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12:02 AM
Hello
Is it possible to find a single ellipse by knowing three points and a center point?
 
Isa
12:56 AM
If A is the upper half plane (without the real axis), then $Fr(A)=\mathbb R$. What is Fr(-iA)? Is it -ki ?
I think it is ki, $k\in\mathbb R$
 
Am I supposed to see something else than this weird code?
 
 
1 hour later…
2:04 AM
0
Q: A question on integrability of derivatives.

Rajesh DachirajuLet $$f : (0,1) \to \mathbb{R}$$ and $$g(x) = |f(x)|^{r-1} f(x)$$$r \in \mathbb{N}$. It is known that $g\in L^2(0,1)$ and the $r^{th}$ derivative, $ g^{(r)} \in L^2(0,1)$. Show that the first derivative $f^{(1)} \in L^2(0,1)$

 
 
2 hours later…
4:03 AM
The new layout of math.se is so bad!
 
@Silent I like it
You'll probably like it after you get used to it.
 
I don't like it ...
but you can remove the left panel in preferences
hi Alex & Silent
 
Hey @TedShifrin
I have to go to class now, so also bye @TedShifrin :P
 
bye, Alex
 
cya
How are you Professor? @TedShifrin
 
4:11 AM
hi user2646, whoever you are
 
Hey Ted!
 
hi Demonark
your summer stuff almost done?
 
Yup, tomorrow is the last day of the program itself and then I just have to finish the paper
 
how quickly time flees
 
This is true. After this is over I'll probably focus on the GRE, hopefully I'll just have to take it only once
 
4:14 AM
practice doing lots of stuff ...
 
Yeah for sure. I have the Princeton Review so I'll go through that and then do a bunch of practice tests
 
OK
is the Princeton Review for the advanced subject test?
 
Yup
Well, that aside, what've you been up to?
 
after much hemming and hawing, I decided to cancel my trip up to Stanford/Berkeley ... too much fire and too much bad air.
you going back to Texas to study, or hanging around Chicago?
 
Heading back to Texas
 
4:26 AM
ah, so you can broil
 
hi @TedShifrin
 
hi Karim
 
The fact that all houses there have central A/C will help a bit with that
 
until there are brown-outs as electricity runs out
 
is T&M good university ?
 
4:27 AM
what's T&M?
 
Do you mean Texas A&M?
 
yeah
 
Not great.
 
I am just recalling from memory
 
Yeah not quite. The good ones in Texas are UT Austin, Rice, and SMU
 
4:28 AM
SMU has no particular math reputation.
 
@TedShifrin I found great people working in geometry at harvard btw hopefully after my PhD I would like to work with someone
there
 
UT Austin is really good, though
Karim: That's like one of the two or three hardest places to get a postdoc :P
 
Oh yeah I meant more in general/for undergrads, for math in particular I've also heard nothing about it
 
@TedShifrin this guy does really awesome things scholar.harvard.edu/artan
what does it take to do a postdoc in Harvard @TedShifrin ?
amazing publication ?
 
Yeah I've considered UT for grad school, though I'm starting to lean against it given what you guys have said about teaching loads for grad students
 
4:30 AM
that's getting bad everywhere, but, yes, we had bad comments in here ... public universities in general are getting bad
 
or amazing PhD thesis ?
 
Karim: I don't have a precise answer. Strong thesis, strong letters, and probably someone who knows someone ...
 
@TedShifrin I see that makes sense
I will go back to work cya all l8ers
 
bubye
 
Catch you around!
Also hey @loch!
 
4:36 AM
Hey @Daminark
 
@TedShifrin Btw I saw couple of videos about people standing against people who say that homosexuals should be killed. Internet is changing a lot of things.
hopefully in next few years there will be actual change
 
except that all the Trompistas want everyone non-white Christian str8 killed ...
heya @loch
 
I think that is just a phase
 
No, this is doing long-term damage
 
Hey @TedShifrin
 
4:38 AM
really ? Is general public being like this ?
 
A very large percentage, yes.
 
That is very dangerous for the U.S as it holds a lot of power.
 
Losing power quickly ... all about Russia and China ...
 
