Mathematics - 2017-06-25 (page 2 of 5)

Secret

10:00

what do elements in $\Bbb{R}/\Bbb{Z}$ look like? I knew for $\Bbb{Z}/n\Bbb{Z}$ the elements has the form $x+mn$ where $m$ is any integer

BAYMAX

I want to draw some drawing

SteamyRoot

That's not the form of an element of $\mathbb{Z}/n\mathbb{Z}$...

BAYMAX

to illustrate this

but how to do that

online

Astyx

They're fibres of the fractionnal part function

BAYMAX

let me do it

SteamyRoot

10:01

Such element has the form $z + n\mathbb{Z}$.

Secret

But I often heard that $x+mn$ can be interpreted as x mod n?

SteamyRoot

That's the form of a representative of the equivalence class.

Astyx

It's not a ring though, it's not compatible with multiplication

SteamyRoot

but the elements are the equivalence classes themselves.

Astyx

Let $$f: \begin{cases}\Bbb R \to [0,1[\\x\mapsto x-\lfloor x\rfloor\end{cases},$$ then an element of $\Bbb R/\Bbb Z$ is of the form $f^{-1}(\{y\})$ where $y\in [0,1[$

BAYMAX

10:07

SteamyRoot

And that is...?

BAYMAX

let me tell

black ones are x

SteamyRoot

$\dot{x}(t) = 1 \pmod 1$ does not imply periodicity in $t$.

BAYMAX

red ones are y

there is some thingy in that problem related to unit circles

like they repeat again

$\dot{x} = 1$

implies

$x(t) = t +c$

and $y(t) = \omega t + d$

SteamyRoot

Yes...

BAYMAX

10:10

so we see the graphs as linear

SteamyRoot

and adding $\pmod 1$ to those differential equations means $x(t) = at + c$ for some integer $a$.

BAYMAX

how

SteamyRoot

Because that's literally what modulo is!

BAYMAX

ok

SteamyRoot

$\dot{x}(t) = 1 \pmod 1$ means $\dot{x}(t) = 1 + m$ for some integer $m$.

BAYMAX

10:12

ok

so $x(t) = at + c$

and $y(t)= \omega t + d$

when x incresing upto 1

then it goes to 0

because of mod 1

SteamyRoot

No...?

BAYMAX

but again it starts repeaing the pattern

ok

SteamyRoot

The derivative being some value modulo $1$ does not imply the function itself has to be taken modulo $1$

$x(t) = at+c$ for some integer $a$. Then $\dot{x}(t) = a \equiv 1 \pmod 1$.

BAYMAX

ok

SteamyRoot

If you "go to $0$ and repeat the pattern" you're creating a disconinuous function

which means there's no derivative defined in $0$, which kind of defeats the purpose of having differential equations in the first place :P

BAYMAX

10:16

but actually

its not discontinuous

but appears to be

here where the circle comes to play

SteamyRoot

Why?

Does the question ever mention you're working on a circle?

BAYMAX

Study the equation x ̇ = 1(mod1); y ̇ = ω(mod1) analytically and what

are the conditions on ω such that

(i) there is a time T such that x(T)= c and y(T)= b

(ii) there is no time T such that x(T)= c and y(T)= b

(iii) analytically solve the map for the following values of ω = 2/5, e, 1/7, 7/1

actually this is the question

if (i) is satisfied then it is periodic

and if (ii) is satisfied then it is quasiperiodic

solution

SteamyRoot

Yeah, well... whatever is meant by "mod" in that question isn't what "mod" means. Nor do I see $b$ or $c$ defined anywhere.

Can't be bothered with this anymore...

BAYMAX

Ok

thanks for looking

Secret

10:46

[Integral symmetries]
$\textbf{Integral conjecture # 001}$ (May be related to Rische Algorithm)
Given any integral where $f(x)$ is any integrable function $$\int f(x) dx$$. If there exists a change of variables $f(x)=g(u)$ such that the integral took the form $$\int \frac{g}{g'}du$$ then no elementary antiderivative exists for this integral without the specific details of $f,g$ being considered

[The notion of bypass]

It is possible that a lot of integrals contains antiderivatives (and definite integrals having closed forms) not because of change of variables, linearity of the integral operator and integration by parts, but because of the various functional equations and identities that the integrand obeys

That means, in a hypothetical mathematical world where all such identities and functional equations are nonexistent, we claimed no integrals except the standard integrals have antiderivatives

In pictures, that is what I think is happening:

Kirill

Hello! I am given a linear projection $\varphi$ and have to show that the trace of $\varphi$ equals to the dimension of $\varphi$. is there any way to estimate the dimension of the kernel of $\varphi$ (evtl. using eigenvalues)?

