This is the Realm of Simply Beautiful Art

Nick

05:15

What's happening here? So glad to see things are well here.

user21820

08:31

Google chrome privacy controversy.

Simply Beautiful Art

18:07

oof xkcd

Holo

Yep

Simply Beautiful Art

So I started by using power reduction formulas to get:

$$8\sin^4(\theta)=3+4\cos(2\theta)+\cos(4\theta)$$

Holo

Sure

Simply Beautiful Art

And since the integrand is even: $$2I(n)=\int_{-\pi}^\pi\sin^4(x+\sin(nx))~dx$$

Holo

Yes

Simply Beautiful Art

18:11

Expanding this out, we need to calculate $$\int_{-\pi}^\pi\cos(2kx+2k\sin(nx))~dx$$for $k=1,2$

I then used $\cos(\theta)=\Re(e^{i\theta})$ and $\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}$ to get:

$$\int_{-\pi}^\pi e^{2kix}e^{k(e^{nix}-e^{-nix})}~dx$$

Substitute $z=e^{ix}$.

$$\frac1i\oint_{|z|=1}z^{2k-1}e^{kz^n-kz^{-n}}~dz$$

One may compute the Laurent series of this

The essentially point here is that $$e^{kz^n-kz^{-n}}=\sum_{m=-\infty}^\infty a_mz^{mn}$$

That is, the powers of $z$ are multiples of $m$.

Multiply this by our $z^{2k-1}$ and you'll see that for $k=1,2$ and for $n\ne\pm1,\pm2,\pm4$, there is no $z^{-1}$ term in the Laurent series.

Hence, by the residue theorem, we may conclude the integral is $0$.

All that's left is $$16I(n)=\int_{-\pi}^\pi3~dx$$

and hence the result.

Holo

Oh god, I got to $\int_{-\pi}^\pi e^{2kix}e^{k(e^{nix}-e^{-nix})}~dx$ but I didn't thought to use $z=e^{ix}$

Simply Beautiful Art

3 mins ago, by Simply Beautiful Art

That is, the powers of $z$ are multiples of $m$.

this should say $n$ instead of $m$

Holo

Yea I understood

Simply Beautiful Art

Hey @Frpzzd

don't read above

Try this problem:

in TheSimpliFire's Chatroom, 1 hour ago, by Simply Beautiful Art

Calculate $$\int_0^\pi\sin^4(x+\sin(nx))~\mathrm dx$$ for all integer $n$ except $n=\pm1,\pm2,\pm4$.

GL HF

Frpzzd

Ha, I've seen this

Simply Beautiful Art

18:19

Dang it lmao

Frpzzd

It's generalizable

Simply Beautiful Art

shrugs can't win 'em all

Frpzzd

in a very cool way

Simply Beautiful Art

Hm, how'd you solve it then?

Holo

I told @SimplyBeautifulArt to generalize this

Simply Beautiful Art

18:20

>:P

It's probably just adding a few more constants inside the sin?

The above method still works for most of those scenarios

Frpzzd

$$\int_0^{2\pi} \text{exp}\big(in(x+f(\sin(mx))\big)dx=0$$

If $m,n\in\mathbb N$ with $m$ not dividing $n$

for arbitrary $f$

Simply Beautiful Art

I don't think you can quite do arbitrary $f$

Frpzzd

Pretty sure it works for arbitrary $f$.

Simply Beautiful Art

:thonk:

Holo

Take $f(\sin(x))=-x$

Frpzzd

18:23

It just relies on the fact that the $\sin(mx)$ inside of it causes it to be periodic, which allows a nice cancellation to happen

Um, but then $f$ is not a function on $(0,2\pi)$

Holo

For $0<x<2\pi$

Frpzzd

arcsin isn

whoops

Holo

- arcsin

Frpzzd

arcsin isn't defined from 0 to 2pi

:P

Whoopsy, my bad. It still doesn't work, but for a different reason.

Simply Beautiful Art

Does it work for $$f(x)=\begin{cases}- x,&x\in[0,\pi)\\x,& x\in[\pi,2 \pi]\end{cases}$$

Frpzzd

18:24

Notice that $\arcsin(\sin(x))\ne x$ for some x

Let's see. I'll hop over to Wolfram

Holo

Correct, we 0 to 2pi is too large

Simply Beautiful Art

Another semi-interesting integral I did recently, though much more straightforward:

$$\int_{-\infty}^\infty \Gamma(1+it)~dt$$

where ofc $\Gamma$ is the Gamma function

Frpzzd

@SimplyBeautifulArt Yeah, it works for your constructed function

Simply Beautiful Art

oof

Frpzzd

Hm. Gamma function.

