Mathematics - 2017-06-25 (page 3 of 5)

Secret

14:00

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

Akiva Weinberger

Ah, OK, got it, I see @BalarkaSen

@Secret No good can come of this

Balarka Sen

It's like a lens space description of S3 I guess

or something

Secret

(and yes, I have actually differentiate it and it is indeed the antiderivative. However integrating this is not my main aim)

My aim is to illustrate the following: (and I guess I learnt a valuable lesson today on how important identities are. This is kinda almost a minimal example where I have used the least number of non polynomial identities to integrate, and we can see how ugly it is)

Balarka Sen

Oh wow that description of $EG$ is obviously the same as the piecewise linear function description

Secret

Contrast this with the first answer of the following:

17

Q: Evaluating the indefinite integral $ \int \sqrt{\tan x} ~ \mathrm{d}{x}. $

I have been having extreme difficulties with this integral. I would appreciate any and all help. $$ \int \sqrt{\tan x} ~ \mathrm{d}{x}. $$

The first answer integrates this much quicker and with a less ugly expression

and he/she did so because of at least 3 uses of trigonometric identities are being exploited

This then brought us to the following conjecture:

Akiva Weinberger

14:05

Oh wow

Hm, how do we write $\cosh^{-1}$ in terms of logs?

'Cause there is a way (using the quadratic formula)

Balarka Sen

log(x + sqrt(x^2 - 1)) something something

Secret

$\textbf{Integral conjecture # 002}$ The length of an integration pathway for an integral that has a closed form is inversely related to the number of function specific identities used

Balarka Sen

something

Akiva Weinberger

I know differential Galois theory is a thing that exists, though I know nothing about it. It looks like it might be related to all this, though @Secret

Balarka Sen

woo my guess was right

Secret

14:09

@AkivaWeinberger I am not sure either (I guess I really have to read that up in detail), the thing is, differential galois theory cannot handle those integrals that have closed form because of the existence of special function identities, it can only handle elementary antiderivatives. Whether there exists a generalisation that can handle special functions I am not sure

Diferential galois theory, if I recall can tell us from the Galois group of the field of functions whethe a given integrand has an elementary antiderivative and if yes, find it

this is analogous to galois theory in solving radicals

Semiclassical

well, special function identities typically come about because they satisfy certain differential equations.

Secret

There's also something called Rische algorithm, though I don't remember which comes first, but it can be derived as a consequence of differential galois theory from one of the online lecture note on the topic that I read some time ago when discussing typhon's periodic constant ODE system

Risch algorithm

In symbolic computation (or computer algebra), at the intersection of mathematics and computer science, the Risch algorithm is an algorithm for indefinite integration. It is used in some computer algebra systems to find antiderivatives. It is named after the American mathematician Robert Henry Risch, a specialist in computer algebra who developed it in 1968. The algorithm transforms the problem of integration into a problem in algebra. It is based on the form of the function being integrated and on methods for integrating rational functions, radicals, logarithms, and exponential functions. Risch...

Semiclassical

but I imagine the place of special functions in diff. Galois theory would be analogous to extending a field by including a particular root in standard Galois theory.

Secret

Hmm, I am thinking about something similar. The issue however is I need to read up how series in general behave in a galois group since a lot of special functions (such as the hypergeometric functions) are basically infinite series, while others are defined in terms of an integral such as erf(x)

Semiclassical

well, hypergeometric functions also satisfy ODEs with three regular singular points

erf is weird, though. so i wouldn't be surprised if that one provides an exception to what I'm saying

Secret

14:19

One thing I notice when Waiting, Cieo, MNCE, robjohn, Lucian etc. when they solve for closed forms of integrals and series is the huge number of identities they exploit in the simplification

The recent test using the $\sqrt{ \tan x}$ seemed to confirm my suspicion that without the existence of these identities, these integrals may be unable to have a closed form

Semiclassical

yeah, there's a huge amount out there. the glories and riches of classical mathematics, you could say

The question, I suppose, is to what extent solving such examples is an art versus a science.

