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21:01
you can actually, just click on "only show top 1" above them, unless it's the last one you were referring to.
it's a secret hat though
oh, there are secret hats :O
I have earned no hats .. (._.)
@Mahmoud just vote on MSE meta ;)
There are secret hats.
And for some reason, my new profile pic isn't showing up.
21:03
@Null Which is .. ?
I just uploaded a picture of a roaring lion about to take a bite out of somebody's face.
I liked your previous one better
Thanks. But I'm really in more of a roaring lion about to take a bite out of somebody's face mood right now.
There are lots of folks who need to get bit.
OOh, Balarka, you're Sherlock Holmes now.
21:05
My hat kind of looks like one of those freaks from the sex party in the movie Eyes Wide Shut.
not a fan of Kubrick
I am, but some of his movies just fell flat. That was one of them.
@Null Do I vote on random posts ?
Anyway, gotta go.
Why am I not behatted?
21:08
Happy Holidays, Merry Christmas, Happy Hanukkah, Happy New Year!
Oh, I am behatted sometimes.
@Mahmoud well, it would be nice if you vote not out of spite, but reasoned, but yes, then you get a hat^^
Hello
Testing.
if $u$ is a solution of a boundary value problem, and satisfy $\lim_{|x|\rightarrow\infty} u(x)=0$ what it means for $u$ ?
21:11
.
@Balarka bad
i used to think i liked 2001, but then i watched solyaris - which happened to be my first exposure to tarkovsky.
starring party on 1.1.17?
solaris has too much going on
@Null Now what ?
21:15
@Mahmoud wait a little, it'll come :d
Make a separate room if you want to do that. No need to clutter this one.
And ..
you get a hat :P
Finally .. :D
Winner hat. $>:)$
21:20
agreed. even tarkovsky himself thinks it's his artistic failure. but yet, the message seems so clear to me - discovering something inherently humane, in an inhumane, alien place far away from the familiar world.
i guess this worked better in stalker but the trouble is there may be some other thing going on there which makes the story more complicated than the story in solyaris
("the zone" is a synonym for "gulag")
no Tarkovsky spoilers please or I'll be left with no plans for the christmas holidays!
i didn't spoil much :)
@BalarkaSen I just mean there's like, five different fundamental conflicts going on. mankind's engagement with God, the discovery of the inhumane, grief, and many other things I've already forgotten
I know, but I was afraid of where your conversation with mike was going
@JessyCat I forget. Did your course talk about solving pdes via conformal mappings at all?
21:23
whereas in stalker it's rather to the point
I find it mildly funny that this chat seems to universally like Tarkovsky.
blame me
or maybe the fear of spoilers will keep me away from the chat and I'll finally be productive
yup, Mike's fault
I need to pack but don't really want to
21:23
Nothing wrong with it, haha.
@MikeMiller I just think the final message in stalker is rather confusing; a mixture of hope and hopelessness. It's as if it's reflecting the director's own confusion about his emotional ties to Russia and estrangement with the USSR.
Ha ha @Alessandro being productive :D
once in a while :P
Hi @TedShifrin
I just did some exercises from G-P
21:28
Heya @Ted.
I just watched an old Renoir movie last night ... such a master
@BalarkaSen You're sounding like an American! You don't need to end on one note.
hi @Fargle, @Mahmoud
But I'm not sending you exercises because I don't have an email, @Alessandro :P
Thoughts on whether $0\in \Bbb N$ or not, can we prove it ? @TedShifrin Sorry for bringing this up again ..
Jesus.
21:31
@Mahmoud: Last word. It can't be proved or disproved. Go away.
Well, I'm not complaining about it - I like it exactly because of this strange confusion I suppose and that's what makes me think about it so much.
Gone way enough.
@Mahmoud gratz^^
Thank you @Null
21:39
apropos only of russian filmmakers
the movie which I want to see at some point is Konchalovsky's Runaway Train
hm, I changed my name on MSE but it's not showing up in chat
testing what? .. I just sent you the stuff.
