Mathematics - 2017-07-05 (page 2 of 10)

Alex K Chen

07:02

Hey @BalarkaSen, did you notice your name is (just saying) an angram of "Blank areas" ?

Balarka Sen

Is it now?

Interesting

Alex K Chen

@BalarkaSen Which one ?

Balarka Sen

I mean I have never noticed that before

Weird!

Alex K Chen

Also Nasal Break :P

Balarka Sen

lol

I like the third one

Alex K Chen

07:05

I like imagining what that mean, though :P

Balarka Sen

a multitude of possibilities

Alex K Chen

Hmmm...

@BalarkaSen How do you think about polynomials ? Visualize the graph ? Imagine the roots on the complex plane ?

Balarka Sen

I don't think I have a special geometric intuition for them.

Alex K Chen

Okay, so how you think about them algebraically ?

Balarka Sen

Just as an expression. They are like the simplest functions.

Alex K Chen

07:10

I mean just like this: Plug a number $x$ in the form $$ \displaystyle \sum a_i x^i $$ and see what happens ?

Balarka Sen

More or less, that's my first thought.

Alex K Chen

Okay, then ?

Balarka Sen

Of course it depends on what I want to do with polynomials.

Do you have a specific problem in mind?

Alex K Chen

Wait a bit then.

Is the problem of given $(a_0, a_1)$ (both integers), determining what numbers would be of the form $a_0 x^2 + a_1 y ^2$ an open problem ?

Leaky Nun

@AlexKChen I don't tend to visualize $x^{15} + 3x^9 - 7x + 1$

Balarka Sen

07:13

Oh by the way for more variables there is a simple answer to your previous question: $x^2 + y^2$. That takes infinitely many prime values, indeed, all the $1 \pmod{4}$ primes.

Leaky Nun

@AlexKChen what do you mean by "determine"?

Alex K Chen

Lol.

Balarka Sen

@AlexKChen It's not very clear to me what the question is.

Not enough to say it's open, at the very least.

Alex K Chen

@LeakyNun I meant does there exists an simple algorithm (you know what I mean, right ?) so that given $(a_0,a_1)$, you can check a number $n$ can be expressible in the form $a_0 x^2 +a_1 y^2 = n$, with integer $(x,y)$ ?

Same question, what happens if we allow $xy, x,y$ along with $x^2, y^2$ too ?

Balarka Sen

These are Pell-type equations aren't they

Leaky Nun

07:16

@AlexKChen what do you mean by "simple"?

Alex K Chen

@BalarkaSen Wha for the extra question ?

Balarka Sen

No I mean the original. I think there are lots of literature on solving quadratic diophantine equations.

Needless to say, I don't know any of it.

Alex K Chen

Also, does there exists a simple algorithm to check whether an integer $k$ appears with the residue class $x^n$ modulo $p$, ($p$ is a prime) ?

OK.

Leaky Nun

@AlexKChen no

Alex K Chen

That's bad.

Leaky Nun

07:22

> The discrete logarithm problem is considered to be computationally intractable. That is, no efficient classical algorithm is known for computing discrete logarithms in general.

Discrete logarithm

In mathematics, a discrete logarithm is an integer k exponent solving the equation bk = g, where b and g are elements of a group. Discrete logarithms are thus the group-theoretic analogue of ordinary logarithms, which solve the same equation for real numbers b and g, where b is the base of the logarithm and g is the value whose logarithm is being taken. No efficient general method for computing discrete logarithms on conventional computers is known. Several important algorithms in public-key cryptography base their security on the assumption that the discrete logarithm problem over carefully chosen...

The factorization of $p-1$ is important though

Secret

> quantum computers

Astyx

Hi chat

Danu

07:42

@BalarkaSen Hey, I remember talking to you about a year ago about special coverings of manifolds with small number of charts.

Balarka Sen

Yeah

Danu

What's that theorem that says you can cover an $n$-manifold by $n+1$ charts

It comes from this other result that says you have a covering where each point lies in at most [something] number of charts

What was the name of that?

