Mathematics - 2019-03-21 (page 1 of 2)

Ted Shifrin

12:00 AM

So what are all 4 numbers and does the equality hold?

Isa

ah

got it

they are equal

Ted Shifrin

Yup.

Isa

thank you Ted!

Ted Shifrin

You're welcome. Always remember to play with examples!

Isa

ok, thanks . Have to go, bye :)

Lucas Henrique

12:06 AM

Hey chat!

Ted Shifrin

heya @Lucas

Lucas Henrique

how you're doing @Ted?

Ted Shifrin

Quite well, thanks, and you?

Érico Melo Silva

hlo folk

user193319

Let $X_n(\Bbb{Z})$ be the simplicial complex whose vertex set is $\Bbb{Z}$ and such that the vertices $v_0,...,v_k$ span a $k$-simplex if and only if $|v_i-v_j| \le n$ for every $i,j$. Prove that $X_n(\Bbb{Z})$ is $n$-dimensional...

Ted Shifrin

12:12 AM

Heya @Eric.

user193319

How is the dimension of a simplicial complex defined?

Ted Shifrin

@user193319: Where did you use $k$?

user193319

Not sure. I just wrote the problem verbatim.

Ted Shifrin

Oh, I see.

Érico Melo Silva

how goes

Lucas Henrique

12:19 AM

@TedShifrin life's giving me a break, so better now. thanks :)

uni is the easiest part of going to uni

Ted Shifrin

LOL

Lucas Henrique

no kidding, my maths is foundations (basic logic but not pedantic), calc 1 which I'm pretty used to work with, analytic geometry and basic linear algebra (by basic I mean matrices and systems of equations only

Ted Shifrin

Sounds too easy for you.

@user193319: So can you show there are no $k$-simplices for $k>n$?

Lucas Henrique

you mean I might be underestimating it?

Ted Shifrin

No, @Lucas. I think the opposite.

user193319

12:23 AM

No, but I can try. So, the dimension has to do with the number $k$-simplices that can be fit into $X_n(\Bbb{Z})$?

Lucas Henrique

well, thank you then. :)

Érico Melo Silva

@Lucas what’s the content of a math degree in Brazil

Ted Shifrin

By dimension they mean the highest-dimensional simplex you can have.

Lucas Henrique

@ÉricoMeloSilva hold on

user193319

Ah, okay. Thanks, I'll think about what you've said.

Lucas Henrique

12:26 AM

@Erico, you can find it here

however, Unicamp's structure is designed to have less obligatory classes and give you more free time to do scientific research

iirc you must do at least 2 in a year after you get into the math course itself (the first 3 semesters are with the physics course so you can choose which one you like the most after this period)

Érico Melo Silva

cool cool

Ted Shifrin

@Eric: So Korean or what for dinner tonight?

Lucas Henrique

"period of time". lmao

non-native speakers suck. :p

Érico Melo Silva

maybe taiwanese porridge

@Lucas are you planning on trying to go to grad school?

Lucas Henrique

sure

my dream is to study at IMPA and Berkeley

Ted Shifrin

12:33 AM

It takes a strong person to succeed at Berkeley.

Dair

so lift weights

Érico Melo Silva

yeah they kind of bury u w teaching

Ted Shifrin

Oh, most places do that, @Eric. That wasn't my issue at all.

Érico Melo Silva

@LucasHenrique i decided not to apply to IMPA for various reasons but I was just at berkeley and met a few cool ppl

@TedShifrin i guess so, among my options it was by far the heaviest teaching load

Ted Shifrin

Huge state school, so sure.

Princeton = spoiled brats.

Dair

12:34 AM

lots of people need lots of teachers

Érico Melo Silva

uh oh am i gonna be a spoiled brat

Ted Shifrin

I think so.

Érico Melo Silva

rip

Dair

@TedShifrin the lack of :P ending that sentence is unsettling

Ted Shifrin

I was referring to undergrads, mostly, but, yeah.

I aim to unsettle.

Érico Melo Silva

12:37 AM

my impression of berkeley from visiting their grad open house was that it’s very easy to get lost there

Ted Shifrin

Yeah, that was more the issue I was speaking to.

