Mathematics - 2016-12-14 (page 4 of 5)

Alessandro

21:01

hi @ted

Ted Shifrin

Oh, I've walked into a punnery.

Hi @Alessandro :)

Null

@TedShifrin how is it to be a cult?

Semiclassical

Hail Shifrin, full of geometries!

Ted Shifrin

A cult?

I take it I've missed something.

Alessandro

look at the starred posts

Ted Shifrin

21:02

LOL, oy vey.

5

Semiclassical

16 hours ago, by Semiclassical

@Brody "Dammit, I leave you guys alone for a few hours and suddenly there's a cult."

Ted Shifrin

Maybe I should beat a hasty retreat, never to return.

Balarka Sen

@TedShifrin Somehow I knew that's exactly the phrase you'd say.

Semiclassical

If real world experience is any sign, that'll just make it worse.

Ted Shifrin

Balarka, DogAteMy will understand :P

Alessandro

21:04

I was expecting more eye rolls

Semiclassical

I'm still proud of Shifrinity as a name :P

MickLH

lol as you should be

Null

@Alessandro in fact Tedillion eyerolls

Ted Shifrin

OK, this is embarrassing.

Where's math?

MickLH

I could spam the channel with some ugly number theory crap I'm working on?

Semiclassical

21:06

Non-math thing, but I got to see a few sun dogs this morning

Alessandro

The only "math" I could do right now is ranting about numerical analysis

MickLH

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?

Ted Shifrin

Are those related to prairie dogs or to dogs in a blanket?

Null

my guess is corndogs

MickLH

@TedShifrin Yes.

Ted Shifrin

21:08

Alessandro, you don't ordinarily rant.

Balarka Sen

Touche, @Mick

MickLH

I can be a Touche-bag sometimes

Ted Shifrin

So, Balarka, I'm assuming you passed your exams and have moved on to the next chapter of life.

Semiclassical

nah, these: en.wikipedia.org/wiki/Sun_dogs

Ted Shifrin

@MickLH: That's Touché !

oh, never heard of those, @Semiclassic.

Semiclassical

21:09

It's really cold here right now and there weren't too many clouds at sunrise, so they were pretty visible

MickLH

@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

Sophie

@MickLH you want the probability that the GCD of all the numbers in the set is prime?

Balarka Sen

I am pretty sure I passed, @Ted. In any case this is a baby-exam that doesn't really matter.

Alessandro

I do when my numerical analysis book looks like a series of examples with ad hoc theorems provided with no proof nor motivation whatsoever :(

Balarka Sen

But I have moved on, yep

Ted Shifrin

21:10

It's really scary for an atheist to find himself likened to the Jehovah's Witnesses :(

Good, @Balarka.

MickLH

@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.

Balarka Sen

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

MickLH

No wait, I butchered that really bad, ouch.

Ted Shifrin

There's nothing wrong with chemistry, @Balarka :P

Balarka Sen

meh bleh

Ali Caglayan

21:12

Hi @TedShifrin

Ted Shifrin

Hi Ali

Sophie

@MickLH I think this is equivalent to the GCD of all the numbers in the set being greater than 1

Mike Miller

@AkivaWeinberger you brought me in here for that?

Sophie

which might be an interesting question to ask, unless I'm missing something

Ted Shifrin

Boo @MikeM

Semiclassical

21:13

An instance of a sun dog:

goo.gl/images/NwydGs

MickLH

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 $1$

Balarka Sen

Those are neat, @SemiC

Sophie

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"

Semiclassical

There's a few more of those on this page here: blogs.mprnews.org/statewide/2013/12/sun-dogs

From Winter 2013 in Minnesota

Alessandro

nice, I didn't even know those were a thing @semi

Mike Miller

21:14

@Sophie You want algebraic independence.

