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16:00
hii do you people know Jacobi method on quadratic forms?
Also, is the plural of emirp "emirps" or "semirp" ?
@Nick 11 is not an emirp
by definition, when the digits are reversed, it should give a distinct prime. otherwise it will just be a pallindromic prime
@BalarkaSen So, the one digit primes aren't emirps either. cool.
:(
Hi @Teddy @Nick @Balarka
16:03
hi Sawarnik
@BalarkaSen Unless you use a different notation than I'm used to, $(H_1,H_2)$ has the same domain as $H_1$ and $H_2$ have.
@DanielFischer ... throws table at himself. you're right.
@BalarkaSen I hope you missed.
@BalarkaSen myself?
@Venus OK. I will. :)
16:05
God, that's a hideous name to have:
"Hi welcome to the funny name convention, I'm John Jacob JingleHeimer Schmidt and you are?"
"Oh, I'm Teddy Nick Balarka"
*John gasps*
math.stackexchange.com/questions/1027392/… if the sum is closed, why n[Delta] is in the image oh H1(q)?
How many tables have you got though?
*of
@Sawarnik no, that'd be incorrect.
^yes right.
16:05
@Sawarnik $2^{\aleph_0}$ many.
@BalarkaSen And where do you store them all?
the real line, of course.
they are bundled over the reals.
nontrivial chair bundle on $\Bbb R$
Oof!
Still no @Chris'ssis :/
16:07
@BalarkaSen Although, I technically don't know what it is. Can I use $\aleph_0$ to mean $\infty$?
@Nick :D
@Nick I assumed it too, lol :D
@Nick $\infty$ is vague. $\aleph_0$ is the cardinality of integers.
so no
@BalarkaSen Have you been to the Salt Lake Stadium?
@BalarkaSen Isn't cardinality of a set the number of elements it has?
what match, @Sawarnik?
@Nick right
16:08
2
Q: How does 2D kriging interpolation work?

BobI have a grid of points Example x 80 82.5 85 87.5 90 y 5 0.5 1.6 1.7 1.7 2.3 2.5 1.6 1.7 1.8 2.1 2.7 0 2.4 2.3 2.6 3.0 3.8 and I want to be able to find any point for example point (81,...

sorry i meant to correctly post the link earlier
@BalarkaSen Psst, does this mean $\aleph$ick can kill Russell's Paradox?
@MattRockwell actually, what you did before is more in keeping with the chat guidelines (note #3)
The whole "Are there more numbers from 0 to 1 than 0 to 0.5?" question. Can I put it to rest using $\aleph$?
How can I write $\aleph$ with my hands? :/
@DanielFischer When your avatar is to the right of mine, it looks like you're about to cough or sneeze on me and I'm trying to get out of the way.
16:12
curly line continued by straightline diagonal and then again a curly line
@Sawarnik It's the mirror image of harry potter's scar divided by the Bizarro's S
LOL @Nick
@RandomVariable Let me see. Post something.
Hahaha @Daniel
@BalarkaSen Superman's S would invert too. Bizarro's S is an inverted superman's S. So, mirror image. yay!
16:14
@DanielFischer sensei have you seen my chat? :-)
@RandomVariable Hahaha .. true
@Venus Sorry, no. What was it about?
I was too late.
Oh no...
@BalarkaSen: You're in contact with smart people like yourself, right?
16:16
@DanielFischer so this exercise in Armstrong wanted me to find out the fundamental group of cylinder and moebius strip. cylinder is $S^1 \times [0, 1]$ so $\pi_1(S^1 \times [0, 1]) \cong \pi_1(S^1) \times \pi_1([0, 1]) \cong \Bbb Z$ but any idea how to approach the moebius strip?
just a hint or some such
@BalarkaSen Do you know anything about deformation retracts yet?
