Mathematics - 2017-01-06 (page 2 of 6)

Kasmir Khaan

09:16

hi chat

can anyone explain to me wierstrass M test for uniform convergent series?

Secret

09:31

[Wheels]
Given $8x/8=x+0*8/8=2*8/8=2+0*8/8=2*0/8$. Find the value of $x^2+3x^3$.
Consider: $x^2+3x^3+0/8=x^2+0/8+3x^3=x(x+0/8)+3x^3=x(2+0/8)+3x^3=x2(1+0/8)+3x^3=2(x+0/8)+3x^3=2(2+0/8)+3x^3=2(2+0/8)+3x^3=4+0/8+3x^3=4+3(0/8+x^3)=4+3x^2(0/8+x)=4+3x^2(0/8+2)=4+3(0/8+8)=4+0/8+24=28+0/8$

Akiva: Actually, on second inspection, I don't really see any advantage conveyed as the zero term 0/8 basically just sits there and not doing anything, and the nonzero term part is basically identical to what happens when you do direct substitution

Secret

09:48

[More wheels] Given $x^5+1=0$ find $x^2+8+7x$

Step one:
$$x^5+1=0$$
$$x^5+1+0/k=0/k$$
$$x^5+0/k+1=0/k$$
$$(x+0/k)x^4+1=0/k$$
= stuck

Hey I already not posted pictures, why is it still span the whole page?

SteamyRoot

Formulas space out alot if you put them in different environments.

Not sure if this works in the chat, but putting all of it in one big align environment would probably fix that.

Secret

Hmm, ok should try that next time

Quick question: Is a klein bottle genus 1 or genus 2?

Balarka Sen

It doesn't have well-defined genus

SteamyRoot

If by genus you mean non-orientable genus, then 2

Balarka Sen

10:04

(which is the number of cross caps, not handles. But yeah some people calls it genus)

Secret

Yeah, which is why I am kinda confused when this guy at 7:14 said the klein bottle is genus 2 and proceed to draw two noncontractile loops on its one sided surface, because by rotating the thing around vertically, the 2nd loop will coincide with the first one thus making them effectively the same loop

Super Bottle - Numberphile

►

By 'rotating vertically' ,I mean a rotation simialr to how one rotates a cylinder along its axes, which for the 4D case will be a rotation centered on the horizontal cross section of the klien bottle (i.e. its circular base) thus in the projection a klien bottle sort of gyrates around its vertical axis without chagning shape

SteamyRoot

Meh. Much easier to write the unit square diagrams to see what the connected sum is :P

I did my differential topology project on the euler characteristic and the connected sum of surfaces

Balarka Sen

$\chi$ is nice

Secret

In that case, it is known a klein bottle is made by two moebius strips glued together along its boundaries, thus it has two cross caps hence nonorientable genus 2.
I am not sure if a connected sum of two moebius strip has a name, though, because such thing will have two boundaries homeomorphic to circles

Balarka Sen

It's a Klein bottle with two punctures

Secret

10:14

ok that makes sense

Btw we also knew in a torus we will describe that "hole" in the middle with a genus since it is one of the things that ensure a horizontal loop drawn on its surface is noncontractile, but a klein bottle also look kinda like a torus. Since the usual genus is not well defined on a klein bottle, how do we describe that "hole" apparently formed by the handle cnnected to the bottle's main body?

SteamyRoot

I always thought it was crazy how you can turn a handle into a $2$ cross-caps if you already have a cross-cap present.a

Secret

that is, what is this hole looking thing in the middle of a klein bottle?

SteamyRoot

Hole looking thing?

The self-intersection?

