Quick question, if we have an uncountable power set $2^S$, and we consider a subset $H\subseteq 2^S$ for which everything in $H$ is disjoint, is it possible for $H$ to be equinumerous with $2^S$?

if you assume choice, there is a choice function $H\rightarrow S$, which is injective by the disjointness condition, so $\# H\le\# S<\#2^S$

@Thorgott I see, thanks

So, here's a thought. If you have a ring $R$, does it make any sense to talk about $\lim\limits_{n\to\infty}R[x^2,x^n]$? Is that notion even well defined?

what's $R[x^2,x^n]$

$R$ adjoin the indeterminates $x^2$ and $x^n$

a) $R[x,y]$ or b) the subring of $R[x]$ generated by $x^2$ and $x^n$?

Ah, I see. It's the subring of $R[x]$

If $x^n$ were not referencing the same $x$ as $x^2$, this would be an easier thing to "answer?"

it would be boring then, cause all those rings would be the same

option a) was silly on my end

Ah, gotcha

anyhow, I don't see a natural map $R[x^2,x^n]\rightarrow R[x^2,x^{n+1}]$

Okay, so the limit would be in terms of homomorphisms? I can see the logic in that.

I guess the larger thing is that I've never encountered the idea of a limit of a sequence of rings

the thing is that just a "sequence of rings" is not enough information to create a sensible notion of limit

you somehow need to "relate" the rings in your sequence

(if you look at a sequence in a topological space, the elements of the sequence are "related" by the given topology - I might be stretching analogy here)

Ah, so we need to set up a metric space where the elements are rings

Or, maybe not even a metric space?

so what you want to do is look at a sequence of rings $(R_n)_n$ and homomorphisms $f_{ij}\colon R_i\rightarrow R_j$ for $i\le j$ such that $f_{ii}=\mathrm{id}_{R_n}$ and $f_{jk}\circ f_{ij}=f_{ik}$ for $i\le j\le k$

It's been a minute since I've thought about more abstract topological spaces

nah, this is something completely algebraic

if you think of it diagramatically, you just arrange the rings in a sequence and have an arrow from one to the next one

Alright, so you can think about the sequence as a sequence of homomorphisms $(f_{1n})_n$

Then consider the images as $n\to\infty$

that is a great idea

for example, take $R_n=R[x_1,...,x_n]$ (where $R$ is some ring), then you have a natural inclusion $f_{ij}\colon R[x_1,...,x_i]\rightarrow R[x_1,...,x_j]$

this is essentially just a tower of rings where you just adjoin more and more indeterminates

the "limit" of this sequence should intuitively be $R[x_1,...,x_n,...]$, the polynomial ring in infinitely many variables

Yeah, alright, I think I get the general idea

but that's not just because of the individual rings, but also because of how we know that they lie inside one another (which is given by the inclusion homomorphisms)

So, if I wanted to somehow talking about an indeterminant of infinitely high power, I would need to have these maps set in place beforehand

So, I could have the inclusion maps $$f_{ij}:R[x^3,x^{2^i}]\to R[x^3,x^{2^j}]$$

superscripts on superscripts

the idea behind constructing the limit object in the general case is to take the disjoint of all the $R_n$ and then identify all the elements that "become the same" under the given homomorphisms (this accounts for how the rings relate to one another under the given homomorphisms)

this works not only for sequences, but more generally for directed sets (or, even more generally, for filtered categories)

Okay, so the limit of my $R[x^3,x^{2^n}]$ idea would be $R[x^3]\cong R[x]$?

I don't see what your inclusion maps are

it seems to me as if the arrows would have to go in the reverse direction

Ah, right, yeah, they would, if they were inclusion

(in which case, you can rediscover the notion of an inverse limit)

I was thinking $f_{12}(x^{2})=x^4$, which wouldn't be inclusion

In fact, thinking about it, I am unsure if that is a well defined homomorphism

how would it act on a general polynomial?

