Mathematics - 2017-04-09 (page 5 of 6)

@Kasmir: In terms of the Laurent series, finitely many negative terms versus infinitely. In terms of behavior of the function, mapping to a neighborhood of infinity versus mapping all over the plane.

Dodsy

I am just saying that I already found those solutions, that's all.

Ted Shifrin

But you've now proved they are the only possible ones, @Dodsy. That's a big deal.

Dodsy

yeah that's true.

Kasmir Khaan

7:21 PM

@TedShifrin Okay thanks ! that was what I wanted in terms of Laurent =p

Ted Shifrin

When you are asked to "solve" an equation (or a bunch of equations), you want an "if and only if" statement, not a "these work but I dunno who else does" statement.

Daminark

@Dodsy Keep this in mind, people screw up these arguments a lot. Might show only one direction of a set inclusion argument, or one inequality, etc

Meow Mix

@Dodsy OK, did you think about the convex functions one?

Ted Shifrin

Zach, leave that alone.

Meow Mix

leaves it alone

Ted Shifrin

7:23 PM

Especially since you're so bad about following up on all the things you're supposed to do for me ...

Meow Mix

LOL so sassy

Dragneel

Can someone help me prove this really quick? I started by getting its contrapositive. I got $(3 \nmid x) \rightarrow 3 \mid(x^2 -1)$

Not sure where to go from here

Daminark

Think about what happens when you square things in mod

Meow Mix

$3 \nmid (x+1)(x-1)$

Daminark

Oh actually that's clever

Meow Mix

7:27 PM

Therefore, $3 \mid x$

Daminark

(The symbol is \nmid)

Meow Mix

Do you see why @Dragneel?

Dragneel

Hmm give me a minute

Ted Shifrin

Demonark: Most likely Dragneel doesn't know modular arithmetic yet.

Dragneel

I do, but haven't had much practice with it yet

Akiva Weinberger

7:30 PM

Hint: any three consectutive numbers contain a multiple of three

Dragneel

Yea I don't really see what happened here

Akiva Weinberger

(ex: 14,15,16 — 15 is a multiple of three)

Ted Shifrin

OK, that's another way to do it, then. But Zach and Akiva are leading you a direct route.

Akiva Weinberger

(another example: 22,23,24 — 24 is a multiple of three)

Why must any three consecutive numbers contain a multiple of three?

Also, hi everybody!

Ted Shifrin

Hi, DogAteMy.

Dragneel

7:32 PM

Because one of them is a multiple of 3

Meow Mix

Hi @AkivaWeinberger

Akiva Weinberger

It contains a multiple of 3 because one of them is a multiple of 3?

Meow Mix

I just had a 40 minute BP game

Akiva Weinberger

By the way, Zach, it's a beautiful day out (assuming the weather hasn't changed since I was last out)

You don't even need a jacket

Meow Mix

Yeah, I went on my bike earlier this morning

Akiva Weinberger

7:33 PM

I'm addressing this to you because I assume you're having roughly the same weather as I am

@MeowMix Nice

Daminark

Lol I should like, actually leave my room

But also I should do this paper that I'm staring at

Meow Mix

I played chess with Nate this morning too

Akiva Weinberger

It's 75 degrees in Chicago, according to Google

Dragneel

I meant because one of them is divisible by 3

Meow Mix

@Dragneel Thats what being a multiple of 3 means LOL

Akiva Weinberger

7:35 PM

But why would it be true for any sequence of three numbers

Dragneel

Okay I don't know how to explain it -_____-

Meow Mix

@Dragneel Well, let's look at it like this

Let's say we have our number $x$ and we want to prove that either $x$, $x+1$, or $x+2$ is divisible by 3

Akiva Weinberger

@Dragneel Does it make sense that it should be true, though?

Meow Mix

In the case that $x$ is, then our argument is finished

Dragneel

Because you have n+1,n+2,n+3 ?

