« first day (5187 days earlier)      last day (19 days later) » 

6:26 AM
@psie yes
7:21 AM
Its interesting that they notify you that your bounty is about to end - like you can do something about it
7:45 AM
@Jakobian It's helpful because if you don't award the bounty manually it will be awarded automatically, possibly to a different answer to the one you'd choose. See meta.stackexchange.com/a/16067/334566 for details. Section "What is automatic awarding?"
8:31 AM
that's if anyone were to answer me in the first place :')
8:52 AM
Got this from the video below 👇🏼
@SoumikMukherjee Nice. Statistics sucks
@SineoftheTime hehe, I want to send your message here
9:08 AM
do it :)
9:31 AM
@SoumikMukherjee statistic isn't, but probability is
and variance/covariance isn't only in the domain of statistics
9:48 AM
@TheEmptyStringPhotographer I’m pretty sure $TREE(3)$ is way bigger than $f_3(3)$ as a lower bound of the latter is only $10^{10^9}$.
10:28 AM
Let $(X,\overline{\mathcal M},\overline{\mu})$ be the completion of $(X,\mathcal M,\mu)$. Let $g$ be $\mathcal M$-measurable, hence also $\overline{\mathcal M}$-measurable. Is it true that $$\int g\,d\mu=\int g\,d\overline{\mu}?$$
Trying to google this is hard.
@psie you read this in Folland already
@Jakobian well, I haven't read it anywhere so far and I'm almost past the integration chapter. It's strange. I guess it is somewhere in there, but I haven't been able to find it.
There's the following proposition:
> Proposition. Let $(X,\overline{\mathcal M},\overline{\mu})$ be the completion of $(X,\mathcal M,\mu)$. Then if $f$ is an $\overline{\mathcal M}$-measurable function, there is an $\mathcal M$-measurable function $g$ such that $f=g$ $\overline\mu$-a.e.
He has also said that we "identify" the spaces $L^1(\mu)$ and $L^1(\overline{\mu})$ due to this proposition.
I guess it follows from this proposition...
It follows that $$\int f\, d\overline{\mu}=\int g \,d\overline{\mu}.$$ Though I still do not see how $\int g \,d\overline{\mu}=\int g\, d\mu$.
10:45 AM
okay, so its more that you didn't think about it and Folland only posed it as a remark
but it was important
@psie let me ask you this, whats the value of both on a simple function
we'll prove something more general, that if $\mu_0$ is an extension of $\mu$, then $\int g d\mu = \int gd\mu_0$
@Jakobian it is $$\sum_1^n a_k\mu(A_k)=\sum_1^n a_k\overline{\mu}(A_k),$$ where $A_k$ are disjoint and partition $X$ and $a_k$ are distinct.
and now whats the value of both on non-negative functions
it was defined as a certain limit of integrals of simple functions... so its the same
so $\int g d\mu = \int gd\mu_0$ holds whenever it can hold, and one expression makes sense if and only if does the other
@Jakobian it's the $\sup$ over all simple functions such that $\sum_1^n a_k\chi_{A_k}\leq g$.
be it $g$ is integrable or not (so that say, we allow $\int gd\mu = \pm\infty$)
ok
10:53 AM
@psie well I guess for this definition the argument is more subtle, but whats matters is both are pointwise limits of increasing sequence of simple functions, which are $\mathcal{M}$-measurable
(since for your definition of $\int gd\mu_0$ we take $A_k$ to be $\overline{\mathcal{M}}$-measurable)
so for your definition we would need to use something like Lebesgue's monotone convergence theorem
Hello, can anyone explain the definition of lift in the topic below?

https://math.stackexchange.com/questions/4709875/bijection-between-lifts-of-a-simple-closed-curve-and-its-cosets-in-pi-1
Does it imply existence of some map $g:S^1 \to \tilde S$ ?
hmm, ok. I'm doubtful about the $\sup$ over $\sum_1^n a_k\mu(A_k)$ is the same as $\sup$ over $\sum_1^n a_k\overline{\mu}(A_k)$. Whatever.
