Mathematics - 2021-10-12 (page 1 of 2)

shintuku

00:45

the proof of $A \preceq B \land B \preceq A \implies A \thickapprox B$ is lengthy

i wonder if some people can do it on demand

leslie townes

schroeder-bernstein? there's a kind of diagram that accompanies the usual argument that is pretty easy to remember

shintuku

yeah that one

oh really? i'll look into it

anakhro

@TedShifrin This was my first meme ever. It hasn't taken off, but when it does, I will be stackexchange-wide famous.

Ted Shifrin

Nested annuli, yup.

leslie townes

Ted Shifrin

00:55

@anakhro Maybe not.

leslie townes

that's how it appears in enderton's set theory book. f is one-to-one from A to B and g is one-to-one from B to A and C_0 = A \ g(B) and C_{n+1} = g(f(C_n))

Ted Shifrin

Oh, no @Leslie. The annuli are way better!

anakhro

@TedShifrin :(((

shintuku

at leslie: i'll check it out thanks a lot!

leslie townes

the annuli are great but i only have an electronic copy of a book that has the ping pong diagram

Ted Shifrin

00:56

I learned the annuli from Munkres. Way superior to a darn logician.

leslie townes

shin: the bijection is defined as f on union_n C_n and g^{-1} otherwise

shintuku

Ted: Munkres' topology book?

Ted Shifrin

Yes

shintuku

i'll look into it too, thanks

I've been using the proof that does ninjutsu with the set $\{C:C\subseteq A \land g"(B\setminus f"C) \subseteq A \setminus C\}$

and I have no clue how the author got intuition for the way he moved in the proof

i'll check out enderton and munkres

leslie townes

so, easy to remember the intuition if you've seen it presented nicely elsewhere. maybe harder to come up something on one's own. even just realizing that it requires proof, and can be done without AC, is kind of a galaxy brain moment

ted: relying on diagrams created by a logician is my way of relying on diagrams without relying on diagrams created by topologists

anakhro

01:04

Can we not pick on people based on their chosen research fields? :(((

This message brought to you by Applied Mathematicians.

leslie townes

anak all of this is a joke based on me being hostile to diagrams and pictures in general. no offense to topologists is intended. even geometers are sometimes OK.

but applied mathematicians, just no. you do have to draw the line somewhere

Avra

Just want to say hello

anakhro

@leslietownes I was making a joke, too, Leslie.

To explain it and ruin it, I was making fun of applied mathematicians.

leslie townes

oh. good.

there's actually a sweet spot to be in. it's being just applied enough that a large portion of your audience doesn't care about proofs, but not being so applied that people actually compare your methods against what people might actually want to use.

if you can strike that balance, i say go for it

shintuku

i've been taking comfort in the fact Ramsey the guy from Ramsey's theorem was an economist, and von Neumann was the guy responsible for general equilibria in economics

anakhro

01:14

Smale and Morse also did stuff in economics, didn't they?

shintuku

didn't know, cool!

anakhro

I think I might be thinking of Smale applying Morse theory in economics.

There was that one amusing story about economists and Kakutani.

> In his game theory textbook,[11] Ken Binmore recalls that Kakutani once asked him at a conference why so many economists had attended his talk. When Binmore told him that it was probably because of the Kakutani fixed point theorem, Kakutani was puzzled and replied, "What is the Kakutani fixed point theorem?"

shintuku

hahaha

Koro

If a function $f:[a,b]\to \mathbb R$ is Riemann integrals then $f$ must be discontinuous at a finite subset $S$ of $[a,b]$.

S also includes an empty set so I the statement seems true. However, I think that there is a function which infinitely discontinuous yet Riemann integrable.

leslie townes

01:33

either f has no discontinuities (take S = emptyset) or it does (in which case the set of discontinuities contains a point p, take S = {p}), done

it's not true that the set of discontinuities of f has to be finite, but the set of discontinuities will certainly contain a finite set

so, file under "weird statement, but true"

anakhro

Heh.

leslie townes

the theorem here is that the set of discontinuities of a riemann integrable function must have "measure zero" (you don't need to define measure in general to define this concept)

but it doesn't have to be finite

Thorgott

the right statement is Riemann-integrable iff bounded + continuous except on a set of measure zero

this is a surprisingly tricky thing to prove

in fact, it is quite non-trivial to give a straightforward argument that a Riemann-integrable function is continuous at any point at all

leslie townes

it's also fairly useless, as results go

along with characterizations of those functions for which FTC holds

Thorgott

well, that's cause the Riemann integral is useless :P

leslie townes

01:37

yeah, i agree

it's always seemed a little weird to me that technical results about a broken integral get more attention (in my view) than simple results about non-broken integrals, at least in some books

anakhro

@leslietownes how would you teach analysis, then?