Oh, we have to clear cache and reload.
received it, thanks a lot! @Ted
21:49
@Alessandro What did you change it to
Did you add a last name?
Hi, DogAteMy.
yep @akiva
Btw, what is DogAteMy supposed to mean ? :o
I hope my algebra professor, who is very active on stackexchange and knows I'm a user too won't notice tho
21:50
then why did you do it, @Alessandro? :D
@Mahmoud My old username was "columbus8myhw" ("Columbus ate my homework")
So Ted made up the name DogAteMy
even though Columbus isn't actually my dog's name
This is going to dog you to the death, DogAteMy :D
through a tractrix?
I should just have a script ready to copy/paste
@TedShifrin Can you make me a name too :P ?
21:51
Ask Fargle about tractrices.
I was considering doing that, and since I shared my last name anyway when sharing my email... Also one only finds an architect or a chess player when googling my name
Zach can be CatAteMy
Someone needs to be Dr. Seuss.
Let $X$ be my made up name, call me $X$ for now until I get somewhat creative and give it a value. :D
21:56
I've met people at conferences before who asked "are you Mike Miller on stackexchange?"
Is the same course followed by graduate and undergraduate students? @Ted just wondering after seeing the exercises you sent me
@Alessandro: Yes, and the undergraduate Honors students had to do advanced problems for graduate credit.
every day the US education system confuses me more and more
good.
Many universities have two tracks of graduate courses, @Alessandro. Masters level (joint with undergraduate courses) and doctoral level (not).
@MikeM: Did it go well after that?
22:00
I see, you need to be a masters student (have completed a Bachelor) to follow masters courses here
@Alessandro: The problem is that at many US universities our best undergraduates are better than many of our graduate students.
@TedShifrin Yes.
Hey guys - just a quick question: if I was not sure whether my answer is correct
makes sense then
for a homework assignment, may I submit the question to stack overflow and ask if my approach was correct?
I don't want to give the impression that I'm trying to make other people do my homework :(
22:04
I was very not happy when I found some of my students posting their homework questions on MSE. ... Well, if someone says it's not correct and tells you how to do it, you're effectively cheating.
But there's no question that it happens a lot.
I see. What are some alternatives to confirming your answer? Because it's the first(well actually second) question in my problemset
Talking with your fellow students or your professor?
which in most cases is usually the hardest, since the next few follow with similar concepts
I'm doing the homework before the course actually starts, because I'm interested in the material
In my problem sets, the first ones were generally easy and the last ones were hardest.
But who knows ...
My professor in UCLA probably is not responding to e-mails right now because it's winter break and he's probably with his family
@TedShifrin you're a professor? :O
22:06
I was.
Woah - hats off to you :) I've always admired math professors
@Balarka has all the hats.
not sure if same case in other universities but math professors in UCLA gets a lot of respect.
I was about to say something along those lines.
Wait your hat is bigger than your avatar... huh
22:07
It would have been "give him the hat instead" or something though
@Departed It's a deerstalker hat. A big one.
Evening everyone, quick question, I'm trying to find a bijection between $\mathbb{Z_+}$ and $\mathbb{Z_+}^n$, but I'm not particularly sure how to construct a function that's surjective to $\mathbb{Z_+}^n$
I don't think the original Sherlock Holmes ever wore one of these in his life.
Well... hmm... If it's not good practice to ask the question on SO, but it's also impossible to contact my professors, but it's also probably not advised to do the rest of the problem set going in blind
then what should I do?
@Perturbative: I don't think I know how to, personally. Try $n=2$.
@OneRaynyDay This is in CS?
22:10
@MikeMiller it's not, I'm a CS major so I'm too used to saying SO, but I actually mean mathematics stack exchange.
do you need an explicit bijection or just a proof that there's one @perturbative
I see. Who's teaching?