Balarka Sen

The name of the number is Lusternik-Schnirelmann category

Danu

Why is the number called "category"... dafuq

Balarka Sen

I am not sure if I knew the theorem you quoted.

@Danu yah no idea

Danu

07:45

Anyways, there was this other theorem about covering a single point only [something] times

it's related

I vaguely remember the word "color" in connection to it

Balarka Sen

huh

Daminark

@BalarkaSen Did you have to look up the name of it to remember how it was spelled?

(I really should sleep but merp)

Danu

@BalarkaSen Yeah, I don't know...

Daminark

But yeah Neves gave a talk on L-S for the REU in finding critical points of a map

Balarka Sen

lol i used to do that, @Daminark

Danu

07:47

Anyone know what I'm talking about with this minimal # of charts covering a single point?

Balarka Sen

Nope

I don't know much about these things

@Daminark Cool stuff

Danu

Lebesgue covering dimension

In mathematics, the Lebesgue covering dimension or topological dimension of a topological space is one of several different ways of defining the dimension of the space in a topologically invariant way. == Definition == The first formal definition of covering dimension was given by Eduard Čech, based on an earlier result of Henri Lebesgue. A modern definition is as follows. An open cover of a topological space X is a family of open sets whose union contains X. The ply or order of a cover is the smallest number n (if it exists) such that each point of the space belongs to at most n sets in the cover...

Muh memory doe

remembering "color"

lmao

visuals too stronk

Balarka Sen

You said chart though.

Lebesgue covering dimension only says you can cover it by n+1 open sets.

Danu

Yeah, that's right.

I didn't remember it correctly.

I could just remember this damn picture, haha

Balarka Sen

pictures +1

Tobias Kildetoft

08:04

Where have fidget spinners been all my life? They are awesome for when I am reading.

Secret

Hi Tobias. Long time no see

Tobias Kildetoft

Hi @Secret

Secret

[To be checked: Other properties of algebraic vs transcendental numbers in rings that can help distinguish them]chat.stackexchange.com/transcript/message/20399274#20399274

Lembik

09:00

if I write $d|x-y$ is that clearly the same as $d|(x-y)$ or do I need the brackets?

Secret

Personally I will add the brackets, but I think d|x and then -y cannot make sense, thus it should force people to interpret it as d|(x-y)

Astyx

First one seems fine to me, although I would also add the brackets

Lembik

does $-$ take precedence over "divides" basically?

Secret

I don't think "divdes" is a binary operator, that is, you don't input d and x and then spit out some number as a result

Lembik

@Secret I think it's true or false

Secret

09:05

"divides" is more like a binary relation telling the property of d

Lembik

@Secret right

@Secret but the question is if I need the brackets :)

I am tempted to write d|(x-y) unless you tell me that is terrible style

Secret

Me and Astyx both agree that putting the brackets is clearer

Lembik

thanks!

I suppose d|x-y is unambiguous but odd

as you can't subtract from a judgement

Secret

indeed

Lembik

09:29

thanks again for your help

Dattier

09:40

Hi, $$\text{Determinate a closed formula for : } u_{n+1}=\frac{u_n^2-2}{2u_n-3},u_0\in \mathbb R$$

gxyd

Is every algebraic extension a simple extension?

I am considering the base of the extension to be $$\mathbb{Q}$$.

And now consider a field $$\mathbb{Q}(\sqrt 2, \sqrt3)$$ which is equivalent to $$\mathbb{Q}(\sqrt 2 + \sqrt3)$$ (is it?), and the extension $$\mathbb{Q}(\sqrt 2 + \sqrt3)$$ is a simple extension, so its equivalent extension should also be simple.

Tobias Kildetoft

@gxyd en.wikipedia.org/wiki/Primitive_element_theorem

gxyd

Are there any computer algebra systems that currently do these type of computations?

Tobias Kildetoft

Probably both Sage and Magma can do them if there are any decent algorithms for it

Alessandro Codenotti

@gxyd think about the algebraic closure of $\Bbb Q$

Dattier

09:56

$$ A_n \text{ the sub-group of order }2^n5^{100-n} \text{ of } (\mathbb Z/10^{100}\mathbb Z,+)\\ \text{ Calculate } \text{card}(\bigcup \limits_{n=0}^{100} A_n )$$

gxyd

That would be $$\mathbb{C}$$.

