Mathematics - 2018-11-28 (page 1 of 4)

Jake Rose

00:00

@TedShifrin how can it make sense everywhere but one point?

Perturbative

Thanks @BalarkaSen, let me just try that approach

Ted Shifrin

@Jake: $f(z) = 1/(1-z)$ makes sense everywhere in the plane except $z=1$ !!!!

@Balarka: I'm really not asking for credit. You can relax about it :) I'm glad to have you back in the chat.

CaptainAmerica16

Das a hole.

Semiclassical

My favorite example of a weird series is f(z)=z+z^2+z^4+z^8+...

Balarka Sen

@TedShifrin Of course you're not asking for it :) But learning multivariable calculus from you was undoubtedly immensely useful for me.

Semiclassical

00:05

Converges inside the complex unit circle, but goes crazy at the boundary

Ted Shifrin

The usual one is $\sum z^{n!}$, @Semiclassic. I guess this one is the same.

Balarka Sen

Yeah, this one works as well

Semiclassical

Yeah. Different gap lengths but same principle

Balarka Sen

Tangentially relevant: There's a thing called Abel's limit theorem out there, which says if $f(x) = \sum a_n x^n$ is a real power series with radius of convergence $r$, and $f(r)$ exists, then $f$ is left continuous at $r$. That is, $\lim_{x \to r^-} f(x) = f(r)$.

Ted Shifrin

That generalizes to the disk in the plane, with an appropriate "cone" of approach.

Balarka Sen

00:09

Aha

Ted Shifrin

I usually did that in the grad complex course.

Érico Melo Silva

hi chat

OneRaynyDay

Hi guys, I've tried searching online for a specific way to prove Baire's Category Theorem for complete metric spaces. Couldn't find the way that E.T. Copson did it in depth online, was hoping to reiterate it here and see if anyone can spot any holes; is this the right type of discussion for this chat?

Ted Shifrin

rehi Eric

Perturbative

@JakeRose Okay so there's this theorem which holds for general metric spaces which says the following, if $X$ and $Y$ are metric spaces and we have $E \subseteq X$ and $f : E \to Y$ and $p$ is a limit point of $E$, then $$\lim_{x \to p} f(x) = q \iff \lim_{n \to \infty} f(p_n) = q$$ for every sequence $\{p_n\}$ in $E$ such that $p_n \neq p$ and $$\lim_{n \to \infty} p_n = p$$ (this is all taken pretty much verbatim from a textbook by Walter Rudin)

Ted Shifrin

00:12

@Balarka: It's the same Abel summation by parts proof.

Balarka Sen

@Ted I think in general it's true that if $f$ is a continuous function on a subset $A \subset X$ of a metric space $X$, $\{f_n\}$ is a sequence of continuous functions defined on the closure $\overline{A}$, and $f_n \to f$ uniformly on $A$, then $f$ extends continuously to a function $\overline{f}$ on $\overline{A}$.

Perturbative

@JakeRose So if we can find two sequences which converge to different points under $f$ then we can conclude that the original limit doesn't exist

Érico Melo Silva

@Ted I responded to u

Ted Shifrin

Hmmm ... not sure, @Balarka.

Érico Melo Silva

thanks for taking the time to read as usual

Jake Rose

00:17

@Perturbative ah i see

@TedShifrin why did you bring up 1/1-z ?

Semiclassical

Do you recognize the sum $1+z+z^2+z^3+\cdots?$

Jake Rose

Oh

I see!

Binomial of 1/1-z

I always forget the basic

Semiclassical

well, or just a geometric series :)

Ted Shifrin

binomial????

Semiclassical

I mean, it's not wrong. $(1-z)^p$ with $p=-1$

Ted Shifrin

00:20

GRR ...

Balarka Sen

@Ted First of all, $f_n$ uniformly converging on $A$ implies it uniformly converges on $\overline{A}$ as well. Say the limit is our candidate $\overline{f}$. Suffices to prove it's continuous: Pick a point $x \in \partial A$ and $x_k \to x$ a sequence of points in the interior of $A$ converging to $x$. $f_n(x_k) \to f_n(x)$ by continuity of $f_n$, for all $n$. $f_n(x_k) \to f(x_k)$ and the speed of convergence is independent of $k$ by uniform convergence of $\{f_n\}$. $f_n(x) \to f(x)$ as well.

