Mathematics - 2024-12-31

pie

04:15

should I ask this question on MO?
https://math.stackexchange.com/questions/661666/upper-bound-for-area-of-polygons

I am sure the sup of area would be impossible to find since its quite hard for quadrangle, one can make a formula for an arbitrary upper bound by using $\max(l_1,l_2, \dots, l_n)$ and finding an area for the n-regular polygon however this is far from what I want, I want a "close-enough " formula to the $\sup$ of the area (and this is the problem of the question I don't know how to state it clearly) what I want want is "as close as it get" to the supremum of the area i.e a bett…

Idk if they would like this question or not, maybe I should re-ask on MSE first? What do you guys think?

Xander Henderson

@pie Do not reask the save question on the same site.

leslie townes

if you could figure out what you mean by "as close as it get" that would make a new question, imvho

pie

@leslietownes Well, That is the problem I can't state it clearly in English.

@XanderHenderson What about MO? do you think this question would be good for that site?

leslie townes

i would regard figuring out what you mean by "as close as it get" as a prerequisite for asking it anywhere

otherwise it's just "hey this old question, any thoughts? anything 'close to' this?"

a question directed at finding out what "as close as it get" ought to mean might be new (i would think more appropriate for MSE than MO), if it set out some parameters that would guide a solution

people who work in extremal problems in geometry have certainly thought about this before

Kripke Platek

04:53

5

Q: Given a polygon of n-sides, why does the regular one (i.e. all sides equal) enclose the greatest area given a constant perimeter?

This doesn't require much more than the title. I just need an explanation, but an algebraic proof would be a bonus. We can demonstrate this for quadrilaterals, a square is best as shown by this graph- the area peaks when both sides are equal at 250.

@pie Maybe this could help you =)

Soham Saha

Hi everyone. I recently got a question about proving the irrationality of tan(1 degree). I thought that if tan(1 degree) is rational, then so is tan(2 degree), using tan(a+b) formula. Similarly tan(3), tan(4),…tan(30) all are rational. But tan(30) is irrational, so tan(1) must be irrational too. Is my thinking correct?

leslie townes

05:11

soham: yes. this argument certainly puts a focus on how one might prove the addition formula for the tangent, or how one would know that tan(30) is irrational, but its logic is solid if you take those things as black boxes.

nickbros123

05:33

Weird, first time got logged out of chat

Jakobian

05:56

@SohamSaha I believe $\tan(\alpha \pi)$ where $\alpha$ is rational can only be rational at few special cases

of course when you say "$\tan(3)$" etc. it should be $3$ degrees there instead

there is a difference between $\tan(3)$ and $\tan(3^\circ)$

(if you don't see the LaTeX on my posts, see description of this channel)

Soham Saha

06:31

@leslietownes thanks for confirming :)

@Jakobian yes I meant 3 degrees. Thanks for pointing it out.

Jakobian

10:20

1

Q: How Can an Open Ball Be Perfect?

In the following proof from Real Mathematical Analysis by Pugh, it is suggested that the open ball $M_{r/2}p$ is perfect. But if we consider a perfect set $S \subset M$, then $S = S'$ where $S'$ is the set of cluster points of $S$. Also based on the proposition below, $\overline S = S \cup S' = ...

real-analysis general-topology metric-spaces

this nomenclature to me, is absolutely horrific

Pugh must have never heard about dense-in-itself sets

Sine of the Time

12:12

A killer question from Japan. Is tan 1° a rational number?

►

@SohamSaha

Soham Saha

15:00

@SineoftheTime yes exactly

I haven't seen their solution yet, though

Hey they've given the exact same solution as mine!

