Mathematics - 2019-02-10 (page 1 of 2)

Akiva Weinberger

00:02

Neat: this page shows how to draw approximate versions of these tilings on a grid

davecomputes.blogspot.com/2018/01/…

Next time I have graph paper in front of me I'll try this

This seems easier than guessing 72-degree inclines freehand

and it seems like it doesn't do much beyond the distortion

Secret

::have a strong urge to compress the above diagram top and bottom::

Akiva Weinberger

Actually, it's not obvious to me that this doesn't mess with the overall structure

How do I know that this version is aperiodic, like the original?

Oh wait I think I see it

Secret

the 5 stars in the figure forming two large pentagons side by side, thus it has to be aperiodic

Akiva Weinberger

The deflation rules are the same

Hm, wait, each inflation/deflation actually changes the shapes of the pentagons slightly

I don't think that should matter, actually

I'll sleep on it

Rumplestillskin

00:19

Hi guys are people familiar with Runge-Kutta (4th order) for systems of DE's and it's implementation in some programming language?

Ted Shifrin

Can always count on DogAteMy to put up serious pictures !!

Akiva Weinberger

53 mins ago, by Akiva Weinberger

It looks so purty

Ted Shifrin

Does it have some particular mathematical construct?

Akiva Weinberger

@TedShifrin IIRC: Take $\Bbb Z^{17}$

Take the plane in $\Bbb R^{17}$ spanned by $(\cos(2n\pi/17):0\le n<17\}$ and $(\sin(2n\pi/17):0\le n<17)$

(By this I mean the vector $(\cos(0),\cos(2\pi/17),\cos(2\cdot2\pi/17),\dots)$ etc)

Note that, under the linear map $e_i\mapsto e_{i+17}$, this plane rotates $2\pi/17$ radians

Take every point in $\Bbb Z^{17}$ that's within a certain distance of this plane and project it onto it

and I think the same for line segments between those points

(That distance might be $1$, I'm not sure)

You get this tiling with rhombuses. It's aperiodic

Ted Shifrin

Ah, interesting.

Akiva Weinberger

00:34

If you do the same in 5D space you get Penrose

I'm not quite sure what happens in 3-, 4-, or 6D space

but I'm guessing you get something periodic

Maybe even just the usual tilings of triangles, squares, and hexagons

Ted Shifrin

These are things I've never contemplated. Thanks. :)

Akiva Weinberger

3 mins ago, by Akiva Weinberger

Note that, under the linear map $e_i\mapsto e_{i+17}$, this plane rotates $2\pi/17$ radians

$e_i\mapsto e_{i+1}$ I meant

where $e_i=(0,0,\dots,1,\dots,0,0)$ in the $i$th place, and addition is mod 17

(Quick puzzle: why are those two vectors perpendicular)

(and I forget what their magnitude is)

(Something involving $\sqrt{17}$ probably)

(Maybe $\sqrt{17/2}$?)

Ted Shifrin

I'm not following quite well enough.

Akiva Weinberger

A simpler example would be, why are $(\cos(0),\cos(72^\circ),\cos(144^\circ), \cos(216^\circ),\cos(288^\circ))$ and $(\sin(0),\sin(72^\circ),\sin(144^\circ), \sin(216^\circ),\sin(288^\circ))$ perpendicular

in $\Bbb R^5$

and I'm 99% sure their lengths are $\sqrt5/2$ each

Also: why does the linear map $(x_0,x_1,x_2,x_3,x_4)\mapsto (x_1,x_2,x_3,x_4,x_0)$ rotate the plane spanned by them by $72^\circ$

Fargle

That just falls out of identities, doesn't it? e.g. cos 72 = cos 288, sin 72 = -sin 288

Akiva Weinberger

00:41

Yeah

Ted Shifrin

Should be a good way to do this with complex vectors.