Secret

not the self intersection, I am thinking this 4 dimensionally, thus a klein bottle is like a mobius stripe with a circular cross section instead of a line cross section (which is justified by it being two moebius strips gloud together at the boundaries

Balarka Sen

@SteamyRoot This is my favorite fact in surface theory. There is also a complex analogue: $(S^2 \times S^2) \# \overline{\Bbb{CP}^2} \cong \Bbb{CP}^2 \# 2\overline{\Bbb{CP}^2}$

Secret

10:22

@SteamyRoot (wikipedia page on klein bottle) Note the construction, the red curve somehow circled a hole looking thing in the middle formed by the handle. That is, the cylidner is rolled up so that its back side glues to the front side, thus overall in 4D there seemed to be something that look like a hole formed by the fact that the cylinder curves around to glue to itself

SteamyRoot

@BalarkaSen That's really neat!

Balarka Sen

I think it's best understood by the blowup construction.

SteamyRoot

A curve on the Klein bottle circling something resembling a hole isn't exactly surprising given the fundamental group of the Klein bottle, or am I mistaken here?

Speaking of fundamental groups, I really should finish this goddamn paper since I'm meeting my co-author from Neuchâtel on monday...

Balarka Sen

What's the paper on

SteamyRoot

Reidemeister spectra of crystallographic groups.

You define an equivalence relation linked with an automorphism $\varphi$ by $\sim_\varphi$ on a group $G$ by $g \sim g' \iff g' = hg\varphi(h)^{-1}$

Balarka Sen

10:33

I have only vaguely heard of the words :)

hmm, ok

Secret

Hmm ok, the wikipedia did mentioend about the klein bottle having a pair of fundemental regions that are mirror images of each other together forming afundemental region of a torus, I guess that's how that hole looking thing is characterised abstractly

SteamyRoot

And then calculate the number of equivalence classes for every equivalence relation

Turns out the fundamental group of the Klein bottle always has infinitely many equivalence classes, whereas the torus can have any number.

Balarka Sen

Huh, very interesting

Secret

The fundemental group of the klein bottle is given by $<a,b | ab = b^{-1}a>$. how to unpack information from that (other than it seems nonabelian by trying to do some quick check via conjugates) besides drawing a table I have no idea as my group theory skill is still preliminary

Balarka Sen

Does any abelian group have finite number of equivalence classes?

SteamyRoot

10:38

Well, taking the identity on any infinite abelian group gives infinitely many equivalence classes.

But every finitely generated abelian group has at least one automorphism for which the equivalence relation has finitely many equivalence classes.

Balarka Sen

Ah, sorry, yes, the latter is what I meant to ask.

SteamyRoot

arxiv.org/pdf/1402.1861.pdf is where my promotor proved that there exist abelian groups (non-finitely generated) for which it doesn't hold.

Balarka Sen

Cool stuff

Secret

10:54

it seems most of the counterexamples tend to arise in groups involving rationals (or maybe I have not read enough)

Anonymous

How to solve the limit : i.sstatic.net/NNb2v.png

Anonymous

I used L hospital rule

Anonymous

And I think the limit does not exist

Anonymous

as I am getting (cos a)^4

Anonymous

11:04

after l hospital

Anonymous

does anyone here know how to solve this ?

Secret

11:23

maybe first try evaluating the integral in terms of a (as powers of trig integrals have analytic formulae) before computing the limit?

Anonymous

@Secret Can't I directly use L Hospital ?

robjohn

@Mystic what is the derivative of the numerator?

Alessandro Codenotti

@Mystic No, L'hospital requires $\lim\frac{f'(x)}{g'(x)}$ to exist and here it doesn't

Anonymous

@robjohn W.r.t a ?

Anonymous

@AlessandroCodenotti I see, so limit does not exist or it exists and I have to calculate it some other way ?

robjohn

11:28

@Mystic yes, as that is the only free variable

mercio

o.o

Anonymous

@robjohn Using Leibniz rule it comes out to be (sin(a))^4

Anonymous

Right?

mercio

all those rules

robjohn

@Mystic yes. What is the derivative of the denominator?

Secret

11:30

I kinda reluctant to l hopital that integral unless i can be shown to unformed converge hence I can use the differentiation under the integral sign.