Alexey Miroshnikov - he's some new professor at UCLA
woah wait a second - you're also from UCLA LOL
@Alessandro, I don't need an explicit bijection, but I would definitely like to find one
I could define $f\left((x_i)_{i \in \mathbb{Z_+}}\right) = (x_i)_{i \in \mathbb{Z_+}}$, but that would just be injective, not surjective
Yeah, I don't know him. Is this modeling or 170A?
22:12
hi @TedShifrin
170A. Probability theory. My intuition do not always match the answer.
@BalarkaSen, it seems you are The Cat in the Hat
I can probably ask Jens for you.
@Mike Who is Jens?
Your TA. :)
22:12
hi all
hi Karim
What the - you know him? :D
We all know each other.
Wow what a small world this is
22:13
Yup, @OneRayny, and a handful of my former students inhabit this site, too ... :P
Well that'd be great! It's about #13 in this problem set: athenasc.com/CH1-prob-supp.pdf , and I already have an answer but not sure if it is correct
@TedShifrin haha this is cool. I'm not always here often, but seems like a paradise for mathematicians
It would almost certainly be fine. If I was TAing I would have no problem with it, especially if you give your thoughts.
On the other hand, you can probably just email him. I'm checking on that.
Is there a grammatical error in this sentence ?
We shall explain the group operation geometrically, then we will use the geometry to proof get a explicit algebraic formula which adds two points on the elliptic curve.
@Alessandro, I could also define $f\left((x_i)_{i \in \mathbb{Z_+}}\right) = ((x_i)_{i \in \mathbb{Z_+}})_k $, where $k \in [1, ..., n]$, but I'm sure that it wouldn't be surjective either
@Adeek I'd call it a run on sentence. If you took out "we will" from the second clause it'd be perfect.
22:15
@MikeMiller Right; but I think some times teachers grade for accuracy of answer
Hello
Read more carefully, @Fargle.
And if I assume my answer was correct for this one question, it's almost certain that if I stretch my assumption to later questions I will get something wrong
@TedShifrin It still seems run-on to me.
The thing to be careful with in probability (I taught the course 2 years ago and had my own issues) is not to cavalierly assume events are equally likely when they're not, @OneRayny. And use conditional probability carefully.
22:16
@OneRaynyDay I poked him on facebook. You should email your question to Jens; my guess is he'll respond.
(Also, shouldn't everything have been due?)
It's run-on because there's a comma separating independent clauses, @Fargle. But there are worse issues.
Ah, I see.
@TedShifrin Right, I see. That makes sense :) but probably requires more experience to notice those things
how can I fix it ?
I need to take an english course :S
@MikeMiller Due last quarter, yeah. I'm taking it next quarter! :) AKA winter quarter
22:17
I will probably take one during the summer.
@TedShifrin Yeah, I'll always mess that up.
I finished the first problem set pretty easily. Set theory at this level doesn't seem painful
Can someone explain why the number of ways you can add up numbers between 1 and 49 to add up to fifty is not a sum of combinations of $\sum_j=1^50\binom{49}{j-1}$?
@Adeek "use the geometry to proof get a explicit..." -> 'a' should be 'an', and 'proof' looks superfluous.
Oh. Yeah, then Jens is probably not going to be TAing the same thing again
22:18
"formula for the sum of two points ..."
@MikeMiller yup. Then I don't really know who is the TA
@OneRaynyDay If it's not even been assigned yet, I see absolutely no problem with asking about your answer on MSE.
Sure :) It's just following @Ted 's argument that the answer shouldn't be disclosed nonetheless
@TedShifrin, There is an explicit injection for the $n = 2$ case, if you define $f(n, m) =2^n3^m$, and it follows from another theorem that if there is an injective map from $f : B \to \mathbb{Z_+}$ then $B$ (in this case $\mathbb{Z_+} \times \mathbb{Z_+}$) is countable
But I suppose if you find no issue with it I'll post it.