Akiva Weinberger

Simplest is $\sin(x)\cos(x)=\frac12\sin(2x)$, average of sine function is 0

Similarly, average of $\sin^2$ and $\cos^2$ are each $\frac12$

(For the perpendicular and lengths)

Oh, I think it actually rotates the plane by $-72^\circ$, if we orient the plane by the order I gave its basis in

Ted Shifrin

and hi @Fargle

Ultradark

the shadow of a 3D cube lattice at certain angles forms a quasicrystal pattern

Fargle

Hey @Ted

Akiva Weinberger

00:46

If you use a 5D lattice you can get 5-fold symmetry easier @Ultradark

If you use 3D I think you get tilings that look like this

$user image$

Ultradark

oh okay, is 5 fold symmetry special?

Akiva Weinberger

No periodic crystal has it

The only symmetries allowed for periodic tilings are 2, 3, 4, and 6

So if you ignore that and try to make a 5-fold symmetric pattern anyway you're guaranteed to either accidentally end up with the wrong symmetry, or make a quasicrystal

Ultradark

I like that one

Akiva Weinberger

1 hour ago, by Akiva Weinberger

^17

I remember someone once saying that quasiperiodic tilings give higher dimensional lattices "something to do"

'cause they seemed kinda abstract, not really predicting any phenomena

but it turns out you can slice and project them and get neat pictures

(Dunno what happens if you try this with, say, the E8 lattice)

Arright, night, I shoulda been asleep three hours ago

Ultradark

See ya

BAYMAX

05:54

woo .. some beautiful pics here :)

Tarun

I have array of characters. I want to balance them in a manner that ever character appears equal number of times. I am allowed to replace any character in the array by any other character. I need to find out minimum number of such replacements required to create such an array.

athos

07:05

hi, may i ask a quick question, what are the primitive root modulo 125?

my computation shows 3,8,13,23,28,33,38,48,53,58,63,73,78,83,88,98,103,108,113,123.

am i right?

athos

07:19

oh these are from 3, from 2 will get another 20 roots.

Akiva Weinberger

07:38

> In July 2017 Michaël Rao completed a computer-assisted proof showing that there are no other types of convex pentagons that can tile the plane.

WHA

Why was I not told about this

I remember when the 15th one was discovered

I thought it was an open problem whether the list was complete or not

quantamagazine.org/…

Akiva Weinberger

07:52

> Rao said he felt disappointed not to have discovered any additional families, but tiling experts say that proving a complete list of 15 is more significant than simply finding a new working example.

Rudi_Birnbaum

@AkivaWeinberger Cool.

So when you have 15 rooms in your house, you can tile the floor in each of them differently :-)

Akiva Weinberger

2

Wolfram MathWorld's page on them still has the outdated information that the list is not known to be complete

(but this wouldn't be the first time Wolfram MathWorld has incorrect data on it)

Rudi_Birnbaum

The "Einstein" also sounds cool.

Rithaniel

I like that type 3 is really just hexagonal tiling.

Zee

Anybody knows how to glue sheaves ?

Akiva Weinberger

08:03

Wait why is type 4 not convex

Oh it's just an error

Rithaniel

1. Acquire glue
2. Acquire sheaves
3. ????

Akiva Weinberger

Here ya go, 4 is fixed

Rithaniel

Actually, in that image type 4 looks for two "stretched" hexagonal tilings overlaid on each other.

Akiva Weinberger

It and 8 look like the Cairo tiling

except that they both have a free parameter and the Cairo tiling doesn't

In fact, I bet that the Cairo tiling is a special case of both 4 and 8, with the right parameters

Buddhini Angelika

Hi, in a direct product $A \times B$ or a semidirect product $A \rtimes B$, is there a particular name by which we can identify A and B? Can we call them the factors of the product?

Rudi_Birnbaum

08:14

@Rithaniel sinc they are periodic, all of them in a way have to show either 2-,3- or 4-fold rotational symmetry.

In crystallography you call the tiles "asymetric units" and the smallest periodic bits "elementary cell"

or "unit cell"

(or 6-fold)

Rithaniel

Fair enough, I can see why that would be true. (Also, I can't immediately discern the unit cell for type 9. What would that look like?)