So, to start $x^{20}=f_{12}(x^2)^5=f_{12}(x^{10})=f_{12}(x^3)f_{12}(x^3)f_{12}(x^2)$, and so $f_{12}(x^3)=x^7$. Checked a few examples, and it seems to be consistent

Wait, I think I found a counterexample

$x^{21}=f_{12}(x^3)^3=f_{12}(x^9)=f_{12}(x^3)f_{12}(x^2)f_{12}(x^2)f_{12}(x^2)=x^7x^{12}=x^{19}$

So my hypothetical map $x^2\to x^4$ is not well defined

@TedShifrin hows it going?

Hi @Faust. What's up?

well my eardrums just blew out

Got my rifle back that my crazy mother stole from my house today

Um, what stupidity did that?

My headphones were apperntly set too high for google so when you pinged me i felt my brain vibrate.

Well, be careful!

^^ been pretty productive last couple days, though i keep having to look up definitions

took me a minute to remember how to do a double integral today. =/

Really happy to have my grandfathers rifle back though ^^

not that i particularly have a use for it as i havent gone hunting since he died.

No gun talk for me.

Fair enough i can understand that with where you live, for me its just a memento of someone who was important to me. Wouldn't matter what it was if it was from him, but i digress. You been keeping busy lately?

are you still doign any teaching?

Nope, no teaching .

Just enjoying retirement then?

Let $K, L$ compacts subsets of $\mathbb{R}$ with usual topology, and $ĸ\cap L = \emptyset$ prove that, exists $G, H$ open subsets of R that $K \subset G, L \subset H$ then $cl(K)\cap cl(H) = \emptyset$ is it just me or does this not make any sense?

It seems liek the way its stated you could just take $H = \mathbb{R}$

Huh?

i dunno its not my question i just don't understand it

They mean $G$ and $H$ have disjoint clisures.

oh ok

But your $H=\Bbb R$ is nonsense.

well the question was nonsense

No.

Reread even with what they have, and what you said is nonsense.

Let $K, L$ compacts subsets of $\mathbb{R}$ with usual topology, and $K\cap L = \emptyset$ prove that there exists $G, H$ open subsets of $\mathbb{R} $ such that $K \subset G, L \subset H$ and $cl(G)\cap cl(H) = \emptyset$

so this is what its supposed to say?

What's the closure of $\Bbb R$?

just R

So how are things disjoint?

they arent which was my concern

But you said to take $H=\Bbb R$; they didn't.

i know i was saying it would be a counter example because of how the question was stated it confused me

No. You need to think more.

?

Anyhow, this is basically normality in topology.

ok well i still don't know what the question is asking the way it was written =P

But thanks for trying to help ^^

Given two disjoint compact (more generally, closed) subsets, they have open neighborhoods with disjoint closures.

oh ok thats what i thought he meant but he wrote the closure of one of the compact sets with the closure of the open set containing the other compact set. which confused me.

Thanks ^^

can use the fact that R is hausdorf and its not too bad to show the result i think.

Why the hell is the word normal always used to describe something that usually doesnt happen in mathematics =\

Not so, but overused for sure.

lol

hi math friends! Ignorant physicist here wondering how to calculate the inverse Laplace transform of 1/(e^s-1) . Been a while since complex analysis and Mathematica is failing me :(

I'm curious, based on the notation $X^Y$ for the set of functions from $X$ to $Y$, do we have $|X^Y|=|X|^{|Y|}$ for finite sets $X$ and $Y$? Because the power set notation $2^X$ comes from the same notation, and finite sets do indeed have power sets of cardinality $2^{|X|}$

@SirCumference Yes, and in fact, the statement also holds for infinite sets, though the behavior starts to become a bit weird.

@TobiasKildetoft yo that's neat

weird i couldn't find anything on it on wikipedia tho

For finite sets, it is a basic exercise in combinatorics

For infinite sets it is essentially by definition

What happened to SteamyRoot ? @LeakyNun

How can i solve $d^2theta/dt^2=mgsintheta/R$ where mg and r are constant

Or does that not contradict ?

What do poles have to do with surjectivity?

I dont know lol just intution

It is with the correctly defined codomain

over R tho?

@Astyx what's steamyroot

ok I have a stupid question:

Can $(e^z)^{\frac{1}{z}}$ where z complex, be simplfiied?

