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2:53 AM
Is there any nice reference on intersection theory to formalize what's done in section four of griffiths harris? It's extremely handwaved it
 
3:23 AM
Learn algebraic topology and Poincaré duality.
 
 
6 hours later…
9:30 AM
@LucasHenrique lol
 
 
2 hours later…
11:44 AM
0
Q: Are there any reinforcement learning benchmarks where the optimal policy is known for each environment?

nbroThere are multiple reinforcement learning (RL) benchmarks (i.e. a set of environments where we can test our RL algorithms), for example, the DeepMind Control Suite. However, given that I am currently not particularly familiar with them or their details, I don't really know if any of them comes wi...

 
 
3 hours later…
2:42 PM
I am trying to show that if $R$ is an integral domain, then $(a)/(ab) \cong R/(b)$. Initially, I thought that ring isomorphism would be $\varphi (ra + (ab)) = r + (ab)$. So, I started proving that it is well-defined and additive, but then I started to prove that it is multiplicative and this is where problems entered...
I could use help finding the right isomorphism.
Hmm...maybe it should just be $ra + (ab) \mapsto ra + (b)$...
I think that fixes the multiplicative issue...
 
@user193319 do an example first
 
Shoot...so $ra + (ab) \mapsto ra + (b)$ doesn't work either? I don't see why...everything seems to check out...
Hmm...it's not clear that it is injective...
 
first, think about why the hypothesis includes $R$ being an integral domain
 
Well, I thought I was using that, because I had to use it in order to show $ra + (ab) \mapsto r + (b)$ is well-defined.
Btw, I made a mistake when defining $\varphi$ in my first post; it should be $\varphi (ra + (ab)) = r + (b)$, but I still don't think it works, although $R$ being an integral domain is needed to show that it is well-defined.
 
no, that works, actually
 
2:51 PM
are you trying to show an isomorphism of rngs (rings without identity)? that won't work, because for example $2\Bbb Z/4\Bbb Z$ is not isomorphic to $\Bbb Z/2\Bbb Z$
you only get an isomorphism of $R$-modules
 
All rings are assumed to be unital...
Oh, maybe I am trying to show they are isomorphic as $R$-modules..
 
I only asked because you talked about multiplicativity
$(a)/(ab)$ is not a unital ring in general
 
Oh, so my professor must want me to show they are isomorphic as $R$-modules.
 
see this is why you always work with an example first
 
It didn't say in the problem statement, so I just assumed it was a ring isomorphism.
 
2:53 PM
@LukasHeger how do you prove the norm-existence theorem?
 
ring isomorphisms should be between rings
 
@LeakyNun I know...you're right...I was just pressed for time, because the assignment is due in a little bit.
Yeah, it didn't occur to me that $(a)/(ab)$ isn't a unital ring.
Oh, if it were, wouldn't that mean $(a)$ contains a unit and therefore $(a) = R$?
 
not necessarily
 
@LukasHeger if $K$ is a non-arch local field and $M$ is an open subgroup of $K^\times$ then there is a finite (abelian) extension $L/K$ such that $M = N_{L/K}L^\times$
 
$(a)$ can be unital with a unit different from the one in $R$
 
2:55 PM
@LeakyNun it's part of CFT, I don't know a simple proof
 
for example, consider the ideal $((1,0))$ in $\mathbb{Z}\times\mathbb{Z}$
 
@LukasHeger because I'm reading Cassels-Fröhlich and they blackboxed it
if (a) is unital with ar being the unit, then aar=a gives a=0 or a is a unit (under the assumption that R is an integral domain)
 
@LeakyNun Sharifi has a proof for p-adic fields here: math.ucla.edu/~sharifi/algnum.pdf
 
if (a)/(ab) is unital with ar+(ab) being the unit, then aar=a+abs, so a=0 or ar+bs=1
 
yeah, the reason is that the unit of a non-trivial proper ideal that is a unital ring must be a non-trivial idempotent
 
3:00 PM
so e.g. 2Z/6Z is unital with 0x4=0, 2x4=2, 4x4=4
@LukasHeger is there a subgroup of $K^\times$ of finite index that isn't open?
I guess the answer is no given LCFT?
 