Alessandro Codenotti

@Zee do you want to glue sections of a sheaf or do you want to glue schemes? I've never heard of gluing sheaves

Leaky Nun

@athos $(\Bbb Z/p^n\Bbb Z)^\times$ cyclic (for $p>2$) of order $(p-1)p^{n-1}$ [1] with $(\varphi(p-1)) (p-1) p^{n-2}$ possible generators. Let $q = \lim_{m \to \infty} i^{p^m}$ where $i$ is a primitive root of $p$. Then $q^p=q$. Then, $(q+pr)^{ap+b} = q^{a+b} + bq^{a+b-1} rp + O(p^2)$, so it should work as long as $r \in (\Bbb Z/p^{n-1}\Bbb Z)^\times$ (if it works then it has to be those).

[1]: combinatorial interpretation: p-1 choice of the ones digit.

what that means for $p=5$ and $n=3$ is that you can compute the primitive roots of $5$, that being $2$ and $3$, and then what you want will be numbers of the form $a \times 5^2 + b \times 5 + \omega^{125}$ where $a \in \{0,1,2,3,4\}$, $b \in \{1,2,3,4\}$, and $\omega \in \{2,3\}$

ÍgjøgnumMeg

@Buddhini yes, and in the semidirect product you call $A$ the normal factor

Leaky Nun

tm = pow(2,125,125)
print(tm)
print(*((tm + 5*b + 25*c)%125 for b in range(1,5) for c in range(5)))

tm = pow(3,125,125)
print(tm)
print(*((tm + 5*b + 25*c)%125 for b in range(1,5) for c in range(5)))

@athos ^ corresponding code

Akiva Weinberger

08:32

@Rithaniel Here's a GIF of type 9, with its parameters varying

Its unit cell is eight tiles (two groups of four)

Leaky Nun

@AkivaWeinberger trippy

Akiva Weinberger

Note that there are two types of groups of four in that image (different orientations)

Find one of each type that's next to the other

(The GIF pushes it into a nonconvex tiling at parts even though this was from the classification of convex tiles)

Leaky Nun

@AkivaWeinberger if you know anything about Lie groups, there's this very important exponential local-isomorphism

which turns out to be also present in the p-adic world

Akiva Weinberger

You mean the map from the algebra (tangent space) to the group?

Leaky Nun

$\exp : p^n \Bbb Z_p \to 1+p^n\Bbb Z_p$ ($n=2$ for $p>2$; $n=3$ for $p=2$)

@AkivaWeinberger yes

$\exp(p^n r) = \sum \dfrac{r^k}{k!} p^{nk} = 1 + p^nr + O(p^{2n})$

oh no, $n=1$ for $p>2$ and $n=2$ for $p=2$

$\exp : p\Bbb Z_p \to 1+p\Bbb Z_p$ for $p>2$

$4\Bbb Z_2 \to 1+4\Bbb Z_2$ for $p=2$

athos

08:42

@LeakyNun thx.

Leaky Nun

no problem

athos

however i don't follow the part that Let $q = \lim_{m \to \infty} i^{p^m}$ where $i$ is a primitive root of $p$

Leaky Nun

those are the Teichmuller representative

very fun to play with

athos

what is this limit for, as m goes to \infity

Leaky Nun

somehow it converges!

to a number $q$ where $q^p=q$

athos

08:45

let p=5, what is q?

Leaky Nun

anyway $\exp$ should descend to a map $p \Bbb Z/p^n\Bbb Z \cong 1 + p \Bbb Z/p^n\Bbb Z$

@athos so if $i=1$ then $q=1$; if $i=2$ then $q=57$

if i=3 then q=68, if i=4 then q=124 = -1

and $\exp : p\Bbb Z/p^n\Bbb Z \cong 1 + p\Bbb Z/p^n\Bbb Z$ should be the proof that $(\Bbb Z/p^n\Bbb Z)^\times$ is cyclic for $p>2$

$p$ has order $p^{n-1}$ so $\exp(p)$ has order $p^{n-1}$ also

so I've found an element of order $p-1$ and an element of order $p^{n-1}$

@AkivaWeinberger might you be interested in this?

