Mathematics - 2017-02-17 (page 1 of 5)

Semiclassical

00:03

This is the image I have in mind, though I can't find a source for it:

(had to make that in paint, oh well)

Ted Shifrin

Yeah, sure.

Semiclassical

I know I've seen that picture done better, but my google-fu is failing me

Skeletor

Given a n distinct points, is there a way to generate a list of all possible betweeness relations between these n points regardless of whether they are true or not? So, just all possibilities.

Daminark

@Semi I like that, Google-fu

Also @Ted what do you think of reading courses?

Zach Hauk

is there a wild @Akiva around here

oh, and hey @Ted

Ted Shifrin

00:08

I have conducted many a reading course. Some were successful, others not.

hi @Zach

Zach Hauk

i need help with an olympiad geometry problem :(

inversion doesn't really help

Daminark

Hmm

Zach Hauk

apparently the book says it has something to do with cyclic quadrilaterals

i found 2 cyclic quadrilaterals but they dont really help me

Ted Shifrin

Oh, usually called concyclic or cocyclic or ....

Daminark

I'm not particularly likely to do a reading course, the main reason the guy I told you about does a lot of reading courses is because his interests are very niche

Ted Shifrin

00:09

I'm not usually clever enough for Olympiad, but go ahead and post it.

Zach Hauk

Ok

so we have quadrilateral $ABCD$

Ted Shifrin

@Daminark: I'm a big believer in socialized learning. So you learn better with people to talk to about the stuff.

Daminark

Yeah, talking things out has helped me a lot

Zach Hauk

and a point $O$ on $AB$, with circle $\Gamma$ centered at $O$ tangent to all of the other sides of the quadrilateral

Daminark

Part of the reason why I'm pretty excited for the bootcamp, after talking around it's apparently more student driven, we give presentations and all

Zach Hauk

00:11

prove that $BC+AD = AB$

Ted Shifrin

Now I have to figure out what all that says, @Zach.

Semiclassical

I'm going to make a Geogebra figure for it

Zach Hauk

yeah

i was going to do that but

i guess semiclassical will

@Semiclassical it should be of use to you to also construct circles centered at $A$ and $B$ intersecting $D$ and $C$ respectively, since the problem is actually asking to prove those tangent

Ted Shifrin

Start with AB and draw the circle. Then put in the other 3 sides.

Zach Hauk

(if they're tangent, then it must be on $AB$ because, well, thats the only line through both radii)

r9m

00:17

Any suggestions for good reference on Topological degree theory? (one that starts from scratch) :)

Skeletor

Does anyone know how to construct such a permutation? I know it's a permutation of the points I just don't know how to construct it for n > 3.

Semiclassical

I'd clean that up more and make it match the labels, but I can't be arsed.

I guess I should've put in the diagonals at least, though.

Oh, wait. BC, AD are the two sides adjacent to AB

Okay, no need for diagonals.

Ted Shifrin

ABCD are the vertices. O is the center.

Semiclassical

Yeah, I wasn't willing to edit the vertices to match.

Ted Shifrin

Interesting question. I'll have to ponder it, @Zach.

Zach Hauk

00:24

its from 1985 IMO

Semiclassical

There's an obvious special case, when AD=AO and BC=BO.

But special cases aren't solutions.

Zach Hauk

again, i think its about proving those circles tangent

Semiclassical

Yeah, I see what you mean.

Ted Shifrin

Proving what circles tangent?

Zach Hauk

the circles with centers A and B intersecting D and C respectively

since that would prove that the lengths add up

to the whole segment

Semiclassical

00:27

Like that.

Ted Shifrin

Aha.

Zach Hauk

@Semiclassical would you mind drawing in the points of tangency for $BC,CD,AD$?

Ted Shifrin

Well, you're way ahead of me ... And I'm going out to dinner with a friend, so I won't think 'til later. :)

Zach Hauk

because i found a cyclic relationship for those points

@TedShifrin alright well, enjoy your meal

Ted Shifrin

@Alessandro: You're turning into Balarka?

Alessandro Codenotti

00:35

Nope, I'm on holiday :P

Ted Shifrin

Having midnight drinks? :)

Alessandro Codenotti

Nah, I was just watching a movie

Adeek

hi @TedShifrin

Ted Shifrin

Hi, Karimm.