Simply Beautiful Art

18:28

oh right mb

I mean

Holo

Ehh, outside of set theory I feel like I need a lot of work. The sad part is that inside of set theory I know how little I really know

Simply Beautiful Art

does your statement even hold for discontinuous functions like 1 on rational, 0 on irrational?

@Holo lol rip

Frpzzd

Yup, it should. Would it convince you if I just showed you the proof?

Simply Beautiful Art

:thonk: but what does the integral even mean in that scenario

Frpzzd

Who knows. The proof doesn't rely on any assumed properties of $f$, other than it being a function.

That's the beauty of cancellation!

Also, what the heck is :thonk:?

Holo

18:31

It is like "thinking hard"

A bit

Frpzzd

Oh XD

Simply Beautiful Art

more of a meme-ish think emoji

Frpzzd

Should I do a proof?

Holo

Yes

Simply Beautiful Art

^

Frpzzd

18:36

Okay. I will try to recall this...

Start with
$$\int_0^{2\pi}\exp\big(inx+in\cdot f(\sin(mx))\big)dx$$

Split it up into $m$ integrals in the form
$$\int_{2\pi k/m}^{2\pi (k+1)/m}\exp\big(inx+in\cdot f(\sin(mx))\big)dx$$
where $k$ is an integer between $0$ and $m-1$, inclusive

Each one is also equal to the following, by making the substitution $x\to x+2\pi k/m$:
$$\int_{0}^{2\pi /m}\exp\big(inx+\frac{2\pi i n k}{m}+in\cdot f(\sin(mx))\big)dx$$

Note that $f(\sin(mx))$ is unchanged

Factor out $e^{2\pi i n k/m}$

and then sum up the terms

That's a sum of roots of unity. It's equal to zero unless $m$ divides $n$.

Or something like that. XD

All good?

Cool corollary:
$$\int_0^{2\pi} \cos^{2n}(x+\lambda \sin(mx))dx=\frac{\pi}{2^{2n-2}}\binom{2n-1}{n}$$
if $m$ divides none of the numbers $2,4,...,2n$

@SimplyBeautifulArt Was that coherent? XD

Simply Beautiful Art

Yeah that was essentially the "simple" solution of mine for $n=3$ in my problem

Frpzzd

Cool!

Simply Beautiful Art

You are, however, assuming the integral exists.

Frpzzd

Oh, blah.

Is that too much to ask of an integral?

Simply Beautiful Art

xP

Frpzzd

18:45

XD

Simply Beautiful Art

idk

it's better than asking for continuity

so I guess not

Frpzzd

haha ok

Holo

@Frpzzd do you want a challenge in set theory?

Frpzzd

Sure!

Holo

In ZFC, for what $\kappa$ is there a partition $P$ of $\Bbb R^+$ such that $|P|=\kappa$ and if $x\in P$ then $x$ is closed under addition? What about ZF?

Frpzzd

18:48

0_0 whaaaaat?

:thonk:?

Simply Beautiful Art

I've another one

Holo

Hmm, what need clarification @Frpzzd ?

Frpzzd

Oh wait. Do I have to prove this by axioms?

Holo

Not need directly by the axioms. The first question is assuming choice and the second is not assuming choice

Simply Beautiful Art

Find an uncountable cardinal $\kappa$ such that for every binary relation $R$ mapping cardinals to $\{0,1\}$, there is a set of $A$ of cardinals less than $\kappa$ such that $|A|=\kappa$ and $xRy=0$ for all $x,y\in A$ or $xRy=1$ for all $x,y\in A$.

Frpzzd

18:52

I think only $\kappa=1$

Holo

@Frpzzd only 1?

Ehhh... "binary relation $R$ mapping" what?

$xRy=0$ hmmm?

Simply Beautiful Art

:thonk: maybe my wording is bad

but xRy = 0 or 1 for every x and y.

Holo

You wording is objectively bad

What does $xRy = 0$ means?

Simply Beautiful Art

Okay, alternatively you could look at it as a function taking two cardinal inputs and outputting either 0 or 1.