Akiva Weinberger

How much of that was addressed to athos and how much was addressed to me

Leaky Nun

@AkivaWeinberger some were addressed to athos and some were addressed to you

so $(\Bbb Z/p^n\Bbb Z)^\times$ has order $(p-1) p^{n-1}$ right

those two numbers are coprime

so to prove that it's cyclic we only need to find something with order $p-1$ and something with order $p^{n-1}$

better, the generators correspond (hey this is the proof for the multiplicativity of totient!)

we have this Teichmuller lifting map $\Bbb F_p \to \Bbb Z_p : [i] \mapsto \lim_{m \to \infty} i^{p^k} =: t_i$ which is a monoid homomorphism (respects multiplication) that sends 0 to 0 and 1 to 1

so that's how you lift the roots of $X^p-X$ from $\Bbb F_p$ to $\Bbb Z_p$

restrict back to $\Bbb Z/p^n\Bbb Z$

and you'll have found a subgroup of order $p-1$ by removing $0$

i.e. $\Bbb F_p^\times \to \Bbb Z_p^\times \to (\Bbb Z/p^n\Bbb Z)^\times$

Akiva Weinberger

Wait is $i$ just a variable here

Leaky Nun

yes

Secret

09:00

sciencedirect.com/science/article/pii/S0370269315001197

hmm...

Reverse casmir effect

I wonder... what it means to reverse infinity...

another thing that is of interest is the following:

For real functions, it seems when its limiting behaviour tends to $\pm \infty$ is very different when it tends to some real number $r$

The latter does not admit functions that increases at faster and faster rate indefinitely, while the former can

Is it the nature of the reals that causes the slowdown as a function approaches $r$?

Is that has something to do with the nested interval theorem being true in the reals?

Akiva Weinberger

@LeakyNun Wait is $\Bbb Z_p$ $\Bbb Z/p\Bbb Z$ here or is it p-adics

Leaky Nun

p-adics

here's a python program for you to play with

Akiva Weinberger

Why does that not depend on the choice of $i$

Buddhini Angelika

@ÍgjøgnumMeg Thanks a lot!

Leaky Nun

ah well

this has to do with lifting the exponent

$v_p\left((i+p)^{p^n} - i^{p^n}\right) = v_p(p) + v_p(p^n) = n+1$

so as $n$ goes to infinity, the difference goes to $0$

at a very uniform rate

Secret

09:17

However, in the projective line, such does not happen. Is there somethingt topologically distinct between a point at infintiy vs a point at r?

Akiva Weinberger

09:30

Thoughts: When we prove the intermediate value theorem, we aren't learning about continuous functions or the real line; we're learning about the definitions and tools we use to study them

We already know that the intermediate value theorem is true. What we don't know is that our definition of "continuous" is the "right" definition (the right way to mathematically model the phenomenon).

So when we show that these tools can prove the intermediate value theorem, we get more confidence that these are the "right" tools.

If we couldn't prove the intermediate value theorem, we would ditch our definitions and get new ones.

(I think they call this Platonism?)

Leaky Nun

@AkivaWeinberger nice

Secret

hmm...

...I need to remind myself again the proof strategy of the intermediate theorem. It has been a long while since I last used ut

hmm ok, so...

we first start with the epsilon delta notion of continuity, and from that using completeness, we can establish the intermediate value theorem, hence our notion of continuity gain confidence as the correct definition

Akiva Weinberger

Note that it's false in $\Bbb Q$

($x^2-2$ is a continuous function from $\Bbb Q$ to itself and attains both positive and negative values but never attains $0$)

(Similarly, $1/(x^2-2)$, which—as a function from $\Bbb Q$ to itself—is continuous and defined everywhere)

Secret

yup, as in this set, some sup/inf can fall outside the set, and hence the epsilon inequality needed to proceed the proof becomes void

Akiva Weinberger

Yeah, this is how you know the proof needs to talk about supremums (or something equivalent to it)

$\sup\{x\in(0,2):x<0\}=\sqrt2$

Secret

09:44

Now to figure out how intermediate value theorem can help to show there exists no continuous functions that increases rapidly to some r whereas such function does exists when the limiting behaviour is $\pm \infty$...