We will see what happens next few years. I have hope that humanity will move forward though, because of the internet makes spread of knowledge happen very quickly.
anyways cya all good nights.
 
See you
 
4:42 AM
Who's to tell the difference between knowledge and disknowledge?
 
hi @Ted
 
hi Leaky
 
@TedShifrin Hello, sir
 
15 hours ago, by Leaky Nun
@Semiclassical @TedShifrin why can we apply the nabla formalism to div and curl even for polar coordinates?
 
@Leaky: It's all chain rule. Easier to do with differential forms than with the classical stuff, but still just chain rule.
 
4:50 AM
also where can I learn about differential forms and de Rham cohomology?
wow jinx
 
there are zillions of places.
 
Where is that preference thing, that you were talking about, to remove left panel?
 
Spivak's Differential Geometry, volume 1.
 
does that also have de Rham cohomology?
 
yes, @Leaky
it wouldn't hurt you to learn some computational stuff, too. For that, of course, I recommend my book/lectures.
 
4:51 AM
Bott and Tu chapter 1 is really good for de Rham cohomology too
 
Bott and Tu and Lee and Sum?
 
@Silent: Go to meta and then click on your name to get profile. There you should get to choices for preferences. (If I remember right.)
I think Bott and Tu is too terse.
wow, what is Mathein doing up so early in the morning?
 
nice, i'm gonna leave the tab of spivak open and never read it, just like every other book I want to read
can I have 36 hours in a day?
 
@TedShifrin I went to bed at a reasonable time for once :P
 
still, university students don't get up at 6 AM :P
 
4:54 AM
it's almost 7AM
 
oh, I forgot you're one hour later than France
 
@TedShifrin Thank you very much
 
I didn't think that was so
 
tag @MatheinBoulomenos
 
morgen @LeakyNun
also retroactively hi @TedShifrin
 
4:55 AM
Hey @Leaky and @Mathein!
 
hi @Daminark
 
I think talking to me replaces "hi," @Mathein :P
 
@TedShifrin how do you even have the time to read a book?
 
I learned most of math by reading books and articles, @Leaky
LOL @Mathein
 
@TedShifrin how many hours do you spend doing math in a day?
 
4:58 AM
Remember I'm a retired bum now, Leaky.
 
eh... used to
how many hours do you used to spend doing maths in a day
 
As a grad student and as a faculty member, I worked pretty hard (including teaching).
 
is it meaningful?
 
I don't know a number.
There was no internet to waste time with ...
9
OK, leaving for now ...
 
should the cross product of two vectors live in the same space as the original vectors?
 
5:00 AM
depends on what you think cross product means
 
Bye @Ted
hi @Alessandro
 
Hi @Mathei
Did I just miss Ted by a couple of minutes?
 
I think cross product is a map $\Lambda V \otimes \Lambda V \to \Lambda^2 V$
and yes you did
 
See you Ted!
 
@LeakyNun Just $\times$?
 
5:02 AM
Hey Alessandro
 
@AlessandroCodenotti that would make it bilinear
 
@Leaky wouldn't that be the wedge product? The cross product is only that modulo applying the Hodge star (which requires a choice of inner product and an orientation on $V$)
 
what is the Hodge star?
0
Q: Argument principle for essential singularities

Kenny LauLet $f : B_0(r) \setminus \{0\} \to \Bbb C$ be holomorphic, where $B_0(r) \setminus \{0\}$ is the punctured disc of radius $r$. Assume that $f$ has an essential singularity at $0$. Is there any way to find $\operatorname{Res}\left(\frac {f'} f, 0\right)$ from the Laurent series of $f$ at $0$?

also I should have asked Ted this :P
 
$\Lambda^n V$ is one dimensional if $V$ is n-dimensional
 
5:06 AM
sure
 
so you have a map $\Lambda^k V \times \Lambda^{n-k} V \to \Lambda^n V$ map
that's equivalent to $\Lambda^k V \to Hom(\Lambda^{n-k}V ,\Lambda^n V)$
if your vector space has an inner product, you can use that to get an isomorphism $Hom(\Lambda^{n-k}V , \Lambda^n V) \to \Lambda^{n-k}V$
composing them you have $\Lambda^k V \to \Lambda^{n-k}V$ (i'm probably missing some conjugates or whatnot if the field is complex..)
 
