00:00 - 19:0019:00 - 00:00

12:32 AM
Oh, this is fun
So $\det A=\sum_{\pi\in S_n}{\rm sign}(\pi)a_\pi$ where $a_\pi=\prod_i a_{i,\pi(i)}$
If $A$ is skew-symmetric, so $A^T=-A$, then $\det A=\sum_{pi\in ECS_n}{\rm sign}(\pi)a_\pi$ where $ECS_n$ is the set of permutations having only cycles of even length

yes!! I have 21 monthly listeners :)

@geocalc33 you get paid for these?

@robjohn yes

cool!

not much cause I only have 21 but it's a start

12:38 AM
How do I get the mathjax stuff to render correctly?

@Bob look at the top right of the webpage. There's a link tinyurl.com/cfqcvpc

@geocalc33 better than a kick in the head

My goal is to ramp up to making 1 complete song everyday
that way if I want to go on vacation for 2 weeks I can have backup songs I can still release

1:10 AM
@peek-a-boo thanks but it does not seem to be working
This is a test $e^$
$e^3$

wait
it is now working

congrats m8

$e^3$
no it is not

wat

1:15 AM
all i sees is e's and threes

Let me try again
$\dfrac{3}{4}$

e^3=(mc^2)^3
e=mc^2

I thought you put a dollar sign around a mathjax expression
and it renders as a math expression

You’re right, Bob. Geocalc is a turkey.

I think the problem is I am missing something

1:18 AM
You have to install robjohn’s link as a bookmark and click on it whilst in the room.

@TedShifrin what do you mean?

geo: bob's testing mathjax, so non-mathjaxxed stuff might be confusing in context.

Figure it out, smart one.

only a caveperson would fail to typeset their mathematics within the dollar signs.

I wasn't trying to confuse him

1:19 AM
You don’t even rise to the level of caveperson, leslie.
WTF were you trying to do?

it is not working for me

@Bob fyi there isnt a need to type new maths to test it. If it works, it renders all mathjax on the screen

Are you on a desktop computer or laptop?

by the way, I have several friends, one of which was a math teacher, who claim to be good in math but none of them know LaTex

so you can scroll back, hit the bookmark, and it should work

1:21 AM
I am a on desktop

They are old.
I learned it in the late 80s.

I am 60

Math teacher at what level ?

high school

Yeah, not surprising at all.

1:23 AM
i have no idea what this (great) arxiv preprint was typed in, but it aint latex

actually I knew somebody who was going for a PhD in math and she did not know LaTex either

When?

2 years ago, maybe more

Maybe more? She’s a fossil. Everyone learns it these days … most undergrads, even.

with three master's degrees
including CS

1:25 AM
She is a serious outlier.

okay

also it could be that she is answering a different question than you think. maybe she can make pdfs, but doesn't know how latex works

It’s been mainstream for 30+ years.

she claimed, a while back, she had never used LaTex / Tex.
She is a big fan of Microsoft Word

Who typed her research papers?

1:28 AM
She did
when I knew her, she was just begining the program

Oh, yuck. OK, Word has LateX built into it as an equation editor.

@CalvinKhor I take it back, i looked in the PDF metadata and found that it was made with MS word 2016

Personally, I want Word burned at every stake.

and she would do her homework with pencil and pen
pencil and paper

virtually everyone does homework with pencil and pen and paper

1:29 AM
and with that

that doesnt really mean someone knows or not latex

I should be going
nice chatting

lol k
cya

bye

But many type it up in Latex. Even 10 years ago, before I retired.

1:30 AM
for homework?

Yes

i typed my homework once. I didnt bother checking the output and all my matrices were wrong. lowest score ever lol

Some faculty require it.

indeed! was easy on the marker for sure
if not easy on their heart
if they still had a heart

Of course, I had to correct bad LaTeX as part of grading (I didn’t take off points). Just taught them that on top of math.

1:34 AM
the way we learned latex in my uni is we had one typed assessed work
after that everyone knew how to at least crawl
some 20 pages i think it was

the idea of using word's equation editor for a math paper makes me want to jump off a cliff.
my wife does some stat stuff in word because it's what her colleagues use.