Meow Mix

7:36 PM

In the case that it isn't, divide it by 3

We'll either get a remainder of 1 or 2, right?

Dodsy

The convex function one seems hard

Dragneel

correct

Meow Mix

If it's a remainer of 1

Then look at $x+2$

That must be a multiple of 3

do you see why?

@Dodsy BTW, a function is convex if it's second derivative is always positive non-zero

Dodsy

@TedShifrin thanks for helping me with that Ted.

Hm alright

Meow Mix

@Dodsy I'd recommend first drawing it out

Dragneel

7:38 PM

I think so. If it has a remainder of 1, then you simply need to add +2 to get a remainder of 0?

Dodsy

And they intersect infinitely many times meaning they represent the same function?

Akiva Weinberger

Yeah @Dragneel

Meow Mix

Draw two curves that have convex graphs and interect countable infinitely many times

No, they can't be equal

Akiva Weinberger

@Dragneel And if it has a remainder of 2…?

Dodsy

Okay that makes sense they just intersect over and over again?

Meow Mix

7:39 PM

@Dodsy Yeah.

Dragneel

You add 1

Meow Mix

@Dragneel So therefore, if we have an $x$, then at least one of the following: $x$, $x-1$, $x+1$ must be divisible by 3

Akiva Weinberger

And if it has a remainder of 0, then it's the multiple of 3

Meow Mix

Correct?

Akiva Weinberger

("It" being the smallest of the three consecutive numbers)

Dragneel

7:40 PM

correct

Akiva Weinberger

So any three consecutive numbers have a multiple of 3.

Meow Mix

So, we have that $3 \nmid (x-1)(x+1)$

Akiva Weinberger

@MeowMix I think I know your convex tangents thing

Meow Mix

Therefore, neither $x-1$ nor $x+1$ are divisible by 3, right?

@AkivaWeinberger Do tell!

Dragneel

which means x is divisible by 3

Akiva Weinberger

7:41 PM

But it won't be the largest convex set inside the loop.

Meow Mix

@Dodsy Also, they have to constantly be growing faster

Dragneel

I think I get it now

Meow Mix

@AkivaWeinberger Yeah I realized that while in my bed, unable to sleep

Akiva Weinberger

For example, the complement of the tangents of a smoothed thickened H-shape is the empty set.

Dragneel

But this isn't true in all cases though right?

Meow Mix

7:42 PM

@Dragneel Whaddya mean?

Dodsy

So they are exponential functions such as x^4 and x^2

Dragneel

Wait nevermind. This is an implication.

Meow Mix

2 things, Nate
1. Neither of those are exponential
2. Possibly?

Dodsy

Well their second derivatives are 2 and 12x^2

Both positive

And they do intersect countably infinitely times

Meow Mix

$12x^2$ isn't always non-zero

Uhh, no they don't

Dodsy

7:43 PM

Oh true

Meow Mix

$x^2$ and $x^4$ intersect once.

Akiva Weinberger

@MeowMix I think it has to do with four facts: (a) that the complement of a union is the intersection of the complements; (b) the complement of a line is two disconnected convex sets; (c) the intersection of two convex sets is convex; and (d) the shape you're looking at (the complement of the tangents) is connected

Well, (d) I don't know how to prove formally

Meow Mix

@Dodsy $x^2$ looks like a good starting point

Akiva Weinberger

but I think that combining them will tell us that the complement of the tangents has to be convex.

Meow Mix

So, draw out $x^2$ on paper.

Akiva Weinberger

7:45 PM

@MeowMix Thrice. $\pm1$ and $0$.

Meow Mix

Oh, heh

Actually, thrice

Akiva Weinberger

Derp edit

Dodsy

no worries I was about to correct you as well :)

Akiva Weinberger

But, whatever, they can't be polynomials

Meow Mix

Uhh

Well, they can't both be

Akiva Weinberger

7:46 PM

'cause if they were, their difference would have infinitely many roots, and the only such polynomial is 0

@MeowMix That's what I meant

Dodsy

so then we just need a similar function which is compressed horizontally by a certain factor

Meow Mix

@Dodsy Noooooo

Dodsy

Oh

Meow Mix

Draw $y=x^2$ out on paper

Dodsy

Done!

