Mathematics - 2022-07-08

Mathematical Emergency

12:10 AM

@geocalc33

Calvin Khor

12:24 AM

@leslietownes lmao

leslie townes

12:49 AM

:)

Alek Murt

1:05 AM

Hey guys how can i show this $\frac{1}{\lambda(B)} \leq H_{d}(Q)$, where $B$ is the unit ball and $Q$ is the unit cube?

$H_d$ is the Hausdorff MEasure and $\lambda$ the lebesgue measure

Mathematical Emergency

1:20 AM

@geocalc33

Dude, I messed up the day we were supposed to code

Time to study some math to recharge mah brain

Koro

Suppose that I have an array of real numbers {$c_{n,k}: 1\le k\le n$}.

Suppose that $\lim_n \sum_{k=1}^n c_{n,k}=1$

Calvin Khor

@Koro im itching to mention toeplitz's theorem on resummation methods

oh its a finite array :')

Koro

Then can I say that there exists a C>0 such that $\sum_{k=1}^n |c_{n,k}|<C$?

@CalvinKhor you got that right : )

Calvin Khor

i suppose you want something independent of $n$ on the right hand side?

Koro

I ask this question because I saw the above two conditions as separate hypothesis in the statement of Toeplitz theorem on the regular transformation of sequences.

Calvin Khor

1:27 AM

siicc

where did you find toeplitz hahaha :D

Koro

and I believe that the limit condition implies that bound condition by C.

Calvin Khor

the answer is no

but i dont know an easy example

Koro

@Calvin: it's in the book 'Problems in mathematical analysis I' as an exercise problem.

Calvin Khor

you can rearrange sum (-1)^k/k to converge to any sum

make it converge to 1. it still doesn't absolutely converge so the upper bound will depend on n

Koro

ahh, you are right!

Calvin Khor

1:30 AM

cool @Koro :) I saw it in Hardy's divergent series, excellent book

Koro

conditionally convergent is not necessarily absolutely convergent.

Calvin Khor

yupp

Koro

But I was doing the following calculation in my head: There is an N such that $n\ge N$ implies $|\sum_{k=1}^n c_{n,k}-1|<1/2$, i.e., $||\sum_{k=1}^n c_{n,k}|-1|<1/2$ so $|\sum_{k=1}^n c_{n,k}|<3/2$ for all $n\ge N$. Now let max$\{|\sum_{k=1}^n c_{n,k}|_{n\lt N}, 3/2\}=:C$

What's wrong with this C?

Calvin Khor

|sum...| ≤ sum |...|?

inequality looks the wrong way

Koro

I used $||a|-|b||\le |a-b|$

Ohh

I think I understood my mistake.

Calvin Khor

1:35 AM

👍

Koro

The calculation above is correct but it has nothing to do with my question. Because the calculation shows $|\sum_{k=1}^n c_{n,k}|$ is bounded but we want to know if $\sum_{k=1}^n \color{blue}{|c_{n,k}|}$ is bounded.

Calvin Khor

yessir

Koro

@CalvinKhor Have you also studied the book Bromvich's infinite series?

Calvin Khor

no

there's a book on infinite series?

Koro

I came to know about this book in Hardy's book on 'Course on Pure mathematics'.

Calvin Khor

1:39 AM

cool

shintuku

from my last adventure in analysis i have come to learn that there are like just two books on infinite series

i guess that one is one of those

Calvin Khor

was it u who was doing harmonic analysis @shintuku

shintuku

the other one is Knopp - The Theory and application of Infinite Series

no i just do math from time to time

very basic stuff

Calvin Khor

kk

Koro

Hardy's 'a course of pure mathematics'

I wrote the title wrong earlier.

