Mathematics - 2024-11-13 (page 1 of 2)

Sha Vuklia

00:01

i will watch it tomorrow! (i'm trying to finish my preparation so i can go to bed :'))

leslie townes

it is a very short clip highly applicable to reaching the end of a math problem in this fashion

enjoy introducing the children (who are the future) to the flux of vector fields

Sha Vuklia

i'll do my best :)

giannisl9

@leslietownes can you take a look at my question maybe?

leslie townes

so many symbols, giannis :D coordinates in a change of basis are usually some matrix product, involving matrices related to the change, applied to the old coordinates. i'm a little nervous about the word "homogeneous coordinates" here, it's suggesting to me that maybe the basic linear algebra will apply to slightly more numbers than you count as "cooordinates" and then the "homogeneous" part will involve doing some other thing.

giannisl9

@leslietownes thanks! Indeed homogeneous is used mainly for translation to include it in the transformation matrix. I think that a translated orthonormal basis in no longer orthonormal.

psie

00:13

I'm getting stuck on another thing while reading the Hahn decomposition theorem. Let $\nu$ be a signed measure that does not assume the value $\infty$ and let $\nu(A)>0$. Then it is claimed that for $C\subset A$ with $\nu(C)<0$ we have $$\nu(A\setminus C)=\nu(A)-\nu(C).$$ Why does this formula hold? As far as I understand, monotonicity does not hold for signed measures.

Monotonicity is what I'd use to prove that formula for positive measures.

leslie townes

nu is additive? nu(A\C) + \nu(C) = ??

i don't remember what goes into folland's definition of 'signed measure'

a number of the questions lately are deeper into the weeds of folland's treatment of this stuff than you might need for applications of this stuff

psie

@leslietownes indeed, but nu(C) could be negative infinity, is that a problem?

Xander Henderson

@psie can it?

psie

Xander Henderson

Isn't one of the requirements of a signed measure that there cannot be sets of both positive and negative infinite measure?

psie

00:17

And in the proof of the Hahn decomposition theorem, we assume $\nu$ does not assume $\infty$.

So it has to assume negative infinity.

Xander Henderson

Dies it HAVE to?

It COULD. Doesn't mean it does.

psie

yeah ok, we don't know :)

leslie townes

psie: if E_1 := A\C, E_2 := C, E_n = emptyset for n >= 3, then E_j is a sequence of disjoint sets in M. doesn't bullet 3 of your definition of signed measures imply that nu(A) = nu(A\C) + nu(C)

Xander Henderson

But the third condition is the one that matters here.

psie

@leslietownes it does, but how do you know nu(C) has finite measure?

to subtract

Xander Henderson

00:19

@psie does it matter?

Can you show that it doesn't?

leslie townes

as a threshold point, following xander, i would work in cases, to see what cases of nu(A) would correspond to that

like up above with what was going on if m as defined in that sup was +oo, what cases would correspond to that

not just "does stuff like this potentially bust extended real arithmetic in this kind of situation" but "would this bust what the theorem is saying about this case"

and maybe it does and maybe it doesn't, but it seems like you're spiralling right into "it does"

can you construct a minimal example where it's a problem

the baseline thing would be like, let's try a two point measure space, or whatever

psie

as I said above, the first sentence has a typo (it should be $+\infty$).

I know it is a problem for positive measures if nu(C) does not have finite measure. But in this very special case, it maybe isn't a problem.

not sure

it maybe follows from the assumption that $N$ is not negative...

leslie townes

okay, so nu(A) has positive [and implicitly non infinite by hypothesis] measure? by the first part of the first sentence of the sentence beginning "Second, if..."

doesn't that and the equality nu(A) = nu(A\C) + nu(C) imply that nu(C) has finite measure

psie

yeah ok :) the way you put it makes more sense now

so nu(A\C)+nu(C) is a finite number...

leslie townes

it's really helpful if to try to test any theorem/argument against some application or example, because then it becomes possible to pinpoint something that is actually going wrong, vs. something that just "feels unaccounted for" in that maybe the phrasing of some proof by some author is just not expressly talking about it

bear in mind that in any application of any of this stuff, the details of follands proof would vanish and you would have, at most, the hypotheses and conclusion of folland's theorem

and maybe in worse cases, "something very similar to the hypotheses" of folland's theorem and "something like the conclusion" of his theorem

psie

00:36

true

leslie townes

good to start with something like folland though where the background assumption that the author paid some attention to what they were doing has a likely basis in fact

maybe less so with random internet sources or even large-publisher-published-textbooks written by random people

psie

@leslietownes so to summarize. Let C be a subset of A. If nu(A) has finite measure, then nu(A\C)=nu(A)-n(C) holds for signed measures (a signed measure by the definition from Folland, where nu can only assume one of positive or negative infinity). For positive measures, at least skimming through Cohn's book, he actually proves this and writes that nu(C) must have finite measure (here nu is a postive measure). Of course, I understand why though.