The integrand is positive thus the integral should diverge as a tend to infinity. Its derivative (assuming i can differentiate under the integral sign here) are trig functions thus it should also diverges unless it is odd

Anonymous

@robjohn 1 ? It comes out to be an oscillating type of limit...

Anonymous

(sin(a))^4/1

Alessandro Codenotti

@Mystic so it might exist or not, l'hopital doesn't tell us enough

mercio

yeah l'hospital's rules doesn't apply like this

robjohn

@Secret it's just the Fundamental Theorem

Secret

11:30

o oops...

Anonymous

@AlessandroCodenotti Ummm, so what other way is there to ensure limit exists or not ?

mercio

have you tried plotting the numerator ?

robjohn

@Mystic oh, this is the limit as $x\to\infty$. Sorry

Alessandro Codenotti

$\displaystyle\int\limits_0^a(\sin(x))^4\mathrm{d}x=\frac38a-\frac14\sin(2a)+ \frac{1}{32}\sin(4a)$

Anonymous

@AlessandroCodenotti Wow! You calculated the whole value of the integral ? That look complicated...anyway if you expression is correct the answer should be 3/8...right ?

Anonymous

11:33

because sine always lies between -1 and 1

mercio

yes

Anonymous

and in denominator it tends to infinity

Anonymous

thanks

Anonymous

i think i got it

Anonymous

:)

Anonymous

11:34

I learnt something new : L Hospital isn't applicable everywhere :)

Secret

yeah that's one of those problem where l hopital does not work because the limit of the d(numerator) does not exist

(deluminator...?)

Alessandro Codenotti

there are some hypothesis that needs to be satisfied for l'hopital to apply

Anonymous

@AlessandroCodenotti 0/0 and infinity/infinity form. And the one you told just some time ago. And anything else ?

robjohn

$$\int_0^{2\pi}\sin^4(x)\,\mathrm{d}x=\frac{3\pi}4$$

$$\int_0^{2k\pi}\sin^4(x)\,\mathrm{d}x=\frac{3k\pi}4$$

Alessandro Codenotti

@Mystic you need the functions to be derivable and $g'(x)\neq 0$ (apart from the point in which you're doing the limit)

user400188

11:37

I was on here in the first few seconds of 2017; but I didn't have the rep to comment then so I couldn't get the Hat. I obtained the rep on the first but more than 12 hours after.

Anonymous

@AlessandroCodenotti gotcha =)..thanks a lot!

Secret

@robjohn easily justified by the diagram, k is the number of lumps you integrated of the $\sin^4 x$ function

so that means, one lump has an area of $\frac{3\pi}{4}$

I wonder if it holds in general for positive periodic functions...?

Hmm...
$$\int_0^{kT} f(x+T)dx=k\int_0^T f(x)dx$$

robjohn

@Secret doesn't need to be positive.

Secret

right (thinking too much of how they will cancel out for odd functions)

When I get to lie groups, I will definitely use them to investigate the symmetry of integrals by treating it as an abstract functional. That might help improve my pattern finding abilities in computing integrals

Anonymous

11:53

@Secret In this formula why is there a T in f(x+T)..what is the meaning of it ? Shouldn't it just be f(x) integration from 0 to kT equals k times the integration of f(x) from 0 to T ?

Secret

that's the period of the periodic function. The above equation does not hold in general if the function is not periodic or the integration domain does not cover an integer number of periods

Anonymous

@Secret No, I get that T is the period of the function. But why did you write f(x+T) ? Shouldn't it just be $$\int_0^{kT} f(x)dx=k\int_0^T f(x)dx$$ ?

Anonymous

I wanted to understand the significance of adding the T in f(x+T)...

Secret

O, that is because I try to emphasise we are dealing with a periodic function here., although you are right that since f(x)=f(x+T), both integrals will give the same results

Anonymous

@Secret Oh I see, I thought you meant something else. I get it :)...are you a college student ?

