Mathematics - 2016-11-19 (page 4 of 10)

Alessandro

09:00

Which part @DHMO?

Brody

@Fargle My intro to higher math never mentioned ordinals, and the only cardinals explicitly mentioned were $\aleph_0, 2^{\aleph_0}$. Might wait 'till grad school then

DHMO

@Alessandro how is it different from $2^{\aleph_0}$?

Fargle

@DHMO $2^{\aleph_0}$ doesn't presuppose that the functions have finite support, if I'm remembering right

DHMO

what does finite support mean?

Alessandro

The function is $0$ everywhere apart from an a finite number of values in the domain

DHMO

09:03

:o ok

Fargle

@Alessandro Hopefully I wasn't terribly wrong in my discussion of ordinals earlier >_>

Alessandro

I didn't see it :P

Fargle

That means it was rock solid.

Balarka Sen

@Alessandro I started with Solyaris too, but Stalker is my favorite. I suggest starting with Stalker.

Brody

@Alessandro You sure didn't!

deletes all dumb comments he made

Alessandro

09:05

I'm no expert, just an interested student, I never had a set theory course (but I'd love to specialize in set theory, we'll see what happens with time)

DHMO

@Alessandro how would you biject between N and w^w?

Fargle

Same here--just wanted a second opinion, haha.

Alessandro

I wrote it down, thanks @Balarka

Fargle

@DHMO I don't know how you'd do that, but it turns out that a countable union of countable sets is countable, and $\omega^{\omega} = \bigcup_{n < \omega} \omega^n$

DHMO

ok thanks

does it require AoC?

Alessandro

09:09

Hm, I don't know if you can write a bijection explicitely, but you can show that every ordinal below $\omega_1$ is countable

Yes (countable choice suffices)

DHMO

how?

why does it require AOC?

Alessandro

If you try to prove it you'll need to pick a function from each set in a countable family of sets of functions at some point (or something similar if your proof is different)

That's just an intuitive motivation, not a proof, but you can construct models of ZF where the reals are a countable union of countable sets

DHMO

ok thanks

Alessandro

Don't ask me how it's done because I have no idea though!

Secret

09:27

[Is bored]
$$\cdots> c0 > 0 > c^{-1} 0 > \cdots$$

Fargle

@Secret I don't quite c what you're up to.

DHMO

what is c?

Secret

just a random element with an inverse

DHMO

ok

Brody

Guys, meet $y=\lfloor x\rfloor \sin x, |x|\le 2\pi$. He's a happy ghost ^.^

DHMO

09:30

wut the hell

Fargle

@Brody pukes

Brody

@Fargle ssshhh His name is ugly, but he's quite adorable

Fargle

It just looks rather...derpy, haha.

Balarka Sen

@Brody yikes man

alan2here

09:45

Can anyone here solve this?

0

Q: For an N by N 2D space with integer coordinates, what is the most unique vectors that can exist that connect M specified pairs?

Take an N by N 2D space with integer X, Y coordinates, for example if N were 3, then there are 9 possible positions/coordinates in this space. Where M points in this space can be marked. Q: What is the most unique vectors that can exist that connect pairs of marked points in this space? Assume...

SteamyRoot

@Brody You're a bit late for Halloween though

3

DHMO

10:57

Here is why the Cantor's diagonal so-called proof is shit:

i dont think i want to explain it

towc

@alan2here there are NxN points, therefore NxN unique set of coordinates, therefore NxN unique vectors...?

unless by unique you mean that in polar form they can't share the same angle or something

ohi @DHMO

DHMO

11:54

@towc hi

Astyx

12:12

Hi chat

Max

12:25

This is driving me crazy... Can anyone help me or give me any clue as to how to prove that this integral is convergent? prntscr.com/d97iq0

I've been trying to find a function whose integral diverges, which is smaller than sin(1/x)/ln(x). I do find sin(1/x)/x, but can't get any further. All functions on the form 1/(x(ln x)^n) seem to not bound the function from below either

Mats Granvik

@Max Have you tried integration by parts? I am no expert on integrals though.

