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1:00 AM
woah
just saw this
"And, given an nxn matrix that does have eigenvectors, there are n of them.
Given a 3 x 3 matrix, there are 3 eigenvectors."
that's... wrong - right
 
Hamilton was supposed to be Huisken there.
Huisken uses Hamilton's methods, so the mistake is forgivable
 
1:17 AM
lol
 
1:28 AM
@EricSilva Anyway, that paper is the first thing anyone should read about MCF
 
1:43 AM
hey @0celo7
can I discuss something with you qucikly in differential forms
because I am teaching something related to that tomorrow ?
it is a very simple thing
 
@orbit-stabilizer Maybe if it's a complex matrix, and you count with multiplicity?
Although I'm not actually sure there
Happy April Fools' slash Easter slash Passover
 
1:58 AM
@AkivaWeinberger can we discuss something small
 
Sure
I'm worried
 
suppose $\omega \in A^k(U)$ i.e differential form on open set U. Then, why is it true that $d(f\omega) = df \wedge \omega + f \wedge d\omega$
I mean why can we even do it ?
oh wait
 
Like, why is it well-defined?
 
yeah
oh wait I see the issue
yeah nvm I understand it now.
 
What happens if we do $d(fg\omega)$
 
2:01 AM
yeah I get it
talking and spelling things out to people makes you think harder of the thing you want to discuss about
 
On the one hand that's $(fg)\omega$, on the other hand that's $f(g\omega)$, on the third hand? that's $g(f\omega)$, hopefully they all give the same answer
 
yeah
 
$d(fg)\wedge\omega+fg\wedge d\omega$ better equal $df\wedge g\omega+f\wedge d(g\omega)$
 
yeah
 
@Adeek maybe'
 
2:04 AM
nvm @0celo7 I see it now
 
ok
 
thanks @AkivaWeinberger for the discussion
 
That's $fdg\wedge\omega+gdf\wedge\omega+fg\wedge d\omega$ on the left and $gdf\wedge\omega+ fdg\wedge\omega+fg\wedge d\omega$ on the right
 
yeah
 
I wonder how much the total fingernail and toenail clippings of humanity, per year, weighs
 
2:07 AM
Many.
 
it's gotta be some insane amount
 
oh
really ?
how much does 1 fingernail clipping weigh ?
 
no idea
 
I mean, there's over seven and a half billion of us
 
I think maximum it would be a ton
 
2:08 AM
but I probably clip about 25 times a year
 
for all of humanity
 
bi-weekly
 
There were only seven billion in 2011 but now there's more
 
I don't think it would be more than that
 
How much does a fingernail grow in a day? Times 10 times 365 times 7.5bil
 
2:09 AM
There's a million grams in a ton
 
Well that's hands
@0celo7 Did you know that a ton of people is like 15ish
 
7.5 billion people clipping 25 times a year
@AkivaWeinberger Yes, it says so on every elevator
For @Adeek's math to work out, a clipping needs to weigh like a microgram
 
My elevator just says "Capacity 2000 lbs.", it doesn't say how many people that is
 
oh, at least in TN all the elevators have human equivalencies
might be a state code
 
This isn't like a public place, it's just my apartment building's elevator
which I am standing next to
 
2:11 AM
elevator codes are determined by the state
 
Goddamn Albany
 
if yours isn't, get out of it
 
it takes 450 fingernail to be a pound
so the math is actually pretty easy
 
are those nails or clippings
 
nails
 
2:13 AM
Mine doesn't say anything about the state, but it mentions the city (NYC)
 
so you're way, way off @Adeek
waaaaaaaaaaaaaaay off
 
yeah
did you calculate it ?
how much is it ?
 
Well, I don't have a nice number for a clipping session
and I don't have a scale
I could collect the clippings and go to a physics lab tomorrow
but that might get me arrested
 
Ooh, Counterstrike season finale is out
*Counterpoint
**Counterpart
 
Let's do a Fermi calculation. According to Google, a nail weighs 1g. Let's assume that a human goes through 5 sets of nails in a year (reasonable?). So that's 1g x 20 x 5 = 100g of nails per person in a year. So times 7.5 billion people, ...a shitload.
750 million tons...
Er
 
2:18 AM
Also, the Chinese space station fell into the ocean today
 
750,000 tons
I can't math
help
 
It's the one that falls out of the sky in the movie Gravity, but like real
 
yeah, 750,000
even if I'm an order of magnitude off, that's insane
 
@Semiclassical
Everyone including blind person has a preferred spot
 
there's so many people
 
2:21 AM
We're well on our way to eight billion
 
dude, 750,000 tons
what's the volume of this
 
How dense is a fingernail
 
I'm googling
Doesn't seem to be readily available
 
In Counteract, there are two universes so like fifteen billion people
I mean Countermand
Counterpart
It's a good TV show
 
Zee
2:58 AM
alpha keratin has a density of 1.3 g/cc
 
is there a good video or resource that shows you how to prove everything regarding natural numbers, integers, rationals, reals, etc?
 