Mats Granvik

15:57

$$\log(2) = \underset{k\to \infty }{\text{lim}}\left(\sum _{n=1}^k \frac{(-1)^n \left(2 \left(\left(1-2^{-2 n}\right) \zeta (2 n) \Gamma (2 n)\right)\right)}{\pi ^{2 n} \left(k^2\right)^n}+\sum _{n=1}^k \frac{(-1)^{n+1}}{n}+\frac{1}{2 k}\right)$$

psie

16:53

Sorry to ask such a basic question, but consider the collection of all sets of the form $$U_1\times\cdots\times U_n$$where $U_i\subset\mathbb R$ is open. Does this collection contain a countable base for $\mathbb R^n$?

Joe

@psie: It is a base for $\mathbb R^n$, but not a countable base – since there are uncountably many sets of those form.

leslie townes

what joe said. it contains a countable base (in fact many of them), but is not itself a countable base.

4

Q: How to show that the set of open balls with rational centres and rational radii form a countable base for $\mathbb{R}^n$?

There are questions on Math.SE that seem relevant, but I will explain why they do not answer my question: 1) $\mathbb R^n$ has countable basis of open balls? (Yes): this question doesn't prove that rational centre/radii balls form a base; it proves that $\mathbb{R}^n$ has such a base 2) Open ba...

general-topology

you can adapt the idea of this

psie

@leslietownes ok, so the link is meant to be an example of a countable base contained in the collection of sets I gave?

Jakobian

$(a_1, b_1)\times ... \times (a_n, b_n)$ where $a_k, b_k$ are rational is an example of a countable base

psie

ah nice, the open rectangles, right?

leslie townes

16:59

well, when n > 1, open balls in the usual metric of R^n are not cartesian products of open subsets of R^1 (this might be a good exercise!). but there are ideas that transfer over

although if you take the metric space generalization that one of the answerers offers up and put a slightly different metric than the usual one on R^n, then the solution provided there will indeed provide you with a countable base of the type you like

but by 'the idea of this' above i meant, the answers suggest that there is significance associated to R^n having a countable dense subset and the possibility of realizing general open sets as unions of ones parametrized by rational numbers

psie

ok

Joe

If we take $n=1$ for simplicity, then we see that $(1,\sqrt 2)=(1,1.4)\cup (1,1.41)\cup (1,1.414)\cup\cdots$. Hopefully this example illustrates the idea behind what Jakobian said.

Jakobian

If $X_k$ are topological spaces, $X = \prod_{k\in I} X_k$, $\pi_k:X\to X_k$ is the projection, $\mathcal{B}_k$ is base for $X_k$ for each $k$, then finite intersections $\pi_{k_1}^{-1}(U_1)\cap ...\cap \pi_{k_m}^{-1}(U_m)$ where $U_i\in \mathcal{B}_{k_i}$ form a basis for $X$

for infinite products we only have finite intersections, but for finite product those are nothing more than sets of the form $U_1\times ...\times U_n$ where $U_i$ are in the individual bases

leslie townes

if you give R^n the metric d(x,y) = max{ |x_1 - y_1|, |x_2 - y_2|, ..., |x_n - y_n|} then the "open balls" in this metric are indeed cartesian products of sets that are open in R^1 and the funny metric topology is the same as the usual one (again a good exercise, all of these are good exercises)

some would say better and more interesting than measure theory :)

Jakobian

another good exercise is to show that a separable metric space is hereditarily Lindelof...

leslie townes

17:11

i want to be hereditarily lindelof when i grow up

Jakobian

An easy exercise shows that a separable metric space is second countable

and conversely, a second countable topological space is separable

it's easy to see that a second countable space is Lindelof

and both being metrizable and being second countable are hereditary properties

so a separable metric space is both hereditarily separable and hereditarily Lindelof

one can also easily see that Lindelof metric space is second countable

in other words for a metric space the following are equivalent: Lindelof, separable, second countable

it's the last property that's the strongest, but separable that's usually the easiest to verify

separable and Lindelof are not comparable to each other

I never saw an argument in which someone goes "oh yeah let's prove this metric space is Lindelof" though