Astyx

That's just a lens effect is it not ? @Semi

Semiclassical

@astyx Heh, no

I saw them with my bare eyes this morning

It's due to refraction off of ice crystals in the air

Astyx

Lucky you

Where exactly, if I may ask ?

Semiclassical

Minneapolis

Mike Miller

Minnesota, but only SemiC can give you the lat/long

Sophie

21:15

@MikeMiller if I use this definition, then is that statement true?

MickLH

@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.

Mike Miller

@Sophie Yes.

But good luck checking algebraic independence.

Balarka Sen

I don't remember the lattitude and longitude of where I live

Semiclassical

It's 5 degrees F / -15 degrees C here, so it's definitely the right conditions for sun dogs

Balarka Sen

It's 23 N 88.2 E off the top of my head IIRC

Mike Miller

21:17

lol

Astyx

Why "dogs" though ?

Semiclassical

and, hey, it'll be about -9 F tomorrow morning, so that'll be even better!

(ugh)

Sophie

@MikeMiller and is the number of solutions the product of the degrees?

Semiclassical

@Astyx No one really knows. Wikipedia has some discussion of that here: en.wikipedia.org/wiki/Sun_dogs#Etymology

Astyx

The french term for this sounds much more poetic :p

"Parhélie"

Semiclassical

21:19

heh

Balarka Sen

oops, 22.57 N. Tropic of cancer is way above.

Mike Miller

@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.)

Sophie

I'm going to pretend I know what that means

Alessandro

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

Mike Miller

Great.

Ali Caglayan

21:20

We have great projections

Mike Miller

@SemiC We should be writing

Semiclassical

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

Astyx

Neither have I. On the other hand I don't tend to look directly at the sun so that might be why

Semiclassical

Grading, in my case

Astyx

Are moon dogs also a thing ?

Mike Miller

21:22

You should probablt be writing too

Ali Caglayan

@Sophie consider the space of all lines going through a point in $\Bbb R^2$

MickLH

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?
$$\forall v_i \in v : \gcd\left(v_i, \prod_{x\,\in\,{v}\setminus\{v_i\}}{x}\right)\neq v_i$$

Semiclassical

I need to grade my last set of lab reports so I can submit lab grades

Astyx

Google seems to say they are

Ali Caglayan

For the most part it looks like $\Bbb R$

Semiclassical

21:23

Yep, moon dogs are a thing

Ali Caglayan

You can think of the lines as projecting onto $\Bbb R$

Mike Miller

I need to do laundry and clean my room

Semiclassical

What I said about a 22 degree halo is related to that

Ali Caglayan

Most of these lines project onto $\Bbb R$

Mike Miller

and also write

Ali Caglayan

21:24

infact we have a 1-to-1 correspondance of sorts

Mike Miller

too much work

Ali Caglayan

only 1 line gives us a problem

Balarka Sen

I want to do math but I also want to be lazy

Semiclassical

I also need to work out a rubric for the problem I'll be grading on the final

Ali Caglayan

the line that is parallel to the real line

Sophie

21:24

@AliCaglayan what does it mean to look like $\mathbb{R}$?

Ali Caglayan

@Sophie we can come up with an explicit example

Mike Miller

@Ali I think this would be hard for anyone who doesn't already know projective space to understand.

Ali Caglayan

Say we are interested in the lines passing through (0, 1)

Astyx

@Semiclassical what topic ?

Semiclassical

@BalarkaSen pretend to do the former by watching the hydra game

Mike Miller

21:25

@Balarka Don't be like me

Semiclassical

Mine is on induction (Faraday's law)

Balarka Sen

@MikeM I don't want to write, no

Ali Caglayan

and then where those lines pass in $\Bbb R$ are our correspodances

Mike Miller

I mean, don't do nothing while feeling bad about it

Sophie

@MickLH ok so let $p=\prod_v$ we want that $\gcd(v_i,p)>v_i^2$ for all $i$

Balarka Sen

21:26

Ahaha, good point

Ali Caglayan

@Sophie here a drew a picture

we have all these lines passing through (1, 0)

they all correspond to a specific number in R

Sophie

@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 $a\in\mathbb{R}$

Ali Caglayan

for every R we can find a line

Balarka Sen

@Semiclassical how much heads have autoplay cut so far?

