Mathematics - 2017-01-23 (page 1 of 4)

TheGreatDuck

00:00

@AkivaWeinberger sorry. Just goofing around. :-)

Akiva Weinberger

Then, we define $\omega+1$ to be the order type of $\{0,1,2,\dots,\omega\}$, with $\omega$ greater than all of the other elements

And $\omega+2$ is the order type of $\{0,1,2,\dots,\omega,\omega+1\}$

(You can think of it as the orderings $\{0,1,2,\dots\}$ and $\{\omega,\omega+1\}$ stuck together; those have order types $\omega$ and $2$, hence, $\omega+2$)

And we can even continue to $\omega\cdot2=\{0,1,\dots,\omega,\omega+1,\dots\}$

and even to $\omega^2$, and $\omega^\omega$, and $\omega^{\omega^\omega}$, and …

$\omega^{\omega^{\omega^{\dots}}}$, with an infinite power tower, can even be defined. It's called $\epsilon_0$.

Because $\epsilon$ usually means a small number.

In any case, it keeps on going, each one representing another well-ordering.

$\{-1,-1/2,-1/3,\dots,0\}$ from before has order type $\omega+1$.

All of the ordinals I've mentioned so far are countable.

There's a natural ordering on the ordinals; the first uncountable ordinal is the order type of all of the countable ordinals.

The Continuum Hypothesis says that it's equinumerous with $\Bbb R$. This has been proven to be independant of ZFC.

TheGreatDuck

@AkivaWeinberger questions regarding the mathematical continuum are best asked to users on sci-fi stack exchange

...or is that the space-time continuum I'm thinking of.

Brody

00:18

@AkivaWeinberger Good lord...

Akiva Weinberger

Yeah I do not endorse this

Brody

You have a fanbase, lol

Simply Beautiful Art

Infinitely many mathematicians walk into a bar. The first says, “I’ll have a beer.” The second says, “I’ll have half a beer.” The third says, “I’ll have a quarter of a beer.” Before anyone else can speak, the barman fills up exactly two glasses of beer and serves them. “Come on, now,” he says to the group, “You guys have got to learn your limits.”

TheGreatDuck

lol

Akiva Weinberger

Three logicians walk into a bar. The bartender says, "Do you all want a beer?" They respond, "I don't know," "I don't know," "Yes."

Brody

00:25

@AkivaWeinberger What's equinumerous with $\Bbb R$?

Akiva Weinberger

The first uncountable ordinal (assuming CH)

Written $\omega_1$

TheGreatDuck

An engineer, a physicist and a mathematician are staying in a hotel. The engineer wakes up and smells smoke. He goes out into the hallway and sees a fire, so he fills a trashcan from his room with water and douses the fire. He goes back to bed.

Later, the physicist wakes up and smells smoke. He opens his door and sees a fire in the hallway. He walks down the hall to a fire hose and after calculating the flame velocity, distance, water pressure, trajectory, etc. extinguishes the fire with the minimum amount of water and energy needed.

@AkivaWeinberger I didn't

Simply Beautiful Art

@AkivaWeinberger Is there anything special about $\omega_2$?

Brody

@AkivaWeinberger nvm, I thought you meant $\omega_0$ for some reason

Simply Beautiful Art

lol @TheGreatDuck

Akiva Weinberger

00:27

It's consistent with ZFC that it's equinumerous with $\Bbb R$ instead, I think

Brody

or just $\omega$

Simply Beautiful Art

@Brody I thought $\omega=\omega_0$?

Akiva Weinberger

I think it's been proven that ZFC proves that $|\Bbb R|\le\aleph_{\omega_4}$

That's the cardinality of $\omega_{\omega_4}$.

Simply Beautiful Art

lol

That's, um... ok then

But is there anything interesting about $\omega_2$?

Brody

@SimplyBeautifulArt right

Akiva Weinberger

00:29

Wait, no, I stated it wrong

Simply Beautiful Art

RIP

Jasch1

01:03

Does Well-Ordering Theorem mean that all sets can become well-ordered or only ordered sets?

Akiva Weinberger

01:18

@Jasch1 It means all sets can be well-ordered. (It doesn't matter if we're talking about ordered sets, because we're giving it a new order anyway.)

Jasch1

ok

Akiva Weinberger

Also, you need the axiom of choice for it.

Fun fact: Without the axiom of choice, you can't prove that $P(\Bbb R)$ (the power set of the reals) can be ordered at all, much less be given a well-order.