I knew the rule $(e^a)^b=e^{ab}$ fails for complex $a,b$

Defined on which domain?

complex numbers $\Bbb{C}$, I think we can take the principal branch?

so not all the complex numbers

yeah, I think there is a branch for $()^{\frac{1}{z}}$ so I don't think there is a way to take it apart without considering cases. Let me think...

@LeakyNun used to be a user in this chat

With blue and yellow icon

quick question, suppose i have $X=u\frac{\partial}{\partial x}+v\frac{\partial}{\partial y}$ and I'm asked to find $X(\omega)$ where $\omega=a\,dx+b\,dy$, how can I do so? I mean what does it mean $\frac{\partial}{\partial x}dx$?

are you sure you don't want to find $\omega(X)$?

@Faust, FWIW, try to think why the proposition is 236% intuitive. What do closed sets look like in $\mathbb{R}$ under the usual topology?

fwiw?

For What It's Worth

@Thorgott i have no clue

maybe you're right actually

Im sorry i just woke up and i am super confused lol

@Thorgott in that case what would it be?

scroll up. the last thing you spoke about with Ted

If your talking about that problem from yesterday

ah

finite subcover

cause hausdorf

sorry really need coffee thats all that comes to mind

well, can you do it for $[0, 1]$ and $[5, 6]$?

$\omega$ is a $1$-form, so it associates to each point $p$ a linear form $\omega_p$ on the tangent space $T_pM$. $X$ is a vector field, so it associates to each point $p$ a tangent vector $X_p$ in $T_pM$. So the "pointwise evaluation" $\omega(X)$ makes sense and is the smooth function mapping $p$ to $\omega_p(X_p)$.

@JoeShmo sorry man i have been off sick mostly in the hospital since april i haven't touched topology in too long i just don

dont remember enough to do anything with it

oh, I'm sorry to hear

eh im pretty happy didn't die and im feeling alot better, i will figure the problem out after breakfast and thanks for the help ^^

@Thorgott i get it, but how to compute?

it's still not clear to me ( i mean the computational part )

review the definitions involved

What is the principal branch of $n^{-z}$?

Anyone remember calc II wanna point out where i messed this up? math.stackexchange.com/questions/3868688/…

@StupidQuestionsInc I'm bothered that your formula for $X$ has $x$ and $y$ as coordinates but $u$ and $v$ as coefficients. Those are functions of $x$ and $y$? ... You also might want to watch my YouTube lectures on differential forms if you're having problems with basics. (See link in my profile.)

@Faust: There's no mistake. $\ln 3$ is a constant.

@TedShifrin no just real numbers

Oh, bizarre notation. Usually letters at the beginning of the alphabet are used for constants.

@TedShifrin sorry for that

So remember that $\partial/\partial x,\partial/\partial y$ and $dx,dy$ are dual bases. What does that mean?

@TedShifrin what I do know is that if $X$ is a vector field $f$ a smooth function then $fX)(p)=f(p)X_p$

@TedShifrin but why doesn't it affect the 9/2 ?

Yes, not sure how that's important for your question.

@TedShifrin oh, $dx^j(\partial/\partial x^i)=\delta_{ij}$

@Faust, OK, so $\frac92\ln 3$ is a constant.

There you go, @Stupid :)

oh ok thanks alot ted

@TedShifrin sometimes i'm just so deluded -_______-

thanks for showing to me that i have way more to go before reaching ideal understanding

in this subject

@Faust: There are sometimes far better disguised "different" solutions. For example, if you try substitution one way or integration by parts the other, you get totally different answers. And yet ... not.

@Stupid: I don't know what level of course you're taking, but, as I said, my lectures might be helpful. (I explain what forms are doing in terms of basics and geometry, no tensor stuff.)

@TedShifrin yeah i haven't dont integrals in a long time and i forgot divided by is the same as minus. doh

@TedShifrin i'm self studying Tu's Intro to manifolds

Um, you mean log of a quotient is the difference of the logs, @Faust ?

@StupidQuestionsInc its a pretty good book, its the one i learned it form

Ah, well, if you get lost in abstract stuff, check out my lectures. Lots of concrete examples.

@TedShifrin yeah, im glad you could hear what i meant not what i said lol

@Faust i'm sure it is

Especially when it gets to surface integrals, Stokes's Theorem, etc.