@BalarkaSen Was that the same armstrong of Armstrong numbers? (just askin)
@robjohn Got it sorry. Unfortunately it is getting little to no attention
(I'd be baffled if it were Neil Armstrong XD)
Lance Armstrong!
16:19
@DanielFischer i am not familiar with the name apparently. what is it about?
@MattRockwell Perhaps no one has seen it who knows what kriging is. I'd never heard of it, and so all I can tell you is what Wikipedia says about it.
@DanielFischer I wanna ask, according to Ramanujan, if
$$f(x)=\sqrt{ab+(a+c)^2+b\sqrt{ab+(a+c)^2+(b+c)\sqrt{ab+(a+c)^2+(b+2c)\sqrt{ab+\cdots}}}}$$
then
$$f(x)=a+b+c$$
How does one prove this one?

Also this one, is it possible the following problem has a unique solution
$$x=\sqrt{i+\sqrt{i^2+\sqrt{i^3+\sqrt{i^4+\sqrt{i^5+\sqrt{i^6+\cdots}}}}}}$$
where $i=\sqrt{-1}$.
2
@Sawarnik He was too busy doping on his bicycle to have ever dealt with numbers... or the moon.
@BalarkaSen Homotopy. The Möbius strip has a pretty easy deformation retract, and that makes computing the fundamental group easy.
@DanielFischer I have nine lives, so I should be OK if I catch something deadly.
16:21
@Nick Haha. true.
@DanielFischer It looks like the second expression is easier than the first sensei :D
@DanielFischer I know what a homotopy is.
I must be the only person who hears "Hippopotamus" every time someone says "homotopy".
oh just googled
you're talking about homotopy relative to A
@Venus Not my cup of tea usually. Looks more like @r9m might know it.
16:23
@DanielFischer Not also the second one sensei?
@DanielFischer You drink math-tea? That must save you fortunes! Give me your formula.
@BalarkaSen Hippopotamus relative to A
@Venus For the second, since $i^4 = 1$, you have a periodic nested radical, so if there is a solution $z$, you can get a polynomial equation for $z$ by un-radicalising $$z = \sqrt{i+\sqrt{i^2 + \sqrt{i^3 + \sqrt{i^4 + z}}}}.$$ I'm not sure how pretty the equation becomes.
@DanielFischer It would be 16 degrees polynomial
@DanielFischer How would you define an anisotropic vector valued function?
@Venus Yes. But with lots of zero coefficients, hopefully.
16:30
@DanielFischer Okay, thank you sensei... :-)
It's nothing to be sorry about. @Hippa sounds like Hippo. It's a fact.
@Nick :/
Hi @alizter, Sarah said hi to me yesterday, lol.
$\text{d}ijk\text{stra}$
@JohnDoe I would first try to find out what "anisotropic" should mean in the context. Is there a bilinear (or sesquilinear) form on the vector space? Then it could mean that $f(x)$ is an anisotropic vector for that form for all $x$.
16:31
@Alizter Yo
@BalarkaSen Hi
@JasperLoy I know! Sarah was one randomly having a go at Dr G.
@Alizter The eccentric German doctor, lol.
@JasperLoy He says his answers are bad because he has to write them really fast.
@Hippalectryon I mean seriously. How offended can you be because of this? I know a guy called Manmeet... everyone calls him "Man-Meat!"... to offend him intentionally but he's always like "Dude, that's my name don't wear it out." If he's cool about his real name being made fun of, why can't you be cool that your username sounds like a fairly decent killer wild animal.
@Alizter He should learn from me, how to write short answers that get lots of upvotes.
16:34
@DanielFischer Could I ask you about something that was posted in chat a couple of weeks ago? It concerns complex analysis.
@JasperLoy No. He shouldn't :P
We need quality!
@Alizter My answers are quality, lol.
@Nick :DDD
I don't know why Ted said I sound pompous that day. I think I don't.
@JasperLoy Sure... but he already does short answers.
With no capital letters to be found in sight.