Adeek

Have next week holiday I can't wait to catch up with everything

@TedShifrin my topology prof likes me a lot

Have exam soon in commutative algebra have to prepare for it

skullpetrol

00:44

No more under graduate courses?

Adeek

@skullpetrol I am not undegrad

skullpetrol

I see.

Adeek

just a week holiday though

Alessandro Codenotti

Now I do have to choose between sleeping and thinking about an interesting question I saw on MSE earlier...

Semiclassical

had to to eat dinner, but lemme edit those in @ZachHauk

@ZachHauk Actually, this is weird. I was adjusting the quadrilateral slightly (as one can do in geogebra) and I'm finding that the circles don't remain tangent.

skullpetrol

00:51

Tangency is very sensitive.

Balarka Sen

@MikeMiller Ah, that's a good point

Ted Shifrin

oh, oh, @Balarka is breaking his sleep schedule still.

Balarka Sen

i woke up...

skullpetrol

Why?

Balarka Sen

dunno. just did

Not a great sleep but I was at least asleep

skullpetrol

00:53

Are you finished your uni course?

Balarka Sen

Can't parse that statement. What uni course?

Daminark

How much do you sleep on average @Balarka?

Ted Shifrin

Don't be parsimonious, then :P

Daminark

Hopefully on normal days it's at least 5-6 hours, right?

Semiclassical

@ZachHauk More importantly, I'm finding that the distances don't add up according to Geogebra.

Ted Shifrin

00:54

Guess again, @Daminark.

skullpetrol

University course. @BalarkaSen

Daminark

sighs

Balarka Sen

@Daminark Pretty much over normal (8-9) but my sleep cycle breaks up into bits and pieces all the time

Semiclassical

It's close, but not exact. So I be confuzed.

r9m

@Simple Hi! o/

Balarka Sen

00:55

Right now it's a single, connected cycle for half the days in the week and two horribly disconnected ones for the other half

Daminark

@Semiclassical Well do you know the de Rham cohomology of your space at least?

Semiclassical

lol

Simple

@r9m hi

Balarka Sen

@r9m Depends on what you mean by that.

Ted Shifrin

hides from @PVAL

skullpetrol

00:57

Hi pal @r9m

PVAL-inactive

Hi @ted

Daminark

Differential forms jokes never get old

Ted Shifrin

Hi @PVAL

Balarka Sen

Good night, @Ted

Ted Shifrin

Well, yeah, they do, @Daminark, but I do more of them than you.

Balarka Sen

00:58

@Alessandro What movie did you watch?

Daminark

Haha, that's fair, my pun game is still pretty elementary

Ted Shifrin

I can't imagine how you thought my comment was wrong, @PVAL :D

@Daminark: Yeah, you're barely past the 20 yard line.

Daminark

Growing slowly but steadily, at least now my friends spend over a third of their time sighing

PVAL-inactive

You thought mine was wrong first.

Ted Shifrin

LOL, @PVAL. I know.

I was just feeling sorry for the poor OP. Otherwise, I wouldn't have tortured you.

Simple

00:59

how do I find the density function of $X/Y$ where $X$ and $Y$ are independent

skullpetrol

Leave him alone he's inactive now :P

Ted Shifrin

I haven't gone back to look at the question, actually.

r9m

@BalarkaSen I shall say the wiki definition is all I am sure of at this point .. (we will start Brouwer's degree theory in following diff geometry class .. so I was thinking if I should give it a read from reference .. ) :)

PVAL-inactive

@Ted Well if you looked at what he was trying to do, it seemed like he was trying to show dA=-I intrinsically

r9m

@skullpetrol Hey :-) How are you?

Balarka Sen

01:00

@r9m Do you have any background on algebraic topology, or differential topology?

PVAL-inactive

Like he was trying to prove that it was true for any choice of charts

or at least thats what I got from it

Ted Shifrin

Well, if I take a chart at $p$ and "the obvious correlated chart" at $-p$, that makes good sense, @PVAL.

skullpetrol

Fine thanks @r9m how are you?

Ted Shifrin

But, anyhow ... :P Sometimes isomorphic things really are equal :P

r9m

@BalarkaSen well I studied bit of algebraic topo last sem .. (no diff topo background whatsoever .. )

Balarka Sen

01:01

Do you know what homology is?

r9m

@skullpetrol I am fine!

