Mathematics - 2017-12-22 (page 3 of 7)

Secret

05:07

Intuition suggests it might be impossible to have skew planes in 4D (note how in whatever direction you move along the blue arrows, there's always an intersection), Now figuring out how to prove that

2

Q: Intersection of 2D planes in 4D space

If I had a four dimensional space, in which I embedded two planes, what possible intersections could they have? Constructing a Plane To give this more context, consider the following. If I had a 4d tuple (w, x, y, z) with no restriction, it can assume any position in 4D space. By imposing one...

Intuition confirmed

I suspect for an n dimensional affine subspace, you need at least n+2 dimensions to have skewed affine subspaces to exist

note the planes can be tilted anywhere as long the tilt is not along the 5th dimension, then there is no way they will intersect

As for the remaining tilting that still will not result in intersection (3D case shown here, note line 1 is tilted towards line 2) , I think my intuition is not strong enough to make a guess

anon

05:23

@Secret two nonintersecting 2d affine planes in R^4 may not be parallel, but they must contain parallel lines

I think that might actually help

Secret

ah right, I focused too much on the perpendicular plane configuration

I wonder if we have enough terminology to distinguish configurations where the planes contains parallel lines but not parallel, and the two planes being parallel in the sense that if the separation is reduced to zero, the planes will coincide

Daminark

05:54

Hey @Mathein!

MatheinBoulomenos

Hi

skullpatrol

The art of mathematical language...

\o

Daminark

How's it going?

MatheinBoulomenos

pretty good, my talk went reasonably well

Just found out the courses they'll offer next semester, some cool stuff, but some unfortunate overlaps

and you?

Daminark

Ah, the one on Galois cohomology?

Also what are you thinking of doing?

MatheinBoulomenos

06:00

yeah

I'll take "Algebraic Number Theory 2" and "Modular Forms 1" in any case. I'll probably do "Local class field theory after Lubin & Tate" as well. I would also like to do "Algebraic Groups", but it overlaps with "Algebraic Number Theory 2" which is really unfortunate imo

Mithlesh Upadhyay

Today is Indian National Mathematics Day "Srinivasa Ramanujan".

Daminark

Damn, that's unfortunate

Re Mathein, not Mithlesh :P

MatheinBoulomenos

Have you decided what you do next quarter? I think commutative algebra would be very useful for the REU you mentioned

Mithlesh Upadhyay

@Daminark :)

Daminark

They won't let me take commutative algebra, since I'd concurrently be doing the intro to ring theory course

MatheinBoulomenos

06:04

ah, that sucks

Daminark

I might try to pressure my ring theory professor into going fast but I dunno if it'll work

MatheinBoulomenos

at my uni, we don't really have any formal prerequisites only recommended prerequisities

but can't you talk with them? You could say that you need this stuff for a REU

Daminark

Oh here the prerequisites are formal, and to get them waived you have to go through the department head

MatheinBoulomenos

damn

Daminark

When I asked him, he said he wasn't too happy with the idea of doing a grad course and a class for which I don't have the prerequisite at the same time

So he was like yeah, I'm gonna say you have 4th year to do commutative algebra, possibly grad, but not now

MatheinBoulomenos

06:06

I see

Daminark

As it stands, I'm doing ring theory, functional analysis, and a course on ancient empires to satisfy the civilization requirement

MatheinBoulomenos

ring theory & functional analysis are really cool

Although introductory ring theory could be quite basic I guess

idk what you're covering

Daminark

In the past that class used to do some linear algebra, plus ring theory a la Dummit & Foote

Now we have a linear algebra class so maybe we'll cover more? I dunno

I'll also have to see which section is likely to be more interesting, since there are two. One of the profs taught my group theory course and was a bit slow for my taste

MatheinBoulomenos

Dummit & Foote covers the basics of ring theory in Part II & III, but they also have some more advanced stuff in Part V

Daminark

Hopefully we'll get to it. I'll see if the other section has prospects of speeding up at all, though I'll want to be able to come back to this section in the spring because the professor is really good for Galois theory

I guess we'll have to wait and see

But yeah I'm still deciding what to do for a 4th class. Since commutative algebra won't be a thing and they likely won't have space in algorithms, there aren't too many good options