more generally, if $G$ is a profinite group that is topologically finitely generated, then all finite index subgroups are open
for p-adic fields $K^\times$ is topologically finitely generated
I think $\Bbb F_q((x))^\times$ has non-open subgroups of finite index, but I'm not sure
 
doesn't profinite already mean all finite index subgroups are open
 
not necessarily
 
but then given LCFT, a f.i. subgroup of $K^\times$ gives you a f.i. subgroup of $G_K^{ab}$ right
doesn't that give you a finite extension by infinite galois theory
 
infinite Galois theory gives a bijection between finite extensions and open subgroups of the Galois group
if $q=p^n$, then $\Bbb F_q((x))^\times=\Bbb Z \times Z/(q-1) \times \Bbb Z_p^{\Bbb N}$, so it suffices to show that $\Bbb Z_p^{\Bbb N}$ has non-open finite index subgroup
this maps surjectively and continuously on $\Bbb F_p^{\Bbb N}$ so it suffices to work with the latter
a construction based on a non-prinicipal ultrafilter for $\Bbb F_p^{\Bbb N}$ is given here: mathoverflow.net/a/82180/117693
 
3:28 PM
@LukasHeger thanks
 
3:51 PM
If $M$ is an divisible module over the PID $R$, is the following a valid way to argue that $M$ is also injective. Let $I = (r_0)$ be some ideal, $f : I \to M$. Then $f(r_0) \in M = r_0M$, so $f(r_0) = r_0m_0$ for some $m_0 \in M$. Let $\varphi : R \to M$ be the $R$-linear extension of $1 \mapsto m_0$. Then a simple calculation shows that $\varphi \circ \iota = f$, where $\iota I \to R$ is the inclusion map. Does this sound fine?
 
4:11 PM
@user193319 I need a small help.
It has to be done either using a program or maybe on paper, but a thing that I require very much.
 
Imagine getting a job at Google, but all you do is make/manage Google Doodles
 
The instructions are like this:
 
"By the way, did I tell you I work for Google?" - said person, in a bar
 
"I do occasional work as an external contractor for a google data center" cit. pizza delivery guy
 
Take a dot matrix of n^2 dots (a square - shaped matrix, actually), choose any one of the corner dots, and using it as a common end of every line segment to be drawn, and then first connect collinear points starting from it and then connecting the remaining dots (the isolated ones).
Then , take a count of the number of line segments formed for each n (the line segment having collinear points are counted as one and no subunits are taken) in such a fashion.
Please try doing this for every n, then take the difference between n^2 and the number of line segments so obtained. Note all these onto a table...
Please do this for as many n's as possible.
I'll be back after prayers..
 
4:26 PM
@BalarkaSen imagine getting a job at google and managing databases
Or doing operations research
Or anything data science
Or training a huge neural network and claiming it understands language
(I'm pretty sure nobody sensible at google claims their models understand language, I just dont like the current glorified statistics in AI)
 
the singularity is coming
Eliezer Yudkowsky will develop the first generation of AI annihilating humans
 
vzn
hopes to be alive for the singularity, but increasingly skeptical of both (life/ singularity) o_O
 
4:41 PM
why do you pop up every time i mention singularity
 
vzn
lol pop up whenever... am a huge AI fan/ geek etc :) ps do you have something against google? :o
 
nah was just a funny bit
 
Howdy
A Balarka. How did your cramming turn out?
 
Fantastic. Did great in the test
 
Not surprised.
 
4:47 PM
I actually like stat after the cramming
It's fun. Nothing deep but fun
 
I wish I knew more. They hide a lot of the geometry that's in the linear algebra.
 
Yeah
 
I should send you the masters thesis I helped direct on that.
 
Yeah please do, I'd be happy to read
 
Cool. In a moment.
Sent.
 
4:54 PM
This is beautiful! Everything here is part of my syllabus. Can I share it with a few friends?
This is really nice. I should just read from here instead of my lecture notes.
 
vzn
Oct 19 at 16:34, by Balarka Sen
@EdwardEvans I have somehow ended up agreeing to give a series of talks on the Riemann zeta
↑ no joke? past, future?
 
i give three talks, i'll give the second one this weekend
 
vzn
huh, a weekend talk? at school?
 
I think my students would be pleased to hear your enthusiasm. I think it's public knowledge, so sure.
 
Great, thanks a lot
@vzn School's online, so
This is part of student seminar
 
5:04 PM
@TedShifrin Is he giving some lecture on a particular topic
 
vzn
ok, congrats on tackling such a formidable topic. is it a chalkboard or powerpoint presentation?
 
writing pad
 
vzn
so do you have a particular take/ angle on it?
 
yeah ill prove prime number theorem
 
vzn
lol ok. worlds funniest mathematician. a standup mathematician.
 