Akiva Weinberger

7:47 PM

You want the two curves to intersect lots of times

Meow Mix

Now, I want you to try and draw a function that intersects infinitely many times with it.

In order to do that, your other function will sort of have to "interweave" between $y=x^2$

So draw that out

And make sure the derivative is everywhere increasing

Dodsy

What about like x^2 + 2 is that not allowed

Meow Mix

That wouldn't intersect infinitely many times

Infact, never at all

Dodsy

Hm

Meow Mix

Draw it out

Dodsy

7:50 PM

So what it will be like a horizontal sine curve or something?

Meow Mix

Hmmm, perhaps

Dodsy

Vertical

Meow Mix

But we want it to interweave with $y=x^2$

How do we make it to that?

Akiva Weinberger

@MeowMix Here's a topology problem, involving path-connectedness and (what I can only assume would he called) polygon-connectedness

The circle $S^2\subseteq\Bbb R^2$ is path-connected, but not polygon-connected,

Meow Mix

@Dodsy I was thinking something like this:

See how it "interweaves"?

Akiva Weinberger

7:53 PM

because no two points can be joined by a sequence of line segments contained in the circle.

Define a space to be polygon-connected if any two points can be joined by a finite sequence of line segments.

@MeowMix So, clearly, the circle example shows that path-connectedness is not the same as polygon-connectedness. Can you show that, for open subsets of $\Bbb R^n$, that they're equivalent?

Meow Mix

So wait

Huh?

Akiva Weinberger

That is, for any path-connected open subset of Euclidean space, you can join any two points with a polygonal path.

Meow Mix

I can join two points in a circle with line segments

Unless you mean the actual circle

Akiva Weinberger

@MeowMix Not line segments that stay inside the circle.

Meow Mix

Not the disk

Akiva Weinberger

7:55 PM

The circle, not the disk

The disk would be polygon-connected

Also, I wrote $S^2$ where I meant $S^1$

Meow Mix

I'm going to think about it during the 5th annual nap of Zach

Akiva Weinberger

That's… not a whole lot of naps

Meow Mix

hmm

One second

trihectosexdecaquintannual nap

Why did I bother finding those prefixes when I could have just said "daily"

Anyways, off for my trihectosexdecaquintannual nap.

SteamyRoot

uhhh

It sounds really weird when you're mixing latin and greek numerals

Astyx

8:12 PM

Hi Zach, Akiva

Alessandro Codenotti

hi @astyx

Astyx

Hi again @Alessandro

Alessandro Codenotti

Can I ask you another probability thing?

Astyx

You can always ask, I'm not sure I'll be able to answer

Alessandro Codenotti

I have a random variable $X$ which has a standard normal distribution and another variable $Y$ defined as $Y=\begin{cases} X\quad \text{ if }-1\le X\le 1\\ -X \quad \text{ if }|X|>1\end{cases}$. I'm asked to show that $Y$ is also a standard normal and then to discuss $X+Y$

I showed that $Y$ is standard normal by showing that $\Bbb P(Y\le t)=\Bbb P(X\le t)$ and that part is fine

however for $X+Y$ I think I get a variable which is $0$ on $(-\infty,-1)\cup (1,+\infty)$ while a sum of Gaussians should be a Gaussian

Astyx

8:18 PM

Is it not : "A sum of independant Gaussians should be Gaussian" ?