Calvin Khor

1:44 AM

well now that i think about it. hardy's divergent series is about a certain special class of infinite series

and generatingfunctionology is about formal infinite series

so we can bump that to 4

shintuku

huh cool

leslie townes

that reminds me of one of my favorite book titles

amazon.com/Functions-Cambridge-Studies-Advanced-Mathematics/dp/…

jean-pierre, we're really going to need a title for that book, what have you got

we need to print this now, no fooling

what's the title, jean-pierre

ehhhhh

shintuku

it's funny because it sounds like it could be a technical term

but it's not quite clear whether it is

and whether they just really said, a bunch of random functions

leslie townes

he does define random series but the "some" is what gets me

he should have threatened the world by putting "Vol. 1" in the title

Calvin Khor

lmao

shintuku

1:52 AM

what could the world expect next?

behold

Vol. 2: Some other random series of functions

leslie townes

Vol. 3: Ooops, we forgot a few! Even more random series of functions

Calvin Khor

vol. 4: random remarks on series of functions

leslie townes

Vol. 5: I don't give a shit anymore just buy the book

shintuku

in the preface, the author would repeat twice that he does not intend to give an exhaustive list of random series of functions. an editing mistake results in this intention being mentioned again just before the preface ends

he is famously quoted as saying, these are just some of them

Calvin Khor

2:07 AM

vol 6: a random series of remarks on functions

Alek Murt

is this measure $\mu_{n}=n \phi(n x) d \lambda$ absolutely continous respect lebesgue measure? $\lambda$ is the lebesgue measure, pls someone help me

Mathematical Emergency

You know your function could have a "backdoor" theoretically?

Since all programs have bugs (usually) what constitutes a mathematical backdoor?

Calvin Khor

@AlekMurt how would you define ac wrt lebesgue?

Alek Murt

If for some funcion $\lambda(f) = 0$ implies $\mu_n (f) = 0$

i forgot to say $\phi$ has support $[-1/n,1/n]$

Calvin Khor

for one function? also you probably want $\phi(nx)$ to have support [-1/n, 1/n], not phi

Alek Murt

2:14 AM

ya, srrry for that u are right

and it is for every function

Calvin Khor

well it is for every "reasonable" function. im gonna guess $L^1(\lambda)$?

Alek Murt

also ya, any space that would fit, no problem with that

Calvin Khor

figure out a way to turn the fs into indicators of measurable sets

Alek Murt

srry but could u be a bit more specific, can't follow the hint

Calvin Khor

reformulate AC in terms of measures of sets, not integrals of functions

then the result is immediate

Alek Murt

2:21 AM

Ok, ya it makes sense, $\mu_n$ would be just a constant times the lebesgue measure

Calvin Khor

a function times the lebesgue measure

leslie townes

a measure of the form f dm is going to be AC wrt m if f is measurable

this is true even for finitely additive measures although you need a different definition of absolute continuity than the one stated above

and there can be measures that are AC in the finitely additive sense wrt m which are not of that form

Alek Murt

what definition are u using

leslie townes

condition 2 in en.wikipedia.org/wiki/… "equivalent definitions" (which are not equivalent for finitely additive measures that are not countably additive)

the one that closely parallels the definition of continuity

i wonder if i can get someone into finitely additive measure theory, it's like indie rock

way better than mainstream measure theory

Calvin Khor

lmao

user10478

3:33 AM

What variable am I missing in order to partition the Monty Hall problem into equally likely outcomes? I still have a 1:1 ratio of winners to losers currently. i.imgur.com/gtnbpiC.png

BAYMAX

4:22 AM

I have given some thoughts to the problem I asked yesterday:

Suppose $A,B \in M_{n}(\Bbb{R})$ such that $A = \left[C_{1}\middle|\frac{I}{0\dots0}\right], B= \left[C_{2}\middle|\frac{I}{0\dots0}\right]$ , where $A$ and $B$ have different first columns (represented as $C_{1}, C_{2}$).

$\textbf{Thus we have $B = A+ \xi e_{1}^T$, where $\xi$ is a $n \times 1$ column vector and $e_{1}^{T} = [1, 0,\ldots,0]$}$.

Let $\lambda_{i}, i=1, \ldots, n$ denote the eigenvalues of $AB^2$. Then we have the condition that $|\lambda_{i}|<1 \, \forall i$.

