Mathematics - 2016-05-31 (page 2 of 3)

Hi everyone, this is likely going to be a dumb question so my apologies :) Would the pattern shown at the following link be a fractal? github.com/MadillJ/InterstingCodeCollection/blob/master/…

Semiclassical

in which case you'd get 10/20 = 2*5/20

Akiva Weinberger

10 mins ago, by MathCubes

@EricStucky I mainly don't get how "3 = a *r^2" then it goes to "a =3/r^2"

Semiclassical

and if you note that 10 and 20 have a common factor of 10, you could simplify the left to 1/2

MathCubes

Yea

Semiclassical

so 1/2 = 10/20. which matches the RHS of what i had before. so it's all consistent.

MathCubes

7:14 PM

True

Akiva Weinberger

In there, we're just dividing both sides by r^2 (remember again that we can switch both sides of an equals sign)

Semiclassical

(just as long as you don't divide by zero. that makes things go wrong immediately)

MathCubes

Oh I think I get it now

So can you give me some questions to do?

Semiclassical

if you search online for some algebra questions, you'll probably find some useful stuff

MathCubes

What is this exactly called?

Eric Stucky

7:16 PM

@Phaeze: Yep.

Semiclassical

elementary algebra, i suppose.

MathCubes

to look it up?

Phaeze

@EricStucky awesome, thank you.

Eric Stucky

two-step equations is a common name, I think, Cubes.

Semiclassical

that'll also include things like solving quadratic equations, which you may not have seen yet but presumably will

Akiva Weinberger

7:17 PM

"Dividing both sides by the same thing"? "Algebra with division"? I'm not sure what they'd call it

MathCubes

I think I have

Akiva Weinberger

Do you have a textbook?

MathCubes

@AkivaWeinberger that is what I mean

Semiclassical

if you've got a textbook, start with that

MathCubes

No

I have learn bit there in there.

I mainy don't get how ab = a/b thing

Semiclassical

7:18 PM

basically, this is high-school algebra

Eric Stucky

Yeah, I remember my ridiculous grade school math terms correctly: two-step equations.

Semiclassical

at least, the symbolic bit is

Akiva Weinberger

Well, "ab = a/b" is in general not true

Semiclassical

the whole 6=3*2 means 6/3=2 is more just arithmetic

MathCubes

What is that called ?

Anyone knows

Semiclassical

7:20 PM

not sure what you'd name it as. it's a sufficiently elementary thing that people tend not to linger on it

MathCubes

Humm...

Semiclassical

Here's a practical example of it.

MathCubes

Okay thanks for the help

What is it again

Ab = a...?

Semiclassical

Suppose someone bakes enough cookies that, in a class of 30 students, each student gets two cookies.

MathCubes

LOL

Akiva Weinberger

7:21 PM

@MathCubes Here's a link that might help: msemac.redwoods.edu/~kyokoyama/math106/Math380pdf/sect2-1.pdf Especially the exercises at the end

Semiclassical

Then I can equally well say that, if there were two such classes, then each class would get a total of 30 cookies.

MathCubes

30Okay thakes

Yea

Semiclassical

so 2*30 = 60 -> 60/2=30

Balarka Sen

@MikeMiller I'm having a little trouble understanding G-P's example of failure of stability theorem for noncompact domains. It says, take a bump function $f : \Bbb R \to \Bbb R$ which is $1$ on $(-1, 1)$ and $0$ on the complement of $[-2, 2]$. Then look at the homotopy $g_t(x) = xf(tx)$ of maps $\Bbb R \to \Bbb R$. Thinking of graphs, $g_0$ is $xf(0) = x$ and $g_t$ for any $t$ is a bump function with a bump on $(-1/t, 1/t)$.

So my homotopy homotopes $y = x$ by damping the "ends" to melt off to $y = 0$, with more and more damping factors, right? And this is not transverse to $y = 0$ for any $t$.

MathCubes

I guess this is called "equivalent equations"?

