Mathematics - 2017-09-20 (page 1 of 5)

Faust

00:00

how do i say first n-1 elements in the right way?

Ted Shifrin

More succinctly, if $\sigma(n)=n$ and $\tau(n)=n$, what is $(\sigma\circ\tau)(n)$?

Faust

n

Ted Shifrin

That's all you have to do ... And inverse.

Daminark

It was re number theory

Ted Shifrin

Oh.

Faust

00:02

Let $\sigma \in H $ if $ \sigma (n) =n $ then $ \sigma^{-1} (n) =n $ as well so $ \sigma^{-1} \in H$?

Ted Shifrin

Yup.

Faust

whats the identity in $ S_n $ called?

( i still havent shown H is non empty )

Ted Shifrin

Use your usual notation. That's fine.

I usually wrote it $\iota$, but ...

Faust

we have diffrent notation depending on the group lol

@TedShifrin its a wonder that humans can remember so much stuff.

Ted Shifrin

Think about speaking multiple languages. There's a lot of vocabulary and grammar/syntax to know/remember.

Math is just another language.

Faust

00:07

eh i have a ridiculous memory

user228700

Hi, everyone :-) I was wondering if anybody could help me out with some multivariable calculus. Specifically, I am extremely confused by the way in which Local Linear Approximations is defined in my textbook.

Faust

i still can't remember what class room i am in or what day it is.... but i can remember a page from a chemistry textbook as an image in my mind that i can read...

Ted Shifrin

I'm leaving shortly, but if it's fast I'll help, @Kaumudi.H.

Jaynot

Okay @TedShifrin I am sorry I still don't understand where this is leading me to solving the original question. (:

user228700

Oh, OK, I'll be quick, then. This is what my textbook says:

Ted Shifrin

00:10

$F'(t)=-2\cos(2/t)+2t\sin(2/t^2)$. Integrate from $0$ to $x$. What does that tell you, @Jaynot?

user228700

"Suppose that $f(x,y)$ is differentiable at the point $(x_o,y_o)$. Then, we can write $f(x_o+\delta x,y_o+\delta y)=f(x_o,y_o)+f_x(x_o,y_o)\times \delta x+f_y(x_o,y_o)\times \delta y$"

Ted Shifrin

Equals missing ... Plus some error. This must be an engineering text :(

That's linear approximation, but there is of course error.

user228700

Right. Using linear approximations, what are we trying to do, exactly?

Ted Shifrin

If $f$ is differentiable, this means that the error goes to $0$ faster than $\sqrt{(\delta x)^2+(\delta y)^2}$.

Faust

@TedShifrin thanks for all your help =)

Ted Shifrin

00:13

Sure, Faust.

user228700

@TedShifrin Right, this is how they have defined differentiability in my textbook.

Ted Shifrin

@Kaumudi.H: The same thing you did with one variable calculus. Use the tangent line to get a good approximation to the function. Now it's the tangent plane instead.

Calculus is about approximations.

user228700

> a good approximation to the function

user228700

What is meant by this?

Ted Shifrin

Physics and engineering use approximations (linear or sometimes quadratic) all the time.

It means that if you don't move far from the original point, you can use the linear approximation to estimate the value of the function.

For example, $\sqrt{1+x} \approx 1+x/2$ when $x$ is small.

You can do that in several variables, too, getting $\sqrt{1+x+y} \approx 1+x/2+y/2$ when $x,y$ both small.

Jaynot

00:17

We get that $F(x) + \int_{0}^{x} 2 \cos \frac{2}{t} dt = \int_{0}^{x}2t \sin \frac{2}{t^2}dt$ @TedShifrin

user228700

Oh, ah, hmm.

Ted Shifrin

So solve for your original question, @Jaynot. And you'll see that $\int_0^x \cos(2/t)dt$ is differentiable at $0$.

Because $F$ is differentiable at $0$ and so is $\int_0^x 2t\sin(2/t)\,dt$ (by the FTC).

You have a typo, btw.

user228700

@TedShifrin I'm afraid I've only just been introduced to the idea; we didn't do this in single variable calculus, I don't think.