Let $f$ be a continuous function in the reals with asymptote $r$

Akiva Weinberger

There's also a neat proof involving ordinals

You basically derive a contradiction by showing that you'd be able to embed $\omega_1$ in the reals

(in an order-preserving way)

Zee

Man, just took a mean dump

Secret

Fix any point $x < r$. Consider the interval $[x,r]$. Then since $f$ is continuous, by intermediate value theorem, there is some $c \in [x,r]$ such that $\min (f(x),f(r))<f(c)< \max (f(x),f(r))$

Now as $x$ tends to $r$, the bounds of this inequality shrinks closer to $\max (f(x),f(r))$

Akiva Weinberger

@Zee I'm proud of you

Secret

and hence $|f(r)-f(c)| < \epsilon$ as $x \to r$

That is thus responsible for the slowdown that must happen as the function approaches $r$

Meanwhile, the same thing cannot happen for $r \to \infty$, because $|f(r)-f(c)| > M$ for any given $x \to r$

Zee

09:52

I wish my mom would say that to me

Secret

and hence for $f$ whose "asymptotes" are at infinity, the function can rise indefinitely rapidly

hmm.. this give me another way to think about infinity: Infinity can be defined as a mathematical object that is roughly invariant when subjected to a relation that is tied to a mathematical notion of size

Or in laymen terms, infinity resists the change in size

That might be a way to unify the different notions of infinity in set theories. For example:

$\omega$ does not change size for any finite element added from the left

Aleph cardinals does not change size when multiplied, added to a cardinal that is smaller than them

Dedekind cardinals, Amorphous cardinals, Russel cardinals can be reduced forever, but will never become finite

Inaccessibles are (... ok I think I get lost...)

ok.... Inaccessibles remains the same no matter how many times they are "subscripted", as long those are not inaccessibly large

And finally, for $x >> y$, |x-y| > M for given $x,y$ (i.e. $\infty$ does not become less infinite)

So that means...

💥(Infinity) =

> Let $S$ be a size relation, let $x$ be a mathematical object of type $\text{Inf}$ and let $y$ be the relation $(S,S)$. Then $x$ is infinite if: $$y(S(x),S(x))$$ has type $\text{Inf}$

So in general, infinity is an object which is preserved under some size relation

Thus for example, given a vector space, a vector $v$ is infinite if there exists at least one linear map $A$ where $\det (A) > 1, \text{Tr}(A)>0$ such that $$Av=v$$

Rithaniel

10:25

Messing around with a subgroup of $S_8$ and I'm finding my mind blown. Specifically $\langle (1234),(1537)(2648)\rangle $. I've not previously encountered a group where commutativity comes and goes so freely.

Secret

Non abelian groups are actually pretty common and important

Rithaniel

Yeah, but this is more on individual elements commuting with each other or not. Like, call the first one n and the second one m. $nm\neq mn $, of course, but $m^2n=nm^2$ while $n^2m\neq mn^2$

Secret

ah, now that's more interesting

I wonder do we have a notion of "power commutativity"...?

Rithaniel

Also $mnmn=nmnm $ and $mnm^{-1}n=nmnm^{-1} $

ÍgjøgnumMeg

Muppet Show - Mahna Mahna...m HD 720p bacco... Original!

►

Rithaniel

10:34

(Should have probabaly gone with h and a for the lols)

ÍgjøgnumMeg

lol

Secret

Commutator subgroup

In mathematics, more specifically in abstract algebra, the commutator subgroup or derived subgroup of a group is the subgroup generated by all the commutators of the group.The commutator subgroup is important because it is the smallest normal subgroup such that the quotient group of the original group by this subgroup is abelian. In other words, G/N is abelian if and only if N contains the commutator subgroup. So in some sense it provides a measure of how far the group is from being abelian; the larger the commutator subgroup is, the "less abelian" the group is. == Commutators == For elements...

You make me curious about the derived series of this group

But trying to calculate that by hand will be a pain

Rithaniel

I was actually working on some more abstract problems involving the commutator subgroup the other day.