Astyx

This looks a bit like Riemann-Lesbegue's Lemma

Max

I tried it a bit, but differentiating 1/ln(x) gives me a ln(ln(x)) term somewhere, so I couldn't get anywhere from differentiating 1/ln(x). Doing the opposite (integrating 1/ln(x)) also gave me a more complicated function

I also tried variable substitution, but that only gives me sin(du/dx), which doesn't help :P

Astyx

12:40

series maybe ?

Max

Like proving that the corresponding series converge?

Astyx

Yeah

Max

I tried that a little bit with the comparison-test (or what it's called), where you divide two functions. That was equally hard as with the integral though. I guess I haven't tried dividing a(n+1) with a(n) though

but then wouldn't I have to simplify an expression with sin(1/(x+1))?

Astyx

Oh actually I read the integral wrong

$\sin{1\over x} \sim {1\over x}$

Alessandro

I'm writing this just after messing around with a graphing software for a few minutes so don't trust it, but your function seems bounded above by 1/x

(After some number around 3)

Astyx

12:45

by $\alpha \over x \log x$ even

For some constant $\alpha$

Max

It is bounded above by 1/x, but 1/x diverges

Astyx

Do you know the series-integral comparison theorem ?

Max

hmm, but it's bounded above right?

is it the theorem that says that both either diverge or converge?

Astyx

Yeah

Max

I know of it

Astyx

12:47

Because your series is decreasing positive and the function converges to 0

Alessandro

Ah, right, my bad

Astyx

Therefore $\int_2^{+\infty} {\sin{1\over x}\over \log x}$ has the same nature as $\sum_{n=2}^{+\infty} {\sin{1\over n}\over \log n}$

And th general term of the RHS is equivalent to $1\over n \log n$

Which does not converge

Therefore your integral does not converges

Max

but doesn't sin(1/x) make it decrease faster?

Astyx

Yes it does

Max

I do see what you mean, but how do you prove that sin(1/x)/x diverges?

as a sum

Astyx

12:50

But not sufficiently faster to make it converge

sin(1/x)/x ?

Max

sorry, *sin(1/x)/ln(x)

Astyx

It's equivalent to ${1\over n \ln n}$

Right ?

Max

isn't sin(1/x)/ln(x) < 1/(xln x)?

since sin(1/x) < 1/x

Astyx

Do you know the comparison with equivalent theorem for series ?

Max

hmm, not quite sure

Astyx

12:53

It states that if $a_n \sim b_n$ then $\sum a_n$ has the same nature as $\sum b_n$

(very very powerful theorem)

Max

oh, so that's the theorem you used?

okay, I do see how it makes sense then

Astyx

yeah

Are you french btw ?

Max

hmm, we haven't learned that one, but it does make sense, which is good :P

no, Swedish

Astyx

Oh, we use ln instead of log too, that's why I asked

Max

yeah, when we write log, we mean log(10)

Astyx

12:55

Here as well

Max

well, actually, what you said makes a lot of sense now that I think of it, since sin(1/x) really becomes closer to 1/x as x -> inf, I suppose?

Astyx

Yes, that's the idea

Max

sweet! thanks a lot by the way :D

Astyx

My pleasure !

Fargle

Why am I still awake? Mathematical answers only please.

Astyx

13:00

4

Fargle

@Astyx Now that I can get behind. Thanks.

Astyx

:)

Mahmoud

Hi !

Astyx

Hi to you too !

Mahmoud

Thanks :D

Can someone please write this for me in the sum notation $y=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\frac{x^{11}}{1‌1!}$

DHMO

13:06

@Mahmoud $y=\sin(x)=\displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$

Mahmoud

The Maclaurin approximation for $f(x)=sin(x)$

Astyx

^

Mahmoud

Thanks @DHMO

:)

DHMO

you are welcome

Astyx

not $\infty$, 5 !