CaptainAmerica16

I was hungry so I ate jalapenos out of a jar. Someone call 911. My tongue.

Balarka Sen

Now it's a $3\epsilon$ trick

Jake Rose

I dont see the sin of that...

Ted Shifrin

You continue to have a death wish, @CaptainAmerica

Érico Melo Silva

00:21

bye everybody

Balarka Sen

cya @Érico

CaptainAmerica16

@TedShifrin I'm not trying to. They've never been this hot before. I'm in so much pain.

Ted Shifrin

eat crackers, white bread, drink milk (NOT water)

CaptainAmerica16

I'm going to have to steal my little brother's milk.

We're out of 2% there's only SOY

Balarka Sen

lol @ CaptainAmerica's profile description

CaptainAmerica16

00:23

brb

Balarka Sen

"The Great Shifrin"

Ted Shifrin

eat bread

@Balarka: This looks like a modified $\epsilon/3$ proof.

Balarka Sen

$3\epsilon$ but yes :3

Ted Shifrin

LOL, I forgot about that thing in his profile.

Balarka Sen

I don't scale my epsilons

Ted Shifrin

00:25

shrug

CaptainAmerica16

I'm back. I ate a piece of cake. It still hurts, but less.

You read my profile?

Ted Shifrin

No, I meant bread/crackers for a reason. Cake doesn't work the same.

CaptainAmerica16

Dang it ;-;

I'm done with life, I burned the stupid chicken we're supposed to have for dinner. I'm just going to make rice and not say anything.

I'm literally sweating. I can't.

The chicken is breaking like crackers. My mom is going to be pissed.

Ted Shifrin

bans CaptainAmerica from the kitchen ...

CaptainAmerica16

I wish my older sis didn't move out. She used to remind me about the chicken ;-;

user330477

00:36

Can anyone look at this question and tell me why near $0$, $f(z)$ is one of the braches of $f(z)=\frac{1 \pm \sqrt{1-4z}}{2}$? I want to say because it is a quadratic in $f(z)$, but still not sure how to make this rigorous. math.stackexchange.com/questions/3011907/…

OneRaynyDay

I ended up just wriitng it here: math.stackexchange.com/questions/3016529/…

CaptainAmerica16

I'll be back in a bit. I have to deal with this and now my stomach hurts from the peppers.

Ted Shifrin

Point by point it's true, by the quadratic formula, @user330477. Now how can you have a continuous (much less analytic) function on some ball centered at $0$?

Jake Rose

If I am caluclating the taylor series of $\ln(1-z)$ about $z=i$

user330477

@TedShifrin So, what I have done there is correct?

Jake Rose

00:41

Is it the same as $ln(1-z)$ expansion but replacing z with $z-i$

?

Ted Shifrin

Not as you've said it, literally, @Jake.

Jake Rose

wdym?

Ted Shifrin

That would be $\ln(1+i-z)$.

Yes, @user330477.

Forget about ln for a minute, @Jake. How do you find the power series for $1/(1-z)$ centered at $z=i$?

user330477

@TedShifrin "Point by point it's true, by the quadratic formula." What does it mean? Is it true that $f(z)=\frac{1 \pm \sqrt{1-4z}}{2}$ by the quadratic formula as I have written in the link.

Ted Shifrin

Why are you asking me this?

Jake Rose

00:45

@TedShifrin

oops

Differentiate a bunch of times and evaluate at i

But thats probably a bad way

user330477

@TedShifrin Because I doubt quite a lot.

Ted Shifrin

That's a horrible way, @Jake

Jake Rose

My taylor series knowledge has gone down the drain

Ted Shifrin

But asking again and again when you've been told it's correct is beyond annoying, @user330477, and we'll all stop answering you.

Your substitution idea is the right track, @Jake.

Jake Rose

Expand it as a geometric series

Ted Shifrin

00:47

$z=(z-i)+i$ ... Maybe even let $u=z-i$ ....

user330477

@TedShifrin I didn't mean to annoy you. It is just that I am not convinced with your answer.

Ted Shifrin

Then ask something very specific, @user330477. Don't just ask "Do you really mean that it's true?"

I actually had a good deal more content in my answer.

If $f(x)^2 = 1$ for all $x$, do you know that $f(x)=1$ for all $x$ or $f(x)=-1$ for all $x$?

