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2:30 AM
@robjohn Ahh, sorry to hear that, not so much fun.
2:51 AM
:(
 
2 hours later…
5:08 AM
@copper.hat why?
5:54 AM
@SoumikMukherjee sry, that was for last year nytimes.com/athletic/4976847/2023/10/19/…
its old news, irl didn't make it
6:44 AM
@BalarkaSen Clearly Gabai's foliation paper is impossible to read. Calegari's book is much clear
 
2 hours later…
8:34 AM
I want to write a paper only 3 people in the world can get lol
I don't get that
Audience should be maximized not minimized
Mathematics is currently a closed field to most of the world. Inaccessibility of journals, unaffordable access to education, elitist attitudes when it comes to dispersing knowledge, it blows my mind.
Wouldn't the world be greater if more people had access to the research
Nope, because then the editors of journals couldn't drive Ferraris or whatever tf floats their boats
Dystopian hell hole
Capitalism is no longer a Nash Equilibrium
psst, haven't you heard "money makes the world go round"
Yes, I've heard that, but it doesn't no longer
Competiton and higher pay drives innovation, but they forgot to include the fact that most of the land / world is already developed. Everyone could be a freakin scientist these days, but we still live like we did 100 years ago.
Also, not to mention all the "low hanging fruit" on the tree of knowledge and technology have already been plucked. So that the remaining fruit is impossibly high up. We can't compete to get it; we must work as one.
In other words, our celebrated Fathers of math such as Gauss, Hilbert, Euler. They ate all the dang fruit! What's left ? Impossibly high up fruit
OTOH they laid strong foundations which are practically the only possible approaches
Steve Jobs had to drop acid and when he did, he bit into an apple of knowledge, hence the bitten apple logo.
In other words, he had to risk his health just to get his foot in the door. So everyone slaves away drinking the caffeine etc, to what end? A paycheck that disappears completely even negatively in one month's time. The other fuckers making decent wage, then criticize street bums for not wanting to work. Everyone loves being productive, it's just that too many people made babies, and once they reach adulthood they're screwed. The world eats them alive.
[/rant]
Oky back to math
I don't mean I hate people making a decent wage. They're all right and work hard. The problem is the harder they work, the less work is available for people starting out. How much work is available for people starting out that don't have the social connects. None.
Freelancer.com: BS, once in a while a living wage is posted. Most the time though, its code this month-long project for < $50. So freelancing is BS unless you know like 20 people giving out work.
All contracts should be posted on a government site. I'm sick of this unfairness / work hoarding.
Also it should be illegal to discriminate against people that do not possess a bachelor's degree.
That's like saying, if you want to work, you got to pay $50k+ for schooling first. That's god damn slavery. Slavery didn't end the day after Lincoln officially ended it. It didn't end the day after that. It just transformed. It still exists. We can't see it because we've habituated to it.
Crap, shouldn't have said all that. Now the adjustment bureau is gonna wipe my mind tonight -_-
Anyone want to answer a homology question I got for free? You'll get unicorn points for it. And a rep that matters nowhere.
Maybe it's better to be a poor mathematician. But they should at least pay us like $500 / month to build this site. Which we have done for free for the past 20 years. Me like 12 years I've been here.
The site maintainers clearly don't even listen to the community's pleas.
The MathJax editor broke about 5 years ago, and was never repaired.
Thanks for letting me vent.
My sincere hope is that the next John Nash of Economics comes around and mathematically proves that the economy could be run more efficiently / productively / progressively. Then perhaps in 50 years after their work has been published and the author has died. The world will wake the fuck up.
The poor people are already aware of all of this. The wealthy are blind to it. Or they have too much to lose to talk shit like a loser as I do.
 
2 hours later…
11:24 AM
Oh f* I wrote all that ... wasn't even st0ned
Time for positivity!
I'm reading about discrete conditioning, i.e. where we have two random variables $X,Y$ and $X\in L^1$ whereas $Y$ is discrete. Then expressions like $E(X|Y=y)$ make sense. I was wondering about when $Y$ is continuous. Maybe I'll get there, but is there such a thing as continuous conditioning? My book very suddenly introduces simply $$f_{X|Y=y}(x)=\frac{f_{X,Y}(x,y)}{f_Y(y)},$$ when $X,Y$ have a joint cont. distribution. I feel like there's some discussion missing.
@psie great observation!
I have no idea. Need to review my probability
Continuity and topology are an extremely powerful tool in math especially when combined with an algebraic structure, such as Furstenberg's topological ring of integers. Which is metrizeable!
E.g. the function that counts the number of twin primes in a growing interval is a continuous function in Furstenberg's topology. See my third to latest post.
