Mathematics - 2021-04-10 (page 1 of 3)

I vividly remember having my head spining in circles when I dealt with the boundary definition copper is talking about the first time

Edward Evans

@HereToRelax I've heard Grand Tour isn't very cool

or whatever it's called

@copper For submanifolds with or without boundary, everything is boundary points. So frontier is just way less confusing.

I am not without reason for my madness.

12:07 AM

@TedShifrin understandable.

@dc3rd using that definition you can see that every point is in exactly one of the interior of $S$, the frontier of $S$ or the exterior of $S$. Since the first & last are open, the remainder must be closed.

visually i see what is happening, in terms of shrinking the size of $\epsilon$ for all neighbourhoods around all my boundary points will give me convergent sequences. Just the solid logic of explaining it is what is missing from me.

if I were to use that definition.

Do we explicitly use these ideas in the upcoming sections @TedShifrin ? I ask because when I did multivariable previously I didn't see these ideas pop up explicitly too much. THey kinda just sat in the background except for defining the usual differentiation and integration ideas, but we didin't use frontier points, limit points, closures, etc too explicitly. I've always wanted to see it in action

12:23 AM

This is not a calculus course. This is a multivariable analysis course. If you study the proofs, they will be used. You will be doing proofs with compactness and integrability ...

Not to mention inverse/implicit function theorems ...

THe course I did was a "middle ground" it wasn't all calculus and wasn't full analysis. We used Folland's "Advanced Calculus"

it was the course for the "inbetweeners" who were not full math specialists and also not math minors.

If you learned that course, you should know most of mine. Hmmm .... Or it could have been taught watered. Down.

"learned" is a stretch..............memorized definitions, theorems, and proofs?.......yes.

but that was my experience because of the weak foundations. But I do recognize all the content you're teaching

also they really didn't do any of the geometry the way you do.

This comes back to the ability to "teach" that you have talked about

HereToRelax

1:04 AM

$$\Demarois$$

HereToRelax

1:19 AM

this person is a legend math.stackexchange.com/a/2353752/33907

apparently someone even tried to post an answer on a sock puppet account to prevent the bounty and got owned

Legend depends on one's universe, evidently

HereToRelax

I thought it was funny :'(

I guess I didn't pay enough attention.

2:33 AM

Perfect timing if you are bored @TedShifrin . I either just finished 9a, or I mixed the results up. What do you think:

i) True/ False: all interior points of $\bar{S}$ are points of $S$?

Claim: False.

Pf: Let $\bar{S} = [a,b]$. Then $b \in S^{f}$. Take $\delta$ nbhds for $b \in S$. If we take $\frac{\delta}{2}$ nbhds we can have $b \in \bar{S}$.

ii) True/ False: If $S$ is open, then all interior points of $\bar{S}$ are points of $S$

Claim: True.

pf: Suppose towards contradiction that there exists interior points $a \in \bar{S}$ such that $a \notin S$. Then $a \in S^{F}$. This means f…

I don't follow a at all. b is wrong.

back to the drawing board...

For a) what I wanted to argue was that the frontier points are also interior points of $\bar{S}$ specifically.

Rover

@robjohn yes, then ...?

@dc3rd check definitioms?

So whatever $\delta$ I have dictating the nbhd around a frontier point of $S$, there exists a smaller $\delta$, this one I said $\frac{\delta}{2}$ that can be chosen to make every nbhd of a frontier point with respect to $\bar{S}$ be fully contained in $\bar{S}$

2:47 AM

Nope.

Alright. I'll go back and read the definitions and proabably check in with you tomorrow about it if you're not around tonight.

The definition of frontier point says nope.

Has anyone heard about this announcement from the CDC?

> Findings of these studies suggest that the risk of SARS-CoV-2 infection via the fomite transmission route is low, and generally less than 1 in 10,000, which means that each contact with a contaminated surface has less than a 1 in 10,000 chance of causing an infection 7, 8, 9.

It's clearly been about aerosol ... for months. What’s your point?

is that a "yes", sir?

2:51 AM

I don't know any new news, and I have no idea what 7,8,9 is there.

sorry, those are the study references

[7], [8], [9]; I guess I'll have to read them

As I say, for months it's been known that it's aerosol transmission that is the grave danger — hence the importance of masks. Hand-washing continues to be a precaution.

"1 in 10,000 after contact" struck me as odd

Surfaces are not a big deal.

Why odd?

Thanks for your attention sir.