Mahmoud

13:07

Oh I forgot the dots :I

DHMO

@Mahmoud then it isn't an approximation right

Mahmoud

He knows what I meant @Astyx

Astyx

^

I know, I'm just teasing

Mahmoud

@DHMO What do you mean ?

DHMO

@Mahmoud I mean it is not an approximation

1 min ago, by Mahmoud

The Maclaurin approximation for $f(x)=sin(x)$

Astyx

13:08

$\sin(x)$ is defined as the limit of this series

DHMO

yes

Mahmoud

Yes, it has to got to infinity, I only got it to work in $[-4,4]$ for the first 5 terms.

It's very clever, I'm amazed.

teadawg1337

13:20

Hello everyone o/

Astyx

Hi

DHMO

hi

heather

hello o/

teadawg1337

How are you all today?

Astyx

Like any saturday : working :p

heather

13:21

don't have school = very happy =D

teadawg1337

I'm super tired, I woke up early today

I suppose that gives me more time to work on mathematics :3

Should I start a bounty on a question of mine that's been unanswered for the 17-ish days it's been up?

Mahmoud

I have a very hard Set theory test Monday, :(

DHMO

13:42

@Mahmoud any interesting questions?

Mahmoud

@DHMO A lot, but it'll be useful if you tell me some tips on Set Theory.

DHMO

@Mahmoud can you share some here?

Mahmoud

Ok.

Mahmoud

13:53

$A,B \in \mathscr P(E)$, we consider the relation $f: \mathscr P(E) \rightarrow \mathscr P(A) \times \mathscr P(B) | X \mapsto (X \cap A, X \cap B)$ Prove that : $ \text{$f$ is Injective} \iff A \cup B=E$ @DHMO

DHMO

@Mahmoud what does the part with $X$ mean?

Mahmoud

The variable is a set @DHMO

DHMO

@Mahmoud $f$ is a function right

Mahmoud

Yes, @DHMO

Alessandro

For one direction: suppose $A\cup B$ is not $E$ and that $z$ is in $E$, but not $A\cup B$, we have $f(E)=f(E\setminus z)$ thus $f$ is not injective

Astyx

14:04

$$f: \begin{cases} \mathscr P(E) \rightarrow \mathscr P(A) \times \mathscr P(B) \\ X \mapsto (X \cap A, X \cap B) \end{cases}$$

DHMO

wait, I thought $f$ is any function, lol

Mahmoud

My bad :I

Astyx

@Mahmoud Do you need help for this or not ?

Mahmoud

@Astyx We solved it in class, but the proof felt like black magic.

Astyx

Haha :P

Would you be able to reproduce it though ?

Set theory exercise are very good for learning to write proofs

Mahmoud

14:09

If I learn it by heart, yes.

I am horrible at writing proofs tho.

Astyx

You should try to understand the concept behind the proof without learning it off by heart

DHMO

I give up, lol

Mahmoud

Exactly, but those kinds of problems really put me down.

The prof looks to me like an alien when he's writing the proof on the board, and I feel like the stupidest person ever for not solving it, repeated situations like this convinced me that I can't prove any statement, even if I know that I can do it if I try hard enough.

Astyx

Huh

DHMO

@Mahmoud well, what's the proof?

Astyx

14:13

Would you like me to guide you through the proof ?

Mahmoud

Yes please.

Astyx

So first let's prove that if $A\cup B \ne E$ , then $f$ is not injective

Mahmoud

Why ?

DHMO

@Astyx I think all steps in your proof are bijective

Astyx

Because $A\cup B \ne E$ implies that $f$ is not injective, then if $f$ is injective, $A\cup B = E$

Which is the indirect way of your equivalence relation

Right ?