Astyx

A linear projection is diagonalisable

And its eigenvalues are 0s and 1s

Kirill

@Astyx with what reasoning?

Astyx

What's your definition of a linear projection ?

Kirill

@Astyx the function equals its' own square

idempotent

Astyx

10:56

So if $f$ is a linear projection, we have $f^2-f = 0$, thus $X(X-1)$ cancels $f$, does this make sense to you ?

Secret

(sorry typo_)

Kirill

@Astyx it is, but: $\varphi(v)(\varphi(v)-1)=0$. How can I know, that there is no $v$, e.g. $9$ such that $\varphi(9)=0$?

Astyx

You're confusing two things, what does $\phi(9)$ even mean ?

Secret

Kirill

@Astyx there is some vector $v$ that gives me 0. That is why I am asking about the kernel. Maybe there are vectors, the function of which gives me zero.

@Astyx 9 is an example if $V=\mathbb{R}$.

Astyx

11:01

No, you really are confusing two things

Kirill

@Astyx will be happy to know what exactly

Astyx

The fact that $X(X-1)$ cancels $\phi$ does not mean that for every $v$ $\phi(v)(\phi(v)-1) =0$, first of all because this makes no sense

You can multiply vectors, and 1 is not a vector

So $\phi(v)^2$ makes no sense and $\phi(v)-1$ makes no sense

Kirill

does 0 make sense?

Astyx

What is means is that $\phi\circ(\phi-id)$ (where id is the identity) is the zero linear application (it sends every vector to 0)

So when you compute the image of any vector $v$ through it, you get $\phi(\phi(v)-v) = 0$

ie $\phi^2(v) -\phi(v) = 0$

It's not th same 0 as before though, here it's a vector

Kirill

you mean I read the square wrong?

Astyx

11:05

The square for linear composition means (and can only mean) function composition

That is, $f^2 = f\circ f$

Kirill

oh God

Astyx

(and more generally $f^{0}=id$ and $f^{n+1} =f\circ f^n$ for any integer $n$)

Kirill

that is a nice fact. Thank you, I didn't know that

Astyx

Multiplying linear maps and vectors is never possible

(at least not without defining clearly what you mean by it)

Kirill

I am used to think about compositions only if I have some space where the composition is defined as a group/space operation.

Astyx

11:08

That is the case for linear maps :

For a vector space $V$, the space of linear maps from that space onto itself (these are called endomorphisms), generally denoted $\mathcal{L}(V)$, is itself an algebra, that is, a vector space with a "multiplicative law" (an operation that behaves nicely with the usual vector space operations)

This multiplicative law is composition

s.harp

on the page for eilenberg maclane spaces it says "Saunders Mac Lane originally spelt his name "MacLane" (without a space)", which makes it seem that Mac Lane originally wrote his name the "space" at the end

Daminark

\('-')/ @s.harp

Astyx

I must go now, see ya

Kirill

@Astyx question: what the $X$ means in your first sentence and what does it mean that the it cancels the function? (not pretty cool in English yet)

Daminark

(Peter May told us about them very recently)

@Astyx darn

Astyx

11:11

Bye/hi s.harp and Dami

Daminark

See you!

Kirill

@Astyx bye, thank you

s.harp

good bye

Astyx

I'll answer your question when I come back @Kirill

Kirill

@Astyx ok

Soham Chowdhury

11:21

Is there a way to prove that localization commutes with finite products and infinite direct sums by universal property?

trilolil

11:52

hello everyone

I have a formula which I don't really understand how to interprete

I am trying to use it to construct a matrix. However I am not sure on how to correctly interprete the formula:

$\chi_i = \mu + [\sqrt{(n+\lambda)\Sigma}]_i$ for i=1...n

$\chi_i = \mu + [\sqrt{(n+\lambda)\Sigma}]_{i-n} $ for i=n+1...2n

source: https://drive.google.com/file/d/0By_SW19c1BfhSVFzNHc0SjduNzg/view page 361/503 chapter 10.10.2

Could somebody show me what form the matrix will have?
It looks to me like those indexes don't make any sense for the second formula:

assume n = 4

if e.g. i = 4+3

Astyx

@Kirill Do you know about polynomials ?