It is continuous because of the topological ring property partially
This genius answerer proved it elementarily
Not sure yet how to use the fact. So was of late looking at what a convergent sequence looks like in this topology and when you apply a continuous function to a converging sequence, then the function will converge to $f(\text{limit of the sequence})$ and so on...
So we have this counting formula for twin primes that fits on a line. But prove that it does not eventually vanish!
12:23 PM
@psie there is, but it's a really subtle issue in general
at least when you try to condition on something with probability zero
ok, I found these notes and they give some perspectives on the different cases; continuous vs discrete conditioning. Interesting stuff.
1:05 PM
@psie $E[X|A] = \frac{E[X1_A]}{P(A)}$ for an event of non-zero probability $A$
however there is other interpretation here
One can consider $E[X|Y]$ which is essentially a function of $Y$ i.e. $E[X|Y] = g(Y)$. Then something like $E[X|Y = y]$ means $g(y)$
yeah, $E[X|Y]$ is a random variable, if I'm not mistaken
it is
@psie the idea here is that after plugging in the integral of $x\mapsto x\cdot f_{X|Y = y}(x)$ as the definition of $E[X|Y = y]$, obtaining some function $y\mapsto E[X|Y = y] = h(y)$, then we can say that $E[X|Y] = h(Y)$.
of course, we're dealing with functions only up to some identification on null sets
ok
(there is no unique function we can define for $E[X|Y]$ either)
in general there is a difference between conditioning on an event vs conditionining on a random variable/sigma algebra
although $E[X|A]$ can be interpreted in a way as $E[X|1_A]$
as you see this is a bit complicated
nonetheless, conditioning is the most important concept in probability so its good to get it right
1:34 PM
yeah, conditioning sounds intricate :) I constantly need to think about the word conditioner when I hear it, which is totally unrelated. That is something for your hair, so let's not mix it up.
it certainly would be odd to bring a conditioner on a probability exam
2:25 PM
Coming from a physics pov, the durac delta serves pretty well for jumping from discrete to continuous random variables
Dirac*
@nickbros123 ...
dirac delta from physics refers to a distribution
but not the one from probability
although, maybe both are used there
Jakobian a while back I think you were mentioning a close connection between completeness and compactness?
Unless I am misremembering and/or misunderstanding what you said at the time. Can't find the comment now
@EE18 not sure if I said closely connected, but they are connected
So I am working an exercise which says for $D \subseteq M$ with $M$ complete, $D$ complete iff $D$ closed
for one, we have theorems such as complete + totally bounded iff compact
IIRC the same is true if I change complete for compact right?
3:25 PM
@EE18 no
oh you mean in both cases?
Ya everywhere you see complete replace with compact
both $M$ and $D$ compact
well, this is true for metric spaces
if we want to be proper, there is some intricacies for non-Hausdorff topological spaces for which the equivalence is not always true
compact no longer implies closed in that case
Ah I see
OK ya I am only seeing metric spaces at the moment
anyway, another reason for why compactness and completeness are connected are compactifications
compactifications can be considered to be type of completion of a space
this is true even for non-metric spaces
although, not every compactification of a metric space is metrizable, so we can't always complete it in the metric space sense
but this can be done by completing it in a uniform space sense
which metric spaces are a special case of (or, er, uniform spaces don't preserve all properties of metric spaces, like isometries type properties)
there is actually some type of characterization of metrizable compactifications by use of Wallman bases of closed sets
Much of the above is well above my head :) but thank you for the interesting extra color here
3:34 PM
I'd say the difference between metric spaces and uniform spaces, is that while uniform spaces have structure more rigid than that of topological spaces, they still don't preserve the "geometry"
3:46 PM
$\sum_1^\infty (-1)^n \cdot \frac{e^{-nx}}{n^2}$ determine the convergence set $I$, leading back to power series through an appropriate substitution. Sequence: $e^-x=t$ therefore $x=-\ln(t) , so \sum_1^\infty (-1)^n \cdot \frac{t^n} {n^2}$, for alternate series test the series converges if the limit of a^n is 0 and the sequence is decreasing.
First point, I used the root criterion, so $|t|<1$
But wolfram said \$ln(t)<1$
Why?¿?¿?¿?¿
Anyway
2 point
I did $\frac{t^{n+1}}{(n+1)^2}\leq \frac{t^n}{n^2}$
How do I proceed from here?
4:03 PM
substituion $t = e^{-x} > 0$ is a good one
You can indeed use alternative series test to show the series converges for $t\leq 1$
for $t > 1$ the series obviously doesn't converge since that's outside of interval of convergence of this power series
so $I = \{x : e^{-x}\leq 1\}$
1.did i use the root criterion correctly?????? 2. how do I get out of the 2nd point
@Jakobian How did you get out of it
4:27 PM
@BinkyMcSquigglebottom hm?
