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12:02 AM
Fine by me, @Jasper. Moderators need to censor him for his insulting behavior toward me. Although he thinks I have harassed him, I've only corrected his math when he was wrong, and he cannot abide that.
If I leave in a fit of anger (and trust me, I'm furious now), I'm just rewarding him for his inappropriate behavior.
At any rate, I posted a correct answer for the OP.
LOL @Pedro: He now downvoted me. That's his mature reaction.
 
@TedShifrin Let's forget about this... so did you assign the problem to show $B(x,y)=\Gamma(x+y)/\Gamma(x)\Gamma(y)$ using polar coords? It is super nice. =D
 
No, it's way off from what is appropriate for these students/this class. I assigned the problem only for the grad student/Honors students, anyhow. Most of the class was lost today when I was doing high school algebra to compute basic stuff with expectation of a function of a random variable. Sigh. :(
I just assigned the integer case with induction.
 
Hehehehhe
@TedShifrin You need to get a whip and make em study.
 
For the most part, the textbook problems are plenty tricky for them. Some of them are plenty tricky for me, too :)
@Pedro: Here's the problem I gave.
 
@TedShifrin Cool!
 
12:11 AM
For your amusement, here are two more problems that everyone has to do:
 
@TedShifrin I don't know expectation stuff. =P
 
Weighted average.
$\sum xp(x)$
 
Nevermind. At the moment I don't want to probability.
 
Fine. :) I'll go disappear, anyhow ...
 
I don't blame @Pedro
 
12:26 AM
@KajHansen I was actually kicked out of my class because I have a pending final exam. =P
 
Thanks a lot @Kaj ...
 
I was willing to sit for the course.
 
@Pedro: This looks a lot like something we were discussing a few weeks ago.
 
Never been super intrigued by probability. Maybe it's just because I haven't been exposed to any?
 
@Kaj: Whose supercool proof of Theorem 14 was that?
@Kaj: It's actually quite more fun than you think. At least a few of my students and I are thinking so :P
 
12:28 AM
@TedShifrin, it's from Dr. Clark's lecture notes he gave us from when he subbed our analysis last week.
 
@TedShifrin Yes. I realized it is actually very trivial. =)
 
Besides, @Kaj, if you had five prizes to try to get (randomly) from Burger King, wouldn't you be curious -- on average -- how many times you'd have to go eat there to get all five prizes? :D
Ah, he couldn't resist writing his own notes even for someone else's class :P
 
I see Doug has been enjoying it. He always seems to have some little quibble with your interpretation of things, though :P
 
Of course he does, @Kaj :D
What was the theorem, @Kaj?
 
Bolzano-Weierstrass in $\mathbb{R}$.
 
12:30 AM
Ah ... Wow, still in chapter 2? :(
I usually do the successive bisection proof (e.g., in 3500) ...
 
Yeah, we're going kind of slow @TedShifrin. It's a little frustrating when I can see roughly how to prove something once it's written on the board, and then we proceed to take up 15 minutes discussing it.
I like being more on the edge of my seat. :P
 
Especially when you're asleep and snoring in the seat :D
 
Hey!
Ugh...
 
@KajHansen You're taking an analysis class?
 
@PedroTamaroff, so they tell me
 
12:35 AM
@KajHansen Is it too easy?
 
@PedroTamaroff, it goes a little slow is all.
It's alright otherwise. The homeworks tend to have some interesting problems.
 
Hit me with some interesting.
 
Consider the set of all irrational numbers in the interval [0, 1] with decimal expansions containing only 4's and 7's. Is this set closed? Perfect? Anywhere dense?
 
Ah, yeah, that's a cool problem :)
 
Plus it gives me another example of a completely disconnected subset of $[0, 1]$ that bijects with $\mathbb{R}$.
(That SHOULDN'T happen!)
"Je le vois, mais je ne le crois pas!" - Cantor
 
12:42 AM
@KajHansen Why not?
 
For me, it's very intuition breaking.
 
I don't know if Pete will do it in topology next term, but you'll have a very cool way of understanding that bijection in terms of a continuous map :D
 
I mean, a subset of a set and the set itself having equal cardinality is just weird to think about.
 
Nah, you're used to that. $\Bbb Z\subset\Bbb Q$
 
It's still weird!
 
12:45 AM
or even $2\Bbb Z\subset\Bbb Z$
 
Still weird!
 
I guess you believe only in finite numbers.
 
I'm very put out that I haven't a good answer for this question.
 
@TedShifrin, he is signed up for it on the course schedule at least.
 
12:47 AM
No, no, @Kaj .. As far as I know, he's teaching it. I just meant I don't know if he'll do the particular discussion (or assign the exercise) I have in mind relative to the Cantor set.
 