Semiclassical

7:23 PM

yeah, that works

MathCubes

OKay again thanks for the help

going to look this up more

Akiva Weinberger

Here's a link to the entire textbook if you want

mathrev.redwoods.edu/ElemAlgText/ElementaryAlgebra.pdf

(The above link was just a chapter)

Semiclassical

two things remain equal if i add, subtract, multiply, or divide* by the same number (*so long as i don't divide by zero)

MathCubes

Thanks I will bookmark it

Semiclassical

dividing by zero is one of those things that's pretty easy to do symbolically if you're not careful

so that warning is more important than you might expect

MathCubes

7:25 PM

Relly

can you give an examable

Semiclassical

yeah. there's a bunch of false arguments one can do to get things like 0 = 1.

MathCubes

Okay thanks for the warning.

Semiclassical

here's a classic one: en.wikipedia.org/wiki/Mathematical_fallacy#Division_by_zero

Akiva Weinberger

0=0 (true)

MathCubes

thanks

Akiva Weinberger

7:27 PM

0*1=0*2 (true)

0*1/0=0*2/0 (neither side is defined)

1=2 (false)

Semiclassical

that works, yeah

MathCubes

Okay

Akiva Weinberger

There, if it were any other number than zero, I could divide by it and cancel it out.

But you can't divide by zero, so you can get a false equation at the end if you do

MathCubes

Yea

So there would be no work arounds?

BRB

Akiva Weinberger

The problem is when variables are involved. Like, you might accidentally divide by "x-x" without realizing that x-x is always 0

Semiclassical

7:30 PM

there'd better not be a workaround, since the place one was trying to reach was definitely false.

Akiva Weinberger

We're, like, 99% sure that you can't prove 1=2!

(Just kidding. You can't prove a falsehood, so we're 100% sure there's no proof of 1=2.)

Balarka Sen

I got 99 problems but 1 = 2 ain't one.

3

Akiva Weinberger

Semi, don't bring up Gödel

Semiclassical

psh

(Godel: proof that the more math you know, the weirder the world gets)

Akiva Weinberger

Gödel is one of those things that's incredibly easy to misunderstand. Along with the ζ(-1) thing.

(1+2+3+… "=" -1/12)

And Cantor.

And .999… .

MathCubes

7:35 PM

LOL

Mike Miller

@Balarka Seems fine.

Balarka Sen

@MikeMiller Alright. Also, I can prove density of Morse functions (although not the Morse lemma, because it's not in there).

Density as in the effective form of density.

Mike Miller

Sure.

user 1618033

7:51 PM

Back.

@AkivaWeinberger did you try my integral on main?

$$\int_0^1 \int_0^1 \frac{\displaystyle(1+y) \log\left(\frac{1+x+y-xy}{1-x+2y+y^2-xy^2}\right)}{(1+y)^3+2x(1+y)(1-y-y^2)+x^2(‌1-y(3-y-y^2))} \ dx \ dy$$

0

Q: Practice for the beginners

A question addressed to the beginners in calculus: Calculate $$\int_0^1 \int_0^1 \frac{\displaystyle(1+y) \log\left(\frac{1+x+y-xy}{1-x+2y+y^2-xy^2}\right)}{(1+y)^3+2x(1+y)(1-y-y^2)+x^2(1-y(3-y-y^2))} \ dx \ dy$$ using simple manipulations with double integrals. EDIT: Then answer is simply $\z...

Also vote (I refer to all) for opening if you see its beauty.

Eric Stucky

No need to bring up godel

Just work mod 1.

Semiclassical

@EricStucky oh, I didn't get a chance to sketch the MO question I had in mind

we established earlier that 1 has the 'Egyptian' representation $$1=\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{43}+\cdots$$

Eric Stucky

43 on that last one, semic?

mkay

Semiclassical

thanks, though that's a bit of an unconscious slip as you'll see in a moment

we could more generally write out the sequence of such representations

$$1 = \frac12 +\frac12 = \frac12+\frac13+\frac16 = \frac12+\frac13+\frac17+\frac1{42}=\cdots$$

so we really have a family of Egyptian representations

Akiva Weinberger

(Same thing happens for my sequence, just with numerators $n$ instead of $1$)

Semiclassical

7:58 PM

now, the nontrivial part is to link this with the paper I linked

(quick question: anyone know the latex symbol for disjoint union? it seems to be an upside-down $\Pi$)

$\coprod$?