Alessandro Codenotti

ohhh, of course

derp

Astyx

:)

Zophikel

I'm having trouble with this consider the sequence: $S_{n+1}=\frac{1}{3}(s_n+1) for \, n \geq$ $S_1 = 1$. I essentially have to prove s_n > \frac{1}{2}

I'm attempting this through induction

I've got so far $P(N) = S_{1+1} = \frac{1}{3}(1 + 1) \, S_{2} = \frac{1}{3}(2)$ $\frac{2}{3} > S_n$

After establishing my base case I attempted do this for $P(n+1)$

$P(n+1) = S_{n(n+1)} = \frac{1}{3}(S_{n+1}+1) for \, (n+1) \leq 1$

^ Is this correct

I Feel like something is missing here

Astyx

Is that a big $S_n$ instead of $s_n$ in the recursion formula ?

Zophikel

@Astyx yes

@Astyx sorry for the bad latex

Astyx

8:22 PM

And third line is $P(2)$ ?

Zophikel

@Astyx yes I belive so

Astyx

What is $P$ ?

Zophikel

$P(N)$ is the base the case

$P(n+1)$ is our extended case

Astyx

What do you mean by that ?

Hi @Liad

Zophikel

@Astyx since $P_n$ is true then it should follow that $P_{n+1}$ is true

That's essentially what I established there

Astyx

8:25 PM

But what is $P_n$ ?

Zophikel

The base case @Astyx do you want me to relatex this

@Astyx see here:

I'm having trouble with this consider the sequence: $S_{n+1}=\frac{1}{3}(s_n+1) for \, n \geq$ $S_1 = 1$. I essentially have to prove $s_n > \frac{1}{2}$

I'm attempting this through induction

I've got so far as $$P(n) = S_{1+1} = \frac{1}{3}(1 + 1) \, S_{2} = \frac{1}{3}(2) \, \, =\frac{2}{3} > S_n$$

After establishing my base case I attempted do this for $P(n+1)$

$$P(n+1) = S_{n(n+1)} = \frac{1}{3}(S_{n+1}+1) for \, (n+1) \leq 1$$

^ Is this correct

Astyx

@ShaVuklia I believe what they meant by this $=\infty$ is simply that the sequence diverges. It is indeed a typo, it should be $\lim_{n\to \infty}ax^{n+1}$. However you said something wrong : we do not need the limit not to equal 0, we need the sequence not to converge (it could converge to 2 and that would make our statement false)

Liad

@Astyx hi

Astyx

@Zophikel I think you are misunderstanding what $P(N)$ is, especially since you are writting $P(n) = S_{1+1}$. First of all what is $n$ ?

Liad

someone here familiar with the $K-$topology ?

Astyx

8:30 PM

No I

Liad

Where is Ted :P

Zophikel

@Astyx sorry for the typo but essentially the for all n, n+1 would be true. also $n \geq 2$ should be in the top of the post sorry

Astyx

@Liad Probably hiding from you ... :p

Liad

hehe

Astyx

That does not answer my question @Zophikel. Can you explain to me how induction works ?

Liad

8:32 PM

well i got to go , see ya @Astyx

Astyx

Bye !

Zophikel

@Astyx yes

Induction works by establish a case where P_n would be true and then it should following that P_(n+1) is true

Speaking form intuition it's like domino's falling

Astyx

And ..?

You're missing something very important

Zophikel

@Astyx what i'm I missing

Astyx

You need the base step

Zophikel

8:34 PM

@Astyx ahh ok

@Astyx I actually established my base case before performing the inductive step

Astyx

So for instance you have a property $P_0$ and you know that for any $n\in\Bbb N$, $P_n \implies P_{n+1}$, induction allows you to conclude that for all $n\in\Bbb N$, we have $P_n$

user243301

@RubénBallester in his web page Fernando Chamizo (Universidad Autónoma de Madrid) has a lecture notes on topology, I hope that it help you. Study and learn mathematics is very difficult. Is not required a response of this message, good luck.

Astyx

So what is your base step here ?