I had a few students over the years who turned in their homework in French. Then I had to correct some of their French, too.

lmao

@leslietownes But math people did that in the 80s.

jumped off cliffs?
today we have pills for that

1:36 AM
Probably.

i was an outlier when i typed my homework as an undergrad in the 1998-2001 time frame. but a number of my fellow students knew latex, they just did not type their homework. in grad school, everyone was all latex, all the time.

Yeah, Bob’s friend must be a shunned person.

i submitted an assignment in law school in latex and got back the remark 'can you do this in word? i need to leave comments on it.' rather than send them a screenshot on how to comment on a PDF, i switched over. i still hate it.

i promise this is good maths

I detest Word.

1:37 AM
i had to typeset a number of equations in something for work a while ago, and it was a nightmare. although word's equation editor does support a subset of latex.

but holy heck its painful

Ugly as hell.
Bad spacing is the least of it.

the most obvious yea

it's also a pain to edit. word is OK at converting basic latex to its own format but there's no keeping it in latex as you edit. as far as i know.

new word has latex

1:41 AM
maybe someone's written a plugin that makes it easier, except anybody with a brain would not be using word, let alone writing plugins to make word less of the hellhole that it is.

@leslietownes right right

shin: it's helpful, but it's a limited subset of latex. and it seems to want to autoformat everything the minute you put it in.
i'm grateful for the minimal latex support. it did allow me to write a few simple equations very easily.

has anyone gotten certified for word or whatever, i recall that being offered at uni. do they actually teach you anything

I’m maximally uncertified.
But certifiable.

I think being carted in excel is probably more useful

1:43 AM
certifiably uncertified

historical person whose name's pronunciation starts with 'h', except homer?

hamel

hannibal, got it thanks geocalc

im gonna quess an excel cert is probably just sql for people who cant sql

who is hamel though

1:44 AM
of hamel basis fame.

hera?

ah i see

nobody went with adolf. i guess that's too easy.
is it a crossword puzzle?

He was notoriously uncountable.

1:45 AM
hamlet?

leslie: no i'm updating some memory palaces
needed something for the russian letter 'x'

@CalvinKhor woah good one too

Helen

$\overset {\Huge 🕶}\smile$

1:47 AM
Helen of Troy, good good

Helen

my daughter is vegging out in front of the TV for the first time in months. she's watching a documentary about polar bears. she began yelling at the screen when they attacked some seals.

keller

A seal sympathizer?

no, she's definitely team polar bear.

1:49 AM
Bully.

i maybe should have vetted this documentary a little better. sometimes nature documentaries show you too much.

saw the last moments of some roadkill ytd. p painful
i'll spare the details

I’ll just visualize crushed cockroaches.

how about someone's name that starts with the sound 'oo' in 'who'

Oona
chaplin

1:52 AM
good one.
uri geller
or does he say yuri

Uzbekistan
idk people are named after countries right

@leslietownes according to wikipedia it's oo, you also get a point
oona is good, evocative because of GoT

@leslietownes yes
@shintuku interesting.

ooiler?
uzman

we'll go with oona, thanks geocalc

1:58 AM
uzumaki

oh wow, that one is also evocative
@CalvinKhor thanks!

12 mins ago, by Calvin Khor
$\overset {\Huge 🕶}\smile$
how are you learnin russian?

whats up gamers

do they teach you cyka blat

have you learned any foreign language before?