I skimmed through Cohn's book because that formula isn't proved in Folland's, probably since it is an immediate consequence of additivity.

leslie townes

00:53

yeah. and generally, there is a lot of give and take between different textbooks about how to define these things, and the static between them is not particularly meaningful in this example. you sometimes do run into situations where one textbook is somehow, by virtue of its definitions or hypotheses, considering a vastly more general context than another, or addressing more cases than another. this does not appear to be one of them.

as far as i've currently looked or read, this looks like a situation of, slightly different definitions leading to slightly different proofs. there isn't going to be, like, "the hahn decomposition theorem," let alone "the proof of the hahn decomposition theorem," that we can put under glass in france and all refer to. you always have this static around the edges.

something i noticed bourbaki doing that i won't mock them for is that in definitions, they tended to seem to be pretty broad, even sometimes losing what one might think of as "the point" of a definition, to give this broad other thing. a definition that is easy to satisfy is a definition that is easy to operationalize and verify in practice.

even if "in practice" is limited to "is what book X saying about Y from their definition of Y actually true, and if so can i prove it from book Z"

bourbaki sometimes seem to be expressly trying to be book Z there

bourbaki isn't usually used as a reference in analysis although some of their analysis stuff is actually pretty good. their stuff got frozen in amber from the lifetimes of some of the more analysis focused contributors

copper.hat

05:42

@psie i have some scanned version of the 2nd ed. thanks for the errata!

leslie townes

copper i think that's the one we had in 2000 :)

copper.hat

06:10

:-). i'm still a little sad after my Moe's visit yesterday, realising that the supply of books is dwindling...

leslie townes

06:21

even 25 years ago i did not think of moe's as a great place for math. cody's on the corner was better. i am not sure that i ever bought math at moe's

copper.hat

no comparison to Cody's. but you could pick up cheap copies of textx

copper.hat

06:34

@psie things that bug me in texts: one of the homeworks in Folland is to prove Lusin's theorem and the hint is to use Theorem 2.26. but the requirements on $\mu$ in 2.26 are stronger than those in the homework problem.

think_meaning_buildß

The good news is the internet archive is back online.

@copper.hat it looks like Mike Tyson is a ~2 to 1 underdog for his bout on Friday.

> Bottom line though, the reason Tyson is a 2/1 outsider in the betting is because it's not the 80s. It's 2024.

copper.hat

09:03

@think_meaning_buildß unfortunately I agree.

would love to see him whup Paul's ass

Claudio

09:46

I talked to my professor. My approach was kind of correct, but I forgot one crucial thing: showing not only that $\exists y_0$ such that $\phi(y_0)=0$, but it must also be unique. Which it’s true since in my case $\phi ‘(y)$ is strictly decreasing

Nobody cared probably but I decided to share this anyways in case somebody actually were interested

Pizza

12:08

hi

Sine of the Time

hi

nickbros123

12:45

i just figured out my method of studying before was extremely inefficient. It took me a while but finally i think i found the method that works best for me

the key i think is to not take math way too seriously. maybe it was obvious but idk

Soumik Mukherjee

13:26

As one of the greatest characters from the goat anime once said "You should enjoy the little detours to the fullest. Because that's where you'll find the things more important than what you want.”

Sine of the Time

13:38

@SoumikMukherjee new WA status :D

Derso

@nickbros123 What do you mean by "too seriously"?

Soumik Mukherjee

@SineoftheTime nice:)

Thorgott

14:46

@Ben I was today years old when I learned that you can also talk about weak $(n,k)$-categories also when $k>n$

Xander Henderson

Gross.