22:20
There's also an explicit surjection $\Bbb Z_+\to\Bbb Z_+^2$, @Perturbative. Try to figure one out.
@Fargle There is a natural way to define a binary operation on E that makes E into a group.We shall explain the group operation geometrically, then use the geometry to get an explicit algebraic formula which adds two points on the elliptic curve.
better ?
Then I would recommend doing the general case by induction, @Perturbative.
@TedShifrin, that's taken directly from Munkres's book, by the way, I didn't come up with that
Karim: I rewrote the end of your sentence up there ^^^^
@TedShifrin, Okay I'll give it a shot
22:21
oh let me check
I guess what I'm trying to ask is, why is a partition not just a sum of a Stars and Bars thing?
It's a common formula, @Perturbative :)
Ok, enough thermodynamics for a night
...why in the world do i have a tab with stuff about A1 homotopy theory open
I think it can be, @JohnDoe, for different numbers of bars, but I have no idea what you wrote up there.
@TedShifrin, is it? That was the first time I came across it in Munkres' book...
22:23
@JohnDoe Do you care about the order of the summands
Yup, you can generalize that injection to $n$ factors by the fundamental theorem of arithmetic, @Perturbative.
@AkivaWeinberger Oh, good point.
I am terrible at combinatorial thinking. I should keep my mouth shut.
Without order you're screwed. But there's still the generating function trick for it, IIRC
Yeah, a sum of stars and bars for every number of bars, with the added mess of dividing every combination by two because of repetitions (1,49) (49,1) (Which makes a mess because it will give 24.5 but I guess just round off to the higher integer)
@AkivaWeinberger
yes
I know an explicit bijection between $\mathbb N$ and $\mathbb N^2$, I'm not sure about $\mathbb Z$ though
Are you only using two summands
22:25
No, I'm trying to calculate every number of summands
Where's the code to run MathJax on this thing?
anyway I'm going now, good night/day everyone!
So then you don't divide everything by 2
Night @Alessandro
night @Alessandro
@BalarkaSen Of course you know the only 'formulas' are infinite series. One also has Euler's inductive formula for its calculation.
22:26
How come? $\binom{49}{1}$ gives 49 when there are 25 possible ways using two numbers (1, 49) up to (25, 25)
Right, but what about when there are three terms
(1,2,47)
That has six copies
@MikeMiller Right, I forgot about the inductive thing. I think that falls out from the generating function.
it's q-series all tjhe way down
Divide by $j!$ ?
Here's an idea: Map $\Bbb Z_+^n\to\{x_0+x_1\pi+\dotsb+x_n\pi^n\mid x_i\in\Bbb Z_+\}$ and then map that to $\Bbb Z_+$ in the unique bijective increasing function @Perturbative
22:29
yeah
Definitely not the intended answer, though
@AkivaWeinberger You need to show that that's discrete, but I like it.
That gives me too low
DogAteMy: We're trying to give a surjection $\Bbb Z_+\to\Bbb Z_+^n$.
Seems to me this shouldn't be too complicated
22:30
hey @TedShifrin does this theorem or its proof is confusing in any way ??
@TedShifrin Inverse. It's bijective
But it is complicated. So. You know.
I have no idea what you intended by your map, DogAteMy, nor why it should be a bijection.
@MikeMiller Yeah, it wouldn't be discrete if it were $\Bbb Z$ rather than $\Bbb Z_+$. That's a thing.
22:31
@TedShifrin: That set he just described is a discrete subset of $[0,\infty)$. The bijection is given by "Take the first element, then the second, then the thrid, then ..."
@AkivaWeinberger I know
@MikeMiller hey mike, I posted the probability question online here: math.stackexchange.com/questions/2066533/…
@MikeMiller It's good you know, because I forgot until you mentioned it
@MikeMiller I'm talking about calculating partitions
Could you let me know if maybe I worded the question improperly or maybe the formatting is bad?
that discreteness could be an issue
22:32
I'd love to get some feedback :)
@JohnDoe Yes. I know. And it's complicated.