@TedShifrin thanks i will do so, i'm also skimming over the book A Visual introduction to differential forms

I don't know Tu's book. I just have known him for a long time, since we were both grad students, basically.

@TedShifrin hopefully i will be back in fighting form in a week or 2

I'm surprised you were interested enough to make it all the way through that integral :)

Its been a long time since doing trig sub so i kinda got into it, anything i remeber doing in recent memory was just a matrix times something or it was in $\Bbb C$ and you find the intergal by takeing the derivative lol

i really like the triangle at the end there, i think its a beautiful problem ^^

so they do cancel out

If $n$ is a natural number and $z$ is a complex number, is it true that $|n^{-z}| \le n^{- Re ~ z}$?

@TedShifrin oh interesting :)

btw one last question

just to check that i'm correct, if $X=a\frac{\partial}{\partial x}+b\frac{\partial}{\partial y}$ and $\omega=xdx+ydy$ then $\omega X=xa+yb$?

seems correct to me

$\omega X: M\rightarrow \mathbb{R}$ is defined by $\omega(X)(p)=\omega_p(X_p)$ , where $\omega: M \rightarrow \coprod_{p \in M} T_p^{*}M$. So locally, yes... it seems correct @StupidQuestionsInc

@mathsresearcher thanks a lot

also recall the useful fact, $\omega \in \Omega^1(M)$ is smooth iff for every $X\in \mathfrak{X}(M)$, $\omega(X)$ is smooth

unrelated to your problem

but it's still useful

and Tu's book is grear

really great

Also, if $a-b \ge 0$, can I say $n^{a-b} \ge n^{0} = 1$?

@user193319 Remember that in general $\log a$ is multivalued, although convention is that it is not when $a$ is positive real. You seem to be inventing your own rules, though. You're saying that $e^{ab} = e^a e^b$? Isn't it the case that $|e^{i\theta}|=1$ for any real $\theta$?

Hey Ted!

Heya Demonark.

Did you deliver your ballot to the unique ballot box in your county, Demonark?

fun fact, I registered to vote apparently on the deadline date in NJ

I haven't done the actual voting yet, though I'm set to do so

That was cutting it close, @JoeShmo.

well, I did register to vote earlier, but they miraculously didn't have me on file. This time I'm on file.

That's the understatement of the century, @Sophie. But the UK isn't far behind us.

Ah, there's a lot of that suspiciousness going on, @JoeShmo.

@TedShifrin I feel like ever since 2010 when everyone got cellphones with social media in their pockets all the time it has been terrible all over the world, but maybe that's because I don't recall much about politics before then

it's a reality-TV circus, @Sophie.

it's certainly framing the politics of the day, I agree

Certainly the internet makes things different from the horse-and-buggy era, yes.

Another fun fact -- this will be the first time I am participating in the democratic process, anywhere.

Wow. My first vote was in 1972. I was proudly from the one state that went for George McGovern (but Nixon was re-elected, not for long, as it turned out). We now have a so-called president who makes Nixon look angelic.

Back to math(s).

not by choice, I would have voted sooner, but I wasn't a citizen here, and you can't vote with an absentee ballot in Israel

Might be of interest to people in functional analysis or calculus of variations:

Interesting. I would guess that there are a lot of Israelis all over the world who are disenfranchised.

In math news, I'm applying for the NSF grad fellowship... fun times

Good luck, Demonark.

Ted, I don't know whether to cry or to laugh..

Thanks. I've finished the personal statement modulo implementing some edits, and yeah I'm gonna spend the day grading to procrastinate on doing the research statement since that's a bit more scary

@NikeDattani This sounds more like engineering than mathematics.

@JoeShmo That's pretty open-ended.

Demonark: It's been a few years since I did that, although I have written recommendations for a number of students over the years.

They apparently are taking impact on minorities very seriously these days, Demonark. So pay serious attention to that.

Well, maybe under Trompolini that's changed. Who knows.

@TedShifrin Are you sure?

Hopefully not, I would imagine that the NSF would be somewhat... perhaps not immune but would have something of a buffer

Since most of it makes absolutely no sense to me, @Nike, I wager yes.