16:35
@RandomVariable Sure. I may not know the answer, but we can only find out if I see the question.
@Alizter Yes, but they look quite ugly.
@JasperLoy Ted is cross at your for saying that Munkres is a bad book.
Wow we are 5 more stars off 10,000 stared messages
@DanielFischer You always know the answer.
@JasperLoy At the end of the movie Kung Fu hustle, in an epic scene, the bad guy kneels before the protagonist and says "Teach me, master!" . In that same gravity of desperation, I asketh of you the same thing.
16:36
@BalarkaSen I think that is an issue he has to work with.
@RandomVariable Selection bias.
@Jasper Me too.
It refers to a anisotropic function a(x,u,\nablau): \Omega \times \mathbb{R} \times \mathbb{R}^{N} \rightarrow \mathbb{R} where u is taken from normed vector space $X$.
@Alizter Oh sorry about that, I'll fix it right away.
NO
Put that mouse down young man!
2
16:38
If I say this ice cream is terrible, will anyone be cross? It just means I don't like the ice cream. So in the same way I say a book is terrible.
@JohnDoe Uh. Try to find the definition in the book.
@Alizter You do know I was euphemistically saying that I'd star more starred posts, right?
@DanielFischer It's a research paper :)
@Jasper Not only do you say that it is terrible, but also say that it is terrible to apparently beginners on topology.
That is not good.
@JohnDoe Still there ought to be some book about the pertinent theory, and it should be defined there.
16:39
Only cookie dough icecream is good.
@BalarkaSen I am entitled to my opinion. And yes, I do know quite a lot about point set topology.
Just sayin'
afternoon
@JasperLoy I have no problem with that. But I think "Artin is terrible" is a totally false statement and a bad advise to algebra students. I believe I am familiar with fundamental algebra.
@DanielFischer Will see if I can source some of the texts online from the reference.
16:41
@Alizter I should be contented with Sarah saying one sentence to me each month lol.
@BalarkaSen Did I say Artin is terrible? I don't recall.
@BalarkaSen He is using terrible in a biased way while you're using it neutrally. He's being opinionated and you're being democratic. Basically, at the end of the day, what either of you think is terrible is not going to matter to the guy that thinks it's great ;D
You are all terrible.
@BalarkaSen I think Artin is great actually, not terrible at all.
oh?
ok. touche.
Even if what was meant was the interval $[-q_{0},q_{0}]$, $\sqrt{\tan^2{q_0}-\tan^2q}$ is not well defined if that line segment is omitted, is it? chat.stackexchange.com/transcript/36?m=18463371#18463371
16:42
The only thing I dislike is him treating only commutative rings. But that book is very good.
@Jasper me too shakes hand
It is one of the very few algebra books that treat the geometry properly.
Hi @MikeMiller, can you tell these fools to stop arguing about books?
no you @Alizter
@BalarkaSen without the accented e, touche sounds like touching the napier's constant inappropriately.
16:43
@Alizter Have you met any interesting girls lately? =)
@Alizter Yo, haters be haters, man.
I have been having many disturbing thoughts these few days, I hope they subside soon.
@Jasper Maybe consult someone?
@JasperLoy I have. He won a beard competition recently.
LOL @Nick
16:46
@BalarkaSen Sometimes, I truly wish I were joking.
@RandomVariable As long as $q_0 \in (-\pi/2,\pi/2)$, there's no problem.
Today I killed a cockroach at home.
@JasperLoy If it involves hurting people, then give me your chest size. I'll send a straitjacket right away!
@Nick I am D cup, lol.
@DanielFischer By "well-defined" I mean well defined on the complex plane.
16:49
@JasperLoy You didn't win a beard competition recently, did you?
@Nick I don't keep one.
@Rafflesiaarnoldii I suspect the question we commented on will not be well-received at academia.SE; it's too broad, its questions are well-answered by searching, etc.