Faust

@leslietownes The only time i have seen the Axiom of choice come up is in topology, the odd broken example; we talked about it when i learned about Hilbert spaces but all of them used the same definition that was on Wikipedia. The guy was just being an asshat i made it clear what definition i was using and he kept saying i was wrong because it wasn't the definition he uses. The definition one chooses to use is irrelevant as long as everyone is on the same page with which on you are using.

I even made it clear to define what definition I was using and explained why i was using that definition. Yet they kept insisting it was wrong, I just wanted to be done with the conversation.

anakhro

@Faust I find that in those sorts of discussions, "Ask, don't tell" is the name of the game.

Faust

I don't remember how the conversation came about in all honesty.

i dont think i was asking them a question, but i may of been.

anakhro

01:54

Well, I just mean that people are less likely to accept the truth if they don't come across it themselves (e.g. through answering a question).

leslie townes

anakh: i would start with the riemann integral, which is fine, but not waste any time on big theorems characterizing riemann integrability

anakhro

@leslietownes Do you think you could adequately highlight how bad the Riemann integral is if you didn't characterize Riemann integrability?

Faust

@anakhro It was just frustrating because it was completely irrelevant to what i was trying to talk about lol.

leslie townes

anakhro: you could say with examples or even just discussion that the riemann integral is not well behaved under various operations and leave it at that

same with the FTC. prove it for the functions that you meet in a calc textbook, do not bother formulating more general FTCs let alone the most general possible

anakhro

@leslietownes What examples would you use in particular?

Faust

01:56

The Riemann integral can be fixed once you understand whats wrong with it but it is still much harder to compute than several other types of integrals.

The one advantage the Reimann integral has is that it has a very nice geometric intuition about what is going on.

leslie townes

anakhro any of the standard ones. for example, pointwise limits of riemann integrable functions do not need to be riemann integrable. a few examples, done, no more about what might or might not be riemann integrable

"here's this notion, let's now abstractly characterize what we can input into it and what its outputs are" is not that useful in teaching introductory analysis. too many pathologies lurk right under the hood

rudin's chapter 2 not withstanding, you shouldn't need to study general point set topology first :)

Faust

One thing the Reimann integral has problems with is when it ends up with a point of discontinuity at infinity as it kind of takes an average which causes it to break. You can fix this with enough analysis applied correctly or the concept of a measure. In the end though even fixing that problem it is generally very rare that it is true that $ \int \Sigma_{n}^{\infty} f(x) dx = \Sigma_{n}^{\infty} \int f(x) dx $

anakhro

@leslietownes do you prefer to introduce the Lebesgue integral with or without the concept of measures (or, e.g. the Lebesgue measure).

Faust

but with something like the lebegue integral its much easier to make such a statement true

leslie townes

anakhro: i don't have an opinion on that. i think after a first exposure to analysis, the choices made don't matter too much

anakhro

02:04

@leslietownes Would you avoid introducing the Lebesgue integral in the "first exposure", then?

leslie townes

yes

Faust

Hmm I am curious i have never seen the lebesgue integral introduces without measures.

at my university its not usually touched on as undergrad.

leslie townes

same at most US universities. i think that this is fine

Faust

Its defiantly cool though ^^

leslie townes

faust: some functional analysis oriented books start with linear functionals on function spaces, pass to a completion, and then define the measure in terms of that

very roughly speaking, you define integrals first, then measures

Faust

02:07

Its not uncommon especially in stats to define measures early

leslie townes

i think i've seen one or two stat books do something similar, where the expectation is given a primary role and the event space is incidental

which kinda makes some kind of applied sense

Faust

It is convenient for stats. Anyway how is everyone doing today? all full of turkeys and sleepy?

Koro

@leslietownes thanks a lot Leslie and Thorgott. This “measure” is introduced while studying Lebegsgue integral?