Daminark

rubs hands Ooh this is gonna be fun

Kirill

@Astyx characteristic ones?

@Daminark this explanation?

Daminark

If we're talking about polynomials and linear algebra this is gonna be very fun

Kirill

@Daminark hope so

Astyx

11:59

No, polynomials in general

Kirill

@Astyx yes. We have had them like $\sum_{i=1}^{n}a_ix^i$ and like the sequences of coefficients, but I am not cool with the last one

back in 2 min

Astyx

@Dami How's it going ?

Kirill

here

trilolil

Does anybody have an idea about my question?

Daminark

Everything's doing alright, how about you?

Astyx

12:04

I guess I'm alright

Kirill

@Astyx my ears(eyes) are all here

Astyx

So you're all ears(eyes) ?

Daminark

Yes, @Kirill's ontology has become merely that of ears(eyes)

Astyx

So you know about polynomials, multiplication between them etc

Kirill

@Astyx Yes, I am the attention myself

Astyx

12:05

The ring of polynomials, euclidean division, degree

roots, the complex field being algebraically closed

Kirill

@Astyx yes

Daminark

(Are you heading the direction of ideals?)

Astyx

Kinda @Dami

You mentionned the caracteritic polynomial earlier, what do you know about it ?

Kirill

@Astyx complex fields being algebraically closed only in theorie. I mean I know this as a theorem, without a proof

@Astyx the zeros of that polynomial are the eigenvalues of $\varphi$

Daminark

You want to hear the proof? eyes light up

Astyx

12:07

Yes, there is another polynomial that is intrinsincally linked to $\phi$, that is its minimal polynomial

Kirill

@Daminark oh that guy. I haven't get him in the lecture.

Daminark

Huh? Which guy?

Kirill

@Daminark the minimal one

Astyx

The algebra of polynomials over a field $\Bbb F$ is usually denoted $\Bbb F[X]$, $X$ is said to be the unknown.

Daminark

Oh lol I thought you were talking about FTA

Astyx

12:10

Formally polynomials are almost zero sequences (that is sequences of elements of $\Bbb F$ of which only a finite amount of terms are non-zero)

Kirill

if $A \in K^{n \times n}$ is a matrix, then the minimal polynomial is the normed polynomial $g \in K[x]$ such that $g(A)=0$. Well,

that is nice, but how one should "cook" it ot get some results in the exercise

oh, I was reading

Astyx

No, I was diverging too far from what you are asking, my bad

Do you know about the kernel lemma ?

Kirill

maybe, but I do not know such a name in German

Astyx

The name is probably specific to french anyway

Daminark

Yeah I've never heard of it

Kirill

12:14

Is that Cayley-Hamilton?

Astyx

No

Kirill

$dim(V)=dim(kern(\varphi)+dim(im(\varphi)$?

Astyx

So it states that if $P_1,\dots, P_n$ are coprime polynomials and if $f$ is an endomorphism, then $$\ker P_1\dots P_n(f) = \bigoplus_{k=1}^n\ker P_k(f)$$

@Kirill That's rank theorem

Daminark

I refer to your "Grassman's formula" as "Rank-nullity theorem"

Kirill

the product of kernels of polynomials is the direct sum of polynomials? Have never heard about it. (also do not know what coprime means)

Astyx

12:17

In $\Bbb F[X]$ there is a unique factorisation of any polynomial as a product of irreductible polynomials

Daminark

And yeah I've never heard of this as the kernel lemma

Astyx

The same way there is one in $\Bbb Z$ : $35 = 7\times5$ for instance

Daminark

Though this is a generalization of what I know

Kirill

@Astyx ok. Like prime polynomials.