Secret

11:57

Currently waiting for my chemistry PhD to start on march, otherwise doing some abstract algebra self study

Anonymous

Chemistry PhD ? :O And studying Maths? :P Why ?

Secret

Because I am interested in it, its really a simple reason

it does help in the end though as in computational chemistry you need a lot of group theory

Anonymous

I see...great!

Krijn

@Secret Represenation theory as well?

Secret

indeed, character tables, though I plan to postpone reading that section until later, cause I want to study groups in general not just its representations in terms of linear oeprators.

Once I am done with lie groups, I will quickly head back to it (I might actually doing that at the same time in my lie group study given the nature of most lie groups)

In my honours year, though, I do learn the basics of character tables and how to use them to characterise molecular orbital symmetries, but I would like to understand deeper about it and groups in general (because their onion structure of sub…

N3buchadnezzar

12:05

@robjohn

Can you shed some light on the transition from eq (6) to (7), in your proof here? math.stackexchange.com/questions/171599/…

I can follow it easily untill that step. Having some problems justifying how one can interchange the concave function $\varphi$ with the integral

Krijn

@Secret This is indeed a good approach I think

Secret

and interestingly for abstract algebra, usually the more abstract it is presented, the easier for me to understand. Possibly because when things are presented abstractly there are not as much wall of text to read and less distraction from specific conditions in the examples

Krijn

@Secret Start with category theory then

Anonymous

Is it possible to solve this question without having to solve a second order differential equation ? i.sstatic.net/8N53O.png

Anonymous

Secret

12:14

@Krijn what are good books or sources on category theory? So far I only know few isolated concepts such as forgetful functor, category of groups, universal property and thus the whole thing is not very coherent. A good book will help organise all thise isolated infomrmation into a coherent whole

Krijn

@Secret I was mildly joking, I don't think you want to start studying category theory until you know a lot of abstract algebra

Neither do I know of books in particular, I so far have only really studied Mac Lane's

Secret

I also noticed rings are frequently discussed in this chat. That is something I intially planned to go into except i truly underestimate the size of group theory compared to other algebraic topics

Roughly 6 months ago, I initially planned my abstract algebra study as follows:
1. Semigroups
2. Quasigroups and loops
3. Monoids
4. Group theory (all) and lie algebras
5. Rings, Semirings, near rings and near semirings
6. Integral domains
7. Fields
8. Modules
9. Lattices and bollean algebras
10. Nonassociative algebras
11. Category theory and universal algebra

however, I am completely off schedule when I reached item 4 (which is where I am when I am not dividing by zero) because I truly underestimate the scope of group theory

SteamyRoot

I'm also surprised you would start with semigroups and quasigroups and such before actual groups.

Secret

because they are more general than groups, my study pattern tend to go for structureless things and add structure one by one as I go

Krijn

You might just wanna go for the basics of group theory, ring theory, and field theory

You can't study all of group theory

SteamyRoot

12:28

^

Krijn

Also, things like semigroups, quasigroups and monoids are not at all that basic to study, in my opinion, when you just want to learn about group theory

Oh, Steamy just said ^

SteamyRoot

Yeah. Because I agree.

Krijn

No, I mean, you just said what I said about semigroups and stuff

So we double agree

robjohn

@N3buchadnezzar Step $(6)$ to step $(7)$ is just integrating over $X$

@N3buchadnezzar was there some question in particular?

SteamyRoot

I understand that going from little structure to a lot of structure may seem a good order of styding, but I really doubt this is the case for abstract algebra.

I'd learn the "nicest" structures first: groups, rings and fields.

Krijn

12:31

Also, Galois theory came much earlier in my algebra studies than most of the later topics

I know really little about Boolean algebras and nonassociative algebras and universal algebra

SteamyRoot

Hmmm... I barely had any Galois theory. And I never really touched any of those either.

robjohn

@Mystic $f(x)=e^{x+1}$

Astyx

Hi chat

Simple Art

:P

N3buchadnezzar

@robjohn How does $$\int_X \varphi\left(\int_Xf(x)\,\mathrm{d}x\right) \,\mathrm{d}x$$ simplify into $$\varphi\left(\int_Xf(x)\,\mathrm{d}x\right)$$?