Point-set and ODE don't sound too appealing, and I'm not ready for the other grad classes

Measure theoretic probability in the stats department would've been good but the prof is apparently quite bad

MatheinBoulomenos

06:27

hmm, from the things you mentioned, I'd probably do point-set tbh

Daminark

Yeah, it's the most sensible option, but I've heard it's rather boring

Might try another reading course

MatheinBoulomenos

you could do a reading course on commutative algebra

Daminark

That could be amusing. I did talk to the guy in charge of the REU on number theory and he seemed to suggest that I had enough background for some of the projects

MatheinBoulomenos

ah, cool

Daminark

Arithmetic geometry probably wouldn't work, but I dunno if I could work myself to having enough background for that no matter what

MatheinBoulomenos

06:34

arithmetic geometry is really demanding

Daminark

They were talking about p-torsion of char p abelian varieties

And one person I know who worked on that project said you kinda need to already know some AG before starting that

Other people talked about monstrous moonshine, modular forms, analytic number theory even

MatheinBoulomenos

one of my friends asked a prof if he could write a bachelor thesis on abelian varieties. He gave him a paper from Tate and said to him "I'm sure that you have no chance to understand any of those proofs any time soon, but if you work hard, you might understand the statements and do something interesting with them"

modular forms are really cool. We're doing that in complex right now

Daminark

I'm gonna ask my complex analysis professor in the spring more details about what she intends to do (I know she'll want to do some analytic number theory), since I want to minimize redundancy, but something along the lines of modular/automorphic forms could be fun to try next

MatheinBoulomenos

Automorphic forms are quite difficult from what I've heard.

Daminark

Hmm

MatheinBoulomenos

06:44

but it depends on the approach, I guess

Daminark

I guess I'll figure out something when the time comes

Alessandro Codenotti

07:09

Hi chat

skullpatrol

heya

Alessandro Codenotti

What are the other options to satisfy the civilization requirement? @Dami (the fact that there is such a requirement looks incredibly weird to me)

MatheinBoulomenos

07:25

Hi @Ted

Daminark

There are quite a few @Alessandro

Though since I'm trying to do them out of order, options narrow down

Hey @PVAL!

PVAL-inactive

God all of those classes sound like horse s.

3

Daminark

Kek

Alessandro Codenotti

I see, the music in western civilization one doesn't sound too bad

PVAL-inactive

If it's by a history professor I doubt there will be any musical content.

but I guess it has a music subject tag nvm.

Salt

07:30

I'm taking Intro to Internet Civilization

PVAL-inactive

Take Intro to Civilization by Sid Meier

Daminark

That's neat @Salt

And re music civ: even the classes for the arts core are more focused on the analysis instead of performance for the most part, so I'm guessing this may be more of a history class with an eye toward the influence of music than anything

skullpatrol

07:49

That's a lot of variety!

Daminark

True

skullpatrol

When's the deadline?

Daminark

For choosing?

skullpatrol

yup

Daminark

So ideally by the time class starts, January 3rd. You have a week where adding/dropping is relatively easy, and then another 2 where it's possible but you have to jump through some hoops

skullpatrol

07:55

right, right

Daminark

After that, you can drop from an honors math class to a regular one for another 2 weeks. Past 5th week, you can only withdraw

skullpatrol

There's also schedule issues...

mathreadler

wow so many planes, but can they fly?

skullpatrol

What's a readler? @mathreadler

is that supposed to be a reader?

Salt

it's a re-adler

skullpatrol

08:03

oh

mathreadler

thats the riddle

what is it really

skullpatrol

Language arts?

mathreadler

just a play on words

you dont have to study languages to do those do you?

skullpatrol

depends

some plays are more difficult than others

mathreadler

if you paardon me, i dont see your point.

i am on math stackexchange, and ask about math, what difference does it make if I am a cat or a lion or a reader?

skullpatrol

08:08

aha, it is reader :-)

mathreadler

actually had to ask one of my professors what a reader was a few years ago. i did not know back then.

skullpatrol

Someone who reads.

mathreadler

they are not called readers where i'm from. i think it is roughly equiv to lektor / lector?

But what do I know, I am just a little housecat on the run trying to solve my homemade least squares.

skullpatrol

yeah, it's a bit of a British term

mathreadler

housecat on british terms?