Jake Rose

@TedShifrin I feel like ive had a stroke and cant remember a thing

Leaky Nun

you "feel like you've had a stroke"?

Jake Rose

@LeakyNun Its called exaggeration for comedic effect

user330477

00:50

@TedShifrin Ok, the question is if $f(z)$ is analytic at $0$ and $f(z)=z+f(z)^2$, then how to we prove that near $0$, $f(z)$ is one of the braches of $f(z)=\frac{1 \pm \sqrt{1-4z}}{2}$?

Ted Shifrin

That's the same question. Did you think about what I just typed to you above?

You really should say "one of the branches of $\frac{1+\sqrt{1-4z}}2$," and not use $f(z)$ both places.

But the main point is what I mentioned about continuity and analyticity. Answer my question about $f(x)^2=1$.

user330477

@TedShifrin This is impossible as if we differentiate both sides at $z=0$, we end up with $0=1$.

Leaky Nun

o_o

Ted Shifrin

How did you get that?

And of course you're still not answering my question, for the fifth time.

@Leaky: I'm gonna put you in charge for a few days. You'll shape people up.

Leaky Nun

@TedShifrin respectfully declines

Ted Shifrin

00:59

Well, cook me a good dinner as a bribe.

Jake Rose

@TedShifrin When making the substitution of $z-i +i$ Im a little confused

Jacksoja

hi

dim V = dim Null T + Dim Range of T

Leaky Nun

weird notation but ok

Jacksoja

i want to prove this and this is how i started

user50393

'dim' v/s 'Dim'

Jacksoja

01:00

let U be a basis for Null T, we extend it to a basis for all of V

Ted Shifrin

@JakeRose Why?

Jacksoja

U has dim = m , and m+r = n dim of V

Jake Rose

I think Im doing it completely wrong

Jacksoja

we need to show that range of T has dim =r

Ted Shifrin

I can't answer that.

Jacksoja

01:02

but how to show that we can find r vectors that span the range of T

Jake Rose

If i just straight up substitute it in the the expansion for $ln(1-z)$ doesnt it still leave me with the same thing?

Jacksoja

and are lin indep

Ted Shifrin

@Jacksoja: A set of vectors doesn't have dimension. You mean $U=\text{Null}(T)$.

Jake Rose

@TedShifrin these are very stupid questions Im sorry

user330477

@TedShifrin "Now how can you have a continuous (much less analytic) function on some ball centered at $0$?" This is the question you are talking about, right? I think that if you expand via power series, then the coefficient of odd terms must be zero by the $f(z)=z+f(z)^2$ condition.

Jacksoja

01:02

@TedShifrin Yes true , sloppy notation by me

Ted Shifrin

So, forget $\ln$, first of all. Let's look at $1/(1-z)$. We can rewrite this as $1/(1-(u+i)) = 1/((1-i)-u)$. Now that you can expand in geometric series. How do you do that?

No, @user330477, I was trying to get to answer a totally different question, which I asked several times. Then you said $0=1$ and now you're saying something else that looks totally wrong. Where are you getting all this?

user330477

@TedShifrin I am not clear on your question. Moreover, how is that related to my original question?

Ted Shifrin

Because it was about understanding how using the quadratic formula pointwise leads to some stronger conclusions with continuity.

CaptainAmerica16

Dinner is #saved because I'm a genius.

Ted Shifrin

But everything else you're saying is just nonsense, I'm sorry.

Where did your $0=1$ come from?

Where is all the odd terms being zero coming from?

Leaky Nun

01:07

@CaptainAmerica16 but is your stomach?

Jake Rose

$1+(u-i)+(u-i)^2/2 ...$?

Ted Shifrin

@CaptainAmerica: We take your genius with many grains of salt ... perhaps boulders.

user330477

@TedShifrin What is quadratic fomula pointwise mean? That $0=1$ came from $f(x)^2=x$ which isnot what you wrote. As for the odd terms being $0$, it was just that the coefficient of $z$ must be $0$.

Ted Shifrin

@Jake: I rewrote it as I did for a reason. Are you ignoring what I wrote?

Jake Rose

I didnt understand how to do it from what you wrote

user50393

01:08

Take out (1-i)

Ted Shifrin

That's false, @user330477. If $f(z)=(1-\sqrt{1-4z})/2$, then the coefficient of $z$ is $1$.