Ted Shifrin

You certainly should have, @Kaumudi.H. Best linear approximation is one of the ideas in calculus.

user228700

:-(

Ted Shifrin

00:20

Are you doing a standard science/engineering science course or a course with lots of theory/proofs?

user228700

The former, I'm afraid :-/

Ted Shifrin

Well, that's OK.

I have taught my share of both.

user228700

I see.

Ted Shifrin

But in multivariable there are lots more subtle things than in single variable, but probably you won't have to worry about them.

Jaynot

Thank you @TedShifrin.

Ted Shifrin

00:21

You'll have to find equations of tangent planes, gradient vectors, etc. It's all cool stuff.

@Jaynot: Pretty cool, eh? Your double angle thing was actually essential.

user228700

@TedShifrin We don't have that stuff, I don't think :-/

Ted Shifrin

@Jaynot: Are you taking an analysis course? This is pretty sneaky for just calculus.

You will, @Kaumudi.H.

user228700

In any case, okay, I will do some more Googling to understand it more. Thanks so much @TedShifrin :-)

Ted Shifrin

Good luck, @Kaumudi.H. I'll be back other days.

user228700

OK :-)

user228700

00:22

Thanks!

Jaynot

@TedShifrin Its a question I found on a qualifying exam

Ted Shifrin

Ohhh ... Whoa. Like for grad school?

Jaynot

Yeah

Ted Shifrin

Then it's certainly fair to expect students to know analysis :P

Bye all.

user228700

Bye :-)

Daminark

00:27

Babai

:P

Dodsy

00:59

What's up yall

usukidoll

Reading

K k finite dimensional is the number of vectors in a basis of a vector space . Gotta repeat this in mai head

Alek Murt

Show that if $||T - I||$ is sufficiently small, then there is an operator S such
that $e^S = T.$. To what extent is S unique?
I already solved the first part, but i having trouble answering the last question, how do i know to what extent is S unique? i believe i'm not understanding the question
Can someone help me please

Mike Miller

01:16

it should be unique more or less up to adding diagonal matrices whose entries are $2\pi k i$

Alek Murt

could u give more details please

Mike Miller

You'd need to tell me how you solved the first part of the problem

Actually I probably don't want to write out more details in any case, I'm feeling lazy

Alek Murt

i used taylor series for log(1+x) and replace x for T-I

the taylor series converge since $||T - I||$ is sufficiently small

Daniel Castle

01:53

Hi guys, does anyone know if it has been proven all numbers of the form $2^a-1$ reach 1 in the Collatz Conjecture?

If not I believe I have a proof

Dodsy

02:17

sigh

Daniel Castle

?

Dodsy

got 75% on my first lab test :(

Daniel Castle

isnt that good?

Dodsy

No, I need 100% on everything!

Daniel Castle

Lol fair enough

You just started uni?

Dodsy

02:20

yes :)

Daniel Castle

Im starting next week for compsci

Dodsy

Wow, that's great :)

Daniel Castle

I didnt even get my first choice uni lol, wait are you in the UK?

Im guessing youre studying maths

Dodsy

No I am in Canada.

Daniel Castle

Ah ok nice

Dodsy

02:22

I didn't get my first choice either, but the school I'm at is considered better in some respects.

Daniel Castle

Same

Dodsy

Did you get into cambridge :o

Daniel Castle

Loool nope just Bath

Its ranked like 5th this year and my first choice (Warwick) was 7th

Dodsy

that's pretty cool.

Daniel Castle

My friend got 100% in almost all his units and got rejected from oxford

Dodsy

02:25

wat why

did he forget to perform miracles?

Daniel Castle

No idea, they only accept like less than 1 in 10

Yup xD

Dodsy

I just found out I have a linear algebra quiz tomorrow

whoops.

they haven't assigned a single reading in that class

and did not tell me about this quiz.

Daniel Castle

Lol

You can always be ill tommorrow

Im gonna be ill a lot next week

Its 3:30 am here imma go bed, cya lol

Dodsy

cya dude

Balarka Sen

02:48

hi chat

Mike Miller

h

Faust

Morning Dodsy

Balarka

Faust

03:13

Morning @KasmirKhaan

Kasmir Khaan

@Faust Good day to you sir

Faust

hows AA going?