Ohhh. Well I guess I'll have to wait and see :)
 
There will be some very good students in that class. I hope he makes it hard :P
 
Who else is taking it? I know Alex and maybe Daniel?
 
Yeah, Gabe Durham and quite a few more.
 
Awesome. I'm looking forward to it regardless. I know Dr. Clark has a bit of a reputation about his 24xx that might carry over...
 
12:53 AM
Well, for the topology course, it's appropriate to aim at a high level. :)
 
When working with LaTeX, how do you produce formatted text within an equation?
I thought it was just \text{insert here} ? Does it need a special library import?
 
No.
 
I use \text{insert here} on StackExchange and it takes it. TeXWorks doesn't like it though :/
$\text{test}$
 
I use it all the time . Hmm ...
 
Test: $S=\{x \in [a, b]: \text{There is a finite subcover of [a, x]}\}$
Very weird! Chat takes it!
 
1:11 AM
I've done it thousands of times in the past 26 years.
 
Beginner's un-luck
 
email me what you're doing ... Including preamble.
 
@TedShifrin, sent. I left out all the extra and just kept the one sentence that was giving problems. My preamble is probably embarrassingly error-full.
And formatting as well, for that matter.
 
Can someone point me in the right direction? I need to prove that all functions with domain $\mathbb{R}$ can be written as a sum of an even function and an odd function, and that this summation is unique.
 
@noahnu, would the functions also need to be continuous?
 
1:22 AM
I believe so. I just read a proof online, I understand now. If you're interested, here's the basic proof: en.wikipedia.org/wiki/…
 
@TedShifrin, I got it working with \mbox{}
@noahnu, it actually doesn't require continuity. Cool proof.
 
Ah, @Kaj, I just sent you two emails. The second was what you just did. The first is what you should standardly do :)
 
Thanks for the advice!
 
@Kaj @noahnu: The decomposition will be continuous whenever the original function is, but there is no need to worry about that.
An even function satisfies $g(x)=g(-x)$, so the obvious way to make an even function out of $f$ is to consider $g(x)=f(x)+f(-x)$, etc.
 
I cannot got believe someone flagged this in the ELU room:

 English Language & Usage: Multi-Layer

Not for the faint of heart or those easily triggered by Englis...
in English Language & Usage, 6 hours ago, by Ice Girl
@WillHunting Your questions in this chat room are really stupid
 
1:33 AM
@skull: I'm still fuming at the crap I got earlier in a comment. I
 
@TedShifrin link please :-)
 
It's gone. I put a link to mods above, but none of them has been by.
 
1:58 AM
Hey guys, does anybody here know how to calculate a trend line for a scatter plot, given n points? I'm trying to find the equation so that I can do it programatically, but I've only been able to find resources that indicate drawing said trend line by hand, with guesses
 
@TedShifrin, my local Lie algebra expert.
 
@GarethParker, you might find more meaningful results by searching for "linear regression".
2
 
Not I, old man @Mike
 
@KajHansen
@KajHansen, thanks for that :)
 
Sure thing @Gareth
 
2:02 AM
@TedShifrin I said local... anyone in this room know more about them than you? :P
 
@Mike: One of your Penn State buddies, who's across the hall from me, was surprised to find out I know you.
 
Juan?
 
Yup.
 
Hope he's doing well
 
So far so good. The one time we played tennis, I beat him a set, so he's pissed :)
 
2:05 AM
Playing tennis with students? Scandalous
 
He's no student of mine.
 
Right... he's interested in number theory, or was a year ago
 
Yeah, who knows ...
What's the LA question?
 
@TedShifrin when is your stress test appointment Professor?
 
Next Thurs morning
 
2:08 AM
you'll do fine
 
@Ted Let $k$ be a CRing. Given a set of polynomials $P_\alpha$ in the coefficients of $n \times n$ matrices $M_n(k)$ write $G(k')$ to be the set cut out of $GL_n(k')$ by $P_\alpha$ for any associative unital $k$-algebra $k'$. He says $G(k)$ is an 'algebraic group' if $G(k')$ is a subgroup of $GL_n(k')$ for all $k'$.
Anyway, then he defines the Lie algebra of an algebraic group to be: $X \in \mathfrak g$ iff $1 + \varepsilon X \in G(k[\varepsilon])$, where $k[\varepsilon]$ is $k$ adjoined with an $\varepsilon^2=0$.
 
I don't have the energy to think ...
 