2:01 AM
mostly reading, a lot of time on discord

@EthakkaappamwithChai 有

I am guessing that is chineese

Ah
I am not sure if Chineese has it, but German has this thing called Case system which ultimately causes like 4 or 6 words for the same idea of "the". I've heard Russian has much more than that which adds to the complexity

2:04 AM
yeah there are two additional cases

Bruh
"Definite and indefinite articles (corresponding to 'the', 'a', 'an' in English) do not exist in the Russian language."
damn

yeah the nouns are declined instead of the articles

that explains why russians dont speak with a or an in english

o
meanwhile in german
while still having definite and indef articles

imagine that moscow becomes moskvye, moskvyoo, moskvey depending on where it is in the sentence
odd thing at first

2:07 AM
Bruh

in chinese theres specific words that are paired with specific nouns for counting. so "dog" is "gou狗" and "one" is “yi一" but "one dog" is "yi zhi gou 一只狗”. One book is "yi ben shu 一本书“. Its like the "bowl of" in "one bowl of noodles", but applies to basically all nouns

last challenge, a person's name or occupation that starts with 'ts', except tsar

tsaikovsky?

it's more tcha

2:10 AM
tsai ing wen
@shintuku yea i thought it looked off

@CalvinKhor woah good one
very cool response

11 mins ago, by Calvin Khor
12 mins ago, by Calvin Khor
$\overset {\Huge 🕶}\smile$

2:40 AM
@Koro hollo!
@Kero hello!

Hi @Calvin !

Ho @Calvon !

I was trying to write two arrows as in “uniformly converges to”.

that's a special symbol
but also without the braces is overset

$f_n\overset {\to}*{\to} f$
oops

2:41 AM
rightrightarrows
$f_n\rightrightarrows f$

$f_n \rightrightarrows f$.
$\ddot \smile$
Shintuku: you’re learning Russian?
@shintuku Heisenberg

yes
oh good one

: )

hello

hi

2:55 AM
@copper.hat ello!

my family is away in Athens atm, house is very quiet here. Reading my Rockafellar for relaxation :-)

Athens where? Greece?

yup

that's a shame. ted has some stuff he needs delivered to his old stomping ground, but can't because of restraining orders.

3:14 AM
That’s quite a long trek … they wouldn’t let you go?

also restraining orders. and maybe more. interpol is involved

Interpol for copper?

yes.
blurry CCTV images of some irish guy in bell bottoms wandering around cory hall during the troubles.

Munchkin probably has several restraining orders filed against her at her advanced age.

yes. miles and ada in her class. she took miles's hat and threw it in the trash. she then did the same with ada's hat.

3:19 AM
Indubitably.

@TedShifrin someone has to pay the bills :-)
just rang my son to wish him happy birthday, he turned 19 today
well, tomorrow here
he's enjoying ouzo, beer etc since he is legal there

Growing up in a hurry.

He's over 6'3" now. His sister is envious.
She's fairly short.
Some minor differences in definitions bug me. For example, $(\liminf f)(x) = \lim_{ \epsilon \downarrow 0} \inf \{ f(y) \mid \|y-x\| \le \epsilon \}$, and others $\lim_{ \epsilon \downarrow 0} \inf \{ f(y) \mid 0<\|y-x\| <\epsilon \}$
Life is tough.

3:59 AM
@copper.hat often, limits don't include the limit point.

Yep, that always throws me a little.

and lim inf is actually a sup.
$\liminf\limits_{x\to x_0}f(x)=\sup\limits_{\epsilon\gt0}\inf\limits_{0\lt|x-x_0|\le\epsilon}f(x)$

yeah, i like that form

@Koro: I wrote a fairly detailed answer. Let me know if it is not clear.

4:29 AM
: math.stackexchange.com/questions/4484581/… "I'm a ring, but then I'm not a ring."

@robjohn thanks a lot : )
I understand the steps but in the end, there is $\frac {x^2}{4\pi}-n\sin \sqrt{4\pi^2 n^2+x^2}\le \frac{6x^4+x^6}{384\pi^3n^2}$.
Shouldn't there have been a mod |.| on the quantity on left to conclude uniform convergence on [0,a]?
For, if there is mod then RHS term (x) can be replaced by $\le$ term (a).

@Koro The quantity is $\ge0$ as shown earlier, but I can express it more definitely.

Okay. Yes.
I understood.
$\sin ()=\sin (()-2n\pi)=\sin \frac{x^2}{(\sqrt{4\pi^2n^2+x^2})+2n\pi}$
now, we can use $\sin t\le t$

4:48 AM
@Koro I have updated the answer
it now shows explicitly that the convergence is from below

5:27 AM
@robjohn yes I have seen it.
I have one question: Is proof for convergence on [0,a], a>0 correct?
(after taking into account your yesterday's comment to the post).
I tried to show that o(1/n) doesn't depend upon x.
(after taking into account your yesterday's comment to the post).
I think that my solution is also correct.