5:07 PM
so zeta has no zeroes on $\Re = 1$ basically
whats funny about the prime number theorem
 
vzn
(oops) thought you were joking about proving it.
 
lol no thats the goal
 
vzn
for ~1½ century.
 
ah i dont mean i will prove it for the first time
that'd probably be hard. although it's a surprisingly easy exercise in a complex analysis -1 course that zeta has no zeroes on $\Re = 1$
 
vzn
are you going to mention its a claymath $1M problem? rereading, forgot about that...
 
5:14 PM
what?
prime number theorem is not the riemann hypothesis
 
vzn
yes understand that.
 
one is about Re = 1, the other is about Re = 1/2
 
@BalarkaSen So you mean one is twice as hard as the other?
 
LOL
 
vzn
Von Koch (1901) proved that the Riemann hypothesis implies the "best possible" bound for the error of the prime number theorem. en.wikipedia.org/wiki/Riemann_hypothesis
so PNT is proven with a "weaker" version? (not an expert on this)
 
5:15 PM
ah i see what you meant
 
@BalarkaSen the 3-4-1 trick?
 
i think i can prove that RH implies pi(x) = x/log(x) + O(x^1/2 log x) or whatever (1/2 + eps?)
it follows from the residue theorem and Perron's formula
i'll see if that falls out of whatever i have in my notes
@LeakyNun ye lol. nuts
just boils down to trigonometry
 
vzn
wow Perron? was recently looking into some of his work. not sure if its the same Perron.
 
I would assume it is the same one. Same as in Perron-Frobenius theorem for positive matrices (and generalizations)
 
yeah should be the same Perron
 
5:19 PM
Damn @Balarka doing number theory takes me back to 2015.
 
no man im not doing NT
 
vzn
"stability of the linear approximation"...
 
i just had these notes i had to make use of somehow
its fun anyway
 
I have used Perron-Frobenius a whole lot in certain papers
 
vzn
In 1929 Perron [1] investigated the question of when the stability of the difference equation x′=Ax (7.1) link.springer.com/chapter/10.1007/978-1-4612-1076-4_7
 
5:21 PM
Lol, enjoy. I am never touching NT hopefully again. Last time I did, I made a mess of myself, "I am going to prove the Riemann Hypothesis".
Jesus what a fucking mess I was
 
vzn
RH will make a mess out of anyone/ everyone.
 
Now, I only believe in infinity categories
I don't know which is worse.
 
sanity check: a connected component of a product space is the product of connected components of the factors, right
If $C\subseteq\prod X_i$ is connected, $\pi_i(C)\subseteq X_i$ is connected, hence $\prod\pi_i(C)$ is connected and $C\subseteq\prod\pi_i(C)$, hence $C=\prod\pi_i(C)$ by maximality. If $\pi_(C)$ were not maximal, then $\pi_{i_0}(C)\subsetneq C^{\prime}$ with $C^{\prime}$ connected, but then$\prod\pi_i(C)\subsetneq\prod_{i\neq i_0}\pi_i(C)\times C^{\prime}$ and the latter is connected, contradicting maximality of $C$.
 
@BalarkaSen I love stats (anyone surprised?). It's often not taught as it is done in practice, which is where things get really interesting. The AI hype is just noise.
 
vzn
AI is (much) more than hype/ noise. :)
 
5:32 PM
@vzn I understand how AI has revolutionized stats. I do not question that. What I do question is seeing the data science students and graduates who think AI can solve every problem.
 
vzn
agreed there is some (over)hype. cf gartner hype curve en.wikipedia.org/wiki/Hype_cycle alas sometimes kurzweil is "part of the problem"...
 
I saw, for example, a post from a data science student who said that Gallup surveying is antiquated data analytics which manipulates data, and that anything you could get from the Facebook Graph API could beat out Gallup surveying.
 
vzn
alas AI doesnt eliminate human stupidity :( :P
facebook is definitely rearranging the political landscape last few yrs, some think 2016 significantly hinged on facebook... and its hard to prove them wrong...
← luvs stats too :)
 
Oh, I don't question that. Not at all. But AI is not a panacea.

Also, quick aside: the things Facebook has done would not get through any IRB. The fact that people don't seem to think that's a problem is a good reason for complacency with regard to anything, including the election.
 
vzn
re AI as not a panacea. wholeheartedly agree. sometimes humans create problems so big even a super AI cant solve em. theres a great/ very well written essay by Chollet on this, can dig it. wish someone would do a documentary on him. (theres more than 1 on kurzweil...)
@Clarinetist IRB= inst review board?
 