Mahmoud

14:17

After re-checking the proof, it does make sense, but it's not the point of this, I want to be better at proving :(

DHMO

@Astyx we call it "contrapositive"

Mahmoud

And yes, you're right.

DHMO

@Mahmoud well, first you need to have an idea about the proof

Astyx

@DHMO I know how we call it

DHMO

@Astyx just want to use formal terms

Mahmoud

14:17

@Astyx Of course you do, btw isn't that two implications ?

So we have to prove it both ways

Astyx

Yes it is

Yep

DHMO

@Astyx I think all steps in your proof are bijective so you only need to do it once

Astyx

Not sure

DHMO

continue then

Astyx

So let's suppose $A\cup B \ne E$. Then there exists $z\in E$ such that $z\notin A\cup B$

Then what is $f(\{z\})$ ?

$A\cap \{z\} = \emptyset$ and $B\cap \{z\} = \emptyset$

Mahmoud

14:21

Yes. :D

Astyx

So $f(\{z\}) = (\emptyset, \emptyset)$

So you have $f(\{z\}) = f(\emptyset)$

And therefore $f$ is not injective

So we proved the first way

Mahmoud

Yes, but how did you know that this is going to work ?

Astyx

Try drawing sets to get used to it

And eventually you won't need to draw them to see what to do

Mahmoud

Thanks.

Astyx

And @DHMO was right in the sense that we only use equivalence, but I personnaly think it's better to structure the proof doing one way at a time (rather than having to justify why it's equivalent each time)

DHMO

14:27

$A \cup B \ne E$
$\iff \exists z \in E: z \notin A \cup B$ (definition of set equality)
$\iff \exists z \in E: z \notin A \land z \notin B$ (use first-order logic to prove)
$\iff \exists z \in E: f(\{z\}) = (\varnothing,\varnothing)$
$\iff \exists z \in E: f(\{z\}) = f(\varnothing)$
$\iff f$ is not injective

The result follows.

prodprod

the last equivalence requires some justification

it is not by definition

Astyx

Exactly

DHMO

@Mahmoud to write a good proof you need to know what each term means

Astyx

And WORDS

DHMO

i don't like words

Astyx

14:29

You should

DHMO

@prodprod you mean $\exists z \in E: f(\{z\}) = f(\varnothing) \iff f$ is not injective?

prodprod

yes

DHMO

@prodprod that follows from the definition of injection

Astyx

I despise proofs such as the one you wrote :p (even though they are right)

prodprod

one direction follows by the definition of injection, yes

but $f$ not being injective only gives us that we have $f(X) = f(Y)$ for some subsets $X,Y$ of $E$

not quite what you're claiming

err, distinct subsets of $E$

DHMO

14:32

$f$ is injective
$\iff \forall X,Y \in E: f(X) = f(Y) \implies X=Y$
$\iff \forall X,Y \in E: f(X) \ne f(Y) \lor X=Y$
$\iff \neg(\exists X,Y \in E: f(X) = f(Y) \land X\ne Y)$
now substitute $X=\{z\}$ and $Y=\varnothing$

Astyx

You're missing some prenthesis right there

DHMO

@prodprod unless we use different definitions of injectivity

@Astyx i don't care as long as it is legible lol

Astyx

Then why not using words which are much more legible ?

prodprod

why are you allowed to substitute anything into that last expression exactly?

Astyx

because of $\exists$

You only need one couple

DHMO

14:36

@prodprod I think it would be better if I cross out the line $\iff \neg(\exists X,Y \in E: f(X) = f(Y) \land X\ne Y)$

substituting things into $\forall$ is better

Astyx

But actually you are right

It's more complicated than that

DHMO

@Astyx why?

Astyx

I was adressing to prodprod x)

Mahmoud

Thanks guys $:)$

DHMO

what is wrong with my proof @Astyx

prodprod

14:37

i do buy the implication $f(\{z\}) = f(\emptyset) \implies f$ is not injective, sure

Astyx

Because the fact that $f$ is injective does not necessarilly mean that there exists a singleton

prodprod

but you're cutting corners with the other implication

Astyx

Not generally

Null

is it normal that i dont see the associated videos on the right on youtube?