To prove that this is indeed a Lie algebra, he first has to show it's a vector space; in the process he says that $P_\alpha(1+\varepsilon X) = P_\alpha(1) + dP_\alpha(X) \varepsilon$. In context the only way one could possibly interpret $dP_\alpha$ is as the total derivative of $P_\alpha$. But I don't see how the claimed equation is actually true unless 1 is the all-1s matrix, and that doesn't make a bit of sense!
It only makes sense for it to be the identity for the analogy with the Lie groups situation.
"total derivative" is the wrong word... I mean $\partial/\partial x_1 + \dots + \partial/\partial x_n$
Alas.
 
No, it's the derivative as a linear map. $1$ should be the identity. What's your issue? This is Taylor's Thm, using $\varepsilon^2=0$.
 
stupid, stupid, stupid
what a bunch of garbage
 
2:18 AM
Yeah, it works algebraically.
 
The problem was with misinterpreting $dP$. It's fine now.
 
Ok :) glad I could in fact help.
 
Thanks. It wasn't anything tough... this is only page 4 :P
Before I nap, what was the comment you linked in your starred message on the right, @Ted?
 
@TedShifrin How can I evaluate the integral of $e^z/z$ around a curve in $\Bbb C-\{0\}$ homotopic to $\partial B(0,1)$?
 
It was René going off on me, telling me he'd told me not to comment on him, and telling me to "go back to stalking little boys." Incompetent homophobe that he is.
 
2:26 AM
Christ.
@PedroTamaroff Convince yourself that integrals of holomorphic functions on contractible curves are 0.
 
@TedShifrin =/
@MikeMiller But that thing is not holomorphic at zero.
 
I don't care.
 
Hey guys. I don't suppose any of you know of resources that would be helpful in understanding Linear Regression Modelling, maybe made for someone who's highest mathematics education was Year 12 some 6 years ago?
 
He better stay away from stuff I know better ...i've had it with him. And he downvoted the answer I then wrote for the OP.
 
@TedShifrin That guy is unbelievable.
 
2:28 AM
The resources I've found go way over my head
 
@TedShifrin Invite Rene into this chat.
 
@MikeMiller You're not helping!
 
@PedroTamaroff I promise you I am.
 
@MikeMiller What is true is the following: for each $\delta >0$, the integral is equal to the integral around $\partial B(0,\delta)$.
 
Use power series for $e^z$, @Pedro? I dunno what you now know.
 
2:29 AM
@TedShifrin Ah, OK.
I can do that.
 
@GarethParker try YouTube Khan Academy
 
@IceBoy Thanks
 
Oh this is residue stuff, right @TedShifrin?
 
Yes, that's true, @PedroTamaroff, but I was suggesting a way to prove that. If a (suitably smooth) curve is contractible in some open subset $U \subset \Bbb C$, then the integral of a holomorphic function (in that open subset) along that curve is 0. $U = \Bbb C \setminus \{0\}$.
 
@GarethParker Thanks for asking, it is a good and easy introduction
 
2:31 AM
@MikeMiller But this curve is not contractible!
Precisely because we have $0$ inside it.
 
Sure, @Pedro.
 
@PedroTamaroff The point of what I was saying is to show that the integral along your curve is equal to the one along $S^1$. But I guess you're taking this as fact.
 
Good suggestion, @skull.
 
@MikeMiller Duh. =)
 
@IceBoy I'll probably be back with questions, but this will be incredibly helpful :)
 
2:32 AM
@GarethParker Have fun :-)
 
2:43 AM
@TedShifrin If a function is entire, does it powerseries expansion around any point of the complex plane have radius infinity?
Ah, nevermind.
 
Yes @Pedro
there's a theorem lurking
 
@TedShifrin I thought so, I haven't thought of the proof.
 
There must be a singularity at at least one point on the circle of convergence.
 
@TedShifrin Oh, Weierstrass'.
 
dunno what you mean ...
 
2:52 AM
I am doing the following exercise "Does there exist a holomorphic function in $B(0,1)$ for which $f(1/2n)=f(1/(2n+1))=1/n$?"
@TedShifrin What you said,.
If there are no singularities on the circle of convergence, we can extend it.
Because nbhd, compactness, yadda.
The proof is in Remmert.
 
Right ...
 
I haven't read it in detail though.
 
I love to assign that for homework :)
 
@TedShifrin What?
 
The proof you just yaddad.
 
2:54 AM
@TedShifrin Ah. And what about my other problem?
I have seen such exercises here around MSE.
Of course one should use the identity theorem.
 
Hi @robjohn
 
You get an infinite sequence of points where some holo fn is $0$, converging to $0$, @Pedro.
 
@TedShifrin So the function must be constant, equal to $0$.
 
Yes, but what function?
 