Balarka Sen

08:21

What is up dramalertnation

mathreadler

I did not know they were employed.

What does a typical housecat contract look like in britain?

Daminark

@BalarkaSen leeeets get riiiiight INTO THE GROUPS

Balarka Sen

Keemstar algebras

Daminark

Amazing

Balarka Sen

I have Keemstar's voice stuck in my head now

mathreadler

08:26

omg are you a pintor balarka sen?

you have a nice painting under that huuge hat

Balarka Sen

Yes, I have a formal training in minimalist paintings. I paint large white boxes as a profession.

I mean no, I am not, that was a joke

My avatar's a scene from the animation "Tale of Tales" by Norshteyn

Daminark

"I paint ironically"

mathreadler

aah

i sometimes wish i could paint

i can only make venn diagrams in ms paint

and stick cats

Balarka Sen

08:41

@Daminark "Irony paints me"

That got a little Rick and Morty level there

Daminark

Lmao

mathreadler

hmm should i go for a brunch or wait for lunch, what difficult questions life throws me.

wait I think my owner left a slice of pizza on the table

lol just kidding, i have no owner, thus no pizza slice on the table to be had :/

Typhon

09:22

0

Q: Verifying an integration method for step function integrals.

I should note that this is a symbolic integration algorithm and therefore intermediate DO appear to be unjustified. The purpose of this question is to justify or reject this algorithm. Suppose we have a function $f(x)$ that has no vertical asymptotes and can be written as some expression $E$ con...

calculus integration algorithms floor-function

can anyone help verify this?

i think the algorithm is much more succinct then splitting the integral onto multiple domains.

TheNotMe

11:38

Why is the following true: $\sum_{k=0}^n \binom{n}{k}\binom{n}{k} = \binom{2n}{n}$?

TheNotMe

12:03

Nevermind. Got it.

math.stackexchange.com/questions/148583/… - if anyone is interested

@anon are you there?

Jarek Duda

13:53

Continuing the graph isomorphism test ...

0

Q: Rotation invariants for higher degree homogeneous polynomials (like Tr($P^m$) for degree 2)?

Treating rotation in $\mathbb{R}^n$ as $x\to Ox$ for orthogonal $O^T O=O O^T=1$, we can easily get complete sets of independent rotation invariants for degree 1 and 2 homogeneous polynomials: Degree 1: $p(x)=\sum_i P_i x_i$ has single independent rotation invariant: $\sum_i P_i^2$, Degree 2: $p...

matrices polynomials rotations multilinear-algebra invariance

LegionMammal978

14:09

I might as well try asking this again: You are given two expressions $P(x)$ and $Q(x)$ consisting of $\sin x$, $\cos x$, $\tan x$, $\cot x$, $\sec x$, and $\csc x$, connected by addition, subtraction, multiplication, and division, such that $P(x)=Q(x)$ for nearly all $x$ (i.e., only a countable number of exceptions); you are attempting to find a substitution path from $P(x)$ to $Q(x)$. Now, the question is, does there exist a minimal set of trigonometric identities that is required to do this?

(i.e., for all $P$ and $Q$, the set of identities, defined on top of standard algebraic identities, should be sufficient to find a substitution path.)

Additionally, forgot to mention that $P$ and $Q$ can contain integer constants, but these can be trivially constructed via $\frac{\sin x}{\sin x}+\frac{\sin x}{\sin x}+\cdots$.

Studentmath

14:28

Question
I have some graph $G$, with $\alpha (G)$ being the size of the independent set.
Assume I colour $G$ with two colours. Any reference to good bounds on the size of the independent set on one of the colours?

must be something better $(n/2) \alpha(G)$

Edge colouring, of course

Mary Star

14:42

Hello!!

We consider the function $f(x)=\tanh x$.

I shown so far the following:
The function is defined for all $x\in \mathbb{R}$, i.e. $D_f=\mathbb{R}$.
The function is strictly increasing.
$\displaystyle{\lim_{x\rightarrow -\infty}\tanh x=-1}$ and $\displaystyle{\lim_{x\rightarrow \infty}\tanh x=1}$. And since $\tanh$ is continuous on the whole $\mathbb{R}$ the range is $(-1,1)$.

