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9:01 PM
hi @ted
 
Oh, I've walked into a punnery.
Hi @Alessandro :)
 
@TedShifrin how is it to be a cult?
 
Hail Shifrin, full of geometries!
 
A cult?
I take it I've missed something.
 
look at the starred posts
 
9:02 PM
LOL, oy vey.
5
 
16 hours ago, by Semiclassical
@Brody "Dammit, I leave you guys alone for a few hours and suddenly there's a cult."
 
Maybe I should beat a hasty retreat, never to return.
 
@TedShifrin Somehow I knew that's exactly the phrase you'd say.
 
If real world experience is any sign, that'll just make it worse.
 
Balarka, DogAteMy will understand :P
 
9:04 PM
I was expecting more eye rolls
 
I'm still proud of Shifrinity as a name :P
 
lol as you should be
 
@Alessandro in fact Tedillion eyerolls
 
OK, this is embarrassing.
Where's math?
 
I could spam the channel with some ugly number theory crap I'm working on?
 
9:06 PM
Non-math thing, but I got to see a few sun dogs this morning
 
The only "math" I could do right now is ranting about numerical analysis
 
Given a set of independently uniformly random natural numbers less than N, what is the probability that every element of the set has at least one prime factor that is unique to the set?
 
Are those related to prairie dogs or to dogs in a blanket?
 
my guess is corndogs
 
@TedShifrin Yes.
 
9:08 PM
Alessandro, you don't ordinarily rant.
 
Touche, @Mick
 
I can be a Touche-bag sometimes
 
So, Balarka, I'm assuming you passed your exams and have moved on to the next chapter of life.
 
@MickLH: That's Touché !
oh, never heard of those, @Semiclassic.
 
9:09 PM
It's really cold here right now and there weren't too many clouds at sunrise, so they were pretty visible
 
@Semiclassical Oh no, exactly what I was afraid of. but I'm still pedantically correct because I chose "Yes" instead of "No" because "Yes" is safe on the technicality that it's related by being in the set of things humans think about
 
@MickLH you want the probability that the GCD of all the numbers in the set is prime?
 
I am pretty sure I passed, @Ted. In any case this is a baby-exam that doesn't really matter.
 
I do when my numerical analysis book looks like a series of examples with ad hoc theorems provided with no proof nor motivation whatsoever :(
 
But I have moved on, yep
 
9:10 PM
It's really scary for an atheist to find himself likened to the Jehovah's Witnesses :(
Good, @Balarka.
 
@Sophie I think something like this? Heuristically I think I could take the GCD of any element with the set, versus the product of the other elements of the set, and if that result is greater than $1$ for every element, then the set satisfies the property I'm interested in.
 
results are yet to come but I hear that I aced at chemistry. I am not sure if I should take that as a compliment or an insult.
but w/e
 
No wait, I butchered that really bad, ouch.
 
There's nothing wrong with chemistry, @Balarka :P
 
meh bleh
 
9:12 PM
Hi @TedShifrin
 
Hi Ali
 
@MickLH I think this is equivalent to the GCD of all the numbers in the set being greater than 1
 
@AkivaWeinberger you brought me in here for that?
 
which might be an interesting question to ask, unless I'm missing something
 
Boo @MikeM
 
9:13 PM
An instance of a sun dog:
 
Well I believe I've severely screwed up that last thing I said, I should have said if the GCD is less than the element, not greater than $1$
 
Those are neat, @SemiC
 
Is there a concept of "linear independence" for polynomials? I'm trying to state a question like "if you have a system of n "linearly independent" polynomials in n variables then there is a finite number of solutions"
 
There's a few more of those on this page here: blogs.mprnews.org/statewide/2013/12/sun-dogs
From Winter 2013 in Minnesota
 
nice, I didn't even know those were a thing @semi
 
9:14 PM
@Sophie You want algebraic independence.
 
That's just a lens effect is it not ? @Semi
 
@astyx Heh, no
I saw them with my bare eyes this morning
It's due to refraction off of ice crystals in the air
 
Lucky you
Where exactly, if I may ask ?
 
Minneapolis
 
Minnesota, but only SemiC can give you the lat/long
 
9:15 PM
@MikeMiller if I use this definition, then is that statement true?
 
