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12:00 AM
So what are all 4 numbers and does the equality hold?
 
Isa
ah
got it
they are equal
 
Yup.
 
Isa
thank you Ted!
 
You're welcome. Always remember to play with examples!
 
Isa
ok, thanks . Have to go, bye :)
 
12:06 AM
Hey chat!
 
heya @Lucas
 
how you're doing @Ted?
 
Quite well, thanks, and you?
 
hlo folk
 
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...
 
12:12 AM
Heya @Eric.
 
How is the dimension of a simplicial complex defined?
 
@user193319: Where did you use $k$?
 
Not sure. I just wrote the problem verbatim.
 
Oh, I see.
 
how goes
 
12:19 AM
@TedShifrin life's giving me a break, so better now. thanks :)
uni is the easiest part of going to uni
 
LOL
 
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
 
Sounds too easy for you.
@user193319: So can you show there are no $k$-simplices for $k>n$?
 
you mean I might be underestimating it?
 
No, @Lucas. I think the opposite.
 
12:23 AM
No, but I can try. So, the dimension has to do with the number $k$-simplices that can be fit into $X_n(\Bbb{Z})$?
 
well, thank you then. :)
 
@Lucas what’s the content of a math degree in Brazil
 
By dimension they mean the highest-dimensional simplex you can have.
 
@ÉricoMeloSilva hold on
 
Ah, okay. Thanks, I'll think about what you've said.
 
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)
 
cool cool
 
@Eric: So Korean or what for dinner tonight?
 
"period of time". lmao
non-native speakers suck. :p
 
maybe taiwanese porridge
@Lucas are you planning on trying to go to grad school?
 
sure
my dream is to study at IMPA and Berkeley
 
12:33 AM
It takes a strong person to succeed at Berkeley.
 
so lift weights
 
yeah they kind of bury u w teaching
 
Oh, most places do that, @Eric. That wasn't my issue at all.
 
@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
 
Huge state school, so sure.
Princeton = spoiled brats.
 
12:34 AM
lots of people need lots of teachers
 
uh oh am i gonna be a spoiled brat
 
I think so.
 
@TedShifrin the lack of :P ending that sentence is unsettling
 
I was referring to undergrads, mostly, but, yeah.
I aim to unsettle.
 
12:37 AM
my impression of berkeley from visiting their grad open house was that it’s very easy to get lost there
 
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.
 
seems to be a problem with large state schools...
 
And many of us on the faculty tried to make sure no one slipped through the cracks.
 
@TedShifrin that’s rlly good
 
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.
 
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?
 
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?
 
No, you don't want \frac.
 
Maybe a copy/paste error? Sometimes copying things into a latex editor can mess it up.
 
When I removed the $, it started working...
 
12:56 AM
Weird. @TedShifrin why wouldn't you want \frac ? /curious
 
You want $a_b$, where $b$ has an overline over the expression.
 
@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?"
 
Ah, I didn't try to psychoanalyze the English.
 
I just wanted the numerator to be on top of the denominator, instead of hanging around on the left side.
 
@DemCode: What are you trying to do?
Oh, OK. So you do want \frac.
 
12:59 AM
That should've read "why I get this". Typo!
Sorry
 
$$\frac{u_{\text{max}}-u_{\text{min}}}{x_{\text{max}}-x_\text{min}}}$$
 
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
 
It was just the fraction.
 
Interesting. I don't know why this isn't compiling.
 
Hm, weird. Are you editing it in stackexchange? Have a link to the question/answer?
 
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?
 
@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
 
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?
 
@ryan Yup, working on it
 
@TedShifrin sorry, $A$ is a proper subgroup of $\mathbb{Z}^n$.
 
So you need to prove it's free abelian.
 
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
 
$\frac{u_{max} - u_{min}}{x_{max} - x_{min}}$
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.
 
There was a \[ missing earlier somewhere in the document, but \] is in there later on. Adding that \[ fixed it (started working with $).
 
So one dollar sign was missing, basically.
 
@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 ${e_i}_{i \in I}$.
 