@TedShifrin Oh, OK.
I see.
 
2:59 AM
Hmm, maybe not right. Maybe realize $f’(0)=0$.
 
@TedShifrin Wait, no.
Actually $f'(0)=2$ right...?
$f(0)=0$ by continuity.
But $f'(0)=\lim\limits_{n\to\infty}f(1/2n)/(1/2n)=2$.
@TedShifrin ?
 
3
Q: Calc III: Volume of the Intersection of two spheres

JDGQuestion: I am not getting the correct answer. How do I get the solution (and why does my solution not work?) Find volume that lies inside both spheres: \begin{align} A: 4 &= (x+2)^2 + (y-1)^2 + (z+2)^2\\ B: 4 &= x^2 + y^2 + z^2\\ \end{align} My solution gives the answer $V \approx 11$, where...

 
@JessyCat HAI
 
My question is where does he get the x=2 from that he integrates over?
@PedroTamaroff, hey!
Brb
There we go. That's better.
Probably the stupidest question ever, but it is what it is.
 
@Pedro: but if $f'(0)\ne 0$, then $f$ is 1-1 in a nbhd of $0$ !!
 
3:11 AM
Hello @TedShifrin
@JessyCat!
 
Hello @BalarkaSen
Long time no see.
 
@TedShifrin Because $|f'(0)|^2=J_f(0)$.
 
Ah @PedroTamaroff is here too
 
@BalarkaSen To your dismay, yes.
 
no, to his delight
 
3:14 AM
Ice ice baby
 
that's me
:D
 
nah, it's dislight. something between dismay and delight
 
Dislight, huh?
 
not sure which one has greater proportion
 
Usually, we frame it in terms of open sets.
So it's a half-delighted interval. Haw haw.
 
3:15 AM
ah drat analysis
 
At least it's delighted at one endpoint.
Yeah, analysis jokes make me happy.
 
Analysis makes me sad altogather
 
GREs make me sad :(
Standardized tests can kiss my _____________
 
Thanks very much for the omission of whatever you meant
 
Haw haw. No problem.
I should get back to studying...
Nitey night, folks.
 
3:18 AM
later pal
 
@PedroTamaroff Let $c = c(n)$ be the least integer such that there is a power of two in $[n^c, (n+1)^c]$
Produce an estimate for $c(n)$. That's what we are working on right now. Fun problem.
Computationally, the best algorithm now is $O(\log^k n)$ using continued fractions.
@Alexander!
 
@TedShifrin i am getting sick of that guy, lemme tell ya what.
@BalarkaSen hi
 
Hi @AlexanderGruber
 
how's the back feeling?
 
3:26 AM
@AlexanderGruber OK, so I told him to remember this is a community and he said "Fuck the community."
I would say he should be kicked out, if that's his mindset.
 
@AlexanderGruber what guy?
 
Rene
 
ah
the severely RH-affected user of MSE?
(RH -- Reputation Hysteria)
 
like I said he should be invited into the chat room };-)
 
what did he do now?
 
3:32 AM
I''ll bring skullparol out of retirement for him
1 hour ago, by Ted Shifrin
It was René going off on me, telling me he'd told me not to comment on him, and telling me to "go back to stalking little boys." Incompetent homophobe that he is.
 
@PedroTamaroff he got 30 days
 
@AlexanderGruber Good.
 
sounds like a troll to me
 
MSE drama is pretty WTF to me
2
 
it escalates roughly by offense: 1 week, then 1 month, then 1 year
with some variation on a case by case basis
what he said was clearly out of line, though. i don't think it's any secret to anyone where his suspension came from.
@Semiclassical You're telling me, man.
 
3:36 AM
i mean, i can understand some level of frustration. putting out a good answer does take some amount of effort, and if that doesn't get recognized it's annouying.
 
anyway i'm gonna cancel the stars on ted's ping, since the situation's been dealt with
 
but rage? idk
 
@Semiclassical Most people deal with it just fine
but the ones who don't tend to be vocal about it (tautologically, i guess)
 
@TedShifrin he got 30 days
 
right
 
3:37 AM
:-)
 
bah. MSE was a math Q&A site last time i looked, not a forum for people to get used to community-based works.
 
Not that anyone asked me, but I find the progression week-month-year to be irregularly spaced. The ratios between consecutive periods are 1:4 and 1:12.
 
@CareBear You're a Care Bear.
Deal with it.
 
lol
 
week-month-quarter-year would be more balanced... and this is how reputation leagues are formed, btw.
2
 
3:40 AM
No, he is officially Thursday, @Pedro
 
Thanks, @Alex. He just cannot be so hateful, for starters, and to think I should be banned from correcting his perfect mathematics ... Sigh. Hell, people catch me in mistakes and I say, "Oops. So sorry. Thank you. I'll fix it." Wth ...
 
mistakes happen
 
@Pedro: So you get the impossibility ...
 