Jasch1

Does the set of perfect squares have the same density as set of naturals?

Sorry for so many questions, I'm just looking back on a paper and I'm confused about some stuff I wrote

Brody

01:37

@Jasch1 How many perfect squares in the first 100 positive integers? How many up to 200 and so on?

Jasch1

i get that the size of the sets are the same

Brody

Sorry, you mean natural density or something else?

Jasch1

natural/asymptotic density

Paul Plummer

The cardinality of the continuum can be pushed to basically any cardinal. The only obstruction is cofinality @AkivaWeinberger

Akiva Weinberger

Oh.

I remember there being some theorem involving $\aleph_{\omega_4}$, though

Paul Plummer

01:42

yah, I think I have seen what you are talking about, but I think there are other things going on and I don't remember what it is

Brody

Do you know the density of $\Bbb N$? @Jasch1

You're not wrong.

Jasch1

Im not?

Brody

What proportion of the first $n$ positive integers are positive integers? And the limit is obvious. :)

Akiva Weinberger

Oh, OK: If $\kappa<\aleph_{\omega}$ implies that $P(\kappa)<\aleph_{\omega}$, then $2^{\aleph_{\omega}}<\aleph_{\omega_4}$

aquire

1.Let f : [0, 1] → R satisfy |f(x) − f(y)| ≤6 |x − y| for all x, y ∈ [0, 1]. Show that f is continuous and that for all ε > 0, there exists a piece-wise constant function g such that sup x∈[0,1] |f(x) − g(x)| <6 ε.
2.For all integers n > 1, let un = $$∫^1_0$$f(t) cos(nt)dt. Show that the sequence (un) converges to 0.

Akiva Weinberger

01:45

That first bit is independent of ZFC, though.

aquire

I have problem with part 2

any hints?

Akiva Weinberger

$u_n=\int_0^1f(t)\cos(nt)dt$?

And we want to show $u_n$ goes to zero?

aquire

yep

Akiva Weinberger

I assume $f$ has to satisfy all of the conditions of the first one?

aquire

yep

Paul Plummer

01:47

Ah yah. It is sort of funny you can have the continuum be $\aleph_1$, $\aleph_{\omega_1}$ but you cant have it $\aleph_\omega$

Akiva Weinberger

Hm. What's $\int_0^1\cos(nt)dt$? It might be relevant

Like, without the $f$

aquire

Brody

apply Cheb. polynomial and integrate?

aquire

$\sin(n)$ @AkivaWeinberger

Akiva Weinberger

So $\int_0^1\cos(nt)dt=\sin(n)/n$, which goes to $0$ as $n$ goes to infinity.

@aquire Off by a factor of $n$

aquire

01:50

i see

Akiva Weinberger

Hm. What if $f$ is positive everywhere?

aquire

Thanks @AkivaWeinberger

Akiva Weinberger

I mean, it probably isn't, but what if

Call its maximum $M$. Then $0\le\int_0^1f(t)\cos(nt)dt\le\int_0^1M\cos(nt)dt=M\frac{\sin(n)}n$?

It has a maximum because it's continuous and defined on $[0,1]$

Oh, wait, that doesn't work

Sorry

The middle inequality doesn't have to hold.

Extrarius

Is how to generate a satisfying tuple from an index (1 giving the lexicogrraphically least tuple, 2 the next, etc) a math.se question? Or would it be more cs?

Akiva Weinberger

$\cos(nt)$ is sometimes negative, so $f(t)\le M$ doesn't imply $f(t)\cos(nt)\le M\cos(nt)$

Socrates

01:54

I try to teach my stuff to a yellow duck on my desk made of plush. Maybe, just maybe, you should do the same?

Akiva Weinberger

So, never mind @aquire

Paul Plummer

Ah finally found the theorem, Easton's theorem @AkivaWeinberger

aquire

we can deal with absolte values @AkivaWeinberger

right?

Akiva Weinberger

Ooh, probably

But we might need to know what $\int_0^1|\cos(nt)|dt$ is.

I doubt that that goes to $0$.

aquire

you are right

Akiva Weinberger

01:58

It's definitely less than $1$, at least

aquire

yep

Akiva Weinberger

Just throwing ideas out there — maybe see what happens at $f(t)=t$?

