Mathematics - 2017-07-25 (page 5 of 5)

Typhon

9:00 PM

The irony if it is one of ted's.

Ted Shifrin

nope, it's not.

Typhon

shrugs

Ted Shifrin

I've never lectured a differential equations course.

Anonymous

Lecture 1 - Introduction to Ordinary Differential Equations (ODE)

►

PVAL-inactive

@Danu I'll be sure to try those if I get a chance.

Anonymous

9:01 PM

Around 21:54

Anonymous

You might find the accent a bit weird

Ted Shifrin

Oh, so the equations he's solving there are exactly the ones I gave you for the envelope. He's using parameter $c$ where I had $t$.

He's thinking of the envelope of the set of general solutions as the singular solution.

I've heard plenty of Indian accents.

Anonymous

@TedShifrin Aha, that makes sense

Ted Shifrin

So if I had written down those lines as solution curves of an ODE, he's saying that the circle (the envelope) would also be a solution, but he's calling it a singular solution.

Anonymous

@TedShifrin Did you have Indian students in your class?

Ted Shifrin

9:04 PM

Some, @Blue, and I've known plenty of Indian mathematicians.

Typhon

which kind?

Ted Shifrin

Exercise for you: Figure out the differential equation that gives our first example.

Typhon

American Indian or Asian Indian?

Ted Shifrin

Asian Indian.

Anonymous

@TedShifrin Sure. Trying!

user193319

9:05 PM

Do there exist any continuous surjections from the integers to the rationals, where both are endowed with the order topology?

mick

Tommy1729 made a prophecy , I would Get 4K Some day !! Now I have 3999 rep aaahhhh :)

Typhon

@mick not anymore

mick

Did you upvote me Typhon ?

Typhon

yup

i was browsing and was in your post when you posted here

mick

Thanks for fullfilling the prophecy :)

SteamyRoot

9:07 PM

So, @mick, decided what you're doing next year?

Alessandro Codenotti

@user193319 yes, plenty, because of silly reasons. Think about it

Mike Miller

@PVAL-inactive I'm also fond of Uigedael which I'm sure I misspelled.

Typhon

@mick no problem.

it was a self-fulfilling prophecy!

mick

Haha

Maybe I am a worthy successor of my master now :)

Typhon

I prophecy that one day you will get to rep amount 6379

mick

9:09 PM

WoW that is alot

Akiva Weinberger

@Typhon No reason to think that's true

Typhon

@AkivaWeinberger D:

are there any counter-examples?

SteamyRoot

Unless you downvote him to make sure he hits that exact amount... :P

Typhon

@SteamyRoot he will hit it exactly and not go down from above. It will be on his first pass.

Akiva Weinberger

It clearly has k=1 repetition because 1 appears more than once (11), it has k=2 repetition because 31 appears twice (…,13,15,… and …,31,…)

Typhon

9:10 PM

I just know this

intuition

mick

Im sure someone Will downvote , So I probably Go over it and Then wobble around it untill I Get it :)

Akiva Weinberger

@Typhon Alternate hint: stop trying to find counterexamples and start trying to prove there are none

Typhon

no i mean that it will be hit before going above

@AkivaWeinberger no thanks

I don't know these sorts of proofs.

I am literally incapable.

user193319

@AlessandroCodenotti Well, I know that if $f : \Bbb{Q} \to \Bbb{Q}$ is continuous, then the restriction of $f$'s domain will be continuous as well; but I need the restriction to be surjective.

Akiva Weinberger

OK, first hint again, pigeonhole

Typhon

9:12 PM

i know what pigeonhole is

mick

Typhon link to your problem ?

Typhon

you miss my point

Alessandro Codenotti

@user193319 which functions $\Bbb N\to\Bbb Q$ are continuous?

mick

I love pigeons :)

Typhon

@mick it's not my problem.

there is no link

SteamyRoot

9:12 PM

If you drill $n+1$ holes in $n$ pigeons, at least one pigeon will have at least $2$ holes drilled into it.