@JohnDoe Another issue is that (2,24,24) only has three doubles
(2,24,24), (24,2,24), (24,24,2)
rather than the usual six
lol so you could in principle get a formula for the number of partitions by finding a formula for the number of partitions into k parts, where the k parts are equal or not depending on a partition of k
thus giving a formula for the partitions of n in terms of... all the smaller partitions
(and a really crappy formula too)
I see
@TedShifrin, I'm not sure if this is correct, but if we define $f : \mathbb{Z_+} \to \mathbb{Z_+} \times \mathbb{Z_+}$ such that $f(x) = ((x, i))_{i \in \mathbb{Z_+}}$, would that be a surjection? The only issue, I think I'm having here is with notation
22:34
Thanks, I didn't know it'd be beyond me
@JohnDoe Look up Euler's pentagonal number formula
@Perturbative: Try to actually enumerate the dots in a giant (infinite) square, starting from $(1,1)$. How would you do it?
@TedShifrin, my idea was pick a single value of $x$ and then have the tuple $(x, i)$, range over all possible values of $i\in \mathbb{Z_+}$, but I was having trouble on how to properly define a function that would do that
That won't work. You never finish and get back to the next $x$.
Draw a picture and see if you can figure out how to enumerate.
I start every problem by telling people to draw a picture
22:37
whence, Mike Miller: A Geometric Approach :D
Not as funny.
we'll see if you can find any geometry in the paper I'm apparently writing
how long have I apparently been writing it?
A few months.
a year
I had a line on my vita that said "in preparation" for over a decade, and finally gave up and removed it.
22:40
This is my first, so I can't afford to do that.
Right.
Do you know Robert Lipschitz?
@TedShifrin, Hmm, well intuitively I would draw the $y = x$ line and fill in everything above the $x$-axis and below the line and do the same for the $y$-axis. Translating that into this problem $f(x) = (x, x)$ would be said $y = x$ line, but I can't seem to define a function to do "fill in" everything in that $x$ by $x$ square
I think so, @MikeM, but not sure.
Doesn't much matter. I'm talking to someone when I get back in Jan about this stuff so I should have the important bits written down.
22:43
@Perturbative: Once you start filling one half, you never finish, so you never get to the other half. Try again.
Why is he relevant, @MikeM?
He's the someone. Co-wrote a paper I'm using extensively.
Ohhhh ...
Eliashberg student ...
hey @Fargle do you see anything confusing in the following proof ?
Yesterday figured out one of the last few things I didn't know how to do, though of course nothing's on paper because me.
Got one piece left before there's nothing I have any reason to do but write.
22:48
I ATC next quarter. I asked Ciprian when I'm supposed to do that and he found out that I hadn't yet.
yes, you should have ages ago, @MikeM
Normally advisors tell people when to do that. :P
I formed my committee the same day. There was a Christmas party.
no, advisers are generally clueless — maybe the graduate secretary tells you.
No, I mean, that actually is how it's done here. They usually tell you.
22:50
I still claim faculty are generally incapable of bureaucratic stuff.
Maybe UCLA faculty are bred differently.
True. But this is just "It's time for you to ATC."
They certainly don't fill out the paperwork.
@TedShifrin, I feel as if you're hinting towards some sort of diagonalisation argument, am I on the right track?
"Diagonals" are the right track ... if you shape them right, @Perturbative.
what diagonal argument @Perturbative ?
@TedShifrin any comments on my proof ?
Karim, I'm not reading.
22:52
oh ok
actually I think aliens won't even talk to us. We as adults think that the size of genitals is irrelevant. Now think what aliens think of being irrelevant. There are very few human thoughts that might be interesting for an alien. But the first contact will be probably with some stoned Teenies.
One thing I really like about the math SO is that people are not pretentious
@OneRaynyDay You haven't talked to me enough, then.
I'm the worst.
Also I think I remember a similar phone problem which was awful.

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