Tobias Kildetoft

@Dattier Where are you copying this from?

@gxyd No, the algebraic closure of the rationals is countable

Astyx

Oh hi @Tobias

Dattier

@TobiasKildetoft : it's a personnal production

Astyx

How are you ?

Tobias Kildetoft

09:57

@Astyx Hi

Alessandro Codenotti

@gxyd no, much smaller

Tobias Kildetoft

@Dattier Then what is up with all of the \text and such?

Fargle

Field theory, one of my many weaknesses.

gxyd

@AlessandroCodenotti BTW how do see the latex written in here?

Fargle

@gxyd there's a link in the top-right, in that link will be a "start ChatJax" link, add that to your bookmarks bar, then click back here and click on the bookmark

Dattier

10:00

@TobiasKildetoft : you don't anderstand the question ?

then sorry my english is bad

where is the understanding problem ?

Tobias Kildetoft

@Dattier I understand the problem fine

I just found the formatting strange

Dattier

Yes I am not professionnal

Leaky Nun

@gxyd Let $x=\sqrt2+\sqrt3$, then $\sqrt2=\frac12x^3-\frac92x$ (from wikipedia), etc.

Liad

someone can take a look here: https://math.stackexchange.com/questions/2347080/differential-geometry-helix-and-frenet-serret-formulas

can't find my mistake.

Leaky Nun

@Dattier usually $A_n$ is the alternating group of order $\frac12(n!)$.

Dattier

10:04

ah

well, $$ H_n \text{ the sub-group of order : }2^n5^{100-n} \text{ of } (\mathbb Z/10^{100}\mathbb Z,+) \\ \text{ Calculate } \text{card}(\bigcup \limits_{n=0}^{100} H_n )$$

gxyd

@LeakyNun wait a moment please, I am still searching for chatJax enabling.

Leaky Nun

the generator of $H_n$ is $2^{100-n} ~ 5^n$, so we are looking for integers whose factorization have at least 100 copies of "2 or 5"

@gxyd it's right in the room description

Dattier

@LeakyNun : the union is not a group

Leaky Nun

@Dattier I never said it is

Dattier

sorry

gxyd

10:13

@LeakyNun thanks.

Secret

I really need someone to say the features that Sage is better than Mathematica

user86234

Hi, if $C$ is a matrix. How do I calculate $\frac{\partial}{\partial C}(C^{-1})$?

Secret

cause so far sage seemed to be in the middle of the strength of most properitery softwares

Sure, sage is open source, but there should be something that maple, mathematica and matlab cannot do and only sage can?

SteamyRoot

@Liad You probably made a mistake with a sign somewhere?

Liad

no idea where. i checked it over and over again @SteamyRoot

SteamyRoot

10:22

Wild guess: when calculating the cross product, you use the determinant trick but forgot that the $y$-coordinate required a minus.

Liad

what do you mean ?

@SteamyRoot amazing.

@SteamyRoot thank you. i did not know that we need to take the minus of the $y$-coordiante

Tobias Kildetoft

@Secret Like "be free"

Dattier

and "be open source"

SteamyRoot

Well, sage is more a bunch of free software thrown together.

Dattier

Another more easy, calculate as quickly as possible, exp(A), with A=[[1,2],[3,4]], matrice 2×2.

SteamyRoot

10:29

It includes GAP, which definitely has routines neither Maple, Mathematica nor Matlab can do

Dattier

how, do you that ?

Tobias Kildetoft

@Dattier diagonalize

Astyx

^

Dattier

Yes I think there is more rapidly (for the big dimension)

Astyx

Compute it as the limit of the remainder of a polynomial modulo the caracteristic polynomial

Doesn't always work though

Dattier

10:32

A step, $$\text{Calculate, } \lim\limits_{n\rightarrow \infty} P_n(x) \mod (x^2+1)^2, \text{ with } P_n(x)=\sum \limits_{k=0}^n \frac{x^k}{k!}$$ this question help to the first

Not the answer but the method to answer it

Astyx

Yeah basically what I said

Dattier

So, how calculate that (the limit of polynomials)

Astyx

Actually no, not what I meant

Dattier

lol

Astyx

Why $(x^2 + 1)^2$ ?