Sha Vuklia

Guys, does anyone know what the commenter in my post means in his last comment? I'm not sure what relations he's talking about, but I feel like our discussion is already getting too long in the comments; math.stackexchange.com/questions/2335694/…

Akiva Weinberger

Speedsolving a Rubik's cube also uses an insane amount of identities.

Seriously, the amount of memorization of what algorithms to use in which cases is insane.

Balarka Sen

The best I can do with a Rubik's cube is color two concurrent faces.

Secret

I am actually starting to wonder, whether the general question on the existence of closed form is whether there exists at least one pathway consists of rewriting expressions and using functional identities that can lead to some finite combination of functions and elements from a field

I knew that feeling, one Rubik cube guy demonstrated to us how the identities are put into action. and one of them is based on one of the commutator of the rubik cube group in the ability to rotate tiles anticlockwise for each 3 moves)

Akiva Weinberger

14:22

I can solve it, but not speedsolve it

Semiclassical

I'm able to color all but the top third of a Rubik's cube correctly

which is not great.

Akiva Weinberger

Mathologer has a good video that describes a way to find your own solution

The secret is conjugation

Balarka Sen

I have done 3 faces but coincidentally. My algorithmic approach just does 2, generically

Oh yes there is a lot of conjugation business popping up

Secret

So far, finding closed form is mostly an art due to no-go results such as nonexistence of antiderivatives (and possibly by extension, nonexistence of antidifferences for series), and the art is basically the intuition of recognising these pathways via experience and other means. But I suspect, if we can map the space of all functional identities and convert that into a manifold somehow, then maybe we will get the science of finding closed forms out of it

Akiva Weinberger

But the point is, solving integrals quickly depends on how many identities you know, even if these identities could all be theoretically be derived from a much smaller set of identities

Balarka Sen

14:25

Technically you can always solve a cube by rotating top third, pushing down front third, ad infinitum repeatedly

but that's not the best strategy :P

Semiclassical

@secret You might find Wilf's A=B interesting: math.upenn.edu/~wilf/AeqB.html

Akiva Weinberger

@ShaVuklia The relations $\rho^3=\sigma^2=1,~\rho\sigma=\sigma\rho^{-1}$

@BalarkaSen That only touches two faces?

Oh, wait, by pushing down from third you mean rotating the whole cube?

Balarka Sen

Yeah.

Akiva Weinberger

Huh, interesting. I didn't know that was a generator

Secret

@AkivaWeinberger I am not sure whether it is an open question on whether the space of all functional identities (which is countably infinite based on discussions between me and Leaky since it still depends on a finite alphabet) forms an algebra with a finite generating set

Balarka Sen

14:28

Just to clarify, it's like: Take the front face, rotate a quarter clockwise. Rotate the cube by pi/2. Take the front face, rotate a quarter clockwise. Etc.

Just win baby

Mar 11 at 16:23, by skill patrol

Straight out of Webster's Note - Authors have not always been careful to use the terms art and science with due discrimination and precision. Music is an art as well as a science In general, an art is that which depends on practice or performance, and science that which depends on abstract or speculative principles. The theory of music is a science; the practice of it an art.

SoumyoB

@Astyx I tried a lot to get how you concluded that $E(e^{tX}) \le e^{t^2/2}$ for all $t\in \Bbb R$ by convexity of the exponential but couldn't get it

could you explain it to me in a bit more detail? Sorry my basics in probability in Markov stuff are a bit rusty

Secret

@Semiclassical Downloaded, it seems it is not just about identities, but approaches on how computers do proofs in general

Semiclassical

@Secret indeed

Astyx

Yeah I skipped the details, sorry @SoumyoB

Semiclassical

14:36

@SoumyoB It may be useful to rearrange that to $E(e^{tX})e^{-t^2/2}\geq 1$

SoumyoB

@Astyx and @Semiclassical is any probability distribution given for $X$?

Semiclassical

Since expectation is linear, you can pull that factor inside $E$. That in turn suggests completing the square.

Secret

@Justwinbaby Don't forget sciart, rare entities that is a harmony between art and science

Astyx

So prove for $x\le 1$ that $e^{xt} \le {1-x\over 2} e^{-t} + {1+x\over 2} e^t$

Semiclassical

Could not tell you, tbh. I'm just following my nose :)

Astyx

14:38

Just that $E(X) = 0$ and $|X|\le1$ almost certainly

Semiclassical

Huh, nice.