Like, what's $\int_0^1t\cos(nt)dt$

Oh, never mind, it's complicated

Brody

What's the simplest example of $f:A\to B,\; A^* \subset A$ such that $f(A)\setminus f(A^*)\ne f(A\setminus A^*)$?

Akiva Weinberger

$\frac{\sin n}n-\frac{1-\cos n}n^2$

@Brody $A^*$ is any subset?

Brody

@AkivaWeinberger Yeah.

aquire

02:03

may be taylor expansion of cos might help

Brody

hmm, I have one that's simple but it doesn't involve $A^*$ empty

Akiva Weinberger

$f:\Bbb R\to\Bbb R,~x\mapsto x^2$, with $A=\Bbb R$ and $B=\Bbb R_{\le 0}$

$f(A)\setminus f(A^*)=\emptyset$, but $f(A\setminus A^*)=\Bbb R$.

I think we need to use the conclusion to part 1 @aquire

About the piecewise function

aquire

makes sense

Brody

Oouu, $\Bbb R_{\le 0}$ is nice notation

Akiva Weinberger

It's the same as $(-\infty,0]$

just like $\Bbb R$ is the same as $(-\infty,\infty)$

Brody

02:06

Yeah, it looks better than some alternatives

imo

aquire

we use $\mathbb{R}^-_0$

Brody

e.g. apparently $\Bbb R^+, \Bbb R_0^+$ can respectively mean $(0,\infty), [0,\infty)$ or vice-versa depending on the text/region/etc.

Akiva Weinberger

Where I wrote $B=\Bbb R_{\le0}$ I meant $A^*=\Bbb R_{\le0}$.

Brody

that terribly confused me for a sec but I had noted :P

and the image of the set difference is a typo too

Akiva Weinberger

It should be $\Bbb R_{>0}$, shouldn't it

Brody

02:13

that

Are you going for the greatest disparity, so to speak?

Akiva Weinberger

The what?

ATaco

I am very intimidated by the math speak here.

I like it.

Brody

By simplest I was thinking simple (co)domain or something like that, e.g. $f:{0,1}\to{0};x\mapsto 0$ and consider the subset ${0}$. The disparity is between $\emptyset$ and ${0}$.

Akiva Weinberger

Oh, the constant function, derp

Brody

as opposed to say between $\emptyset$ and $\Bbb R$, which though (no more or less) unequal than the above are indeed the most unequal in some sense @Akiva

Akiva Weinberger

02:18

That's a good example

Brody

domain $\{0,1\}$, latex derp

Akiva Weinberger

How about the function from robbers to the banks they rob?

The function on everyone but Robert isn't necessarily the same as all the banks but the one Robert robs

say, if someone else also robs the same bank.

Brody

barring that a robber may rob multiple places

Akiva Weinberger

Stopping Rob won't necessarily save his bank, essentially.

@Brody Assume spherical cows

Brody

I don't know how common that kind of serial robbery is irl. the cows perhaps needn't be that round

oh, and "Robert". lol

Akiva Weinberger

02:24

Little Bobby Tables, we call him

Brody

quite an unfortunate path he's gone down D:

aquire

$$

We can say ${-1\over n}\le\int_0^1 \cos nt dt\le {1\over n}\\\Rightarrow \left|\int_0^1 \cos nt dt\right|\le{1\over n}$

That will do Thanks a lot @AkivaWeinberger

bjb568

03:22

TIL my nose is a wave.

Latin is weird.

Brody

Is it possible to have an endofunction on $\Bbb R$ such that a subset and its complement both have images $\Bbb R$? Don't think so but could be wrong...

Akiva Weinberger

What's an endofunction @Brody

Brody

I just mean $\Bbb R\to \Bbb R$

Akiva Weinberger

Is it required to be continuous?

Brody

no restrictions

Akiva Weinberger

03:31

Oh, $x\sin x$. Positives and negatives.

Doesn't matter which one you put $0$ into.

Actually, any cut into $(-\infty,a]$ and $(a,\infty)$ would work.

(And same for switching open and closed.)

Brody

nice, thanks

this is an example where $f[A] \setminus f[A^*]=\emptyset$ and $f[A\setminus A^*]=\Bbb R$

Semiclassical

04:10

hi chat

Daminark

@Akiva So you have what's called an "endomorphism"

Which is a morphism from an object to itself

If you're in the category of sets with your morphisms being functions, then you get an endofunction

But you could also make that case for a homomorphism from a group to itself, continuous function, and so forth, depending on the type of object

Akiva Weinberger

What are the sine curves that pass through $(2,\sin2)$, $(8,\sin 8)$, and $(14,\sin14)$ other than $y=\sin x$

$\sin\left(\dfrac{2\pi}6(2-x)+x\right)$ works.