One has to be a very aggressive graduate student to make sure to get attention.

At UGA we pampered the grad students, by comparison.

Dair

seems to be a problem with large state schools...

Ted Shifrin

And many of us on the faculty tried to make sure no one slipped through the cracks.

Érico Melo Silva

@TedShifrin that’s rlly good

ryan

12:51 AM

Anybody know if there's a question similar to this out there? or resources that may help https://math.stackexchange.com/q/3156156/445911

I checked around but couldn't find anything (among the many "recommended" posts) that quite fit.

DemCodeLines

Anyone who is familiar with Latex, can you tell me I get this?

When I do

$\frac{u_{max} - u_{min}}{x_{max} - x_{min}}$

Why does the numerator decide to be on the left instead of top?

ryan

Looks like you messed up somewhere, the first 'u' is not in math mode

$\frac{u_{max} - u_{min}}{x_{max} - x_{min}}$ should work?

Ted Shifrin

No, you don't want \frac.

ryan

Maybe a copy/paste error? Sometimes copying things into a latex editor can mess it up.

DemCodeLines

When I removed the $, it started working...

ryan

12:56 AM

Weird. @TedShifrin why wouldn't you want \frac ? /curious

Ted Shifrin

You want $a_b$, where $b$ has an overline over the expression.

ryan

@TedShifrin I think DCL wants the corrected version, where it is actually a fraction. Unless I'm mistaken

Presumably "can you tell me I get this?" should've been "can you tell me why I get this?"

Ted Shifrin

Ah, I didn't try to psychoanalyze the English.

DemCodeLines

I just wanted the numerator to be on top of the denominator, instead of hanging around on the left side.

Ted Shifrin

@DemCode: What are you trying to do?

Oh, OK. So you do want \frac.

DemCodeLines

12:59 AM

That should've read "why I get this". Typo!

Sorry

Ted Shifrin

$$\frac{u_{\text{max}}-u_{\text{min}}}{x_{\text{max}}-x_\text{min}}}$$

ryan

Anyway, I would assume if removing $ fixes it, then you probably have an open math expression somewhere before it, meaning you didn't close it with $ earlier. What's the full expression you're trying to get? If it's just the frac, then your code should be fine

DemCodeLines

It was just the fraction.

Ted Shifrin

Interesting. I don't know why this isn't compiling.

ryan

Hm, weird. Are you editing it in stackexchange? Have a link to the question/answer?

abuchay

1:03 AM

This is my first time chatting here in Math Stack Exchange. So I am not sure if this is frowned upon but just a quick question: I am trying to prove that a proper subgroup of $\mathbb{Z}^n$ is isomorphic to $\mathbb{Z}^k$, where $k \le n$. So we must have $rank(A) = rank(\mathbb{Z}^k)$ , right?

ryan

@DemCodeLines I still find it odd that the first u is not in math mode but the rest is... maybe if you figure out why that's occurring it may solve the problem

Ted Shifrin

Yeah, that's not my problem, though.

Oh, I left out a {.

$$\frac{u_{\text{max}}-u_{\text{min}}}{x_{\text{max}}-x_{\text{min}}}$$

@abuchay: What is $A$?

How do you know the subgroup has to be free?

DemCodeLines

@ryan Yup, working on it

abuchay

@TedShifrin sorry, $A$ is a proper subgroup of $\mathbb{Z}^n$.

Ted Shifrin

So you need to prove it's free abelian.

ryan

1:09 AM

@DemCodeLines I'm not quite sure why it's not working for you. However, if you post the answer/question and put the link to it in here, someone will certainly propose an edit that fixes it for you.

Anyway, hope you get it resolved. I'm off now

Ted Shifrin

$\frac{u_{max} - u_{min}}{x_{max} - x_{min}}$

I literally copied your text and pasted it, @DemCode, and that's what I got.

For mine, I put the subscripts in text mode, because I hate the math mode with text.

DemCodeLines

There was a \[ missing earlier somewhere in the document, but \] is in there later on. Adding that \[ fixed it (started working with $).