one can also show that a space is hereditarily Lindelof if and only if every open subspace is Lindelof

but that's besides the point

either way the application here is of course that if you take an open set $U$ of a second countable space, and $U = \bigcup_{i\in I} U_i$ is some union of open sets, then $U = \bigcup_{i\in I_0} U_i$ for some countable $I_0\subseteq I$

and the same thing happens in any hereditarily Lindelof space

why? Because $U_i$ for $i\in I$ form an open cover of $U$, which you can then extract countable subcover, because $U$ is Lindelof

so this is really about showing a separable metric space (such as $\mathbb{R}^n$) is hereditarily Lindelof, and this follows from separable + metrizable $\implies$ second countable $\implies$ hereditarily Lindelof

but I think the equivalence of all three properties for metric spaces is just important to remember

nickbros123

17:35

@Jakobian me neither

@Jakobian I have, i think seen a counter argument technique using lindelof though, to show that a metric space is not one of the more important things: 2nd countable or separable

by showing its not lindelof

ok perhaps I hallucinated. I was thinking about ell_infinity, but to show its non-separable, we can look at the set $G$ of sequences that contain 0s and 1s and directly prove inseparability: If $E$ is any countable dense set, we consider all $\varepsilon<1/3$ balls around every point of $E$ which is supposed to cover all of $G$. From pigeon hole principle, one such ball must contain 2 points of $G$, but the distance between 2 distinct points of $G$ in the sup norm is 1,

but the ball is of a smaller radius than 1/3

raises a contradiction

Jakobian

18:02

well, yes, for a metric space, extent equals to density

this is a bit different from the Lindelof number, but equal to it as well

Xander Henderson

@leslietownes That would require growing up, and my impression is that you don't want to do that.

ILikeMathematics

18:28

How does $\alpha = 0$ follow from this?

This seems like a contradiction. $\sum_{i=1}^n |a_i|^2 < \sum_{i=1}^n |a_i|^2$.

Thorgott

this reads like nonsense, unless you've omitted context

Jakobian

rule 1: Always omit context

@Thorgott I mean, if they didn't omit context, would it not read like nonsense?

I feel like what's written is written... no

ILikeMathematics

@Thorgott Well, Sheldon Axler responded in a comment and said this was how he imagined the solution. math.stackexchange.com/questions/1452559/…

Thorgott

maybe I'm being blind, but I have no clue where that first equality comes from

Jakobian

I mean it kind of makes sense but those strict inequalities are only really strict if $\alpha$ is non-zero and $n > 1$

ILikeMathematics

18:36

The problem is this: Let $B = \{v_1, \dots, v_n\}$ be an orthonormal basis of $V$ and $(V, \beta)$ an euclidean space. Let $w_1, \dots, w_n \in V$ with $$||v_i - w_i|| < \frac{1}{\sqrt n} \quad \text{for all $1 \leq i \leq n$.}$$ Show that $\{w_1, \dots, w_n\}$ is a basis of $V$.

Jakobian

@Thorgott this is norm of $\alpha$ squared

Euclidean norm

Thorgott

nvm, I was just being blind

ILikeMathematics

@Jakobian Hm, so we make them non-strict

But then we obtain $\sum_{i=1}^n |a_i|^2 \leq \sum_{i=1}^n |a_i|^2$.

That doesn't say anything

Or is there something in-between that delivers the proof?

Jakobian

@ILikeMathematics one of those inequalities is strict if $\alpha$ is non-zero

but definitely if $\alpha = 0$ then all of those are non-strict

so this is really a proof by contradiction but the author was lazy

ILikeMathematics

Ah, assume $\alpha \neq 0$, then this is a contradiction

Thorgott

18:41

there is no need to phrase this as a contradiction, but yeah

ILikeMathematics

Thank you!