ILikeMathematics

For the construction of $\mathbb R$ using Dedekind cuts, my professor defined $L$ as a left subset and then $L + L' = \{s + s' \mid s \in L \wedge s' \in L'\}$. Afterwards, he defined $L \cdot L' = \{s \in \mathbb Q \mid s < r \cdot r' \text{ with $r \in L$, $r \in L'$ and $0 < r$, $0 < r'$}\}$. But isn't it supposed to include $r \cdot r'$?

Jakobian

@ILikeMathematics depends on your definition

ILikeMathematics

@Jakobian It just seemed inconsistent, $L + L'$ includes the rightmost element, $L \cdot L'$ doesn't

Xander Henderson

@ILikeMathematics What is your definition?

ILikeMathematics

14:53

@XanderHenderson Of $L$?

Xander Henderson

Yes.

What is a "Dedekind cut"?

ILikeMathematics

A subset $L \subset \mathbb Q$ with $L \neq \mathbb Q$ is a left subset if $x \in L \wedge y < x \implies y \in L$ and $\forall_{x \in L} \exists_{y \in L}: x < y$.

Ben Steffan

@Thorgott what's a weak $(n, k)$-category?

Xander Henderson

That definition doesn't seem to include right endpoints.

Jakobian

Yeah. And $L+L'$ won't have one either

ILikeMathematics

14:56

Oh, the second condition prevents right endpoints

Thanks @XanderHenderson and @Jakobian

Jakobian

If they didn't then rationals would have two corresponding Dedekind cuts, something you don't want

Xander Henderson

So much notation... the intuition is that if something is in a "left subset", then everything to the left of that thing is also in the set (the first condition); and that if something is in the set, then there is also something to the right which is in the set (the second condition).

That second condition pretty much explicitly says that there is no greatest element.

Anywho, time to teach.

ILikeMathematics

@Jakobian Which two?

Thorgott

@BenSteffan I'm not quite sure yet!

For $k=1$, I believe what I want is $(\infty,1)$-categories whose mapping spaces have vanishing homotopy in degrees $\ge n$

Jakobian

Personally I like the definition of Dedekind cut as a pair of sets as here

@ILikeMathematics $(-\infty, r]$ and $(-\infty, r)$

Thorgott

15:00

but I'm just using "weak" philosophically to mean "not strict"

cause a strict $(n,k)$-category for $k>n$ does not really make sense (or rather, it's no different than an $(n,n)$-category)

Ben Steffan

yeah

that's why I'm asking

ILikeMathematics

@Jakobian Ah

Thorgott

but a weak $(0,1)$-category is essentially the same thing as a preordered set lol

in general, an inductive definition should probably do the trick: a weak $(n,k)$-category is an $(\infty,k)$-category whose mapping categories are $(n-1,k-1)$-categories

Ben Steffan

@Thorgott a thin $\infty$-category

Thorgott

yeah, homotopically thin

Bml

15:28

Hi everyone, I have a question. If a user has offered a bounty for an existing answer, but this same user has been temporarily suspended, what happens to the bounty?

Jakobian

Its a question for the moderators, but I believe its automatically given at random if the user doesn't give it to anyone

But maybe I'm wrong and it just gets wasted

Bml

@Jakobian OK, then I await a response from a moderator :-)

Jakobian

$Mathematics$ Math Mods' Office

For informal chat with the site moderators about moderation, s...

1

Await it here and mention Xander

Bml

15:48

@Jakobian Thanks.

Sha Vuklia

@leslietownes lol that was very fitting

leslie townes

:D

ILikeMathematics

16:07

@Jakobian With this definition, where exactly did we close all the gaps in $\mathbb Q$? How come just because we take this set of left subsets as $\mathbb R$, we get no holes? And how do we identify $L$ with a real number as we know it?

Soham Saha

Hi! Does anyone know what this kind of graph transformation is called (img below)? Something like splitting a node into 2 connected nodes, while keeping the rest of the connectivity of the graph intact?

1,2,3,4,5 are edges to the rest of the graph from node A in the original graph G

ILikeMathematics

Ah, we identify $a \in \mathbb R$ with the left subset of $a$

leslie townes

it's hard to answer questions like that without just diving into axioms and definitions and proofs. you have to balance any intuition you might have for Q itself plausibly being 'the number line,' with the fact that for many purposes Q just won't do

soham: how do you decide which edges go with which new vertex? before i look for a candidate name for this operation, i'm not sure i understand what the operation would be.