Ted Shifrin

So one dollar sign was missing, basically.

abuchay

@TedShifrin correct me if i am wrong; isn't $A$ free abelian group since it is a subgroup of $\mathbb{Z}^n$ and it can be spanned by ${e_i}_{i \in I}$.

Ted Shifrin

It is a theorem, @abuchay, that a subgroup of a free abelian group must be free. It is non-obvious.

abuchay

1:21 AM

@TedShifrin Alright, i can prove that. Thanks.

Fargle

Heya chat.

Érico Melo Silva

sup @Fargle

Fargle

Nothing much, how have you been?

Érico Melo Silva

mostly good, agonizing over grad decisions but that’s not a real problem. hbu?

Ted Shifrin

Wow, it's a @Fargle.

Fargle

1:26 AM

Just hanging around. Been reading mostly.

Ted Shifrin

Actual literature?

anakhro

mr. shifrin

Q: why is the isomorphism $T_p(T_pM)\cong T_pM$ natural

Ted Shifrin

You're just asking about $T_x(V) \cong V$ for any vector space $V$.

You have a global chart.

anakhro

Yes, but how is that iso natural

Ted Shifrin

How is it not?

anakhro

1:32 AM

Doesn't it depend on the basis you use?

Ted Shifrin

Hell no.

Fargle

@TedShifrin No, math. :P

anakhro

W H A T

Ted Shifrin

Blah, @Fargle :P

Fargle

I do read other things. Just haven't lately.

Érico Melo Silva

1:33 AM

the iso literally looks like (p, v) maps to v

what could be more natural

anakhro

The proof I know of $T_pV \cong V$ for a vector space $V$ begins with "take a basis of $V$..."

Ted Shifrin

One definition of natural would be that if you have an iso $\phi\colon V\cong W$ it induces an iso $T_xV \cong T_{\phi(x)}W$?

Let's actually look at the definition of natural.

anakhro

pulls out Mac Lane

Ted Shifrin

For me, I'd show $T_p(V) \cong T_0(V) = V$.

As Eric suggests, I don't see why you need any basis.

Érico Melo Silva

@Fargle pshhhh nerd

Ted Shifrin

1:37 AM

I suppose it depends on which definition of tangent space you use.

anakhro

Which one are you using?

Ted Shifrin

I use whichever one I feel like.

Here I'll use equivalence classes of curves.

Fargle

@ÉricoMeloSilva :(

Érico Melo Silva

me too tho

anakhro

Is it not natural for any definition? I don't see why it would matter.

All definitions are equivalent, maybe one is easier?

Ted Shifrin

1:38 AM

I think you are using a naïve definition of "natural," i.e., independent of choices. But there is a definition more like the one I said above.

I.e., it corresponds right under morphisms.

anakhro

Alright, but I just want to be sure that there is no possible way for my linearisation to be anything but the one I define it to be.

So I want there to be no choice whatsoever for the identification of $T_pM$ with the tangent space to the fibre.

However, it's not clear to me how or that I can do this.

Ted Shifrin

I've made several remarks.

anakhro

This is what "natural" was meaning to me.

Ted Shifrin

Well, no one will protest that $T_p(V) \cong V$ is not natural. You should just learn it.

anakhro

I am struggling to show it for myself.

MrAP

1:43 AM

For four proper fractions $a, b, c, d$ X writes $a+ b + c >3(abc)^{1/3}$. Y also added that $a + b + c> 3(abcd)^{1/3}$. Z says that the above inequalities hold only if a, b,c are positive.
(a) Both X and Y are right but not Z.
(b) Only Z is right
(c) Only X is right
(d) Neither of them is absolutely right.

Ted Shifrin

Translate curves through $p$ to curves through $0$.

anakhro

Normally I would set up a chart which is just an isomorphism $V\cong\mathbb R^n$ (so pick a basis).

MrAP

Someone please help me with this.

anakhro

@TedShifrin so just take $\beta(t) := \gamma(t) - p$

Ted Shifrin

Yes, of course.

@MrAP: I disqualify the question entirely on the grounds that "neither" in (d) is totally incorrect.