"everything"
 
well yeah, the usual definitions, how things are typically proven, that sort of thing
like tao's analysis except minus the "do this as an exercise instead" spiel
 
3:15 AM
You probably want an intro number theory text and a intro analysis text I'm guessing
Maybe something like this? scribd.com/doc/86664082/…
 
@orbit-stabilizer I believe he means proving basic things from the axioms
@user525966 firstly, what is your foundation
 
I have a question about terminology
 
@LeakyNun Graduated high school already (up through calc BC), familiarity with certain number theory concepts, going back and re-learning everything from the ground up so i can fill in all my gaps and really understand how everything works
 
@user525966 by foundation I mean foudnational theory...
you need to have something if you want to prove things
 
not sure what that means
 
3:19 AM
things are proved from other things
 
Is an arbitrary intersection of compact sets compact? Here by "arbitrary" do they mean a finite intersection of compact sets or is the case of infinitely many also included.
 
but ultimately, things are proved from axioms
 
Infintely doesn't work
wait
 
yes
 
it does
 
3:20 AM
@user525966 so you need to have axioms before you can prove things :)
are you going to build it from ZFC or something
 
waiiiit
 
for example peano's axioms for natural numbers, definitions of addition -- and then from there proving all the various properties and attributes we naturally understand about addition (such as associativity, commutative property, all that stuff), number ordering, so on
 
right
 
so far been using tao's analysis which is great and all
 
24
Q: Intersection of finite number of compact sets is compact?

John BuchtaIs the the intersection of a finite number of compact sets is compact? If not please give a counter example to demonstrate this is not true. I said that this is true because the intersection of finite number of compact sets are closed. Which therefore means that it will be bounded because the in...

 
3:21 AM
but i hate that it skips things intentionally
i'm not a student so that stuff is more of a hindrance for me
i'd rather just have the answers so i can check my work against it and make sure i am doing it right
(as I lack a teacher/prof who can check my work or teach me stuff, etc)
 
yeah I mean I've already proven stuff like that
was just using that as an example
 
I mean that site
 
oh, less helpful for me because it's not structured in any way
i have to hop around a lot
unlike a book or class that presents things in a logical order
where one thing builds from the stuff before it
 
I see
 
3:24 AM
@LeakyNun Proof 1...
 
i do use that site though to cross-check certain concepts though which is helpful
but it's super time consuming
to piece together crap from numerous sources
 
@0celo7 one could define N categorically :P
or type-theoretically
def N (X : Type) := X -> (X -> X) -> X
 
Question
Question in real analysis.
Question in real analysis.
A set is called clompact if it has the property that every closed cover admits a finite subcover. Closed cover basically is the same as open covers, just with closed sets.
Describe all clompact subsets of R.
 
Zee
Open sets
 
why not closed sets
 
Zee
3:38 AM
Let me think am drunk
So closed and bounded sets are compact by hein Borel
Actually it’s the closed and bounded sets , hein Borel is an IFF statement
Oh god , never mind I thought you said compact
But it’s probably the same except with open bounded sets , just use de Morgan laws
 
@LeakyNun, Suppose $a,b,c\in G$ where $G$ is a group. If a and b do not commute, is it always true that $ac$ and b do not commute?
 
@orbit-stabilizer Line with two origins. Take $[-1,1]$ with the first origin, and $[-1,1]$ with the second origin…
 
@Silent no
@PiyushDivyanakar hint: singletons are closed
 
So,
9 hours ago, by Silent
How do we know that $(ml)(jk)$ does not commute with $\sigma$ here in last line??
@LeakyNun
 
Zee
@LeakyNun that’s only true in atleast a hausdorrf space
 
3:48 AM
@Zee R
 
Zee
Yaaaa
 
@LeakyNun @Zee
 
@Zee get back here when you're not drunk
 
Zee
But that’s not as fun :(
I have HW am gonna be up till 7 AM , am just using you guys to sober up
 
Hi, does anyone know how to get the LaTeX rendered in chat?
 