SteamyRoot

12:37

@robjohn Quick meta-question, if you don't mind. I'm offering a bounty on the following (ex-tumbleweed) question. I tried to solve it myself before, but didn't manage; though I did get quite a few steps.

Given that my partial answer is rather very long to fit in comments, should I post the incomplete answer; or post a separate question which my answer does cover completely; and answer that one myself? I've seen both suggested on meta posts...

Krijn

@SteamyRoot Then maybe Galois theory is less standard than I thought

SteamyRoot

3

Q: Given that $G$ is subdirect product of $A\times B$ and $G/G'$ is not subdirect product of $A/A'\times B/B'$ this group contains $G'$?

Given a direct product $AB=A\times B$, a subgroup $G\le AB$ is a subdirect product of $A$ and $B$ if $AG=BG=AB$. Let $R=G'(B\cap G)\cap G'(A\cap G)$. Apparently it is easy to see that if $G/G'$ is not a subdirect product of $A/A'$ and $B/B'$ then $R>G'>1$. As it is easy to see I suspect I am ov...

robjohn

@N3buchadnezzar $$\int_X1\,\mathrm{d}x=1$$

N3buchadnezzar

@robjohn Yes?

I know that =)

robjohn

@N3buchadnezzar That is given in $(1)$

N3buchadnezzar

12:39

Gah, thanks =)

just a big fartbrain

SteamyRoot

@Krijn Maybe... But my professors were much more into (commutative) rings and groups research-wise. I imagine if you are taught by someone who does research in fields, you'd get a lot more Galois theory.

robjohn

@SteamyRoot If you have asked the question and answered it partly, just include your work. If you are asking a different question, then make a new post.

SteamyRoot

@robjohn It's a question by someone else.

robjohn

@SteamyRoot Oh, then if the question has not been fully answered, and your answer gets part way there, just answer that question, but mention it is only a partial answer.

SteamyRoot

@robjohn Okay, thanks!

Anonymous

12:47

@robjohn How did you get it ? Did you solve the second order differential equation ?

Anonymous

Or is there any shortcut ?

robjohn

@Mystic I differentiated both sides

Krijn

@SteamyRoot I do think Galois theory was one of the most fun and beautiful undergrad courses I took, so I'd advise people to take it

Alessandro Codenotti

@Krijn Galois theory isn't a mandatory undergrad course at my university (most people take it though)

SteamyRoot

It was about 2/5th of my Algebra II course, which has been deprecated completely since this academic year. Students going into their masters have never heard of Galois Theory :P

Krijn

12:57

@AlessandroCodenotti Most of the mandatory stuff at my university is just the basics, nothing specialised

@SteamyRoot :(

Alessandro Codenotti

Same here, you have mandatory courses in the first 2 years of undergrads and then you're (mostly) free to choose what you want in the third

Secret

I only know that galois theory expains why we cannot have closed forms for roots of polynomial equation of deg > 5. Anythign else I am still learnign as part of my self study schedule above

Pichi Wuana

Can someone give me a hint of how to solve $(-1)^n + n(-1)^{n-1} = 2$? I got to $(-1)^n = \frac{2}{1-n}$.

Krijn

$(-1)^n$ is either -1 or 1

Depending on n

Simple Art

@PichiWuana assume odd. Assume even. Solve

Krijn

13:01

Now consider cases where $n$ is odd and even

Pichi Wuana

So I just check for both cases?

Krijn

I think your formula is wrong though

SteamyRoot

Is it?

Krijn

Check it for $n = 4$ or so, gives $1 - 4 = 2$?

Simple Art

@PichiWuana yeah pretty much

Krijn

13:03

@SteamyRoot Is your image the Wina-emblem?

Simple Art

How I feel you should tackle problems like this.