Faust

oh i forgot its not thanksgiving in the US

leslie townes

koro: a lot of the time yes

ha, my daughter was just chanting 'canada, canada, canada' today because they had something about thanksgiving at her day care.

Faust

02:11

@Koro Usually, but you can come across it without it.

I remember something blowing my mind when i saw it the first time. I think it was showing that the measure of the set of open balls around every rational number having measure 0 was what blew my mind.

I mean i had known for a long time that the integral from 0 to 1 over Q was 0 but that is alot more points than just Q and it was still 0.

anakhro

@leslietownes Soon she will become one of us.

Faust

It seems much more natural to have an introduction to topology before learning about the lebesgue integral imo

Koro

For now, I perceive “measure” of a set as “length” of the set. For example: Cantor set has measure zero for it does not have any line segment in it.

[0,1] has measure 1-0=1

anakhro

@Faust To define the Lebesgue integral without mentioning the premeasure or measure, you first define the integral of a step function. Then define a Lebesgue integrable function as a function f such that there is a sequence of step functions f_n such that (i) $\sum\int|f_n| <\infty$ and (ii) $f(x) = \sum f_n(x)$ for every x such that $\sum |f_n(x)| < \infty$. Then the Lebesgue integral of f is defined by (i).

Of course there are a few technical results to show (e.g. that it doesn't depend on the family f_n)

But by far, this avoids having to define anything remotely close to a sigma algebra.

Faust

02:27

@Koro It depends on what cantor set your talking about some of them have a measure. somehow cantor found the perfect thing to describe what was going on in some sense

anakhro

However, it does abstract away the notation of measures, so that might be undesirable.

Faust

@anakhro thats intresting

Koro

@Faust I see. I was however referring to the cantor set one gets when they keep on trisecting [0,1] discarding middle thirds.

Faust

Care to take a guess what happens if you discard the middle 1/4's instead of 1/3's.

Koro

In that case, do we get a set that’s not perfect and has a line segment in it?

Faust

02:34

well both cases have line segments in them but the 1/4 case you end up with measure 1/2 instead of measure 0

at least line segment in some sense of the word

Koro

@Faust I see. I have yet to see how measure is defined.

But in case of cantor set (obtained by trisecting and discarding middle thirds), there is no line segment in it.

Faust

its very similar to the idea of the cantor set

Koro

I’ll read about that in more detail, when I cover Lebesgue integrals.

I have noted many concepts recently which use Lebesgue measure .

Faust

basically, if your talking about the length of the interval (1,0) your set will have some measure between 0 and 1 and the complement of your set will have measure so that the sum of the measure of the original set and its complement added together equals 1.

In the cantor trisecting example you basically prove that all of the measure in that interval is contained in the Complement of the cantor set.

You still have alot of points in the cantor set, in fact what seems like many, many more points than Q has.

zed111

05:31

I am seeing two definitions of a reducible representation, 1. if there exists a non trivial G-invariant subspace and 2. if the matrix can be written into upper triangular form

How are these two connected

leslie townes

if U is an invariant subspace and you choose a basis of U, and then extend it to a basis of the whole space, with respect to that basis, the image of the rep will be block upper triangular

and vice versa, if you can turn something into block upper triangular via a change of basis, that gives you an invariant subspace

Nano Miratus

05:56

hi! has anybody of you knowledge on this topic: ftp.jssac.org/Editor/Suushiki/V18/No1/V18N1_102.pdf ?

(cyclic polygons and their circumradius)

i just can't believe that it's that hard to find a formula for this

337,550,051 terms and a polynomial of degree 38 just for a heptagon, ???

can't find any more recent papers/research/work on this specific topic

ah, just found another one from 2018: jssac.net/Editor/CJssac/V03/V3_101.pdf

still madness

"For example, the expansion of P¯28 took 371 days of CPU time in total (with 182 jobs, on Machine B described in Subsection 4.1)" :$

i don't understand why exactly it's that difficult

leslie townes

06:13

i guess the lesson is that the intuition that this problem ought not to scale horribly is wrong :)

difficult to know what to make of that figure without knowing more about the CPU and whether anybody bothered too much with how fast to make that code. maybe not too surprising that large amounts of exact arithmetic with large numbers of things could take a while.