Ali Caglayan

@Sophie that restricts you though

how to you talk about the line x = 0

Sophie

21:29

I just chose the origin because I'm lazy, you can do that for any other point

GFauxPas

Hi friends

Ali Caglayan

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 $\Bbb R$

GFauxPas

Does anyone know a straightforward way to tell if a contour is simple?

Balarka Sen

RP^1

GFauxPas

Mine winds itself into a tight loop getting arbitrarily tight as it goes near the origin

Ali Caglayan

21:31

@GFauxPas googling simple contour just gives make up tutorials

GFauxPas

$Ali I know :(

Ali Caglayan

@Sophie do you see what I am talking about?

For each $a$ you describe we have a line

Sophie

@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

Ali Caglayan

so there is some copy of $\Bbb R$ sitting inside $\Bbb {RP}^1$ which is the space of all lines going through a point

GFauxPas

So my question is equivalent to how can I check if a contour's winding number is 0

Sophie

21:34

then when you get the limit as $a\to\infty$ you "approach" the line $x=0$, but what happens for that exact line?

GFauxPas

Or does a winding number only make sense if the curve is closrd

Ali Caglayan

@GFauxPas construct some homotopy

MickLH

@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.

Ali Caglayan

that shrinks your contour to a point

Sophie

@MickLH this is definitely false for sufficiently large $N$

Ali Caglayan

21:35

@Sophie the exact line sits in $\Bbb {RP}^1$ which is the space of all lines

Balarka Sen

@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)

Ali Caglayan

but we also have a copy of R in RP1

Balarka Sen

and indeed, RP^1 is just a circle.

GFauxPas

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

Ali Caglayan

@GFauxPas is there an exact set up?

MickLH

21:37

@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?

GFauxPas

I posted the graph earlier today but I'm not on my computer, let me see if I can find it

MickLH

and then I could just directly evaluate the definition

Sophie

I'm very confused. If you post it as a regular question I'll think about it

GFauxPas

Sophie

@AliCaglayan what's the motivation for thinking of this as a circle?

GFauxPas

21:39

So @ali

Null

@GFauxPas still the one from yesterday?

GFauxPas

I don't remember what i asked yesterday

Sophie

I know a bit about harmonic bundles and such

Null

the graph i mean^^

Ali Caglayan

@GFauxPas ah wait I thought the contour was jordan

GFauxPas

21:40

@Ali, if the curve is simple, I can choose a branch cut to have the curve be analytic on a simply connected set

Ali Caglayan

yeah then I am useless

sorry

@Sophie its like we have a point at infinity if you will

GFauxPas

It's .... half a jordan curve?

Why does it need to be jordan

Ali Caglayan

@GFauxPas it doesn't need to be, but the point still stand about me being useless

GFauxPas

We can still be friends :)

Ali Caglayan

@TedShifrin is here he knows the answer

GFauxPas

21:42

He's sick of my obsession with contours

Ted Shifrin

@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 $t=T$.

GFauxPas

Yes now I'm not talking about integrals anymorw

Ali Caglayan

@Sophie lets look at your line $y=ax$

GFauxPas

I'm asking if there's a branch cut that allows that curve to be analytic on a simply connected set

Ted Shifrin

"Curves" are not analytic.

GFauxPas

21:43

Not analytic, smooth

I meant

Sophie

I'm bad at geometry. I think I'm going back to number theory

Krijn

@Sophie Feel you.

Null

@TedShifrin I learned from you making a hash of sth, and being bogged down

Ted Shifrin

@GFauxPas: Nothing to do with branch cuts.