\coprod? $\coprod$

that works

$\sqcup$

\sqcup. That's uglier.

Semiclassical

yeah

Akiva Weinberger

8:03 PM

$\amalg\coprod$ \amalg\coprod

Semiclassical

anyways, suppose you have a group $G$. then it has 'groupoid cardinality' (see paper for the definition) $|G|=\frac{1}{\#G}$ where $\#G$ is the order of the group

Akiva Weinberger

And for groupoids you replace that with the automorphism group of a dot (groupoid elements starting and ending at the same dot) and add over the components

Semiclassical

and if you take a disjoint sum of groups G,H then the resulting groupoid $G\coprod H$ has cardinality $|G\coprod H|=|G|+|H|=\frac{1}{\#G}+\frac{1}{\#H}$

Eric Stucky

mmm I think I see where this is going

Semiclassical

so you have groups of appropriate order, you can build an egyptian fraction out of them in a fairly brute-force way

for example, you could take G=Z/nZ to get a contribution of 1/n

Now, going back to the Egyptian reps from earlier

Using them, I can make groupoids with cardinality 1 in a pretty brute force way:

Akiva Weinberger

8:09 PM

Are groupoids essentially categories?

Oh, categories don't need things to have inverses.

Balarka Sen

Groupoids are a kind of categories.

Semiclassical

$$1 = |\mathbb{Z}/2\coprod \mathbb{Z}/2|=|\mathbb{Z}/2\coprod \mathbb{Z}/3\coprod \mathbb{Z}/6|=\cdots$$

right?

thing is, this is a pretty brute-force construction. it may exist, but it's hardly natural

Akiva Weinberger

I think there's an AlgTop book somewhere that deals with the fundamental groupoid a lot

Semiclassical

so here's the question: is there a 'natural' sequence of groupoids $\{G_k\}$ such that the desired Egyptian representations arise from the definition of groupoid cardinality?

Akiva Weinberger

where it's just paths, not necessarily loops

Semiclassical

8:13 PM

which I forget to cite, and which I don't really appreciate the meaning of

Balarka Sen

Ronnie Brown.

Who else?

Semiclassical

$|G|=\sum_{[\bullet]\in G}\frac{1}{\#\text{Aut}(\bullet)}$ where $[\bullet]$ is a component of $G$ and Aut is the automorphism group of $\bullet$

anyways, that be the question I have in mind.

No idea if it's got a chance of being true, but I think the intention is clear.

Balarka Sen

Hi @Clarinetist

Eric Stucky

Yeah, that seems plausible

that was a heck of a typo

Andrew Thompson

Hi @MikeMiller, @BalarkaSen

Balarka Sen

8:18 PM

Hello.

Andrew Thompson

How's life?

Semiclassical

Yeah

Balarka Sen

So-so, Andrew. Too much schoolwork, trying to get some time to do math.

Semiclassical

It amounts to some claim about automorphism groups and the like, I suppose

Clarinetist

This is probably a long shot, but I might as well try: does anyone know anything about time series? I would like to show that $\text{AR}(p)$ is a linear filter. Let $\tilde{z}_t = z_t - \mu$, for some constant $\mu$. Then the $\text{AR}(p)$ process is given by
$$\tilde{z}_t = \phi_1 \tilde{z}_{t-1} + \phi_2 \tilde{z}_{t-2}+ \cdots + \phi_p \tilde{z}_{t-p-1} + a_t$$ where $a_t \sim \mathcal{N}(0, \sigma^2_a)$. A linear filter is just a linear combination of the $a_i$ (plus, if applicable, a constant).

Hi @BalarkaSen

Semiclassical

8:20 PM

is that basically just a recurrence relation?

Mike Miller

Hi @AndrewT. I want to take a nap.

Clarinetist

@Semiclassical It seems to be, but you would think... there has to be an initial condition somewhere. IDK.

Semiclassical

Hmm.

Andrew Thompson

@MikeMiller Do it. Helps with productivity afterwards.

Balarka Sen

@Clarinetist I learnt a thing or two about index numbers.