3:52 AM
Consider $\sigma=(mj)(kl)$. Then $(jk)$ does not commute with $\sigma$, but $(ml)(jk)$ does, so your final claim is incorrect. — Akiva Weinberger 19 secs ago
 
@LeakyNun @Zee
So a singleton can be regarded as clompact, and so can be closed sets. As a finite union of closed set is closed, closed sets can be clompact. But open sets can be contained in closed sets so are all sets clompact??
 
your question has been addressed by the god
 
16 mins ago, by Piyush Divyanakar
Question in real analysis.
A set is called clompact if it has the property that every closed cover admits a finite subcover. Closed cover basically is the same as open covers, just with closed sets.
Describe all clompact subsets of R.
 
@LeakyNun Wha
 
@PiyushDivyanakar use the definition of clompact and the fact that singletons are closed
 
3:53 AM
You saw my comment and posted it here so quickly
 
@AkivaWeinberger ?
well someone was asking me that a second ago if you scroll up
6 mins ago, by Silent
9 hours ago, by Silent
How do we know that $(ml)(jk)$ does not commute with $\sigma$ here in last line??
 
I know but still, that was fast
 
@AkivaWeinberger I was looking at the question because I was also thinking
 
@AkivaWeinberger, @LeakyNun, thank you!
 
Zee
@PiyushDivyanakar without even having to think , that sounds way to nice to be true , just take your question , translate into open sets using de Morgan and apply the last century worth of knowledge to solve it
 
3:54 AM
@PiyushDivyanakar kindly ignore the guy above me ^^
 
Hey @Zee
 
Zee
My man
Haha , can’t believe your here
 
Does anyone know how to have latex rendered in chat? All im seeing is dollar signs
Lol yeah, i came in here on a whim!
 
@NicholasRoberts see room description
 
Zee
Ya, I was just at the house we were at last year
Drunk and have to do my HW , FML
 
3:57 AM
What house?
 
Zee
The one by my house with the dog ?
 
@NicholasRoberts >>>^^^
Room info box in the top-right of the screen
 
Thanks @Ak
@AkivaWeinberger
And nice, those were good people!
 
Zee
With the lady who gave me her Snapchat and was hitting on me
 
Is anyone interested representation theory here?
 
3:59 AM
For a second I thought that said "reputation theory"
 
Zee
puke algebraist
 
and I was thinking "yes I attend a high school"
 
@LeakyNun, can you suggest other proof that 'center of $A_n$ trivial, if n>3'?
 
Zee
Just take the centralizer
Can’t anybody solve problems here ?
 
4:00 AM
Just to think, @Zee was contemplating taking intro to Algebraic geometry last week...
$\sigma$
 
1 min ago, by Zee
puke algebraist
(jk)
 
Zee
That’s only because I have a weak spot for Grothendieck
 
Dude, are you gonna be good to drive to school tomorrow?
 
Zee
Ya man , we gonna take AT together too
 
Word, I hope you actually take it lol
 
4:02 AM
@Silent An is simple, and center is normal subgroup, and An is non-abelian
 
@LeakyNun is that true in general? Or just $n > 3$>
*?
 
@NicholasRoberts A_n is simple iff n >= 5
 
Right, but he said n > 3
 
that leaves only n=4 unproved
which one can check by hand
 
Oh gotcha
@Zee how was the rest of JJ's?
 
Zee
4:06 AM
That guy was into bitcoins and stuff , he said he wanted me to teach his son to play chess and that we should discuss hiring me into his startup , but maybe that was drunk talk...
 
That guy was crazy. Whats his startup?
 
Zee
Bitcoins
Idk the details
 
@Silent How about this
 
Zee
Anyway , seeing you here sobered me up, I gtg do HW , don’t let these guys know who I am
 
Given a $\sigma$ (other than the identity) I'mma find something that doesn't commute with it
Case 1: $\sigma$ is not a cycle
so it contains at least two cycles
Call them $C_1$ and $C_2$
Choose $a\in C_1$ and $b,c\in C_2$
Then $(abc)$ does not commute with $\sigma$
Proof: $\sigma(abc)$ maps $a$ to something in $C_2$ while $(abc)\sigma$ maps $a$ to something in $C_1$.
Case 2: $\sigma$ is a cycle
Choose $a$ in that cycle
Say $\sigma$ maps $a$ into $b$
and let $c$ and $d$ be anything other than $a$ and $b$ (we can do this 'cause $n>3$)
Then $(bcd)$ does not commute with $\sigma$
Proof: $\sigma(bcd)$ maps $a$ into $b$ while $(bcd)\sigma$ maps $a$ into $c$
@Silent It occurs to me that the proof for Case 2 works for Case 1
so let me actually simplify this proof a lot
New proof: Since $\sigma$ is not the identity, $\sigma$ maps some element $a$ into $b$ with $a\ne b$. Choose $c$ and $d$ not equal to $a$ and $b$ (which I can do because $n>3$). Then I claim $(bcd)$ does not commute with $\sigma$. Proof: $\sigma(bcd)$ maps $a$ into $b$, but $(bcd)\sigma$ maps $a$ into $c$.
Therefore, no $\sigma$ other than the identity commutes with every element of $A_n$; therefore, the only element of the center of $A_n$ is the identity. QED.
 