Astyx

You have to developp from the definition of $f$ here

@Null no

Mahmoud

14:38

But please clarify what this means $P(A) \times P(B)$

Astyx

It's a couple of parts of respectively A and B

for instance $(\{a,b,f\}\times\{1,3,4\})$ is a couple of $P(A)\times P(B)$ where A is the alphabet and $B =\Bbb N$

DHMO

@prodprod where?

Mahmoud

But $P(A)$ is different from $A$

prodprod

you're saying that $f$ is not injective is somehow equivalent with $f(\{z\}) = f(\emptyset)$ for some $z \in E$

Astyx

yes

Otherwise a couple would be $(a,4)\in A\times B$ for instance

DHMO

14:40

@prodprod that isn't what I said

prodprod

but $f$ is not injective only gives us that we have $f(X) = f(Y)$ for SOME distinct subsets $X,Y$ of $E$

Astyx

@DHMO this is exactly what you wrote on the last line

DHMO

@Astyx let me clarify

Mahmoud

$P(A)=\{\emptyset , \{a\}, \{b\} ...\}$

Astyx

@Mahmoud Yes

DHMO

14:41

$A \cup B \ne E$
$\iff \exists z \in E: z \notin A \cup B$ (definition of set equality)
$\iff \exists z \in E: z \notin A \land z \notin B$ (use first-order logic to prove)
$\iff \exists z \in E: f(\{z\}) = (\varnothing,\varnothing)$
$\iff \exists z \in E: f(\{z\}) = f(\varnothing)$
$\iff \exists X,Y \in \mathscr P(E): f(X) = f(Y) \land X \ne Y$
$\iff \neg(\forall X,Y \in \mathscr P(E): f(X) \ne f(Y) \lor X = Y)$
$\iff \neg(\forall X,Y \in \mathscr P(E): f(X) = f(Y) \implies X = Y)$
$\iff f$ is not injective

Mahmoud

For instance, $\{a,b\} \times \{1\}$ Makes no sense for me @Astyx

DHMO

@Mahmoud $\{a,b\}\times\{1\} := \{(a,1),(b,1)\}$

Mahmoud

@DHMO Nice.

DHMO

@prodprod @Astyx any problem?

Astyx

You're still assuming line 5 to 4 is obvious

But to prove it you need to explicit what $f$ does

Otherwise that's not generally true

prodprod

14:44

how do you prove the implication $\exists X,Y \subset E \colon f(X) = f(Y) \land X \neq Y \implies \exists z \in E \colon f(\{z\}) = f(\emptyset)$

DHMO

@prodprod you are right

how should i proceed then

Astyx

You should explicit the definition of $f$ (ie say what line 5 means without writting $f$)

DHMO

@Astyx I don't understand your objection. Could you quote?

Astyx

$\exists z \in E: f(\{z\}) = f(\varnothing)$
$\iff \exists X,Y \in \mathscr P(E): f(X) = f(Y) \land X \ne Y$

should be replaced by

$\exists z \in E: f(\{z\}) = f(\varnothing)$
some work here
$\iff \exists X,Y \in \mathscr P(E): X\cap A = Y\cap A, X\cap B \ Y\cap B \land X \ne Y$
$\iff \exists X,Y \in \mathscr P(E): f(X) = f(Y) \land X \ne Y$

prodprod

@DHMO i guess you could prove the implication by looking at some $z$ that is in exactly one of the sets $X,Y$, but admittedly i haven't thought about it too much

i would not try to do the whole proof in a bijective fashion tbh

Astyx

14:48

^

DHMO

@Astyx so same objection

Astyx

Yes

Null

can someone tell me what's wrong?

Astyx

About Youtube ?

