Mathematics - 2017-10-12 (page 6 of 9)

Secret

15:02

So if $c$ is rational, a finite string in $c$ must be fixed by countably many multiplications of $2$ on $c$, that is, the number of fixed points of $c$ under the action of $2\times$ is finite

MatheiBoulomenos

note that we also need to use that $g$ and $h$ are of degree $>1$, since $f$ has no roots in $K$

Secret

But the the question is, how does one check the orbit induced by countable actions of a magma on a set

MatheiBoulomenos

or else this argument could fail

Leaky Nun

@MatheiBoulomenos hast’i mehr aoyfgaben?

MatheiBoulomenos

interesting spelling

Leaky Nun

15:07

yiddisch

Secret

I wonder what function will allow me to zoom into the next layers of a continued fraction

MatheiBoulomenos

@LeakyNun try to contruct a finite extension that is not simple

Well, maybe start with a finite extension that is not separable

Secret

Hmm....

$c=[i;a_1,a_2,...]$

$\frac{1}{c-i}=[a_1;a_2,a_3,...]$

So the required function is:

$$\frac{1}{c-\lfloor{c}\rfloor}$$

Lucas Henrique

yo.

Secret

Therefore, given $c$, $\frac{1}{\cdot - \lfloor \cdot \rfloor}^n (c) = c_n$

Lucas Henrique

15:22

How do I prove that $|\{(x,y): (x,y) \in \Bbb Q^2 \land x^3 + y^3 = 9\}| = \infty$?

Secret

where $c_n$ is the nth layer of the continued fraction

So for rationals, $n$ is finite

hmm...

Let $k=\frac{1}{e^e-\lfloor e^e\rfloor}$

$\ln k = -\ln (e^e-\lfloor e^e\rfloor)$

$\ln k = -\ln (e^e - 15)$

$ke^e - 15k -1 = 0$

Let $k=\frac{1}{e-\lfloor e\rfloor} = \frac{1}{e-2}$

$\ln k = -\ln (e-2)$

$ke-2k-1=0$

...

Let $k_0=e, k_1 = \frac{1}{e-\lfloor e\rfloor} = \frac{1}{e-2}$

$k_1e-2k_1-1=0$

Let $k_2 = \frac{1}{k_1-\lfloor k_1\rfloor} = \frac{1}{k_1-1}$

$k_2k_1-k_2-1=0$

$k_2 (\frac{-e+3}{e-2})=1$

ok this is getting nowhere, clearly thinking about this mechanically is not going to work, I need to think about this more globally...

Faust

15:53

taco

Semiclassical

@LucasHenrique that'd be equivalent to $x^3+y^3=9z^3$ having infinitely many integer solutions. kinda interesting that that's doable but $x^3+y^3=z^3$ isn't.

(it's not equivalent, since 9 isn't a perfect cube)

Faust

Morning Semi

TastyRomeo

I can spot (2,1,1)... hmmm

Semiclassical

morning

Faust

or Semi Morrning?

TastyRomeo

15:59

There should just be some easy family that satisfies it, I guess

Faust

I swear doing too much math makes you insane

Semiclassical

either

Oskar Tegby

Okay. I have a nooby one for ya'll. If you have a bookshelf with 4600 centimeters of books, and the width of the books is normal distributed as N(1.8,0,7). What is the probability that you have more than 2500 books?

Semiclassical

that sounds like a central limit theorem question

Faust

i dont undestand the notation FTW

Oskar Tegby

16:01

Really?

Faust

EGG NOG IS SO GOOD

Semiclassical

should that be N(1.8,0.7) ?

Oskar Tegby

The average width is 1.8 centimeters, and the standard deviation is 0.7, yeah.

Faust

that makes more sense

Oskar Tegby

Hence, we consider N(1.8,0.7), or am I wrong?

Semiclassical

16:02

is that std dev = 0.7 or var =0.7 ?

wikipedia has the notation as $\mathcal{N}(\mu,\sigma^2)$

TastyRomeo

So there's a tiny chance of having a book with negative width :^)

Faust

i havent done stats in 10 years

Oskar Tegby

Oh.. Yeah, 0.7 is the std dev.

Leaky Nun

@MatheiBoulomenos of what field?

Oskar Tegby

So, rather N(1.8,0.49) then.

Semiclassical

16:03

right

Faust

@LeakyNun good morning

Semiclassical

worth knowing if you have a reference text

Oskar Tegby

Yeah. Well, we haven't solved it quite yet

Semiclassical

right

MatheiBoulomenos

@LeakyNun any field you want

But not all fields have such extensions

Oskar Tegby

16:06

So if we have $Y=X_1+\dots+X_n$ then where each $X_i\sim N(1.8,0.49)$ then $Y\sim N(1.8n,0.49n)$, am I right?