@Alizter Instead of not arguing about books, people should not argue with Jasper. Not because he's right or wrong, but because nobody will convince the other, and it will be a waste of the time of everyone involved.
@JasperLoy Ah well, glad to know that wasn't you. So, how did you kill the roach?
@DanielFischer Maybe we're talking about the same thing.
@MikeMiller Yar.
16:51
@Nick I stepped on it with my shoe on.
@RandomVariable Ah, you mean the integrand. You have branch-points wherever $\tan q = \pm \tan q_0$, so you need to choose the domain somewhat carefully.
@Alizter We are not really arguing, just talking nicely.
@BalarkaSen My calculus students take priority. Also, these basic questions you should really fight through without hints; I believe the value gained from the struggle exceeds the time lost from the same. So I will not help you on these questions for a while, and for your own good, you should try not to seek help either - you'll learn more if you don't.
That's harsh, @Mike. snif
No, it's kind.
16:54
Well. OK.
I'll try to solve them by myself afterwards.
@JasperLoy That's not as bad as what I did to one. One day, a flying cockroach landed on my head and held me keyboard as hostage. He was a brave and valiant fighter but when he got serious, I was left with no choice but be a sissy and throw a spoon at him. The spoon decapitated him but he didn't die. One part of his body squirmed to one end of the room and the other head part amazingly also moved. The goop in between the two cockroach segments traced out a perfect sinusoidal wave.
@BalarkaSen Back then, did you mean this: $g(m) \sim h(m)$ implies $g(m) = h(m) + o(h(m))$ and thus $$\displaystyle \lim_{m\to \infty} \frac{ \prod_{n \leq g(m)}f(n)}{\prod_{n < h(m)} f(n)} = \lim_{m\to \infty}\frac{\prod_{n \leq h(m) + o(h(m))}f(n)}{\prod_{n < h(m)} f(n)} = \lim_{m\to \infty}\prod_{n \leq o(h(m))}f(n)=1.$$ ?
@Nick I try to kill them quickly to minimise its suffering. I also pray it has a good rebirth.
@VincenzoOliva yes.
@JasperLoy I don't do the whole "Na'avi one with Gaya" sort of thing but I do try kill them off humanely. Except that one guy, may he be a Viking in the future past :D
17:00
@balarka
@BalarkaSen
As I haven't encountered little o yet, is this because $o(h(m)) \to 0$ as $m \to \infty$, for any $h$??
just a ? * , uff, hating this keyboard
@VincenzoOliva $f(x) = o(g(x))$ as $x \to \infty$ iff $\lim_{x \to \infty} f(x)/g(x) = 0$.
@VincenzoOliva That keyboard will always be a keyboard, it won't transform into an iPhone just because you hate it.... hmm, that gives me a brilliant idea.
@BalarkaSen I see.
@Nick I hate iPhones too.
@JasperLoy: Remember that guy in church story I said earlier. Yeah, that was seriously misquoting that story. That's not the story at all. Sometimes memories get all fuzzy in our heads,. It was from a movie I watched a few years ago. Let me dig up that link. Possibly with subtitles so that you'd understand.
@DanielFischer If you scroll down a bit, he says that it can be evaluated by integrating above and below the branch cut on $[-1,1]$ and using the residue at infinity. (He probably meant to say $[-q_{0},q_{0}]$.) But unless I'm horribly mistaken (which is always a possibility), the integrand is not well-defined outside of the contour, and doesn't have a Laurent expansion at infinity.
17:09
@VincenzoOliva iHate takes care of that for you. Why hate when you can outsource that job to an Indian Call center open 24/7 ;D
tagline: Why hate when you have iHate ;D
@BalarkaSen: One day, can you prove that 37 and 73 are the only Sheldon Cooper Primes?
Hi @trig lol.
@RandomVariable I guess he transformed the integral, setting $z = \frac{\tan q}{\tan q_0}$ to get an integral of $f(z)\sqrt{1-z^2}$.