@Sophie Ok that is almost what I'm looking for, but it's "inverted" in a way. Like it's checking for a related condition but not the same one.
 
@Sophie Yes.
But good luck checking algebraic independence.
 
I don't remember the lattitude and longitude of where I live
 
It's 5 degrees F / -15 degrees C here, so it's definitely the right conditions for sun dogs
 
It's 23 N 88.2 E off the top of my head IIRC
 
9:17 PM
lol
 
Why "dogs" though ?
 
and, hey, it'll be about -9 F tomorrow morning, so that'll be even better!
(ugh)
 
@MikeMiller and is the number of solutions the product of the degrees?
 
@Astyx No one really knows. Wikipedia has some discussion of that here: en.wikipedia.org/wiki/Sun_dogs#Etymology
 
The french term for this sounds much more poetic :p
"Parhélie"
 
9:19 PM
heh
 
oops, 22.57 N. Tropic of cancer is way above.
 
@Sophie You would have to count multiplicity, and it's not obvious to me what that means for more than one polynomial.
(You also need to projectivize.)
 
I'm going to pretend I know what that means
 
Parelio in Italian @Astyx I wonder which atmospheric conditions are needed to see them, I was well below 0°C for months in China and never saw anything like this
 
Great.
 
9:20 PM
We have great projections
 
@SemiC We should be writing
 
I should keep an eye out tonight: if the moon is clearly visible, it's cold enough that we should see a good 22 degree halo
 
Neither have I. On the other hand I don't tend to look directly at the sun so that might be why
 
Grading, in my case
 
Are moon dogs also a thing ?
 
9:22 PM
You should probablt be writing too
 
@Sophie consider the space of all lines going through a point in $\Bbb R^2$
 
Ok @Sophie here's a more careful description: Given a vector ${v}$ of natural numbers less than $N$, what is the probability that the following statement holds?
$$\forall v_i \in v : \gcd\left(v_i, \prod_{x\,\in\,{v}\setminus\{v_i\}}{x}\right)\neq v_i$$
 
I need to grade my last set of lab reports so I can submit lab grades
 
Google seems to say they are
 
For the most part it looks like $\Bbb R$
 
9:23 PM
Yep, moon dogs are a thing
 
You can think of the lines as projecting onto $\Bbb R$
 
I need to do laundry and clean my room
 
What I said about a 22 degree halo is related to that
 
Most of these lines project onto $\Bbb R$
 
and also write
 
9:24 PM
infact we have a 1-to-1 correspondance of sorts
 
too much work
 
only 1 line gives us a problem
 
I want to do math but I also want to be lazy
 
I also need to work out a rubric for the problem I'll be grading on the final
 
the line that is parallel to the real line
 
9:24 PM
@AliCaglayan what does it mean to look like $\mathbb{R}$?
 
@Sophie we can come up with an explicit example
 
@Ali I think this would be hard for anyone who doesn't already know projective space to understand.
 
Say we are interested in the lines passing through (0, 1)
 
@Semiclassical what topic ?
 
@BalarkaSen pretend to do the former by watching the hydra game
 
9:25 PM
@Balarka Don't be like me
 
Mine is on induction (Faraday's law)
 
@MikeM I don't want to write, no
 
and then where those lines pass in $\Bbb R$ are our correspodances
 
I mean, don't do nothing while feeling bad about it
 
@MickLH ok so let $p=\prod_v$ we want that $\gcd(v_i,p)>v_i^2$ for all $i$
 
9:26 PM
Ahaha, good point
 
@Sophie here a drew a picture
we have all these lines passing through (1, 0)
they all correspond to a specific number in R
 
@AliCaglayan I get that you can parametrize, say the set of lines in $\mathbb{R}^2$ in a one to one correspondence with $\mathbb{R}$, say the lines going through the origin can be written as $y=ax$ then you variate $a\in\mathbb{R}$
 
for every R we can find a line
 
@Semiclassical how much heads have autoplay cut so far?
 
@Sophie that restricts you though
how to you talk about the line x = 0
 
9:29 PM
I just chose the origin because I'm lazy, you can do that for any other point
 
Hi friends
 
we have this extra point
but yes your correspondance is the right idea
This extra point is what make the real projective line $\Bbb {RP}^1$ different to $\Bbb R$
 
Does anyone know a straightforward way to tell if a contour is simple?
 
