Mathematics - 2022-12-25

Gwyn

00:04

Asking Wolfram Alpha wolframalpha.com/input?i=plot+%7Cx%7C%281-y%5E2%29<%3D0 to plot the inequality $|x|(1-y^2)\le0$ it gives back only the two half planes $y \ge 1$ or $y \le -1$. But since for $x=0$ the inequality is true for any $y\in\mathbb{R}$ and for $x>0$ the inequality is equivalent to $1-y^2 \le 0$ shouldn't it be the union of the two half planes $y\ge 1$, $y \le -1$ and the vertical line $x=0$?

Ted Shifrin

You're never going to see a single line on Wolfram. But, yes, it's the union of the three sets.

Goku カカロット

@TedShifrin Ted, I think I found something you might like. Turkish Kebab filled with water. Not dry, as you liked

Gwyn

Thank you Ted:) moreover, can I say that $A=\{(x,y)\in\mathbb{R}^2 \text{such that} \ |x|(1-y^2) \le 0\}$ is closed because it is finite union of those three closed sets? And it is not convex because $(0,0), (1,1) \in A$ but the segment of endpoints $(0,0)$ and $(1,1)$ does not belong to $A$ (for instance, because the segment contains $(1/2,1/2)$ but $(1/2,1/2) \notin A$)?

Ted Shifrin

@Gokuカカロット You don't seriously mean "filled with water." You mean not overcooked and succulent? :)

@Gwyn It is closed, yes. No disconnected set (this one has three connected pieces) can ever be convex. Maybe they want you to notice that each of the three pieces is convex, but the union can never be.

Goku カカロット

@TedShifrin Being overcooked is probably better than being undercooked isn't it?

Ted Shifrin

00:14

Not to me. I like most meat rare to medium rare. Even pork needn't be murdered to avoid trichinosis (and freezing the meat for a few weeks kills any worms that might have been there).

Lamb, in particular, is fabulous medium rare and like sawdust when cooked well-done.

The taste is totally different.

Gwyn

@Ted Unfortunately I have to justify it in a "more elementary" way, because I have not studied topology yet and we have been given only the definition that for any pair of points in the set the whole segment joining the points must be contained in the set; thanks again, by the way)

Ted Shifrin

lecture over

Goku カカロット

@TedShifrin medium rare is....rare

Ted Shifrin

@Gwyn: Yes, that's fine. Just go from one of the three pieces to another and ... no convexity. I was just stating a general principle for purposes of your intuition.

Gwyn

that was indeed useful for intuition) thanks again and happy Chritsmas!

Ted Shifrin

00:21

You too.

BAYMAX

01:14

2

Q: A pen-and-paper proof for a matrix implication.

Suppose $A = \begin{bmatrix} x & 1\\ y & 0\end{bmatrix}, B = \begin{bmatrix} z & 1\\ w & 0\end{bmatrix}$, for $x,y,z,w \in \Bbb{R}$. I have observed by considering many examples of $x,y,z,w$ that: If all the eigen values of $A^2B$ and $AB^2$ are less than one in absolute value $\implies$ $\det(A...

Akiva Weinberger

05:05

So

1

Q: What functions from $\Bbb N\to\Bbb N$ can be made from $+$, $-$, $\times$, $\div$, exponentiation, and $\lfloor\cdot\rfloor$?

What functions from $\Bbb N\to\Bbb N$ can be made from $+$, $-$, $\times$, $\div$, exponentiation, and $\lfloor\cdot\rfloor$? Call this class of functions $\mathcal Flex$ (for "floor and exponentiation"). The $\rm mod$ function $a\operatorname{mod}b$ can be defined by $a-b\lfloor a/b\rfloor$. In ...

exponential-function diophantine-equations recursion computability ceiling-and-floor-functions

Joe Shmo

06:34

@TedShifrin maybe question for tomorrow: $\exp(\frac {z^2 - 1} z)$ is analytic, but I am getting that via a left method $\exp(\frac {z^2 - 1} z) = \exp(z) \exp(- \frac 1 z)$, and then using the formula for the coefficients of a product of two laurent series (can be deduced from Cauchy's integral formula) gives you that the negatively indexed terms vanish. This is not surprising, since the numerator approaches $0$ faster than the denominator. Is there a better way to see this?

