Mathematics - 2017-06-15 (page 7 of 9)

@TheGreatDuck If $(X,\tau)$ is a topological space, then it's homeomorphic to the diagonal (with subspace topology) of $X \times X$ (with product topology)

TheGreatDuck

uuh

sounds nice

Akiva Weinberger

@TheGreatDuck Ale said it's possible for the diagonal of $X\times X$ not to be homeomorphic to $X$. This turned out to be false.

Astyx

@SteamyRoot Is that always true ?

TheGreatDuck

umm...

ok

Akiva Weinberger

@Astyx Yeah

TheGreatDuck

8:08 PM

(I have no clue what any of that just meant)

Astyx

(Sorry if I don't follow the conversation too well, I can't see from time to time)

TheGreatDuck

@AkivaWeinberger here's a weird concept. Noneuclidean matrices.

grids of numbers on surfaces other than flat planes

like a number cube but only the surface

or a number sphere

Alessandro Codenotti

@AkivaWeinberger that's what happens when you check only one example and it's wrong :P

Astyx

@TheGreatDuck You can already do this in some sense with Manifolds (things that locally look like $\Bbb R^n$)

user379685

There are three bus companies A, B and C that transport 2000 people from X to Y. Company A takes 30% of them, B takes 35% and C the rest. How many routes does the company A need to do so that the chance of rejecting the client is less than 1%. Buses of company A can hold 40 people.

Astyx

8:12 PM

If that's what you mean

TheGreatDuck

no i meant matrices

the same algebra as matrices

but shaped like spheres or cubes or surface geometries.

Astyx

I'm not sure I follow

TheGreatDuck

you have a matrix

SteamyRoot

That sounds like restriction a tensor to one of its indices being either $0$ or maximal

TheGreatDuck

now imagine one that was a sphere of numbers

Astyx

8:14 PM

Ah ...

So instead of a table of numbers you have a dice or something else ?

TheGreatDuck

well dice with several numbers on each side

but yeas

@SteamyRoot what about a circle of numbers?

Akiva Weinberger

You'd need a good way to spread your numbers onto the surface of the sphere

Maybe vertices of Platonic solids or some such

Astyx

That would be an map from a partition of your object to the natural numbers, I don't know of anything like it though

TheGreatDuck

hmmm

SteamyRoot

A circle of numbers just sounds like a vector with indices mod $n$

TheGreatDuck

8:16 PM

@SteamyRoot not a ring of numbers. A filled circle.

Akiva Weinberger

Or a function from $\Bbb Z_n$ to $\Bbb N$ I guess?

SteamyRoot

A disk, you mean.

TheGreatDuck

yeah

Akiva Weinberger

In the same sense that a usual vector is a function from $[n]$ to $\Bbb N$

user379685

hello, anyone?:c

Akiva Weinberger

8:16 PM

@user379685 What have you tried?

Astyx

Or $D \to \Bbb N$ or $\text{anything} \to \Bbb N$ more genrally

SteamyRoot

Dunno, you'd have to specify how these numbers lie with respect to eachother

Akiva Weinberger

I'd probably start by seeing how much "30% of them" is @user379685

TheGreatDuck

@user379685 you have provided no context. Therefore, we will ignore you.

Akiva Weinberger

Well, not context per se, but at least tell us what you've done so far.

user379685

8:18 PM

Well, I tried using the Central Limit Theorem but i got 1600 which is waay to many

Akiva Weinberger

Ohh

TheGreatDuck

you have a bad history of no context posts

why should we help you?

Akiva Weinberger

Oh, you know what, I'm gonna duck out of this one

I don't actually know statistics

Sorry!

TheGreatDuck

i probably could help but i refuse to help on principle.

sorry!

Akiva Weinberger

You might as well type out what you've tried with the Central Limit Theorem

Incidentally, do you know how to turn on MathJax?

That's the sort of thing that turns $a^b$ into actual exponents for us

(There's a link in the room info on the top-right corner)

tinyurl.com/cfqcvpc

user379685

8:28 PM

$P(S_n > 600) < 0.01$
$P((S_n - n*0.3)/(sqrt9(n)*sqrt((3/10)*(7/10)))>((600-n*(3/10))/((sqrt(n)*sqrt((3/10)*(7/10))) < 0.01$
$P(Z>x_o) < 0.01$
$1-P(Z<x_o) < 0.01$
$0.99 < P(Z < x_o)$
$0.99 < Φ(x_o)$
$2.33 < x_o$
$((600-n*(3/10))/((sqrt(n)*sqrt((3/10)*(7/10))) > 2.33$
$n=1849$

TheGreatDuck

use English

Akiva Weinberger

Let me fix the formatting

TheGreatDuck

@user379685 where is your explanation of those lines, hmm?

Akiva Weinberger

What's sqrt9(n)?

user379685

whops that was a mistae

Akiva Weinberger

8:29 PM

$\sqrt{9n}$ or $\sqrt[9]{n}$?

user379685

theres no 9

Akiva Weinberger

Oh, OK

TheGreatDuck

please revise your post before posting....