Dattier

10:41

I resume the situation, and the link between the first and second questions, if you have P polynomials with P(A)=0, then if you calculate $\lim P_n(X) \mod P(X)$ we are an answer, but how calculate that ?

@Astyx for augment difficulty

Astyx

But $P(A) \ne 0$ here is it ?

Dattier

Calley-Hamilton, or anoter method

Astyx

We have $A = \begin{pmatrix}1&2\\3&4\end{pmatrix}$ right ?

Dattier

right

Astyx

Its caracteristic polynomial is $X^2 -5X -2$

Dattier

10:44

you can derteminate {Id, A,A^2} linear relation

Astyx

It's not $(X^2+1)^2$ (which is of degree $\gt2$ and doesn't cancel $A$)

Dattier

So, the calculate of $$\lim P_n(X) \mod (X^2-5X-2)$$ give the answer

if P(A)=0

I: "Not the answer but the method to answer it"

Astyx

But why $(X^2+1)^2$ then ?

Dattier

I : "@Astyx for augment difficulty"

Astyx

What does that mean ?

So you have $X^2 \to 5X + 2$

Dattier

10:48

for augment the difficulty (by google translator) Pour augmenter la difficulté

Astyx

So if you write $X^k \to a_k X + b_k$

"To make it more difficult" is more understandable imho

You get $X^{k+1} \sim a_kX^2 + b_kX \sim (5a_k + b_k)X + 2a_k = a_{k+1}X + b_{k+1}$

Dattier

$$X^{k+1} \sim a_kX^2 + b_kX \sim (5a_k + b_k)X + 2a_k = a_{k+1}X + b_{k+1}$$

Astyx

I meant $\equiv$, not $\sim$

Dattier

yes

Astyx

So you're left with the recursive formula for two sequences $$a_{k+1} = 5a_k + b_k\\ b_{k+1} = 2a_k$$

Akiva Weinberger

10:53

The exponential is gonna have roots and stuff, no?

Astyx

With $a_0 =0$ and $b_0 = 1$

Tobias Kildetoft

@Astyx Why are we not just diagonalizing?

Dattier

@Astyx
And we go back to a problem of diagonalization

Astyx

No clue

Tobias Kildetoft

this ends up requiring the same calculations anyway

Dattier

10:53

@Tobias yes

Astyx

He wanted another method

Akiva Weinberger

I'm curious how this one ends up being solved

Dattier

a quick method

Akiva Weinberger

A *quick method

Astyx

First step would probably to decompose that fraction

Dattier

10:56

@Akiva, here the trick : math.stackexchange.com/questions/2114710/…

Kasmir Khaan

hello

Dattier

hi

Kasmir Khaan

i have a question about the solution used in the book

a+b+c+d+e+f <10

i need to count the number of solutions to that , we only use non negative integers

Akiva Weinberger

@Dattier Wow, cool!

Kasmir Khaan

my idea was to find solution to a+b+c+d+e+f =9 then 8 then 7 ect

Akiva Weinberger

11:06

Did we ever determine if there existed a similar solution to $u_{n+1}={u_n}^2+1$?

Tobias Kildetoft

@KasmirKhaan It is not a bad idea, but unfortunately, it gets more complicated than it needs to

Kasmir Khaan

but the way they did it was a+b+c+d+e+f+G = 9 , claiming that they have the same solutions

Akiva Weinberger

Oh, that's clever

Kasmir Khaan

@TobiasKildetoft yes with larger numbers it will be hard to do

Tobias Kildetoft

@KasmirKhaan Indeed, since the sum being less than 10 means that the sum can be written as 9 - G where G is a non-negative integer

Kasmir Khaan

11:08

but how adding 1 more variable and making it equal to 9 does the trick?