3:05 AM

Just unraveling the definition a bit more I see what you're getting at with respect to i). To explain it informally. If I take an interior point $a \in \bar{S}$ then there exists a nbhd fully contained in $\bar{S}$. Equivalently it means $a \in S$ or $a \in S^{F}$. We are only concerned with the $S^{F}$ case. If $a \in S^{F}$ then by definition every nbhd of $a$ will contian points not in $S$, negating the definition of an interior point

....... So then that means i) is true.....just have to write it out

3:24 AM

No, it's still not correct. You're not quite catching the subtlety yet. Think about more examples.

Alright will do

The subtlety yo're hinting at has to do with the interior of the set of frontier points I gather.

Not quite.

hmmmm....interesting.

I don't actually know what you mean by that.

Well I was thinking about the set of frontier points exclusively and disregarded the point of $S$ because I thought it was trivial.

3:39 AM

That seems correct, but you went the wrong way down the road.

@dc3rd Pick a point $x$. Then exactly one of the following three things is true: 1. there is an open set containing $x$ that is entirely in $S$. 2. there is an open set containing $x$ that is entirely in $S^c$. 3. neither 1 nor 2 hold.
Now note that if 3. holds, then no open set containing $x$ can be a subset of $S$ or a subset of $S^c$. In other words, any open set containing $x$ must contain points of both $S$ and $S^c$.

Sorry, I've been saving that since before I started dinner.

The issue is to imagine different scenarios with where frontier points are located. Let's not give it away.

@geocalc33 there?

The room is surprisingly lively despite its dearth of tools. I think that is due to you @Ted.

3:56 AM

Is that good or bad?

And whom are you calling tools?

@copper.hat yeah, who da tools ?

:-)

We all can be tools of the prolific spread of mathematics I guess

Tool is also slang.

Perhaps dated slang.

nah it's still relevant. prbably has greater emphasis now because of its age as a term

4:01 AM

As slang, it's rather derogatory. Whence my original question.

Well, if it's slang it could mean 1 of 10 things, and 80% of us will fall into a tool category

Therefore, I think it can be replaced with "people"

Whence my original question.

i just wanted to use the word dearth.

dearth vader

LOL ... or death

dear'th

4:04 AM

a few years ago my daughter gave me a present of "vader's little princess"

He used a bad word
Dearth of t-word, could mean stuff
But who cares, time's little

a Haiku I composed

I might have some questions in just a bit. I'm working through this #-theory book at the moment.

how did hash become number?

lol

LOL, another victim of the primes

Leonhard Euler

@zacts burton?

4:06 AM

@LeonhardEuler I'm looking at that and another text too.

# has always been number!

@zacts what's the q?

illustratedtheoryofnumbers.com

[#] looks like a rubik's cube to me

Math is indeed an infinite-dimensional rubik's cube phenomenon

Just as is coding

@StudySmarterNotHarder well, I'm kind of stuck with some of the problems. maybe my first question could be what's a good study strategy for these texts? Also, might analysis be a useful way for me to get started before these texts, like with Tso's Analysis for example.

4:08 AM

i think i have only seen it used for number in the us, at least a few decades ago.

@zacts what part do you first get stuck on in the book?

@copper.hat what is your etymological meaning of #?

the interweb is homogenising.

@StudySmarterNotHarder I'll link a particular problem

Seems also like you need to know some Python from that page

4:09 AM

@StudySmarterNotHarder no python is needed for the actual text of an illustrated theory

that's like extra supplement for the text

Oh, cool, either way I could teach Python 3.x / SymPy coding if that's what you wanted

i could easily have forgotten, but i believe it was used to indicate a number as a label.

as opposed to having some order. similar to a modern SKU

#:

$\# A$ is an alternative to $|A|$ when $A$ is a set

that is where the cs usage came form

4:11 AM

@copper.hat ah cool

Ok, let me link a question from this text by burton

k

it is called an octothorpe

Prove the following facts concerning triangular numbers: (a) A number is triangular iff it is of the form n(n+1)/2 for some n>=1.

Another way to phrase that is they want you to prove that $T(n) = \dfrac{n(n+1)}{2}$

The information given is: "This led the ancient Greeks to call a number /triangular/ if it is the sum of consecutive integers, beginning with 1."

4:14 AM

is that a definition?

where $T$ is the sequence of triangular numbers

It's actually a very well-tought proof, what are you having trouble with? It's an obvious candidate for inductive-style proof because $n + 1$ occurs directly within it and that's how induction proceeds.