RP^1
 
Mine winds itself into a tight loop getting arbitrarily tight as it goes near the origin
 
9:31 PM
@GFauxPas googling simple contour just gives make up tutorials
 
$Ali I know :(
 
@Sophie do you see what I am talking about?
For each $a$ you describe we have a line
 
@MickLH no clue, I bet you can find asymptotic estimates as $N\to\infty$ but I don't think there exists a closed form for the probability that a randomly (uniformly, etc...) has the property you described
 
so there is some copy of $\Bbb R$ sitting inside $\Bbb {RP}^1$ which is the space of all lines going through a point
 
So my question is equivalent to how can I check if a contour's winding number is 0
 
9:34 PM
then when you get the limit as $a\to\infty$ you "approach" the line $x=0$, but what happens for that exact line?
 
Or does a winding number only make sense if the curve is closrd
 
@GFauxPas construct some homotopy
 
@Sophie I could get by just with proving when it's $> 2/3$, if you have any suggestions on how to attack it. So far I've just been able to write a Euler product for it, but it was flawed so I should try that again I guess actually.
 
that shrinks your contour to a point
 
@MickLH this is definitely false for sufficiently large $N$
 
9:35 PM
@Sophie the exact line sits in $\Bbb {RP}^1$ which is the space of all lines
 
@Sophie The deal is such lines are parametrized over $\Bbb R \cup \{\infty\}$ instead of $\Bbb R$, where $\infty$ is the extra point. You can also think of this as being a circle (which is what you get when you glue the two "ends" of the real line to a point at infinity)
 
but we also have a copy of R in RP1
 
and indeed, RP^1 is just a circle.
 
I have a homotopy that moves it into a line segment but that assumes the homotopy doesn't carry the contour across a branch cut
 
@GFauxPas is there an exact set up?
 
9:37 PM
@Sophie The trouble is, I'm interested in fairly "small" $N$
Though I just got a crazy idea... if I restrict $N$ to a primorial power, then maybe the partial product converges?
 
I posted the graph earlier today but I'm not on my computer, let me see if I can find it
 
and then I could just directly evaluate the definition
 
I'm very confused. If you post it as a regular question I'll think about it
 
@AliCaglayan what's the motivation for thinking of this as a circle?
 
9:39 PM
So @ali
 
@GFauxPas still the one from yesterday?
 
I don't remember what i asked yesterday
 
I know a bit about harmonic bundles and such
 
the graph i mean^^
 
@GFauxPas ah wait I thought the contour was jordan
 
9:40 PM
@Ali, if the curve is simple, I can choose a branch cut to have the curve be analytic on a simply connected set
 
yeah then I am useless
sorry
@Sophie its like we have a point at infinity if you will
 
It's .... half a jordan curve?
Why does it need to be jordan
 
@GFauxPas it doesn't need to be, but the point still stand about me being useless
 
We can still be friends :)
 
@TedShifrin is here he knows the answer
 
9:42 PM
He's sick of my obsession with contours
 
@GFauxPas: Did you see my ping yesterday? You're totally making a hash of that question. The path is just the portion of the real axis from $t=\epsilon$ to $t=T$.
 
Yes now I'm not talking about integrals anymorw
 
@Sophie lets look at your line $y=ax$
 
I'm asking if there's a branch cut that allows that curve to be analytic on a simply connected set
 
"Curves" are not analytic.
 
9:43 PM
Not analytic, smooth
I meant
 
I'm bad at geometry. I think I'm going back to number theory
 
@Sophie Feel you.
 
@TedShifrin I learned from you making a hash of sth, and being bogged down
 
@GFauxPas: Nothing to do with branch cuts.
 
i mean the meaning of the words, of course ;)
 
9:45 PM
Good, @Null. God forbid you should learn any mathematics from me :D
 
If a curve crosses a branch cut surely it's not smooth?
 
You're confusing a branch cut of a function with a continuous curve. Totally not involved.
There's a question about whether the integral of your function over the curve makes sense.
 
So my contour is well defined even if it's not simple?
 
hey @TedShifrin
I aced my geometry exam
 
Good, Karim :)
@GFauxPas: In that problem you keep drawing, the contour is just a segment of the real axis.
 