Daminark

(The version I heard is when the product of the $p_i$ is the minimal polynomial)

Astyx

12:19

Coprime means that if you take the gcd of $P_i$ and $P_j$ (with $i\ne j$) it's 1

Kirill

ok

Astyx

Now you can always decompose a polynomial (modulo going in an algebraically closed field containing $\Bbb F$) as a product of degree 1 polynomials $P = \prod_{k=1}^p(X-\lambda_k)^{\mu_k}$

Product indeed

brb

LegionMammal978

I'm going to repost an earlier question of mine in case someone here can answer it:

18 hours ago, by LegionMammal978

Is $\sin(r\pi)$ algebraic for all $r\in\Bbb Q$? If not, what would be a counterexample?

It's reducible to the case where $r$ is a unit fraction, but I'm not sure where to go from there

Astyx

Back

So this means that $\ker P(f) = \bigoplus_{k=1}^p \ker(f-\lambda_k id)^{\mu_k}$ (we consider the $\lambda_k$ to be distinct)

So if $P$ cancels $f$, we have $$V = \bigoplus_{k=1}^p \ker(f-\lambda_k id)^{\mu_k}$$

Kirill

@Astyx Why do you use $\mu$? What does it mean (besides power)

Astyx

12:25

$\mu_k$ is the algebraic multiplicity of $\lambda_k$ in the polynomial $P$

Kirill

ok

but I am thinkg about the formula

Astyx

Which one ?

Kirill

P cancels f means $P-f=0$, right?

Astyx

Oh no, i should have emphasized on that

So you know $\mathcal L(V)$ is an algebra, this means we can apply polynomials on it

For a polynomial $P = \sum a_k X^k \in\Bbb F[X]$, we define $P(f)$ to be : $$P(f) = \sum a_k f^k$$ for $f\in \mathcal L(V)$

Daminark

@Legion I think the answer is yes

The logic is this

$\cos(\pi/n) + i\sin(\pi/n)$ is a root of unity

Astyx

12:28

Note that this is an element of $\mathcal L(V)$

Kirill

@Astyx I can hardly follow starting from "back"

Astyx

We say "$P$ cancels $f$" when $P(f) =0 \in \mathcal L(V)$, that is when $P(f)$ is the zero endomorphism

Do you follow my last 4 messages (they are not dependant on the rest) ?

Daminark

So that expression is a solution to $z^{\frac{1}{2n}} - 1$, and then playing with conjugates should do it

Kirill

@Astyx I do not know what an algebra is. But I can be ok since I know that if it is an algebra, it is also a vector space. And, I am confused about setting polynomials in polynomials.

Astyx

setting polynomials in polynomials ?

It requires a lot of abstraction to grasp this concept, but it really is worth the effort

An algebra is a space in which you can add, multiply elements together, and multiply by scalars

trilolil

12:33

Does anybody have an idea about my question?

Astyx

Examples are $M_n(\Bbb F)$, $\Bbb F[X]$ etc

trilolil

Hello

To make a long story short, I have a formula I am trying to use to construct a matrix. However I am not sure on how to correctly interprete the formula:

$\chi_i = \mu + [\sqrt{(n+\lambda)\Sigma}]_i$ for i=1...n

$\chi_i = \mu - [\sqrt{(n+\lambda)\Sigma}]_{i-n} $ for i=n+1...2n

source: https://drive.google.com/file/d/0By_SW19c1BfhSVFzNHc0SjduNzg/view page 361/503 chapter 10.10.2

Could somebody show me what form the matrix will have? e.g.:

[ 1,1 1,2 1,3 1,4 1,5 1,6 1,7 1,8 1,9;

Kirill

@Astyx You said $P$ is our polynomial that consists of prime polynomials. Now $f$ is a linear map, right? So, yes, setting linear maps in polynomials, not polynomials in polynomials

Astyx

I said we defined $P(f)$ as $\sum a_k f^k$

Do you understand what this means ?

Akiva Weinberger

@trilolil I have no idea @Astyx What are you doing?

Kirill

12:35

the value of $P$ at $f$ is the sum of WOW no

Astyx

@Akiva Explaining some concept about polynomials and vector spaces

Daminark

Ideally, he intends to talk about the minimal polynomial

Kirill

@Astyx the sum of compositions multiplied with scalars?

Astyx

Yes

For instance if $P = 1 + 2X + 5X^3$ then $P(f) = id + 2f + 5f^3$

trilolil

@AkivaWeinberger ok thanks.

Does anybody else have any idea?