Ah, OK. Let me type up a thing with images:

imgur.com/a/bPiMr

When you take the dots $(n,\sin n)$ for integer $n$ and graph them, they form a wave… unless you squish the $x$-axis, after which they seem to form lots of waves.

Akiva Weinberger

04:51

It might be more apparent for $(n,\sin cn)$ for some $c$ — that is, for $y=\sin cx$.

Everything was graphed in Desmos.

Yeah, it very much becomes more apparent for $c\ne1$.

Kimberly James

05:32

hi is anyone good with first-order-logic here?

John P

You might want to just ask.

Martin Sleziak

07:22

Since I am trying to get the bounty room going, perhaps a reasonable thing could be to promote it here. So here is link to the relevant meta post and here is the room.

9

DHMO

10:28

@AkivaWeinberger how do you plot dots?

Akiva Weinberger

11:09

@DHMO You could type $([-50...50],\sin([-50...50]))$ into Desmos. The $[m...n]$ format makes a list.

Alternatively, you could type in $(n,\sin n)$ and $n=[-50...50]$.

DHMO

@AkivaWeinberger I see, thanks

Akiva Weinberger

11:23

@DHMO Here it is for $\sin(2n)$ rather than $\sin(n)$.

imgur.com/a/4Evgg

DHMO

ok

DHMO

12:15

As n increases, the ratio n^n : (1+n)^n goes to 1:e.

Can we prove that 1/2 + 1/12 + 1/30 + 1/56 + ... = ln 2?

Akiva Weinberger

12:32

That's $(1-\frac12)+(\frac13-\frac14)+\dotsb$, so use the well-known $\sum(-1)^{n+1}\frac1n=\ln2$.

@DHMO

DHMO

@AkivaWeinberger can we prove it without going through a conditional-convergent series?

Mahmoud

Hi.

DHMO

@Mahmoud wa-alaykom salam

Mahmoud

@DHMO $:)$

I always like to say that math is not about computation, but when I come along something like this, I get both confused and disappointed.

DHMO

why?

Akiva Weinberger

12:38

@DHMO Well, it's easily shown to be convergent, by comparison to $\sum1/n^2$.

Mahmoud

$\lim_{x\to 3} \frac{\sqrt{x^2-8}-2\sqrt{3x}+5}{\sqrt{x+1}-\sqrt{x+6}+1}$

Computation at it highest limits, unless someone comes up with a miracle to avoid it.

@DHMO Any ideas not involving multiplication by the conjugate ?

Akiva Weinberger

@Mahmoud That's clearly $\infty$.

Divide top and bottom by $\sqrt x$.

Mahmoud

@AkivaWeinberger It turns out to be $24$.

Akiva Weinberger

Oh, wait, $x\to3$, sorry

I didn't see that

Hm. I would substitute $u+3=x$ so that we could get the limit to be at $0$.

Mahmoud

So $\lim_{u\to 0}$ ? First time I see it.

@TedShifrin I apologize, you were right, after I read those notes with the concrete example of $\lim_{x\to 3} x^2=9$, I began to understand it, please forgive my haste. $$\text{Concrete}\rightarrow \text{Abstract}$$ Is a good rule to keep in mind.

We'll get $0$ in the denominator @AkivaWeinberger

Akiva Weinberger

13:00

You're right. Hm…

Oh, OK, I think I see what to do. Rewrite the denominator as $(\sqrt{x+1}-\sqrt{3+1})+(\sqrt{x+6}-\sqrt{3+6})$

and do conjugates on that

Mahmoud

Just on the denominator @AkivaWeinberger ?

Akiva Weinberger

Also the numerator, but I haven't worked it out yet

The denominator becomes $\frac{x-3}{\sqrt{x+1}+2}-\frac{x-3}{\sqrt{x+6}+3}$

Which is $x-3$ times something that goes to $\frac14-\frac16=\frac1{12}$. Hopefully, the numerator also is $x-3$ times something that goes to some finite value.

In that case, they'd cancel out.

Mahmoud

Yes, @AkivaWeinberger.