SteamyRoot

Or, with the formula you ended up with, you could just check: for which values of $n$ is $\frac{2}{1-n}$ either $1$ or $-1$

And then check if those values are a solution

Pichi Wuana

Oh so it's similar to the cases where I get rid of the square root. I need to check when I finish...

SteamyRoot

@Krijn yup

Simple Art

Nice idea @SteamyRoot

DHMO

13:21

What do you guys think about the Diophantine equation n!=a!b!?

SteamyRoot

Ummmmm...

Sophie

I think it's big solutions are tremendously unlikely

SteamyRoot

Hard to say what I "think about it", but I like that there are (semi-trivial) solutions of the form $(m!)! = m! ((m-1)!)!$

DHMO

@Sophie why?

mercio

um

$(m!)! = m!(m!-1)!$

SteamyRoot

13:28

Oh, woops, right.

mercio

i have no idea if there should be a finite of infinite amount of nontrivial solutions

DHMO

I would appreciate a probabilistic argument from the density of the prime numbers

SteamyRoot

Have you posted a question on MSE about this?

mercio

I wouldn't be surprised if the ABC conjecture was useful somehow

SteamyRoot

There's a lot more expertise (and willingness to spend a lot of time on questions) there than just among us chat regulars.

Akiva Weinberger

13:32

> 5. Two sides of a triangle are $x=3$ and $y=4$, and the included angle is $\theta=\pi/3$. To a small change in which of these three variables is the area of the triangle most sensitive? Why?

mercio

o.O

rrrw

DHMO

@mercio there is a paper that the ABC conjecture implies the existence of finite non-trivial solutions

mercio

of at most finitely many nontrivial solutions ?

DHMO

yes

mercio

not surprising o.o

Ramanujan

13:35

@AkivaWeinberger theta ?

SteamyRoot

Well, in that case, wait until people finally figured out Mochizuki's (attempted?) proof :P

mercio

i also think it's theta

DHMO

@AkivaWeinberger A = 0.5absinθ, find ∂A/∂a, ∂A/∂b, ∂A/∂θ.

Ramanujan

Area of triangle=1/2 * x*y*sin w (w is the angle)
(When x=3 y=4 w=pi/3)
d(Area)/dx=1/2*y*sin w
=1/2*4*sqrt(3)/2=sqrt(3)
d(Area)/dy=1/2*x*sin w
=1/2*3*sqrt(3)/2=3sqrt(3)/4
d(Area)/dw=1/2*x*y*cos w
=1/2*4*3*1/2=3
As u can see d(Area)/dw is larger the change in area is more sensitive to change in angle

mercio

differentiating $\log A$ is slightly easier

Akiva Weinberger

13:37

@Ramanujan I actually was just about to write, "@TedShifrin, is it $\theta$?" before I lost internet connection

Ramanujan

Copy paste from Yahoo

Akiva Weinberger

(I was trying to verify my answer)

DHMO

∂A/∂a = 0.5bsinθ = sqrt(3)
∂A/∂b = 0.5asinθ = 3sqrt(3)/4
∂A/∂θ = 0.5abcosθ = 3

∂A/∂θ is the largest, so the answer is θ.

Ramanujan

Why largest?

DHMO

because 3 > sqrt(3) > 3sqrt(3)/4

SteamyRoot

13:40

Also, if anyone is willing to try some group theory, there's a bit of rep up for grabbing if you can finish my incomplete answer to this question

DHMO

this is essentially the answer you just copied

Secret

A group AB that is as large as one of its subgroup direct product with its "consituents"? does that mean G somehow generates A or B?

Akiva Weinberger

@DHMO Yes, that's how I did it.

SteamyRoot

@Secret It's bad notation, with $AB$ they mean $A \times B$.

Akiva Weinberger

I didn't mean to post it here to ask the chat how to do it, I meant to post it so Ted would know to which question I was referring. (Ted wrote the problem.)