5:38 PM
@vzn Yes
 
vzn
hey they called out zuckerberg in front of congress what more can you ask for? and now dorsey/twitter (next in line "pandora box") is being accused of all the worlds evils...
 
Until laws get passed, it's just complacency IMO
I don't see the US doing anything like the GDPR anytime soon
 
vzn
we havent passed much meaningful laws in years. except the ones that give (BIGLY) handouts to corps :(
 
Right. That's why I can't take it seriously. It's just all a show.
 
vzn
 
5:42 PM
And it's really tough for me, because, well, I teach data science. The things I've seen of COVID are identical to what I've had to deal with professionally.

People doubting the data all of the time. People doubting the metrics. People disagreeing what to do with the metrics. People thinking there's not a problem at all.

I have to point out the problems, but as faculty, I'm not allowed to tell them specifically how I view the problem, because it gets political.
 
vzn
yes scientific illiteracy is rampant in the age of science. not following why talking about your pov about science illiteracy is political. oh, yeah, can see how it can be misconstrued. it seems even scientific literacy is a partisan issue now.
however, in defense, science is a fundamentally human enterprise... warts and all...
 
All I could say is that in the situations I've been in, buy-in from the top is necessary. I did not elaborate beyond that, but anyone who was actually absorbing the material could figure out what I would suggest.
 
vzn
handouts windfalls Corporations reap windfalls from coronavirus tax breaks axios.com/…
@Clarinetist so are you a math prof?
 
I'm a data science adjunct instructor at a CC, and work as a data analyst full-time
 
vzn
@Clarinetist what buy in are you looking for? teachers are already teaching...
@Clarinetist cool what kind of data/ prjs?
 
5:51 PM
@vzn To elaborate on "buy-in," for any initiative to be successful, the people at the top levels need to agree on what data to collect and how to act on it. That is clearly not happening with COVID.
The course I teach is basically a one-semester introduction to how to do automated data analysis: cleansing, visualization, reporting, etc. Not much on the actual analytics.
 
vzn
in some defense, there is not really a strong scientific consensus on how to deal with a raging pandemic. the last one of this scale was almost exactly over 1 century ago.
it is hard to even point to a single paper that (unequivocally) proves mask efficacy.
 
@vzn IMO it doesn't matter what experts say. That has never been the case in any organization I've worked in. If the top people can't make up their minds, it's all useless.
 
vzn
science can help, but it can also be rather powerless, seeing both (extremes) play out.
 
Yes, I am well aware about the lack of evidence of mask efficacy, even though people seem to think they're quite effective.
 
vzn
ofc top people would have trouble making up their mind if experts cant make up their minds either.
 
5:54 PM
Not necessarily. Just throw a bunch of money at something and you'd be surprised how quickly execs would be willing to do something, even if it makes no sense
 
vzn
but yeah can agree US overall handling of crisis is abysmal for many reasons, leadership at top of list...
 
I've been on the advisory situation of that end. Warned my boss at the time that that a consultant's model was going to be invalidated, because I knew of the multicollinearity existing in it and the final decision was based on the parameter estimates. It was a few million bucks. I resigned. Boss had to deal with the PR afterward.
 
vzn
alas though some of the leadership vacuum seems to be due to cultural aspects... ie leaders nearly as chaotic and characterless as the "unwashed" masses...
@Clarinetist quite a story but hard to imagine a project/ initiative failing because of ("mere") bad curve fitting o_O :P
 
That was, in fact, how the model was sold
Basically, the idea was that you could measure a student's expected growth taking into account all of their demographics
using prior data, compared to their actual score
 
vzn
how about 2008? the quant models were flawed, but in exactly such a way that they gave corps the "message they wanted to hear"... ie a green light for risky/ reckless speculation...
 
5:59 PM
Now, the thing is, I was taking a graduate-level linear models class at the time, and as I progressed in the class, I started asking more technical questions
So I got from the consultant one day that pseudoinverses were being used to calculate the parameters, which were being used for whatever they were calling expected growth
and I was just thinking... you're going to screw us over
 
vzn
glad youre into scientific/ modelling integrity but theres a lot more to initiatives than that...
 
At that job, my job was just reporting the numbers
The initiatives happened outside of my organization
Clearly no one believed me, but the local media sure picked up on it. I've seen my former boss' name in the papers, blaming it on "noise" from the statistics
 
vzn
huh a dramatic case. what was the initiative about?
 
I'm not going to elaborate much besides this, but I took this seriously, as the numbers I reported could cause schools to close down.
To get an idea, I once received a phone call from a superintendent - about this very same number, actually - who said to me that if he couldn't explain how this number was obtained, he'd have to resign.