Kasmir Khaan

Well iv seen better days , we did yesterday homomorphism bettwe G/H

and G

Faust

oh shig

thats bad

Kasmir Khaan

was a little comfusing and the natation was the part that made me understand none ><

Anyway one advice is to get the concept of hom asap =p

what chapter are you on now?

Faust

03:15

oh i did that in my last course of AA

im just starting after that alittle bit in my second class

Kasmir Khaan

you do AA in 2 classes?

like a and b ?

Faust

yeah its two semesters

ones 212 the others 312

Kasmir Khaan

><

we doing this in less than 2 months

Faust

it used to be 233A and 233B

Kasmir Khaan

have exam 23 oct

we gonna start with sylow soon

Faust

03:17

we cover up to fields in one semester

Kasmir Khaan

Ted told me the course we doing is done for grad students and am an undergrad ><

Faust

and some stuff with complex numbers and some other crap after

Kasmir Khaan

We cover up like everything in 14 lectures

lucky you ><

Faust

then we go deeper into it in 312

more so in 412 and 413

Kasmir Khaan

thats a lot better for the student

its crazy how we doing it

Faust

03:18

lol

Kasmir Khaan

the problem is like 5 % are understanding the stuff

and am being generous

with that 5%

Faust

yeah its a haard but veyr useful class

Kasmir Khaan

I dont even get the chance to be familiar with the new concepts and i get a whole new list on new concepts

yepp =p

Faust

i did my hw assinment that i sent you in about 3 hrs but it is all review from 212

Kasmir Khaan

Oh we had few same questions there

I think they taking them from the same bank

Faust

03:20

any chance 5a was the same?

i rember and forgot how to do it 3 times now

Kasmir Khaan

what is the Q

or let me check

Faust

Prove that if G is abelian

then H is a subgroup

$ H= \{ g \in G | g^2 = e \} $

i know someone explained how to do it to me

and then i forgot

Kasmir Khaan

Yeah i remeber this question you posted

that I missunderstood it and was talking about what is a subgroup ><

anyways

Faust

i remember it being obvious

Kasmir Khaan

you have to assume G is abelian

ab=ba for a ,b in G

and conclude that H is a subgroup

first prove H is a subset

non empty subset

then let a, b in H prove that ab also in H

Faust

03:24

well e

that wont hold

Kasmir Khaan

then a^-1 is in H for all a in H

e wont hold?

Faust

if a, b in H

then ab also

that wont hold less its abelian

Kasmir Khaan

well if we assume G is abelian

a^2 =e

b^2 =e

(ab)^2 = e

aabb= e = abab

Faust

so $ab= a^{-1} b^{-1}$

Kasmir Khaan

H is the subgroup that have that property g^2=e for all the lements

I think what they want you to conclude here

that the subgroup inherit the group structure

You take a group , you assume its abelian, and the condition on the subgroup H is exactly the condition for abelian, g^2 =e means that ab=ba

and in part b) if you did not have an abelian group to begin with, you would conclude that the subgroup cant be abelian

like i told you =p show x ,y in H implies xy in H

Faust

03:32

yeah i got it from my last line cause it implys order two for ab

Kasmir Khaan

and x in H , implies x^-1 is in H

Faust

but thats not rue unless its abelian

Kasmir Khaan

(x^-1 ) ^2 = e also

yes

but your arguments have to follow this direction

abelian ==> that H is a subgroup of an abelian group

anyway I have to go =p

5.30 am here

getting ready for school :D

Faust

have a good one ^^

Kasmir Khaan

we can talk later about more problems :D

Faust

03:33

have a good day!

Kasmir Khaan

thanks and see you ! :D

u2:)

Secret

04:21

Apr 2 '15 at 7:37, by David Wheeler

If $f''(0) = 2a_2$ then $a_2 = \dfrac{f''(0)}{2}$.

That's nice Taylor's theorem derivation

Hmm

[Random]

Let $f(x)$ be some known function

consider the following formal power series:

$$\sum_{n=1}^{\infty} a_nf(nx)$$

Besides Fourier series and Taylor series, I wonder what other approximation type power series are possible...

Balarka Sen

@Secret It's not actually :P

The derivation is begging the question; it's assuming the function was analytic in the first place.

It's easy to find out the coefficients of the power series if the function is analytic. The main power of Taylor's theorem is that it holds for all $C^k$ functions whatsoever, with an additional exact error term.