Astyx

12:36

I don't @trilolil

trilolil

@Astyx ok thx

Sha Vuklia

Guys, say we consider $D_3$ (set of congruences that leave a regular triangle invariant). Say we first reflect the triangle (say with $\sigma_{\pi/3}$), and then we rotate it by $2\pi/3$ degrees (so we apply $\rho_{2\pi/3}$). I tried to construct the according permutation to this rotation.

If we use a triangle that hasn't been reflected (with the vertices numbered 1,2,3 in the counter-clockwise direction), we get $(132)$. However, if I consider the permutation for the reflected triangle, I get $(231)$, so that can't be right. Any ideas on how to find the according permutation when there has already been applied a permutation?

Astyx

I have to go, I'll be back in an hour at most I think @Kirill

Kirill

@Astyx I have to go, too. I will be here in 6 hours, will be glad to continue if you will be here!

Akiva Weinberger

@BalarkaSen I think we can make a ring of formal power series in two noncommutative elements $x$ and $y$

so that each element is a formal series of $ax+by+cx^2+dxy+eyx+fy^2+\dotsb$

Like, a term for each possible expression of the form $cx^my^nx^oy^p\dotsb$

In any case, suppose we have the quotient of some ring like that.

Theorem: If $w$ commutes with $x^2+2x+1$, then it commutes with $x$.

Proof: We can write $x+1=\sqrt{(x^2+2x)+1}=\sum_{n=0}\binom{1/2}n(x^2+2x)^n$. So anything that commutes with $x^2+2x$ commutes with $x+1$; everything commutes with constants, so QED.

(I like that the Taylor series of the square root still works in the formal power series setting)

Secret

12:50

This is SOOOO UGLY, Now to check whether it actually recovers the question when I differentiate it...

$$\int \sqrt{ \tan x} dx=\frac{1}{2\sqrt{2}}\left(-\ln \left((\sqrt{\tan x}+\frac{\sqrt{2}}{2})^2+\frac{1}{2}\right)+2\sqrt{2}\tan^{-1}\left(\sqrt{2 \tan x}+1\right)+\ln \left((\sqrt{\tan x}-\frac{\sqrt{2}}{2})^2+\frac{3}{2}\right)+\frac{2\sqrt{2}}{3}\tan^{-1}\left( \sqrt{\frac{2}{3}}\sqrt{\tan x}-\frac{\sqrt{3}}{3}\right)\right)+C$$

Akiva Weinberger

O_O

Astyx

@AkivaWeinberger What set is this in ?

Akiva Weinberger

@Astyx In a quotient of that sort of ring

Astyx

Oh I don't know how to read

Akiva Weinberger

in which we can have any formal power series in $x$, and in which for any $y$, we can write $y\sum a_nx^n=\sum a_nyx^n$

I believe that commuting with $x^2$, though, does not necessarily imply commuting with $x$.

The difference being that $\sum_{n=0}^\infty\binom{1/2}n(x^2-1)^n$ doesn't exist—the constant term would be an infinite sum

Astyx

12:58

So if $w$ commutes with $x^2+2x$ really

Akiva Weinberger

Yeah

I mean, commuting with $x^2+2x$ is the same as commuting with $x^2+2x+1$

Astyx

I don't know why but that feels like cheating to me

Akiva Weinberger

shrug

Astyx

Is it always true that $\sqrt{y} = \sum_{n=0}^{\infty} {1/2\choose n} y^n$ ?

Akiva Weinberger

Yeah, as long as that sum exists

Oh, wait

You want $\sqrt{y+1}$, or $(y-1)^n$

You essentially only need to prove that $\sqrt{x+1}=\sum_{n=0}^\infty\binom{1/2}n x^n$. The rest comes from lemmas about how composing functions in the ring of formal power series works

Astyx

13:04

Yeah +1 my bad

So through formal comutation, the resulting formal series of the product is the series consiting only of $y$ ? Interresting

Secret

13:21

Oh crap, careless mistake...

$$=\int\frac{-\frac{1}{\sqrt{2}}u}{(u+\frac{\sqrt{2}}{2})^2+1-\frac{1}{2}}+ \frac{ \frac{1}{\sqrt{2}}u}{(u- \frac{\sqrt{2}}{2})^2+1+!!! \frac{1}{2}}du$$

that + should be -

Akiva Weinberger

Copy/paste \frac{4}{\pi }\sum _{n=0}^{40}\frac{\sin \left(\left(2n+1\right)x\right)}{2n+1} into Desmos