Mike Miller

8:21 PM

I have something to read on the plane in an hour, but yeah, too tired right now

Semiclassical

Main thing I notice is that the product terms in that sum are a convolution

Clarinetist

@Semiclassical The text states:
> It is not difficult to see that the $\text{AR}(p)$ is a special case of the linear filter model. For example, we can eliminate $\tilde{z}_{t-1}$ from the RHS by substituting $$\tilde{z}_{t-1} = \phi_1 \tilde{z}_{t-2} + \phi_2 \tilde{z}_{t-3} + \cdots + \phi_p \tilde{z}_{t-p-1}+a_{t-1}$$ Similarly, we can substitute for $\tilde{z}_{t-2}$, and so on, to yield eventually an infinite series in the $a$s.

Semiclassical

well, the other way to get it is to think of it as a big matrix equation

Andrew Thompson

You travel a lot, Mike. Common for gradstudents?

Clarinetist

@Semiclassical Hmmmm, I'll have to try that when I get home

Semiclassical

8:23 PM

i.e. you've got a column vector $Z$ such that $Z=\Phi Z+A$

Clarinetist

@BalarkaSen I know nothing about index numbers. Looks like economics

Semiclassical

and $\Phi$ has some structure. Toeplitz with finite bandwidth, I think?

Clarinetist

@Semiclassical Lol, that's gibberish to me

Semiclassical

heh. Toeplitz just means it's a matrix with constant values along diagonals

Clarinetist

@Semiclassical How bizarre! I've definitely seen matrices like these before, but didn't know there was a name for them!

They're used all of the time in stats

Semiclassical

8:25 PM

and finite bandwidth just means that there's a finite range of diagonals that are nonzero

Balarka Sen

@Clarinetist Sort of. It's a way to compare two given data sets, just like central tendency and variation.

Semiclassical

Additionally, $I-\Phi$ is itself Toeplitz by construction

Mike Miller

@AndrewThompson I just go wherever someone will pay me. I'm not traveling this summer.

Balarka Sen

But more interesting from a practical point of view.

Semiclassical

and $Z=(1-\Phi)^{-1}A$, so the main concern is how to invert this toeplitz matrix

i think it's also a lower (or upper, depending on how you order the column vector) triangular matrix

Andrew Thompson

8:28 PM

@MikeMiller Pay you for what? Do you not get paid by UCLA?

Mike Miller

Pay for my travel.

Semiclassical

and if memory serves there's a lot that can be said, especially since the main diagonal of $I-\Phi$ is just ones

Andrew Thompson

Ah, I see.

Just begged my department for travelmoney, hope I get it.

Clarinetist

@Semiclassical I will have to check this out later! Gotta leave work. Thank you!

Semiclassical

plus, you formally have $(I-\phi)^{-1}=I+\Phi+\Phi^2+\Phi^3+\cdots$

which seems right for a time series

np, glad to be helpful

Eric Stucky

8:29 PM

Same, Andrew :P

Semiclassical

@EricStucky Anyways, I think I'll try to draft a problem based on that sketch.

Eric Stucky

Sounds good :)

Mike Miller

Where are you trying to go?

Andrew Thompson

I'm going regardless. Conference in Copenhagen.

Mike Miller

on?

Andrew Thompson

8:34 PM

Various. Can send a link.

gecogedi.dimai.unifi.it/event/316/1

Obliv

Anyone know how I might go about proving: if $n \geq m$ then the amount of $m-$cycles in $S_n$ is given by $\frac{n(n-1)(n-2)...(n-m+1)}{m}$?

Mike Miller

Very nice. I don't know why I didn't apply this year, that looks like an excellent event.

Obliv

also when it says $m-$cycle, does that mean $(1~2)(3~4)$ is a $2-$cycle?

Balarka Sen

@Obliv So you want to prove that the number of $m$-cycles in $S_n$ is $n!/(n - m)!$. Can you tell me what an $m$-cycle means?

Obliv

or is it strictly singular $m-$cycles

Balarka Sen

8:40 PM

I mean, what does it do?