11:19 AM
@Jakobian yes, the creator probably put that as a joke that there is no categorical definition of covariance
@psie I just explained this. Why are you getting mad
Regarding my post, in the link they define a lift of a curve $\alpha: S^1 \to \tilde S$ as a lift of the map $\alpha \circ \pi: \Bbb R \to \tilde S$, where $\pi: \Bbb R\to S^1$ is the usual covering.
24 mins ago, by Jakobian
so for your definition we would need to use something like Lebesgue's monotone convergence theorem
you can directly use that a non-negative $\mathcal{M}$-measurable function $f$ is an increasing pointwise limit of simple functions $f_n\to f$ with $f_n$ being $\mathcal{M}$-measurable. From Lebesgue's monotone convergence theorem, $$\int f d\overline{\mu} = \lim_n \int f_n d\overline{\mu} = \lim_n \int f_n d\mu = \int f d\mu$$
Do we need the MCT? Let $S$ be the set of $\sum_{k=1}^na_n\mu(A_k)$ where the sets $A_k$ are $\mathcal M$-measurable. Let $\overline S$ be the set of $\sum_{k=1}^na_n\mu(A_k)$ where the sets $A_k$ are $\overline{\mathcal M}$-measurable. Evidently $S\subset\overline S$ and since for every $s=\sum_{k=1}^na_n\mu(A_k)\in\overline S$ we can find sets $B_k\in\mathcal M$ such that $B_k\subset A_k$ and $\mu (B_k)=\overline{\mu} (A_k)$.
Then $s=\sum_{k=1}^na_n\mu(A_k)=\sum_{k=1}^na_n\mu(B_k)\in S$. So also $\overline S\subset S$
Sure I wasn't sure if elements of $\overline{\mathcal{M}}$ are unions or unions and differences of elements of $\mathcal{M}$ with a null set
But what you are saying is true
11:31 AM
@psie Meeh, sorry, but some of the $\mu$'s here should be $\overline{\mu}$, my bad
in particular, when I write $s=\sum_{k=1}^na_n\mu(A_k)\in\overline S$
I only looked at the idea of the proof
ok 👍
 
3 hours later…
2:17 PM
Have y'all ever called a mathematician before
I need a mathematical hotline
 
1 hour later…
3:28 PM
Can this matrix problem be formulated as an ILP? Given an n by n binary matrix, I want to find the smallest number of bits that need to be flipped to reduce the rank of the matrix over the field of integers mod 2. I don't think there is a fast algorithm so I was hoping it could be formated as an ILP problem. But I am not sure if the rank restriction allows that.
Reduce the rank to some given bound, or just reduce at least by $1$?
@VladimirLysikov by at least 1
Joe
Joe
4:11 PM
I'm trying to think of examples of spaces that are path connected but not locally path connected. If we take a circle in the plane and remove a point, this is an example, right?
@Joe no
you need more pathology than that
@Joe Warsaw circle
the punctured circle is in particular homeomorphic to $\mathbb{R}$, a very nice space :)
Joe
Joe
Oh, I see where I went wrong now.
Thanks to you both.
4:29 PM
👋
Do mathematicians have any feelings about this? mindmatters.ai/2021/03/…
@SineoftheTime But for this "Criteria of exactitude for differential forms C⁰" its mean to write the characterization theorem of exact forms? Is that all or is there something else to write?
It's an article with Gregory Chaitin
About science and beureocrecy
How it's all being held back
I don't even want to read that
4:34 PM
$\int_\gamma \frac{e^{5z}}{(z-i)^3}dz$ $\gamma=C_2(0)$
it's worth being suspicious of any academic or technologist making broad claims about "science" vs. "bureaucracy." i think my default position is that serious people do not do things like that. at a high level i would certainly agree that science is often at odds with bureaucracy. the sky is also blue
Can you confirm that I have to do the derivative?