AbstractSpacecraft

Made a modular arithmetic answer on MO and somebody DV'd it! :(

-1

A: Are there finitely many primes $x$ such that for a fixed odd prime $p$, $n=x^{p-1}+x^{p-2}+\dotsb + x+1$ is composite and $x \mid \phi(n)$?

Quoting a comment above: I'm getting: $\phi(n) = n \prod_{p \mid n} (p-1)/p$ so that if $\phi(n) = 0 \pmod x$, then that implies $n \prod_{p \mid n} (p-1) = 0 \cdot \prod p = 0 \pmod x$ and of course $n = 1 \pmod x$ so we get, if $x \mid \phi(n)$ then $x \mid \prod_{p \mid n} (p - 1)$ and of co...

David Choi

06:40

Guys, I have a pretty simple question:

I was asked to expand $\sqrt{1+x}$ to two terms plus remainder

Real simple; I busted out the expansion:

$1 + \frac{x}{2} -\frac{x^2}{(1+\theta x)^{\frac{3}{2}}}$

But then my textbook says the expansion should be

$1 + \frac{x}{2} -\frac{1}{4(1+\theta x)^{\frac{3}{2}}}$

I made a typo for the first equation

It should be $-\frac{x^2}{8(1+\theta x)^{\frac{3}{2}}}$

I don’t see at all how to obtain the textbook’s expression for remainder

Btw I apologize for any formatting errors; I’m typing this on my phone

David Choi

07:04

I made a post abt this: math.stackexchange.com/questions/4274301/…

porridgemathematics

07:16

could anyone with knowledge about the dirichlet problem/greens functions help me take a look at this please? math.stackexchange.com/questions/4274327/… , I am thoroughly confused

actually it isnt even necessary, just basic knowledge about harmonic functions

robjohn

@DavidChoi that looks wrong.

porridgemathematics

i think it is possible that the assumption I made about regularity of the domain is just wrong.. and so I can't solve the dirichlet problem with those boundary values because of the horziontal slit, this makes intuitive sense to me, but the result of ransford does indeed claim that the dirichlet problem can be solved in this type of domain

David Choi

@robjohn do you mean the textbook?

robjohn

07:48

@DavidChoi yes.

I get your second error term, with the $8$ in the denominator.

Koro

10:25

I claim that Dirichlet function is algebraic.

That is, $f:\mathbb R\to \{1,0\}$ defined by $f(x)=\begin{cases} 1; x\in Q\\ 0; x\in Q^c \end{cases}$ is an algebraic function.

Consider the polynomial $P(z)=(z-1)(z)(z-x)$. Clearly for all $x\in R$ we have $P(f(x))=0$

It follows that function $f$ is a root of the polynomial equation $P(z)=0$.

Therefore, by definition $f$ is an algebraic function.

Is my understanding correct? Thanks.

robjohn

@Koro what is $x$ in $P(z)$?

Koro

Sorry, I meant $P(z,x)$, which means a polynomial in $z$ whose coefficients are polynomials in $x$ (variable). That is, $P(z,x)= a_1(x)z^m + a_2(x)z^{m-1}+...+a_m(x)z + a_{m+1}(x) $

where $a_i$'s are polynomials.

robjohn

I still don't see how that says that $f(x)$ is algebraic

Koro

10:41

professor Rob, what I think is that since $z=f(x)$ satisfies the polynomial equation $P(z,x)=0$, that's why $f$ is algebraic. We have $P(f(x),x)= (f(x)-1)(f(x))(f(x)-x)$ so for rational $x$, we have $P(f(x)=1,x)=0$ and for irrational $x$ also, since $f(x)=0$, we again have $P(f(x)=0, x)=0$

professor Rob, math.stackexchange.com/questions/4081438/…

I had discussed the same in the above linked question under comment section of the answer.

But today, I noted through someone that in Terrence Tao's Analysis 1, Dirichlet function has been called transcendental and so I got confused now.

Professor Rob, I think that this is the possibility: The definition of algebraic function: We call a function $y=f(x)$ an algebraic function if there exists a polynomial $P(y,x)=0$ such that $P(f(x),x)=0$ for all $x\in R$.

However, on chapter 9, page 218 professor Terry Tao says: "This function is not algebraic (i.e. it cannot be expressed in terms of x purely by using the standard algebraic operations of $+,-$ , square root, division etc.; we will not need this notion in this text".

