definitions of extremally disconnected, $P$-space, basically disconnected (under the name $\sigma$-complete), $F$-spaces, $F'$-spaces, and relations between them, see [3]. Some of them to note here are: $$\text{ext. disconnected}\implies\text{basic. disconnected}\implies F\text{-space}\implie...

Alessandro Codenotti

A cozero set is just the complement of the zero set of a continuous function right?

Jakobian

20:05

here I prove that this is not the case, in fact, it can fail one of the weakest properties I know that generalizes extremally disconnected spaces

@AlessandroCodenotti yes

At the end I've included a question which I'm not sure about. I'm pretty sure it must be true though. If an extremally disconnected space can have a subspace which is not an $F'$-space

There are theorems about compact $F$-spaces embedding into extremally disconnected spaces. Although my space seems to have weight which is relatively large for me to use those

nevertheless, the example is really simple so there's a good chance that the space $Y$ can be embedded into an extremally disconnected space

Jakobian

20:23

never mind. I see no one is even remotely interested in what I do

Joe

21:40

If $A$ and $B$ are commutative rings, is every surjective ring homomorphism $\varphi:A\to B$ finite? I think yes, since $B$ is generated as an $A$-module by $\{1\}$, right?

Thorgott

21:59

yes

leslie townes

22:33

@ILikeMathematics this is an example of a function that is continuous on all of R and has limits at infinity and -infinity (both 0). any such function is uniformly continuous. if you don't have a theorem that could apply in that situation like a black box, see something like math.stackexchange.com/questions/705961/… (which like your example is where the limits at both infinities not only both exist but are both 0)

skullpatrol

\o @robjohn

robjohn

@skullpatrol hey, there. How are you doing?

leslie townes

if you have more general tools (e.g. a notion of compactness in metric spaces more general than R) you might be able to get this out of some black box-ish theorems. any f with such limits extends to a continuous function on a compact metric space that contains R (sometimes called the "two-point compactification"), a continuous function on a compact metric space is uniformly continuous, and the restriction of a uniformly continuous function to a subspace of its domain is uniformly continuous.

skullpatrol

@robjohn fine thanks, how are you?

leslie townes

you are generally right to have an instinct that one does not want to dig too deeply into the specific definitions of functions to prove uniform continuity, when you have it you can often get it from more general properties of the function than whatever its specific formula is. it is more common to find oneself dealing with formula specifics when a function isn't uniformly continuous

robjohn

22:43

@skullpatrol Dealing with a lot of stuff irl, but I hope I can spend more time online soon.

skullpatrol

Nice to see you back.

:-)

robjohn

@ILikeMathematics Have you checked if the derivative is bounded?

that is sufficient, but not necessary.

ILikeMathematics

23:00

@robjohn Yeah, it is. Thanks

@leslietownes Thanks!

Let a_n and b_n be sequences so that lim n -> infty a_n b_n = 0. Show that a_n or b_n has a subsequence converging to 0

What would be the idea to prove this?

robjohn

Suppose neither has a subsequence converging to $0$

Alessandro Codenotti

23:19

@Jakobian I had to leave earlier, but I don't really know what $F$ or $F'$ spaces are

I stick mostly to continua nowadays, haven't really looked at weird spaces in a while

I'm thinking about weird subsets of the reals though

leslie townes

23:51

ilike: personally i find robjohns idea to be the most intuitive approach. if you did not like thinking in terms of contradiction, i might first convince myself that because min(|a_n|,|b_n|)^2 <= |a_n b_n| the hypothesis implies [via some reasoning that you would need to supply, but simple reasoning] that min(|a_n|,|b_n|) goes to 0. at least one of the sets {n: |a_n| = min(|a_n|,|b_n|)}, {n: |b_n| = min(|a_n|,|b_n|)} is infinite (maybe both of them are).

and an infinite one of those subsets would index a subsequence that has the desired property for whichever sequence it corresponds to.

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