MrAP

1:45 AM

Why is that?

Ted Shifrin

Because "neither" refers to precisely two.

Érico Melo Silva

that’s what ted said about showing the natural isomorphism to T_0V

MrAP

I was thinking it would be either (a) or (c).

Ted Shifrin

You can't rule out $Y$ if $d=1$.

MrAP

(a) seems more likely to me.

Ted Shifrin

1:46 AM

"more likely" isn't mathematics.

anakhro

So that shows that any curve through $p$ with tangent vector $v$ at $p$ corresponds to a curve through $0$ with tangent vector $v$ at $0$.

Ted Shifrin

Then you have to give an example where they're not all positive.

anakhro

So translation is an isomorphism

Ted Shifrin

Oh, I can give lots of examples where they're not all positive.

anakhro

Well, the first you suggested.

Ted Shifrin

1:49 AM

BTW, @MrAP, don't you know that the GM/AM inequality can be an equality sometimes?

MrAP

Yes. I know.

Ted Shifrin

Anyhow ... I'm out of here.

anakhro

:( ted i luv u

Érico Melo Silva

@TedShifrin tchau tchau

MrAP

@TedShifrin, will it be (d)?

If all variables are substituted with 1, the inequalites do not hold.

And I think the "neither" in (d) should "none".

anakhro

2:04 AM

@MrAP as Ted said, if it is (d), then you should be able to find a counterexample

MrAP

It is 1

If all the variables are 1 then the inequalities do not hold.

@anakhro

anakhro

>proper fraction

A proper fraction is one of the form $a/b$ where $a < b$ and $a,b\in\mathbb Z$

MrAP

Oh I forgot about that.

Its 1/2

If all the variables are 1/2, then the inequalities do not hold.

I think.

anakhro

Do better than just think: calculate.

MrAP

In fact, if all the variables are equal proper fractions then the inequalities do not hold.

I had calculated before writing that.

The first inequality becomes $3/2>6$ which is false and the second inequality becomes $3/2>3(16)^{1/3}$ which is false.

Correct?

anakhro

2:17 AM

well you are right that the inequalities do not hold, but your calculation is wrong.

Set $a=b=c=d$ and re-write your inequalities.

MrAP

Oops. I am totally out of my mind. It would be $3/2>3/2$ and $3/2>3/(16)^{1/3}$.

anakhro

1/16 in the last cube root

MrAP

So its option (d).

anakhro

If you say so.

MrAP

What do you mean by that?

anakhro

2:24 AM

I mean that if that is what you have shown, then that's what it is.

Not to cast you into doubt.

MrAP

You don't say so?

anakhro

No, I haven't looked at the question

Like the other options

MrAP

2:38 AM

Ok. Fine.

Buddhini Angelika

3:57 AM

Hi, what is meant by when you say, "a group $H$ embeds in to $Aut(C_{p}^2)=GL(2,p)$, where $C_{p}^2$ denotes the (non-cyclic) abelian group of order $p^2$ and $p$ is a prime?

Can we think about the structure of H by this, if we know the order of H too?

Ted Shifrin

4:40 AM

Well, you know (or can find) the order of $GL(2,p)$, so the order of $H$ would need to divide that.

Buddhini Angelika

4:55 AM

Yes, @TedShifrin the order of $GL(2,p)$ is $p(p+1)(p-1)^2$. But I found this on a classification of groups of order $p^2qr$. There order of $H$ should be $qr$ and it is present as $G = C_{p}^2 \rtimes H$. I want to know that whether we can know the structure of $H$ that can be present?

Like can we think $H=C_q \times C_r$ or something like that from the given data?

When we say it embeds into $GL(2,p)$ does that mean we can say $H=C_q \times C_r$? or $H=C_q \rtimes C_r$? or should we consider all possibilities?

Buddhini Angelika

5:13 AM

Please help me with this question. Thanks a lot in advance.

Tobias Kildetoft

8:39 AM

Man, how did it become so popular to group together windows from the same application that desktops stopped offering the option to not do that?