oops, this should be surprising, because the numerator doesn't approach $0$ at all, as $z \to 0$, and $\exp (\frac 1 z)$ has an essential singularity at $z = 0$

But I can't do the same thing with $\exp \left ( \frac 1 {z - \frac 1 z} \right)$

I could write $\frac 1 {z - \frac 1 z}$ in a partial fraction decomposition, but that's only valid for $|z-1| < 2$ (say when I expand in $(z-1)$)

Joe Shmo

07:17

^^actually, the negatively indexed coefficients don't vanish (unless I am making another mistake), and there's infinitely many of them; so that $\exp \left ( z - z^{-1} \right )$ has an essential singularity at $z=0$, commensurate with intuition.

also, yeah, $|z-1|<2$ for partial fraction decomposition shouldn't be a problem.. ^^

DIRAC1930

09:32

If I have a projective unitary representation with $U(G_1) U(G_2) = \pm U(G_1 G_2)$, it this a statement that I can find a representation with elements $F$ such that $\{ F(G_1), \imath F(G_1)\} \{ F(G_2), \imath F(G_2)\} = \{F(G_1 G_2), - F(G_1 G_2)\}$

The $U$'s are elements of $U(2)$ which from a projective unitary representation of $O(3)$

shintuku

12:47

there are no pure rhymes for fifth

Thorgott

13:47

twentyfifth

shintuku

ok, there are no pure rhymes for *fifth, where * ranges over the natural numbers

nvm that's a dumb statement

but i won't delete it because it's the 25th

hell is having no syntax for a semantic compound

Koro

14:03

I have one confusion: Let $\mathbb Q=\{q_1,q_2,\cdots\}$. $\{q_i\}$ is closed in $\mathbb R$ so $\mathbb R-\{q_i\}$ is open in $\mathbb R$. Arbitrary union of open sets is open so $\cup_i \mathbb R-\{q_i\}= \mathbb R-\mathbb Q$ is open, which is false. So what did I do wrong here?

shintuku

hi @Koro

Koro

Hi @shintuku!! How are you?

shintuku

maybe it is this. let $i$ be given. then $\cup_i \mathbb{R} - \{q_i\}$ is open

but you can give no $i$ which makes $\cup_i \{q_i\} = \mathbb{Q}$

i'm good, and you?

Koro

By that notation, I mean $\cup_{i\in \mathbb N} \{q_i\}=\mathbb Q$.

I'm good. Thanks :)

@shintuku we are taking union over all i running through natural numbers.

shintuku

hm, i'm still not sure that equation is true

can you prove it?

both the naturals and the rationals are countable, yes

but this just means there exists a bijective function between the two, i'm not sure it implies the above mentioned equation

Koro

14:19

hmm, you are right. The union is not correct.

We can only say $\mathbb R-\mathbb Q\subset \cup_i \mathbb R-\{q_i\}$.

RHS is basically $\mathbb R$.

:-)

shintuku

here is a proof: suppose $x_i \in \mathbb{Z}$, let $q_i = x_i/x_i$. then clearly the union is false

i think we should note, to even properly state that union, we need $\mathbb{Q}$ as a set of equivalence classes of integer pairs

we need a couple of sets to do this

hm there is probably a simpler way to think of this

Yai0Phah

14:40

It seems to me that $\mathbb R\setminus\mathbb Q=\bigcap_{q\in\mathbb Q}\mathbb R\setminus\{q\}$, not the union.

shintuku

it has the same problem, you cannot get $\mathbb{Q}$ by iterating $\{q\}$

in fact: there is a theorem that states that the union of finite sets is finite @Koro

this is a simpler way to think of it

!!: it seems that a finite union is qualitatively different than an infinite union

Koro

@shintuku no no, that's wrong.

@shintuku $Q=\cup_{i\in \mathbb Q}\{q_i\}$ is true indeed. The other union was problematic.

@Yai0Phah that's true. :)

For example: $\cup_{n\in \mathbb N}\{n\}= \mathbb N$ is union of finite sets and is not finite. Perhaps, you meant finite union of finite sets.

shintuku

it is not obvious to me that $\bigcup_{n\in \mathbb{N}}\{n\} = \mathbb{N}$ :/

Koro

Anyways, I managed to show that an uncountable set with cofinite topology is separable. :)

Yai0Phah

14:56

What do you mean by separable?