4:40 PM
@Jakobian I can't do 0 calculations like you
5:33 PM
hmm... yes, I did skip the calculation of the radius of convergence of this power series, which you can essentially do in your head
or on paper, either is fine
5:45 PM
@Jakobian hi. u didnt reply in the other chat. have u changed ur mind about Big Bang?
6:15 PM
@RyderRude as in?
hi
6:32 PM
hello
7:05 PM
What (other) sites do you use for math discussions? I have seen math groups on discord.
Discord is ok.
Any social media platform can be used.
Btw I am asking everyone, so anyone can answer if they want.
7:38 PM
@SoumikMukherjee none
I've tried posting on mastodon before. But I thought it was pointless
as fun as it sounds to have math discussions, eventually you have people that don't like you and start looking at you the wrong way
I was wondering, the theorem that states that if $f$ is a nonnegative measurable function, then there exists an increasing sequence of (positive) simple functions converging towards $f$. Does this theorem also hold if $f$ is not nonnegative, i.e. real-valued? I have rarely seen it used, but I think it holds in some fashion.
Or even complex-valued...
@psie in what sense do you want it to hold
obviously a complex valued function is made up of real valued functions which are made up of non-negative valued functions
well, I was thinking we might no longer have that the sequence of simple functions are increasing towards $f$, right?
I believe by simple function you mean a finite-valued measurable function?
imagine a function $f:\mathbb{R}\to\mathbb{R}$ unbounded from below
does there exists any lower bound for $f$ which would be a simple function?
hmm, well, if you allow simple functions valued in the extended real line than I believe such analogue might be possible though
as for complex functions, what does increasing even means
7:58 PM
ok, I think we'd still have $0\leq |\phi_1|\leq|\phi_2|\leq\ldots\leq|f|$ or something like that, where $\phi_n$ is the simple function
that can be easily achieved if you just break a complex function into real pieces, then that into non-negative pieces and apply the theorem for each
ok, will look into it
i.e. $f = (f_1-f_2)+i(f_3-f_4)$ where $f_1, ..., f_4\geq 0$
now applying the theorem there exist non-negative simple functions $\phi_i^{(n)}$ for $i = 1, 2, 3, 4$ which increase to $f_i$
Letting $\phi^{(n)} = (\phi_1^{(n)}-\phi_2^{(n)})+i(\phi_3^{(n)}-\phi_4^{(n)})$ we have our desired sequence of simple functions
ok, silly question maybe, I see the inequality $|\phi^{(n)}|\leq |\phi^{(n+1)}|$, but why are they all smaller than $|f|$ in absolute value?
well actually there is a subtlety here that I didn't focus on properly
8:05 PM
I guess it is just because $\phi^{(n)}$ converges to $f$
you want $f_1, f_2$ be chosen in such a way that for each $x$, either $f_1(x) = 0$ or $f_2(x) = 0$
similarly $f_3, f_4$
otherwise what I said doesn't quite hold, but if you do this then it does
$f_1 = \text{Re}(f)_+, f_2 = \text{Re}(f)_-$ and so on
so that for fixed $x$, you have either $f_1(x) = 0$ or $f_2(x) = 0$. Say the latter. Then $\phi_2^{(n)}(x) = 0$ for all $n$.
you see how this makes the inequalities hold?
yeah, more or less
8:39 PM
@leslietownes For any topological space $X$, there exists a Tychonoff space $Y$ and continuous function $\tau:X\to Y$ which induces an isomorphism of rings $\tau':C(Y)\to C(X)$, where $C(Z):=C(Z;\mathbb{R})$. You do this like this, you let $Y$ be equivalence classes of relation $x\sim y$ iff $f(x)=f(y)$ for all $f\in C(X)$. Then you let $C'$ to be the set of those functions $g:Y\to\mathbb{R}$ for which $g\circ\tau$ is continuous, and topology on $Y$ is least such that $g\in C'$ are continuous.
Now the interesting part that if in this construction you let $x\sim y$ be $f(x) = f(y)$ for all $f\in A$ instead, then this construction with Stone-Weierstrass theorem can be used to show that any closed subring including all constants $A\subseteq C^*(X)$ is of the form $A \cong C(Y)$ for some compact space $Y$ (here $C^*(X)\subseteq C(X)$ are bounded continuous functions)
so closed subrings including all constants of $C(X)$ (say $X$ is compact) are precisely rings of the form $C(Y)$ for some compact $Y$
@Jakobian the space $Y$ obtained using $A$ in this construction is precisely the one for which $A\cong C(Y)$ (after replacing $X$ by $\beta X$)
9:04 PM
@Jakobian ooh I haven't heard of this one before
its basically an alternative for twitter but there is some split in terms of communities so you can technically post some math

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