Thorgott

I'm not sure how they obtain the last inequality tho

Jakobian

Jensen inequality

Thorgott

ah, has to be one of those I always forget

Jakobian

I don't know maybe there's another way to see this

Thorgott

18:42

it's not obvious when you just multiply it out (which is what I was doing), so probably not

but I'm baaad with inequalities

Jakobian

Generalized mean

In mathematics, generalized means (or power mean or Hölder mean from Otto Hölder) are a family of functions for aggregating sets of numbers. These include as special cases the Pythagorean means (arithmetic, geometric, and harmonic means). == Definition == If p is a non-zero real number, and x 1 , … , x n {\displaystyle x_{1},\dots ,x_{n}} are positive real numbers, then the generalized mean or power mean with exponent...

there's a general statement like this

ILikeMathematics

Can't we say $$\left(\sum_{i=1}^n|\alpha_i|\frac1{\sqrt{n}}\right)^2<\left(\sum_{i=1}^n|\alpha_i|\right)^2\leq\sum_{i=1}^n|\alpha_i|^2$$

Jakobian

if you take square roots and mingle with the constant term, it's basically arithmetic mean $\leq$ quadratic mean

@ILikeMathematics definitely not

ILikeMathematics

Oh wait

Jakobian

all those inequalities between means are basically applied Jensen inequality/Holder inequaity too so its not like I am saying much

ILikeMathematics

18:45

It should be the other way around in the last step, yeah

That doesn't work

Jakobian

there should be a lot of different ways to justify this

Quasi-arithmetic mean

In mathematics and statistics, the quasi-arithmetic mean or generalised f-mean or Kolmogorov-Nagumo-de Finetti mean is one generalisation of the more familiar means such as the arithmetic mean and the geometric mean, using a function f {\displaystyle f} . It is also called Kolmogorov mean after Soviet mathematician Andrey Kolmogorov. It is a broader generalization than the regular generalized mean. == Definition == If f is a function which maps an interval I {\displaystyle I} of the real line to the real numbers...

there's also this cool thing, but it seems like it can't provide you maximum of numbers, so they need to get some kind of approximated functions instead

also cool that for $p\to 0$, $M_p$ approaches geometric mean which corresponds to logarithm, so in some sense $\lim_{p\to 0} x^p = \log(x)$

@Jakobian yeah because those are constant on a huge portion

ILikeMathematics

@Jakobian For $\alpha_i \neq 0$, why is $$\left|\left|\sum_{i=1}^n \alpha_i(e_i- v_i)\right|\right|^2 < \left(\sum_{i=1}^n |\alpha_i| ||e_i - v_i||\right)^2?$$

Jakobian

8

Q: Characterization of the quasi-arithmetic mean

The $f$-mean, where $f$ is a continuous monotonically-increasing function, is defined as: $$ M_f(x_1, \dots, x_n) = f^{-1}\left( \frac{f(x_1)+ \cdots + f(x_n)}n \right). $$ For any $f$, $M_f$ has the following nice properties: Continuous; Monotonically-increasing in each argument; Symmetric ...

real-analysis means

@ILikeMathematics have you heard of this one cool thing called like... uh... triangle something?

Thorgott

that's the one I know!

ILikeMathematics

this is the triangle inequality?

Jakobian

18:59

yes, triangle inequality and homogenity of the norm

ILikeMathematics

Ah

Jakobian

I suppose I should say "absolutely homogenity" if I want to be completely precise

@Jakobian I think I might read this

it's actually pretty interesting result

actually I don't like how they are doing things for when $f$ is defined on $\mathbb{R}$ since I think the $\log$ case is very important

copper.hat

19:44

happy new year to all :-)

Xander Henderson

@copper.hat Oh, is that happening?

Or did it happen?

copper.hat

19:59

just getting ahead of myself

not even drinking...

Xander Henderson

@copper.hat I realized this week that I have not had an alcohol since Thanksgiving.

I should have an alcohol...

copper.hat

$3^4 \cdot 5^2$. Rather pleasing for some reason.