Now I want to show that $\tanh$ is uniformly continuous on $\mathbb{R}$ (without using differential calculus). Could you give me a hint how we could show that?

philmcole

15:59

Hello! Quick question. $\mathbb{C} \setminus((-\infty,0]\times\mathbb{R})$ corresponds to the right part of the complex plane?

Balarka Sen

16:13

Suppose $N_1, N_2$ are $n_1$ and $n_2$-dimensional submanifolds transversely intersecting at $p$ inside $M^n$. I want to prove that there is a chart $U$ at $p$ that admits a diffeomorphism $f : U \to \Bbb R^n$ such that $f(U \cap N_1)$ and $f(U \cap N_2)$ are two transverse $n_1$ and $n_2$-dimensional affine subspaces of $\Bbb R^n$ intersecting at $f(p)$.

Can I give $M$ a Riemannian metric such that $N_1$ and $N_2$ are totally geodesic submanifolds of $M$? In which case the exponential map at $p$ would be the inverse of the desired diffeomorphism.

I have had this idea about proving a local modelling theorem for transverse submanifolds for a while, but I didn't get around to proving it.

Given a set of submanifolds, can you always make a Riemannian metric on your ambient space such that those submanifolds are totally geodesic? Or are there restrictions?

TheNotMe

how is $\sum_{k=0}^n \binom{n}{k} k (k-1)$ equal to $2^{n-2}n(n-1)$?

Mike Miller

16:38

@BalarkaSen To do your original goal, first pick a chart in which one submanifold is flat; say V inside W. Then consider the map from the other submanifold as a map to W/V. That you were transverse means that this is an immersion

So you can use the implicit function theorem to fiddle with W/V to make the second submanifold flat inside W/V.

Semiclassical

@TheNotMe start by figuring out what $\sum_{k=0}^n \binom{n}{k}x^k$ would be

TheNotMe

binom

success in k experiments out of n experiments

Semiclassical

right. so that's...?

there's a simple closed form expression.

TheNotMe

nx

Semiclassical

No.

Balarka Sen

16:43

@MikeMiller Ah, that seems to be a more hands-on approach.

TheNotMe

well look

Mike Miller

@BalarkaSen I got lazy with what I was writing. I think you need to do some work still - lift this to a splitting of W asV + W/V so that the second submanifold lives in W/V. Then finish the job.

TheNotMe

I solved it a bit differently.

Mike Miller

But that should still be easily achievable (maybe a 3-step problem in the end)

Semiclassical

when $x=0$, you should have $\sum_{k=0}^n \binom{n}{k}x^k=1.$ But $nx=0$ when $x=0$.

TheNotMe

16:45

$\sum_{k=0}^n \binom{n}{k} k (k-1$ is like choosing a group of any size out of a group of $n$ people and then choosing two represenatives from that chosen group. You can also express it like this: $\sum_{k=2}^n n (n-1) \binom{n-2}{k-2}$. ie you choose first the two representatives and then choose the rest of the k-2 people

and the latter sum is easy to calculate: $n(n-1)2^{n-2}$

Mike Miller

I wonder if there are global obstructions to making a single submanifold geodesic

Semiclassical

sure, that works.

TheNotMe

But I am interested in your way

It's almost like a mean value of binomial.

Except it is missing the $(1-x)$ factor and $k$

Semiclassical

it's the generating function way, really

Mike Miller

The inverse function theorem says there are no local obstructions if I'm not trying to preserve anything on the submanifold or larger manifold

Semiclassical

16:46

but, again, what's the closed-form expression for $\sum_{k=0}^n \binom{n}{k}x^k$?

Balarka Sen

@MikeMiller True. I think it's workable

Semiclassical

it's not $nx$.

you were on the right track with binomial...

TheNotMe

As I said, its close to binomial.... but its not the expected value of it

Balarka Sen

Actually I think what you said was my initial approach a long time ago

Semiclassical

It's close to the binomial what

Balarka Sen

16:47

But I abandoned it because it looked messy

Semiclassical

maybe it's easier to see with a particular case, though: When $n=3$, that's $\sum_{k=0}^3 \binom{3}{k}x^k=1+3x+3x^2+x^3$

TheNotMe

beats me... sorry

Balarka Sen

@MikeMiller I feel like we have wondered about the question on other categories before. For example, given two transverse subs, extend a triangulation given on both to a triangulation on all of the ambient dude

Semiclassical

think less probability, more algebra.

the first few cases (n=1,2,3,...) are $1+x,1+2x+x^2,1+3x+3x^2+x^3$,...