Let $\beta_{i}, i=1, \ldots, n$ denote the eigenvalues of $A^2B$. Then we have the condition that $|\beta_{i}|<…

Mathematical Emergency

5:28 AM

-2

Q: In general what's the max. $y$ such that $\sqrt{x + y}\# = \sqrt{x}\#$ do we just look at the floor of the square roots, or can we say more? I rhymed

Let $x\# = \prod_{ p \text{ prime} \\ p \leq x} p$ be $x$ "primorial". Determine $\max y \in \Bbb{Z}$ such that: $$ \sqrt{x + y}\# = \sqrt{x}\# $$ Go! Attempt. Take $x = 36$. It's square root is $6$ and we can keep adding to $36$ until we reach $49$. In terms of $x$ that's $(\lfloor\sqrt{x}\rfl...

-2 son

Yes, anything math related... -2 votes immediately. If it doesn't have a Kevin Bacon degree of less than 3 from Galois theory, it's garbage.

Calvin Khor

5:39 AM

@MathematicalEmergency I cannot 'let [something] be "primorial" ' if I do not know what this word means. If you mean to say you are defining the "primorial", then I think you should edit. In addition MSE is not a race track, I think you can remove "go!"

leslie townes

I upvoted because I couldn't understand what the downvotes were for but I also agree with Calvin

Calvin Khor

also what does rhyming have to do with anything lol

Mathematical Emergency

...

fixed

It's not my fault MSE gives me the impression that people race to post

like they have to be the first one

or something. >.>

<_<

Calvin Khor

well that is also my impression sometimes

but i think calling attention to that may bring out some goblins

Mathematical Emergency

.I think there should be a "open problem stack exchange" where people can collaborate

On unsolved mysteries of math

Calvin Khor

5:50 AM

also did you actually rhyme i cant find a rhyme

Mathematical Emergency

title

Calvin Khor

floor and more?

ok

Mathematical Emergency

Yes. What's more, we used floor. We feel that the ceil version is non-standard.

Calvin Khor

whaaat you have unsung hero lol

Mathematical Emergency

What's that

Calvin Khor

5:53 AM

this thing

Mathematical Emergency

Is that Japanase?

Unsung

Am I a warrior?

Calvin Khor

nein

if it wasnt english my next gues would be korean

japanese doesnt have this ung i dont think

Mathematical Emergency

Yes, I have that badge. I don't know why...

Calvin Khor

because people dont upvote u lul

Mathematical Emergency

That sux. I guess they can focus on their manifold cohomology theories, and I'll stick over in my corner messing with some blocks of wood.

it was a joke.

Calvin Khor

5:57 AM

u say that but im the one answering the basic analysis questions and ur up in weirdo algebra land .-.

dang im only at 9% unsung, but you need to be 25% unsung for the badge

impossible

Mathematical Emergency

lol newb. You have to be more esoteric

:D

leslie townes

it seems kinda dickish for MSE to have a badge for that.

Calvin Khor

i tried but i just keep finding grateful people

maybe i need to be colder and leave more things as exercises for the reader

Mathematical Emergency

I think i deleted that. Anyway, I just think my brain is hurting from doing math and I was like here MSE you solve it and even though I didn't say that directly, the dogs sniffed it out and downvoted. They were right. I should make more effort to solve it.

Calvin Khor

yeah they didnt really think through the incentives well @les

leslie townes

6:02 AM

there at least ought to be a category for badges where, if you qualify, you get to choose to adopt them.

otherwise it just seems shitty to me.

Calvin Khor

@MathematicalEmergency tough luck

Mathematical Emergency

Have you seen papers written from an Indian institution that say "We've adopted these codes to handle this algorithm." I think they meant adapt, but adopt sounds so funny. I think they should not spell check. Ever!

Calvin Khor

@mathem is down to 22.5% now. @les is 1.4% unsung lol

leslie townes

huh. is that a standard usage? the thing about taking issue with indian english is... it's a substantial portion of the world's english.

as in, significantly more than england's english.

Calvin Khor

you can adopt a procedure, not sure about objects like code

leslie townes

6:08 AM

calvin i feel like that number should be lower.

1.4. who are my haters.

Calvin Khor

its the lowest nonzero percentage possible given the number of accepted answers you have

i.e. you have 1/71 accepted answers with a score of 0

leslie townes

ok. i'm not happy, but i will accept this.