Jakobian

Xander reminiscing the days of alcoholism

copper.hat

@XanderHenderson this year had been a disaster for my drinkiing

one friend got gout, another was laid off and depressed so his wife told me that we are not supposed to go drinking

a lot of folks have got the "any alcohol is bad for you" messaging from the temperance society

Xander Henderson

@copper.hat Yeah, those people are nutty.

@Jakobian I mean, I've never drunk to the level of an alcoholic. I'm typically in the two or three drinks per week category.

copper.hat

20:04

truly a disaster. and my 21 & 23 yo kids have an inherited trait that limits their intake.

Jakobian

@XanderHenderson how much categories do you do per week?

copper.hat

i'm not a heavy drinker, but i do enjoy drinking a bottle of wine or two with friends

Xander Henderson

@Jakobian None.

copper.hat

:-)

Jakobian

oh, so you're an abstinent

Xander Henderson

20:06

@Jakobian Indeed. Categories are dumb.

My doctor warned me away from them.

Jakobian

I would disagree if this weren't true

copper.hat

there was a temperance thing in Ireland (linked to religion) when i was growing up, i didn't drink until i was 21 (excepting an accidental mouthful of something called shandy).

you wore a pioneer pin (not sure if that was what it was called then).

ILikeMathematics

How do you pronounce $\overline z$ of a complex number in German? "Das konjugierte von $z$", "Das Konjugat von $z$"?

Xander Henderson

"zee bar". In any language.

copper.hat

it is zee bar, a place you got to drink

Xander Henderson

20:09

Because all languages are really just English with a funny accent.

Jakobian

not in Polish

sprzężenie

Xander Henderson

That looks like "zee bar" to me.

copper.hat

of course, it should be zed bar

ILikeMathematics

Yeah, "$z$ quer" works I guess, if you explicitly want to say that this is the conjugate of $z$, is "Konjugat von $z$" right?

Xander Henderson

@copper.hat That's just zee with a funny accent.

copper.hat

20:11

ah, i zee :-)

2

Jakobian

I haven't had enough alcohol for this discussion yet

why does everyone speak German

copper.hat

unfortunately i only speak English. And I am Irish

Jakobian

as someone who is bilingual, speaking in any language is just like breathing

you don't even notice

copper.hat

i wish

Jakobian

I meant it to be kinda a positive message? Like it wouldn't actually be much different

ILikeMathematics

20:20

Ok, seems it's "die Konjugierte von $z$"

copper.hat

my memory is hopeless. trying to synthesize sentences taking into account conjugations, declensions, etc is mostly beyone me

Jakobian

I don't even know what half of those words mean

copper.hat

when my kids ask me to explain a word i use i am often unable to explain, which is scary.

Thorgott

@ILikeMathematics if you wanna be explicit about it being the conjugate, yes, otherwise "z quer" (as you said)

copper.hat

20:36

i will hit z liquer in a while

leslie townes

wheres grandad

monoidaltransform

21:56

If X is a metric space an A is an anr is it true that the closure of A is an anr?

Thorgott

22:52

no, take X=R and A={1/n,n in N}

copper.hat

grandad went for a dip in the pool

Jakobian

23:27

@Thorgott I stumbled on an interesting concept

A Tychonoff space $X$ is called an $F'$-space if for any disjoint cozero sets $U, V\subseteq X$, $\overline{U}\cap \overline{V} = \emptyset$

An $F$-space when $U, V$ can be separated by zero sets instead

$F$-spaces are interesting because every countable subset of them is $C^\ast$-embedded

example 1.10 here msp.org/pjm/1983/108-2/pjm-v108-n2-p03-s.pdf provides an example of a locally compact $F'$-space which is not an $F$-space, and is a subspace of an $F$-space

I haven't really gone through this example yet. It'd be interesting to find an example of a subspace which isn't even an $F'$-space

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