@JasperLoy: Ah yes, found it! Apparently it's from "My Name is Khan: I am not a terrorist"
I am sad that Chris has left this chat. I hope she returns soon.
6
I am sad that Jonas left this chat long ago. But he has better things to do than chat.
17:15
@JasperLoy Hello. Just popping by.
@JasperLoy And that compares to losing your son "Sam" how?
I think Jonas might win the Fields medal one day.
@DanielFischer Perhaps that it is what he did. He does say "an integral" and not "the integral".
@JasperLoy I'm sure he'll give you some credit :D You believed in him!
@Nick I need to believe that I can get well soon.
17:20
@JasperLoy Did you not listen to the song right after the sad dialogue in the video I just linked?
@Nick This video is not available in my country.
@JasperLoy good god, what country is that? Have you tried youtubeunblocker
@Nick It is probably some copyright issue in this case.
@JasperLoy Not available in this country (Ireland) either.
About 100 sites are blocked in my country, mostly porn sites.
17:26
@JasperLoy Tsk. How do you know porn sites are blocked? Have you been looking for them? ;)
@JasperLoy the unblocker didn't work?
@Nick I did not try it, nvm, thanks, lol.
Aww, you deserved to hear it out.
@Nick I heard the previous one, so good enough.
@TRiG Yes, I like them.
@JasperLoy yeah but that was all gospelly and not pop enough.
17:28
@Nick Oh, gospel music is nice too.
I am going to sleep, good night.
@JasperLoy Yeah, but I linked to the original version of the song. That song has been much changed since its inception. The blocked video I linked provides for the most memorable version of "We shall overcome" that I know.
@JasperLoy Goodnight Jasper :D
I believe in you, get well soon :D
@DanielF I don't quite see how möbius transforms apply to that question.
@MikeMiller Cuz you don't see the right side of the question
side. Möbius. pun. >.>
HI @DanielFischer If you have a chance can you see if its possible to conclude that $g \in L^{1}$ in the post
Möbius did more than one thing, @Hippa, and hence you should be able to make more than one joke about him.
By the way, I don't know if I said this before, but elegant solution to your nilpotent matrices problem. I was definitely working too hard - my attempted proof would have given the result from the paper I linked for all fields of characteristic 0.
17:42
@MikeMiller All holomorphic automorphisms of the unit disk are Möbius transformations. Now if you know Möbius transformations, you could look at them to see that conjugating $z \mapsto z+1$ with a biholomorphic map between the unit disk and the upper half-plane fits the bill.
@MikeMiller Thanks :)
@Nick LOL

I got suspended, was someone offended by the short true story of a poor little iPhone? :0
Thanks @Venus
@DanielF I knew the former, the idea that eluded me was to take that transform of the upper half plane (which is indeed a cute idea).
@MikeMiller Some things are easier on disks, others on half-planes. The trick is to recognise which applies.
17:47
Right. I was thinking topologically; I axtuallt think the map I was thinking of is probably the same as yours. The trick is that I couldn't write mine down, just visualize it.
Anyone know Jacobi's method for diagonalizing a quadratic form? I know the proof I'm just hoping for a sanity check on it's logic, and to talk about how it compares to Lagrange's method and the eigenvalue method :D
Hey, silly question. In a CFG rule, what does a vertical bar mean?
A -> AB | A
Is that "or?"
@Moshe Yes. A can be replaced with either AB or with A. (The latter is kind of pointless, however.)
I suck at speaking english, especially combinatorics =( math.stackexchange.com/questions/1029565/…
@Venus either this answer or this answer can be adapted to answer that.
18:01
1
Q: complex analysis $\int_0^{\infty} \frac{\sqrt{x}}{x^2+2x+5} dx$

alicehow do i compute $\int_0^{\infty} \frac{\sqrt{x}}{x^2+2x+5} dx$ with complex analysis? I feel like im calculating the residue wrong and I cant get to the answer correctly. I tried to branch cut the real $0 \rightarrow \infty$ but I feel like im doing it wrong. any help is appriciated.