It seems that Terence Tao uses a different definition of algebraicity. The definition that I am using (the one stated above) is more general. One should keep in mind that not every polynomial can be solved explicitly (for example there is no formula for finding $x$ in $7x^{10}+787 x^7+63x^6+19x^5+17x^4+11x^3+131x^2+100x+43=0$) but that doesn't mean that its roots are transcendental.

professor Rob, is my understanding correct? Thanks.

Koro

11:19

professor Rob, en.wikipedia.org/wiki/Algebraic_function first para seems to be in line with my understanding.

Thorgott

11:33

Wikipedia seems to agree with my common sense that says one shouldn't call discontinuous functions algebraic

Koro

@Thorgott where is it mentioned that a discontinuous function can't be algebraic?

Thorgott

"In more precise terms, an algebraic function of degree n in one variable x is a function {\displaystyle y=f(x),}{\displaystyle y=f(x),} that is continuous in its domain and satisfies a polynomial equation"

Koro

??

@Thor: I am using the definition as stated in the text -A course of Pure mathematics by GH Hardy.

The definition nowhere mentions continuity.

Thorgott

12:00

don't give me a "??" when I'm quoting the Wikipedia article you yourself have linked

Hardy's text is outdated by over a century and doesn't even contain a rigorous definition of the notion of a function, that's not a place to take as reference

Koro

@Thorgott yes, I saw it. Somehow, I overlooked that in the article.

So my understanding now is: based on the definition that I am using, Dirichlet function is algebraic.

Based on the definition using continuity, of course Dirichlet function is not algebraic.

Thorgott

I think a definition allowing discontinuous functions to be algebraic is absolutely useless mathematically, but you do you

porridgemathematics

if I have some spaces universal cover $U $, the covering map $p : U \rightarrow X$, and I have two (path-connected) open sets $W_1,W_2$, such that $\{\gamma(W_i) \}_{\gamma \in \Gamma}$ are disjoint and their unions are $p^{-1}(V_{i})$ respectively, for some evenly covered (path-connected) open sets $V_i$ (respectively), then is it true there are at most finitely many $\gamma$ s.t. $\gamma(W_1) \cap W_2 \neq \emptyset$?

here $\Gamma$ is the deck transformation group

Koro

@Thor, I don't know as of now whether allowing discontinuous functions as algebraic functions is useless or not. You probably know that better :). I'm just learning. Also, encyclopediaofmath.org/wiki/Algebraic_function does not seem to use continuity to define algebraic functions.

i'm not willing to debate definitions right now. For now, I only wanted to know whether my understanding based on the definition of algebraic functions (stated earlier) is correct or not. :(

porridgemathematics

12:19

or perhaps, if what I said is incorrect, how does one show that there are only finitely many $\gamma \in \Gamma$ s.t. $\gamma(K) \cap K \neq \emptyset$ for $K \subset U$ compact, the reason I asked the former is because I can show this for small enough open sets, and i'm trying to generalize that result to the compact case by the usual open covering stuff

on the way to doing that , I realized for how I want to prove the compact case, I need to prove the former claim

Thorgott

eh, I think what you're taking away from that article is misleading

if you read the latter parts, they wanna think of algebraic functions as algebraic elements over a rational function field

which is a legitimate thing to do, but those are only functions in a formal sense, not concrete functions like what you wanna talk about

robjohn

@Koro So you're saying that if $f(x)(f(x)-1)=0$ then $f$ is an algebraic function? That means any function whose range is $\{0,1\}$ is algebraic? That is just not right.

Later in the Wikipedia article it says: "In more precise terms, an algebraic function of degree $n$ in one variable $x$ is a function $y=f(x)$, that is continuous in its domain and satisfies a polynomial equation..."

porridgemathematics

12:35

oh, I seem to have solved it... although im not too solid on my proof, in fact under my assumptions I think there is at most one such deck transformation sending a $W_1$ to $W_2$, since it would have to send the translate of $p^{-1}(p(W_1) \cap p(W_2))$ sitting in $W_1$ to the translate sitting in $W_2$, and there is only one such deck transformation

Koro

12:45

@robjohn yes because that's how the definition was introduced to me.

math.stackexchange.com/questions/4081438/… as I mentioned here also. :(

robjohn

16:09

@Ted: good morning!

Ted Shifrin

Hi @robjohn

robjohn

It was really cold here this morning.