Like G/N is abelian if and only if N contains the commutator subgroup, if I recall correctly. Though I feel that might be incorrect, for some reason.

Secret

Feb 6 at 11:59, by Rithaniel

Alright, homework question: "If $N$ is a normal subgroup of $G$, show that $G/N$ is abelian if and only if $N$ contains the commutator subgroup of $G$." I'm having some difficulty coming up with a good approach to this. (I think I need to have that if $n\in gN$ then $n^{-1}\in gN$ as well, but I can't figure out how to show that either.)

Rithaniel

Yeah, that's the one.

Secret

10:41

yeah, the commutator subgroup and derived series in a sense, measures how nonabelian a group is

I think it might be possible to teach 1st year undergraduates about commutator subgroups using the rubik cube rotational group as an intuition

this is because many moves of solving a rubiks cube are actually commutators that do not vanish

Thus a group where its derived series is of length 2 will be like an alien rubiks cube where say the move LRRL (each of these being commutators) is not the same as the identity

Eran

11:00

How can I show that a countable union of sigma algebra is an algebra? let A,B in the union then both A and B are in some sigma algebra A_k and A_n , how can I make sure their union is in some sigma algebra?

Secret

you can verify that the 3 axioms of sigma algebra is satisfied. The sets in each sigma algebra, in particular, will be preserved by countable unions

Eran

It doesn't work

the sequence has to be an increasing sequence

If the union does not work with 2 it'll not work with countably many

Akiva Weinberger

11:47

@Rithaniel So you want to figure out when $xy=yx$ in $G/N$, yeah?

When is element of $G/N$ equal to the identity element?

Doesbaddel

12:12

I want to calculate $$\lim\limits_{x\to 0}\frac{\ln(1+2x)}{x^2}$$

using L'Hospitals rule: $$\lim\limits_{x\to 0}\frac{\ln(1+2x)}{x^2}\overbrace{=}^{L'Hospital}\lim\limits_{x\to 0}\frac{\frac{2}{1+2x}}{2x}=\lim\limits_{x\to 0}\frac{4x}{2x+1}\to \frac{0}{1}=0$$

The solution from my lecture is that $\lim\limits_{x\to 0}\frac{\ln(1+2x)}{x^2}$ doesn't exist. What am I missing?

Rudi_Birnbaum

12:30

$\frac{\frac{a}{b}}{c}=\frac{a}{bc}\ne\frac{ac}{b}$

Doesbaddel

oh thx

Rudi_Birnbaum

;-)

Doesbaddel

:D

Im dumb xD

Rudi_Birnbaum

don't say that, Akiva is around ...

Doesbaddel

Oh, sry xD

Mary Star

12:53

Hello!! Could someone of you take a look at my question about LU decomposition with column pivoting?

0

Q: LU decomposition of matrix using column pivoting

I want to find the LU decomposition of the following matrix $A$ using Gauss algorithm and column pivoting. $$A=\begin{pmatrix}6 & 4 & 3 & 1\\ 1 & 1 & 0 & 2 \\ 2 & 3 & 1 & 6 \\ 1 & 3 & 7 & 3\end{pmatrix}$$ $6$ is the largest element of the first column, so we don't have to change something and th...

Akiva Weinberger

13:35

Hm

Astyx

How do you generate those images ?

Akiva Weinberger

Connecting all the rosettes (five fat rhombuses surrounded by five skinny rhombuses forming a decagon)

The original I found here preshing.com/images/penrose.jpg

On an iPhone there's a feature that lets you draw on your images

@Astyx Did you see this thing from last night

Astyx

I did

Akiva Weinberger

I described there how it was obtained (to the best of my memory)

Astyx

In the transcript ?

Akiva Weinberger

13:39

Yeah

A plane cuts through 17D space, and then points and squares from the integer lattice that are close to the plane get projected onto it

Astyx

That's the integer lattice ?

Akiva Weinberger

13 hours ago, by Ted Shifrin

Does it have some particular mathematical construct?

