Mathematics - 2022-08-27

user4539917

12:23 AM

I'm not really sure if geometrical wars fought in cyberspace have crypto bounties that are blockchainable.

leslie townes

back to axie infinity for me, i guess

user4539917

:D

Koro

4:32 AM

@copper.hat sure copper, thanks a lot. I think in Folland's there is an exercise which deals with what you are suggesting.

I'll be doing that exercise shortly.

If upper sum = lower sum, then the function is Riemann integrable.

In measure theory, it should translate to: outer measure = lower measure then the function is measurable.

Also, what is geometric measure theory?

Ted Shifrin

@Koro This sounds very rare.

Geometric measure theory deals with generalizations of manifolds to non-smooth settings.

Koro

4:48 AM

I should have said upper/lower integral.

Koro

5:08 AM

$f: \mathbb R^n\times \mathbb R^m\to \mathbb R^p$ is bilinear. Then $\lim_{(h,k)\to (0,0)} \frac{|f(h,k)|}{|(h,k)|}$ is to be calculated.

We have $|f(h,k)|=|f(\sum_{i=1}^nh_ie_i, \sum_{j=1}^mk_ie_i')|=|\sum_{i=1,j=1}^{i=n,j=m}h_ik_jf(e_i,e_j')|\le \color{blue}{\max\{|f(e_i,e_j)|, 1\le i\le n, 1\le j\le m\}}|\sum_{i=1,j=1}^{i=n,j=m}h_ik_j|$

So $|f(h,k)|\le \color{blue}M |\sum_{i=1,j=1}^{i=n,j=m}h_ik_j|\le M\sqrt{\sum h_i^2}\sqrt{\sum_j k_j^2}\le M(\sum_{i=1}^nh_i^2+\sum_{j=1}^m k_j^2)$

$|(h,k)|=\sqrt{|h|^2+|k|^2}=\sqrt{\sum_{i=1}^nh_i^2+\sum_{j=1}^m k_j^2}$, using $l_2$ norm. Therefore, …

I think that my solution is correct.

user4539917

$Mathematics$ Advanced Set Theory Study Group

This chat is to find and organize people interested in studyin...

copper.hat

5:48 AM

@Koro I would let $M= \max_{\|x\|\le 1, \|y\| \le 1} |f(x,y)$, then you have $|f(x,y)| \le M { \|x\| \|y\| \over \sqrt{ \|x\|^2 + \|y\|^2} }$.

Urrg, The $ { \over \sqrt{ \|x\|^2 + \|y\|^2} }$ should be on both sides.

@Koro That is the correct idea, the basic concept is that if $\infty > \mu X = \mu^* E + \mu^* E^c$ you can find measurable $D \subset E \subset F$ such that $\mu^* E \setminus D$ and $\mu^* F \setminus E$ are $ < \epsilon$.

one potato two potato

@TedShifrin this can be a measure theory I can enjoy

Koro

6:20 AM

@copper.hat I have one confusion though. Given E (a subset of X), I can find a $B\supset E$ such that B is in $\sigma $- algebra (the smallest sigma algebra containing algebra $\mathfrak A$). This is fine. Now, $E\cup E^c=B\cup B^c$.

From here, can I deduce that $\mu^*(B^c)=\mu^*(E^c)$?

The problem is that: $\mu^*(E\cup E^c)=\mu^*(B\cup B^c)=\mu^*(B)+\mu^*(B^c)\le \mu^*(E)+\mu^*(E^c)$

I forgot to add that: B also has this property that $\mu^*(B)=\mu^*(E)$.

So I can get: $\mu^*(B^c)\le \mu^*(E^c)$. How do I get rid of inequality?

Oh wait. That's easy. $\mu^*(E\cup E^c)=\mu^*(X)=\mu^*(E)+\mu^*(E^c)$ is given.

I conclude that: $\mu^*(B\setminus E)=0$.

$E\setminus B=\emptyset$ and $E\triangle B= E\setminus B \cup B\setminus E$

So $\mu^*(E\triangle B)=0$. Now from this, I'm done.

thanks a lot @copper.