Ted Shifrin

Going down to the 40s tonight, I think.

robjohn

Yeah, it was 44° here when I was in the park this morning.

Ted Shifrin

Nice!

robjohn

16:13

But your temp is moderated a bit more by the ocean.

AMDG

17:42

Could someone verify for me please if $n + \lfloor\log_2(x)\rfloor \leq \lfloor\log_2(y)\rfloor$ for $\lbrace n,x,y \rbrace\in\mathbb{N}_0$ holds?

leslie townes

is there some relation between x, y, n

other than the stated one, i mean

AMDG

Well, not in this context as far as I'm concerned.

leslie townes

consider x = y = 1, n arbitrary, or x = 2^f(n), y = 2^g(n), n arbitrary and f, g arbitrary positive-integer-valued functions, or what the thing ought to mean if x or y is 0

AMDG

Although I'm pretty sure the whole inequality that I'm interested in is actually $n+\lfloor\log_2(x)\rfloor\leq\lfloor\log_2(y)\rfloor\leq 1 + n + \lfloor\log_2(x)\rfloor$.

If correct, then it describes the largest power of two for which I can multiply x such that $2^n x \leq y$.

This of course is just a modified form of the first inequality I posted.

AMDG

18:02

@leslietownes Interesting, I see what you're saying. Assume in this context, then, that the natural logarithm and its inverse being used here are defined for the entire set of $\mathbb{N}_0$.

copper.hat

18:37

we are moving to beyond logarithm as the natural one produces too much methane.

Ted Shifrin

19:02

So cows have to go back to the abacus?

anakhro

No, only the slide rule.

Ted Shifrin

slide rules are nothing but logarithms!

anakhro

We are getting into things that only Ted and Leslie know about. These are dangerous waters.

Ted Shifrin

more likely Ted and copper. Leslie isn’t old enough.

surely even you have seen how slide rules work?

leslie townes

yeesh, i'm not that old. although i have a slide rule somewhere. i never had any reason to use it.

Ted Shifrin

19:11

I willed my circular slide rule from 1969 to the math club at UGA when I retired.

i remember bicycling to the Green Line and going to the MIT Coop to get it one Saturday in high school. Big trip!

Leaky Nun

20:26

if any finite extension of Qp is solvable, then where is the quintic formula for Qp?

leslie townes

20:44

26

Q: Solubility of the quintic?

Over the p-adics, every Galois group is solvable. Does this imply that the quintic (and higher-order polynomials for that matter) can be solved by radicals over $\mathbb{Q}_p$? EDIT: The original place I learned that the p-adic galois groups were solvable was in Milne's Algebraic Number Theory ...

galois-theory

kinda depends on what you mean by 'solvable.' this comes up in the case of Q too, a little bit.

Derivative

god damn the linear algebra professor is going on about proving things about dual spaces and adjunct transformations and all my classmates ask on TA office hours are trivial computation problems

AMDG

I believe quintics have about 0 solubility in any known chemical element.

@copper.hat Need to make it produce CO2 instead in order to compute $\log(0)$.

Speaking of, something I've wondered about is why $\log(0)$ isn't defined in spite of the fact that it seems to approach negative infinity for its real part as x approaches zero for $\log(x)$.

leslie townes

21:03

nothing stopping you from defining it any way you want, i guess

AMDG

I think it would make the most sense to first define the result of division by zero before defining log(0). If I were going to do that, I'd probably define division by zero as a function that maps from reals to some new set.

Leaky Nun

21:17

@Derivative be the change and ask difficult problems in office hours

Ted Shifrin

21:44

My experience in 40 years of teaching was that too often students came to office hours unprepared, often not having read the homework problems — let alone having tried to do them.

Plenty of exceptions, of course.

shintuku

21:57

is there a way to formalize the idea of a function $f: \mathbb{N} \times \{-1,1\} \to \mathbb{N}$ such that $f(x,r) = x + r$, such that we randomly get either $r=1$ or $r=-1$ for each successive $x$?

Semiclassical

@TedShifrin yeah, that happens a lot in tutoring

shintuku

it might be a function that maps to the integers and not to the natural numbers, and that's among one of the things I'd like to figure out

Semiclassical

@shintuku i feel like you'd do that by having a family of functions $F(x,s)$ where $s$ is some sequence of $\pm1$'s, and then somehow randomly choosing one of them

that said, it's probably best to just think of the simpler version first: how would you formalize a 'random' sequence

Ted Shifrin

You gave a formula, so I don’t understand what you’re talking about.