Astyx

So when taking the expectancy you get $E(e^{tX})\le\cosh(t) \le e^{t^2/2}$, directly from comparing the terms in the taylor series

Just win baby

@Secret cool

Secret

[Integral symmetries] Btw, below are what I suspected to be no-go forms:

$$\int \frac{f(x)}{f'(x)}dx, \int \left(\frac{f'(x)}{f(x)}\right)^n dx, \int \frac{f''(x)}{f(x)} dx$$

and for that, I formulated a conjecture # 001 which states that any integral that can be brought into these forms (and possibly more) cannot have a non infinite series closed form if suitable functional identities are not used to manipulate the integrand

That A=B and other books on special functions may help me to check this conjecture

SoumyoB

@Astyx be right back from the market

Secret

14:44

in particular, when evaluating integrals, I especially hate to see this form:

$$\int \frac{f(x)}{f'(x)}dx$$

because it often signals simple integration techniques (IBP, partial fractions, change of variables) often don't work at least for all the integrals I have done so far in the past, like in my studies, readings etc.

If conjecture #001 is true, then it will mean that whenever a change of variable give you this, it signals you to use functional identities

Akiva Weinberger

@Secret Counterexample, $f=\sin$

Or any polynomial

Secret

For $\sin$, well I guess $\tan x=\frac{\sin x}{\cos x}$ is pretty much a definition more than an identity, For polynomials, yes........

O and also $f(x)=e^{ax}$ as well, because of the special role that $e^x$ is an eigenfunction under the integral operator

Akiva Weinberger

I suppose it depends on your definition of closed-form. Are you allowing the use of any constants? Or only the ones expressible in terms of integers and elementary functions?

Ivan Ivanov

Hello if I know vertex cover is NP complete and I want to show that Set cover is also NP complete, I should reduce Set cover to Vertex cover or in other words finding Set cover is the same as finding Vertex cover of a graph ?

Secret

@AkivaWeinberger My idea of a closed form is slightly more general than the notion of finite combination of elementary functions. For me, a closed form is either a number (for series and definite integrals) or a finite combination of special functions and elementary functions (thus there will be numbers e.g. in linear combinations)

user84215

14:54

Why is any compact and connected surface in R^3 with constant Gaussian curvature a sphere ?

Secret

So in general, no infinite series will be considered a closed form unless it happens to solve some (at least 3?) functional equations (and thus often make them have a name)

similarly, no expresisons contains integrals will be considered a closed form unless it is famous enough to get a name

s.harp

In TeX I want to write $ff$ where both fs are close to each other so that it looks like one symbol, how can i do that?

Secret

actually in light of this, I think I will consider $\tan x=\frac{\sin x}{\cos x}$ a functional identity, since if you imagine replacing all the functions with e.g. $f,g,h$ then $f=\frac{g}{h}$ does not always hold, making it quite specific to trigonometric functions

Liad

@BalarkaSen hi

Akiva Weinberger

@s.harp How about f\!\!f $f\!\!f$?

Secret

15:03

As for polynomials, I am not sure, since if you go into the way of considering its various properties as identities then suddenly a lot of algebraic identities become specific only to polynoamials

s.harp

@AkivaWeinberger nice!

Liad

@AkivaWeinberger hi

Akiva Weinberger

Test: $f\!\!\!f$

s.harp

thats too close

Akiva Weinberger

Oh, no, that's too far

@Liad Hi

$\fortissimo$

Oh, that's too much to hope for

Liad

15:06

im trying to find $f:X \to Y$ where $X,Y$ are path connected and the induced homomorphism is not trivial nor the identity.

user84215

Why is any compact and connected surface in R^3 with constant Gaussian curvature a sphere ?