It is a theorem, @abuchay, that a subgroup of a free abelian group must be free. It is non-obvious.
 
1:21 AM
@TedShifrin Alright, i can prove that. Thanks.
 
Heya chat.
 
sup @Fargle
 
Nothing much, how have you been?
 
mostly good, agonizing over grad decisions but that’s not a real problem. hbu?
 
Wow, it's a @Fargle.
 
1:26 AM
Just hanging around. Been reading mostly.
 
Actual literature?
 
mr. shifrin
Q: why is the isomorphism $T_p(T_pM)\cong T_pM$ natural
 
You're just asking about $T_x(V) \cong V$ for any vector space $V$.
You have a global chart.
 
Yes, but how is that iso natural
 
How is it not?
 
1:32 AM
Doesn't it depend on the basis you use?
 
Hell no.
 
@TedShifrin No, math. :P
 
W H A T
 
Blah, @Fargle :P
 
I do read other things. Just haven't lately.
 
1:33 AM
the iso literally looks like (p, v) maps to v
what could be more natural
 
The proof I know of $T_pV \cong V$ for a vector space $V$ begins with "take a basis of $V$..."
 
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.
 
pulls out Mac Lane
 
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.
 
@Fargle pshhhh nerd
 
1:37 AM
I suppose it depends on which definition of tangent space you use.
 
Which one are you using?
 
I use whichever one I feel like.
Here I'll use equivalence classes of curves.
 
@ÉricoMeloSilva :(
 
me too tho
 
Is it not natural for any definition? I don't see why it would matter.
All definitions are equivalent, maybe one is easier?
 
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.
 
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.
 
I've made several remarks.
 
This is what "natural" was meaning to me.
 
Well, no one will protest that $T_p(V) \cong V$ is not natural. You should just learn it.
 
I am struggling to show it for myself.
 
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.
 
Translate curves through $p$ to curves through $0$.
 
Normally I would set up a chart which is just an isomorphism $V\cong\mathbb R^n$ (so pick a basis).
 
Someone please help me with this.
 
@TedShifrin so just take $\beta(t) := \gamma(t) - p$
 
Yes, of course.
@MrAP: I disqualify the question entirely on the grounds that "neither" in (d) is totally incorrect.
 
1:45 AM
Why is that?
 
Because "neither" refers to precisely two.
 
that’s what ted said about showing the natural isomorphism to T_0V
 
I was thinking it would be either (a) or (c).
 
You can't rule out $Y$ if $d=1$.
 
(a) seems more likely to me.
 
1:46 AM
"more likely" isn't mathematics.
 
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 $0$.
 
Then you have to give an example where they're not all positive.
 
So translation is an isomorphism
 
Oh, I can give lots of examples where they're not all positive.
 
Well, the first you suggested.
 
1:49 AM
BTW, @MrAP, don't you know that the GM/AM inequality can be an equality sometimes?
 
Yes. I know.
 
Anyhow ... I'm out of here.
 
:( ted i luv u
 
@TedShifrin tchau tchau
 
@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".
 
2:04 AM
@MrAP as Ted said, if it is (d), then you should be able to find a counterexample
 
It is 1
If all the variables are 1 then the inequalities do not hold.
@anakhro
 
>proper fraction
A proper fraction is one of the form $a/b$ where $a < b$ and $a,b\in\mathbb Z$
 
Oh I forgot about that.
Its 1/2
If all the variables are 1/2, then the inequalities do not hold.
I think.
 
Do better than just think: calculate.
 
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?
 
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.
 
Oops. I am totally out of my mind. It would be $3/2>3/2$ and $3/2>3/(16)^{1/3}$.
 
1/16 in the last cube root
 
So its option (d).
 
If you say so.
 
What do you mean by that?
 
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.
 
You don't say so?
 
No, I haven't looked at the question
Like the other options
 
2:38 AM
Ok. Fine.
 
 
1 hour later…
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?
 
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.
 
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?
 
5:13 AM
Please help me with this question. Thanks a lot in advance.
 
 
3 hours later…
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?
 
