« first day (2399 days earlier)   

3:00 AM
A funner example: is the set $\left\{\frac{1}{n}\;:\;n \in \Bbb N\right\}$ compact?
 
Now, can you show that, for every $x$, there is an open interval $(a,b)\ni x$ such that $f$ is bounded on $(a,b)$?
Sure — choose an epsilon, and make the interval $(x-\delta,x+\delta)$ so that $f$ can't vary more than $\epsilon$. It's surely bounded there.
 
oh ya :P
 
(Since it is always between $f(x)+\epsilon$ and $f(x)-\epsilon$)
@ZachHauk So now let our cover be the set of all intervals on which the function is bounded.
Clearly, this is a cover, since we just prove every point is in one of these intervals (using the fact that the function is continuous).
So use the fact that $[0,1]$ is compact!
This gives us a finite collection of open sets on which $f$ is bounded!
Do you see how to conclude? @ZachHauk
 
But yeah so the way the direct proof that $[a,b]$ is compact goes is basically like this
So your open cover is $O$
 
This will also explain why it fails on intervals like $(0,1)$ (that aren't compact). An example unbounded function is $y=1/x$.
 
3:04 AM
Let $A$ be the set of $x\in [a,b]$ such that $[a,x]$ is covered by finitely many sets in $O$
 
It's clearly not empty because $a\in A$
 
sorry what
 
@AkivaWeinberger Or $\tan \pi x$.
 
i was playing a gane
let me scroll up and read
 
3:05 AM
And it's bounded above by $b$
So there's some supremum, call it $c$
 
This just makes me want to read Munkres again. goes off searching
 
@Daminark This supremum is either in A or it isn't.
Both of which lead to it not being the supremum very quickly…
…unless the supremum is $b$.
Yes?
 
We know that $c\in A$ because if we take an open set containing it, then you have $(c-\epsilon, c+\epsilon)$
Which is in that open set
But then take $c-\frac{\epsilon}{2}$
 
Consider the quadratic function $\mathbf{\frac{1}{2}x^TGx+b^Tx}$ in four variables where $$\mathbf{G}=
\begin{bmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{bmatrix}$$
and $\mathbf{b}=(-1,0,2,\sqrt{5})^T$. Apply the conjugate gradient method to this problem from $\mathbf{x_0=0}$ and show that it converges in two iterations. Verify that there are just two independent vectors in the sequence $\mathbf{g_0,Gg_0,G^2g_0,\dots}$.
In this case, what is $\mathbf{g_0}$?
 
We have that $[a,c-\frac{\epsilon}{2}]$ is covered by finitely many sets in $O$ since it's less than $c$
But then take that finite collection and stack on that open set around $c$ we were just talking about
So $c\in A$
Now, assume that $c < b$
 
3:08 AM
But then also $c+\epsilon/2$!
Is finitely covered by the same cover!
 
Then that $c+\frac{\epsilon}{2}$ screws us up
So $c=b$
 
Thus, $[a,b]$ is compact
 
I was gonna say that because it's bounded at every point we conclude that it's bounded but I don't know why that argument wouldn't work for open sets
and hopefully I'm not making myself look stupid here
 
@ZachHauk It's bounded on each individual set of the cover, right?
 
3:10 AM
Then you use Tychonoff's theorem to get that any product of compact sets is compact
 
And there's finitely many
 
(o v e r k i l l)
 
well yeah
 
So $f(x)<M_1$ on the first set, $f(x)<M_2$ on the second, $f(x)<M_n$ on the last, say
 
yep
 
3:11 AM
Since there's only finitely many, we can take the maximum of them.
 
mhm
 
So $f(x)<M:=\max(M_1,\dots,M_n)$ on the entire interval, so it's bounded.
 
yep
I understand that
 
QED. On $(0,1)$, though, things mess up.
Take $f(x)=1/x$
then the set of open intervals on which it's bounded is a cover has no finite subcover
 
Actually one problem on my pset this week is to prove that if the dimension of a vector space is infinite, the unit ball is necessarily not compact
 
3:12 AM
so you can't take the maximum of the bounds because there's always infinitely many, and the maximum becomes infinity.
@Daminark Hm. Maybe do something with the vertices of the $\infty$-octahedron?
The unit points on the axes
 
That's the geometry for sure
But the thing is
 
oh, good point!
 