TheNotMe

$(x+1)^k$?

Semiclassical

16:51

(1+x)^n

TheNotMe

for a particular k I meant

oh sorry, yes.

$(x+1)^n$

Semiclassical

kk

TheNotMe

I see where this goes.

Semiclassical

which is really just the binomial expansion: $(1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k$

Mike Miller

@BalarkaSen Yes, I remember that

TheNotMe

16:52

indeed. and then you want to split the original sum to 2, and then one of them is this and the other is something simpler.

Yeah, I get it

Semiclassical

well, you can do that. but you can also differentiate twice and evaluate at $x=1$

on the RHS you get $x^k\mapsto k(k-1)$. on the LHS you get $(1+x)^n\mapsto n(n-1)2^n$

and bam

Balarka Sen

@MikeMiller Say $N$ is a compact subfold of $M$. The tubular neighborhood of $N$ admits a metric such that $N$ is totally geodesic, right? (because it has the structure of a vector bundle over $N$, which can be obtained as gluing pieces of the form $U \times \Bbb R^k$ where $U$ is an open in $N$; give each of those pieces the product metric and partition of unity to glue those up). Then in turn use partition of unity to extend that metric on the tubular neighborhood of $N$. Does that work?

Mike Miller

@BalarkaSen I think vector bundles have metrics so that the zero section is totally geodesic but I'm not entirely sure how we get it. I guess we need a metric and compatible connection on the vector bundle?

Balarka Sen

Yeah maybe that's not so easy. Ignore my goofy parenthesized explanation.

Mike Miller

@BalarkaSen Seems we maybe just need a metric on the base manifold and the bundle.

at v, a vector in E_x, T_v E has E_x as a summand. pi: T_v E/E_x -> T_x M is an isomorphism

Oh hm but I need to choose a lift of T_x M

Balarka Sen

17:04

That's where the connection comes in?

Mike Miller

I guess so

TheNotMe

@Semiclassical, do you have the time to help me with another probability question?

Mike Miller

@BalarkaSen Suppose the splitting scales like v in E does. Then the R-action of scaling preserves the metric

So the R action sends geodesics to geodesics. I think that implies the zero section is totally geodesic

Yeah it does

Balarka Sen

Oh yeah fixed point locus of isometries are geodesics

Semiclassical

@TheNotMe You can make the connection to the binomial distribution rather explicit, btw. Suppose $X\sim B(p,n)$. Then (formally, at least) you have $\mathbb{E}[z^X]=\sum_{k=0}^n \binom{n}{k}p^k (1-p)^{n-k}z^k =(1-p)^n\sum_{k=0}^n \binom{n}{k}\left(\frac{pz}{1-p}\right)^k$

...dangit latex

TheNotMe

17:08

haha latex got confused

Semiclassical

there we go

Mike Miller

Really nice idea Balarka

TheNotMe

Sweet!

Thanks for the different point of view. In the end I ended up writing the other combinatorial solution

Silent

"Suppose $x$ is rational, and $x^2>3$, and $\delta=\frac{x^2-3}{2|x|}$, and $y\in(x-\delta,x+\delta)$ and $h=y-x$. Since $|h|<\delta$, we get $y^2=x^2+2xh+h^2>x^2-2|x|\delta=3$" In this argument, I can't understand how did we get $x^2+2xh+h^2>x^2-2|x|\delta$.

TheNotMe

Do you have the time for an extra question that is also confusing me?

Semiclassical

17:09

in particular, when $p=1/2$ we have $\mathbb{E}[z^X]=2^{-n}\sum_{k=0}^n \binom{n}{k}z^k$

sure

TheNotMe

Can we go to a separate chat room? it is a tiny bit longer as I need to explain a preliminary

Semiclassical

my main point now is just that, if you differentiate the LHS with respect to $z$ and evaluate at $z=1$, you get (for instance) $\mathbb{E}[X]$

TheNotMe

Yup, I noticed that

I have never seen this approach before. Is it common?