3 hours later…
8:26 AM
is it true that $1 + x + x^2 - y + y^2 + xy > 0$ $\forall x,y$
I saw it in DESMOS and it kind of covers the all of $xy$ plane
$x,y \in \Bbb{R}$

8:46 AM
Expand $\frac{(x+y)^2+(x+1)^2+(y-1)^2}2$
So it seems to be $\ge0$ $(x=-1,y=1)$

Nice idea! thank you1

9:25 AM
@robjohn yes i think it can be generalized too
In a $2 \times 2$ matrix
$A = [x,1;y,0]$ then det(A) $= 1 + x + x^2 + y^2 -y + xy > 0$
a $n \times n$ how we can show indeed that det$(A) > 0$

10:00 AM
@BAYMAX what does $A=\left[-\Delta\middle|\frac{I}{0\dots0}\right]$ mean?

Oh ok, $A$ is a $n \times n$ matrix with the first column as $-\Delta$

If by $A=[x,1;y,0]$ you mean $A=\begin{bmatrix}x&1\\y&0\end{bmatrix}$, then $\det(A)=-y$

oh yes, am sorry for that notation
@robjohn That's right
@robjohn I think it must be true for $n \times n$ matrix $A$ of the form $A = \left[ -\Delta \middle | \frac{I}{0 \ldots 0}\right]$
as I have tried but could not find a contradiction

$\det\left(\left[-\Delta\middle|\frac{I}{0\dots0}\right]\right)$ is $(-1)^n$ times the bottom element of $\Delta$.

Oh ok, but how you concluded it? quick!!
suppose we consider $3 \times 3$ matrix
Say $B = \begin{bmatrix} a,1,0\\ b,0,1\\ c,0,0\end{bmatrix}$
then $det(B) = c$

10:12 AM
Every time you swap to rows, the determinant is negated. to the the bottom row to the top takes $n-1$ row swaps. The negative on the $\Delta$ makes that $(-1)^n$. Then, expanding on the top row you get the bottom element of $\Delta$.
@BAYMAX Yes, your $\Delta=\begin{bmatrix}-a\\-b\\-c\end{bmatrix}$ and so my statement is true. $(-1)^3(-c)=c$

oh nice
yes suppose $c$ is the last element of $\Delta$

then the determinant would be $-c$

then $det(A) = (-1)^n c$

$\det\left(\left[\Delta\middle|\frac{I}{0\dots0}\right]\right)$ is $(-1)^{n-1}$ times the bottom element of $\Delta$.

yes for a general $n \times n$ matrix
cool!
So $\det((\left(\left[\Delta\middle|\frac{I}{0\dots0}\right]\right))^2) = (-1)^{2(n-1)}$ times the bottom element^2
yes but this mightnot help as we are thinking of $det(I+A+A^2)$

10:20 AM
$(-1)^{2(n-1)}=1$

Right!

@BAYMAX $\det(A+B)\ne\det(A)+\det(B)$

yes therefore I thought det$(A^2)$ mightnot help
seems like a series though $I+A+A^2$

If $A-I$ is invertible ($1$ is not an eigenvalue of $A$), then $I+A+A^2=\left(A^3-I\right)(A-I)^{-1}$

Cool analogy from geometric series!
so, $\det(I+A+A^2) = \det(A^3 - I) \det((A-I)^{-1})$
$\det(A^3 - I) = c^3-1$
where suppose $c$ is the last element of $\Delta$
$\det(A-I)^{-1} = \frac{1}{c-1}$
so, $\det(I+A+A^2) = 1+c+c^2$
but can we conclude if $\det(I+A+A^2) > 0$ for any real entries in $\Delta$
Is that correct? @robjohn

10:48 AM
@BAYMAX no
29 mins ago, by robjohn
@BAYMAX $\det(A+B)\ne\det(A)+\det(B)$

Oh I thought it followed from $\det(I+A+A^2) = \det(A^3 - I) \det((A-I)^{-1})$
I think $\det(A^3 - I) = c^3-1$ is wrong

@BAYMAX yes, that is the comment I linked to

Oh okie
Do you also have an intuition that $\det(I+A+A^2) >0$ ?