I'd rather not make that isomorphism $(\Lambda^{n-k}V)^\ast \to \Lambda^{n-k}V$
 
not sure what you mean
but anyway, that final isomorphism is the hodge star operator
 
ok
 
@Mathein do you know some interesting/presentable results of Kronecker-Weber?
 
5:22 AM
Hello again
 
I guess in particular I am trying to see if there are some corollaries I could state that are more identifiable as number theoretic
Hey @AlexClark!
 
@Daminark every abelian algebraic rational number can be expressed as a linear combination of roots of unity
 
How are you doing @Daminark
 
Lmao Leaky
Everything's doing alright, how about you?
 
Good now, finally got a place to live in :P
 
5:24 AM
$\sqrt 5 = \zeta_5 - \zeta_5^2 - \zeta_5^3 + \zeta_5^4$
I'm serious @Daminark
 
Had to sleep on a couch in one of the math building rooms a few times
 
@AlexClark sounds like every mathematician ever
 
@LeakyNun Apparently if I got caught doing it again, I would have my after hours access removed :P
 
really :o
 
I mean that's kind of a restatement of the result, and the specific expression you mentioned requires not only knowing that it lies in some quadratic extension, but which one
But say, I dunno, my mentor sorta outlined for me the other day how one could go about proving quadratic reciprocity using ramification instead of Gauss sums, I guess that's the sort of thing I'm looking for
 
5:27 AM
I think it’s important to know that $\sqrt p \in \BbbQ (\zeta_p)$ for $p=4k+1$
and the other result for 4k+3
 
@Daminark wait, the abelianization of the absolute Galois group is not immediately identifiable as number-theoretic?
 
I guess I just haven't seen much of the Galois theory <-> NT interaction aside from quadratic reciprocity
 
Come back and give me a music recommendation @BalarkaSen
 
So as of now it's one of those things where the nature of the fields makes it make sense that they're linked, and I know that they are, but I don't know that many theorems in NT which are proven via Galois
 
@AlexClark Listen to some Volbeat while you wait
 
5:32 AM
@Daminark yeah I know what you're asking for, I said that in jest (though I met someone who said that to him, number theory is about certain Galois groups and diophantine equations are more geometry than number theory)
 
Hey @Tobias!
 
@Daminark Hi
 
Hi @Tobias
 
Hi @MatheinBoulomenos
 
@TobiasKildetoft I'll try it
 
5:33 AM
@Mathein ah lmao
Yeah I've had multiple grad students say something to the effect of "Modern NT is the study of the absolute Galois group of Q"
 
@Daminark I think the "number-theoretic" bit of Galois theory is really about where Frobenius sits
In that sense, Kronecker-Weber isn't that number-theoretic per se
but if you can use it to help deduce Artin reciprocity or something, then it's number theoretic
 
@user31415 that is a good point
 
Personally I found "Modern NT is the study of the absolute Galois group of Q" extremely mysterious for a long time
until I see Joel Bellaiche's answer here: mathoverflow.net/a/162733/31814
which makes more sense, and explained why we care about Galois representations for example
 
the proof that Artin reciprocity => quadratic reciprocity is well-known, but I'd have to think how you get the special case of CFT from Kronecker-Weber
 
(as a tool to understand where Frobenius is - since very often just knowing when Frobenius = 1 is great)
@MatheinBoulomenos I'm not too sure if that can be done off the top of my head actually :/
 
5:37 AM
@user31415 I remember reading in some historical overview of CFT that something like that was done, but maybe you need more than just Kronecker-Weber
 
I imagine you'll need some way to get the reciprocity for cyclotomic fields
 
Find $\operatorname{Res}(\sin(\tan z), \pi/2)$ :p
 
but with Kronecker-Weber it probably means you can reduce Artin reciprocity to that case
(with some sort of functoriality of the reciprocity map)
 
I see
 
6:00 AM
so the laplacian equation is something like a function without local maximum/minimum
 
6:21 AM
@MikeMiller I'm very thick
$\frac {f'} f$ doesn't need to have a Laurent series if $f$ has an essential singularity at $0$
The zeroes of $e^z + e^{1/z}$ has an acculumation point at $0$
so $(e^z + e^{1/z})^{-1}$ doesn't have a Laurent series
 
 
1 hour later…
7:43 AM
@TobiasKildetoft I thought that the new Group Theory room might be of interest to you. (But maybe you already know about it.)
 