Calvin Khor

SEDE will have it at 0 next week

unless u screw it up by posting a boring answer that someone accepts

leslie townes

this will interfere with my planned campaign of shitposting, but maybe i'll hold off on that.

for one more week

Calvin Khor

its gonna go down so badly when someone comments "hey why is this true" and then I reply "ive edited to address this", and my edit is "The proof of this is in Exercise 5. [...] Exercise 5: literally their comment"

polite proofs

6:40 AM

math.stackexchange.com/questions/4487906/… Could someone give me a hint please?

I feel as though I have to use the given integral inequality, but I'm not sure how

Calvin Khor

whats your definition of integrable?

regulated? riemann?

polite proofs

We say $f$ is integrable on $[a,b]$ if for every $\varepsilon > 0$, there exists a partition $P$ such that $U(f,P) - L(f,P) < \varepsilon$

Where $U(f,P) = \sum_{i=1}^n \sup f(x)_{[t_{i-1},t_i]}(t_i - t_{i-1})$

Assuming $P = \{ t_0,t_1,\dots,t_n\}$ with $a = t_0 < t_1 < \dots < t_n = b$

Calvin Khor

ok

polite proofs

And similarly for the lower sum $L(f,P)$

Actually we could have two partitions $U(f,P_1) - L(f,P_2) < \varepsilon$

But it's a quick consequence that this would imply $P$ containing both partitions would give us what I said

Calvin Khor

I feel like you're not using that you have step functions, whose integral is written explicitly as a finite sum

polite proofs

6:49 AM

Well, it's kind of hard to use them lol

It's easy to show that if $f$ is integrable, then there exist $s_1$ and $s_2$ which satisfy that, and define the $s_i$

Calvin Khor

you can probably, fix epsilon, this gives you s1 and s2, which come equipped with their own partitions whose L and U sums are exactly equal to the integral

and then argue using a partition that refines both these partitions

polite proofs

I tried to do that I suppose? But it is a lot harder to do than it seems

I'm just not convinced by my own argument

Calvin Khor

on each part say [a,b] of this partition you should have s2 > f > s1 so the integral on [a,b] of s2 will be larger than any U(f,[a,b]) which is larger than the integral of s1 on [a,b]

you probably want to squeeze U ≥ L ≥ ∫s1 in there

then you sum over all the pieces and essentially get the result i think?

i havent actually tried

polite proofs

Okay I'll try again, thank you Calvin Khor

Calvin Khor

gl!

Gwyn

7:47 AM

Is it correct to say that $A \subseteq B \implies int(A)\subseteq int(B)$ because of the following reasoning: it is always true that $int(A) \subseteq A$, by hypothesis $A \subseteq B$ and so $int(A) \subseteq B$. Moreover, for any set $X$ I know that $int(X)$ is the "biggest" open set contained in $X$ and so, since $int(A)$ is open and I shown that $int(A) \subseteq B$, it must be $int(A) \subseteq int(B)$ exactly because $int(B)$ is the "biggest" open set contained in $B$.

Koro

8:06 AM

It looks correct to me.

Gwyn

10:01 AM

Thank you Koro!

polite proofs

10:33 AM

FYI I finished my proof a while ago, would appreciate if someone could have a look: math.stackexchange.com/questions/4487906/…

Farhad Rouhbakhsh

11:11 AM

0

Q: Is this question even correct?

Suppose that $X$ is a topological space which is path-connected. Let $x_0∈ X$. Show that the function $ t_{x_0} : {X} \to {X×X}$ where $t_{x_0}(x)=(x,x_0)$ is an embedding and image of $X$ under $t_{x_0}$ is closed in ${X×X}$. I know it is trivial that $X$ and its image are homeomorphic. My ques...

general-topology path-connected

need urgent help. I'd be happy if someone could answer it asap.