Buddhini Angelika

9:20 AM

When considering finite groups $G$ of order, $|G|=p^2qr$, where $p,q,r$ are distinct primes, let $F$ be a Fitting subgroup of $G$. Then $F$ and $G/F$ are both non-trivial and $G/F$ acts faithfully on $\bar{F}:=F/ \phi(F)$ so that no non-trivial normal subgroup of $G/F$ stabilizes a series through $\bar{F}$.

And when $|F|=pr$. In this case $\phi(F)=1$ and $Aut(F)=C_{p-1} \times C_{r-1}$. Thus $G/F$ is abelian and $G/F \cong C_{p} \times C_{q}$.

In this case how can I write G using notations/symbols?

Is it like $G \cong (C_{p} \times C_{r}) \rtimes (C_{p} \times C_{q})$?

Buddhini Angelika

9:33 AM

First question: Then it is, $G= F \rtimes (C_p \times C_q)$. But how do we write $F$ ? Do we have to think of all the possibilities of $F$ of order $pr$ and write as $G= (C_p \times C_r) \rtimes (C_p \times C_q)$ or $G= (C_p \rtimes C_r) \rtimes (C_p \times C_q)$ etc.?

As a second case we can consider the case where $C_q$ acts trivially on $C_p$. So then how to write $G$ using notations?

There it is also mentioned that we can distinguish among 2 cases. First, suppose that the sylow $q$-subgroup of $G/F$ acts non trivially on the sylow $p$-subgroup of $F$. Then $q|(p-1) and $G$ splits over $F$. Thus the group has the form $F \rtimes G/F$.

Akiva Weinberger

9:52 AM

So why does the Schrödinger equation look the way it does

(Also anyone else notice that ö looks like a shocked face)

Ü

Ultradark

$\vec F_{\Bbb Q^2}=(x,y)$

$\vec F = \Bbb Q^2 \to \Bbb Q^2$

Silent

10:51 AM

I am having hard time understanding 'unique upto associates' in the definition of unique factorization domain.

I am using this definition:

please help!

Alessandro Codenotti

You want to say that $\Bbb Z$ is a UFD even though $6=(-2)(-3)=2\times 3$ has two distinct factorization, because they are different in a way which shouldn't matter

Hi @Balarka

Balarka Sen

Hi!

ÍgjøgnumMeg

Hey @Alessandro @Balarka et al

Balarka Sen

Hey @ÍgjøgnumMeg

ÍgjøgnumMeg

waddup

Alessandro Codenotti

11:04 AM

Hi @ÍgjøgnumMeg! How are you?

ÍgjøgnumMeg

Hey @Alessandro, I'm alright, working atm and trying to prepare for my interview lol

How about you?

Alessandro Codenotti

When will it be?

ÍgjøgnumMeg

This monday :(

Alessandro Codenotti

@ÍgjøgnumMeg I finished my exams so now I'm free until the new semester begins in April, I'm looking forward to it

ÍgjøgnumMeg

Nice :)

Alessandro Codenotti

11:06 AM

Oh, that's soon! What are you interviewing for?

ÍgjøgnumMeg

You're ditching alg. geo?

It's for a full DAAD Scholarship

which I've been shortlisted for

Balarka Sen

@Alessandro Teach me geometric group theory lmao

Silent

@AlessandroCodenotti Thank you

Alessandro Codenotti

@ÍgjøgnumMeg Yeah, I wanted to go to the lectures without doing the exam at the end, but it overlaps with a logic course

Balarka Sen

now that ur free i mean

Alessandro Codenotti

11:07 AM

Ah, that's great, good luck for the scholarship!

Balarka Sen

Good luck from me as well

Alessandro Codenotti

I think you already know more GGT than me :P

Balarka Sen

Nah

I don't know anything precisely

Alessandro Codenotti

I'm free as in having no lectures and no exams, I'm still preparing a seminar talk for the next semester

Balarka Sen

Oh what's it on

Alessandro Codenotti

11:09 AM

Kunen's inconsistency theorem and the large cardinals I1, I2, I3 (they ran out of cool names I guess) on the verge of inconsistency

Balarka Sen

Yikes

runs away

Alessandro Codenotti

lol

Balarka Sen

Have you looked at the proof that word problem is solvable for hyperbolic groups?