I borrowed a book from my college library and after I brought it home, I noticed 'Reference book (not to be taken out of the library)'. My conscience is telling me to return it immediately.

But that also begs a question why they issued it to me in the first place when it wasn't supposed to be taken outside.

copper.hat

6:40 AM

They may have lifted the reference requirement but cannot remove the stamp.

leslie townes

it could be an old stamp. sometimes a book gets put on reserve for a course in one year and then later it's not like that anymore.

there should be a stamp for that too, but maybe somebody missed it.

user4539917

6:53 AM

To ease your conscience @Koro can't you just phone the library and ask them?

leslie townes

be sure to use an app that electronically disguises your voice

user4539917

and call from an unlisted number

leslie townes

and also ask about a different book that definitely isn't on a reserve list

user4539917

Make that^ a couple of different books.

leslie townes

also call a different library

user4539917

7:00 AM

Try to sound like a tourist?

leslie townes

maybe say you're trying to settle a bet with a friend about whether a given book is on reserve or not

user4539917

✅

leslie townes

go steal another book from the library without checking it out, so you can have a guilty conscience with probability p = 1

user4539917

Make sure the value of that^ book far out weights the original book.

BAYMAX

7:23 AM

not[idx_] := Complement[Range[NN], idx]

SC[A_, idx_] := A[[idx, idx]] - A[[idx, not@idx]] . Inverse[A[[not@idx, not@idx]]] .
A[[not@idx, idx]]

Any idea what the above code does in Mathematica??

Koro

7:36 AM

@leslietownes haha

HelpMeToUnderstandContours

7:52 AM

My friend gave me a wrong proof of $1 = i$ and asked to find what's wrong.
Here's what my friend did:
$$i^2 = -1\implies i^4 = 1 \implies i^4 = 1^4 \implies i = 1$$

I think the mistake is that $i^4= 1$ has 4 solutions i.e. $i = 1, -1,i , -i$. And considering only $1$ is kind of partiality. Is it right?

leslie townes

sure. a^4 = b^4 doesn't always mean that a = b.

same thing in real numbers, with -1 and 1.

HelpMeToUnderstandContours

@leslietownes But since $i^4 = 1 \implies i = -1, 1, i, -i$. How to decide what to choose?

leslie townes

why do i need to choose? why can't i^4 just be 1?

other things^4 can also be 1.

HelpMeToUnderstandContours

$i$ is imaginary unit iota.

leslie townes

if all i tell you is that b^4 = 1, you don't know what b is. as you point out, if you know
only that b is complex, there are 4 possibilities. if you know that b is real, two possibilities.

either way, you don't know what b is.

HelpMeToUnderstandContours

8:00 AM

Ohh yes.

copper.hat

there's also mathjaxcoin

why would $a^4=b^4$ mean that $a=b$?

leslie townes

yeah, the general phenomenon here is that if f is a function and a and b are inputs, f(a) = f(b) doesn't tell you that a = b

it's sort of the rule, and not the exception

copper.hat

Note that $a^4-b^4 = (a^2-b^2)(a^2+b^2)=(a-b)(a+b)(a-ib)(a+ib)$.

so lots of juicy choices

maybe your friend is real and positive?

Adil Mohammed

8:42 AM

Goodmorning everyone

I am in a dilemma

Okay dilemma resolved, bye

robjohn

9:23 AM

Glad we could help

@leslietownes sort of the definition of injectivity

user4539917

12:38 PM

A study group for people interested in set theory using Jech, Kunen or a similar text.

5

one potato two potato

1:24 PM

2

A: Existence of pullback tensor

You know that we can choose local coordinates on $M$ and $N$ so that $F(x^1,\dots,x^n,x^{n+1},\dots, x^m)=(x^1,\dots,x^n)$. Write $$\omega = \sum_I a_I(x)dx^I + \sum_{I',J} a_{I'J}dx^J\wedge dx^{I'},$$ where $I$ is a $k$-multiindex involving only $1,\dots,n$ and $J$ is a $j$-multiindex involving ...