Guess I'm confused of the goals of these subjects in a way
Or is someone like Musk just privileged/ lying to hope to advance society?
Idk maybe it's a skill issue lol I just put in so many hours over the last 8 years and havent seen a payoff
@ZacU. Honestly, I got maybe three or four paragraphs in before it felt like b*llsh*t. There seems to be a certain uncritical fawning over rich technologists (Musk and Wolfram, in particular) which I find really off-putting.
These men are snake oil salesmen.
Oh
So is it a matter of if you can't beat them, join em?
4:42 PM
I think its a matter of throwing out popular names for visibility
Well I mean for somebody who wants to do what those guys claim is their aim
Seems like slippery slope
Or maybe just dull
@ZacU. I've called my prof at 1 am once. Asked him if 0 is a natural number
He blocked my number
for realises tho don't call mathematicians to discuss math, more fruitful would be to book some office hrs with them
@Pizza there are different criteria
Even better is some chance encounter at the cafeteria
4:57 PM
You're right and I see why now. Dang I thought I had life all figured out and then whoossh
You had life figured out? What protein does the following RNA sequence correspond to? AAGAUUAGAA
@SineoftheTime Could you give me some examples, because I can't find them
Well I mean as a research scientist mathematics hybrid I could theoretically have skirted by the tedious natures but now I see a bit differently
Like bro I was gonna be the Greatest.
Like no one ever wass
5:14 PM
@nickbros123 that's an RNA sequence? sounds similar to the language I used to speak when I was 1 or 2 years old
@SoumikMukherjee I may or may not have keyboard mashed :)
usually RNA sequences come in multiples of 3 so I think this is fundamentally wrong
ok that may also be wrong. Ill stop with the misinformation now lol
Talk about ANR, this is a math chatroom
(jk) no restriction on what you wanna talk about
@SoumikMukherjee solve riemann hypothesis
@SoumikMukherjee sure let's talk about absolute neighborhood retracts
I was actually refering to them
5:18 PM
lol I read that as ANT xD
something is wrong with me atm
You sure it's only atm? :p
all the moments
@nickbros123 "pirate struggling to withdraw money"
or an irish guy :)
@Pizza closed form implies exact on a simply connected domain
@Gian'sPizzeria you have to use Cauchy formula for derivatives
5:23 PM
(chanting of "de Rham, de Rham" echo in the background)
this may be the antihistamines talking but I rly like numerical analysis lol
@SineoftheTime But isn't this true for class C¹ functions?
Did you see a theorem like this: $\omega$ is exact iff for all $x,y\in A$, for all $\gamma, \phi$ $\int_{\gamma} \omega=\int_{\phi}\omega$
where $\gamma,\phi$ are curves that start at $x$ and end at $y$
@Pizza right, you need $C^1$
an another theorem is: $\omega$ exact if $\oint_{\gamma} \omega=0$ for all curves in $A$
@SineoftheTime yes
This should have the things you wrote
yes
that's the theorem I was referring to
5:37 PM
Yes, so I only have to write this, is there nothing else?
I saw that there is also the proof
@Pizza I don't know what you did in class so I can't say if you did other theorems
I only found this related to the question to answer
that's the standard one as far as I know
@nickbros123 smart guy
In analysis usually $0\notin \Bbb N$
5:47 PM
> Exercise: Let $(X\times Y,\mathcal L,\lambda)$ be a complete measure space. If $f$ is $\mathcal{L}$-measurable and $f=0$ $\lambda$-a.e., then $f_x$ and $f^y$ are integrable for a.e. $x$ and $y$, and $\int f_x\,d\nu=\int f^y\,d\mu=0$ for a.e. $x$ and $y$.
I have worked towards a solution to this exercise, but the solution doesn't use the fact that $f$ is $\mathcal L$-measurable. Why is this needed?
Here $\lambda$ is the completion of $\mu\times\nu$.