Akiva Weinberger

Numerator is $(\sqrt{x^2-8}-\sqrt{3^2-8})-(2\sqrt{3x}-2\sqrt{3(3)})$

I mean, $(\sqrt{x^2-8}-1)-(2\sqrt{3x}-6)$

So do conjugates on that as well?

First term looks annoying, though.

Krijn

@arctictern Your bird just came up on the tv show QI that I'm watching

logical123

13:11

Can we have a bijection from a countable union of finite sets to an uncountable set?

Krijn

What an interesting bird

arctic tern

mine is an allusion to a joke on whose line is it anyway (US version)

@logical123 a countable union of countable sets is countable

Krijn

Oooh I like that show

Scenes from a hat, mostly

logical123

@arctic

yeah that's what I read but does that imply the bijection can't exist?

arctic tern

@logical123

Krijn

13:12

@krijn

arctic tern

@logical123 there are no bijections between countable and uncountable sets

logical123

it must be very early, I know this stuff lol

thanks for confirming

Tobias Kildetoft

@Krijn QI is great

Krijn

Yeah and Sandy is a great replacement

Akiva Weinberger

@Mahmoud I have to run. Good luck, though.

Tobias Kildetoft

13:15

Yep, she is doing great

And as a fellow Dane, I appreciate her Randy Scandy segment

Krijn

It's a bit pushy at times

Alan is much funnier this season

Mahmoud

Bye @AkivaWeinberger

Adeek

hey @TobiasKildetoft could I ask some quadratic forms stuff ?

There is a proof I want to make sure my understanding for it is correct.

Akiva Weinberger

13:35

@Mahmoud Yeah, that plan should work. The numerator becomes $\frac{x^2-9}{\sqrt{x^2-8}+1}-\frac{12x-36}{2\sqrt{3x}+6}$

which is $\frac{(x-3)(x+3)}{\sqrt{x^2-8}+1}-\frac{(x-3)(12)}{2\sqrt{3x}+6}$

Mahmoud

And we divide by $(x-3)$

Akiva Weinberger

So it's $x-3$ times something that goes to $\frac62-\frac{12}{12}=2$

So the numerator was $x-3$ times something that goes to $2$, the denominator was $x-3$ times something that goes to $\frac1{12}$

Cancel out the $x-3$, and we get $\frac2{1/12}=24$.

Which is what you said the answer was.

Ramanujan

@Mahmoud اَلسَّلاَ مُ عَلَيْكُمْ وَرَحْمَةُ اللهِ وَبَرَكَا تُهُ

:P

Adeek

nvm I get it

Mahmoud

وعليكم السلام ورحمة الله تعالى وبركاته

@Ramanujan

Ramanujan

13:42

How are you doing,you just started limits?

Akiva Weinberger

Shalom aleichem

Ramanujan

@AkivaWeinberger aleikhem shalom

Mahmoud

Yes @Ramanujan.

I'm a bit confused, I need to re-order my thoughts and make sense of everything.

Ramanujan

Limits is deception and we need to get correct answer…

(as I think)

Akiva Weinberger

14:00

I don't know why limits are called that.

Ramanujan

Because firstly everything looks indeterminate form,but from some where we get answer

Semiclassical

14:27

hi chat

Typically, something which 'looks indeterminate' ceases to be so when you actually plug in numbers which are near the value of interest.

What you get may be surprising or counter-intuitive, but hardly deceptive.

DHMO

14:41

@Mahmoud why would you not multiply by the conjugate?

Simply Beautiful Art

15:30

@DHMO :|

Because I am the conjugate

Simply Beautiful Art

15:41

@Mahmoud Just use binomial expansion?

eurocoder

Define $A: l^2 -> l^2$ by $Ax = (\frac{1/i}{x_i})_{i \in \mathbb{N}}$ for $x = (x_i)_{i \in \mathbb{N}}$. Then the $A$ is self-adjoint and injective but not surjective. I can't see why it is not surjective?! Anyone know why?!

Akiva Weinberger

15:53

@SimplyBeautifulArt That's an infinite series, for $\sqrt{x+a}$. I doubt he's learned that yet.

Semiclassical

@eurocoder I don't see an answer off the top of my head, but evidently you need a sequence $x$ such that $\sum_i \frac{1}{(i x_i)^2}$ is convergent but $\sum_i x_i^2$ isn't.

Which, in fact, seems pretty easy to arrange.

Transcript for

$Mathematics$ Mathematics