Zophikel

@Astyx the base step is this:$$P_1 = S_{1+1} = \frac{1}{3}(1 + 1) \, S_{2} = \frac{1}{3}(2) \, \, =\frac{2}{3} > S_n$$

Astyx

No, we don't have $P_1 = S_{1+1}$

In fact $P_n$ is the property $S_n \gt {1\over 2}$

Balarka Sen

8:38 PM

Hi, bye, hi @Akiva.

Zophikel

@Astyx all right

Astyx

^

So what is $P_1$ ?

Zophikel

@Astyx I think I see the initial mistake

@Astyx would it be $P_1 = S_{n+1} = \frac{1}{3}(1+1)$

?

Astyx

No, reread what I wrote two messages ago

Zophikel

$P_1 = S_{2} = \frac{2}{3} \, > S_n$

@Astyx sorry latex error

Astyx

8:42 PM

Still not

Zophikel

@Astyx what is it

Astyx

Just that $S_1 = 1 \gt {1\over 2}$

Zophikel

@Astyx how did you get that ?

Astyx

Get what ?

Zophikel

$S_1 = 1 \gt {1\over 2}$

Astyx

8:45 PM

Well you wrote that $S_1 = 1$. And obviously $1 \gt {1\over 2}$

Zophikel

ok nvm sorry ]

Ted Shifrin

Salut, @Astyx.

Astyx

Salut Ted

Liad te cherchait quelques messages plus tôt

Pour parler de K-topologie ou quelque chose comme ça

Ted Shifrin

Quoi que ce soit ...

Astyx

Sinon tu vas bien ?

Ted Shifrin

8:48 PM

Il ne faut plus me demander ça ...

Astyx

Comment ça ?

Ted Shifrin

Parce que je suis toujours malade ... :(

Astyx

Tu as vu ton médecin ou toujours pas ?

Ted Shifrin

Oui. On m'a dit d'attendre ...

Astyx

C'est assez bon signe non ?

Ted Shifrin

8:51 PM

Beaucoup de monde me disent qu'il faut attendre 2 ou 3 mois ... :(

Astyx

Ne laisse pas ça abattre ton moral en tout cas ! On est toujours là nous :)

Zophikel

@Astyx for reworking it out $P_{n+1} \, => S_{n(n+1)} = \frac{1}{3}(S_{n+1}+1) for \, n \leq 1$ S_{n+1} > \frac{1}{2}

Ted Shifrin

LOL, merci.

Mike Miller

Isn't Liad in a point set topology class?

Ted Shifrin

yup

Astyx

8:53 PM

Tu passes toujours en Juin ?

Ted Shifrin

Any luck finding contact solution, @MikeM?

Mike Miller

Why the hell are they doing K-spaces?

Ted Shifrin

Oui, @Astyx.

Mike Miller

Nope. RIP contact.

Ted Shifrin

No, no. I assume this is the compact complement topology or something, @MikeM.

Mike Miller

8:54 PM

I had to squint at some of the talks today.

Astyx

@Zophikel That's not the way to go. You need to suppose $P_n$, not $P_{n+1}$ for a start

Zophikel

@Astyx note that the n in $n \leq 1$ should $(n+1) \, \leq 2$ and finally this initially implies that $S_{n+1} > \frac{1}{2}$

Mike Miller

@Ted Ok, much better.

Zophikel

@Astyx all right sorry my mistake

@Astyx so to fix my solution we start from $P_n$ and then do $P_{n+1}$ correct sorry for the mistakes I have a hard time chatting about this

Astyx

Super @Ted

Balarka Sen

8:55 PM

I can't see the blackboard from any magnitude of distance unless I'm smashed to it

but eh, whatever. glasses are heck of a pain

Ted Shifrin

I realized when I got to UGA at the age of almost 30 that I was getting headaches trying to read the blackboard during someone else's lectures. So I went to an opthalmologist and got glasses.

Balarka Sen

i don't wear mine... maybe i should. don't want my eyes to break or something

probably

Astyx

The less you use your glasses the better your eyes stay, IMHO

Balarka Sen

orly?

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$Mathematics$ Mathematics