Semiclassical

hmm

that feels backwards, unfortunately.

it's not that you know how many books you have and you want to get the predicted height

Oskar Tegby

Okay. So if $Y=4600$, we want to find out the probability that $n\geq2500$.

Semiclassical

you've got the height and you want to predict the number of books

Leaky Nun

@Faust good morning

Faust

|Get enough sleeps?

Oskar Tegby

16:07

Right.

Faust

i hope u like a camel

Leaky Nun

sure

Faust

built it all up when u were a kid or something

Oskar Tegby

So, where do we start instead?

Semiclassical

that's what I'm not seeing. it seems like you'd need a probability distribution on the number of books

Faust

16:08

how old is leaky?

Oskar Tegby

It can't be this one. Here, we let $n$ go to $\infinity$.

$\infty$

lol

Semiclassical

i suspect i'm missing something obvious

Oskar Tegby

Law of large numbers

In probability theory, the law of large numbers (LLN) is a theorem that describes the result of performing the same experiment a large number of times. According to the law, the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed. The LLN is important because it "guarantees" stable long-term results for the averages of some random events. For example, while a casino may lose money in a single spin of the roulette wheel, its earnings will tend towards a predictable percentage over a large...

Me too.

Semiclassical

what i'm thinking is about the problem in reverse

you start with 4600 cm's of books, and you start pulling them off the shelf

Oskar Tegby

Yeah? We know the number of books, but not the length.

Okay.

Faust

16:10

Alright i have to go do that think at the school with the learning and stuff. ill see you guys laterz =)

Semiclassical

so the first book you remove would lower it on average by 1.8, the next by 1.8 again, etc

Oskar Tegby

Keep it real, @Faust

Yeah

Semiclassical

until you eventually pull all the books off the shelf

Oskar Tegby

Dead sure

Semiclassical

what I find a bit uncomfortable is the end of that process

i.e. you need the length of the last book you remove to be a normal random variable as well

Oskar Tegby

16:12

Yeah, so this method isn't what we want.

Semiclassical

well, is it true that $4600-\sum_{n=1}^{N-1} X_n$ is a normal random variable if the $X_n$'s all are?

i should think it is.

Oskar Tegby

Well, what are the criteria for a random variable? I think they hold.

Semiclassical

yeah. what i'm not sure is the variance/mean

mean should be 4600-1.8(N-1), I guess

bleh. this doesn't seem to work either

how I'd want this to proceed is to say

keep removing books until the sum exceeds 4600, and take the number of books removed as the random variable

I'm not convinced that gives the answer perfectly right, but it shouldn't be far off the mark

Oskar Tegby

This feels like a pretty standard probability question, there must be some thread on MSE about it. I can't find any, but still.

MatheiBoulomenos

@LeakyNun I'll give you a hint: characteristic $0$ won't do it. Finite fields won't either

Oskar Tegby

16:19

Perhaps I should start one. Worst case scenario is that it's a duplicate.

Semiclassical

I vaguely feel like the problem doesn’t make sense as written

Leaky Nun

@MatheiBoulomenos :o

Oskar Tegby

Okay. I can quote it word by word.

Semiclassical

Like it should be that the width doesn’t exceed 4600

But that seems wrong too

Oskar Tegby

"John has 46 bookshelf meters of books in his library. The width of the books can be seen as independent random variables with expected value 1.8 cm and standard deviation 0.7 cm. Determine approximately the probability that John has more than 2 500 books!"

Semiclassical

16:22

Hmm. “Approximately” is a big hint there, I think

Balarka Sen

Hi @Ted

Oskar Tegby

Yeah, me too.

Semiclassical

Hmm

Maybe one way to start here: how would One formulate the (absurdly small) probability of him having one book?

Oskar Tegby

Yeah

Semiclassical

Then you’ve just got one random variable and it’s easy

How do things change if we make it two...

Oskar Tegby

16:28

How, exactly, do we compute the probability that he has one book?

We have the random variable for the book. What's the probability of it being exactly 4600 centimeters wide?

Semiclassical

right. so that'd be (4600-1.8)/0.7 = many many standard deviations from the mean

Oskar Tegby

Why?

Semiclassical

I think the issue is really this: If $\sum_n X_n=4600$, are these random variables independent?

Oskar Tegby

Yes

Semiclassical

yes that's the issue, or yes they're independent?

Oskar Tegby

16:31

They are independent, sorry for the unclear answer.