From there^

@Astyx Pieces of $\Bbb Z^{17}$ that got projected onto the plane

Astyx

That's cool

Akiva Weinberger

Kinda like that^

Astyx

Yeah I figured

Akiva Weinberger

13:43

This one (preshing.com/images/penrose.jpg) is called the Penrose tiling, and it was discovered from messing around with pentagons. People later discovered you could get it from higher-dimensional lattices

Astyx

I guess you need a rational slope or something to get something nice

Akiva Weinberger

Rational slope would be periodic

Irrational slope would be quasiperiodic, and that's the interesting case

The Penrose tiling is quasiperiodic

It never repeats, but every finite piece of it appears infinitely many times in the tiling

Astyx

Hmmm

Akiva Weinberger

These two tiles actually force aperiodicity

They tile the plane, but they cannot tile the plane periodically

(The shapes of the notches aren't so important, it's just to ensure that the rhombuses only get put together in certain ways)

@Astyx There's also something called the einstein problem

"Einstein" means one stone

The problem is to find a single tile that forces aperiodicity on its own (instead of the set of two that we have)

There's a weird nonconnected tile called the Socolar–Taylor tile that works

but it's still an open problem for connected tiles

It's known it can't be convex

Alessandro Codenotti

There are some connected single tiles that can tile the plane aperiodically but also periodically as well iirc

Akiva Weinberger

13:51

@AlessandroCodenotti Just use a domino

Take a square tiling, cut each square into two dominoes, but randomly choose whether to cut each one horizontally or vertically

Alessandro Codenotti

I was thinking of the L shaped one, there is a substitution rule producing an aperiodic tiling

Akiva Weinberger

Ah right yeah that's a neat one

You can't force it to do that with notches and tabs though I think

or maybe you can but only if you use multiple tiles (each with their own configuration of notches)

Lucas Henrique

14:25

@Ted: I did the proof myself and I understood it. I just don't get motivation, i.e, how did you see the relation between the constraint equations and $\mathbf{C}(A) \subseteq \mathbf{N}(A^T)^\perp$

Hi @Akiva, @Alessandro

Alessandro Codenotti

Hi

Lucas Henrique

You're kinda addicted to tiles, aren't you, @Akiva?

Akiva Weinberger

briefly reconsiders life choices

Maybe

Lucas Henrique

Chill, it's ok. We all have something to disappoint our parents... for example, I plan to major mathematics. :P

Rudi_Birnbaum

@AkivaWeinberger: Did you have a look at the paper?

@AkivaWeinberger I would die for it if anyone could explain to me how to get the Fourier-transform of the dots in on the black line ....

The thing is the FT of the 2D lattice is simple. Also in principle its not too difficult to get a strip of finite thickness and arbitrary angle in the FT. But I somehow always failed to do it properly.

And its not clear to me how substantial it is to have the projections, or if it a finitely thick stripe does also the trick (of quasiperiodicity).

Rudi_Birnbaum

15:17

@AkivaWeinberger I mean the idea is that taking a strip of the 2D lattice corresponds to a multiplication with a kind of step-function (one direction step the other just a constant). And then multiplication Fourier-tranforms to a convolution operation.

So actually it should be a piece of cake ...

Derek Adams

15:51

Hello gamers

Did you know you could make fractals in MS Paint?

Akiva Weinberger

@Rudi_Birnbaum I did but I didn't really get it

Rudi_Birnbaum

@AkivaWeinberger good to hear that I'm in good company

Jack Don

gamers rise up

Eran

Let $(A_n)_{n=1}^\infty$ be a sequence of measurable sets. A sanity check, is it correct that , $\liminf A_n \subseteq \limsup A_n$?

famesyasd

16:06

what is limsup

Lucas Henrique

@famesyasd $\operatorname{sup} A_n$ as $n\to \infty$

Eran

Nope

Lucas Henrique

What so?

Alessandro Codenotti

@Eran What do you think?

@LucasHenrique $\bigcap_{n=1}^\infty\bigcup_{k\geq n} A_k$

Eran

$\limsup A_n = \lbrace x\in \mathbb{X} : \forall N \in \mathbb{N}, \exists n > N s.t x \in A_n \rbrace $

Lucas Henrique

16:10

That's a new thing to me. Thanks.