@psie not needed
although, depending on your definition of $g = h$ a.e. you might need it
but for your definition I'm assuming its just that $\{f \neq 0\}$ is contained in a set of Lebesgue measure zero, which makes it Lebesgue measurable
yeah, it's baked into "complete measure space"
oh. This is not the Lebesgue measure
well, what I said doesn't need that
psie just in the background, while this kind of hypothesis chasing is necessary in the textbook, in many "real life" applications of measure theory you will stay in the world of measurable functions if you start in it, so it is common for people to tack that kind of hypothesis on just as a security blanket because that's what they're going to be confronting even if it isn't crucial to really anything at all in the argument
so it wouldn't upend my understanding of what an author is trying to do if they added in a measurability hypothesis that seemed superfluous or even was provably superfluous
some people define symbols like "$g = h$ a.e." only when $g, h$ are a priori measurable
5:58 PM
yeah, I don't find it very disturbing that the author put it there, but it's reassuring that you say it is not needed, then we're on the same page
with lebesgue measure in particular, once you deal with functions having any kind of structure there are a lot of 'automatic measurability' results where, if you do your homework, anything that satisfies whatever you are talking about is going to be measurable, but the proof and argument style used for the proof is just so far from whatever else you might be doing that people ignore it and throw in the superfluous assumption instead
you should be sensitive to completion issues whenever you deal with product measures and i would be generally concerned with measurability concerns if you are doing anything resembling an 'infinite product' construction, people get lazy and sometimes f up in that setting
but not in just like doing calculus or PDE or whatever on R^n
ok 👍
6:20 PM
Its important though. Sometimes you want to restrict to Borel sigma algebra
7:11 PM
Consider the above proposition. I'm trying work a proof of another theorem where this proposition is needed. I'd like to make two assumptions, one of which I feel confident about, the other which I'm doubting.
1. If $f\geq0$ in the above proposition, I can assume without loss of generality that $g\geq0$. I understand this, because we could just consider $h=|g|$, which will be nonnegative and still measurable if $g$ is. Also, we still have $h=f$ $\overline{\mu}$-a.e. if $f\geq0$.
2. The second assumption I'd like to make is that if $f\geq0$, we can assume without loss of generality that $g\leq f$. Is this possible? In other words, if we have $g>f$ on the $\overline{\mu}$-null set, can I redefine $g$ so that it still satisfies all the conditions in the proposition and we get $g\leq f$?
I thought about simply defining $h=g\chi_{E}$, where $E^c$ is the $\overline{\mu}$-null set, though I'm not sure if this function is $\mathcal M$-measurable.
I’m thinking of writing a paper about Hilbert’s hotel and the fast-growing hierarchy
which will certainly include limit ordinals
7:27 PM
@SineoftheTime In Gian's exercise he has to use : $\frac{1}{2\pi i} \int_\gamma \frac{f(z)}{(z-z_0)^{n+1}} dz = \frac{f^{(n)}(z_0)}{n!}$?
8:09 PM
Oh okay
@Pizza did you see Residue theorem?
8:31 PM
@SineoftheTime nop
@SineoftheTime but shouldn't we use the residue theorem?
Anyway, for now I'm doing analysis 2, I'm not really focused here either
Anyone know what happens to the point $(0,-3)$ when $f(x)=-3\cdot\cos(45(x))$ becomes $f(x)=-3\cdot\cos(45(x-2^{\circ}))$? My thinking is that since it deals with trig functions that $-2^{\circ}$ is the same as $\frac{2}{2\pi}$ or just $\frac{1}{\pi}$ & that this may be useful supposed to be dealing in degrees the whole time.
@Gian'sPizzeria Cauchy theorem is a special case of the residue theorem
do you know how to compute the residue at a pole?
@sine see my big brain simplifying move Qh2+ :"/
8:40 PM
the knight is pinned :(
would the points on the graph just be equal to radians in a sense? 3.14... is denoted as $\pi$ so wouldn't subtracting 2 degrees just be a shift to the right by 0.0349066 radians?