Secret

Also, $e^x$ is pretty much the only family of functions that satisfy $f'=f$, so I think we can consider it as a functional identity

I think the conclulsion is that we can relax or modify our condition so that the only counterexample to conjecture #001 are the polynomials

More rigorously, functional identities are relations that holds only for soem functions f,g,h

That way, $f=f$, $0f=0$ is not considered a functional identity since it holds for all functions and thus is due to the underlying algebraic structure

This together with the above definition for closed forms, should make it specific enough

Liad

so i took $X= S \ ^1 , Y = S \ ^ 1 \times S \^ 1$ and $f(x) = (x,-x)$ im trying to show that the induced homo' is $f(n) =(n,-n) $ for $n \in \Bbb Z$

Akiva Weinberger

@Liad Induced homomorphism on the fundamental group?

s.harp

@aminliverpool there should be a theorem that a surface of genus $g$ cannot be embedded into $\Bbb R^3$ with curvature $≤0$ everywhere

Liad

15:07

@AkivaWeinberger yea. i know that $\Pi_1(S \ ^ 1 ) = \Bbb Z $

@AkivaWeinberger any chance you know how can i show that the induced homo' is indeed $n \to (n,-n)$ im a bit stuck on that

s.harp

@aminliverpool no compact negative curvature surfaces in $\Bbb R^3$: "throw planes at your surface from very far away. At the point of first contact, your plane and the surface are tangent. But the surface is everywhere saddle-shaped, so it cannot be tangent to your plane without actually piercing it, contradicting first contact."

SoumyoB

@Astyx just understood your steps, that was very clever I must say

Astyx

It is tricky, and this is why I'm surprised it works so well

Akiva Weinberger

@Liad I'm not sure, maybe use the covering space of the torus

s.harp

15:40

are path spaces $C([0,1],Y)$ Hausdorff? google only tells me things about path connected spaces that are hausdorff

Balarka Sen

Surely you need $Y$ to be Hausdorff?

Pretty sure path space of the, say, Sierpinski space is not Hausdorff.

s.harp

The path space of an indiscreet topology cannot be Hausdorff, it is also indiscreet

so it is likely that you want $Y$ to be Hausdorff

Balarka Sen

Right.

s.harp

im trying to see whether the evaluation map defined as a map $Y^I\times I\to Y$, $(\omega,t)\mapsto \omega(t)$ is continuous. With the exponentiation law this is super easy, but that requires $Y^I$ to be Hausdorff

Akiva Weinberger

What's the topology on $Y^I$?

Is this homotopy classes of paths?

Balarka Sen

15:47

Compact-open.

I think if $Y$ is Hausdorff $Y^I$ is Hausdorff but I'm a little slow and lazy today. Maybe Akiva would want to help

Akiva Weinberger

I forgot what compact-open means

s.harp

its true and hte proof is trivial my bad^

Akiva Weinberger

looks it up

s.harp

@Akiva if $K\subset X$ compact and $U\subset Y$ open then $K(X,U) = \{f:X\to Y\mid f(K)\subset U\}$ defines a subbase

this subase gives the compact open topology

to see why the proof is trivial let $\gamma\neq\omega$, so you have a point $a$ with $\gamma(a)\neq\omega(a)$, choose two disjoint neighbourhoods of $\gamma(a)$ and $\omega(a)$ and call thenm $U,V$. You have that $K(\{a\},U)$ and $K(\{b\},V)$ are disjoint and one contains $\gamma$ the other $\omega$

the first set is the set of all maps that send $a$ into $U$, the second set is all maps sending $a$ into $V$, since $U$ and $V$ are disjoint it is obvious that the two sets must be disjoint too

SoumyoB

MAGIC^^

s.harp

15:54

?

SoumyoB

empty line

SteamyRoot

Balarka Sen

how did you do that

i'm cofnsued

SteamyRoot

Soft hyphen doesn't render

alt-0173

s.harp

you submit a unicode character that doesnt hve width or height or something

SteamyRoot

15:55

It's a common way to troll on the internet :P

Balarka Sen

ah

SoumyoB

@Astyx I saw the remaining steps you had posted earlier, I didn't get how you arrived at the last step

Astyx

Which one is that ?