Secret

13:45

Yeah but they said it is a subdirect product if $A \times B=A \times G=B\times G$. Since all of them are finite groups, AG will be as large as AB despite G < AB, so it wonders me whether G somehow generates B and A. (Well I might stuff up somewhere in my knowledge of direct products, let me read again...)

SteamyRoot

Nono, they mean $G \leq A \times B$ and $A \times B = G(A \times 1) = G(1 \times B)$

Balarka Sen

@AkivaWeinberger Definitely does sounds like a Ted problem :)

Null

if an equationsystem can have none, one or infinite solutions, depending on one variable, what is the most likely outcome if I choose the variable at random?

i guess no solution

mercio

it depends on everything

SteamyRoot

^

Secret

14:09

> Suppose $A' \neq 1$, i.e. $\exists a_1,a_2 \in A$ such that $[a_1,a_2] \neq 1$

But having no elements such that the commutator is the identity is not enough to show A' is trivial, A' could be a nontrivial abelian group?

SteamyRoot

$A'$ is the commutator subgroup. It's generated by all commutators in $A$.

Secret

ok makes sense

so If A'=/=1, it means we are assuming A is nonabelian continue reading proof

Mary Star

Hello!! Can a reducible polynomial in $\mathbb{Q}[x]$ is irreducible in $\mathbb{F}_p[x]$ ?

SteamyRoot

@Secret Exactly: $A$ is abelian if and only if $A' = 1$.

Null

lol, i wrote horrible vid under some video and the creationer said thank you

mercio

14:20

awkward

Null

is an element of $A\times A\times A$ (a,a,a) or ((a,a),a)?

^ what about different Sets

like $A\times B\times C$

mercio

i think it's better to write it $(a,a,a)$

unless you talk about $(A \times A) \times A$

Secret

$A\times A \times A$ should be a 3 tuple, not a two tuple, ((a,a),a) is a 2 tuple (an ordered pair) of (a,a) and a

Null

@Secret fair enough. altho I wish there was a way to define that unambigiously

SteamyRoot

In the category of sets they're all isomorphic :P

Secret

14:24

@SteamyRoot Are all primed labelled groups in that question commutator subgroups given how we seemed to quotient them away in the later part of the statement?

SteamyRoot

Yes, they are indeed

Null

is $\mathbb{Z}\times\mathbb{R}\times\mathbb{C}$ a thing somewhere? Or is $A^n$ much more common?

SteamyRoot

You could define a situation where it's a thing. Never seen it pop up, though.

Mary Star

When an irreducible polynomial $f$ divides the polynomial $g^n$, does $f$ divide also $g$ ?

Astyx

I think so

Alessandro Codenotti

14:30

I think you need them to be polynomials over an integral domain at least

otherwise $g^n$ might be $0$ with $g\neq 0$ and then you have a trivial counterexample

Null

is $g^n$ not some $h\in R[x]$?

Mary Star

We are in $\mathbb{F}_p[x]$

Balarka Sen

@Alessandro You probably need it to be in a poly. ring over a UFD

Mary Star

Is $\mathbb{F}_p$ a UFD?

Balarka Sen

Or at least, that's sufficient

Alessandro Codenotti

14:32

@Balarka yeah I was thinking about that after writing integral domain because it looked suspicious, UFD might be required

@MaryStar every field is an UFD

because there aren't many irreducible elements so "unique factorizations" are trivial

Mary Star

I see!!

Thanks!!

Balarka Sen

@Alessandro Oh, btw, did you understand the Sard's theorem from yesterday?

Alessandro Codenotti

I understand the statement but I have no intuition about it. It's in G-P's book though

Balarka Sen

It's one of the genericity results in the smooth manifolds that doesn't hold in any sense for topological manifolds. Do you see why it implies there are no space-filling curves?