Oh shit I was the one explaining that derivation. RIP me from 2 years back

Secret

04:40

I see. I think I would except the Taylor series approximation for $C^1$ function would then have very large errors since the error term will be 2nd order in (forgot)

Balarka Sen

depends on the function

Leaky Nun

The taylor series approximation for $|x|x$ is quite nice.

Secret

$|x|x$ is an interesting function. It is kinda like crossing a cubic with a quadratic together

and it shares the nice properties of both

Balarka Sen

Hm, if $X$ is a Riemann surface, I can describe analytic continuation of a germ $g \in \mathscr{O}_{p, X}$ as the monodromy homomorphism $\varphi: \pi_1(X, p) \to \mathscr{O}_{p, X}$ given by lifting a loop $\sigma$ at $p$ to $\tilde{\sigma}$ in the espace etale $|\mathscr{O}_X|$ starting at $\tilde{\sigma}(0)=g$, and sending $\varphi(\sigma) = \tilde{\sigma}(1)$

Alternatively, it's an unbranched holomorphic map $p : Y \to X$ of Riemann surfaces with a function $f \in \mathscr{O}_Y(Y)$ such that $f(q) = p$ and the germ of $f$ at $q$ sends to the germ $g$ at $p$ under the isomorphism between the stalks that comes from the local biholomorphic nature of $p$.

This kind of looks like the fundamental group-covering space correspondence. This should just be a bijection between $\mathscr{O}_{p, X}$-representations of $\pi_1(X, p)$ and unbranched holomorphic "covers" of $X$ with germinal information ("the basepoint") $g \in \mathcal{O}_{X, p}$

The "maximal representation" should be the one given by lifting loops to the etale space $|\mathscr{O}_X|$ and on the other side of the dictionary there should be the "maximal unramified cover" of $X$.

Alessandro Codenotti

05:48

@BalarkaSen self sniped?

Balarka Sen

looooool

Little Rookie

07:57

How do i write the formal version of "No student likes every theorem in every course." ?

Leaky Nun

08:15

@LittleRookie is it (no student likes every theorem) in every course or no student likes (every theorem in every course)?

Little Rookie

i dont know.

should be second

Leaky Nun

$\neg \exists x [ Sx \land \forall c \in C [ \forall t \in c [L x t]]]$

not sure what your system is

@LittleRookie

AnlamK

Hey is anyone here

Can someone help me show why if cosh x / K = |f(x+iy) |, then f(z) can't be analytic. z = x + iy

sorry it should be K / cosh(x)

Little Rookie

Not all student likes every theorem in every course.
$\exists$ student $x$, $\exists$ theorem $t$, $\exists$ course $c$, $x$ does not like $t$ in $c$

Is my formal version (2nd line) of first line correct?

user308168

08:52

I have a question. Is there anyone?

Tobias Kildetoft

@WDNWBM That is more a question of philosophy than math

user308168

I do not distinguish philosophy, math and physics from each other.

Tobias Kildetoft

@WDNWBM You should

user308168

In my view they are not distinguishable.

Tobias Kildetoft

@WDNWBM That is because your view is not detailed enough. Anyway, if you don't distinguish, then you will be unlikely to be able to ask meaningful questions for us to answer

Balarka Sen

09:00

Well the rest of the world would think your view is garbage, and hence would not answer your question.

user308168

The tangent space of an n-dimensional differentiable manifold is an n-dimensional vector space?

Balarka Sen

Yes.

user308168

even for $C^k$-manifolds?

Balarka Sen

Yes, whenever you have $k \geq 1$. (For $k = 0$ tangent space doesn't make sense)

user308168

But in my textbook it has been mentioned that the tangent space is infinite dimensional in that case.

Balarka Sen

09:09

I think this depends on how you define the tangent space. If I recall correctly there's some subtlety with using the derivations definition (is this how your book does it?)

user308168

Yes, algebraic definition

Balarka Sen

Right, in that case there's trouble.

I usually do it by looking at curves passing though a specific point and tangent vectors of those curves at that point.

In which case the dimension anomaly doesn't happen, I believe

user308168

But it has also been mentioned that the algebraic and geometric definitions are equivalent.