4:19 AM
@AkivaWeinberger, I am so thankful to you! Why don't you add that as an answer thare?
@LeakyNun Oh! just saw this approach in Dummit and Foote. Thank you!
@AkivaWeinberger Are we taking $b\ne c$?
 
@Silent Yeah
0
A: The center of $A_n$ is trivial for $n \geq 4$

Akiva WeinbergerI will show that, for every $\sigma$ other than the identity, there is something in $A_n$ that does not commute with it. Since $\sigma$ is not the identity, $\sigma$ maps some element $a$ into $b$ with $a\ne b$. Choose $c$ and $d$ not equal to $a$ and $b$ (which I can do because $n>3$). Then I c...

@Silent Good idea: I posted it ^
 
Confusion about seperated sets?
The definition of separated sets states that $A,B$ are separated if $\bar{A} \cap B = \emptyset$ and $A \cap \bar{A}=\emptyset$.
By this definition I don't see how $(1,2)$ and $(2,5)$ are separated as $\bar{A} = (-\infty, 1] \cup [2, infty)$ and $\bar{B} = (-\infty, 2] \cup [5, infty)$ now $\bar{A} \cap B = B$ and $\bar{B} \cap A = A$.
What am I doing wrong here?
typo: $\bar{B} \cap A = \emptyset$
 
@PiyushDivyanakar I think $\bar A$ there means the closure of $A$, not the complement of $A$
 
oh okay
 
4:25 AM
People use it to mean both things, it's annoying
You kinda just need to know from context
 
@AkivaWeinberger thanks
 
So $\overline{(1,2)}=[1,2]$.
 
We know $(b\,c\,d)\in A_n$ because, $(b\,c\,d)=(b\,d)(b\,c)$, right?
@AkivaWeinberger
@LeakyNun About that 'clompact' question, I think every subset of $\Bbb R$ is clompact, because arbitrary intersection of closed sets is closed. Am I right?
 
@Silent no
 
oh!
 
4:38 AM
@Silent Yeah
Yeah, $k$-cycles are even iff $k$ is odd. (Which is an annoyingly confusing consequence of the terminology…)
 
@LeakyNun Finite sets and empty set are the only sets that are clompact?
 
@Silent the former includes the latter, and yes
 
ok :)
@AkivaWeinberger thx
 
not sure why you said "finite sets and empty set"
since empty set is finite
 
@LeakyNun I am also not so sure, i think to emphasize that empty set is clompact, too. I have this feeling that we do not give enough respect to $\varnothing$.
 
4:47 AM
are open sets perfect?
no right
 
Well $\Bbb R$ is perfect and open
but like perfect sets are closed so other than that (and the empty set) you're good
 
i find it extremely strange how empty set and $\mathbb{R}$ behave.
 
Intersections of closed sets are closed. Intersection of $[1,2]$ and $[3,4]$ is the empty set. Therefore…
Similarly, finite intersections of open sets are open, and the intersection of $(1,2)$ and $(3,4)$ is the empty set
and similar stuffs for $\Bbb R$
I mean yeah the empty set is weird but if nothing were weird then math would be boring
 
5:44 AM
Yo so this problem is slick
So, recall that a hypergraph is some $H = (V,E)$ where $E\subset \mathcal{P}(V)$. Basically, a graph but we're not requiring the edges to just be two-point sets.
 
@cello what do you need
 
Now, a legal coloring of $H$ is some map $f:V\to [k]$ such that no edge is monochromatic, ie $(\forall \mathcal{E} \in E)(|f(\mathcal{E})| \ge 2)$
And the chromatic number $\chi(H)$ of a hypergraph is the fewest colors needed in a legal coloring
So the problem is this, define $\mathcal{H}_n = ([n],\{\{a,b,a+b\}\subset [n]: a\ne b\})$. Show that $\chi(\mathcal{H}_n) = O(\log(n))$
It's pretty short but I dunno, I really liked it
Hmm, quick test: $\mathcal{V}$
@Akiva I think you'd like this one
 
5:59 AM
@0celo7 I don’t understand your question.. Did I said anything before...?.
 