I found a solution using Feynman's integration. Should I post it or it's irrevelant as he's requiring complex analysis? The residue theorem method is definitely faster
@DanielFischer I think between the two of us, we've answered the questions :-)
What's the most natural way to map the interior of the unit disk $D^n$ in $\mathbb{E}^{n+1}$ to the boundary?
@BalarkaSen sine and cosine?
@DanielFischer Thanks!
@UserX You mean differentiation under the integralsign ?
18:04
@N3buchadnezzar translating by a matrix?
@N3buchadnezzar yes
@UserX how?
Introduce an $a$ in the root; $\sqrt{ax}$
It flows naturally after that
Should I post it?
@UserX Not quite sure if I follow your reasoning?
@UserX $$f'(a) = \frac{1}{2\sqrt{a}} \frac{\sqrt{x}}{x^2+2x+5}$$
$\frac{1}{2\sqrt{ax}}$
18:12
@UserX Yeah I used $\cfrac{x}{\sqrt{ax}} = \cfrac{\sqrt{x}}{\sqrt{a}}$
Chain gang rule
@robjohn I really don't want to talk about this anymore, but would I at least lose the badges if I deleted that answer?
Then you're missing an $x$ in the numerator of the second fraction
@RandomVariable You would keep them, but the next time an answer of you qualifies, you don't get new badges. At least, it used to be so.
@UserX You are differentiating with respect to $a$...
18:16
Yea. The result is $\dfrac{x}{2\sqrt{ax}(x^2+2x+5)}$
$$\frac{\mathrm{d}}{\mathrm{d}a} f(a) = \frac{\mathrm{d}}{\mathrm{d}a} \frac{\sqrt{ax}}{x^2+2x+5} = \frac{\sqrt{x}}{x^2+2x+5} \frac{\mathrm{d}}{\mathrm{d}a} \sqrt{a} = \frac{1}{2\sqrt{a}} \frac{\sqrt{x}}{x^2+2x+5} $$
@UserX How is that easier to integrate? =)
@UserX Since you could pull out the $\sqrt{a}$ from the integral directly, you can't use that to obtain the integral by differentiating under the integral sign.
@DanielFischer I can't. That differentiation is wrong. Trying to find where
I'm confident I'm correct as my result agrees with numerical integration on mathematica exactly
Can anyone help explain Chomsky Normal Form?
I'm not getting it entirely.
Hmm. If I don't pull out the constant $x$, with the chain rule it goes $(ax)'\times \frac{1}{2\sqrt{ax}}=x\times \frac{1}{2\sqrt{ax}}$
18:26
I get that there's replacing the start variable, and then removing any RHS variables from LHS of rules, but that's it.
@DanielFischer I'm assuming you know the answer I'm talking about. Perhaps I'm acting a bit nuts. But I don't want rep points and badges that I don't rightly deserve. It feels like I cheated the system.
Hi, can someone please explain to me why this is true?: if $H$ is a Hilbert space and $A$ an operator such that $\langle Af, f\rangle \ge 0$ that implies $A \ge 0$.
@RandomVariable It was the $\frac{x^4+1}{x^6+1}$ integral, I think? You knew the trick yourself, didn't you? So what makes you think you don't deserve the rep and badges just because somebody else also knew the trick and posted it earlier? It would be different if you saw the other answer and intentionally copied the way without giving credit, but you didn't. Just don't take these things so seriously.
@N3buchadnezzar I think you can't break the root into two roots
You don't know what x,a are
I think
Otherwise I can't tell why I got the correct answer
@Twink That means $\langle Af,f\rangle \geqslant 0$ for all $f\in H$, presumably? That's usually the definition of $A \geqslant 0$.
18:32
@UserX $x$ is positive. Also your chain rule is wrong
How?
What's the name of the backwards 3?