How can I use Cartan's formula for Lie derivative to show the definition is independent of $p$? (Last sentence) @TedShifrin

Ted Shifrin

4:49 PM

@onepotatotwopotato Move $p$ with a vector field.

Koro

6:15 PM

I know that (0,1) can be written as countable disjoint union of open intervals.

this can be proven also.

But I find this statement contradictory to the definition of connected sets.

The set S in metric space X is said to be disconnected if it can be written as $S=A\cup B$, where $A,B\subset X$ are open in X such that $A\cap B=\emptyset$.

So if $(0,1)$ can be written as countable disjoint union of open intervals, then noting that any union of open intervals is open set, we can write $(0,1)=A\cup B$, where A and B are open in R and are disjoint.

This shows that (0,1) is disconnected.

which is a wrong conclusion.

leslie townes

isn't there a nonemptiness condition on A and B

Koro

oh yes, I forgot to add that A and B are non empty above.

leslie townes

when you say (0,1) can be written as a countable disjoint union of open intervals, you have to be allowing writing (0,1) = \bigcup {(0,1)} as a union of a single open interval

Ted Shifrin

Leslie took the words right out of my mouth

leslie townes

for more or less the reasons you state, its connectedness implies that it is not a union of two or more disjoint open intervals

Ted Shifrin

6:23 PM

Indeed, the only way to do $(0,1)$ — as opposed to an arbitrary open subset — is as a single element union.

Koro

ok

let's take $\mathbb R$ instead of (0,1).

leslie townes

ted and i are on the same page today. let's see how long this lasts.

Koro

:-)

Ted Shifrin

Take Munchkin to the ducks.

leslie townes

she wanted to go to the 'spinny swings' park today, which doesn't have ducks.

Ted Shifrin

6:24 PM

Ted is grumpy — he has been ordered to lose weight and lower his triglycerides.

Does she like getting dizzy on the spinny swings?

Koro

Leslie, I want to show you a bird that I saw for the first time. I think you may recognize the bird.

leslie townes

she doesn't actually like the swings very much, so it's weird that she named the park with them.

Ted Shifrin

So what does she like about the park?

leslie townes

it has some structures she can climb on. kinda like cargo nets, but more well supported.

koro: R itself is an 'open interval' here. i think 'interval' has to mean something that's closed under betweenness for this verbal formula to work. so that 'open interval' includes "half-open" infinite intervals (-oo, a) and (a, +oo) and the fully infinite crazy interval (-oo, +oo)

Ted Shifrin

OK, cut out that oo stuff. You will be thrown in Chat jail.

Koro

6:30 PM

Leslie: By open interval, I mean (a,b), where a, b are real numbers such that a<b.

robjohn

Looks a bit like a heron

can't tell the coloring, though

leslie townes

yeah, heron or egret or something in those families. i don't know shore birds very well.

Koro

the bird is camouflaging so well.

leslie townes

around here some of the juvenile herons look like that, but they might not be as tall.

Ted Shifrin

Hard to judge the neck versus body, but it seems very necky.

leslie townes

6:33 PM

koro: ok, i think that your definition of 'interval' will fail to make R a countable union of disjoint open 'intervals.'

a lot of herons can do a thing where they zoop their neck up and down. it's pretty funny.

zoop being the technical term.

i have some photos of my daughter menacing a black-crowned night heron, but i don't have any of her menacing a juvenile black-crowned night heron with a zooped neck, which might be the closest i could get to that look with local birds.

Ted Shifrin

I've never thought about that point, but leslie may be right.

robjohn

Looking closely and comparing, it might be an American Bittern

Koro

@leslietownes I see.

leslie townes

if R = \bigcup (a_n, b_n), with the union disjoint and the a_n < b_n sequences of real numbers, you might ask, what interval on the right hand side does b_1 (which is certainly in R) lie in, and how can that interval be disjoint from (a_1, b_1).

robjohn

they can stretch the neck out long like that

Ted Shifrin

6:38 PM

@robjohn Is that related to the Angostura Bittern?

Koro

But usually, by open intervals we mean the definition that I said above. (oo, a) etc. are called open rays.

robjohn

@TedShifrin I am not sure. Could be.