@MartinSleziak Thanks. I did see it, but so far there has not been much activity there
 
Yes, that true. (Which is why I used this also as a possibility to advertise the room here. Although I believe that the room owner will mention the room occasionally in order to get some initial interest.)
 
is $\frac 1 r \in C^\infty(\Bbb R^3 \setminus \{0\})$ irrotational / solenoidal?
Also what is the de Rham cohomology of $\Bbb R^3 \setminus \{0\}$?
 
8:50 AM
So, that guy is gonna be $S^2$
(I'm very much cheating here by using many theorems)
So we know that $S^2$ is simply connected, so $H_1$ is $0$, meaning $H^1 = 0$ (that's in $\mathbb{Z}$ coefficients but I think it should translate over to $\mathbb{R}$, and then de Rham's theorem)
 
yes it does translate by universal coefficients
since the cell decomposition of $S^2$ is very simple, it's also easy to work with cellular cohomology with $\Bbb R$ coefficients directly. you still use a lot of theorems (deRahm=singular=cellular)
the cellular chain complex of $S^2$ with the standard decomposition is just $\Bbb Z \to 0 \to \Bbb Z \to 0 \to 0 \dots$, then you apply $\mathrm{Hom}(-,\Bbb R)$ and take the cohomology of that
 
Haven't ever really learned that kinda business. I think I know the definition more or less of singular homology but that's about it. I'm aware of de Rham and know how to compute it for the circle and I know the $[X,K(G,n)]$ and axiomatic business
 
Hello!!

I want to find the values of $a$ such that $a^2=a^{10}$ in $\mathbb{Z}_{11}$. So it holds that $a^8=1$ in $\mathbb{Z}_{11}$. We have that $\left (a^4\right )^2=1$, which means that $a^4=\pm 1$ in $\mathbb{Z}_{11}$, or not?
So we get $a^4=1$ and $a^4=-1\equiv 10\pmod {11}$.
Is this correct so far? Do we continue in that way till the power is $1$ ?
 
cellular is great if you actually want to compute things
 
I guess we can compute $[S^2,\mathbb{CP}^{\infty}]$ and use universal coefficients
:P
(I should probably learn algebraic topology at some point)
 
9:06 AM
@MatheinBoulomenos @Daminark Could you maybe use a function to demonstrate that the second exterior derivative is non-trivial?
(probably wrong terminologies)
 
9:40 AM
Maybe. I'm not really sure how to compute de Rham cohomology explicitly
I've only ever seen one example and it was really contrived
 
$\Bbb R^3$?
 
Punctured plane
 
I thought you would know about $\Bbb R^3$
since the derivatives are just grad curl div
 
I mean that's not contrived, it's just nullhomotopic
 
1 min ago, by Daminark
I've only ever seen one example and it was really contrived
 
9:42 AM
Well, I don't really phrase things often in terms of grad/curl/div
I guess replace "one example" with "one example which isn't stupid"
 
could you tell me about the punctured plane?
 
Okay so, first fact is, two curves are homotopic in the punctured plane iff they have the same winding number, defined as $\int_{\gamma} \omega_0$ where $\omega_0 = \frac{-ydx + xdy}{x^2 + y^2}$ is the angle form
 
hi @MatheinBoulomenos
 
Actually I'm lazy to type
 
what is a closed 1-form?
 
9:47 AM
One whose exterior derivative is 0 (1-form, here just in the plane, is given by $adx + bdy$)
 
I don't see why they don't just work on $S^1$ if their argument is just $S^1$ anyway
 
Well, we appeal to $S^1$ to essentially get that the fundamental group is $\mathbb{Z}$, but then we hadn't talked about forms on the circle
 
so if I'm understanding correctly, there should be a closed 2-form that is not exact on $\Bbb R^3 \setminus \{0\}$?
 