I did check in the news after: he did indeed resign.
 
vzn
huh interesting. what state was this?
 
6:06 PM
I'm going to identify myself too easily if I reveal that.
 
vzn
cambridge analytica is a case where statistics can go bad/ blow up in massive ways... congressional investigation etc... 2008 quant research (justifying/ rationalizing/ "covering up/ whitewashing" real estate subprime loans) is another case study...
 
All I can say for now is I no longer live in that state, and I still look at what's going on over there occasionally with disappointment.
Yeah, you know what's funny: I worked in real estate after. Not very long, but long enough to know how things worked in a brokerage.
 
vzn
seems public education system in US is being "ghettoized"... or turned into/ strengthened into a caste system...
 
It's so, so much more complicated than that
 
vzn
huh... whats the short take on it?
 
6:11 PM
You'd have to look at NCLB and the implications it has on assessment and accountability. That's where it all really began.
 
vzn
yeah NCLB was a big deal/ "gamechanger"... once again suspect it was heavily Corp driven... (behind the scenes...)
 
Well, yeah, there's a reason why ACT and the College Board profit so much from those implications
 
 
vzn
exactly yeah testing companies seemed to be involved in NCLB somehow...
 
Consider question 4b, specifically finding the highlighted conditional expectation. The solution says the conditional distribution is Bin(5,0.3), and so the expectation is 5*0.3. However, isn’t the factor 1/4 missing as given in the density?
 
6:16 PM
@schn Conditioned on $X \geq 0$, observe that $\sum_{x=0}^{5}f(x) = \dfrac{1}{4}$. Divide $f$ by this value to obtain the conditional density given $X \geq 0$, and you will notice then that $\binom{5}{x}0.3^x(0.7)^{5-x}$ is the probability mass function of a $\text{Bin}(5, 0.3)$.
 
@Clarinetist Thanks for the reply. My definition of conditional density is the joint density (of $X$ and $X\geq0$) divided by the density of the condition ($X\geq0$). What is the joint density in this case?
 
@schn In this case, it would be $\dfrac{f(x)}{\mathbb{P}(X \geq 0)}$.
@schn It's kinda unfortunate, because you're not really told how to deal with these sort of "compound" conditionals. You're only told the density given some other fixed thing.

I'm actually taking measure-theoretic probability right now, and I suspect there's more to conditional stuff than what I know currently from computations.
 
@Clarinetist Yes. I interpret the condition as $X=k$ for some $k\geq0$...?
@Clarinetist What is the density of $X\geq0$?
 
@schn There's no such thing, as far as I know. It's just the measure of that event.
 
vzn
@Clarinetist not following exactly, what "reason" are you referring to?
 
6:28 PM
@vzn Basically, NCLB is clearly corp driven, and you see it with the ACT and College Board. Nothing more than that.
 
vzn
:(
 
@Clarinetist And the measure is 1/4, or?
 
@schn Yes
 
vzn
this is fineâ„¢
 
@Clarinetist So there really is no joint density (between $X$ and $X\geq0$) either, right? It only is $f(x)$.
 
6:38 PM
I want to solve this problem using Tonelli's theorem
But that \infty bothers me
I learned Tonelli's theorem on $\mathbb{R}^n\times\mathbb{R}^m$
 
@Clarinetist Anyway, thanks for your replies! Clarified it.
 
Or can we consider \int_0^\infty as \int_{[0,\infty)}?
 
well, that's the same thing
 
@schn Sorry, was in a meeting. Yes, that's correct
 
@Thorgott Well then it makes sense. just consider $\int_0^\infty$ as $\int_{[0,\inty)}$ so that all defined on Euclidean space
 
6:59 PM
just want to check my understanding here: it is not always true that given a covector $\omega$ we have that $\omega\wedge\omega=0$, the only way this can be proven is trough the anticommutativity property of the wedge product, and we need $\omega$ to be a $k$-covector where $k$ is odd, but for instance if you we deal with $dx_1\wedge\cdots\wedge dx_n$ and we have $dx_i=dx_j$ for some $i\neq j$ then it becomes immediately zero since each $dx_i$ is a $1$-covector (hence odd),
did i say anything wrong here?
 
@Thorgott I wonder why the condition $f$ is integrable is necessary isn't it enough to say that $f$ is measurable to use Tonelli's theorme?
 
yeah, you don't need integrable
 
Oh, thanks
One more, $\int_{\mathbb{R}^d} f = \int_{\mathbb{R}^d} f(x,y) dxdy$?
 