Yes, factor out $1-i$. As a warm-up exercise, give me the power series for $1/(2-z)$. @Jake

CaptainAmerica16

@LeakyNun Not really, but I drank some water. I'll see what happens.

@TedShifrin I took off the really burnt pieces and made some wet vegetable stuff. I'm going to throw it all in a pot with the rice so you can't tell it's burned. I think it'll work.

user330477

@TedShifrin I don't what I am doing now. But what does quadratic formula pointwise mean?

Ted Shifrin

For one $z$ at a time.

Now we go back to the extra question I asked you so many times. Would you please find it and think about it?

I'm about to leave. I'm really running out of patience.

Jake Rose

$(1-i)^-1 (1+u/(1-i) +u^2/(1-i^2)2...$

?

Thats for the first question sorry Id already started working it out

Ted Shifrin

01:11

Parentheses issues, but, yes, you're doing it right ...

Jake Rose

Im sorry for how dreadful tht latex looks

Ted Shifrin

So it turns into the geometric series in $u/(1-i)$.

That's how you do this. Now you can get $\ln$ from that in one simple step.

Jake Rose

So why doesnt straight substitituion like the way I did it the first time work?

user330477

@TedShifrin "If $f(x)^2 = 1$ for all $x$, do you know that $f(x)=1$ for all $x$ or $f(x)=-1$ for all $x$?" How did you get $f(x)^2 = 1$ for all $x$ in the first place.

Ted Shifrin

I made that up, @user330477. It's to teach you a lesson in an easier setting. That's how you understand mathematics. Call it $g(x)$. Nothing to do with your $f(z)$.

I don't even remember what you did, @Jake. It's nuts in here. Do you give me a power series in $u$?

Prashin Jeevaganth

01:14

Bob

Here in the chat room, when somebody types in an expression with $ should I see what he types or should that be processed as MathJax

Ted Shifrin

It needs to be enclosed in $ .... $, @Bob, to be processed as MathJax.

darn, enclosed in dollar signs

Prashin Jeevaganth

Could someone help verify this solution? I got $y^{2}=-1+Ce^{-2x}$ instead

Ted Shifrin

Does your answer work? @Prashin ... Easy enough to check.

Bob

When I write $\frac{3}{4}$ I do not see a fraction.

Ted Shifrin

01:15

You need to use the LaTeX in chat link up on the right ....

Bob

do I need to install software or something?

Ted Shifrin

not software ... a page

user330477

@TedShifrin My question is why can we use quadratic formula for $f(z)^2-f(z)+z=0$ for every point?

Leaky Nun

@PrashinJeevaganth wolfram alpha

Ted Shifrin

Because you're solving a quadratic equation $y^2-y+z=0$ for some number $z$.

Bob

01:17

I do not see it

Ted Shifrin

Are you on a desktop or on a mobile?

user330477

@TedShifrin Does this quadratic equation give us that $f(z)=\frac{1 \pm \sqrt{1-4z}}{2}$ for every point $z$?

Ted Shifrin

@user330477: I've said yes to this five times. I'm done. Maybe someone else will help. I have no more patience.

user50393

six times*

Nvm, I lost count.

Ted Shifrin

smacks @user50393

Jake Rose

01:19

Once ive integrated is u my new z?

@TedShifrin forget my previous quesiton

Ted Shifrin

Once you've integrated, you remember that $u=z-i$ and you get a power series in $z-i$.

Leaky Nun

@TedShifrin I have to say, I'm impressed by your patience

Ted Shifrin

Well, Mike wanted me fired for being too patient. Maybe he has a few points.

Bob

@TedShifrin Are you talking about the page that says create/edit a book mark?

Ted Shifrin

Luckily for me, I'm going to cook dinner and then go play bridge for the evening, so I'm leaving momentarily.

Yes, you have to create a bookmark, Bob.

Prashin Jeevaganth

01:21

@LeakyNun ok my solution according to Wolfram Alpha is wrong... do you mind telling if there's a need to evaluate $d/dx( e^{2x} z)$

Bob

I did that but it is not working

Jake Rose

@TedShifrin final question. WHy do you have to do this method and not just simply replace every z with z-i?

Ted Shifrin

Once you're in here, you have to click on the bookmark, Bob.