Secret

ok, so far the proof makes sense. Now to the part where is stuck. Might have to think about it. Some rather sketchy venn diagrams seemed to suggest it is true but I need to think carefully on why it is true (because drawings are representations on R^2 thus not always reliable)

Ted Shifrin

14:39

DogAteMy: I didn't check computations, but the discussion was right. That's meant to be a routine standard problem. Eschewing icemageddon and heading home from ATL 3 days early :(

Secret

mercio

woah lol

Ted Shifrin

Hey @Balarka @Alessandro.

Balarka Sen

Hi @Ted

Alessandro Codenotti

Hi @Ted

Ted Shifrin

14:40

You don't need Sard for that. It follows directly from the mean value inequality and a little work.

Balarka Sen

Of course, that's in your book @Ted. It's my favorite approach.

Ted Shifrin

Exercise in my book — tee hee.

Oops, too slow.

Balarka Sen

:)

mercio

@SteamyRoot is the question saying that when $A=B$, the diagonal $D = \{(a,a) \mid a \in A\}$ is a subdirect product of $A \times A$ ?

Balarka Sen

But I am trying to get Alessandro to understand it via Sard because that'd get him a bit ahead in G-P

SteamyRoot

14:42

@mercio Yeah, that's indeed the case.

mercio

because I would have thought that subdirect products would be things like $G \times H$ where $G$ and $H$ are subgroups of $A$ and $B$

so this makes me confused

the name is unfortunate

SteamyRoot

I guess you could interpret "subdirect" in that sense, yeah... :/

Secret

yeah it seems natural to interpret it that way before reading its definition. Glad it is provided in the problem

R > G' suggest we cannot have R = G', I wonder what happens if we assume R=G'...

SteamyRoot

That's what I tried too, but I didn't get anywhere.

mercio

why do they assume that $G/G'$ is a subgroup of $A/A' \times B/B'$ ?

SteamyRoot

14:47

May have missed something, so be sure to let me know (or answer the question!) if you come up with something

Krijn

Has there ever been any attempt to replace grading homework with some better approach?

To give feedback to students

Grading is so mindnumbingly boring

SteamyRoot

@mercio Again, that's really wacky to me. Unless $G' = A' \times B'$, it's not a subgroup in any "natural" way, I think.

Secret

(I forgot) is it always true that for finite groups the quotient of a subgroup H < K H/H' < K/K'?

SteamyRoot

But perhaps there's a good way to identify $G/G'$ with a subgroup if $G$ is a subdirect product?

mercio

they never say that $G$ is a subdirect product

Ted Shifrin

14:49

On-line stuff, like WeBWork, WebAssign, etc. @Krijn

mercio

$G$ is a subgroup of $A \times B$

and $G'$ is a subgroup of $(A \times B)' = A' \times B'$

SteamyRoot

Oh, I see your confusion.

They do explicitly mention that $G$ is a subdirect product of $A$ and $B$ in the paper where the question is from

I should probably edit the question to state this more clearly.

mercio

so $G/G'$ injects into $A/A' \times B/B'$

er

no

not an injection

Alessandro Codenotti

Hmm, I can't justify why a space filling curve needs to have a positive measure set of critical values @Balarka

Jessy Cat

Just wanted to throw out there that I'm offering a 100 point bounty for a completely detailed solution

Balarka Sen

14:57

What does a differential of a map $(0, 1) \to (0, 1)^2$ look like?

Jessy Cat

2

Q: Direct product of two nilpotent groups is nilpotent and direct product of two solvable groups is solvable

Let $G = G_{1} \times G_{2}$. I need to prove the following two things: If $G_{i}$ is nilpotent of degree $n_{i}$, $i = 1, 2$, then $G$ is nilpotent of degree $n = \max \{ n_{1}, n_{2} \}$. If $G_{i}$ is solvable (some people call them soluble) of degree $n_{i}$, $i = 1,2$, then $G$ is solvabl...

abstract-algebra group-theory normal-subgroups solvable-groups

Alessandro Codenotti

it's a row vector with $2$ elements

Ted Shifrin

No, column.

Alessandro Codenotti

right, I agree

Balarka Sen

What is the domain/range of it, as a linear map?

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