Balarka Sen

Only for $C^\infty$, I think. Otherwise in lower smoothness conditions there exists derivations which do not come from tangent vectors on the manifold.

These ares subtle issues I haven't seen dealt before. I'm pretty sure there exists question on MSE answering these thoroughly

user308168

I think it has been stated that for all dimensions they are equivalent.

Balarka Sen

09:15

The reason people don't usually care is because if you have a $C^k$ manifold for $k \geq 1$, then you can find a $C^\infty$ atlas compatible with the $C^k$ structure, so you can define the tangent space without any trouble whatsoever anyway.

@WDNWBM That'd be interesting.

user308168

However, I can not see the use of $C^\infty$ property of the manifold in the proof of the fact that the tangent space is an n-dimensional space.

Balarka Sen

Check out this paper.

That seems to have a proof.

Also yes, I was right: the tangent vectors of curves definition is the right definition for $C^k$ manifolds. It's the same dimension as the manifold, and is pretty easy to check.

I just worked it out

user308168

Thanks, But I think that paper has used algebraic geometry tools.

Balarka Sen

It has used basic ring theory.

user308168

"The algebraic tangent space is canonically isomorphic to (I/I^2)*"

Balarka Sen

09:32

Yes, this is equivalent to the derivations definition.

user308168

That is an algebraic geometry concept.

Balarka Sen

No, it is not.

Algebraic geometry simply generalizes the idea.

user308168

I want to know why we need $C^\infty$ property to show that the tangent space is an n-dimensional space.

user308168

09:50

Yes, I think you are right. The algebraic and geometric definitions are equivalent when the manifold is $C^\infty$.

user308168

However, I still have the following problem:

user308168

13 mins ago, by WDNWBM

I want to know why we need $C^\infty$ property to show that the tangent space is an n-dimensional space.

Waiting

@Semiclassical here is a typical example of a problem where one single simple observation makes the big difference and brings you very close to the final result.

19

A: A difficult logarithmic integral ${\Large\int}_0^1\log(x)\,\log(2+x)\,\log(1+x)\,\log\left(1+x^{-1}\right)dx$

\begin{align} & \int_0^1\ln(2+x)\,\ln(1+x)\,\ln\left(1+x^{-1}\right)\ln x\,dx\\ & \quad=\frac{71}{36}\,\ln^42+2\ln^32\cdot\ln3+4\ln2\cdot\ln^33-7\ln^22\cdot\ln^23-\frac23\,\ln^32-\frac23\,\ln^33-\ln^22\cdot\ln3\\ & \quad \quad +6\ln^22+3\ln^23-12\ln2-\frac{\pi^4}{216}+\pi^2\!\left(\frac{49}{36}\,...

$$\int_0^1\ln(2+x)\,\ln(1+x)\,\ln\left(1+x^{-1}\right)\ln x\,dx$$

@Semiclassical think well of it and do the right step. I want you to confirm what I said above is perfectly right.

I let you think well over it.

One single and simple step! I return to my work, there is so much to do here.

Felix.C

How could I quickly check whether cos(kx^2) is absolutely integrable?

user308168

I think I have gotten it.

Secret

10:06

...how?

user308168

The tangent vectors are defined on the set of $C^\infty$ functions.

Secret

Ah ok

user308168

Thanks for your answers. However, why is the following true?

user308168

53 mins ago, by Balarka Sen

The reason people don't usually care is because if you have a $C^k$ manifold for $k \geq 1$, then you can find a $C^\infty$ atlas compatible with the $C^k$ structure, so you can define the tangent space without any trouble whatsoever anyway.

Secret

[To be checked] whether a series where the summand is a polynomial of generalised harmonic numbers will also be a polynomial of generalised harmonic numbers

meqnwhile, I knew almost nothing about diff geom

Kasmir Khaan

10:48

@LeakyNun Hello :D

Jasper

@Waiting Sometimes, I wonder if you and Cleo are the same person, lol.

mick

11:28

0

Q: $f " (x) = f(x) f " (x-1) $

I know the equation $ f ' (x) = f(x) f ' (x-1) $ is solved by $f(x) = C$ or by tetration ( $ f(x+1) = exp(f(x)) $). So I wonder What are the solutions to $$ f " (x) = f(x) f " (x-1) $$

Leaky Nun

12:35

@KasmirKhaan hi

Kasmir Khaan

@LeakyNun ohhoo was waitin on you:D

@LeakyNun I got 3 question i want to them with u

Leaky Nun

what is up

Kasmir Khaan

like discussion them and see how to think :D

nothing is up :D

Leaky Nun

ok

Kasmir Khaan

all righty ill type first Q

what is the order of 888 in Z/ 1000

in Z/1000 the operation is addition

so the order of elements in Z/1000 are 1000 only for the numbers relativly prime to 1000 right?