Because replacing $z$ with $z-i$ is looking at $f(z-i)$, not $f(z)$. @Jake ... Do simple examples.

user330477

@TedShifrin Then, is it now obvious that near $0$, $f(z)$ is one of the braches of $f(z)=\frac{1 \pm \sqrt{1-4z}}{2}$ as $f(z) is analytic at $z=0$. As for the patience thing, I really thank you for being patient.

Bob

It is working

I like this $\frac{3}{4} = 0.75$.

Thank you very much @TedShifrin

Ted Shifrin

01:22

You're welcome, Bob.

Prashin Jeevaganth

@TedShifrin from Wolfram Alpha my solution doesn't work. Do you mind telling what is the wrong assumption I'm making? Usually I don't evaluate $d/dx(ze^{2x})$ and I only integrate the right hand side from the resulting differential equation after substitution, which is $d/dx(ze^{2x})= (-2x)(e^{2x})$, so now I lost the x and the half term from my solution

Fargle

01:51

y'all wild lmao

Bob

02:03

good night

user330477

02:16

Where is a branch of the square root function analytic? This branch of the square root function is not analytic over any disk containing $0$. Also, I know that along the negative real axis, square rooot function is discontinuous.

Ultradark

02:50

how do I come up with a differential equation based on how a node is updated in its position on the coordinate plane

Daniel ML

Hey someone can help me with the proof of the signum function is integrable on $I:=[-1,1]$

Sank

03:12

1

Q: Central Limit Theorem to approximate probability

I have a probability question using CLT: George and Julia work at the campus coffee shop. The management wants to award a prize to the quicker worker. They will each be set the task of making 200 consecutive double decaf skim milk lattes and, for each, the total of the 200 independent t...

would like some feedback :)

Daniel ML

03:36

no one? :(

SMA.D

05:13

Hi. Is there any specific name for a function which has a single maximum and is decreasing around it (e.g. Gaussian function)?

Semiclassical

@SMA.D unimodal?

Akiva Weinberger

@TedShifrin I didn't even know there was a Ben Gurion University. (And I haven't been to the Negev this year so far anyway.) So I don't think it's from me.

ninja hatori

06:22

0

Q: about dedekind domain

Let $A$ be a Dedekind domain, $S$ a multiplicatively closed subset of A. Show that $S^{-1}A$ is either a Dedekind domain or the field of fractions of $A$. Attempt: A is an integrally closed Noetherian domain of dimension one. Then we know that $S^{-1}A$ is integrally closed and noetherian. $S^{...

commutative-algebra dedekind-domain

Prashin Jeevaganth

06:42

0

Q: Sequencing events of Truth or Lie in possibility Trees

Clarence is known to speak the truth two out of three times. He throws a fair, six-sided die and reports that it shows a 6. What is the probability that the number shown on the die is really 6? I will explain my issues on this problem, forgive me if I use the wrong terminologies as I’m not that...

probability conditional-probability decision-trees

If anyone's interested in looking at my question, it's quite long so I didn't ask it here

Akiva Weinberger

06:59

$\oiint$ doesn't work :(

$\oint$ does

$\oint\!\!\!\int$

Fargle

$\require{esint}\oiint$

hmm

Akiva Weinberger

I noticed from this question

Balarka Sen

09:24

@AkivaWeinberger You can make it by hand: $$\bigcirc \!\!\!\!\!\!\!\int\!\!\!\!\!\, \int$$

Alessandro Codenotti

09:42

ikeaTex

ÍgjøgnumMeg

Mornin' all

Alessandro Codenotti

hi @ÍgjøgnumMeg

ÍgjøgnumMeg

@Alessandro how's it going

Alessandro Codenotti

I like to complain about AG whenever I'm asked that, but actually fairly well!