Prove it first for $n = 1$

I think my confusion was with the definition for /triangular/.

$1(1 + 1)/2 = 1$, done

Oh, cool

triangular means "The sum of consecutive integers, beginning with 1"

so if I can prove that n(n+1)/2 is the sum of consecutive integers, beginning with 1,, I have proven it's triangular, right?

iff that is

I guess because you can form a "pyramid" with 1 object at the top, 2 objects on the row below that, 3 on the row below it and so on, and the sum of all row counts is equal to the area approximately (as $n$ grows) of a triangle.

4:17 AM

Show $T(1) = 1$ and $T(n+1) = n+1+T(n)$.

If part is proof the formula $T(n) = \dfrac{n (n+1)}{2}$, and only if would be simply by definition I think

The integer n is a triangular number iff 8n+1 is a perfect square.

I wonder if I should start with a different text first.

I also am looking at this Analysis I text by Tso.

Ok either I'm thinking too hard or I'm on to something. But I've been sitting here thinking about the statement of

"either $a \in \bar{S}$ or $a \in \bar{S}^{c}$ or $a \in \bar{S}^{F}$"........should I unravel the idea from there?

For that you're asked to solve: $8 T(n) + 1 = m^2$ for $m$, for any given $n$. And the converse part is for all $m$ such that $m^2 = 8 k + 1$, we have that $k = T(n)$ for some $n$.

but let me post a cool looking question from the other #-theory (pardon my use of #) book.

4:23 AM

Nice, let's hear it

*see it

@dc3rd i'm not sure what you are trying to do.

Either $a \in A$ or $a \notin A$ for all objects $a$ in the universe of thought, for any given set $A$.

try a few examples. what is the frontier of $\mathbb{R}$?

Draw a spiral to demonstrate that 100 = 10 + 2(9) + 2(8) + 2(7) + 2(6) + 2(5) + 2(4) + 2(3) +2(2) + 2(1).

What does the spiral look like?

4:26 AM

I don't know.

What's the general pattern they are implying?

it's got to be stacked somehow

like stacked dots

Then I would try it with a smaller number than 100 maybe

ok

yeah

25 maybe

4:27 AM

that's kind of what is going on in the ch for sure

I don't think this would be too difficult now that you mention it.

let's see if I can sketch a pic for that.

25 = 5 + 2(4) + 2(3) + 2(2) + 2(1)

Look at each term, can you tell me the general formula for a any square $n^2$, $n \in \Bbb{N}$?

That's a good question @copper.hat ....I'm drawing it now and I'm asking what IS the frontier of $\mathbb{R}$..only thing that comes to mind is $-\infty$ and $+\infty$ and those don't mean much.

The frontier of $\Bbb{R}$ is the great wide open $\Bbb{C}$-plane, son!

J/k

@StudySmarterNotHarder n^2 = 2n(n+1)/2

wait

I was wondering if it was $\mathbb{C}$ or not.

4:31 AM

I was speaking poetically though

Lets stick with reals for now.

It is not the complex plane.

$n^2 = n + ?$

what points can be approached from the complement?

sorry just a sec..

Use \sum_{k = 1}^{n-1} ?$ notation

4:33 AM

if a point is $x \notin \mathbb{R}$....hmmmm....I'm truly stumped

would we have to include higher dimensions?

Noooooooo

$\mathbb{R}$ is the whole universe here.

Well maybe I shouldn't search too hard because all I got is the empty set beyond $\mathbb{R}$

exactly the complement is the empty set

@zacts I'm not sure about what spiral it should be

or look at it this way, for any $x$ you can find a open set containing $x$ that is a subset of $\mathbb{R}$.

4:36 AM

but the empty set has no points in it, so how can I have a frontier point at "the end" of $\mathbb{R}$?

if $x$ was a frontier point then any open set containing $x$ must contain points in $\mathbb{R}$ and $\mathbb{R}^c$.

Peas porridge hot, peas porridge cold

however, there are no points in $\mathbb{R}^c$.

ha.....which it cannot because there are no points in $\mathbb{R}^{c}$

from which we conclude that the frontier of $\mathbb{R}$ is ....

4:38 AM

imgur.com/zR3piku

remember that the frontier is a set.

^ I think it's got to be something like this stacked dots idea

ooooohhhh..............the set is empty

exactly

how about the rationals as a subset of the reals.

what points $x$ can be approached arbitrarily closely by rationals and irrationals?

See the rationals I used as an example of a frontier set.