@LeakyNun also got Z/12 is isomorphic to Z/4 x Z/3

@LeakyNun and need to show that there is no isomorphism of Z/128 and Z/64 x Z/2

Jester Tran

12:52

@Vrouvrou No inverse occurs at $t \in (-1, 1)$

Leaky Nun

@KasmirKhaan ok

Kasmir Khaan

@LeakyNun okay i think if i do the isomorphisms first its better because the first one looks simple =P

Leaky Nun

ok

Kasmir Khaan

So the idea here

Z/12 is cyclic

need to find an element that generated Z/4 X Z/3

and that element is simple (1,1)

Leaky Nun

ok

Kasmir Khaan

13:02

now I only need to find a hom map

that is bijective

Z/4 X Z/3 = < (1,1) >

but hmm

how do we do that map

a to (a,a) ?

Leaky Nun

5 to ?

Kasmir Khaan

(5,5) ?

Leaky Nun

is it an element of Z/4 x Z/3?

Kasmir Khaan

oh ><

Ehm 5 to

(1,2)

@LeakyNun hmm how to write that?

Leaky Nun

think.

Kasmir Khaan

13:07

because am gonna be having to take first component mod 4

and other component mod 3

okay

Leaky Nun

right

Kasmir Khaan

I meant like a map to be well defined

Leaky Nun

or just (1,1)^a, lol

@Secret hi

Kasmir Khaan

oh that an easiar way

like more clean direction =p

Leaky Nun

you literally found the generators for two groups

you just need to map powers of the one to powers of the other

Kasmir Khaan

13:09

Yeah I did other question like this before =p

need just to show it is a hom

and bijective

Leaky Nun

ok

Kasmir Khaan

let f : (1,1)^a ==> a

okay

f(a) = (1,1)^a

where 0<= a < 12

f(ab)= (1,1)^ab = (ab,ab ) = (a,a) + (b,b)

so thus hom

@LeakyNun here?

Leaky Nun

f(2x3)=?

Kasmir Khaan

@LeakyNun the notation comfuses me now

Leaky Nun

ok

Kasmir Khaan

13:22

well

(1,1)^a

since 3|12 and 4|12 this is a well defined map

because if a = 12

(4x3 , 3x4 ) = ( 0,0)

hmm what is am trying to do now ><

okay i got the list

(1,1) -> 1

(2,2) -->2

(3,0) --> 3

(0,1) -->4

(1,2) -->5

ect

@LeakyNun .

@LeakyNun all right , solved it =p

need to do one product mod4

and other mod3

Dodsy

13:45

wat is a binary vector

Kasmir Khaan

@Dodsy never heard of that term in math

Dodsy

me either :S

Kasmir Khaan

I got a Q

if ( a mod 3, a mod 4) = ( b mod 3 ,b mod 4 ) how to prove a=b

Balarka Sen

@WDNWBM Nontrivial theorem of Whitney. I don't know the proof.

gian

Hey chat.

Balarka Sen

13:52

Hi @gian.

Kasmir Khaan

Guys who can help with that?

gian

Hey @BalarkaSen.

@BalarkaSen, when computing the simplicial homology of $S^1$, would I just use three 1-simplices that form a triangle since it's homeomorphic to the circle?

Balarka Sen

Yup, that's it.

Semiclassical

@Dodsy vector whose entries are all 0 or 1

Balarka Sen

Well, you can make it a bit simpler by considering a bigon instead; just two edges glued to each other along the corresponding endpoints. This is NOT a simplicial complex, but it's a $\Delta$-complex (see Hatcher), which can also be used to compute simplicial homology.

Jasper

13:59

Baygon is the name of an insecticide, lol.

And a mosquito is a vector.

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$Mathematics$ Mathematics