ahh

Tobias Kildetoft

so you move over the G and get a new equation that you can solve instead

Kasmir Khaan

9-G so that does cover all cases =p

Tobias Kildetoft

right

Kasmir Khaan

okay thanks guys :D

Akiva Weinberger

$x<N\iff\exists n:x+n=N-1$ when everything's a nonnegative integer

Kasmir Khaan

11:09

i thought it was wierd that it was only equal to 9 , when it could have been any number from 0 to 9

but now i see why :)

thanks again @AkivaWeinberger @TobiasKildetoft

Dattier

@Akiva, I try it, but now nothing

no solution

Tobias Kildetoft

It is one of those tricks that you pretty much just have to get shown

Dattier

@Kasmir there are a trick

$$x_1+x_2+x_3+...+x_n=k$$

Kasmir Khaan

I wish one day that I could find soemthing like that by myself =p

Dattier

$$\text{coeff}(\frac{1}{(1-X)^n},X^k)$$

In math there is always a simple trick but either it is hidden, or it is known

Tobias Kildetoft

11:13

@Dattier That is only true for exercises you get given

Akiva Weinberger

Oh yeah that's a binomial thing

Dattier

I don't think

@Tobias

Akiva Weinberger

It's kinda like the hockey stick identity

Dattier

Even the garnished chiefs have their secrets

Et j'aime divulgué les secrets (dans les sciences utiles)

Astyx

Depends what you call a trick really

Dattier

11:16

And I like disclosed the secrets (in the useful sciences)

trick : a simple idea that facilitates understanding

Tobias Kildetoft

@Dattier So what was the trick for Fermat's last theorem?

Dattier

and solving

Astyx

Yeah but your idea of simple changes as you know your subject better

Dattier

In the first, I do that for the "little" problem, des grand concours d'oraux ENS-X

And that's not finish

by, my enigmas, I can see, even a simple idea is difficult to found, if you don't know it,

@Kasmir, the idea (of this problem) is simple ?

now

?

Kasmir Khaan

@Dattier Yes! the part of 9-G was the part that I missed =p

9-G covers all cases i did at start ( the long way )

Akiva Weinberger

11:23

If $f(x)=\frac{ax+b}{cx+d}$ and $\underline2(x)=2x$, then what's $f\circ\underline2\circ f^{-1}$?

Kasmir Khaan

a+b+c+d+e+f =9 then 8 then 7 ect can be replaced with adding extra variable

Tobias Kildetoft

@KasmirKhaan Note that this then gives a proof of an otherwise non-obvious identity between a sum of binomial coefficients and a single binomial coefficient

Dattier

@Akiva : here I use homographie and $$x^2$$

Kasmir Khaan

@TobiasKildetoft Yes I should look in into that , thanks :)

Akiva Weinberger

@Dattier Oh, is that what it's called?

Dattier

11:32

@Akiva, I don't understand

Geralt of Rivia

hello, does a course in differentiable manifolds use much abstract algebra ? analysis? (like measure theory analysis)
And vice versa?

Akiva Weinberger

@Dattier I've never heard the word "homography" before

s.harp

@Geralt no. almost no measure theory or powerful algebra is used

Dattier

$$h(x)=\frac{ax+b}{cx+d}$$

Akiva Weinberger

So doing $h\circ(x^2)\circ h^{-1}$?

Dattier

11:36

here yes

Akiva Weinberger

What does that simplify to?

Dattier

But, the trick work, even h is not inversible

user84215

Does human's mind not have any primary axiom?

Dattier

$$u_n=\frac{a\times x^{2^n}+ b}{c\times x^{2^n} +d}$$

@Akiva

@aminliverpool no, I don't think

The same functioning perhaps

user84215

if it has no, then why did it start to move or work?

Tobias Kildetoft

11:41

@aminliverpool how is that a math question?

user84215

it is because math is a consequence of mind's performance

Tobias Kildetoft

No. Just...no.

user84215

why ?

lattice

Hello chat

Any hint on how to show that the underlying topological space of a noetherian scheme is noetherian?

Dattier

@aminliverpool : step by step, this question is too simple for me (so too difficult !)

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$Mathematics$ Mathematics