SteamyRoot

Like, If I ask an MSE question and I just put that char down a few times

SoumyoB

the one where you said you used the Bernoulli law

SteamyRoot

15:56

Then the question title will be empty, so no link to click on :D

Astyx

Oh you linked it

Akiva Weinberger

Astyx

I'm not too sure anymore. You'd probably have to apply Bienaymé-Tchebychev to that inequality to get bound it from above by $nV(Y)\over \epsilon_n^2$ where $Y+p$ follows Bernoulli

Akiva Weinberger

@AkivaWeinberger Magic

Astyx

Magic indeed @Akiva

Secret

16:04

$U+2028$

ok I have no idea how to type unicode

Balarka Sen

@BalarkaSen Hi, future

@BalarkaSen Hi, past

Astyx

So you get ${n p(1-p)\over \epsilon_n^2} \le {2\over \sqrt n}$ so $$\epsilon_n \ge \sqrt{n^{3/2}p(1-p)\over 2}$$

Lemme double-check that

Balarka Sen

@Akiva Fixed.

Akiva Weinberger

:38357941 Taking a gamble

Balarka Sen

gamble some more

Astyx

16:07

Oh actually I guess you could get a better bound using what's just been done

Balarka Sen

47 boo

Akiva Weinberger

Gah.

How are these numbers even generated?

Balarka Sen

@AkivaWeinberger Hi

Akiva Weinberger

Let's see

Balarka Sen

Hey, 64

Akiva Weinberger

16:09

Woo! Wait

Balarka Sen

But it didn't work

Astyx

So you want to bound from above $P(\left|{\sum{X-p}}\right| \ge \epsilon_n n)$

The X-p satisfy what's just been done

Akiva Weinberger

@BalarkaSen Edit it and do nothing?

Balarka Sen

Done.

Astyx

Now we make a case distinction wether $p\ge 1/2$ or not

Akiva Weinberger

16:10

@BalarkaSen It still gets that marker next to it that says it's been edited :(

Balarka Sen

Yeah I don't like it

Akiva Weinberger

:00000001 Test?

Hrm.

Balarka Sen

Maybe that's not from this room

Akiva Weinberger

@ChaoXu Test!

Astyx

If it is, then $P(|\sum X-p|\ge \epsilon_nn) \le 2\exp\left({-\epsilon_n^2n^2 \over 2np^2}\right)$

Akiva Weinberger

16:12

Woo

Astyx

So you want $-{1\over 2}\ln n \ge {-\epsilon_n^2 n\over 2p^2}$

Balarka Sen

Voyage to past.

Akiva Weinberger

Oldest message!

:9999999 Final message!

Balarka Sen

Probably that won't be in the room either.

Astyx

So $\epsilon_n \ge p \sqrt{\ln n\over n}$ which is a way better bound

I'm surprised

That can't be right

Oh wait it can

That's the law of big numbers for bernoulli variables

I'm confused by how good this result is

@SoumyoB

Abcd

16:20

Hey!

Astyx

Hi

Abcd

In cross product of vectors through components we only get a magnitude because $icap. icap = 1 (same for j and k)$ so how is that possible that we ain't getting any direction?

Please let me know if you don't understand the question.

Astyx

You might want to add some $ there

Abcd

@Astyx I don't know how to represent i cap through math jax

Astyx

$\cap$

\cap

But I was reffering to your "$sameforjandk$"

Balarka Sen

16:25

Hi @Daminark

Astyx

If two power series $\sum a_k x^k$ and $\sum b_k x^k$ coincide on any open set $U$ inside of the intersection of their disk of convergence, are they necessarily equal ? (even if $0\notin U$)

How do you prove that efficiently

Abcd

In cross product of vectors through components we only get a magnitude because $\hati . \hati =1$ (same for y and z compoents). How is it possible that we ain;t obtaining any direction?

@Astyx I don't want this cap.

Astyx

@Abcd Check this

Can you ping yourself ?

Abcd

@Astyx Why?

@Abcd check this. //Test message

No we can't @Astyx.

Astyx

How do you do that ?

Okay answers my question then

Thanks

Secret

16:51

WTF?

Integral of inverse functions

In mathematics, integrals of inverse functions can be computed by means of a formula that expresses the antiderivatives of the inverse f − 1 {\displaystyle f^{-1}} of a continuous and invertible function f {\displaystyle f} , in terms of f − 1 {\displaystyle f^{-1}} and an antiderivative of f {\d...