Yeah, I think $H^2_{dR}(S^2) = \mathbb{R}$ (based on a conjecture of mine that if your integral cohomology is $\mathbb{Z}^n$, then your real cohomology is $\mathbb{R}^n$ (and probably $k^n$ for char 0 fields? idk))
 
10:37 AM
I need some help with the completing the task given in the last sentence of this old exam question.

"The system
$$
x+y+z&=6 \\
x^2+y^2+z^2&=14
$$
is satisfied at the point $(1,2,3)$. Show that $x$ and $y$ can be solved in a neighborhood of $(1,2,3)$ as a function of $z$. Calculate also $x'(3)$ and $y'(3)$, where $x$ and $y$ are regarded as functions of $z$.".

Any clues?
The formatting was screwed up. Sorry!
 
$$\begin{cases} x+y+z&=6 \\ x^2+y^2+z^2&=14 \end{cases}$$
 
Thanks
I've tried to solve the equations for $x$ and $y$ as a function of $z$ respectively, but I don't succeed. I get mixed terms.
 
I'm probably picking on words now
but if you just need to show that $x$ and $y$ can be solved
without explicitly calculating the inverse
then you could just cite the inverse function theorem (theorem 3.1)
 
I have shown that. I just need to complete the instruction "Calculate also $x′(3)$ and $y′(3)$, where $x$ and $y$ are regarded as functions of $z$.".
 
then I believe you can use implicit derivative
well firstly you would need to find $x$ and $y$ first
given $z=3$
 
10:44 AM
That's my problem.
 
we have $$\begin{cases} x+y&=3 \\ x^2+y^2&=5 \end{cases}$$
try squaring the first equation
wait no
we are given the point $(1,2,3)$
so $x=1$ and $y=2$
 
Yeah
 
so we're done?
 
We need the deriative.
 
use implicit differentiation, as I said
$$\begin{cases} x+y+z&=6 \\ x^2+y^2+z^2&=14 \end{cases}$$ becomes $$\begin{cases} x'+y'+1&=0 \\ 2xx'+2yy'+2z&=0 \end{cases}$$ becomes $$\begin{cases} x'+y'+1&=0 \\ 2x'+4y'+6&=0 \end{cases}$$
 
11:23 AM
Okay. I'm wondering if there's any way to check the veracity of the result.
 
38 mins ago, by Leaky Nun
try squaring the first equation
 
I meant for the derivatives.
 
if you solve the equation then you can differentiate it by hand
$$\begin{cases} x+y &= 6-z \\ x^2+y^2 &= 14-z^2 \end{cases}$$
 
Okay. So that's another way to do it?
 
(1)^2 - (2): $$2xy = (6-z)^2 - (14-z^2) = 22-12z+2z^2$$
well that's the ugly way to do it
 
11:26 AM
I feel.
 
(2)-(3): $$(x-y)^2 = (14-z^2) - (22-12z+2z^2) = -3z^2 + 12z - 8$$
we're in a neighbourhood of $(1,2,3)$, so $y>x$
$$y-x = \sqrt{-3z^2+12z-8} = \sqrt{-3(z-2)^2 + 4}$$
((1)+(4))/2: $$y = 3-0.5z + 0.5\sqrt{-3(z-2)^2+4}$$
((1)-(4))/2: $$x = 3-0.5z - 0.5\sqrt{-3(z-2)^2+4}$$
valid when $4-3(z-2)^2 \ge 0$, i.e. $(z-2)^2 \le \frac 4 3$, so $|z-2| \le \sqrt{\frac43}$
 
Thanks!
 