Let $h$ be an analytic function on the unit disc. If there exists a constant $C > 0$ such that $|h'(z)| \le C$ for every $z$ in the unit disc, then there exists a constant $M > 0$ such that $|h(z)-h(w)| \le C|z-w|$ for every $z,w$ in the disc....If $z$ is some arbitrary point in the disc, using the definition of complex differentiability at $z$, I can show that $|f(z) - f(w)| \le M$ for all $w$ in a nbhd of $z$...How do I get that it holds for every w in the disc?
 
I don't know what that latter is supposed to mean
$f$ is a function of $d$ variables, not of two
 
7:13 PM
Oh, I mean $\mathbb{R}^d =\mathbb{R}^{d_1}\times\mathbb{R}^{d_2}$ and $x\in\mathbb{R}^{d_1}$ and $y\in\mathbb{R}^{d_2}$
 
then aren't those just two different notations for the same thing
 
@Thorgott hey! can u quickly look at what i wrote above?
 
Maybe I need to use Cauchy's integral formula
 
maybe. The book didn't specify
 
Nah, that won't work.
 
7:22 PM
@StupidQuestions if $x_i$ are supposed to be the coordinate functions, you won't have $dx_i=dx_j$ for $i\neq j$ and if they're not supposed to be the coordinate functions, I suggest you don't call them $x_i$
but yes, in general, $\omega\wedge\omega=0$ is only true if $\omega$ is of odd degree
 
@Thorgott my mistake sorry i meant $dx_i=dx_j$ for $i=j$
thanks for the confirmation, that thing bugged me for like 3 hr xD
 
If some sort of mean value theorem held for analytic function, then I'd be done...hmm...
 
for example, $dx_1\wedge dx_2+dx_3\wedge dx_4$ doesn't wedge to $0$ with itself (as forms on $\mathbb{R}^4$)
 
thank u @Thorgott , one last question, do u know any book like "Counterexamples in Analysis" but for higher-dimensional analysis and differential geometry/manifold theory?
 
Maybe Cauchy's integral formula would help...hmmm
 
7:29 PM
I believe counterexmples in analysis does have some higher-dimensional stuff, but otherwise no
@user193319 what statement are you trying to show exactly, your notation's over the place
 
Using Cauchy's integral formula, I believe one can show that $|f(z)-f(w)| = \frac{1}{2 \pi} |\int_{\gamma} \frac{f(\xi)(z-w)}{(\xi - z)(\xi - w)} d \xi | \le \frac{|z-w|}{2 \pi} \int_{\gamma} |\frac{f(\xi)}{(\xi - z)(\xi - w)} | |d \xi|$.
@Thorgott Let $f$ be an analytic function on the unit disc. If there exists a constant $C > 0$ such that $|f'(z)| \le C$ for every $z$ in the unit disc, then there exists a constant $M > 0$ such that $|f(z)-f(w)| \le M|z-w|$ for every $z,w$ in the disc
The only problem is that that last integral depends upon $z$ and $w$...
And I don't see how to connect it to the derivative of $f$.
 
this is a weird exercise
cause this has nothing to do with analyticity
well, I guess it makes the bound simpler
but weird regardless
 
Is only continuity needed?
I'm basically trying to show that $f$ is Lipschitz, right?
 
not every continuous function is Lipschitz-continuous, no
 
But analyticity implies Lipschitz?
 
7:41 PM
well yes
continuous real-differentiability + bounded derivative implies Lipschitz on convex sets
 
Hmm...
What about complex differentiable? How do I show that?
Are $Re \circ f$ and $Im \circ f$ complex differentiable?
 
pick $f=\operatorname{id}$
 
Oh, so $Re$ isn't differentiable...hmmm...
I wasn't thinking I could reduce it to the real case.
 
not complex-differentiable, yes
 
Hmm...so how should I attack this problem?
 
7:44 PM
well, yes, complex-differentiable implies continuously real-differentiable
and the norm of the complex derivative equals the operator norm of the Jacobian, so boundedness of the derivative translates into one another
 
I wonder why $\int_{0}^\infty \chi_{x:|f(x)|>\alpha} d\alpha = |f(x)|$
 
hmm...Is that helpful? I'm not familiar with that fact. I know that $|f'(z)|^2 = det J_f$, where $J_f$ is the Jacobian matrix (I might have the formula a little off).
 