The Fourier series of the square wave looks so cool

Its derivative as well, wow

\frac{4}{\pi }\sum _{n=0}^{10}\cos \left(\left(2n+1\right)x\right)

Wait, oh my god, the derivative is explicitly summable!

$$\frac4\pi\sum _{n=0}^N\cos ((2n+1)x)=\frac2\pi\frac{\sin((2N+1)x)}{\sin(x)}$$

Oh, wait, duh, of course it is -_-

But that means that the $N$th Fourier approximation of the square wave is $\displaystyle\frac2\pi\int_0^x\frac{\sin((2N+1)t)}{\sin(t)}\operatorname d\!t$.

Semiclassical

@akiva mmm, dirichlet kernel

Just win baby

13:43

hi chat

Astyx

Hi

Akiva Weinberger

@Astyx Note that $\sum_{n=0}^\infty a_n(x^2+2x)^n$ always exists (since every coefficient ends up being determined by a finite sum).

Astyx

Formally yeah

Balarka Sen

@AkivaWeinberger This is interesting.

Akiva Weinberger

Right. You can screw with radii of convergence all you want ($a_n=n!$ and stuff)

@Astyx But I think the general rule is, $f(g(x))$ exists when $g(0)=0$.

So, in formal power series, $e^{e^x}$ is problematic, but $e^{e^x-1}$ isn't (interpreting $e^x$ as $\sum x^n/n!$).

(Which means, by the way, that the Taylor series of $e^{e^x-1}$ has all rational coefficients.)

(Which implies that the Taylor series of $e^{e^x}$ has all coefficients in $e\Bbb Q$.)

Balarka Sen

13:47

Depending on what your formal power series ring is over.

If over R, e^e^x exists no

Akiva Weinberger

The whole point of formal power series is that you don't want to have to deal with issues of convergence.

So the moment that you realize the constant term is defined by an infinite sum, you're dead.

Astyx

Or consider formal formal series

Akiva Weinberger

(In this case you could just define $e^{e^x}=e\cdot e^{e^x-1}$, I guess, if you really needed to)

Astyx

And you don't care about the convergence of the sums defining the constant terms

Balarka Sen

@AkivaWeinberger Which makes sense only if the power series ring is over R

Akiva Weinberger

13:49

@BalarkaSen OK, sure

@Astyx I think you'd end up with horrible stuff like $0=(1+1+\dotsb)-(1+1+\dotsb)=(1+1+\dotsb)-(0+1+1+\dotsb)=1+0+0+\dotsb=1$

Astyx

That statement is obviously correct

Balarka Sen

I forget what join of spaces is

$X \times Y \times [0, 1]$ mod squishing $X \times pt \times [0, 1]$ on one side, and squishing $pt \times Y \times [0, 1]$ on the other side I think

Akiva Weinberger

I know a circle join a circle should be a sphere, I think

Balarka Sen

S^3, I think

Akiva Weinberger

Like, a torus but with a meridian and a longitude line squished

Balarka Sen

13:55

I think that's smash

Akiva Weinberger

Oh.

Balarka Sen

$X \times Y/X \vee Y$

Akiva Weinberger

Oh, you're right

Right, this is the thing what turns a line and a line into a tetrahedron.

Line segments.

Balarka Sen

Right, yup

Mike Miller

yikes

Balarka Sen

13:57

HUH. So $EG$ can be written as a direct limit of iterative join of $G$'s.

Akiva Weinberger

I'm having trouble verifying that circle join circle is indeed S3

s.harp

Does the identity theorem (holomorphic function on connected domain is uniquely determined by accumulation point) hold in banach spaces/ banach algebras also?

Balarka Sen

@AkivaWeinberger It's like the tetrahedron with two pairs of concurrent faces identified.

Akiva Weinberger

Oh, wait, OK

Balarka Sen

Cut along the tetrahedron along a small square from the middle, glue each "piece of pie" individually to get solid torii, and then glue the solid torii back to get S^3.

Akiva Weinberger

13:59

For some reason I thought that joining the first pair would give us something weirder than it does

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