ÍgjøgnumMeg

10:08

@Alessandro nice! I'm quite excited for AG... should I not be? lol

Alessandro Codenotti

You're probably going to like it more than I do to be fair

SMA.D

10:45

@Semiclassical Thank you

mick

12:02

2

Q: About $ F(u) = \int_{-\pi/2}^{+\pi/2} \ln(g(x) + u) dx $

We know for $ u > 1 $ $$ \int_{-\pi/2}^{+\pi/2} \ln(\sin(x) + u) dx = \pi \left(\ln\left(u + \sqrt{u^2 -1}\right) - \ln(2)\right) $$ Usually this is shown by using differentiation under the Integral sign or contour integration. This made me wonder : Consider the log-transform For a given r...

calculus representation-theory closed-form integral-transforms inverse-problems

4

Q: A Thue-Morse Zeta function ( Generalized Riemann Zeta function and new GRH )

Consider $t_n$ as the Thue-Morse sequence. Let $m$ be a positive integer and $s$ a complex number. Odiuos Number Now consider the sequence of functions below $f(1,s)=1+2^{-s}+3^{-s}+4^{-s}+...$ This is the zeta function valid for $Real(s)>1$ $f(2,s)=1-2^{-s}+3^{-s}-4^{-s}+...$ This is the ...

reference-request special-functions zeta-functions riemann-hypothesis

Julius

12:51

Hi. If $S_{x,r}$ denotes a sphere with radius $r$ centered at $x$, how come the last inequality is true in tinypic.com/view.php?pic=ilf14n&s=9? Reducing $rh_n$ to $rh_n/2$ is clear, but why replacing $x$ by $z_j$?

ÍgjøgnumMeg

@Alessandro thankfully I can take Alg. Top. 1 in Heidelberg for the masters (I think this is usually an undergrad course) and this apparently has an introductory section on point set topology so I don't miss out lol

Alessandro Codenotti

13:08

You don't really need much point set topology to do AG, you'll be fine

Knowing commutative algebra well (which I don't) is way more important

That's how it is in the course I'm taking at least

Akiva Weinberger

Book: "To produce a model explicitly, we need a Yang–Baxter solution. Deus ex machina, here it is."

That's certainly one approach to motivating your results

Kinda like "rabbit out of a hat"

ÍgjøgnumMeg

@Alessandro well I can take up to two undergrad courses to fill out my background so AlgTop1 is definitely gonna be one and then the other can be algebra or diff.geo.

Akiva Weinberger

13:31

I want to write a book or paper with the line, "Unfortunately, the proof is left to the reader"

or "I regret to inform you that this is left as an exercise" or some such

Alessandro Codenotti

13:56

I see, sounds good @ÍgjøgnumMeg

Balarka Sen

You should give a proof-sketch instead, and state some fact in the sketch to prove which, you refer back to some earlier proposition which in turn uses, say, lemma 5.8.78, which you should state is just a trivial corollary of theorem 3.5.10 and is left as an exercise

Those epic trolls are the most annoying

Alessandro Codenotti

I'm doing two undergrad courses this semester and I might do one more next semester

Nice internet at uni today, took only 8 minutes to send a message

ÍgjøgnumMeg

@Alessandro the algebra II course in Heidelberg would be a masters course in the UK though, now I understand why @Mathein is @Mathein

lol

@Alessandro Ah yeah, which courses ?

Akiva Weinberger

14:13

How to turn your nonassociative algebra into an associative algebra in one easy trick

(and then hopefully there's a way to simplify $\left.\overline{b\;\overline c\rceil}\right\rceil$, like there is with quandles)

Alessandro Codenotti

@ÍgjøgnumMeg Set theory, because it's the only set theory or logic course offered this semester, and topology I, because they do two algebraic topology courses in the bachelor here while I had none in my bachelor

ÍgjøgnumMeg

@Alessandro I had no topology at all in my bachelor :(

Akiva Weinberger

That's not typeset-able, is it

ÍgjøgnumMeg

@Alessandro or number theory.. or in fact any algebra.. or geometry... we just did a little bit of analysis and then a tonne of applied crap

Akiva Weinberger

Aw

What kind of applied crap

Alessandro Codenotti

14:18

I had a great point set topology course, but as far as AT goes we only saw what's a fundamental group

ÍgjøgnumMeg

Like.. statistics

hahaha

Alessandro Codenotti

While here they do homology and cohomology and then have two courses on homotopy theory for the masters

ÍgjøgnumMeg

and I'm awful at stats, I essentially dumped all of my stats knowledge after passing the exams so I couldn't even speak meaningfully about it

@Alessandro damn

My university was unfortunately mostly a choice for people whose A-Levels didn't go well

for whatever reason

Akiva Weinberger

So it had a statistically insignificant effect on you

ÍgjøgnumMeg

couldn't tell you

lel

also we spent a lot of time learning to solve some specific PDEs in the 3rd year

but no general methodology

or any theory

such a waste of time

Akiva Weinberger

14:25

How come Boolean algebra seems so separate from abstract algebra

They're both algebras but I never find them in the same context

ÍgjøgnumMeg

@Akiva there are Boolean rings

Akiva Weinberger

What's a Boolean ring?