3 hours later…
2:08 PM
@BAYMAX Do you know the eigenvalues of $A$? If $\lambda$ is an eigenvalue of $A$, can you find an eigenvalue of $I+A+A^2$?

2:25 PM
Good morning @robjohn

2 hours later…
4:28 PM
@geocalc33 Hi there. I am just back from walking the dog.

@AkivaWeinberger are you proving that $\det(A)=0$ for skew-symmetric matrices of odd dimension?

4:51 PM
Let $S_{\infty}$ be the group of finitely supported permutations of $\Bbb{N}$. I am trying to argue that $S_{\infty}$ is not residually finite. I feel like a stronger claim is true, namely that any homomorphism into a finite group is trivial, but I don't know how to argue this. I could use some hints.

i strongly believe the sign is a non-trivial homomorphism $S_{\infty}\rightarrow\mathbb{Z}/2\mathbb{Z}$

Ah yes and the kernel is $A_{\infty}$, the finitary alternating group.

I hate to agree with @Thor, but ...

I missed this! :P

Okay, that's good to know...so the stronger claim isn't true...Hmm...back to the drawing board...

4:58 PM
@Thorgott Just remember that next time you disappear for months and months. :P

5:12 PM
I shall

Okay, maybe the "bad" element (i.e., the element which shows non-residual finiteness) is some element outside of $A_{\infty}$.

also, I have an inkling suspicion that $\mathrm{sgn}\colon S_{\infty}\rightarrow\mathbb{Z}/2\mathbb{Z}$ might actually be the profinite completion of $S_{\infty}$, but I don't have an argument

Oh, I see. And that map definitely can't be injective, so it can't be residually finite.

5:36 PM
If $f$ is a $C^{\infty}$ function and $U$ is a open subset of $R^n$, what does $C^{\infty}(U)$ mean? More generally I've seen $C^{\infty}(C^n)$, $C^{\infty}(R^n)$ etc. Is it the set of all $C^{\infty}$ functions whose domain is $U$?

@ShikiRyougi Yes.

Thank you very much! :))

@Thorgott I think I got it. Because $A_{\infty}$ is an infinite simple group, any finite-index subgroup of $S_{\infty}$ must actually contain $A_{\infty}$. Indeed, if $N$ is a normal finite-index subgroup of $S_{\infty}$, then $|A_{\infty} : A_{\infty} \cap N| \le |S_{\infty} : N| < N$. Infinite simple groups do not have proper finite-index subgroups, so $A_{\infty} = A_{\infty} \cap N \le N$.
Hence, the intersection of all finite-index normal subgroups of $S_{\infty}$ will contain $A_{\infty}$ which is certainly not trivial.
So, $S_{\infty}$ is not residually finite. This means there exists an element $g \in S_{\infty}$ which is killed by every group homomorphism from $S_{\infty}$ to a finite group...interesting...I wonder what that element looks like.

6:26 PM
nice, right
this argument does show the sign map is the profinite completion of $S_{\infty}$
so the elements getting killed by every group hom to a finite group are precisely $S_{\infty}\setminus A_{\infty}$
more precisely, two elements can be distinguished by their images in a finite group if and only if they are in different $A_{\infty}$-cosets

6:52 PM
i can define a gradient of a functional $f$ on a Hilbert space $H$ via reisz representation theorem like this right : given $f:H\to\mathbb{R}$, the Frechet derivative of $f$ is an operator $Df:H\to \mathbb{R}$, i.e $Df\in H'$, so by reisz representation we can associate to $Df$ a unique $v\in H$ such that $Df(u)=\langle u, v \rangle$. We call $v$ the gradient of $f$.
Is that generally the idea?

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