Thorgott

11:34 AM

why so pressed for time

Mad Spaces

12:57 PM

How does an element of the exterior Algebra look like? "not the kth exterior algebra, rather the exterior one that is defined to be a direct sum of all others" $\alpha \in \Lambda \: V $

shintuku

jesus, the news these days

Thorgott

1:16 PM

it's a sum of elements from the various $k$-th exterior algebras

Xander Henderson

@shintuku Don't pay attention to the news. It will only make you depressed.

shintuku

my field of study forces me to keep a close watch on what's going on in the world

thankfully i get revitalized by bad news

it is the Spirit of Humanity developping itself, awkwardly and violently, calling upon Mankind to participate in it to improve its self-actualization

Koro

1:44 PM

Suppose that $(a_n)$ is defined recursively as follows: $0<a_1<1, a_{n+1}=a_n(1-a_n)$, then it is to be shown that $\lim na_n=1$.

I proceeded as follows: I claim that $a_n\in (0,1)$ for all $n\in \mathbb N$. The assumption is true for m=1. Assuming it to be true for m=n, $a_{m+1}\le (\frac {a_m+1-a_m}{2})^2=\frac 14<1$. This proves the claim. $\frac{a_{n+1}}{a_n}=1-a_n<1\implies (a_n)$ is a decreasing sequence. Since the sequence is bounded from below, it must converge to some $L$ such that $0\le L\lt a_1$

Now, let $b_n:=na_n$. The ratio $\frac {b_{n+1}}{b_n}=\frac {n+1}n (1-a_n)\to 1-L<1$

whence, it follows that (using ratio test) $\lim b_n=0$

But the statement asks to prove $b_n\to 1$.

What step went wrong above?

Ah, I see. L=0 is problematic.

Infact, L=0

Thorgott

2:08 PM

indeed, $L=0$

Prithu Biswas

2:34 PM

$\sup\{L(f,P)\} = \inf\{U(f,P)\} \\$
$\text{This means that for each}\ \epsilon > 0 \ \text{there are partitions} \ P', P'' \ \text{with}
\\$
$U(f,P'') - L(f,P') < \epsilon$

I do not understand this part of a proof from Spivak Calculus. Can someone help me with this?

Xander Henderson

3:01 PM

@PrithuBiswas Where, precisely, are you confused?

Prithu Biswas

This part:

$\text{This means that for each}\ \epsilon > 0 \ \text{there are partitions} \ P', P'' \ \text{with}
\\$
$U(f,P'') - L(f,P') < \epsilon$

@XanderHenderson I don't know how they got P' and P''.

shintuku

what page

@PrithuBiswas

Xander Henderson

If $\inf A = c$, then for any $\varepsilon > 0$, you can find an element $x$ of $A$ such that $x < c + \varepsilon$, yes?

That is, while there may not be an element of the set which achieves the infimum (i.e. there may not be a minimal element), there are elements of the set which are arbitrarily close to the infimum.

Prithu Biswas

@shintuku page number 239.

Xander Henderson

Do you see where I am heading with this argument?

Prithu Biswas

3:10 PM

@XanderHenderson What if A = {1,2,3}?

Xander Henderson

Choose $x = 1$.

(That set has a minimum.)

polite proofs

Can I try to answer, Xander?

I'm currently reading Spivak as well

Xander Henderson

For any $\varepsilon > 0$, we have $1 < \inf\{1,2,3\} + \varepsilon$, yes?

@politeproofs Sure.

Prithu Biswas

@XanderHenderson Yup that seems correct.

@XanderHenderson Is the proof of that lemma something like "if it weren't true, then we can construct a new lower bound which is greater than c."?

Xander Henderson

Something like that.

If there is some $\varepsilon > 0$ such that $x \ge c + \varepsilon$ for all $x \in A$, then $c + \varepsilon$ is a lower bound of $A$, and so $c$ cannot be the greatest lower bound (since $c + \varepsilon > c$ and $c+\varepsilon$ is a lower bound).

Prithu Biswas

3:22 PM

@XanderHenderson Exactly :D

So where shall we proceed with this argument?

Xander Henderson

Okay, so what is the next step?

Prithu Biswas

Oh okay I will try to guess.