Alessandro Codenotti

Boolean algebras are mostly of interest in set theory and related areas

ÍgjøgnumMeg

A ring $R$ in which $x^2 = x$ for all $x \in R$

Akiva Weinberger

Hm

Alessandro Codenotti

14:27

Every element has order two @Akiva

ÍgjøgnumMeg

and every element is additively self-inverse

Akiva Weinberger

@AlessandroCodenotti That makes it sound like it's nilpotent rather than idempotent

@ÍgjøgnumMeg Oh, wait, that's a consequence isn't it

ÍgjøgnumMeg

@Akiva yis

Alessandro Codenotti

@AkivaWeinberger you're right, order was not the right word here

Akiva Weinberger

'cause $x+x=(x+x)^2=x^2+x^2+x^2+x^2=x+x+x+x$

and then we can subtract $x+x$

Alessandro Codenotti

14:30

Actually my previous comment was probably wrong. All boolean rings are $\Bbb F_2$ algebras though

MatheinBoulomenos

@ÍgjøgnumMeg lol

Akiva Weinberger

@ÍgjøgnumMeg That's a bit weird 'cause in Boolean algebra you have both $x\cdot x=x$ and $x+x=x$

It's basically $\land$ and $\lor$ in different notation

ÍgjøgnumMeg

$x + x = 0$

Akiva Weinberger

In Boolean algebra, not a Boolean ring

MatheinBoulomenos

there's a 1-1 correspondence between boolean rings and boolean algebras

ÍgjøgnumMeg

14:35

@Akiva I see, sorruy

was not reading

am covertly mathing while at work

Akiva Weinberger

What's work?

(As in, "what work are you doing", not "what does the word 'work' mean", to be clear)

ÍgjøgnumMeg

@Akiva I'm working at my university's IT department to save money before I go for my masters lol

Akiva Weinberger

So you fix teachers' projectors

MatheinBoulomenos

Given a boolean algebra $(A,\land,\lor,\lnot)$, you can define a boolean ring structure on $A$ by setting $a \cdot b = a \land b$ and $a+b=(a\lor b) \land (\lnot a \lor \lnot b)$ (basically "exclusive or")

ÍgjøgnumMeg

@Akiva haha nah, there's a "fleet refresh" project going on so I'm replacing all of the university laptops and desktop PCs

Akiva Weinberger

14:39

Arright

@MatheinBoulomenos Ah OK

MatheinBoulomenos

and conversely every boolean ring gives rise to a boolean algebra and the constructions are inverse to each other

Akiva Weinberger

@ÍgjøgnumMeg So basically you've got a lot of computers with faster processors and newer OSes and stuff?

@MatheinBoulomenos I see

ÍgjøgnumMeg

@Akiva right

Akiva Weinberger

Neat

All Linux, I assume

(/s)

ÍgjøgnumMeg

rofl

Akiva Weinberger

14:43

I'm kidding, I know they're probably actually Raspberry Pis

ÍgjøgnumMeg

\s

$\mathfrak{S}$

Leaky Nun

@MatheinBoulomenos mumbles something about clopen subsets

MatheinBoulomenos

@LeakyNun this is far easier than the Stone representation theorem

Leaky Nun

sure

@MatheinBoulomenos I was wondering whether the sigma algebra generated by $S$ is just countable unions of countable intersections of elements of $S$ and their complements

and if so, why I have never heard of this concrete definition

Balarka Sen

No

Leaky Nun

14:56

hmm?

Balarka Sen

You have to do that a transfinite amount of times to get the full sigma algebra generated by S

Alessandro Codenotti

The borel hierarchy would be really boring if that were the case

Leaky Nun

Surely $(\bigcup_i \bigcap_j s_{ij})^c = \bigcap_i \bigcup_j s_{ij}^c = \bigcup_{f:\Bbb N \to \Bbb N} \bigcap_k s_{f(k) k}^c$

I see

the problem is that $\Bbb N^\Bbb N$ is uncountable

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$Mathematics$ Mathematics