Adeek

@TedShifrin I love your later chapters btw

I am reading them in my spare time

r9m

@BalarkaSen nope :P sorry .. we only learnt stuff about $\pi_1$'s

PVAL-inactive

Well I might define the smooth manifold S^2 as the standard sphere, that wouldn't be the data I would consider it as.

Ted Shifrin

Thanks, I guess, Karim :)

Alessandro Codenotti

01:02

@BalarkaSen lost in translation, by Sofia Coppola

PVAL-inactive

If you defined it as a slightly morphed sphere, it wouldn't affect me mathematically

Ted Shifrin

@PVAL: I think that to have the antipodal map well-defined, we're stuck with the usual sphere. :P

Balarka Sen

@r9m Ahh, ok. I would have recommended you Hatcher's "Algebraic Topology", chapter 2. I think you should be able to dive into it if you want though... chapter 2 doesn't, technically speaking, need any of the stuff from the previous chapters as prereqs (but you say you know fund. group. so you shouldn't need a lot of that either)

Adeek

@BalarkaSen did you see this ? youtube.com/watch?v=TDY-4X8NJfA

Ted Shifrin

@Alessandro: How cool. She's the daughter of the famous director and vintner Francis Ford (whose wines I buy regularly).

Balarka Sen

01:04

Chapter 2 section 2 computes degrees of stuff.

Ted Shifrin

He has a whole line of wines named after her. :P

Balarka Sen

The actual geometric intuition really comes from differential topology though. Milnor's diff. top. book is a quick and dirty intro

PVAL-inactive

Well my tangent space definition doesn't depend on the existence of an embedding

Balarka Sen

@AlessandroCodenotti I have heard of that movie

Ted Shifrin

@PVAL: Of course, I understand that. But the mapping he's looking at very much does. So ... enough already.

r9m

01:06

@BalarkaSen 'kay thanks! :) I'll look into it .. (I only studies algebraic topology or whatever we had in our topology course last sem from Munkres .. never opened hatcher .. )

Ted Shifrin

Hi @r9m

Alessandro Codenotti

Oh, interesting, I didn't know about the wine

r9m

@TedShifrin G'morning professor :)

Ted Shifrin

It's actually very good wine, @Alessandro. I get a couple of cases a year (but it ain't cheap).

Balarka Sen

I really need to brace myself up and watch Videodrome sometime this week

Alessandro Codenotti

01:07

@BalarkaSen it's undoubtely a good movie, I quite liked it

Videodrome has been on my list of movies to watch for at least a year, but I never feel like actually watching it

Anyway I should sleep now, good night/day everyone!

Ted Shifrin

Night, @Alessandro :)

Balarka Sen

Night

Adeek

I am also going to sleep I have been staying up past 2 days

r9m

@Adeek holy Cow! :P last time I stayed awake for 72 hrs I couldn't sleep after that even if I tried to .. :P (I needed stuff to put me to sleep .. )

Ted Shifrin

Night, Karim!

Adeek

01:12

I had so much to do

Ted Shifrin

Funny that they say "holy cow" in India ... :D

Adeek

assignments + TA work + other stuff but I am glad I have break for 1 week to catch up and catch up on more stuff @r9m

nights @TedShifrin everyone

r9m

@Adeek good night ..

Daminark

Hehe

And yikes that's pretty rough

I don't think I've ever stayed up for 72 hours

Like, I've only pulled a couple all nighters

r9m

@TedShifrin you'd think it's just cows .. there's too much holy s*** lying around :P

Balarka Sen

01:16

@TedShifrin Cows are indeed holy.

poetryfoundation.org/poems-and-poets/poems/detail/54163

Ted Shifrin

I know, @Balarka. I wonder who started the expression.

G'night @MikeM

Interesting that you linked to Allen Ginsberg, @Balarka. Very gay, iconoclastic, beat poet. :P

Mike Miller

@BalarkaSen It's not even true that things of a given noncompact type are open in the leaf space of a codimension 1 foliation, I dont think.

Balarka Sen

@TedShifrin Mhm, I am aware. I haven't read much by him but that one is great.

Ted Shifrin

Night, all ... Or, good day, @Balarka.

skullpetrol

Are you finished with Edgar Allan Poe?