It is certainly not Hausdorff.

Koro

@shintuku Take any $n$ in $\mathbb N$. Then $n\in \cup_{i\in \mathbb N} \{i\}$.

@Yai0Phah By a space X being separable, I mean X has a countable dense subset.

shintuku

15:15

oh, I got where I was going wrong

thanks Koro

Koro

:-)

shintuku

16:38

@Koro I think our initial thought that $\bigcup_{i \in \mathbb{N}}\{q_i\} = \mathbb{Q}$ is incorrect was wrong

consider the following proof

Recall that $\bigcup_{i \in \mathbb{\mathbb{I}}}F(i) := \bigcup\{F(i):i\in\mathbb{I}\}=\bigcup\{F(i_1), F(i_2), \cdots \}$ Since $\mathbb{Q}$ is countable, there exists at least one bijection $\mathbb{N}\to\mathbb{Q}$, let it be $f$.

Notice that $\bigcup_{n \in \mathbb{N}} \{f(n)\}=\{f(0), f(1), \cdots\}$ is a subset of $\mathbb{Q}$.

Let $x \in \mathbb{Q}$. Then, since $f$ is surjective, there is an $n^* \in \mathbb{N}$ with $f(n^*)=x$. But notice $\{f(0), \cdots, x(=f(n^*))\}$ is a subset of $\{f(0), f(1), \cdots\}$, and therefore, $x \in \bigcup_{n \in \mathbb{N}}\{f(n)\}$.

Koro

@shintuku This is correct. The proof is same as that in the example on N.

The incorrect thought was $\mathbb R-\mathbb Q=\cup_i(\mathbb R-\{q_i\})$, where $\mathbb Q=\cup_i \{q_i\}$

shintuku

alright, we remain on the path of truth

Koro

It is true for any set X that $X=\cup_{x\in X} \{x\}$, whether X is countable or uncountable.

shintuku

right, and your very first thought, that you can do $\mathbb{Q} = \bigcup_{i \in \mathbb{N}}\{q_i\}$ was right

Simd

If I have this function

w(u)=1/u for 1 <= u <= 2,
(uw(u))'=w(u-1) for u>2.

how can I compute w'(u) for a given u >= 1?

shintuku

16:50

@Simd find the derivative of w at x for an arbitrary x

Simd

@shintuku right.. but how?

any help hugely appreciated

shintuku

first do the derivative of 1/x

Simd

that I can do

shintuku

oh, nvm the second part is harder

maybe someone else can help

Simd

thanks and I hope so

shintuku

16:55

for the second function, have you tried applying the integral on both sides of the equation?

Simd

I haven't... can we maybe just solve for w'(u)?

shintuku

that implies taking the integral of both sides of the equation, simplifying to get w(u), and then applying the derivative, I think

Simd

can we get 1/u + u w'(u) = 1/(u-1) ?

shintuku

ah nevermind this is a differential equation, i don't know much about them

Simd

the suggestion is that we can get the equation that starts 1/u directly somehow

then we are done

but I am not sure how....apparently it's simple :)

@shintuku does (u w(u))' = w(u) + u w(u)' ?

shintuku

17:01

yeah, but that's still a differential equation

can't help you much with those personally, but someone else might

Simd

ok thanks

Yai0Phah

17:13

@Simd For any $n\ge2$ and any $x\in[n,n+1)$, you integrate the second equation on $[n,x]$, obtaining $xw(x)-nw(n)=\int_n^xw(u-1)du=\int_{n-1}^{x-1}w(u)du$.

Thus $w(x)=x^{-1}(nw(n)+\int_{n-1}^{x-1}w(u)du)$.

Then you can inductively on $n$ determine the function $w$ on $[1,n]$.

shintuku

hm, how do you evaluate that integral?

Yai0Phah

Inductively

That integral only depends on the value of $w$ on $[n-1,n]$, thus already known.

Simd

what do you get for u = 3?

shintuku

@Yai0Phah ahhhh, swell

Simd

the answer should be close to 1 at u=3 I believe

oh maybe I mean close to 0

Ted Shifrin

17:36

You get $uw(u)-1 = \ln(u-1)$ for $2\le u\le 3$. So $w(u) = \dfrac{1+\ln(u-1)}u$. But on the next interval, you cannot integrate explicitly.