Prithu Biswas

3:40 PM

Let $\sup\{L(f,P)\} = c$ and $\inf\{U(f,P)\} = d$
Given $\epsilon > 0$
we will find $L(f,P')$ such that $L(f,P') > c - \epsilon/2$
we will find $U(f,P'')$ such that $U(f,P'') < d + \epsilon/2$
or $U(f,P'') < c + \epsilon/2$ [Because c = d]

So Now we have:
$U(f,P'')-L(f,P')$
$< c + \epsilon/2 - (c - \epsilon/2)$
$= c + \epsilon/2 - c + \epsilon/2$
$= \epsilon$

@XanderHenderson Is this a correct reasoning?

Xander Henderson

Sure. But you know that the sup and inf are the same, so why use two different letters?

Also, why "we will find" and not "there is"?

Prithu Biswas

@XanderHenderson Yea, I could have used the same letters at the start.

We could also write "there is" or "There exists".

Xander Henderson

And "There is a partition $P'$ such that $L(f,P')$..."; it is about specifying the thing that you have control to choose.

But I am being pedantic.

Prithu Biswas

Oh now I understand why I wrote "we will find". Because you wrote "you can find". So my brain caught on that phrase =P

@XanderHenderson That is a lot more accurate.

Anyway, Thank you Xander for solving my dilemma =)

Xander Henderson

No problem.

Koro

3:53 PM

$0<a_1<\pi$, $a_{n+1}=\sin a_n$, then $\lim \sqrt n a_n=\sqrt 3$

robjohn

@Koro this answer might be of use here.

Koro

Book solution: $\color{blue}{(\frac 1{a_{n+1}})^2-(\frac 1{a_n})^2=1/3}$ so $\frac 1{na_n^2}=1/3$ whence $\lim \sqrt{n} a_n=\sqrt 3$. I understand this. But this is very tricky. The blue colored part is not 'guessable' easily. Is there any systematic way?

robjohn

@Koro did you look at my answer? It uses that $\lim\limits_{n\to\infty}\frac{a_n-a_{n+1}}{a_n^3}=\frac16$. That looks like $\frac{\mathrm{d}a}{a^3}=-\frac16$ which hints that $\frac1{a^2}\sim\frac n3$.

That is why it then looks at $\frac1{a_n^2}-\frac1{a_{n+1}^2}$

Koro

@robjohn looking

Instead of $(3)$ in your answer, the book considers the limit $\lim_{x\to 0} \frac{x^2-\sin^2x}{x^2\sin^2x}=1/3$

so the trick is to consider such limit.

robjohn

4:12 PM

@Koro Since $\frac{\sin(x)}x\to1$, they are the same

@Koro that limit is actually in $\text{(6a)}$

hansika

6:13 PM

Hi! Can anyone please help me with this:

>If $f:(0,∞)→(0,∞)$ satisfy $f(xf(y))=x^2y^n$, $(n\in R)$ then
Number of solutions of $2f(x) = e^x$ is___

robjohn

9:21 PM

@hansika can you compute $n$?

This looks like an exam or homework question

Ted Shifrin

9:51 PM

Or competition-style question ….

user4539917

10:03 PM

it could be self-made

jay

10:18 PM

how do I show that hamiltonian dynamics i.e $$\frac{dq}{dt}(t,x,v)=p(t,x,v)$$ $$\frac{dp}{dt}(t,x,v)=\nabla V(q(t,x,v)) $$ $$ q(0,x,v)=x,p(0,x,v)=v $$ are reversible,

in the sense that $q^{-1}(t,x,v)=q(-t,x,v)$ and $p^{-1}(-t,x,v)$

or is this even true?

robjohn

10:56 PM

@user4539917 I really doubt that. The whole thing revolves around $e\lt2\sqrt2$ and that seems too tight to leave to chance.

robjohn

11:55 PM

0

Q: Functional equation $f(xf(y)) = x^2y^a$

If $f:(0, \infty)\to(0, \infty)$ is an into function satisfying $f(xf(y)) = x^2y^a, (a\in\mathbb R)$, then find the value of $a$ and the number of solutions of $2f(x) = e^x$. My approach: $f(x(f(y))) = x^2 y^a$ $y=x$ $f(x(f(x))) = x^2 x^a$ $f(xy) = x^2 x^a$, (as $f(x) = y$) $y = 1$ $f(x) = x^a ...

functional-equations

Transcript for

$Mathematics$ Mathematics