Daminark

01:21

See you around @Ted!

skullpetrol

Cya.

Balarka Sen

@MikeMiller Hmm. I don't know of an immediate counterexample though.

The easiest example (Reeb foliation) has that

@skull Poe is good. I have read a nontrivial fraction of his works. I am not sticking around with him anymore these days.

skullpetrol

cool

Mike Miller

@BalarkaSen Is my total space allowed to be noncompact?

Balarka Sen

Sure, I have no problem with that

Maks

01:26

Hey, can someone enlighten me with the name of these kind of integrals ? $ \int_c^{f(x)} g(t) dt $ The ones that are bounded with a function ??
I want to know how to solve then but I cant find them

Mike Miller

@BalarkaSen Pick a fibered knot in $S^3$ such that the fiber has genus $g$. Foliate the knot complement; you "turbulize" the obvious foliation of the circle bundle so that it becomes tangent to the boundary. The leaves are open of genus g. Glue this to the Reeb foliation of the torus, and puncture a point on the torus.

I mean, shit, you could just puncture the Reeb foliation.

skullpetrol

:O

Mike Miller

Then the leaf space is S^1 sqcup S^1 sqcup *, and the only open set containing * is everything.

Balarka Sen

@MikeMiller Sorry, you mean Reeb foliation on the torus or the solid torus?

Mike Miller

solid torus

Semiclassical

01:33

"turbulize" is a fun word.

Balarka Sen

And puncture from the interior or the boundary or anywhere?

Mike Miller

I meant the Reeb foliation on $S^3$ induced from the Heegaard splitting. Then puncture on the torus.

Balarka Sen

Ahh, alright. I agree with your description of the leaf space (the * corresponding to the punctured torus leaf)

But a neighborhood of a generic leaf (one of those twisted paraboloids) still consists of diffeomorphic leaves so I am a bit confused what this is a counterexample to

Leaves of the same topological type are still open in the leaf space

Mike Miller

Nah, * is not open :)

But ok, I still don't think it's true but I don't have an easy counterexample.

Balarka Sen

On the other hand your previous fibered knot example gave a counterexample to that I believe. Turbulizing makes it tangent to the boundary, which in turn says a neighborhood of the boundary leaf can contain drastically different leaves

@MikeMiller Ohh, I see

I was just thinking of the R^2 leaves

skullpetrol

01:44

@Semiclassical to make turbulent? So another way of saying "stir." :-)

Dragneel

Afternoon everyone. Does anyone know what this question is asking for? Find the relation between $n$ and $k$, $(k \leq \lfloor \frac{n}{2} \rfloor)$ such that $(\binom{n}{k+1})$ three times the previous entry, in the same row of Pascal's triangle?

Mike Miller

4.1.14 might give a c/e

Balarka Sen

@SemiC it's just this picture.

4.1.14 is interesting.

Aha, that indeed sounds like an example

logical123

02:43

Asked this yesterday, and have worked on it from multiple directions since then to no avail...
If the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is to enclose the circle $(x-1)^2 + y^2 = 1$, what values of a and b minimize the area of the ellipse? I am reviewing lagrange multipliers and this problem has me stuck. I seem get stuck in an algebraic loop everytime, after solving for my lagrange multiplier.

Semiclassical

What've you written out so far?

logical123

i can share a pdf i have of some stuff typed, but my most recent attempt was to parametrize each curve

Semiclassical

Why? Seems like it's easiest to do it in plain old Cartesian coordinates.

logical123

That was my original thought hahah

I restrict to quad I, use $\frac{\pi}{4}ab$ as my function to be minimized

i noticed too, to write an inequality in $y^2$ to keep roots out of the problem

Semiclassical

Eh, I feel like this is overcomplicating it.

logical123

02:48

Which part is an overcomplication?

Semiclassical

Well, I feel like the thing to do is to find the condition on a,b such that the ellipse is tangent to the circle on the outside.

And then minimize the area with respect to that a,b condition.

logical123

ooo that would def work

Dragneel

If (x = odd) and (y = odd), then x-y = even. I believe this statement is true, I can't think of a counter example. Can you guys think of any?

Semiclassical

x,y odd -> x=2m+1, y=2n+1 -> x-y = 2(m-n) -> x-y is even.