Epsilon
(Dumb question of the day, having a moment.)
Thank you.
21 mins ago, by N3buchadnezzar
@UserX Yeah I used $\cfrac{x}{\sqrt{ax}} = \cfrac{\sqrt{x}}{\sqrt{a}}$
18:34
I know @N3buchadnezzar I can see what u did...
This problematizes me a lot. If it's right you're wrong, if you're wrong I can't spot a mistake.
Thank you, @UserX, and @DanielFischer
HI @DanielFischer do you think it is possible to show that $g$ is in $L^{1}$ by using $|g| \leq f$ and some partition? With $g$ given as in this post?
@UserX We get the same friking thing, and it does not help to solve the integral. As Fisher said you can just pull the a out. It scales linearly with the integral
@DanielFischer If you can see a simpler way can you just give me a hint?
@Alex The domination tells you $x \mapsto g(x,u)$ is in $L^{1/(1-\varepsilon)}$ for every $u$ [I do hope that Caratheodory functions are at least separately measurable]. If $\Omega$ has finite measure, that implies $g(\,\cdot\,,u) \in L^1(\Omega)$.
18:44
This question is so lame.
3
@UserX I'd post it for verification, if it's not too involved
@N3buchadnezzar my battery died and on my way to home to charge it I realised how stupid I am
I evaluate the antiderivative of the first integrand(without a) at some point
@RandomVariable I don't know... badges are sticky, but they might go.
@DanielFischer The finite measure of $\Omega$ allows you to dominate it by finitely many $L^{1}$ functions and then the result follows?
So I just evaluate it normally, and multiply/divide by a constant factor
18:48
@Alex The finite measure of $\Omega$ gives you $L^p(\Omega) \subset L^1(\Omega)$ for all $p \in [1,+\infty]$.
@UserX Indeed. Wanna look at a question I asked? :p
Why not
@PedroTamaroff True. One can shake one's head about what a certain somebody objects to, however.
@DanielFischer Oh no I see, I was reading that the $\sup$ was taken over $x \in \Omega$, so I was complicating it but I see it's simple.
@DanielFischer I didn't know, thanks
18:50
@DanielFischer The answers are mostly "let me rub all my math pet peeves on yer alls."
Bbl my mother's out of surgery
@PedroTamaroff Although the one with the $\mathbb{S}^n$ with the quote from wolfram was indeed shocking.
@DanielFischer What did you mean by 'hoping Caratheodory are separately measurable', separately in what sense?
@DanielF I think this is likely a peculiarity of Coxeter. I am not so shocked that there are strange mathematicians.
@Alex Well, we only need that for fixed $u$, the function $x\mapsto g(x,u)$ is measurable here. Separately measurable means measurable in each variable when the other variables are kept fixed. That is weaker than being jointly measurable.
@MikeMiller Strange is normal. But there are limits to what I would expect.
18:57
@DanielFischer Okay I see, but by definition of Caratheodory functions if you keep $x$ fixed then $g(x,u)$ is continuous for almost every $x \in \Omega$, so I guess it is separately measurable.
@Alex We're more interested in the behaviour for fixed $u$ here, however.
@DanielFischer It is measurable for fixed $u$ by definition also.
@DanielFischer I didn't know there were separate conditions of measurability for multi-variable functions.
@DanielFischer I think you should just write down the answer to the auto-of-$\bar{\Bbb D}$ question.
@DanielFischer I don't feel that it matters at all that I knew the trick myself. I don't deserve the rep and the badges, and I don't want them. But I won't bother you about this. Seemingly everyone is asking you questions.
@RandomVariable How many Gummibears can you buy with the rep and badges? It is that meaningful.
19:07
Hats are tied to rep and badges. Hats is life and life is hats.