4:41 AM

so what is the frontier?

sorry I used the integers

what real numbers can be approached arbitrarily closely by rationals?

the integers would serve as a frontier set

not sure how you're getting that.

in a metric space, a point is in the frontier if (and only if) it can be approached arbitrarily closely from the set and its complement.

the number ${1 \over 2}$ is a rational that is far away from the integers.

@zacts I think you take a square, say $n^2 =25$ and draw an actual square of 5 x 5 units

Cross out the top row (5)

and fit in $2(k)$ for each $k \lt 5$

in a spiral-like pattern

4:46 AM

well to your question of what real numbers $x$ can be approached by rationals. I would say the irrationals.

@StudySmarterNotHarder would it be $n^{2} = 2(n-1)n/2$?

oops

that is not all

$0$ as well

well, any rational

$n^2 = n + \sum_{k = 1}^{n-1} 2k = n + 2 \sum_{k=1}^{n-1} k = n + 2 T(n-1)$

4:47 AM

$n^{2} = n + \frac{2(n-1)n}{2}$?

yes

I left in the 2 there to reflect the grouping in the original problem

But the spiral they ask for is starting with an $n \times n$ square.

and then vice versa for the irrationals towards rationals as well.

@dc3rd given any real $x$, and an open set $U$ containing $x$. Does $U$ contain rationals?

4:49 AM

@StudySmarterNotHarder oh, so I'm drawing a spiral within the square?

like draw a line within the dots kind of idea?

But the spiral for case $n$ has in it the spiral for case $n - 1$ which has in it the spiral for case $n - 2$, and so on down to $1$.

I would use grid squares

Where each grid box is a unit in area

Yes it does.

Does $U$ contain irrationals?

Definitely does

So what is the boundary of the rationals?

4:52 AM

Was just going to type it too.....the boundary is the reals

correct. good.

551133
552233
552233
554444
554444

what is the boundary of the irrationals?

@zacts where the labeling just reflects what it index it corresponds to

sorry i meant frontier. old habits die hard.

4:56 AM

boundary is the rationals.

Nope.

lol....ted with the RKO out of nowhere......

what real numbers can be approached arbitrarily closely by irrationals?

i'm going to ask the same questions again...

Don't worry: I'll disappear as quickly as I appeared.

@StudySmarterNotHarder is the spiral formed from the actual dots themselves within the grid?

4:58 AM

The spiral is centered approximately at $1 \ 1$

You then double 2 to get: $1 \ 1 \\ 2 \ 2 \\ 2 \ 2$

That always gives you a height or width that is $3, 4, 5, \dots$

The next is 3 so:
1 1 3 3
2 2 3 3
2 2 3 3

The point is that it's always of the form $k(k-1)$

real numbers that can be approached by irrationals are the irrationals

that is a different answer to the same question when we were looking for the boundary of the rationals.

can 2 be approached by irrationals?

yes it can

so, what real numbers can be approached arbitrarily closely by irrationals?

the integers

but then the rationals contain the integers.....

5:05 AM

lets back up a bit

pick any real $x$. Pick any open set $U$ containing $x$.

Does $U$ contain irrationals?

was just doing that too.... but yes $U$ contains irrationals

Does $U$ contain rationals?

@StudySmarterNotHarder thanks. I'm still trying to visualize this, but that looks like a cool solution.

yw

Yes $U$ contains rationals.

so the reals would serve as a boundary for the irrationals...

given that the irrationals are viewed as a subset of the reals.....very important

5:12 AM

correct.

note that the boundary of the rationals and the boundary of its complement (the irrationals) is the same.

@TedShifrin Poof!

i mean frontier

@copper.hat depending on the book you use

the subtlety in this is keeping note of what is a subset of what. I understand what you mean by boundary, no worries

what is the frontier of $(-\infty,0]$ as a subset of the reals?

@robjohn Ted wants me to use frontier in current discussion so i am complying :-)

5:14 AM

@StudySmarterNotHarder just to be clear, but was my solution to the $n^2$ problem correct?

@copper.hat Yes, master.

@dc3rd It is important, since complements are involved.

i am very obedient :-)

$n^2 = n + 2[\frac{(n-1)n}{2}]$

@zacts yes

ok, thanks

5:16 AM

@zacts $n^2=2\binom{n}{2}+\binom{n}{1}$

the frontier would be $(0, \infty)

nope

what $x$s can be approached from inside and outside the set?