9:47 PM
there was 14 questions in the exam prof told us to solve 5 questions
I end up solving all 14
haha
 
Seriously, all 14, Karim? That's a LOT of questions.
 
yeah @TedShifrin I solved all of them.
 
Correctly? :D
 
yeah haha
 
Yeah I had that problem
The correct part
 
9:48 PM
That's not so unusual, Ali.
 
Hello quick question, i am learning some vectors and i am calculating the lengths of i and j vectors, im given the bearing of 240 degrees and a magnitude for vector v of 5.2ms

So when calculating the lengths of my triangle for example length a , why do i have to do sin theta = a / -5.2

How can you have a negative for a length of a triangle =/
 
Forgot what my integrand is, my contour is what I'm going to ontegrate over, for example
 
@TedShifrin I am taking today off then from tomorrow I will go over how I should be next semester. Since, this semester I end up like studying one subject and neglecting others.
 
@WDUK what do you mean by length of a triangle?
 
You can't, @WDUK, but in trig you work with signed lengths (when the angle is obtuse, for example).
 
9:49 PM
for my vector im making a right angled triangle from it to calculate the lengths of i and j vectors
 
@WDUK are you sure you aren't mixing up degrees and radians?
 
its definately degrees
 
@GFauxPas: I still think you're totally confused.
 
@TedShifrin we will use atiyah macdonald for ring theory.
 
240 degrees is the same as -120 degrees
 
9:50 PM
for next semester
 
It's a very terse book, Karim, but it's got good exercises.
 
i know ali my question is why am i doing sin a = opposite / - hypotenuse though
 
You may want to look at Eisenbud's book, too.
 
okay cool. @TedShifrin I will check it out. I want to use many sources. I don't want to say oh I understand this in class and not study.
 
@Ted why do you think I'm confused? I have a contour on the plane, parameterized. Why shouldn't I be allowed to integrate something on it
 
9:52 PM
A-M was too hard for me but I should have another stab at algebra next time I feel like it
 
one of my friends told me A-M is really awesome. But, I will look at Eisenbud and also allufi it has a lot of nice stuff about ring theory.
I will take a break from math today I am tired
 
Take a break for several days. Go hiking. Go out to a nice dinner with your girlfriend.
 
yeah
 
@GFauxPas: I just want to be sure you are doing what you're supposed to be doing.
 
that is what I will do.
 
9:55 PM
My idea of a break is very different, but I don't recommend it.
 
I never found any algebra book I liked very much.
But I'm not an algebraist.
 
@ZachHauk nice answer
 
(So why am I only doing algebra lately?)
 
What algebra are you doing
 
Infty categories
 
9:56 PM
Even I did some algebra in a few papers, @MikeM. Who'd have dreamed?
 
Sanity check: I cannot parametrize the circle with a single map because that would give a diffeomorphism from an open subset of $\mathbb{R}$ to the circle so in particular an homeomorphism from a noncompact space to a compact one @balarka
 
@TedShifrin I'm just trying to understand contours. My book doesn't deal with this kind of contour
 
Correct, @Alessandro. True for any compact manifold.
 
What Ted said :)
 
@Null which answer? :P
hey @TedShifrin :)
 
9:57 PM
The point is, @GFauxPas, that for most integrals that are interesting in complex analysis, you can deform contours homotopically, provided you don't go through singularities of the function. So why mess with a ridiculous contour?
 
Hi @Zach.
 
if two curves on the complex plane are homotopic over a region in which $f$ is holomorphic, then their contour integrals are equal
precisely because they encircle the same singularities
hmm, should i say "homotopically equivalent"
 
Nah, homotopic is the right word
 
I'm a bit confused by this definition requiring smooth manifolds to be locally diffeomorphic to an open subset of $\mathbb{R}^k$ because I would have expected an open ball instead
 
9:58 PM
Mumbles "closed $1$-form" ... mumbles.
 
@Alessandro You can check that doing that with open ball changes nothing
 
have you guys watched Miss Peregrine home for Peculiar kids ?
I am considering watching it
 
They're equivalent, @Alessandro.
 
Ah, interesting
 
@Adeek watch it and tell us if its good
 
9:59 PM
yeah @AliCaglayan
 
But make sure to do open ball instead of all of $\Bbb R^k$. Of course, they're equivalent for smooth manifolds, but dangerous if you do something more general!
 

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