Derek Adams

Is there a way of rendering tex in the chat?

Eran

@AlessandroCodenotti it's a yes for me, just checking with you guys

.

Derek Adams

* rendering someone else's TeX

Alessandro Codenotti

@Eran I mean do you have an argument supporting that feeling or is it a guess?

(It is true indeed)

famesyasd

what does this have to do with measurability?

Alessandro Codenotti

16:12

Nothing

Eran

every x that is in A_n for all n>N for some N is definitely infinitely many times

Alessandro Codenotti

But usually one encounters such limsups and liminfs in measure theory, for example in the Borel-Cantelli lemmas and stuff like that

Eran

just from the definition I said

Alessandro Codenotti

Yeah I agree it's just a matter of checking the definitions

Eran

thank you

Alessandro Codenotti

16:15

@LucasHenrique The intuition is that to be in the liminf you need to be in all but finitely many $A_n$ while to be in the limsup you need to be in infinitely many $A_n$

@DerekAdams tinyurl.com/cfqcvpc

Derek Adams

Many thanks

What's everybody studying on this sunday?

Alessandro Codenotti

16:31

Partition relations and partition cardinals here

Derek Adams

Care to explain?

(in laymen's terms preferably)

Vrouvrou

any help for this:math.stackexchange.com/questions/3107573/…

Alessandro Codenotti

Uhm they are some kind of large cardinals but it's hard to tell much about them without going into details

Derek Adams

I see. Sounds more interesting than what I'm doing. I'm currently working through Halmos' Finite-Dimensional Vector Spaces

Alessandro Codenotti

I'm not familiar with it, but if it's written by Halmos I'm willing to bet it's good

Derek Adams

16:47

Very. Have you read his text on Set Theory?

Alessandro Codenotti

I did, it's an extremely readable introduction!

Tobias Kildetoft

Hi @AlessandroCodenotti

Ultradark

I'm studying

Ted Shifrin

@Lucas: Remember that $N(A)$ tells you what linear combinations of the columns give you the zero vector. Thus, $N(A^\top)$ tells you what linear combinations of the row vectors give you the zero vector. That's what is encoded in the constraint equations: Think about how they arrive algorithmically.

hi italic @Alessandro, @Tobias

Tobias Kildetoft

Hi @TedShifrin

Ted Shifrin

16:51

So you failed them all, huh? :)

Tobias Kildetoft

Well, 8 of them in total

Ted Shifrin

Oh, that's not a big percentage.

Tobias Kildetoft

no, it was fine.

Alessandro Codenotti

Hi @Ted

Hi @Tobias

Tobias Kildetoft

For one of them it was more than a fail though, as it was the third attempt, which means the last unless you can get a dispensation from higher up (which requires a good reason)

Ted Shifrin

16:52

Meanwhile, none of my high school kidlets in my AoPS calculus class has done his/her homework. None. I'm sorta tired of teaching.

Well, @Tobias, I stopped several people from graduating during my career.

Ultimately, most of them should probably never have gotten out of first or second year, but too many faculty gave them "generous" grades in calculus and linear algebra, etc., that they definitely did not deserve.

Tobias Kildetoft

Yeah, I feel fine about this one. They had enrolled in the class again, but did not do any homework this time (the homework is mandatory in order to take the exam, but only first time you do the course)

The resulting exam was not even close to a passing grade

Ted Shifrin

Yes, so of course I won't do exercises to try to learn the material ... because I already know I don't know it.

No sympathy whatsoever.

Mike Miller

This is my shortest ragequit yet.

Ted Shifrin

LOL, there was a ragequit?

Mike Miller

For about 12-18 hours.

Alessandro Codenotti

16:55

We still missed you

Ted Shifrin

I don't think I'm to blame for this one :P

Mike Miller

Nope, you're safe. Anyway.

Ted Shifrin

Last conversation we had was about circle-valued $1$-forms :P

Transcript for

$Mathematics$ Mathematics