5

mick

Lol

Akiva Weinberger

@mick Gon' copy/paste it, one sec

mick

K

Akiva Weinberger

> Call a real number repetitive if for every k you can find a string of k digits that appears more than once in its decimal expansion. The problem is to prove that if a real number is repetitive then so is its square.

Typhon

@AkivaWeinberger if all numbers are repetitive, then what is the point of mentioning squares?

mick

9:14 PM

Similar to one of my OWN questions

Akiva Weinberger

@Typhon Please delete this

Typhon

@AkivaWeinberger why?

Akiva Weinberger

Don't spoil it please

user193319

@AlessandroCodenotti I am not sure. I would guess just constant functions--and that wouldn't have my cause!

Typhon

we literally just did two seconds ago

mick

9:15 PM

If a number is normal , Its square is normal .... right ??

Typhon

besides, I'm not spoiling. I don't know if it is or isn't.

Akiva Weinberger

Yeah but now other people are thinking about the problem

Typhon

im guessing

Akiva Weinberger

OK fine whatever

Alessandro Codenotti

@user193319 Why just constants? Pick the immersion of $\Bbb N$ into $\Bbb Q$, does this map seem continuous to you?

Typhon

9:15 PM

for all I know, there is a counter-example. XD

either way, proving it for all real numbers would be way harder than proving the implication.

user193319

@AlessandroCodenotti By 'immersion,' do you mean 'inclusion' or 'embedding'? If so, I would say this is continuous, but it isn't surjective.

Alessandro Codenotti

ok, but don't worry about surjectivity now

so not only the constant functions are continuous

user193319

@AlessandroCodenotti I agree.

Alessandro Codenotti

what's an example of a discontinuous function $\Bbb N\to\Bbb Q$?

Akiva Weinberger

What's an example of literally any discontinuous function with domain $\Bbb N$

Alessandro Codenotti

9:20 PM

that's where I was headed

Akiva Weinberger

Why don't you just say the thing

SteamyRoot

Well, if we're not working with a predetermined topology... ;)

Akiva Weinberger

Order topology was specified

Alessandro Codenotti

because I wanted him to work it out :P

user193319

I am blanking! I only think about continuous functions!

SteamyRoot

9:21 PM

What does continuous mean?

mick

Simplify cuberoot ( 1 + sin )

Akiva Weinberger

@user193319 Name a random function with domain $\Bbb N$

user193319

That the preimage of every open set is open in the domain space.

SteamyRoot

And what are the opens in $\mathbb{N}$ with the order topology?

Alessandro Codenotti

which subsets of $\Bbb N$ are open?

Akiva Weinberger

9:22 PM

and then ask yourself if it's continuous or not

user193319

All of them.

Alessandro Codenotti

Steamy is starting to snipe too

Typhon

@AkivaWeinberger I'm thinking that the square of a number with repeated n decimals has repeated n-1 decimals.

or ast least...

I think that is easy to prove.

user193319

Because the order topology is the same as the discrete topology on $\Bbb{N}$.

Typhon

but im not equipped to do so

sorry

Alessandro Codenotti

9:23 PM

@user193319 Ok, so pick a function $\Bbb N\to X$, with $X$ your favourite topological space, take a random open set in $X$, is its preimage open in $\Bbb N$?

user193319

Oh...Every function is continuous?

Akiva Weinberger

Yeah

Literally every function with domain $\Bbb N$ is continuous

user193319

And just to verify, $\Bbb{N}$ is locally compact, right? Because $\{n\}$ is compact and open?

Akiva Weinberger

In fact, every function with whose domain is discrete is continuous

Alessandro Codenotti

it's a standard question to classify continuous function with discrete domain, discrete codomain, trivial domain and trivial codomain

Akiva Weinberger

9:25 PM

@user193319 Yeah should be

Alessandro Codenotti

@user193319 yep

user193319

Then choose any $f : \Bbb{N} \to \Bbb{Q}$ is that surjective, $\Bbb{N}$ is locally compact but $f(\Bbb{N}) = \Bbb{Q}$ is not.