@love_sodam Isn't that literally the same thing as the exercise you posted above?
 
it's helpful, because of the above general fact
 
I was trying to use Tonelli's theorem and during that, I think it's not obvious that the above equality holds
 
7:51 PM
Hmm...Is there a more elementary way? I don't think our professor is looking for a solution that uses the operator norm of the Jacobian.
 
What's being sliced by Fubini on the RHS is the indicator function of the undergraph of $f(x)$
 
no, this is completely elementary
it's just an application of the mean value theorem
 
@love_sodam It is integration by parts using the Riemann–Stieltjes integral
 
nvm
 
@Thorgott how does the mean value theorem work over $\Bbb C$? I'm only familiar with the case over $\Bbb R$
 
7:57 PM
@Thorgott How is it an application of the mean value theorem?
 
It's an application of the real mean value theorem
 
To what real function?
 
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Q: Measure theory and existence of set integral

orientablesurfaceLet $(X,S,\mu)$ be a measure space. Suppose $f\in \mathcal{L}^p$. Let $\epsilon>0$. Show that there exists a set $E \in S$ with $\mu(E)<\infty$ such that if $F\in \mathcal{S}$ and $F\cap E=\varnothing$ then $||f\chi_{F}||_p<\epsilon$ My idea was to do something like: $f\in \mathcal{L}^p$ implies ...

 
@AlessandroCodenotti Oh then it's just $m((0,|f(x)|))$ right?
 
@Thorgott Can the function be of a real variable but complex-valued? E.g., does the mean value theorem apply to $g: [0,1] \to \Bbb{C}$ given by $g(t) = f(tz + (1-t)w)$?
Where $z,w$ are fixed points in the unit disc?
 
8:04 PM
$$\int_0^\infty\raise{2pt}{\chi}_{\{x:|f(x)|\gt\alpha\}}\,\mathrm{d}\alpha =-\int_0^\infty\alpha\,\mathrm{d}\raise{2pt}{\chi}_{\{x:|f(x)|\gt\alpha\}}$$ where $\mathrm{d}\raise{2pt}{\chi}_{\{x:|f(x)|\gt\alpha\}}$ is the negative of the measure of $\{x:\alpha\le|f(x)|\lt\alpha+\mathrm{d}\alpha\}$ and $\alpha$ is essentially $|f(x)|$.
 
the one-variable mean value theorem doesn't apply, but your idea is very good
try applying the FTC to $g$
 
hi chat
 
@Thorgott this version? mathonline.wikidot.com/…
So, in our case, $\gamma (t) = tz + (1-t)w$ and $F = f$?
 
no, I'm saying apply the FTC to $g$
not integrate something else over $g$
 
@robjohn I don't know why the first equality holds (I'm in introductory course on Lebesgue integral theory). I think as a function of $\alpha$, $\chi_{E_\alpha}=1$ if $|f(x)|>\alpha$ so that $\int_{0}^\infty chi_{E_\alpha} d\alpha = m((0,|f(x)|))$
 
8:24 PM
y0
 
GREETINGs
 
8:40 PM
greetingS
 
We mathematicians like our duals
 
8:58 PM
Hi Ted
 
9:14 PM
Was Ted here?
 
My browser told me he was
 
Weird.
 
9:48 PM
@Astyx i read as brother at first and I was very confused
 
this question got me thinking, is there a good way of doing this without invoking like the rank theorem: math.stackexchange.com/questions/3884626/…
 
10:01 PM
@Thor: The composition of immersions is an immersion.
 
submanifold means embedded submanifold here
and while it's easy to see that the composition of embeddings is an embedding, the fact that the image of an embedding is an embedded submanifold (in the sense of admitting adapted charts) relies on the rank theorem/immersion theorem
 
The topology isn’t the challenge. It's the local analysis of the inclusion mapping. We know that a composition of homeomorphisms is a homeomorphism.
The immersion normal form is far easier than the rank theorem, but yes it's a special case.
 
I think I want something like "every chart of a submanifold can, at least locally, be extended to a chart of the manifold", but I'm not seeing an elementary argument for that
 
10:17 PM
That's the local immersion theorem. shrug
 
Yeah that's exactly the local immersion theorem
The inclusion of the submanifold is an immersion. Then pass to coordinates such that it becomes the standard immersion $\Bbb R^k \to \Bbb R^k \times \Bbb R^n$ near a point
 
10:34 PM
@love_sodam that is integration by parts. The boundary terms are $0$ If $f\in L^1$.
 
hmph, I guess you're right
 
Hey all
 
Hi @Krijn!
 
Are you by any chance bored, Bala?
 