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 $G \cong (C_{p} \times C_{r}) \rtimes (C_{p} \times C_{q})$?
 
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 $F \rtimes G/F$.
 
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
 
$\vec F_{\Bbb Q^2}=(x,y)$
$\vec F = \Bbb Q^2 \to \Bbb Q^2$
 
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!
 
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
 
Hi!
 
Hey @Alessandro @Balarka et al
 
Hey @ÍgjøgnumMeg
 
waddup
 
11:04 AM
Hi @ÍgjøgnumMeg! How are you?
 
Hey @Alessandro, I'm alright, working atm and trying to prepare for my interview lol
How about you?
 
When will it be?
 
This monday :(
 
@ÍgjøgnumMeg I finished my exams so now I'm free until the new semester begins in April, I'm looking forward to it
 
Nice :)
 
11:06 AM
Oh, that's soon! What are you interviewing for?
 
You're ditching alg. geo?
It's for a full DAAD Scholarship
which I've been shortlisted for
 
@Alessandro Teach me geometric group theory lmao
 
@AlessandroCodenotti Thank you
 
@ÍgjøgnumMeg Yeah, I wanted to go to the lectures without doing the exam at the end, but it overlaps with a logic course
 
now that ur free i mean
 
11:07 AM
Ah, that's great, good luck for the scholarship!
 
Good luck from me as well
 
I think you already know more GGT than me :P
 
Nah
I don't know anything precisely
 
I'm free as in having no lectures and no exams, I'm still preparing a seminar talk for the next semester
 
Oh what's it on
 
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
 
Yikes
runs away
 
Have you looked at the proof that word problem is solvable for hyperbolic groups?
Sounds up your alley
 
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
 
What's a Dehn presentation
 
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
 
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
 
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!
 
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
 
11:35 AM
@Alessandro thank you :) I am just relearning a lot of the stuff from my dissertation so I can waffle about it
 
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 $\pi_1(S)$
 
Heya
 
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
 
 
2 hours later…
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?
 
@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 $p_k$
 
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.
 
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.htm‌​l?file=math.uh.edu/~dlabate/settheory_Ashlock.pdf
 
@ParasKhosla, sorry to derail the discussion. Which book are you using for complex analysis?
 
@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
 
@Silent I have yet to find a really good book on complex analysis
 
@anakhro Is there no link for hardcopy? maybe on Amazon or elsewhere
 
2:25 PM
But Visual Complex Analysis is probably one of the better.
@ParasKhosla afaik these are strictly online notes.
 
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
 
You could just print it out.
Probably cheaper than buying a book
 
@ParasKhosla OK! Thank you very much for suggesting this! seems that it covers many topics
 
Yeah that's still an option
What? @Silent the course?
 
@anakhro in india, it is way cheaper to by an indian edition of book, better, if you buy used copy! :)
 
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.
 
@ParasKhosla yeah. it covers conformal mapping, residue theorem etc! the core and cool topics
 
Yeah it's really great. Anyways are you a math student in India? @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.
 
Oh cool @Silent
 
@ParasKhosla, so that book does not seem be be either on set theory or analysis! Which subject does it cover?
 
2:36 PM
Basic algebra
 
Yeah perhaps that abstract algebra.
rings and groups in further chapters
 
@anakhro oh!
 
@Silent did you do an undergrad in math?
 
@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.)
 
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 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.
 
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!
 
Can a triangle on a surface
 
@anakhro Thank you very much for all this information!
@anakhro How did you guess that? :)
 
automorphism groups are basically where group theory developed from.
Automorphisms, and then symmetries.
Graph theory makes heavy use of both of these concepts.
 
ok!
@ParasKhosla, how can i increase speed in coursera video? can i see those vids in youtube, as we can do in edx?
 
Can you tell the geometry of a surface based on how
A triangle is curved on it
 
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?
 
3:04 PM
hi chat
 
@Ultradark you will have to define "geometry".
 
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 :)
 
@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.
 
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 $a,b,c,d$.
 
@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
 
Thank you so much!
 
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