12:00 PM
not A is so hopelessly huge that don't try to plot it
While all anti A are not A, not all not A are anti A
 
how do you have so much time to help in here and stay caught up with your courses at uni?
 
is that question to leaky?
 
you also
 
well uh, I put most of my work in scripts, and I am mostly lurking really, doing various things in the background
Compact sets are super useful to confine infinity into a limited space so you can study it
Ezekiel is a goat with
an infinite horn, but
he thinks of himself
as mostly a zebra.
infinite chicken is so scary...
Infinity physics
The world of infinity is ruled by things which are themselves inexhaustible sources and rearrangement.
As long you still have countable things intact, you are able to fix all problems.
But seriously hammerwood would be better off to just brush his teeth with a countable brush, because having to pull your teeth around in the mouth with rubber bands every so often is so painful just from the description of it
one of the reasons why you may not want a world ruled by infinty: Crowding is VERY COMMON
Hyperbolic spaces are very compact containers for infinite objects
Ok, now try an uncountable farm :D
 
1:01 PM
 
1:29 PM
@Secret how does putting most of your work in scripts save you time?
 
Because all you need to do is to execute the script, and some hours later check the results. Thus you ca do other things while waiting
 
doesn't writing the script take much effort?
 
I done most of that hard work in my 1st year
 
what do you mean?
 
All that job submission scripts, error checking, data collection etc. is coded back in 1st year
 
1:42 PM
what about learning the chemistry
 
During the beginning of the year, I read some literature, otherwise I am analysing the data in recent days
 
so you're an applied chemist?
 
2:01 PM
Isn't it amazing how symbols so small can tell stories so great such as that of science?
 
example please
 
2:26 PM
I am a computational chemist
(though I do have lab experience back before PhD)
 
I see.
Sorta like ACM, you're always on the computer.
 
2:48 PM
ᕦ(ò_óˇ)ᕤ ᕦ(ò_óˇ)ᕤ ᕦ(ò_óˇ)ᕤ👎👎😡
@user2646 there is a huge difference. Secret is a nice guy and ACM not
 
3:28 PM
@LeakyNun I am not convinced. I think you can have that phenomenon with Laurent series that go infinitely negative, even with e^{1/z}: that has zeroes at -i/(2 pi n).
Remember great Picard says values will accumulate at an essential singularity.
 
3:54 PM
@MikeMiller yes, but most of them have accumulated zeroes
a bi-infinite function that misses zero would be exp(exp(z)+exp(1/z))
and I have yet to figure its argument principle
 
@LeakyNun I gave an example of a function with accumulating zeroes and a Laurent series, yes?
 
Complex analysis is supposed to be beautiful and work well, stop this atrocities!
 
4:12 PM
@AlessandroCodenotti If f and g are meromorphic in the entire complex plane, let $f_w$ be the set of points for which $f(z) = w$. If $f_w = g_w$ for five values $w$ (possibly including $\infty$), then $f = g$
 
That's some high level of arcane magic, let me think about it
 
4:30 PM
@MikeMiller oh sorry i wasn't thinking
but how does e^(1/z) have any zeroes?
 
Hm I have no idea why this holds
I tried to see what happens if we use entire functions for simplicity and got nowhere
 
more importantly, where did you get 5 from
 
@LeakyNun Oops, subtract 1 :)
 
still it's oddly specific
 
4:50 PM
I meant from e^(1/z)
The theorem is definitely about 5values
Show that e^z and e^(-z) share the values 0, infty, 1, -1
 
vzn
@Secret where are you getting those screenshots? the (schwartz) link on the page seems to be wrong it points to bookstore.ams.org/chel-366-h a different book by Artin-Tate...?!?
 
5:19 PM
@MikeMiller oO
I didn't know that theorem
 
5:35 PM
Hi! I discovered that in Math.SE the zoom trigger of MathJax is a bit ugly (Prob. 1). Also there is a problem with some equations (Prob. 2). Please see the image below:
I think the two problems are in fact one, but I don't know how change the configuration to restore back the old style. Any ideas? Where can I report this issue or how can I solve it?
 
@mercio @LeakyNun it's from the field of value-distribution theory, and requires some heavy lifting to set up
 
:o
 
o:
 
ö
 
Hello my friends
 
5:48 PM
@MikeMiller do you have like a link or a book?
 
I learned this from a book by Lee Rubel but I think there are better sources
Here are good notes
it should follow from the second main theorem but I am having trouble recalling the details
 

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