Replace $xRy$ with $R(x,y)$ if that's clearer

Holo

No, because you said binary relation, it is not binary

Also, if you use = it implies that the relation is a function

Simply Beautiful Art

18:58

oic nvm lol

Holo

The correct way to write it is $R(x,y,0)$ or $R(x,y,1)$

Simply Beautiful Art

So either $xRy$ for all $x,y\in A$ or we don't have $xRy$ for any $x,y\in A$.

Holo

I see

Now, what does "there is a set of A of cardinals less than κ such that |A|=κ " means?

Do you have one extra "of" there?

Simply Beautiful Art

yeah :P

there is a set A of cardinals less than κ such that |A|=κ

Holo

@Frpzzd what do you think about $\kappa=2$?

@SimplyBeautifulArt So $A$ is a set of cardinals?

Simply Beautiful Art

19:01

yeah

Holo

If so we do assume choice

Simply Beautiful Art

:thonk: you may need to assume a lot more than just choice for the whole thing

and this may or may not just be some large cardinal

>.>

TheSimpliFire

oh, set theory :P

Holo

Wait, if $k$ is a set of cardinals, then $|k|\le \sup k$, denote $\sup k=A$, now assume that every ordinal till $A$ is in $k'$, then $k'\cup \{A\}=A+1$, so $|k'\cup \{A\}|=|A+1|=|A|=A$, and $|k|\le |k'|$ sp $|k|\le A$

@SimplyBeautifulArt are you sure this is the question?

Simply Beautiful Art

yeah I'm sure

also wdym by $\sup k=A$ and $|k|\le A$?

wait, what is $A$?

Holo

19:09

My k is your A

Simply Beautiful Art

.-. y u be backwards

Holo

Let me rewrite it:

If $A$ is a set of cardinals, then denote $k=\sup A=\bigcup A$.

Now, let $A'$ be the set of all ordinals till $k$

This is clear that $A\subseteq A'$

Simply Beautiful Art

mhm

Holo

Then $A'\cup \{k\}=k+1$

So $|A'\cup \{k\}|=|k+1|=|k|=k$

So $|A\cup \{k\}|\le k$

Thus $|A|\le k=\sup A$

This means that $\kappa$, if exists, need to be very big

I think that $\aleph_{\varepsilon_0}$ is the first cardinal it can possibly be

No wait

It is still too small

You follow @SimplyBeautifulArt ?

Simply Beautiful Art

@Holo yeah

@Holo huehue

Holo

19:22

This is the first time I said $\aleph_{\varepsilon_0}$ is too small

Hmm, we need a cardinal $\kappa$ that $|\{\aleph_a\mid a<\kappa\}|=\kappa$ no?

No, wait

The problem is $|A|=\kappa$, so we need $\sup A=\kappa$...

If so we need a fixed point of the aleph numbers...

So now we need to show that for any relation there exists $A$ like that....

Simply Beautiful Art

19:39

>.>

Holo

I feel like the first aleph fixed point is an answer

Simply Beautiful Art

nope

Holo

So we need something larger?

Simply Beautiful Art

19:56

Yes

A wee bit larger (more like way way larger :P)

Holo

Large cardinal... So we are assuming they exists... This is why you said we need to assume more

The heck, do you really need large cardinals for this simple question!

Simply Beautiful Art

lmao

Yes

uwu

Holo

>.<

Can I throw the names of the large cardinals I know :/, I don't know a whole lot about large cardinals

Simply Beautiful Art

20:17

lol

sure

Holo

20:30

Is is inaccessible cardinal? or maybe $k=\aleph_k$-inaccessible?

amWhy

20:52

@SimplyBeautifulArt, @Holo $\Huge {\left(\ddot\smile\right)^{\Huge /}}$

Sorry for the interruption!!

Holo

@amWhy Hi! $ \tiny{\left({\Huge\ddot\smile}\right)^{\Huge /}}$

Simply Beautiful Art

o\

lol

@Holo it is

but which one?

amWhy

@Holo Wow, you're a slimmed down version of me!! :)

Holo

@amWhy ;), so how are you today?

amWhy

@SimplyBeautifulArt You're hanging upside down, waving!

Simply Beautiful Art

20:55

:P

Holo

@SimplyBeautifulArt i would say [1-]inaccessible cardinal because $k=\aleph_k$-inaccessible is to random

Simply Beautiful Art

@Holo but which one?

Holo

which inaccessible?

amWhy

I'll let you both continue without my nonsense! $\ddot \smile$.