That's the counterexample I needed.

Akiva Weinberger

@AlessandroCodenotti Anything, locally constant (constant on connected things), constant(?), anything

Wait the third one is confusing

Alessandro Codenotti

yeah the third one is not as nice as the others

Akiva Weinberger

Yeah no it doesn't need to be constant

mick

9:27 PM

Goodnight friends :)

Adeek

@Daminark I was wondering what book are you using for complex analysis ?

user193319

An even easier counterexample would be the identity between $\Bbb{Q}$ and itself, where the domain is endowed with the discrete topology, and the range is endowed with the usual topology.

Akiva Weinberger

Wait no

Never mind

I had that backwards

@user193319 You'll find that that's the same example, actually :P

$\Bbb Q$ with the discrete topology is homeomorphic to $\Bbb N$ with the discrete topology

SteamyRoot

Any two discrete spaces with the same cardinality are trivially homeomorphic :P

user193319

@AkivaWeinberger Yes. You are right. Thanks everyone for the help!

Akiva Weinberger

9:32 PM

@AlessandroCodenotti For trivial domain, I want the thing to be constant up to indistinguishables I think

Alessandro Codenotti

hm, yeah I think it should be constants only with some mild assumptions on the codomain

Akiva Weinberger

What the hell are these

Matroid

In combinatorics, a branch of mathematics, a matroid is a structure that abstracts and generalizes the notion of linear independence in vector spaces. There are many equivalent ways to define a matroid, the most significant being in terms of independent sets, bases, circuits, closed sets or flats, closure operators, and rank functions. Matroid theory borrows extensively from the terminology of linear algebra and graph theory, largely because it is the abstraction of various notions of central importance in these fields. Matroids have found applications in geometry, topology, combinatorial ...

Weird generalizations of independence in vector spaces

Mike Miller

also of graphs

Alessandro Codenotti

never heard of them before

where does one encounter them?

Mike Miller

combinatorics when you're going real wet and wild

2

Alessandro Codenotti

9:38 PM

this question is how I discovered I have no idea whatsoever of what combinatorics actually is

Mike Miller

me neither but i understand it better than representation theory

Akiva Weinberger

@AlessandroCodenotti Combinating?

Hm, nice picture from that article:

Danu

There is this crazy stuff relating matroids to complex geometry, and particularly to some kind of analog of Kähler geometry

Akiva Weinberger

If $S$ is a subspace, then the subspaces $T$ covering $S$ (meaning they contain $S$ and there are no intermediate subspaces between $T$ and $S$) partition the complement of $S$

Gaurav Agarwal

What to do about a question that wasn't a duplicate but marked as one?

Danu

9:47 PM

Recently covered by Quanta magazine

Akiva Weinberger

(Proof: Straightforward)

@GauravAgarwal I'm not sure. What's the question?

Gaurav Agarwal

-3

Q: A polynomial intersecting the x-axis while not intersecting the x-axis?(Complex Numbers)

I know three questions (that gained momentum) that have been posted asking a question which seems the same, but answers to none of them answer the following very well. Please jump to point 2 & 3 for immediate addressing to the problem. Knowledge that I currently have: I was introduced to imagi...

Daminark

@Mike lol this reminds me, one of the lecturers in atop said that we only really know how to do linear algebra and kinda combinatorics, so in atop we try as hard as we can to reduce to those

Mike Miller

that's an extremely common catchphrase

Daminark

@Adeek officially we're using Titchmarsh Theory of Functions, but I'm kinda meh on that book. I've been meaning to look at either Narasimhan or this one book called Berenstein and Gay

Which kinda devotes chapter 1 to some material like differential forms, homology, partitions of unity, etc, which I'm hoping to pick up

Gaurav Agarwal

9:56 PM

@AkivaWeinberger Thanks I believe it was you who up-voted. I do not undesrtand however, that how do they become a 4-D thing

Akiva Weinberger

No that was not me

@GauravAgarwal First, why are graphs of real functions 2D?