Semiclassical

it's the end of the day and i can feel my brain leaking out

Ted Shifrin

22:03

you want a random variable mapping $\Bbb N\to{\pm 1\}$

Semiclassical

it's easy if you only deal with a subset of N, since then there's finitely many sequences and thus you can just pick one of them at random

shintuku

oh, so it would actually be $f:\mathbb N \to \mathbb Z$ s.t. $f(x) = x + R(x)$, where $R$ denotes a random variable

Semiclassical

i'm forgetting how one formalizes this for N as a whole tho

Ted Shifrin

Can’t edit. I hate chat on tablet.

shintuku

also didn't know you could shorten mathbb to Bbb, thanks

although the above mentioned $f$ wouldn't exactly be a function

Ted Shifrin

22:05

LOL, my inability to edit served a purpose.

Semiclassical

editing chat on the mobile site is possible but it be annoying

Leaky Nun

they would need to be iid $X_1, X_2, \cdots$

where each $X_i$ has image in $\{\pm1\}$

and then the "function" would be $f(i) = i + X_i$

Semiclassical

as an example of how this kind of thing works, think of probability questions involving random coin flips

shintuku

at Leaky Nun: does idd stand for independent and identically distributed?

Leaky Nun

and each $f(i)$ would be a random variable

yeah

Ted Shifrin

22:07

Who cares about identically distributed?

Semiclassical

e.g. repeatedly flipping a fair coin and asking for the probability of seeing HTH before TT

Leaky Nun

ok then just independent

Semiclassical

i never did manage to figure out how you'd solve problems like that from countable additivity

which irritates me

shintuku

so $f$ wouldn't exactly be a function, right? since $f(4)$ has $(1/2)^4$ possible values

Semiclassical

you really wouldn't think of the randomness being in $f$ itself

you'd have $f$ be one particular realization of the $\pm$ sequence

to get a set of all such $f$'s

and then pick one at random

Ted Shifrin

22:10

@shintuku Huh?

Semiclassical

to put it differently: functions aren't random, but the choice of a function can be random

shintuku

@TedShifrin for $f(n)$ each iteration is either $n + 1$ or $n-1$. so 4 iterations in we have a couple of possible branches

oh you're right, I meant $2^4$ possible values

Leaky Nun

so you want $Y_i := \sum_{j=1}^i X_j$ basically

Semiclassical

mmm, random walks

Ted Shifrin

I still say you’re wrong.

shintuku

22:14

oh, right @LeakyNun, I think it would be $\sum_{j=1}^i 1 + X_j$

Ted Shifrin

There are 2 choices of value.

Semiclassical

if you had some kind of recursion going on then i could see that summation

but i don't see where that's happening

Ted Shifrin

I give up on following.

Semiclassical

your case is more like taking the identity function and adding a random noise

amWhy

@TedShifrin What region of the US gonna be that cold? I know it'll be in WI in no time! But not yet!

Ted Shifrin

22:16

southern Cal!

shintuku

oh, right! there's no recursion, $f(1) = 1 + 1$ or $f(1) = 1 - 1$, then $f(2) = 2 + 1$ or $f(2) = 2 - 1$, etc.

Semiclassical

@TedShifrin wow

surprising to see it drop that much this early

Ted Shifrin

i love it.

shintuku

thanks to all for the help

Semiclassical

it's 64 here. a bit cooler than it's been but pleasant

Ted Shifrin

22:17

Robjohn chillier than here.

Semiclassical

though if i wasn't wearing sleeves i might be saying a different story

amWhy

@robjohn Are we talking CA? Sorry, I'm late to the game. And, sorry to say, I don't, and never did, own a slide rule.

leslie townes

i don't think we're supposed to go below the low 50s. i assume we are talking night time lows.

Semiclassical

hmm

i wonder what the oldest analog computer would be. the slide rule is a candidate

leslie townes

i don't want to turn the heat on. :(

amWhy

22:18

Thanks, @Ted!

leslie townes

do abacus-like things count? those go way back.