Dragneel

I see.

This would even work with negative numbers. Anyway, thanks for the reassurance.

logical123

02:56

semiclassical, am i guaranteed that if i have a tangent between those two curves, the ellipse lies outside the circle?

Semiclassical

No, but you're guaranteed that it's either inside or outside.

Actually, I take it back.

There will be a few different possible tangencies.

logical123

thats what i was thinking

and thinking for an arbitrary circle placement, very much so

Semiclassical

Hmm.

I still think this can be made to work, but it requires some care.

Oh, but something that helps a lot

logical123

That's part of my problem; as it is, doing lagrange multipliers where x and y seem to serve as dummy variables is something new to me

Semiclassical

The circle is still on the x-axis.

logical123

03:00

yes

Semiclassical

In that case, there's only two ways for the ellipse to possibly be tangent.

One of them encloses, the other doesn't, and the one that does has the larger x-value as the point of tangency.

logical123

copy that!

should do it, ill start scribbling away and report back in a while

thanks

Semiclassical

np

logical123

i do want to figure out why my polar method goes wrong though :P

Semiclassical

sure.

Oh, hah, it's even easier than what I was saying. The circle intersects the origin.

So the only way for an ellipse centered at the origin to be tangent is to enclose it.

logical123

03:04

no thats not true

let a = 2

Semiclassical

?

logical123

all those have tangents at the edge of the circle

Semiclassical

Hmmm.

Point.

logical123

(2,0)

Semiclassical

Heh.

Actually, if I think of it geometrically.

I have a sneaking suspicion that the minimal case will have $a=b=2$.

logical123

03:06

the soln is (3 root(2) / 2, root(6)/2)

Semiclassical

That $a=2$ is needed is pretty obvious by now. What's not so clear is $b$.

logical123

confirmed numerically

Semiclassical

...well, shoot.

logical123

via a bruteforce search in python

Semiclassical

Right. So the points of tangency aren't on the x-axis in that case.

logical123

03:07

literally coding that took me 12mins, but doing the work pen and paper

Semiclassical

lol

logical123

try number 6 begins

Semiclassical

That's what said parameters give, so it seems sound.

logical123

Yep

Semiclassical

Anyways. Doing things more carefully, it looks like you can show via Lagrangian multipliers that tangency occurs when either $a=2$ or $a(b)=b^2/\sqrt{b^2-1}$.

The first one won't enclose unless $b$ is sufficiently large. Not sure how to eliminate it as a possibility, but evidently it won't work. For the other, all one does is minimize $a(b)b$ w/r/t $b$.

logical123

03:16

Copy that

Did you take partials wrt x and y or and b?

* a and b

Semiclassical

x and y alone.

logical123

Aha, I have been doing wrt a and b alone

Semiclassical

Ah. Yeah, I was still going with the "I want the ellipse to be tangent" logic.

Simple

$Y=\begin{cases}
X,\;\;&|X|<a\\-X,\;\;&|X|\geq a \end{cases}$ with $X$ has $N(0,1)$ and $a>0$, do $X$ and $Y$ bivariate normal?

logical123

Was tangency your constraint or are you saying doing Lagrange multipliers implies tangency

Semiclassical

03:20

The latter.

logical123

Right

So the constraint is the eqn of the ellipse, circle, or both?

Semiclassical

I was doing the circle as the constraint.

logical123

Gotcha

Semiclassical

One approach to checking the case where $a=2$ and they're tangent at $(2,0)$ is to try to solve for the intersection between the two curves. If an intersection exists, then it doesn't enclose.

BAYMAX

Hi guys , i have a doubt --- If $x<y$ then can we always say $log(x) < log(y)$, if not under what conditions?

Semiclassical

03:25

Is $x>0$? It presumably should be, since otherwise $\log x$ isn't well-defined.

BAYMAX

yes , $x \geq 0,y \geq 0$

logical123

log is strictly increasing

so yes

Semiclassical

^

RE60K

Can anybody help with this statement: "Accordingly consider the set of all natural numbers of the form ax + by
with x, y in Z. The set is not empty since, for instance, it contains a and b;
hence there is a least member d, say. Now d = ax + by for some integers x, y,
whence every common divisor of a and b certainly divides d. Further, by the
division algorithm, we have a = dq + r for some q, r in Z with 0 ≤ r < d; this
gives r = ax' + by' , where x' = 1 − q x and y' = −qy. Thus, from the minimal

i cant get because of the use of the word :whence

logical123

y need not be positive though

log(x) when x<1 is less than 0

Semiclassical

03:26

? $y>x>0$.