Gaurav Agarwal

@AkivaWeinberger there is only one variable, and we define y as the output of the functions, thus making it 2-D

and hence, 'plottable' on the cartesian plane

Akiva Weinberger

The domain is $\Bbb R$, the set of real numbers, and the codomain is also $\Bbb R$. So we can describe every value of the function as a pair of two real numbers, like $(x,f(x))$

Now, for complex numbers,

we can represent every value of the function by a pair of complex numbers.

You know how you can visualize the real numbers on a line, pairs of real numbers as a plane, and triples of real numbers as 3D space? And you can represent complex numbers as a plane?

You can represent complex numbers as a plane because you can represent each one by a pair of real numbers

Now, a pair of complex numbers would be represented by a pair of… a pair of real numbers. Or a quadruple of real numbers.

Which we can represent geometrically on a 4D space.

In other words: Two dimensions are for the input (x-value), and two dimensions are for the output (y-value). @GauravAgarwal

Gaurav Agarwal

Okay, I seem to have wrapped up my mind about it. After hours, this is the first thing that has been easy to understand! Thank you very much @AkivaWeinberger

Akiva Weinberger

This means complex functions are very difficult to visualize. We have to use something other than just graphing them, because we can't draw or visualize 4D space

One way is to think of a function as "doing something" to the plane

Like, $2x$ stretches the complex plane. $ix$ rotates it.

Lucas Henrique

10:06 PM

@AkivaWeinberger is there a way of doing this without continued fraction?

Akiva Weinberger

$x^2$ does a weird foldy-over thing

Lucas Henrique

It looks... really weird to me

Akiva Weinberger

@LucasHenrique Not sure

Mike Miller

bro if you're talking about the complex plane you need to say $z$

Lucas Henrique

I'd never have this idea

Mike Miller

10:06 PM

you'll just confuse me

Lucas Henrique

@MikeMiller lol what

Daminark

Usually people use $x$ to denote real numbers and $z$ for complex

Changing that is off-putting

Lucas Henrique

well ok. ikr, but IMO it's arbitrary

Akiva Weinberger

@LucasHenrique Found this but didn't read it yet math.stackexchange.com/a/207775/166353

Lucas Henrique

(well, in fact every naming is arbitrary, but I don't feel weird using x as a complex)

Daminark

10:08 PM

True, it's arbitrary, but it's one of those things where you just get used to a convention so changing it messes with you

Gaurav Agarwal

@Daminark I understand that... Will keep in mind from next time

Akiva Weinberger

Tradition is tradition

I think that was in response to me @GauravAgarwal

Daminark

"Let $\underline{\sum_{n=1}^{\infty} x^n}$ be the category of all k-simplices"

Or whatever

Mike Miller

lmfao

Daminark

That phrase almost surely doesn't make sense but like, homotope it to something which does, then that notation is just evil

Gaurav Agarwal

10:10 PM

@AkivaWeinberger that was for the tagged person for the x and z confusion. I was going to reply to you by follow up posts....

@AkivaWeinberger I have come to understand the ways of depicting a $-d surface by colors in a 3-d space, but I fail to understand the two plane method. In-fact the quadruple real numbers, I guess I will only understand after going to college.

Akiva Weinberger

Re: $z^2$ is a foldy-over thing,

there's a nice visualization over here at around 6:55 in

Gaurav Agarwal

3-D not $-D

Akiva Weinberger

(the rest of the video is interesting but kinda unrelated)

In fact, check out the rest of that channel, it's really good

3Blue1Brown

Gaurav Agarwal

@AkivaWeinberger I had watched the video about a month ago, but failed to understand how does it co-relate to four dimensions

Infact, if you want you could also see the series : Imaginary numbers are real by Welch Labs

Very good explanation of complex numbers

Imaginary Numbers Are Real [Part 1: Introduction]

►

Akiva Weinberger

@GauravAgarwal Yeah the video doesn't use four dimensions

The graph of $z^2$ would be 4D, so nobody ever draws the graph

and people hardly even think about it

arctic tern

10:26 PM

riemann surfaces & algebraic geometry over C?