@TedShifrin Ah, got it

Ted Shifrin

LOL ... I was wondering.

Koro

@robjohn humans can do that too to some extent. haha

robjohn

@TedShifrin I need to do more drinking ;-p

leslie townes

6:41 PM

koro: i don't know that there is standard terminology in this area. i've definitely seen books that define 'intervals' to be sets that are what i loosely referred to as "closed under betweenness" above. meaning J is an interval iff for all a, b, c, if a, b in J and a < c < b then c in J). this embraces "finite" intervals (including ones with one or both endpoints included), as well as what you would call 'rays' and all of R.

the 'open' then removes endpoints from this collection.

anyway, i think this notion is the one that's used in at least some intro analysis books that use order topologies as opposed to metric spaces as the "one level up" generalization of R.

Ted Shifrin

Huh?

@robjohn That was going to be my next comment :P

Koro

@TedShifrin I wrote $\mathbb R=\cup_{n\in \mathbb Z}(n,n+1)$ wrongly. I realized that and deleted the comment. :(

Ted Shifrin

I know. That's why I said Huh?

I was not going to bother deleting my Huh?

Koro

yeah, and when I deleted my comment, your comment looked empty so I replied to it to fill in the deleted comment.

But I still don't understand how allowing rays as open intervals solves the problem.

Ted Shifrin

What precisely is the problem now?

leslie townes

6:48 PM

all of R is also an 'open interval' here, koro. i don't know what this 'ray' thing is.

i mean, i do, but the notion of 'interval' i'm talking about is not introduced simply to include them. it includes anything that satisfies the definition, which would also include R.

Ted Shifrin

Try not to be stung, leslie.

leslie townes

they have rays you can pet at the aquarium. my daughter's really into it. i'm less into it.

Koro

260

Q: Any open subset of $\Bbb R$ is a countable union of disjoint open intervals

Let $U$ be an open set in $\mathbb R$. Then $U$ is a countable union of disjoint intervals. This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as poss...

leslie townes

look at the definition of 'interval' in the 16-upvoted answer.

the answers that do not define 'interval' appear to be implicitly using that definition.

Koro

@leslietownes Ah I see. R is also an interval in the theorem statement.

leslie townes

6:52 PM

e.g. in the 66-upvoted answer "which, as a union of non-disjoint open intervals (each I contains x), is an open interval subset to U". that's allowing +infty and -infty as endpoints. has to be.

Koro

That is, end points are not supposed be considered in the open interval definition in theorem statement.

Ted Shifrin

Sometimes stubbornness is not so much a virtue, Koro.

copper.hat

i think of an interval as any convex set of the reals.

of course.

Ted Shifrin

Of course. Oh, have I got a convexity one for you, copper.

This person sort of drives me crazy, but ...

Koro

I understood now. Thanks a lot @leslietownes and @TedShifrin.

copper.hat

6:55 PM

I run when I see the word manifold and there is no car in sight.

ObjectsMorphisms

clicks link to check that it's not me

@Koro hey :)

Ted Shifrin

You don't need to worry about manifolds. Just a smooth surface.

Koro

@ObjectsMorphisms hi :)

ObjectsMorphisms

It's EnjoysMath (old name)

Koro

I know.

ObjectsMorphisms

6:56 PM

Now I'm Objects Morphisms because category theory, etc

Ted Shifrin

I didn't know. Long forgotten.

And if you morph into category theory, you'll be forever forgotten to me.

ObjectsMorphisms

Categories are a good way to organize your code with OOP

copper.hat

@TedShifrin Its beyond my notational complexity limit.

ObjectsMorphisms

I'm making a CAS for homological algebra, first of its kind

C++ of course. I tried out Nim, but lost fight to the compiler

Ted Shifrin

I get it, copper. It may be the first time I've seen the second and third derivatives of a vector-valued function appearing, too.

ObjectsMorphisms

6:59 PM

Anyway, I'm ratproofing my parents deck steps today with rocks and concrete.