@TedShifrin there are clearly some sort of emotional issues at work over there...
 
there should be a minimum level of respect in MSE
 
3:43 AM
Oh yeah. Presumably related to why he couldn't keep a job. @Alex But he may hate me. It doesn't matter. He can't be around here and act abusively or god-like.
 
@CareBear after being a mod this long i'm inclined to say week month year is perfect
 
@AlexanderGruber he has a point about balance
 
it's not easy to get us to suspend you for something. By the time you've managed to do it twice, you need to go away for a while.
 
Well weak-month-quarter-year would be much healthier @AlexanderGruber
 
Any fun math lately, @Alex?
 
3:44 AM
;)
I asked @TedShifrin. Just graphs =(
 
@TedShifrin i've been working on queueing networks.
 
Like I said.
 
Ah ... Some cool applied homology :)
 
@BalarkaSen embarassing confession: I read that as "much heathier," like a Health bar, for the entire time he had that. I only figured it out after he changed it. I am an idiot.
 
@AlexanderGruber True, repeat offenders are showing malicious intention.
 
3:45 AM
@AlexanderGruber haha
 
weak
week
 
I think he's unstable and depressed, @skull. But he can't go off on me in public when I say politely that there is some confusion .... In the field of my expertise! And slandering me on top of it?
 
Agreed.
 
@IceBoy Yeah. first time, maybe you got heated and stepped over the line- i get it. 1 week to cool, fair enough. Second time, you've showed the first time wasn't an accident, but we're not ready to say you don't deserve to be here at all. Third time, you're a menace with no intent to change.
 
@AlexanderGruber he did it even before?
 
3:48 AM
Not the homophobic slander, but yes.
 
that's just great
 
Sigh. I'm too old for this ...
 
he's just mud slinging
 
No, he resents me.
 
jealousy?
 
3:49 AM
@TedShifrin chill out. have some water.
 
@TedShifrin Do you know him outside of MSE?
 
He'd never admit it, but, clearly?
 
hmmm...
 
wait @TedShifrin so you DO know him outside MSE?
*of
 
3:50 AM
No. I've never heard of him. Although I'm tempted to ask my friends at his ph.d. Alma mater ...
 
@TedShifrin I cannot solve this problem.
 
Really? @Pedro
 
Let $f:\Omega \subseteq \Bbb R^n\to\Bbb R^m$ be $C^1$, and $K$ compact
 
oh, a different one.
 
Then there is $\delta >0$ and $c>0$ such that $|f(x+h)-f(x)|\geqslant c|h|$ for every $x\in K$ and $|h|<\delta$, provided $Df(x)$ is injective at points of the compact set $K$.
 
3:52 AM
@TedShifrin Who do you know over there? one of my very good friends is doing his PhD at boulder
 
I am not sure if the hypothesis is that for each $x\in K$, $Df(x)$ is injective, or $Df$ is injective in $K$.
@TedShifrin
 
Surely the former, @Pedro. The latter doesn't really make sense.
@Alex: Jeff Fox and Jeanne Cleland
 
@TedShifrin ah i see. Different areas than my friend. (I've heard of Jeff Fox.)
 
@Pedro: There may be other ways, but I'd redo the proof of the Inverse Function theorem, locally.
 
@TedShifrin Ah, I see.
 
3:55 AM
Fox and I were undergrads together, @Alex.
 
He needs to be told that the next one is a year.
It straightened out Bill.
 
It is actually a consequence of the rank theorem, @Pedro, too.
 
@TedShifrin What's the rank theorem? I think it is in Rudin.
 
Ugh, :)
 
3:59 AM
A $C^1$ map of constant rank $r$ looks (in appropriate coords) like linear inclusion followed by linear projection, or some such.
With your dimensions, it looks locally like inclusion of $\Bbb R^n\hookrightarrow \Bbb R^m$.
 
@TedShifrin Yes, I noted that $n\leqslant m$ here.
 
ok, g'night, all
 
Because of the rank-nullity theorem in usual linear algebra.
Bye
 
Yup @Pedro
 
 
2 hours later…
6:19 AM
@Ice
@IceBoy: Thanks for the tip earlier. I'm actually understanding how it all works now
@IceBoy I should be able to implement this in my program
 
6:31 AM
hi all
what sup
 
hi
 
hi
@GarethParker you are very welcome, thanks for asking :-)
@PedroTamaroff
 

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