Null

Mahmoud

14:50

Oh Vsause !

Astyx

have you tried turning it off and on again ?

Null

@Astyx my pc or the site?

Astyx

Your browser

Null

ah ok, i try :)

SteamyRoot

Try ctrl-F5 to clean cache on the page, worth a try too

Astyx

14:51

Can the 24 on the top right have anything to do with it ?

prodprod

@DHMO yeah, just pick some $z$ in $X \setminus Y$ (or $Y \setminus X$, one of these is guaranteed to be nonempty) and note that since $f(X) = f(Y)$, we must have $f(X \cap \{z\}) = f(Y \cap \{z\})$

DHMO

@Mahmoud any more question?

Mahmoud

@DHMO A lot.

Null

@Astyx i try to disable all addons, then see. but thanks i wouldnt have thought this way, i thought youtube changed :O

DHMO

@Mahmoud can you share?

@prodprod and then?

Mahmoud

14:53

Oh, yes :D

prodprod

from this you get $f(\{z\}) = f(\emptyset)$

SoumyoB

so I was wondering in a theoretical world, where there are infinite people, if I went on killing 1 person, then 2 persons, then 3 persons and so on, till infinity, would I be considered a murderer for killing infinite people or would I be considered God for giving life to $\frac{1}{12}$ persons?

Astyx

@Null Tell me if it worked

DHMO

@SoumyoB ... misconceptions about mathematics #314159

Astyx

Misconceptions ? Really ? :D

DHMO

14:54

the result $1+2+3+\cdots=\dfrac1{12}$ depends on what + means

Astyx

More like on what $\dots$ means no ?

SteamyRoot

actually, in your case, @SoumyoB, you'd be dead before you reach infinity

DHMO

and you can be assured that it is not the usual +

SteamyRoot

But, either way, you'd never get $-1/12$

DHMO

@Astyx no, the $\cdots$ still mean $\Bbb N$

Astyx

14:55

Yeah but you have different approaches to "go to infinity"

And that's kind of the point I think

SteamyRoot

Because what you're killing is $\lim_{n \to \infty} \sum_{i = 0}^n i$ people

DHMO

@SoumyoB the formal statement is that "if the summation method is linear and stable, it implies that $1+2+3+\cdots=\dfrac1{12}$ by the summation method"

where we are exploring different summation methods

SteamyRoot

wut

SoumyoB

sorry guys I meant $\frac{1}{12}$ persons given life, which means I'm implying $1+2+3+... = -\frac{1}{12}$ @DHMO

Astyx

$-{1\over12}$ is what you get at $-1$ with the analytical continuation of the $\zeta$ function

SoumyoB

14:56

^

Mahmoud

$A,B \subset E \land A \subset B$ Solve in $\mathscr P(E)$ $X \cap B=X \cup A$ @DHMO :)

DHMO

@SoumyoB whatever, just replace every occurrence of $\dfrac1{12}$ by $-\dfrac1{12}$ in what I said above

Null

@Astyx it was an addon very likely, altho not the adblocker, but some download addon. at least now its back. strange stuff...

DHMO

@Astyx that is another way of stating it

@Mahmoud $\Bbb P(E)$ means $\mathscr P(E)$?

Astyx

@Null Ok glad i helped :)

SoumyoB

14:57

I didn't get what you were asking @DHMO

SteamyRoot

@DHMO what are you saying? That any linear and stable summation method sends $1 + 2 + \cdots$ to $-1/12$?

DHMO

@SteamyRoot no?

Mahmoud

Yes.

DHMO

@SoumyoB I wasn't asking anything. I was trying to clear up your misconception.

@Mahmoud hey you changed $\nand$ to $\land$??

Mahmoud

Yes @DHMO I'm still learning MathJax

DHMO

14:59

but they are opposite things lol

Null

@DHMO i like the video of vsauce :)

40 haha

Transcript for

$Mathematics$ Mathematics