Semiclassical

hmm

I think my concern is how $4600-\sum_{n=1}^{N-1}X_n $ is distributed

Oskar Tegby

I don't get why you got (4600-1.8)/0.7 from. We were supposed to get a probability.

Semiclassical

that's the number of standard deviations 4600 cm is from the mean value of 1.8 cm

since 0.7 cm is the standard deviation

Oskar Tegby

All right

Semiclassical

main point is that, while that answer is well-defined, the resulting probability is obviously way too small to be relevant.

Oskar Tegby

16:34

Yeah, but few books are 46 meters wide indeed.

Semiclassical

yup

so now we go to the two book case.

and here's where I get unsure. suppose X1 ~ N(1.8,0.7^2).

If X1+X2=4600, then X2 = 4600-X1

Oskar Tegby

Right

Semiclassical

the mean value of X2 is then <X2> = 4600<1>-<X1> = 4600-1.8 and the variance is <(X2-<X2>)^2> = <(X1-<X1>)^2 = 0.49

so X2 isn't distributed as N(1.8,0.7^2) if X1 is.

that makes me realize another issue, come to think of it.

suppose we have a single random variable X~N(0,1). what's the probability that x=1 ?

Oskar Tegby

0

Semiclassical

right.

we never talk about the probability of a single value, but the probability of a range

Oskar Tegby

16:41

I've thought about that as well.

Semiclassical

The interpretation I want to give this is that we should be asking for the probability that X1 is at least 4600

Oskar Tegby

That's not what they are asking.

Maybe, we can try to find that first.

Semiclassical

Hence why I say I want to say that :/

Oskar Tegby

Okay. Let's go for the easier one first.

Semiclassical

kk

Suppose we've got $N$ random variables $X_n \sim \mathcal{N}(1.8,0.7^2)$. What's the probability that $\sum_{n=1}^N X_n\geq 4600 ?$

by the central limit theorem, this sum should be distributed as $\mathcal{N}(1.8N,0.49N)$

Oskar Tegby

16:46

Yes

Semiclassical

so 4600 would correspond to $z=(4600-1.8N)/(\sqrt{0.49N})$ std deviations

and we'd want to compute $\int_z^\infty \Phi(z)\,dz$

...hrm. I'm not liking where this is heading

Oskar Tegby

Integrating the distribution function for the normal distribution, many attempts die that way.

Semiclassical

I think what I should do is take $N=2500$ specifically

not sure i'm right on that, but let's at least see what that gives

Oskar Tegby

Sure!

Semiclassical

z=(4600-1.8*2500)/sqrt(0.49*2500) = 2.857

Leaky Nun

16:51

Poll: "combinatorical interpretation" vs "combinatorial interpretation"

Semiclassical

second

though I think i'd do "combinatoric interpretation", so no -al

Oskar Tegby

2

I'll be back soon. I've got to change place in uni, they're closing the library those fuckers.

PVAL-inactive

All of those options seem immensely popular.

Leaky Nun

@Semiclassical google search doesn't like "combinatoric"

PVAL-inactive

all 3 break arxiv.

user2154420

16:56

If a cover is compact, is it finite sheeted?

Balarka Sen

yes

preimage is a discrete set; discrete subsets of compact spaces are finite

user2154420

Always? I assume we need some assumptions on the base space...

Simply Beautiful Art

Hopefully this isn't a repost

13

Q: $a_n$ is the smallest positive integer number such that $\sqrt{a_n+\sqrt{a_{n-1}+...+\sqrt{a_1}}}$ is positive integer

An infinite sequence of pairwise distinct numbers $a_1, a_2, a_3, ...$ is defined thus: $a_n$ is the smallest positive integer number such that $\sqrt{a_n+\sqrt{a_{n-1}+...+\sqrt{a_1}}}$ is positive integer. Prove that the sequence $ a_1, a_2, a_3, ... $ contains all positive integers numbers. ...

sequences-and-series

Balarka Sen

I guess you need Hausdorffness of the base

MatheiBoulomenos

@LucasHenrique I did some searching regarding your question. It seems to be related to elliptic curves. Note that a rational solution to $x^3+y^3=9$ is equivalent to a integer solution to $x^3+y^3=9z^3$. In this post there is a method to generate infinitely many solutions starting from a single solution: math.stackexchange.com/a/475562/348926

Balarka Sen

16:58

'cuz you want points to be closed to take preimage and say the preimage is closed

MatheiBoulomenos

you just need an initial solution, you can take $x=1$ $y=2$ $z=1$

Balarka Sen

closed discrete subsets of compact spaces are finite :)

MatheiBoulomenos

if you only need closedness of single points, then $T_1$ is sufficient

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