If you tell me the underlying Banach spaces that are the domains, maybe it'll be more mathematical.

that is preposterous. I had no idea Nixon won practically unanimously.. how does it feel to have been right, Ted?

LOL, @JoeShmo, I'm still proud of my vote.

wth I've never seen a map so red. What was it? The Soviet threat?

In large part, yeah, and this was the middle of the Vietnam war, too.

man there too many wars in this world

Too many evil, evil people, @Faust.

was the Vietnam war popular at that point?

Hell no.

oh you're saying in spite of Vietnam

Many of my friends and I were ready to escape to Canada if our numbers came up too low on the draft lottery (educational deferments had stopped).

No I think Nixon's appeal was getting us out of Vietnam

Well, it was popular among the Nixon supporters, yes.

Or am I misremembering

oof

I think you're misremembering.

McGovern was the peace candidate for sure.

I can't imagine a republican being anti war

in that day, anyway

@TedShifrin its easy to say that but i would like to think people aren't born good or evil; life perhaps forges evil people due to circumstance and inequality.

@TedShifrin Since it has the DFT tag, they are functionals of density functions rho(r) which are real-valued functions of x, y, and z Cartesian coordinates. here "r" is a vector (x,y,z)

This doesn't make it of interest to mathematicians, @Nike. You're not enlightening me.

haha

but it's a vector

@TedShifrin I was just answering your question about the domain

You didn't tell me the Banach spaces.

Anyhow, I'm not objecting to the existence of the website. I'm saying it doesn't look like it's of general interest to mathematicians.

@TedShifrin What exactly do you want to know about the Banach space of the domain?

Yeah looking it up it seemed like Nixon's deal was that he wanted to pull out without admitting defeat by basically throwing resources at South Vietnam and saying "Alright now it's on you glhf"

Is anyone here familiar with drawing hmm diagrams?

I looked at a bunch of tutorials and pdfs, but couldn't get a solid answer for the following question. How and where is the initial probability encoded into an hmm diagramm?

I know how to draw out the transition, emission probabilities. But not sure how to deal with the initial probability.

No earthly idea what you're talking about, sorry.

By hmm, i mean hidden markov models.

yes and ... what exactly are you trying to draw..?

if you have a verbatim problem statement, that would help

Well in the simple case of a person using a fair vs loaded coin, I have an initial probability of whether the person starts with a fair coin or loaded coin.

How do I incorporate that into an hmm diagram.

For example, let's say a person has one fair and one loaded coin, with P(H=0.5) and P(H=0.8) for the two coins respectively.

And they flip it 3 times. And the probability that he'll use a fair coin given he used it in the previous turn is 0.8.

Similarly the prob that he'll use a loaded coin given he used it in the previous turn is 0.9.

All this is veyr straightforward and I can draw the transitions, nodes, etc.

But there's one piece of information that I do not know what to do with and that is the initial probability. Let's say there is a 75% probability he starts with a fair coin.

How and where does this get encoded?

Obviously my diagram will look something like the one on page 5 of this pdf: cbcb.umd.edu/confcour/Fall2012/hmms.pdf

But even this pdf doesn't show how to use the initial probability.

start with an initial "null state" that has an arrow to the fair and loaded coins

Oh wow, that's a simple and elegant way to handle it, thank you!

@joe

Why it feels differential-equations are harder than real and complex analysis?

@Stupidquestioninc what kind of DE's?

@Faust ordinary

nfi they are generally alot easier than most other things

even easyier than complex analysis and that saying something

without proof I find everything hard

i mean real analysis is pretty hard

much harder than complex anyway

do you have a specfic question?

I find rudin book a lot easier than ordinary diff equ book

what book lol?

@Faust Nah I found I forgot most of the diff equ but didn't forgot real analysis

James c Robson intro to diff equ

any ok that makes sense i forgot almost everything about DE's right away

ah*

I can't even remember pitchfork bifurcation

nfi if i see a problem in DE's though i always figure it out pretty quick but i never remember the names of shit

yeah same

analysis though its pretty hard to forget its just kind of "how would a mathematician think about this?"

if u cna think like one thats 90% of analysis

However its usually pretty hard to get to that point for most people