Simply Beautiful Art

yeah

:P

Holo

20:57

Bye @amWhy

Well, I have absolutely no idea :-]

Simply Beautiful Art

These are weakly compact cardinals :P

Holo

Sigh, there is no way in hell I could answer this question alone... Well can you tell me the full solution, how do you prove that accessible cardinals can't have this?

$◕\ddot‿◕$

Simply Beautiful Art

oh I've no idea on that one

:thonk:

if I did

Maybe I would've had my "beyond weakly compact"-OCF made already

Holo

So what do you know to prove here?

Simply Beautiful Art

nothing lul

xD

Holo

21:08

WHAT

Simply Beautiful Art

Well I mean

Holo

This is a question from discord?

Simply Beautiful Art

I can show some smaller cardinals can't satisfy it

no this straight from the definition of weakly compact

Holo

Well, I can show that all cardinals beneath $k=\aleph_k$ can not satisfy it

Nothing more really

Where did you hear about this question?

Simply Beautiful Art

:P

I made it up

look up weakly compact cardinals xD

Holo

21:14

κ is defined to be weakly compact if it is uncountable and for every function f: [κ] 2 → {0, 1} there is a set of cardinality κ that is homogeneous for f. In this context, [κ] 2 means the set of 2-element subsets of κ, and a subset S of κ is homogeneous for f if and only if either all of [S]2 maps to 0 or all of it maps to 1.

Simply Beautiful Art

So essentially I was asking you to find a weakly compact cardinal

Holo

Yes, just the wording was horrible :)

Simply Beautiful Art

;P

Holo

Also, what you said is not exactly correct, because $f(x,x)$ is undefined, but in your question it does

Simply Beautiful Art

rip

/shrug

Holo

21:17

But I would not be able to answer it at all, maybe if I would have gone through wiki page of inaccessible sets

Simply Beautiful Art

lol

Inaccessibles and Mahlos are much more intuitive to me as far as size

Holo

Inaccessibles are indeed the easiest to understand in large cardinals

@SimplyBeautifulArt do you know what weakly even/ weakly odd sets are?

Simply Beautiful Art

no

Holo

A set $A$ is called weakly even an union of disjoint sets of size 2 can create it

weakly odd is the same but "union of disjoint sets of size 2"+a singleton can create it

Hmm, is my explanation clear?

Simply Beautiful Art

yeah

Holo

21:27

Do you know what dedekind finite set is?

Simply Beautiful Art

Probably not

Holo

A set $A$ is dedekind finite(d-finite) if for all $B\subsetneq A$ we have $|B|<|A|$

Simply Beautiful Art

ic

Holo

Now, assuming choice d-finite set is equivalent to finite set, but without choice they are not. The question is

How to prove that d-finite set can not be weakly odd and weakly even at the same time

Simply Beautiful Art

wait so

It is weakly even if you can partition it into two sets of equal size?

Holo

21:32

No, this is strongly even

Assuming choice they are equivalent but without choice weakly even is strictly weaker

Simply Beautiful Art

Okay then what does size 2 mean? lol

Holo

For example, $\Bbb N=\bigcup_{n\in\Bbb N}\{2n,2n+1\}$

Or $[6]=\{0,1\}\cup \{2,3\}\cup \{4,5\}$

If you want: $A$ is weakly even if there exists partition of $A$ such that all of its elements are of size $2$, and weakly odd if there is partition of $A$ such that all of its elements apart from one are of size $2$ and the one other element is of size $1$

Simply Beautiful Art

okay

well I'm not good with proofs like this yet so x.x

Holo

I don't know the proof

I heard it is true but I didn't prove it yet

because there exists infinite d-finite sets which are hard to deal with

Simply Beautiful Art

oof

kek

interesting

Holo

21:40

Yea, those are the weirdest sets

Well, there are weirder sets

For example:

Amorphous set

In set theory, an amorphous set is an infinite set which is not the disjoint union of two infinite subsets. == Existence == Amorphous sets cannot exist if the axiom of choice is assumed. Fraenkel constructed a permutation model of Zermelo–Fraenkel with Atoms in which the set of atoms is an amorphous set. After Cohen's initial work on forcing in 1963, proofs of the consistency of amorphous sets with Zermelo–Fraenkel were obtained. == Additional properties == Every amorphous set is Dedekind-finite, meaning that it has no bijection to a proper subset of itself. To see this, suppose that S is a set...

Simply Beautiful Art

owo

Holo

Yep yep

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$Mathematics$ This is the Realm of Simply Beautiful…