Semiclassical

oh

Antikythera mechanism

The Antikythera mechanism ( AN-tih-kih-THEER-ə) is an ancient Greek hand-powered orrery, described as the oldest example of an analogue computer used to predict astronomical positions and eclipses decades in advance. It could also be used to track the four-year cycle of athletic games which was similar to an Olympiad, the cycle of the ancient Olympic Games. This artefact was among wreckage retrieved from a shipwreck off the coast of the Greek island Antikythera in 1901. On 17 May 1902 it was identified as containing a gear by archaeologist Valerios Stais. The device, housed in the remains of a...

Ted Shifrin

I needed one for all my chem and physics courses in high school and at MIT. HP calculators started to show up later in my college career, but they cost huge fortunes.

Semiclassical

that beats a lot of them

Derivative

@LeakyNun easy way to get the TA and the other students to hate you

amWhy

22:19

@leslietownes Just hope it's not a repeat of Texas's saga from last winter!

leslie townes

my dad had stories of seeing 'the engineers' on his college campus with slide rules. i don't think he ever used one in high school.

amWhy

I've seen one, a slide rule, in a museum once, I think (@Ted) ;P

Ted Shifrin

You’re too young.

Semiclassical

i like thinking about the history of stuff like this, even though it's impossible to answer, b/c it puts into perspective how much of human history is simply not within reach

like, how tf did human language emerge

Ted Shifrin

Derivative. Tough. If you’re asking for help with lecture and not showing off, that’s their job.

Semiclassical

22:22

can't have history without language

amWhy

@Semiclassical Exactly! Or what did humans do before flushable toilets! I mean, those are details rarely discussed in history books!

leslie townes

semi there is also the huge issue of making tangible things that are durable. if you stored information via plant fiber that wasn't wood, or wood and not metal/stone, good luck. particularly depending on climate.

robjohn

@TedShifrin The HP-35 came out when I was in high school. It was $400.

Ted Shifrin

Outhouses?

Semiclassical

water closets

robjohn

22:24

Then when each new one came out, it was $400.

Ted Shifrin

right, robjohn, like 1973?

amWhy

@TedShifrin Buckets

Leaky Nun

@Derivative then there's nothing that can be done, so you don't need to worry about it

robjohn

@TedShifrin that sounds about right.

Semiclassical

so, this is a sentence i stumbled on via wikipedia's page on billiard-ball computers

leslie townes

22:24

my father in law began his career with a curta calculator.

Semiclassical

"Logic gates based on billiard-ball computer designs have also been made to operate using live soldier crabs of the species Mictyris guinotae in place of the billiard balls."

robjohn

@Semiclassical Crabby slave labor

amWhy

Just so you don't think I'm all that young: my first computer programming class, when I was 12, was done via punch cards...

Ted Shifrin

That was my college days for sure.

leslie townes

'computer programming class' already marks you as young.

Ted Shifrin

22:27

so I gave up on learning to program.

Semiclassical

math profs turn coffee into theorems, biologists turn living species into computational automata

neat

Ted Shifrin

nah, I took a course in high school in 1969.

leslie townes

you do wonder if anyone was protesting outside the lab for that.

Ted Shifrin

PDP11 with tape

Semiclassical

soldier crabs aren't cute enough for that

leslie townes

22:28

here's where i realize i forgot that our country used to have a functioning school system.

amWhy

@leslietownes Hah!

Ted Shifrin

And antiwar protests instead of the president overthrowing democracy

Semiclassical

the fact that there's a nonzero chance of him running again and winning sorta looms like a specter for me

Ted Shifrin

Yup, and the governors have it set up for him to win even when he doesn’t… democracy’s about done.

Semiclassical

especially with how many thumbs are being placed on scales in advance of the next election

yeah

Ted Shifrin

22:32

Time for me to die soon.

Semiclassical

governers and sec's of state and election commissioners....it's loathesome all the way down

copper.hat

politics is never pretty at anytime or place.

amWhy

@TedShifrin I might seriously think about relocating; but "where" is the question.

@copper.hat You just answered me! No doubt. But there is bad, worse, worser, and a few vying for worst.

Ted Shifrin

@amWhy Yup. Not likely for me at this point with messed up back, etc.

copper.hat

i still have my beautiful faber castell slide rule. i found my dad's circular financial slide rule a few months ago while going through my mum's stuff.

Transcript for

$Mathematics$ Mathematics