Semiclassical

well, $(12)(34) \neq 1$. so if $(12)(34)(12)(34)=1$ then it's a 2-cycle

where 1 is the identity cycle, whatever that's typically denoted as

Eric Stucky

Semic, some authors distinguish 2-cycles and transpositions.

Obliv

well it permutes $m$ elements, right? @balarka

Mike Miller

I like Randall-Williams as a researcher and a speaker, and homological stability is big in Copenhagen and also very cool (which is what Wahl will be talking about).

You're probably more interested in Akhil's talk. I have no opinions on that.

Semiclassical

hm. i'd think (12)(34) would be a 2-cycle but not a transposition

Eric Stucky

8:42 PM

^.– I meant to write 'involution' whoops

Semiclassical

every 2-cycle being a product of pairwise-distinct transpositions

Eric Stucky

yeah, I agree it's not a transposition.

Semiclassical

anyways.

Balarka Sen

@Obliv Yes, an $m$-cycle is precisely a permutation of a set of $m$ elements from $\{1, \cdots, n\}$.

Andrew Thompson

I am, but homological stability looks fun as well.

Have you been at YTM previous years?

Obliv

8:44 PM

under this definition,then, disjoint cycles of $m$ length are not considered a single $m-$cycle? @balarka

Mike Miller

I have not.

Eric Stucky

(12)(345) is not a 5-cycle, though ;)

Mike Miller

I want someone to teach me Madsen-Tillman-Weiss at some point.

Balarka Sen

So, first, can you count for me how many set of $m$ elements you can choose from $\{1, \cdots, n\}$, not caring about the order?

Obliv

they must be unique? @balarka

Balarka Sen

8:47 PM

What do you mean by that?

Obliv

like if $\{1,2,3,4,5\}$ and $m = 3$ can I choose $\{1,2,3\}$ and $\{2,3,4\}$?

Balarka Sen

Yeah, sure, why not? Just any m element set from an n element set.

Can you count the number? Do you know what it's called?

Obliv

if the sets must be unique from each other but the elements can be in multiple sets is it $m!$?

m factorial**

Balarka Sen

Um, no.

Obliv

no

lol

Mike Miller

8:51 PM

@Andrew but at the same time, travel is tiring, so I guess I'm glad I'm not doing any more travel this summer - despite there being a few very good events

Obliv

i'm thinking of difficult examples. for $\{1,2,3\}$ and $m=2$ then the amount of sets of $m$ elements is 3? but you said not to care about order so I can pick 6 @balarka

is this true

Semiclassical

@BalarkaSen that's not quite what he said. his would be that divided by $m$

Andrew Thompson

@MikeMiller Yesh. Just 1h30min from my city, thankfully.

Semiclassical

( (123) = (231) and all that)

Balarka Sen

@Semiclassical Right, typo. Whoops.

@Obliv OK, if you don't recognize this, let us start from scratch.

user 1618033

8:53 PM

@EricStucky I'm not sure what you mean by the comment you let. Are you challenging me somehow?

Balarka Sen

Suppose I have a set $\{1, \cdots, n\}$ of $n$ numbers. I want to pick a set of $m$ numbers from it, this time caring about the order (by that I mean e.g. {1, 2}, {2, 1} are distinct).

Eric Stucky

No

Did you not see JMP's edit?

It was atrocious

user 1618033

@EricStucky "Does the OP know what ζ means? Who knows?"

Mike Miller

@Andrew Well, I just mean that one doesn't get any of their own work done and a lot of math in a short period of time can make you tired.

Eric Stucky

Come on, u16.

You know me, I know you

We both know that you know what zeta is

Obliv

8:55 PM

again then for $\{1,2,3\}$ $m=2$ the obvious is $\{1,2\},\{2,1\},\{1,3\},\{3,1\},...,$ ending up with $6$ sets of $m$ elements. i suppose it is represented by $n!$ @balarka

Eric Stucky

But, read JMP's edit from an outsider's perspective.

You will see that, if someone didn't know you

they would have no way

to tell if you knew what $\zeta$ was.

Obliv

ah I see then since in cycle notation we only care about one representation so then divide by $m$ which is 2 in this case

user 1618033

@EricStucky OK OK

Balarka Sen

@Obliv No, be very careful about making such assumptions.

Obliv