Garfield>gummibears
Oho I just found a weird pun
Some mathematician should be called Mr. Gar
Then he could make Gar Fields
:D
65
Q: Most ambiguous and inconsistent phrases and notations in maths

fvelWhat are some examples of notations and words in maths which have been overused or abused to the point of them being almost completely ambiguous when presented in new contexts? For instance, a function $f$: $f^{-1}(x)$ can be an inverse and a preimage and sometimes even $\frac{1}{f(x)}$. $f^...

Can someone explain me what Git Gud means on his 7?
27 mins ago, by Pedro Tamaroff
This question is so lame.
He's whining that $=$ is being used to denote definition rather than equality. It's impossibly pedantic.
just like the other 20-whatever things in that answer
19:14
There's also an answer, near the bottom, that complains about 1992=100.
@MikeMiller That answer is pretty fudging stupid.
@UserX [text](link)
much better^ :-)
hi @PedroTamaroff
@user2179021 Who are you?
The only things I find reasonable are the trig identities thing, because $\sin^{-1}$ should probably not be $\arcsin$ when $\sin^2$ isn't $\sin \circ \sin$, and the fact that Coxeter thinks $n$-spheres are $n-1$-spheres, which come on, no.
19:16
@MikeMiller Convention.
@MikeMiller don't you find 1 reasonable?
@PedroTamaroff Lembik.. my name gets ruined somehow by chat
@PedroTamaroff I agree that there is no ambiguity there. I think it is poor notation nonetheless.
well turned into user2*
it's a bit random.. doesn't happen in all *.SE sites
19:17
I think 1 is silly. It's difficult to be confused by it.
Bad notation is something that isn't immediately clear in context. That's not bad notation.
the latter looks doable by a good probabilist
the former I am not so sure how to attack
@user2179021 Your main account is on StackOverflow, and that's where your username is user2179021. Change your name there to Lembik and it will be Lembik in chat.
@user2179021 Not really, no.
19:20
@MikeMiller maybe you're so used to it it's intuitive to you. If I go ask my whole class that took something similar as AP calc in my school about 1, they'll say it's a tautology, or f(x) is more correct.
@DanielFischer It was a pretty big deal when people started upvoting my answers and leaving nice comments. It made my really crappy life feel just a bit less crappy. So I do take this more seriously than most people.
What's a tautology? There's no logical statement there.
Someone called someone else "armpit" and got flagged.
Although I don't understand what @DanielFischer means with that comment. $f\colon \Bbb R\to C(\Bbb R, \Bbb R)$. The function $f(x)$...
because an "armpit" stinks
19:23
@MikeMiller I thought tautology meant these two things are a trivial restatement
hahah that's great
the domain of $f$ is a space of functions, so instead of being a real number, $f(x)$ (where $x$ is a fixed real number) is a function
@PedroTamaroff Pedro, ssup?
so $f$ is a function and $f(x)$ is a function for all $x$
@JayeshBadwaik Hello.
Semester end approaching?
19:27
Yes.
You?
Cool, yup, exams start in a week.
@RandomVariable The valuable reputation is the one expressed in the nice comments. The number in the top-bar is not entirely unrelated, but the important thing is that you give helpful answers and (some) people know that, and associate your (user) name with good quality posts.
@PedroTamaroff was no, not really in reply to whether anyone found the questions interesting
@MikeMiller thanks.. oddly it does work in some places
@user2179021 It was my personal opinion.
I'm not speaking for everyone here.
then put "imo"
19:34
@PedroTamaroff ok :) Do you mind if I ask why ?
Pedro speaks for me.
@PedroTamaroff it's a genuine question.. I am interested in what people find interesting :)
Pedro speaks at me.
@DanielFischer It's funny how this all came about because I was editing some old answers that I felt weren't of good quality because they had bad formatting.
19:54
So does asking questions no longer give rep?
this guy has almost 500 questions and only about 150 rep math.stackexchange.com/users/66906/xin-wang
oh he bounties it ok...
@PedroTamaroff Are you there?
@SwapnilTripathi Yes.
@DanielFischer

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