All they want is $w'(u)$, but you still have to know $w(u)$, because $(uw(u))' = w(u) + uw'(u)$.

Where did this question come from?

copper.hat

18:15

happy holidays to all!

Gwyn

it is correct to say that $\lim_{\|(x,y)\| \to \infty} x$ does not exist because, put $f(x,y)=x$, the points $(x,0)$ are such that $\|(x,y)\| \to \infty$ when $x \to \infty$ or $x \to - \infty$ but $f(x,0) \to \infty$ when $x\to\infty$ while $f(x,0) \to -\infty$ when $x\to -\infty$ contradicting the uniqueness of limit?

leslie townes

probably yes. i would reword some of that but you have the idea. you can fiddle with the limiting value of f(x,y) depending on how (x,y) approaches "infty" and that is something that does not happen if the limit exists.

@copper.hat i see you have been appropriated by the woke mafia. we used to say Merry Christmas.

copper.hat

18:31

@leslietownes i did send merry christmas cards to all my jewish friends though...

leslie townes

well, that's a level of anti semitism that i would not aspire to, but you're irish, so we sort of imagine you as a bull in a china shop to begin with.

copper.hat

wokeness has hit ireland, was corrected by a neice & nephew for asking where someone was really from

i am happy to report that ostensibly ireland is a much more diverse place than when i was growing up

leslie townes

i hear you're a racist now, father

should we all be racist now? what's the church's position?

copper.hat

have got that a few times.

leslie townes

:D

copper.hat

18:34

i did bring & attend mass with my 94 yo aunt, with whom i just had a conversation

somethings are generational.

my aunt (who is nominally broad of mind) commented "a lot of non nationals"

what she means is folks for whom english is not a first language

slippery slope

leslie townes

my mom's priest is irish, and "non national" applies to him

we're all very suspicious

copper.hat

we had a lot of irish priests who went on the missions to africa and were much more interesting on return

i have found that modern priests fall into two categories

(i) toe the lines and (ii) people i like to have a drink with.

but that said, my social calendar, apart from family, has been remarkably barren for the last few months

was feeling a bit mopey this morning until i had a chat with my sister this morning who made me laugh

listeing to only you by the flying pickets does not help the mood

leslie townes

haha

copper.hat

my 94 yo aunt has of this moment polished off more food that the other 19 guests.

not a big person, still graceful

the entire family will be going to a place outside rome for a family wedding in march. that should be interesting.

Goku カカロット

18:51

I tend to eat a lot too

@leslietownes NORAD is tracking Santa again

copper.hat

i really believed santa existed until i was around 10

my folks would like out a sideplat with fruitcake crumbs and a few drops of whiskEy in a glass

Goku カカロット

Wait....he doesn't exist???

Ted Shifrin

19:23

Happy holidays to grinches one and all!

Tell Munchkin to have a good swim with the ducks.

This “mark all read” crap does NOT work.

Koro

@TedShifrin it works but is very annoying!

leslie townes

yes, felices fiestas to all. we are celebrating christmas this morning, i have a bunch of texts going from which i understand chanukah (which may not be the hugest deal in the grand scheme of jewish things but is at least happening now) is still going. and also happy sunday to people who don't celebrate but might not be working.

Goku カカロット

It doesn't work for me either. I hate the new inbox. Who asked for this again??

leslie townes

i try to mark all read, but it remains unread. i'm thinking this is SE's attempt at some kind of spiritual instruction.

Goku カカロット

Or a late and horrible April fools joke

Ted Shifrin

19:29

@leslietownes I need to count on my fingers, but I think last night was the last night of Chanukah.

I’m cooking rack of lamb (French) and Indian appetizer today.

leslie townes

i hear the best way of cooking lamb is to really char it. you want to see the char on all sides.

@TedShifrin munchkin woke me up very early this morning so it still feels like yesterday.

Ted Shifrin

It’s been nestling in its mustard, garlic, fresh herb coat for a day. It will be blasted in the air fryer and cooked only to medium rare.

@leslietownes Send her to cavort with the ducks.