:p

Akiva Weinberger

$P(S_n > 600) < 0.01$
$P\left(\dfrac{S_n - 0.3n}{\sqrt{n}\sqrt{(3/10)*(7/10)}}>\dfrac{(600-(3/10)n}{\sqrt{n}\sqrt{(3/10)*(7/10)}}\right) < 0.01$
$P(Z>x_o) < 0.01$
$1-P(Z<x_o) < 0.01$
$0.99 < P(Z < x_o)$
$0.99 < Φ(x_o)$
$2.33 < x_o$
$\frac{(600-n*(3/10)}{(\sqrt n*\sqrt{(3/10)*(7/10)}} > 2.33$
$n=1849$

Alessandro Codenotti

@Akiva take an $n$ dimensional topological vector space $V$, are all $n$ dimensional subspaces of $V\times V$ homeomorphic to $V$? It sounds reasonable

user379685

8:32 PM

the second line the probabilty is of the whole thing not just the numerator

yeah that's correct

Akiva Weinberger

$P(S_n > 600) < 0.01$
$P\left(\dfrac{S_n - 0.3n}{\sqrt{n}\sqrt{(3/10)\cdot (7/10)}}>\dfrac{600-(3/10)n}{\sqrt{n}\sqrt{(3/10)\cdot(7/10)}}\right) < 0.01$
$P(Z>x_0) < 0.01$
$1-P(Z<x_0) < 0.01$
$0.99 < P(Z < x_0)$
$0.99 < \phi(x_0)$
$2.33 < x_0$
$\dfrac{600-(3/10)n}{\sqrt n\sqrt{(3/10)\cdot(7/10)}} > 2.33$
$n=1849$

user379685

well obviously that's wrong, could you help?

Akiva Weinberger

Where did 2.33 come from?

@TheGreatDuck (I hope your moral qualms are quelled since they posted their work) Do you know this stuff?

TheGreatDuck

@AkivaWeinberger i don't actually know statistics at this level. I'm in high school!

user379685

from tables for the N(0,1)

TheGreatDuck

8:38 PM

:-P

Akiva Weinberger

Yeah, OK, I don't really know this stuff either

Sorry!

TheGreatDuck

sometimes I read about things and talk about them to sound smart, but tbh I don't know much advanced math.

Astyx

Why is probability so ... tricky ?

Alessandro Codenotti

I'll have a probability exam in July... definitely not looking forward to that one

Akiva Weinberger

@Astyx Provability > probability :P

(Almost every time I try to write "provably" on my phone, it autocorrects to "probably")

TheGreatDuck

8:53 PM

@Astyx because it is evil incarnate along with calculus and geometry.

Akiva Weinberger

Hey, what's wrong with calculus?

TheGreatDuck

idk

i just don't like much math really

Astyx

Great choice of chat then :p

I'll go to bed now, and hope a night's sleep will solve my eye problem

Cya

Akiva Weinberger

What time is it where you are, 10pm?

Astyx

11

Akiva Weinberger

9:05 PM

Oh

Astyx

I wish I could go to bed at 10 :(

Akiva Weinberger

Yeah sleep

TheGreatDuck

@AkivaWeinberger i mostly come here because I have nothing better to do.

Akiva Weinberger

Eat, drink, piss

TheGreatDuck

that's not better

Daminark

9:26 PM

Hai everyone!

5

Fargle

Sup nerd

Semiclassical

@LeakyNun this turned out to be a prophetic comment

Akiva Weinberger

@Semiclassical Want me to tell you the flat ordered pair thing? Or do you still want to think about it

Daminark

How goes it @Fargle?

Semiclassical

Not right now

Fargle

9:29 PM

@Daminark Alright. Listening to the Hotline Miami soundtrack so I feel lucid and yet strangely violent. You?

Daminark

I just woke up like, basically now

Semiclassical

Actually I was curious

Daminark

shrugs This sleeps schedule is like, not sustainable for long

Semiclassical

How sure are we that type is equivalent to rank in the senses described?

Fargle

yeah I woke up about an hour ago with some serious sleep paralysis

Daminark

9:31 PM

Kek

Semiclassical

I need to avoid taking naps during the day

I tend to end up feeling really tense/anxious after

Akiva Weinberger

How is type defined?

Daminark

Eek

Initiating homotopy type theory

Semiclassical

I dunno. But that's the word used in the Wikipedia bit on Quine-Rosser ordered pairs

Akiva Weinberger

Oh, so you saw how it's defined?

Semiclassical

9:35 PM

I saw it described as being type-level

I didn't really follow the definition though

I'm mostly paranoid about whether type-level implies flat in your sense

Akiva Weinberger

What's your definition of type-level though

Daminark

This sounds odd but, I've been running wild and free

Can I ransack someone's city?

Fargle

@Daminark Yeah sure, I've been needing an excuse to move

Akiva Weinberger

Wat

Semiclassical

Again, I don't know. I am literally saying that, since the portion on Wikipedia is not in terms of rank, I can't assess whether or not the Quine-Rosser definition is flat

Daminark

9:41 PM

Thx

Semiclassical

All I can tell from Wikipedia is that it's type-level. That sounds like flatness, but I can't draw such a conclusion just from that

Daminark

Fargle, being ransacked

Fargle

Temba, his arms wide.

Akiva Weinberger

Bill Wurtz?

@Semiclassical It's hard to draw conclusions from words you don't know their meaning