Astyx

Nice proof without words

Secret

Integration using parametric derivatives

In mathematics, integration by parametric derivatives is a method of integrating certain functions. For example, suppose we want to find the integral ∫ 0 ∞ x 2 e − 3 x d x . {\displaystyle \int _{0}^{\infty }x^{2}e^{-3x}\,dx.} Since this is a product of two functions that are simple t...

That, I guess, will only work if you can differentiate under the integral sign

Abcd

17:09

Could someone please explain why $ \hat{i} - \hat{j} $ lies in the third quadrant?

According to me it should lie in the 4th Quadrant.

BAYMAX

if $\hat{i}$ is in x direction and $\hat{j}$ is in y directon thenit must lie in 4th quadrant

Abcd

@BAYMAX My book said it's in 3rd Quadrant therefore I got confused.

BAYMAX

ok

Akiva Weinberger

You want \hat{\imath} $\hat\imath$ and \hat{\jmath} $\hat\jmath$ (curly braces optional), by the way

\imath $\imath$ and \jmath $\jmath$ are like $i$ and $j$ but without the dots.

Daminark

17:44

Hey @Balarka!

Astyx

Yo Dami

Dodsy

my alergies are so bad today :C

Paweł Kusz

Hello :) $A$ is compact,$ B$ is closed $ \Rightarrow A \cap B $ is compact ?

Daminark

How's it going @Astyx?

Darn @Dodsy

Astyx

Kinda good, you ?

Daminark

17:51

Yes @Pawel

Paweł Kusz

@Daminark Thanks :)

Daminark

Do you know that a closed subset of a compact set is compact?

Dodsy

@Daminark I think I'm gonna study Serge Lang's Basic Mathematics before I go to school, what do you think?

Daminark

And I'm doing alright, thanks! @Astyx!

Dodsy

if I have enough time I'll study spivak's calculus too

Paweł Kusz

17:54

@Daminark Yes, I know, but Is the true in every topology ?

Dodsy

Anyone else know about serge langes basic maths?

or does anyone have better recomendations?

I'd like to gain a better foundation.

s.harp

recommendation for what?

Dodsy

books to read before university

s.harp

How would you describe your math level?

are you bored when you hear stuff like uniform continuity?

for analysis my favourite book eva is "Foundations of Modern Analysis" by Dieundonné

Secret

[The Weird Integral manuscript] this is so weird, I guess I need to wrote more definitions to clarify...

s.harp

18:03

its not for a beginner, but if you are already a little bit acquainted with analysis its greatg

Abcd

Can someone please tell me how to approach this problem (not the entire solution)?

Not a homework question ( I practice problems ...) . Secondly, it might be a basic question but I am new to this topic...

Astyx

What have you tried ?

Abcd

@Astyx Distance covered = 24 m

Astyx

If a particule is at vector position $\vec{OM}$ at some point and it goes with velocity $\vec v$, where is it after some time $t$ ?

Abcd

18:19

@Astyx velocity vector*t

Astyx

Not quite

Abcd

@Astyx $v_{avg} = (r_2 - r_1)/ \triangle t$

Astyx

Not what I asked

Kirill

@Astyx hello Astyx! Do you wish to continue?

Astyx

Make a drawing

Abcd

18:21

@Astyx I have it.

Astyx

@Kirill Sure, where was I at ? What did you understand/misunderstand ?

@Abcd Show it to me

Kirill

@Astyx let me summirize it in TeX for a minute

what was the symbol for $L(v)$ in Tex?

Dodsy

@s.harp I feel like I have problems simplifying sometimes.

Astyx

\mathcal L(V)

Kirill

ok

Dodsy

18:24

I know how to make both sides equal in an equation to solve for x most of the time

but that's a little bit of a hang up

I'd like to review trig

and geometry

Astyx

You can use $L(V)$ just as well, I just enjoy fancy typefonts :p

Kirill

@Astyx So, what I got: there are many mysterious things :) But, first of all there is a space $\mathcal{L}(V)$ of linear functions, where composition is defined as a group/space operation. Then, we can define algebra on it. Then we spoke about the minimal polynomial but I couldn't get it all together.