There were < 10 packrat nests underneath - they even sealed it off which will make the deck wood start to rot

Koro

I wish to learn OOP one day.

ObjectsMorphisms

@Koro if you know math, you already know a lot

Ted Shifrin

All I have to say is ... OOPs

copper.hat

@ObjectsMorphisms Other than hermetically sealed (jk) I have never seen a rodent proof deck or roof. I have found the best focus is to remove whatever it is that they want.

@Koro OOP is a much overblown concept.

ObjectsMorphisms

I'm removing the soil, they want the shelter

copper.hat

7:01 PM

Shelter is hard to remove.

ObjectsMorphisms

The rock in sedona is only a few inchest deep, so rock + concrete will do

I'll be hauling rocks from a dry bed by two buckets because wheel barrels were > $150 at Ace

Should be good exercise

copper.hat

@Koro Think of it as one way of factoring code.

ObjectsMorphisms

It will take a few days, but in exchange my parents are paying back my long-forgotten student loans

copper.hat

please don't get me started

ObjectsMorphisms

So I do stuff around yard for $20/hr

to pay them back

copper.hat

7:04 PM

it seems i will be paying back lots of student loans whether i like it or not

ObjectsMorphisms

C u guys l8er

copper.hat

have a good afternoon!

@Koro A bit late, but I found this: math.stackexchange.com/questions/2354595/…

Koro

Thanks a lot @copper.hat.

user10478

8:13 PM

Doing something wrong here...I start with 100 units and wager fraction $f$ of bankroll per flip on a series of $n$ I.I.D. weighted coin flips with 60% probability of success per flip. i.imgur.com/vxIIPka.png If I bet half of bankroll, $f = .5$, I should expect 110 units after $n = 1$, regardless of whether the formula is supposed to yield an arithmetic mean or geometric mean (they should be the same after one period). What am I doing wrong?

copper.hat

8:45 PM

What does wager mean? If you bet $x$ and you win you get $2x$ with 0.6 probability or $0$ otherwise?

user10478

Yes

copper.hat

Your formula should be something like $E v_n = (2fp+(1-f))^n v_0$. Where $v_0$ is the initial capital.

user10478

Why is this formula for $r$ (en.wikipedia.org/wiki/Kelly_criterion#Proof) not applicable?

It feels like I should be able to plug that $r$ into the geometric growth formula, $v_n = (r + 1)^n$, which is exactly how I got my formula, and I used the case $a = b = 1$.

copper.hat

9:10 PM

I am not sure what you are trying to do. You need to elaborate a little more.

Tiny Tim

9:29 PM

Is there a chance someone could talk me through what it means for polynomials to be relatively prime?

I've never used this chat before so I hope I'm using it the right way.
Basically, I have an equation that looks like $p_{1}^{a_{1}} \times f(x) = p_{2}^{a_{2}} \times g(x)$ where the p's are prime numbers, the a's are integer values of at least 1, and both f(x) and g(x) are functions that will yield an integer solution of at least one for all x. Is there a way for me to show that f(x) and g(x) are coprime?

I can share the original equation if that's helpful.

user10478

I am just trying to apply the linked formula, $r = (1 + fb)^p (1 - fa)^q$ to a real scenario (ultimately applying it to the case of multiple sequential wagers), but I am getting counterintuitive results even in the single wager case.

PM 2Ring

9:55 PM

This went onto the Hot Network Question list. But it got removed within ~30 minutes. Ok, it could potentially attract under-informed opinions &/or start a religious war, but that didn't happen.

4

Q: Do we draw a distincton between a number as an element of the reals, and an element of the naturals?

I see in some explanations of attempts to formalize numbers such as Von Neumann's ordinals like in this rather philosophical question that we can draw a distinction between a real number '1' and a natural number '1', obviously in mathematical contexts I have encountered, we describe these as one ...

logic set-theory real-numbers ordinals natural-numbers

leslie townes

disaster averted.

PM 2Ring

I guess being kicked off the HNQ isn't as bad as being locked or deleted. But I don't think a pre-emptive strike was necessary.