Semiclassical

Yeah, pretty common. this function $G(z)$ defines the probability-generating function

you also see variations on it, for instance $\mathbb{E}[e^{i t X}]$

that's the characteristic function

(maybe without the i, I forget)

nice thing in that case is that it's equivalent to a fourier transform, which is...fun

I'm not sure I can help right away with another problem, but invite me to another room and write your question out

TheNotMe

ok great

Silent

17:38

28 mins ago, by Silent

"Suppose $x$ is rational, and $x^2>3$, and $\delta=\frac{x^2-3}{2|x|}$, and $y\in(x-\delta,x+\delta)$ and $h=y-x$. Since $|h|<\delta$, we get $y^2=x^2+2xh+h^2>x^2-2|x|\delta=3$" In this argument, I can't understand how did we get $x^2+2xh+h^2>x^2-2|x|\delta$.

Please help me with this!

Erik Humphrey

How do you get 3[21 + 108 + 0 - (27 + 72 +0)] - [36 + 0+ 7 - (24 + 0+ 9)] from matrices 3[7, 6, 3 // 4, 1, 2 // 9, 0, 3] - [3, 0, 1 // 7, 6, 3 // 4, 1, 2]?

I don't get where the 108 and 72 come from.

it's just the sum isnt it

No, that would be too easy

user193319

17:59

Are there are any metrization theorems for Hausdorff spaces?

Alessandro Codenotti

19:23

The Nagata-Smirnov and Urysohn metrization theorems are results in this direction, but they require Hausdorff and regular (and other stuff)

Clarinetist

This question went WOOSH over my head

11

Q: Homology groups of a Pokeball

I tried to discover the singular homology groups of a simplex, but I found some dificulties and want to know if someone knows what I've got wrong. The simplex is the following (0-cells are in blue, 1-cells in red and 2-cells in yellow): As it looks like a Pokeball, I called this space $PB$. F...

homology-cohomology

Balarka Sen

lol

there's nothing special about the pokeball there. good joke title though

Semiclassical

19:48

Reminds me of how a certain Feynman diagram looks like the lines on a basketball

\|/

+++

/|\

No idea how that looks btw since I’m doing this on mobile

TheNotMe

@Semiclassical it didn't work in the end. Couldn't get rid of the $0.5$ factor.

Semiclassical

Eh, not great

@TheNotMe strange

Akiva Weinberger

@Silent $x^2+2xh+h^2\ge x^2+2xh\ge x^2-2|xh|$

Clarinetist

Good ol' UMVUEs

Akiva Weinberger

${}=x^2-2|x||h|>x^2-2|x|\delta$

Clarinetist

19:52

(stats people and their words...)

Semiclassical

@Clarinetist did you figure out that entropy problem?

Clarinetist

@Semiclassical The continuous one? Yeah, Ted helped me out. Definitely not intuitive, but it makes sense

Semiclassical

Hmm

The integral version or the countable one?

Clarinetist

@Semiclassical Sorry, the integral one is what I mean by "continuous one"

Semiclassical

Kk

I think I see how the countable one (fails to) work

Studentmath

19:55

Someone familliar with "weakly induced" ramsey theorem?

Semiclassical

Suppose you’ve got n outcomes of equal probability 1/n. Then the total entropy is $n\cdot -\frac1n \log(1/n)=\log n$

Leaky Nun

ugh, physics

Clarinetist

@Semiclassical Yeah, and then its suprenum is $\infty$

Semiclassical

Right

Erik Humphrey

In a matrix with characteristic equation λ(1 − λ)(λ − 7) = 0, how do I know 0 is an eigenvalue?

Semiclassical

19:58

So in the countable case there’s no max entropy case

Akiva Weinberger

@ErikHumphrey λ=0 is a root of that, isn't it?

Clarinetist

@Semiclassical I also wonder if by "max entropy," people actually mean $\infty$. It's misleading, I agree.

@Semiclassical It also makes me wonder, is the uniform distribution the ONLY distribution which could yield an infinite entropy?

Erik Humphrey

@AkivaWeinberger Maybe! It's been awhile, how do I tell?

Semiclassical

Nah

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$Mathematics$ Mathematics