Astyx

It is the space of linear functions from $V$ to $V$

Kirill

@Astyx what I do not understand is the way from projection to the diagonalization, eignvalues and why did you speak about the minimal polynomial.

Liad

Astyx

18:29

That second part is important

Liad

why does fact that $c $ not belonging to $A_0,B_0$ contradicts the fact that $[a,b] = A_0 \cup B_0$ ?

Kirill

@Astyx it was a direct sum of the kernels of prime polynomials, right?

Astyx

And what $P(f)$ is when $P$ is a polynomial and $f$ an endomoprhism right ?

That second question is more important

Kirill

@Astyx yes, though I do not have a formual now.

Astyx

$P(f) = \sum a_k f^k \in \mathcal L(V)$

Kirill

18:33

and $f^i \in \mathcal{L(V)}$, too?

Astyx

Yes @Kirill

Kirill

ok

Astyx

It's $\underbrace{f\circ f\circ \dots\circ f}_{i \text{ times}}$

Which is a linear application from $V$ to itself (check it if you doubt it)

Ideally, you should be confident enough about this for it to seem obvious to you

Daminark

Hey, I'm back

Kirill

@Astyx linear applicaton $\equiv$ linear map?

Balarka Sen

18:36

rehi

Daminark

@PawełKusz yeah it is, try proving it

Astyx

So now the minimal polynomial of $f$ is the unique nonzero unitary polynomial of minimal degree such that $P(f) = 0$ (we know it exists because $\{P\in \Bbb F[X]\mid P(f) = 0\}$ is what we call an ideal (ie a group that's absorbant for multiplication) ... and we'll see it's not trivial later)

@Kirill Yes

Daminark

@Dodsy I don't know that Lang book much, Spivak is quality

Akiva Weinberger

@BalarkaSen Got anything clever upon your sleeve?

Although, actually, that's a strange way of putting it

I don't have much on my sleeves

Astyx

I don't have sleeves

Daminark

18:38

Anything upon your sheaves then?

Balarka Sen

No, but I have something beneath the fold of my collars.

Kirill

@Astyx ok

Akiva Weinberger

Where does the king keep his armies?

Military bases.

Daminark

Anti joke

Balarka Sen

"I don't need an army; I have an atom bomb on my arsenal."

Akiva Weinberger

18:39

…On?

Daminark

Yep, right on top

Balarka Sen

Easy to dispatch.

Astyx

Now the kernel lemma states that if $P_1, \dots P_n$ are coprime (their pairwise gcd is 1), then $$\ker P_1\dots P_n(f) = \bigoplus_{k=1}^n \ker P_k(f)$$

Akiva Weinberger

(The word you're looking for is "pairwise")

Astyx

Thanks

Abcd

18:41

@Astyx Nevermind...

Balarka Sen

@AkivaWeinberger I'm wondering if we can relate $\pi_n(X, A)$ with $\pi_n(X/A)$.

Astyx

Found the solution ? @Abcd

Balarka Sen

But that's not particularly clever.

Secret

(Prepare for chat explosion)

[Integral symmetries: Attempt to formalise]

Definition

$\textbf{Elementary functions}$: Given $f \in \textrm{L}^p(\Bbb{F})$,$f$ is elementary if it can be expressed as a finite combination of $\textrm{exp}$, $\log$, $c\in \Bbb{F}$, powers and $\frac{p}{q},p,q \in \textrm{P}(\Bbb{F})$.

$\textbf{Standard integral functions}$: Given $f \in \textrm{L}^p(\Bbb{F})$, $a, a_k,b \in \Bbb{F}$, $f$ is a standard integral function if it take only linear arguments $a*\textrm{id}+b$ and is any of the following and its derivatives:

$\sin$, $\cos$, $\tan$, $\csc$, $\sec$, $\cot$,

Dodsy

@Daminark thanks.

Astyx

18:44

@Kirill Got it ?

Kirill

@Astyx not actually, I am thinking about the right question

Astyx

The right question ?

lattice

Hey

Daminark

Yeah Astyx the question you were talking about was wrong. It has a truth value of z e r o

Kirill

@Astyx I think that is it: what is a kernel of a polynomial if we already put $f$ in it? I know about the kernel of a linear map, but here - not yet.