Sounds up your alley

Alessandro Codenotti

Kunen's inconsistency basically says that there is no nontrivial elementary embedding of the universe $V$ into itself

Yep, you have to show that they admit Dehn presentations

Balarka Sen

What's a Dehn presentation

Alessandro Codenotti

11:16 AM

A presentation $\langle S\mid R\rangle$ is a Dehn presentation if for some $n\in\Bbb N$ there are words $u_1,\cdots,u_n$ and $v_1,\cdots, v_n$ such that $R=\{u_iv_i^{-1}\}$, $|u_i|>|v_i|$ and for all words $w$ in $(S\cup S^{-1})^\ast$ representing the trivial element of the group one of the $u_i$ is a subword of $w$

If you have such a presentation there's a trivial algorithm to solve the word problem: Take a word $w$, check if it has $u_i$ as a subword, in that case replace it by $v_i$, keep doing so until you hit the trivial word or find no $u_i$ as a subword

There is good motivation for such a definition here

Balarka Sen

Let me see if I can interpret the Dehn presentation in terms of the van Kampen diagram or something.

It's some kind of cancellation condition; if my word in $F(S)$ is trivial in $G$ then there is some large subword $u_i$...

Well, if $w$ contains $u_i$ as a subword then in $G$ length of $w$ gets reduced

Because I just replace $u_i$ by $v_i$ in $w$, as $u_i = v_i$ in $G$.

Oh it's some shortening procedure for geodesics

Alessandro Codenotti

Yeah the proof that hyperbolic groups have Dehn presentations is basically that you can take shortcuts with local geodesics but the details are quite ugly, they're in the paper I linked above

(also this is something I know nothing about but hyperbolic groups are automatic groups as well which is a much bigger class of groups with solvable word problem)

I'm leaving for lunch now, bye!

Balarka Sen

Cya! I'll think about this a bit

@AlessandroCodenotti This is equivalent to finding a geodesic representative for the word, right? The "final word" after all these reductions would be a length minimizer

ÍgjøgnumMeg

11:35 AM

@Alessandro thank you :) I am just relearning a lot of the stuff from my dissertation so I can waffle about it

Balarka Sen

So I don't know how to do it precisely for hyperbolic groups, but if $S$ is a surface of genus $g \geq 2$, to get a geodesic representative for a class $[\alpha] \in \pi_1(S)$ where $\alpha$ is an embedded loop, one lifts it to $\widetilde{\alpha}$ in $\Bbb H^2$ by the locally isometric universal covering, and then the deck transformation corresponding to $[\alpha]$ is an isometry of $\Bbb H^2$ which preserves the embedded arc $\widetilde{\alpha}$

It has to be an isometry fixing a geodesic $\gamma$ with endpoints at the boundary being the same as the endpoints of $\widetilde{\alpha}$.

Consider the homotopy of $\widetilde{\alpha}$ to $\gamma$ by straightline homotopy, but straightlines being the hyperbolic geodesics. This is $\pi_1(S)$-equivariant, so projects to a homotopy of $\alpha$ and the image of $\gamma$ (which is a geodesic in $S$) downstairs, and you have your desired representative

I don't know how to interpret this coarsely in $\pi_1(S)$

Akiva Weinberger

Hi

Balarka Sen

Heya

Akiva Weinberger

Say we take the pairs (0,100), (1,99), etc, (50,50)

Pairs of integers adding to 100

Let's do the sieve of Eratosthenes

2

We're left with (1,99), (3,97), (5,95), etc, (49,51)

Yeah?

Now 3

Notice that (1,99) dies ('cause of the 99). (3,97) doesn't die just because we don't sieve out the prime itself, but (7,93) dies and (9,91) dies

Does that make sense so far?

So we started with 51 pairs, then went to 24

and now we have (3,97), (5,95), (11,89), (17,83), etc, (47,53)

so that's 9 pairs

I realize 1 never gets sieved out by Eratosthenes, so if instead of 100 we had some number that was 1 more than a prime, it would never get sieved out

@BalarkaSen Am I making sense?