I forgot that, (fortunately my opponent forgot that too)
that happens a lot of time
I should read about psychology in chess
@VulpesInculta Using the same letter to mean multiple things is generally a bad idea. You have a function $f$, defined by the formula $f(x) = -3\cos(45^\circ x)$. You then define a new function, say $g$, by the formula $g(x) = -3x\cos(45^\circ(x-2^\circ))$.
Observe that $g(x) = f(x-2^\circ)$. So the graph of $g$ is identical to the graph of $f$, but translated to the right by $2^\circ$. This implies that the point $(0,-3)$ (corresponding to $f(0)=-3$) is translated $2$ units to the right, to the point $(2,-3)$ (corresponding to $g(2)=-3$).
@XanderHenderson okay thank you. this makes sense
8:47 PM
It should be a straightforward exercise to confirm that, indeed, $-3\cos(45^\circ(2^\circ-2^\circ)) = -3$.
@XanderHenderson it seems pretty obvious now the graph was represented using fractions of $\pi$ so I guess I just assumed I should do everything else in radians lol
The units don't actually matter here.
I assumed that $x$ was measured in degrees, because otherwise, that factor of $45$ seems like a bit of a non sequitur, but it doesn't actually matter.
The key idea is that $g(x) = f(x-2)$. So $g$ is a right-translation of $f$, by $2$ units, whatever those units happen to be.
(Degrees, radians, hogsheads, whatever).
@XanderHenderson so far the most part i don't need to worry about the types of units & just work based off the integers?
@VulpesInculta That question is way to broad to be answerable. It depends on context.
In this particular problem, the units don't matter.
@XanderHenderson I suppose just in this context then of trig graphs
9:00 PM
@SineoftheTime I want to learn about this double blindness in chess.
I hope my professors are blind when I make mistakes😩
@SineoftheTime Really? I would hope that they wouldn't be blind, and that they would correct my mistakes.
yep, joking
9:24 PM
Erase few words and it becomes evil
"I hope my professors are blind"
Is $\log 0$ undefined even under surreal numbers?
Nevermind; surreal numbers don't admit division by zero either.
danny: separate from that, if it were defined one way vs. another, what would you do?
@leslietownes Cry.
rejoice
9:40 PM
Even though wheel theory is a thing, it ruins topological properties such as Hausdorff-ness, so not interesting whatsoever.
that's a weird criterion to discard wheels on
but yeah, wheels are a bit of a joke
I never knew Rachmaninov also composed sacral works
@BenSteffan Huh... I thought Wheel Theory was a punchline. Still not sure what the joke is...
@BenSteffan Oh, muchly!
@XanderHenderson The joke is that they figured out how to divide by 0, finally, and then nobody cares (as far as I'm aware)
not being able to divide by 0 is a critical problem in mathematics, even to the point that mathematics as a whole is a scam, y'know (according to... you know the kind)
Shh...! You might summon the dude who insists that zero is not an even number.
not that I disagree that mathematics as a whole is a scam but :^)
@XanderHenderson if it were an even number you could divide by it...
9:50 PM
@BenSteffan SHHHHHH!
@leslietownes yes
10:27 PM
Folland says the Lebesgue measure $m^n$ on $\mathbb R^n$ is the completion of $m\times \cdots\times m$ on $\mathcal B(\mathbb R^n)$, or equivalently, the completion of $m\times \cdots\times m$ on $\mathcal L\otimes \cdots\otimes\mathcal L$. Why is this an equivalence? Are both these measure spaces incomplete and their completion is $\mathcal L(\mathbb R^n)$?
10:55 PM
I think these spaces are both not complete. I feel quite shaky by all the surprises one finds when examining product measures on product spaces, so I'm even a bit unsure if $\mathcal B(\mathbb R^n)\subset \mathcal L\otimes \cdots\otimes\mathcal L$.

« first day (5187 days earlier)      last day (19 days later) »