Sort of. Why do you ask
 
10:41 PM
I have a trivial question that I can't solve
 
Feel free to ask. No promises I can help
 
Okay, so let $q$ be any prime power, doesnt really matter, and $n,m$ are integers, (and we may assume $n \leq m$ as everything will be symmetric
Define $\Gamma(z,k) = \prod_{i = 1}^k (q^z - q^{i-1})$
and then $W_{n,m}(k) = \Gamma(m,k)* \Gamma(n,k) / \Gamma(k,k)$
Then my computer confirms that $\sum_{k = 0}^n W_{n,m}(k) = q^{nm}$
But everything I throw at it is not working :(
 
How many $k$-subspaces does $\Bbb F_q^z$ have again?
 
Yeah, it comes from that, and that's how I know that last formula should work
 
$\Gamma(z, k)/\Gamma(k, k)$ maybe
 
10:46 PM
But I also want to show it by manipulating formula's
 
Oh but why
That's so annoying
 
Because it would really strengthen the argument I am trying to make
What we basically count is matrices $\mathbb{F}_q^{n \times m}$
 
yeah that makes sense
$k$ is rank
 
And then $W_{n,m}(k)$ are the matrices of rank $k$
you're so quick
So clearly summing over all ranks should yield the $q^{nm}$ matrices
SO WHY CANT I SHOW IT
 
It's pure geometry. $M_{m \times n}(F)$ is stratified by rank $k$ matrices, which is a Grassmannian more or less
Dunno man who can do arithmetic
Nice formula though
 
10:49 PM
Yeah the formula is pretty
The geometry is also clear
but arithmetic, meh
 
yeah
haha
 
I'll ask it as a question, maybe some combinatorics peepz have nice tricks up their sleeves
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Q: Showing a summation of powers of a prime power $q$

KrijnI calculated that the number of rank $k$ matrices in $\mathbb{F}_q^{n \times m}$ (where we may assume $n \leq m$) for a prime power $q$ is equal to $$W_{n,m}(k) := \prod_{i = 1}^k \frac{(q^m - q^{i-1})(q^n - q^{i-1})}{(q^k - q^{i-1})}$$ And so summing all $W_{n,m}(k)$ from rank $k = 0$ to full ra...

Time for some tea
 
What do you get if $q \to 1$? Some binomial coefficients summing to $mn$
Ah no you're counting elements so I don't think that's interesting
 
Yeah it seems really related to Gauss binomial coefficients
(it is!)
but that's not helping me
 
Yeah this is a little strange
 
11:10 PM
is the formula correct?
I'm only counting 6 matrices for $n=m=q=2,k=1$
 
(1,0|0,0), (1,1|0,0), (0,1|0,0) - flip rows and you get 3 more. Then (1, 1|1, 1), (1, 0|1, 0), (0, 1|0, 1)
3+3+3 = 9
@Thorgott ??
 
ah, so i am too stupid to count
 
Stop doing so much category theory
It's clearly not helping
 
dw, im done with higher category theory
 
did you move on to topos theory now?
 
11:24 PM
move to TOPOS theory
 
I moved on to reviewing complex analysis in preparation for my Riemann surfaces course this semester
 
let's read a good proof of the riemann mapping theorem someday
the montel magic sucks
 
Could someone help me evaluate $$\int P_n(x) P_m'(x) dx $$ where $P_n$ and $P_m$ are Legendre polynomials for $n \ne m$ (n = 8 , m = 11) I tried: a) Using the relation obtained for Pn'(x) from the Legendre equation. b) The orthogonal property c) Integration by parts, but nothing worked
 
I wanted to go to sleep, but I just noticed that The Hirsch Effekt published a new album in May this year and now I don't know whether I want to listen to it more than I want to sleep
 
I don't know any proof of it as of currently
 
11:43 PM
Problem: If $f \in H^2$ is cyclic, where $H^2$ is a Hardy space, then $f$ does not vanish on the unit circle. Proof: $f$ being cyclic is equivalent to $\{pf \mid p \in \Bbb{C}[z]\}$ being dense in $H^2$. If $f$ were a vanishing function on the unit disc, then $pf = 0$ for every polynomial $p$. But this means that $\{pf \mid p \in \Bbb{C}[z]\}$ can't possibly be dense in $H^2$...QED?
Is it really that simple?
 
11:54 PM
@Bala A german guy named Georg Landsberg solved in in three lines for me in 1894 :-)
 
No, that can't be right...it's too easy. I wonder if "does not vanish" means it has no zeros on the unit disc...I wonder how you solve that problem.
 
@myself, nevermind that, my German sucks
 

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