You don't a priori know you have a Schauder basis
 
So no octahedron. :(
 
Well I guess that might not mess it up
It uses a theorem called Hahn-Banach
 
3:14 AM
What's a a Schauder basis
 
Basically it's a basis where you can take infinite linear combinations and talk about convergence
 
hmm
i wish my parents would do stuff like that for me, @Akiva
 
How do you define the topology, @Daminark
 
So you take a set $\{e_n\}_{n=1}^{\infty}$
 
maybe i sound whiny but i really do.
 
3:15 AM
@ZachHauk So Heine-Borel says that a subset of $\Bbb R^n$ is compact iff it's closed and bounded
@ZachHauk Maybe some day you'll be able to visit my house in Brooklyn and I could let you take home a few books
 
Such that for any $x\in X$ ($X$ is your Banach space), $x = \sum_{n=1}^\infty \alpha_ie_i$. This representation is unique
 
@ZachHauk Alternatively: Look up some of Ian Stewart's popular math books.
(Popular math meaning math intended for a layperson audience)
 
And actually we're dealing with Banach spaces in particular
 
I was about 20 minutes away from you today (was in the city)
 
So you have a vector space, over $\mathbb{R}$ or $\mathbb{C}$ with a norm
And then $d(x,y) = \|x-y\|$
It's Banach if it's complete under this metric
 
3:18 AM
@Daminark I never studied Banach spaces
but they sound interesting
 
I mean I just started recently
I will say, it would've been somewhat nice if we went the more standard route
Which is to do measure theory first
Now we're stuck dealing with $\ell^p$ spaces
While a lot of books do it for $L^p$ spaces
The former is a special case of the latter by considering $\mathbb{N}$ with the counting measure
But yeah, so the proof of the unit ball in infinite dimensional Banach spaces not being compact relies on Hahn-Banach
 
@AkivaWeinberger well if I see you at mathcamp next year...
 
Basically, given some linear functional $f:X\to\mathbb{R}$, where $X$ is a Banach space, you can define its operator norm $\|f\|_{op}$
 
too bad i'm not the best with social situations lol
 
Which is $\sup_{\|x\|\le 1} |f(x)|$
Now, a functional is called bounded if its operator norm is finite
 
3:23 AM
@ZachHauk Lol, me neither
 
@Akiva, @Zach: I thought this just came with the territory.
 
So, if you take some closed subspace of a Banach space, say $V\le X$
 
Staring at the globe in my fourth grade classroom instead of interacting with my other classmates during recess did make me semidecent at geography, though
 
And you have some functional $f$ defined on some $V$, Hahn-Banach allows you to extend $f$ to all of $X$ while preserving the operator norm.
 
Hiding in the stairs instead of playing outside was probably less useful
 
3:24 AM
It's ridiculously overpowered
 
good night
have school tomorrow
 
Alright, see you @Zach
 
@AkivaWeinberger same
 
Lol I'm like, bad at social situations but with a desire to engage
Meaning, I like to talk a whole lot, but my social skills are meh
 
EFFING YES BOY.
 
3:33 AM
What
 
Wut @MickLH
 
algorithm breakthrough, 50% resource requirement
 
Dayum
 
and it's the most scarce and important resource that's cut in half
 
Woo
What are you algorithming
 
3:34 AM
crypto
 
@MickLH Donuts?
 