Gaurav Agarwal

@AkivaWeinberger no, it does, it uses colour coding the surface to represent the fourth dimesion

Imaginary Numbers Are Real [Part 10: Complex Functions]

►

start from watching this 10 part, uptill 13 and you'll get to see the fourth dimesion, in the two plane method

it is very intuitive

(actual thing statrs in part 11)

starts*

Akiva Weinberger

@GauravAgarwal I meant the 3Blue1Brown video

@arctictern Oh derp sorry

Daminark

@Akiva I've got a better get rekt song: youtube.com/watch?v=lXMskKTw3Bc

Gaurav Agarwal

@AkivaWeinberger ohh

Akiva Weinberger

@GauravAgarwal Oh that's a nice video

Akiva Weinberger

10:51 PM

@Daminark Here's something interesting

Called Blakey's secret-sharing scheme

I have a secret, which I encode as a real number

I choose a random point in 3D whose first coordinate is that number

Now I choose a bunch of random planes through that point and give everyone in this chat one of the planes

You all have no idea what the secret is (assuming no plane is horizontal).

If two people get together and share their planes, assuming I did this smartly, they still don't know what the secret is

(I just need to make sure that the intersection of two planes is never a horizontal line)

We need three people together to recover the secret.

And any three people can recover it (assume general position of the planes blah blah)

So I have shared my secret among a bunch of people in such a way that no two people can recover it but any three people can recover it.

Another idea, Shamir's scheme, chooses a random quadratic equation whose first coordinate is the secret, and gives people random points on it.

(Or degree $n-1$ in general)

Apparently secret sharing can also be done using the Chinese remainder theorem somehow

Daminark

Huh, that's interesting

All I know along the lines of coding and whatnot is RSA

arctic tern

@Danu Doesn't Sp(n) contain Sp(1)?

Akiva Weinberger

11:08 PM

@Daminark Theoretically if one person has three accounts and I don't know that then they could figure it out

You guys aren't all alt-accounts of each other, right? :P

Daminark

Heheheheheh AHAHAHAHAHAH

Akiva Weinberger

cowers in fear

Akiva Weinberger

11:40 PM

"I didn't do A because B" is ambiguous

It could mean I did A, but B was not the cause; or it could mean I didn't do A, and B was the reason for that.

Daminark

This is a true statement

Akiva Weinberger

The first could be reworded as "I did A not because B [but rather C]", I guess

The second one is harder

Daminark

B is the reason I did not do A

Akiva Weinberger

Maybe "The reason I didn't do A is B" or something

@Daminark That too

In any case, I didn't bring this up because of anything we were discussing before

(Oh, that sentence doesn't actually look ambiguous… the "anything" seems to force it be the first one)

I didn't bring this up because of an earlier conversation.

Daminark

Makes sense, I often do that too, like yo btw this is neat!

Semiclassical

11:48 PM

Hmm. I wonder how one would do this problem without using the obvious trick: math.stackexchange.com/q/2371738/137524

(The answer might be "you wouldn't")

Akiva Weinberger

Huh, I guess the equation $(a+b)^2=a^2+2ab+b^2$ is easy enough to visualize in terms of areas of rectangles, but I don't know of any good way to visualize it for the complex case

Complex multiplicative can be visualized (read: represented geometrically) by scaling and rotating

If there were a way, we could try to shove that problem through it and see what we get @Semiclassical

Semiclassical

Hmm.

Akiva Weinberger

What happens when we apply the squaring map to the circle $|z-1|=1$?

Transcript for

$Mathematics$ Mathematics