I'm doing Goldbach here

So we have a lower bound of 50(1/2)((3-2)/3) pairs surviving I think

(which works 'cause that's 8⅓ and we have 9 pairs)

Then what, we sieve out the 5s

(5,95) and (35,65) die and I think everything else stays

We're left with (3,97), (11,89), (17,83), (23,77), (29,71), (41,59), and (47,53)

That's 7 pairs

and we can get a lower bound of 50(1/2)((3-2)/3)((5-2)/5)=5 which still works

@BalarkaSen So I have someone on Reddit who is really bad at English (and communicating in general) who is convinced he has a proof of the Goldbach conjecture

and his argument is basically, do the sieve of Eratosthenes on the pairs $(k,2n-k)$, give a lower bound for the surviving pairs at each step, and show that the lower bound is never 0

like I started doing above

So I haven't really thought it 100% through yet

but right now my task is to figure out why this doesn't work

user2723984

1:39 PM

hi everyone, a quick question: is there a name for the "inverse" of the commutant? i.e. the problem of having a sub algebra and trying to find if it can be seen as the commutant of another algebra?

Akiva Weinberger

@BalarkaSen OK so I see the main problem:

once we've sieved out all the multiples of 2,3,…,$p_{k-1}$ in the range $[0,n]$, we assume that roughly $1/p_k$ of the remainder are multiples of $p_k$

and, through numerical experiments, that seems to be true

but the problem is, the numbers in the range $[0,n]$ that aren't multiples of 2,3,…$p_{k-1}$ aren't necessarily evenly distributed

so it's theoretically possible that a much higher than expected proportion of them is a multiple of $p_k$

Silent

2:12 PM

I don't understand in above proof why p
must be associate to one of the irreducibles occurring either in the factorization of a or
in the factorization of b.

Paras Khosla

Hi! I was looking for books on analysis and set theory. I saw this link but I'm not able to figure out which book this chapter is a part of. Does anyone know?

chrome-extension://cbnaodkpfinfiipjblikofhlhlcickei/src/pdfviewer/web/viewer.htm‌l?file=math.uh.edu/~dlabate/settheory_Ashlock.pdf

Silent

@ParasKhosla, sorry to derail the discussion. Which book are you using for complex analysis?

Paras Khosla

@Silent I'm not using any book. I took a course from Coursera offered by Wesleyan University.

I'm currently looking to research on what book to use. I found this chapter of a book which seems to be good but I can not seem to figure out which book it is

anakhro

@ParasKhosla eldar.mathstat.uoguelph.ca/dashlock/math4140

@Silent I have yet to find a really good book on complex analysis

Paras Khosla

@anakhro Is there no link for hardcopy? maybe on Amazon or elsewhere

anakhro

2:25 PM

But Visual Complex Analysis is probably one of the better.

@ParasKhosla afaik these are strictly online notes.

Paras Khosla

Oh. There's a different feeling while studying from an actual book. I guess I'll have to look for other resources. Although thanks a lot

anakhro

You could just print it out.

Probably cheaper than buying a book

Silent

@ParasKhosla OK! Thank you very much for suggesting this! seems that it covers many topics

Paras Khosla

Yeah that's still an option

What? @Silent the course?

Silent

@anakhro in india, it is way cheaper to by an indian edition of book, better, if you buy used copy! :)

anakhro

2:29 PM

@Silent I am sure there are budget printers who allow you to print even cheaper.

They have to make money off the books somehow.

Silent

@ParasKhosla yeah. it covers conformal mapping, residue theorem etc! the core and cool topics

Paras Khosla

Yeah it's really great. Anyways are you a math student in India? @Silent

Silent

@anakhro Well, they print in bulk, and on really cheap paper, almost transparent and very thin, and offset machine is really cheaper per page than a printer, you know, but you should be printing in bulk, its all economy of scale.

@ParasKhosla Yes, I am Indian, and trying to get in some good masters progam in math.