Goku カカロット

@TedShifrin is it dried, just like the way you enjoy?

Ted Shifrin

Goku, like that you’ll never get invited!

Goku カカロット

@TedShifrin You dodged a bullet, Goku is known for devouring food like a black hole

leslie townes

19:35

we might actually go to see some geese later.

Ted Shifrin

I have cooked geese for xmas dinner before. But don’t tell them.

leslie townes

i last ate goose in christmas of 2001. it was delicious.

Goku カカロット

If geese is plural for goose then, is cheese plural for choose?

Ted Shifrin

Choose rhymes not with goose.

But meese is the plural of moose.

Goku カカロット

Okay fine, if mice is plural for mouse, then hice is plural for house

Ted Shifrin

19:42

Agreed.

Koro

@TedShifrin How long will it take usually to cook ?

1 hour at max?

leslie townes

1 hour at 750F will kill everything.

Koro

goose? I never ate that.

@leslietownes does it taste like Chicken?

Goku カカロット

Goose. Reminds me of Top Gun

Ted Shifrin

More like 16 minutes. It’s not a giant leg.

leslie townes

19:54

yes, if only because everything does.

Koro

@Gokuカカロット the Su vs F fight movie?

leslie townes

non-farm-raised non-chicken birds are often fattier and gamier than chicken. but it's all roughly the same experience.

Goku カカロット

@Koro I hate how you described that but yes, the prequel to that movie

Simd

20:20

how can I do \int _3^t (1/s) log(s-1) ds ?

for t > 3

Ted Shifrin

You can’t.

Simd

hmm

how come?

Ted Shifrin

It does not have an expression in terms of standard functions.

Simd

that's a shame

I just tried wolfram alpha and it gives me an answer it terms of Li_2

Ted Shifrin

Right.

Simd

20:27

but I can't really interpret that

Ted Shifrin

I asked you an hour ago where you got the question.

Simd

oh sorry I missed that

it's a function used in number theory

Ted Shifrin

Never seen such a thing, but that doesn’t make me think it should be an elementary function.

Simd

integral_3^t log(s - 1)/s ds = Li_2(1 - t) - Li_2(-2) + log(t - 1) log(t) - log(2) log(3) is definitely not elementary

I don't even understand when the polylogarithm is real

when is Li_2 real?

Ted Shifrin

20:55

Polylog is defined by an integral like yours.

Oliver Díaz

@leslietownes Us Jews (religious and otherwise) have to thank the catholics for preserving the history of Hanukkah (the book of Maccabees) Protestant bibles don'r include those text in their cannon. In any event happy holidays, I'll be celebrating with a meal in a Chinese restaurant and playing Dreidel with the kids.

Simd

@TedShifrin so it should always be real?

Ted Shifrin

If you put in a real value in the domain, yes.

Simd

thanks

I will always do that

Akiva Weinberger

21:19

I just put a bounty on an old (August) question of mine

5

Q: How many ways to arrange $n$ points in $(\Bbb F_q)^2$ with no three collinear?

How many ways are there to arrange $n$ points in the finite field plane $(\Bbb F_q)^2$ with no three of the points collinear? An easy upper bound is $(q^2)^n=q^{2n}$, but of course it's less than that. (Of course, if I asked the same question over $\Bbb R^2$, it would be infinite.) The collineari...

combinatorics geometry finite-fields inclusion-exclusion mobius-inversion

Interesting discussion in the comments

Joe Shmo

21:42

I'm surprised it's not a known question (with known answers)

I see a lot of that sorta stuff from the Israeli school, Terry Tao/Netz Katz type of question

leslie townes

22:10

sounds like a government agency interview question.

akiva is a cop

Ted Shifrin

Go flock with Munchkin's geese!

Goku カカロット

22:27

I wanna own 3 hice

leslie townes

22:48

ted: honnnnnnk

Ted Shifrin

23:21

Ah, much better.

@Gokuカカロット Isn't one houce more than enough?

Goku カカロット

@TedShifrin nah, one house for me, one for my wife and one for my kid. So we can all live comfortably

Ted Shifrin

Oh, of course.

And if there are four kids? Then three more hice?

Goku カカロット

@TedShifrin absolutely

Transcript for

$Mathematics$ Mathematics