@dc3rd are you using the definition $\partial S=\overline{S}\cap\overline{S^C}$?

for any $y>0$ i can find some $r>0$ such that $(y-r,y+r) \subset (0, \infty)= (-\infty, 0]^c$.

no closures on them @robjohn

5:18 AM

@dc3rd then that would be empty

@StudySmarterNotHarder @robjohn thanks

@dc3rd loosely, for 'nice' sets, the frontier is the 'edge' of the set.

what is the edge of $(-\infty, 0]$ ? In the context of the reals?

@copper.hat that would seem to be what I wrote...

in this case then $0$ is the frontier

@zacts in your proof you have to state that $\sum_{k = 1}^{n-1}k = T(n-1)$

5:20 AM

@copper.hat $\partial S=\overline{S}\cap\overline{S^C}$

@robjohn would you mind explaining in a nutshell how that notation works?

@robjohn i think @dc3rd is using the defn that $x$ is in the frontier iff any open set containing $x$ contains points in the set and its complement.

@zacts which notation?

which is, of course, equivalent,

{ n \choose r }

5:21 AM

and the closure definition is one i prefer.

the bionomial coefficient notation parenthetical (n/2) (n/1)?

binom{n}{m}

with a slash

backslash

$binom{n}{m}$

verbatim the definition is: "$a \in \mathbb{R}^{n}$ is a frontier point of S if every neighbourhood of $a$ contains both points in $S$ and points not in $S$"

$\binom{n}{m}$

5:22 AM

@zacts right-click on the mathjax?

ok

enthu

@TedShifrin I have complex shapes of geometries which makes it hard to integrate over the area, I am thinking to use the theory to make things more analytical and reduce integrations. do not know though how things may work or succeed.

\binom{n}{m}

@robjohn no, can you conceptually explain how bionomial coefficients work in a nutshell, not how to write it in $\latex$.

ah, okay

5:23 AM

Although, I do apologize as I'm still learning LaTeX at the moment as well

I do have another venue for a practice pad too which I'll use.

@dc3rd in my example. pick 3 different classes of points. $x<0$ $x=0$ and $x>0$.

@zacts They are the coefficients of $(1+x)^n=\sum\limits_{k=0}^n\binom{n}{k}x^k$

for each class. ask if it can be 'approached' from in the set and from outside the set.

well I was typing something, but I'll save it. I think I'm going to conclude the same thing, but it is worth the practice.

@zacts They can also be computed in Pascal's Triangle

5:26 AM

@dc3rd you need to work examples involving intervals open closed half inifinite and become comfortable with the reasoning.

love_sodam

Can anyone describe a simply connected covering space of a 2-sphere with a circle intersects with two points?

@zacts also $\binom{n}{k}=\frac{n!}{k!(n-k)!}=\frac{n(n-1)\cdots(n-k+1)}{k!}$

@dc3rd as i mentioned, in 'nice' cases the frontier is what you might expect the 'edge' of the set to be.

i should not have started with the rationals.

love_sodam

I don't know how to describe when a portion of circle is inside of sphere

a not so great circle?

5:28 AM

@zacts I am not sure what you mean by "how they work"

Got it. with regards to 'approached', I take it you mean that it can "eventually" get extremely close to the point?

@robjohn how would you think of that notation conceptually within the context of the $n^2$ problem, perhaps in terms of how things are grouped together? if that makes sense. I mean you wouldn't be thinking in terms of the factorial formula given above right?

@dc3rd well, more precisely, any open set contains points from the set and its complement.

@zacts No. I was simply considering exactly the formula you stated: $\frac n1+2\frac{n(n-1)}2$

but thinking of it as binomial coefficients allows it to be manipulated using things used to manipulate the binomial coefficients

such as the hockey-stick identity

Or Vandermonde's Identity

what would $\binom{n}{3}$ look like?

5:34 AM

@zacts $\frac{n(n-1)(n-2)}{3!}$

8 mins ago, by robjohn

@zacts also $\binom{n}{k}=\frac{n!}{k!(n-k)!}=\frac{n(n-1)\cdots(n-k+1)}{k!}$

so, would $\binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{4!}$

ok

@zacts yes

in $\binom{n}{k}$ I only consider $k$ being a non-negative integer. $n$ can be any real (or complex even)

so you were directly notating my original solution to the $n^2$ problem in terms of binomial coefficient notation, which then allows you to use operations specific to binomial coefficients.

yes

cool, thanks.

5:37 AM

@zacts there is a nice interpretation as the number of paths to get from the top to a point in Pascal's triangle.