 
1 hour later…
7:16 AM
@nitsua60 Hi ! Are you free right now?
 
How do I solve this question?
 
8:19 AM
hello does someone know where to find Concrete Mathematics book with better typographic layout, original one is very hard to read
 
8:37 AM
1 hour ago, by Tanuj
@nitsua60 Hi ! Are you free right now?
I guess he's forgetful right now
 
8:48 AM
[insert joke about forgetful functor here]
 
9:02 AM
God I had forgotten how great of an album Fear of Music is.
 
9:31 AM
Let $\mathcal{A} = \{ H \leq G \ |\ A \subseteq H \}$ and $\mathcal{B} = \{ H \leq G \ |\ B \subseteq H \}$. Since $A \subseteq B$, we have $A \subseteq H$ whenever $B \subseteq H$; thus $\mathcal{B} \subseteq \mathcal{A}$. By definition, we have $\langle A \rangle = \cap \mathcal{A}$ and $\langle B \rangle = \cap \mathcal{B}$. We know from set theory that $\cap \mathcal{A} \subseteq \cap \mathcal{B}$.
I can't understand how we got $\cap \mathcal{A} \subseteq \cap \mathcal{B}$ from $\mathcal{B} \subseteq \mathcal{A}$, it seems to me that containment should be in reverse direction. @BalarkaSen
 
I don't know the answer; maybe someone else does.
 
@Silent let $x \in \bigcap \mathscr A$, so $\forall H, A \subseteq H \to x \in H$.
We want to show that $x \in \bigcap \mathscr B$, i.e. $\forall H, B \subseteq H \to x \in H$
Indeed, if we have $H$ such that $B \subseteq H$, then $A \subseteq H$, so $x \in H$.
therefore, $\bigcap \mathscr A \subseteq \bigcap \mathscr B$
 
@LeakyNun Thank you!
So, derivation $\mathcal{B} \subseteq \mathcal{A}$ is wrong, correct?
 
@Silent I can't answer that with "yes" or "no" without disambiguity
but $\mathscr B \subseteq \mathscr A$ is correct
 
oh
@LeakyNun :)
 
9:44 AM
simple example, let $\mathscr B = \{S\}$ and $\mathscr A = \{S,T\}$, then $\mathscr B \subseteq \mathscr A$, but $\bigcap \mathscr A = S \cap T \subseteq S = \bigcap \mathscr B$
 
now its clear!!
 
!!
 
9:59 AM
0
Q: Closed form for $\frac{\int_0^1 Ei(x)^5 ln(x) dx}{\int_0^1 Ei(x)^3 ln(x) dx} $?

mickConsider the expression $$\frac{\int_0^1 Ei(x)^5 ln(x) dx}{\int_0^1 Ei(x)^3 ln(x) dx} $$ A) Can we rewrite this with a single integral sign ? B) Do we have a closed form for this expression in terms of hypergeometric functions ? C) Is there a closed form without hypergeometric functions for ...

Any ideas ?
 
10:09 AM
that moment when you think that the forgetful functor Top->Set is just forgetful but it turns out that it is also free
@BalarkaSen
 
I will forgetful your functor up
 
10:37 AM
@BalarkaSen every function from discrete topology is continuous; every function to indiscrete topology is continuous
so the indiscrete functor kind of "forgets" that it's a set (with distinct elements)
 
11:16 AM
@Daminark $a+b$ mod $n$?
 
 
1 hour later…
12:16 PM
@cello you’re on the star board...
 
12:50 PM
if a function f:M->M' satisfies f(x+y)=f(x)+f(y), where M and M' are monoids, does it follow that f(0)=0?
 
since you're writing the monoid operations as addition, should we assume that both monoids are commutative?
 
hi there,
do you know similar websites like Paul's Online Math Notes? I'm looking for a complete a thorough treatment in any math field.
 
@Semiclassical hmm, let's say they are
and see what happens
 
I guess what I'd note is that in the case $y=0$ then the above becomes $f(x)=f(x)+f(0)$, so $f(0)$ acts like the identity element for everything in the image $f(M)$.
 
[PhD stuff] So..., after almost 1.5 months of manually crawling through the data, the most important conclusion I get is:
$$\phi(L1,L2,L3)=\phi(L1)\phi(L2)\phi(L3)$$
 
12:56 PM
@Semiclassical thanks, that helped me
 
and $$L3 > L2 >> L1$$
 
$f:\Bbb Z \to \Bbb Z^2 : x \mapsto (x,0)$
 
now the interesting things will begin soon
 
@LeakyNun in that case it seems pretty simple
 

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