Astyx

18:46

@Dami :'(

The kernel of a polynomial isn't a thing, I'm talking about the kernel of "the polynomial evaluated at $f$"

Kirill

@Astyx or, could you please tell the formula in words?

:)

Astyx

"The kernel of the endomorphism $P_1\dots P_n(f)$ is the direct product of the kernels of $P_i(f)$ for all $i$"

Kirill

but it is a multiplication on the left side, right?

Astyx

What is ?

Abcd

@Astyx Yes.

Astyx

18:49

Glad to hear it

Kirill

@Astyx $\prod_{i=1}^{n}P_i(f)$? Or $(\prod_{i=1}^{n}P_i)(f)$?

Perturbative

Hey everyone

Astyx

Oh right, by product I mean product of spaces

Abcd

@Astyx Still one confusion. We are dealing with graphs right? How can we have- time, velocity, position - all in one graph?

Astyx

Actually I mean sum, sorry I'm tired

Daminark

18:52

In the context of linear algebra you'd be better off calling it a direct sum of spaces

Kirill

@Astyx I do not know the product of spaces

Astyx

Do you no about direct sums of subspaces of $V$ ?

Kirill

@Astyx yes, a bit

Astyx

@Abcd A graph isn't necessarily 2d

Abcd

Oh yes.

Astyx

18:53

So "The kernel of the endomorphism $P_1\dots P_n(f)$ is the direct sum of the kernels of $P_i(f)$ for all $i$"

user193319

I am trying to find a counterexample to the following claim but am having trouble: Let $f : X \to Y$ be a continuous function. If $A \subseteq$ has limit point $a$, and $f(a)$ is a limit point of $f(A)$, then $f$ is injective.

Astyx

Sorry for the confusion

Kirill

@Astyx I still do not get the left side

Astyx

You agree $P_1\dots P_n$ is a polynomial right ?

Kirill

@Astyx yes, but why an endomorphism?

Astyx

18:55

(God did I really write no instead of know ? :()

Because $P_1\dots P_n(f)$ is a linear map $V\to V$

That is an endomorphism of $V$, ie an element of $\mathcal L(V)$

Kirill

@Astyx let me smoke about it

Astyx

Smoking is bad for your health

Kirill

@Astyx yes, it is like Don Juan answered to Castaneda in his books :)

Abcd

@Astyx I multiplied the velocity with time in that question. How is that possible on a graph?

Akiva Weinberger

@Secret Do you have any idea on the $E^{1/2}$ problem?

Astyx

18:57

I don't get it @Abcd

Kirill

@Astyx Castaneda said he had a problem. Don Huan said that he has already smoked about it

Akiva Weinberger

I actually discussed an important element of the solution earlier today

Astyx

Heh :p

Sha Vuklia

Guys, I'm trying to generalise the Chinese remainder theorem. A couple of months ago when I tried this, I wrote down the following, and never really came back to it:

"Consider $(\mathbb Z/n_1\mathbb Z)\times\dots\times(\mathbb Z/n_{t+1}\mathbb Z)=[(\mathbb Z/n_1\mathbb Z)\times\dots\times(\mathbb Z/n_t\mathbb Z)]\times(\mathbb Z/n_{t+1})$. Denote $N=\prod_{i=1}^t n_i$.

We know that $(\mathbb Z/N\mathbb Z)\cong(\mathbb Z/n_1\mathbb Z)\times\dots\times(\mathbb Z/n_t\mathbb Z)$. So there exists an isomorphism $f\colon(\mathbb Z/N\mathbb Z)\to(\mathbb Z/n_1\mathbb Z)\times\dots\times(\mathbb …

Abcd

@Astyx How does time* velocity yield displacement vector on a graph?

Astyx

18:58

That's a lot of $\Bbb Z$

Sha Vuklia

lol it is

Astyx

By graph do you mean drawing ? @Abcd

Kirill

@Astyx really, I need two minutes, as I was sure we were speaking about polynomials, and now it suddenly reveals to be a linear map

Abcd

@Astyx Yes. How do I do this on a graph and obtain the answer that I obtained while working on paper?

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$Mathematics$ Mathematics