Null

i mean the meaning of the words, of course ;)

Ted Shifrin

21:45

Good, @Null. God forbid you should learn any mathematics from me :D

GFauxPas

If a curve crosses a branch cut surely it's not smooth?

Ted Shifrin

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.

GFauxPas

So my contour is well defined even if it's not simple?

Adeek

hey @TedShifrin

I aced my geometry exam

Ted Shifrin

Good, Karim :)

@GFauxPas: In that problem you keep drawing, the contour is just a segment of the real axis.

Adeek

21:47

there was 14 questions in the exam prof told us to solve 5 questions

I end up solving all 14

haha

Ted Shifrin

Seriously, all 14, Karim? That's a LOT of questions.

Adeek

yeah @TedShifrin I solved all of them.

Ted Shifrin

Correctly? :D

Adeek

yeah haha

Ali Caglayan

Yeah I had that problem

The correct part

Ted Shifrin

21:48

That's not so unusual, Ali.

WDUK

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 =/

GFauxPas

Forgot what my integrand is, my contour is what I'm going to ontegrate over, for example

Adeek

@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.

Ali Caglayan

@WDUK what do you mean by length of a triangle?

Ted Shifrin

You can't, @WDUK, but in trig you work with signed lengths (when the angle is obtuse, for example).

WDUK

21:49

for my vector im making a right angled triangle from it to calculate the lengths of i and j vectors

Sophie

@WDUK are you sure you aren't mixing up degrees and radians?

WDUK

its definately degrees

Ted Shifrin

@GFauxPas: I still think you're totally confused.

Adeek

@TedShifrin we will use atiyah macdonald for ring theory.

Ali Caglayan

240 degrees is the same as -120 degrees

Adeek

21:50

for next semester

Ted Shifrin

It's a very terse book, Karim, but it's got good exercises.

WDUK

i know ali my question is why am i doing sin a = opposite / - hypotenuse though

Ted Shifrin

You may want to look at Eisenbud's book, too.

Adeek

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.

GFauxPas

@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

Balarka Sen

21:52

A-M was too hard for me but I should have another stab at algebra next time I feel like it

Adeek

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

Ted Shifrin

Take a break for several days. Go hiking. Go out to a nice dinner with your girlfriend.

Adeek

yeah

Ted Shifrin

@GFauxPas: I just want to be sure you are doing what you're supposed to be doing.

Adeek

that is what I will do.

Balarka Sen

21:55

My idea of a break is very different, but I don't recommend it.

Mike Miller

I never found any algebra book I liked very much.

But I'm not an algebraist.

Null

@ZachHauk nice answer

Mike Miller

(So why am I only doing algebra lately?)

Ali Caglayan

What algebra are you doing

Balarka Sen

Infty categories

Ted Shifrin

21:56

Even I did some algebra in a few papers, @MikeM. Who'd have dreamed?

Alessandro

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

GFauxPas

@TedShifrin I'm just trying to understand contours. My book doesn't deal with this kind of contour

Ted Shifrin

Correct, @Alessandro. True for any compact manifold.

Balarka Sen

What Ted said :)

Zach Hauk

@Null which answer? :P

hey @TedShifrin :)

Ted Shifrin

21:57

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?

Null

@ZachHauk math.stackexchange.com/questions/2059094/…

Ted Shifrin

Hi @Zach.

Zach Hauk

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"

Balarka Sen

Nah, homotopic is the right word

Alessandro

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

Ted Shifrin

21:58

Mumbles "closed $1$-form" ... mumbles.

Balarka Sen

@Alessandro You can check that doing that with open ball changes nothing

Adeek

have you guys watched Miss Peregrine home for Peculiar kids ?

I am considering watching it

Ted Shifrin

They're equivalent, @Alessandro.

Alessandro

Ah, interesting

Ali Caglayan

@Adeek watch it and tell us if its good

Adeek

21:59

yeah @AliCaglayan

Mike Miller

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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$Mathematics$ Mathematics