ObjectsMorphisms

OOP is just a way to organize your app, and do some code-reusage. And works like this. You'd have a class called Ring and a class called Field that derives from Ring. For example. Because naturally every field is the same thing as a ring, minus one axiom @Koro

PM 2Ring

@TinyTim I assume that's connected with your recent question: math.stackexchange.com/q/4519831/207316

ObjectsMorphisms

If you can say A is a B, then you can usually apply "subclassing, inheritance, type derivation" each of those vocabulary terms mean the same thing

Except in Python they call it subclassing and in C++ inheritance

You can inherit from multiple classes, except in the malfangled language called D

A class is just a user-described type (user = coder here). So you have like < 10 builtin types of a language like int, char, string, etc. In C++ there are different sized ints and in Python every int has arbitrary precision meaning you're not limited to just 64-bits or 2^64 different values

So then you can create a class called PairOfInts which has two private members: left & right

that are the builtin type int

That's basically all OOP is in a nutshell

Polymorphism is this. If you have that example of a Ring/Field in mind and you have a function that you pass an instance of a Ring into, then it can also be passed an instance of a field (instance = sometimes called object)

And a Field might override a method in the definition of a Ring, so if your object is a field, the Field's method will be called. So you can have a list of all rings some of which might be fields, and the appropriate method gets called

@Koro

If you want to learn a prog lang. The best way imho is to pick a goal project. Then it's just simply googling around and learning how to solve each subproblem. The union of all your solved subproblems equals your app

leslie townes

10:12 PM

@PM2Ring my own vibe is, not necessary, but as an exercise of discretion/judgment, it makes sense to me.

ObjectsMorphisms

I think the hardest part about coding is getting your dev environment up and running

leslie townes

qualifying this with the fact that i have no idea of how things normally arrive on, or depart from, the HNQ list

PM 2Ring

It makes sense to me too. I just think it's a little premature.

The HNQ gets criticised because it boosts questions that aren't necessarily representative of the site. The distortion is especially severe for Math.SE because any question title containing MathJax is automatically disqualified from the HNQ.

copper.hat

The hardest thing about coding is unraveling the mess left behind.

@PM2Ring a bit random and out of place but have you ever used a src control called fossil?

PM 2Ring

This? en.wikipedia.org/wiki/Fossil_(software)

I think I've vaguely heard of it, but I've never used it.

copper.hat

10:26 PM

@ObjectsMorphisms At some point, Apple were going to unleash a beautiful (to me) language called Dylan, essentially a C-style syntax on Lisp underpinnings with OOP (real OOP with multiple dispatch & call/cc, etc, not the C++ style).

@PM2Ring Yep, I'm looking for an answer to a rather peculiar question (I'm changing a CVS based setup to a fossil based setup).stackoverflow.com/questions/73495844/…

For some reason, many of my interactions on SO are vaguely negative.

@user10478 It think you have to look at the assumptions behind the Kelly bet (infinite plays).

@PM2Ring Thank you.

PM 2Ring

No worries. Some people don't realise that questions about programming tools are on-topic on SO.

I've done almost nothing on SO for a few years, apart from updating some old answers when they get upvotes. In the old days, I rarely got downvotes, but the last 3 or 4 answers I posted all got downvotes. I think I might've annoyed someone on meta SO, because I got 3 downvotes within an hour, 2 on SO, 1 on a very high scoring answer on Astronomy.

user10478

@copper.hat I have arrived at your formula now, but I still cannot figure out how to relate it to the formula in the Kelly article. Pattern matching your formula with this one (en.wikipedia.org/wiki/Exponential_growth), the growth rate $r$ would be $2fp - f$. Is this the same as the expected geometric growth rate $r$ from the Kelly article? Do I have to somehow take a geometric mean of the former to get the latter?

copper.hat

@user10478 The Kelly formula is maximising the expected geometric growth rate.

user10478

right

Surely this expected geometric growth rate relates to growth rate of the geometric growth formula you posted, unless none of the words in statistics make any sense at all.

$Mathematics$ Advanced Set Theory Study Group

Transcript for

$Mathematics$ Mathematics