Soham Saha

@leslietownes That is not fixed, any of the original edges can be 'transferred over' to one of the two child nodes

Does that sound reasonable, or am I making a mistake in defining the transformation?

leslie townes

OK, so for a given vertex in a given graph we generally have something like a family of operations, as opposed to one particular kind of transformation.

Soham Saha

16:15

Yes

leslie townes

it sounds perfectly reasonable, i just think it lowers the likelihood of there being a name for it :)

Soham Saha

oh...

Derso

@ILikeMathematic But we only know left subsets of rational numbers, right? We don't know what the left subset of an arbitrary $a\in \mathbb{R}$ might be, if we still trying to define $\mathbb{R}$... What am I missing?

Soham Saha

@leslietownes Actually I was trying to prove non-planarity of a given graph. I believe this transformation preserves non-planarity, and so am looking for a name.

Derso

Actually, we don't even define left subsets "of $a\in \mathbb{Q}$". We just define left subsets in general, using rational numbers. They're not associated (yet) with rational (or real) numbers... (even though we want to write them as $(-\infty, a)$)

leslie townes

16:21

3

Q: edge contraction and planar Graphs

Let $G=(V,E)$ be a graph and $e\in E$ be any edge. Let denote $G/e$ denote the graph obtained by contracting the edge e. Proof: If G is a planar graph, so is G/e While this is intuitively clear, I don't know how to formulate it. Would appreciate any help

graph-theory

@SohamSaha there's no answer to the question i just posted, but there are some links and terminology. do they help? your G would be an "edge contraction" of G'

Jakobian

I like to define Dedekind cuts as pairs (A, B) because it has the advantage of being symmetric

and by advantage I really mean what I find way nicer

leslie townes

Edge contraction

In graph theory, an edge contraction is an operation that removes an edge from a graph while simultaneously merging the two vertices that it previously joined. Edge contraction is a fundamental operation in the theory of graph minors. Vertex identification is a less restrictive form of this operation. == Definition == The edge contraction operation occurs relative to a particular edge, e {\displaystyle e} . The edge e {\displaystyle e} is removed and its two incident vertices, u...

appears to be a kind of inverse to the family of operations you are thinking about

Soham Saha

@leslietownes Ah, that question is exactly what I need I think. If planarity of $G'$ implies planarity of $G$, then we can conclude that reverse edge contraction preserves non-planarity

Derso

@ILikeMathematic How do we approach the gap thing? First of all, we need to say what we mean by "gaps", I suppose. The thing is, I believe you just build the whole $\mathbb{R}$ and its operations using the cuts. So, for example, you'll need to show that there is some cut $L$ such that $L'+L=L'$, for every $L'$. Then you'll name this guy "zero". Show then that there is a cut that behaves as "one". Things are commutative, associative, bla bla bla

Soham Saha

@leslietownes yes exactly

Derso

16:25

And then conclude Dedekind cuts have all the things we want from $\mathbb{R}$. :D

leslie townes

en.wikipedia.org/wiki/Wagner%27s_theorem seems to be the elephant in the room. and the notion of a 'minor' of a graph

cool ideas

Xander Henderson

@leslietownes How did the elephant get into the room?

The door is so narrow, and short...

leslie townes

they contracted some of its edges

Soham Saha

then the faces would kinda bulge out

Derso

@leslie townes The door or the elephant?

leslie townes

16:27

it's wild to me that graph theory is so 20th century of a thing. wagner of "wagner's theorem" for example was younger than both of my grandfathers and lived long enough that i could have met him

it feels like humanity should have gotten to that stuff sooner

Derso

@leslie townes Math "is" trivial after it's done.

leslie townes

@Derso the elephant

the sermon on the mount is really about edge contraction

derso: well, yes, but to me, elementary graph theory seems particularly trivial after it's done, if that makes any sense

i sort of understand why it took until the 20th century for people to introduce the lebesgue spaces L^p for example, although those aren't particularly counter intuitive either

Xander Henderson

@leslietownes I'm not sure that there was much interest in it until the 20th century because there wan't any obvious application, and a lot of the now-obvious applications require a lot of computing time to do anything useful.