Paras Khosla

Oh cool @Silent

Silent

@ParasKhosla, so that book does not seem be be either on set theory or analysis! Which subject does it cover?

anakhro

2:36 PM

Basic algebra

Paras Khosla

Yeah perhaps that abstract algebra.

rings and groups in further chapters

Silent

@anakhro oh!

anakhro

@Silent did you do an undergrad in math?

Silent

@anakhro, its my first time that I saw graphs in algebra text. I knew that graphs are closely related to (equivalent to?) topology. How does algebra relate to graph?

@anakhro well, not formally. (I regret that.)

anakhro

Graphs are largely a combinatorial object. Here they seem to be just adding an additional topic of graphs at the end, despite it not really pertaining that closely with the rest of the material.

That being said, algebra sees many applications in graph theory.

Algebraic graph theory

Algebraic graph theory is a branch of mathematics in which algebraic methods are applied to problems about graphs. This is in contrast to geometric, combinatoric, or algorithmic approaches. There are three main branches of algebraic graph theory, involving the use of linear algebra, the use of group theory, and the study of graph invariants. == Branches of algebraic graph theory == === Using linear algebra === The first branch of algebraic graph theory involves the study of graphs in connection with linear algebra. Especially, it studies the spectrum of the adjacency matrix, or the Lap...

I can probably guess that they are using symmetries and permutation groups on graphs in this course.

For example, orbits and studying the automorphism groups of graphs.

Silent

2:43 PM

@anakhro I have heard really good thing about Palka. Also, if you do not worry about little sacrifice of rigor (e.g. counterclockwise orientation based on your intuition, rather than, on winding numbers, etc.), Howie's Complex analysis is good. It is teeming with typos here and there, but you will be fine, i think. Also, thisbook contains all the solutions in appendix!

Ultradark

Can a triangle on a surface

Silent

@anakhro Thank you very much for all this information!

@anakhro How did you guess that? :)

anakhro

automorphism groups are basically where group theory developed from.

Automorphisms, and then symmetries.

Graph theory makes heavy use of both of these concepts.

Silent

ok!

@ParasKhosla, how can i increase speed in coursera video? can i see those vids in youtube, as we can do in edx?

Ultradark

Can you tell the geometry of a surface based on how

A triangle is curved on it

chandx

2:50 PM

Got a simple question: I gotta find kernel of linear transformation $F(P)=xP^{''}(x) + (x+1)P^{'''}(x)$ where $F: \mathbb{R}_3[x] \to \mathbb{R}_3[x]$, so I think it would be just $\ker (F) = \{ ax+b : a,b \in \mathbb{R} \}$ since only polynomials of degree at most 1 would give zero polynomial in this case

am i right?

Lucas Henrique

3:04 PM

hi chat

anakhro

@Ultradark you will have to define "geometry".

vzn

in The h Bar, 17 hours ago, by vzn

Quantum Machine Appears to Defy Universe’s Push for Disorder / quanta mag

anakhro

Hi @vzn

vzn

↑ hi all really jazzed about this new discovery, anyone into dynamical systems theory? has a lot of neat new math, looks breakthru or even revolutionary :)

Lucas Henrique

@chandx you're looking for all the $G = P''$ such that $xG + (x+1)G' = 0$; if $G \neq 0$ you can solve the DE to get $G'/G = -x/(x+1) = -1 + 1/(x+1) \implies \ln G = -x + \ln(1+x) \implies G = (1+x)e^(-x) + C$ which is obviously not a polyonomial, so $G = 0$ and thus $P = ax + b$

could you suppose that $\operatorname{deg} P \geq 2$ and show that you wouldn't have nonzero polynomials? Sure.

but... yeah, idk why i did this.

anakhro

3:15 PM

@chandx the kernel is all degree 1 polynomials, si.

But you can just calculate it easily.

Let $P(x) = a + bx + cx^2 + dx^3$, find $P''$ and $P'''$, and then find $F(P)$.

Then let your expression for $F(P) = 0$, then solve for $a,b,c,d$.

Paras Khosla

@Silent @Silent Yes you can speed up the video look at the panel on the bottom, right side maybe has the tool to modify speed

Silent

Thank you so much!

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