RE60K

I only don't get the last line

BAYMAX

@logical123 then $log$ will not be well defined

logical123

no

just remember

negative exponents give you POSITIVE fractions

Semiclassical

That doesn't really help. Log isn't defined as a real function over the negative reals.

logical123

i was saying log takes on negative values in its image...

log(0.1) = -1

Semiclassical

03:29

...and?

logical123

10^-1 = 0.1

Semiclassical

still doesn't make log(-1) well-defined over the reals.

logical123

I never said it did lol

Semiclassical

@logical123 ummm

I'm guessing you meant "log(y) need not be positive", then, rather than "y need not be positive"

logical123

logx < 0 for x btw 0 and 1

BAYMAX

03:30

Yes $log$ can give us negative values but $log $ cant take negative values!!!

Semiclassical

I think you made a typo, because you did say that y doesn't need to be positive...

logical123

which is why i said y need not be positive, baymax

Semiclassical

???

logical123

the image of log(x) = y does not need to be positive

he restricted the range of log(x) to be greater than or equal to 0

which i was correcting

Semiclassical

...considering y is already used as the argument of log in the inequality he used, you really shouldn't be using it for y=log x.

logical123

03:31

just graph log(x) lol

ohhhhhh im so sorry

Semiclassical

just a miscommunication, then.

logical123

interpretational mishap, i thought of y after the initial line as the range of log

BAYMAX

seee @logical123 i have taken y in the domain of $log$ , hence i defined $log(y)$ but here $y$ must be postive , yes the value of $log(y)$ may or may not be positive !!!!!

logical123

yeah derp and a half

Semiclassical

yeah, @baymax. miscommunication: he was thinking of $y$ as in the $y$-axis of a $y=\log x$ plot.

BAYMAX

03:33

yes @Semiclassical

logical123

but i interpreted it correctly at first, just the x≥0,y≥0 line made me think otherwise

BAYMAX

so , conclusion then ?

logical123

the same hahaha

BAYMAX

oh sorry

Semiclassical

(Admittedly, I think it's clearer to write $0<x<a\implies \log x<\log a$. using $y$ as something 'like' $x$ is just a bit annoying)

logical123

03:34

log is strictly increasing

BAYMAX

$x>0 , y>0$

$0$ is not in domain of $log$

logical123

therefore if, in your symbolage, $x < y$, then $log(x) < log(x)$

Semiclassical

That shows both of the smallest ellipse possibilities.

logical123

orange def has less area even visually

pilko

Man, I won't care trying to find good answers to not so finely asked questions anymore. The post get removed and I get deranked. I'm learning new SE pitfalls by the day.

logical123

03:36

im surprised my algorithm converged so rapidly to be honest

im always surprised when they dont end up in a local minimum and get stuck

Semiclassical

Yeah. The orange has $ab=3\sqrt{3}/2\approx 2.6$ whereas the green has $ab=2\sqrt{2}\approx 2.8$.

BAYMAX

@Semiclassical is the figure related to my question or any other>?

Semiclassical

Nah, not to yours.

pilko

it's pretty though

BAYMAX

ok

Semiclassical

03:39

It is rather nice.

logical123

I need to modify my code

so i can put the circle anywhere

see if it still converges as well

Semiclassical

I can see that making the Lagrangian approach a lot harder, since it gets rid of symmetry.

Hmmmmm

BAYMAX

@pilko it will take some time ,post your question before searching them in MSE and also post what have tried so far , as it will help you better!

logical123

you would have to break it into cases i think

Semiclassical

I think that's not what @pilko was getting at.

Rather, they submitted an answer to a post of questionable quality; when the question was removed, so was the answer.

BAYMAX

03:41

ohh @Semiclassical

ok

Semiclassical

Which, yeah. Pretty annoying.

Though what's worse is if the author removes it voluntarily. "Oh, thanks for the answer. kaybai."

pilko

yep, I like whinning a little, then I move on.

BAYMAX

he he .. yeah..

nice @pilko

Simple