*2nd most important resource
 
Computational time?
(I'm not sure how much it was using to start with)
 
Actually that's already good, better than factorization schemes at least, it's the size of the ciphertext itself that is reduced
 
Noice
 
3:38 AM
SIHT KCARC MHTIROGLA RUOY NAC TUB
 
With a subtle re-arrangement, I can floor divide away a ton of information and reconstruct it on the other side with a best guess, and the corruption is isolated purely into the randomized part, so the message is in-tact
@AkivaWeinberger no but I have another one that can
 
WhAt AbOuT tHiS oNe
See, the thing about that one is, whatever tries to read it gets cancer
 
lol
 
1337 sp33k: 1
All the efforts of modern research: 0
5
Also hey @arctic
 
hey
 
3:43 AM
use generating function to show that for a symmetric random walk $P(S_1S_2,\dots,S_{2n}\neq 0)=P(S_{2n}=0)$ for all $n\geq 1$
 
Hey @TedShifrin ! Thz for the answer of vector spaces
 
-1
Q: Prove that the group ring $\mathbb{Z}(\mathbb{Z}_{n})$ can be generated by a single element

ALannisterI need to prove that the group ring $\mathbb{Z}(\mathbb{Z}_{n})$ can be generated by a single element, but I'm not really sure how to begin. I know that the group ring $\mathbb{Z}(\mathbb{Z}_{n})$ consists of all formal sums $w_{1}[z]_{1} + w_{2}[z]_{2} + \cdots + w_{k}[z]_{k}$ where the $w_{i}$...

 
@ALannister Any guesses what element might be a generator? (Hint: what's special about the group $\Bbb Z_n$?)
Also, what would you say "ring generated by a single element" means?
 
3:58 AM
@DHMO How would one write the "gwoo" sound in Spanish? Güu, perhaps?
Just trying to fill in the sets "ja ge gi jo ju / ga gue gui go gu / gua güe güi guo ??". Could be that it just doesn't exist, though
 
 
1 hour later…
5:16 AM
Hi guys..
Any kind of Question/problem sets available for Schwarzs lemma , application of Liouvilles theorem ,open mapping theorem, as these can simply make many complicated looking questions in complex analysis as simple as possible , any site or book for these will be very good!!!...thank you
 
Well, I don't know much offhand about this sort of thing
 
means about complex analysis??
@Daminark
 
But I can give general books on complex analysis that I've heard are good
And for which there are pdfs
They'll likely have this result
Stein-Shakarchi, I know is good, that's what they'll use in complex next quarter
Ahlfors is not particularly modern, and also kinda terse, but I've heard is quality
 
that would definitely help, .. but i was looking for those which have examples of these concepts as using these results is like using LAW80... :) @Daminark
Thank you !!
 
I mean these are reasonably comprehensive books
So in all likelihood they'll have what you're looking for
 
5:28 AM
ok ! got it !!
Actually if one want's to read basic of complex analysis then book by Dennis Zill is awesome ...
@Daminark
 
I'll look into it for sure, thanks!
My path through complex was bizarre, I will need to eventually sit down and learn about this stuff
The bootcamp should have some, done out of Titchmarsh I think
 
ewwwwww
 
bootcamp out of Titchmarsh ??
 
Partially
Oh @Eric not fond?
Oh lol @Baymax it's this program that my analysis prof holds in the summer
 
No I think Titchmarsh is bad and outdated.
I get what he's trying to do by assigning it, but I won't ever be reading it again.
 
5:33 AM
Ohh .. Titchmarsh is a programme ?
 
Lolol, do you know why Schlag uses it? When he actually taught complex he used Silverman's translation (likely bastardization) of Markushevich
 
It's a book on complex analysis that an analysis professor at our school uses for this summer program he started running last year
 
@Baymax It's a book, it's just that the program I'm hoping to do is going to use it
 
I did the program last year, Daminark may do it this year
 
Ohh .. both @Eric and @Daminark you are in the same class ??
 
5:35 AM
He's a year ahead of me wrt classes
But we're in the same school and everything
 
Which book would you recommend? @Eric
 
Ohh..nice..@Daminark ...
 
He uses it to "give you technical fluency" which is fair I guess, but it's just so dry and hard to read.
Stein and Shakarchi is really good in my opinion
The entire four volume series in analysis by stein and shakarchi is quite good
That's for basics, beyond that there are so many directions with complex that you can go that no one book is "comprehensive" anymore.
 
How can I show the Cauchy point gives the minimizer of $$m_k=f_k+\mathbf{g_k^Tp}+\frac{1}{2}\mathbf{p^TB_kp}\quad \text{s.t.}\quad \vert\vert p \vert\vert\le\Delta_k$$
along the direction $-\mathbf{g_k}$
 
Ok guys..I will definitely look into it ,but there exists no supremum to books or materials i think , hence i am searching .. please you too search as i was telling ..these concepts are bullets .. so if anyone finds any nice material share it ...Thanks...All the best..
 
5:39 AM
Here, the Cauchy point is $$\tau_k =
\begin{cases}
1 &\quad\mathbf{g_k^TB_kg_k}\le0 \\ \min(\frac{\vert\vert\mathbf{g_k}\vert\vert^3}{\mathbf{\Delta_k g_k^T B_k g_k}},1) &\quad \text{otherwise}
\end{cases}$$
 
sorry to ask but it belongs to which topic @ozarka
 
This is relating to convex optimization, @BAYMAX.
 
Ok@ozarka , thanks!
 
@Eric Are the other books somewhat better at least?
 
uhhh well he got rid of the one by Arnol'd right?
I have no idea if Brin and Stuck is good
 
5:48 AM
Fair
 
And you can just ask Ted directly if his book is good l o l
 
Oh right yeah, diffgeo
I mean surely
So what does that leave? Probability?
 
I would expect Schlag will use it again since it is the perfect length
oh yeah i forgot about that because I just didn't do any of the problems
It was Sinai and I have no idea if it's good or not.
 
Oh speaking of probability, a bunch of people are kinda bummed out that Fefferman's special probability class isn't happening next quarter
At least 2 people I know offhand were gunning at it and it's not being offered
 
Oh I didn't know he taught one
Isn't he teaching just regular analysis
 
5:52 AM
Yeah, I think it only so far happened last year
And yes
 
6:05 AM
Would anyone be willing to help me with convex optimization. I am struggling with a problem.
 
 
1 hour later…
7:11 AM
In my book there was a phrase saying that. "x is 2 times less than y " what does it mean,,,man!!
I am confused in this English phrase ,I can't convert it to maths
Hey @Daminark please say mr
 
Hello
 
Hey @satyatech
No need to call me mr, I'm like, probably around your age
So that registers for me as $x = \frac{1}{2}y$
But I'm not terribly fond of that way of saying it
 
7:38 AM
Hi @Daminark, @Alessandro
 
Hey @Balarka!
 
Hi @Balarka!
 
What's up
 
Not much, I had a lazy weekend. I'm a bit disappointed by some of the oscars
 
Oh those happened?
 
7:43 AM
I didn't see which got them
 
But I'll have an algebraic topology lecture in the afternoon, that's going to be the highlight of the day probably
@Daminark yep, tonight
 
Eek they mixed up the winner of best picture
 
Huh
 
Hello.
 
Damn it's 2, I should get some shuteye lmao
Well, see you guys around!
 
7:54 AM
Night
 
8:06 AM
Cya
 
 
1 hour later…
9:10 AM
@AkivaWeinberger One wouldn't. Just leave the gap there.
 
9:49 AM
So I'm trying to minimise $f(x)=x^8+x^6-x^4-2x^3-x^2-2x+9$ . I differentiated it and factorised it to $f'(x)=2(x-1)(4x^6-4x^5+7x^4-7x^3+5x^2-2x+1)$. I've got no idea how to solve for the other roots or prove absence of any (there are none) apart from numerical methods. Any ideas?
 
nope nope nope nope
 
10:35 AM
@CompulsiveMathurbator $f(x) = (x^8-2x^4+1) + (x^6-2x^3+1) + (x^4-2x^2+1) + (x^2-2x+1) + 5$
 
10:56 AM
@DHMO Thanks
 
@DHMO How do you even notice that?
 
@MateenUlhaq completing the square...
 
o_O
 

« first day (2399 days earlier)