Soham Saha

@leslietownes I was trying to prove the non-planarity of this graph (below), by contracting edges and going till $K_5$. But I need to show that the inverse retains non-planarity, so that I can prove that the original graph is non-planar

And I have no idea about "minor", need to read up about it.

leslie townes

@XanderHenderson it had an obvious application to street planning in konigsberg

Xander Henderson

16:37

@leslietownes Not really. That was just a fun puzzle.

leslie townes

haha i like the lighting and erased portion of that image, it looks like a ransom note

Xander Henderson

Not a real application. :P

Soham Saha

@leslietownes Ha

leslie townes

do you know for a fact that the graph isn't planar? living in 2024 there is software that should be able to answer this quickly for a small number of vertices and low investment of time in input into that software

Xander Henderson

@leslietownes We hAve YOur $K_5$. GiVE us ThE monEY, & nO 1 GETs huRt.

leslie townes

16:38

@XanderHenderson what about that one town with three houses and three utilities. that felt real to me

Xander Henderson

@leslietownes Utilities live in three dimensions, not two.

Soham Saha

@leslietownes I had no idea about software for that, gonna look online. The question was posted on MSE, but deleted due to low quality

leslie townes

there's a mildly fun browser game that involves detangling a planar graph so it's planar jasondavies.com/planarity

i'm not sure if that's the OG version of the game (which was called "planarity"), just the first google match

Soham Saha

Hey that is quite cool, thanks!

@leslietownes For the lighting just click a pic of any page with some text on it and click AutoCorrect in MSOffice Photos

No idea why it does that though...

leslie townes

i guess it's trying to widen the visual range of color information from a given input, which you might do if the image conceivably carried a lot of color information but was taken under circumstances that washed it out

which would be more about photos of people and places than stuff drawn on paper

Soham Saha

16:45

@leslietownes yeah maybe

@leslietownes confirmed non-planarity:

leslie townes

OK. do you believe the picture? does it show or suggest a sequence of edge contractions that lead to K_5? if you assume wagner's theorem, such a picture would be a proof that the original graph was not planar

Soham Saha

So, since "building a minor" contains edge-contraction as a valid transformation, so if the contraction steps are all right, then I have successfully proved non-planarity right?

leslie townes

as a matter of the logic of everything, yes. so it might just be a question of how you visually present this information to whoever would be evaluating your work

Soham Saha

Why didn't I see this Wagner before? I was only trying to work with Kuratowski's theorem, and that doesn't mention minors.

@leslietownes Thanks a lot. You're a lifesaver. (No one to evaluate my work, just took up the problem as a fun puzzle :))

leslie townes

17:02

:D

Soumik Mukherjee

@XanderHenderson Sending our K9 to the rescue

Jakobian

@Derso only seemingly because most often we see simple cases

ILikeMathematics

@Derso Thanks

Let $\mathbb K$ be an ordered archimedean field and $I = (a,b) = \{x \in \mathbb K \mid a < x < b\}$ a non-empty interval in $\mathbb K$. To show that there exist rational numbers in $(a, b)$, why can't we just pick $\frac{a+b}{2}$?

Vladimir Lysikov

17:18

Because $a$ and $b$ are not necessarily rational

Ben Steffan

Why would $(a + b) / 2$ be rational?

ILikeMathematics

@VladimirLysikov Oh

Thanks!

@Derso I think we can actually associate a left subset to each real number by taking suprema of subsets of $\mathbb Q$

Xander Henderson

@ILikeMathematics That is circular.

The idea is that $\mathbb{R}$ is the set of all "left subsets".

@ILikeMathematics What is a "rational number" in $\mathbb{K}$?

I feel like there are steps missing here...

Derso

@ILikeMathematics I see. So, for example, $\sqrt{2}$ is associated to the left subset of rational numbers $q$ such that $q^2<4$. We need to show it indeed agrees with that formal definition of left subsets etc

So, the $\sqrt{2}$ "gap" is solved.

However, to speak of "gaps" in general, I guess we need to talk about Cauchy sequences of left subsets. And show that every such sequence converges (in the set of left subsets).

In particular, to talk about Cauchy sequences, we need to define a metric on the set of left subsets. In short, you'll need to define everything we want from $\mathbb{R}$ only using the cuts.

Nasser

17:36

some books use horrible math notations. They use   /  for fraction instead of real \frac{}{}.  How would you read this (image below). Do you read it as

       t^2/( y * (1+t^3) )

or as

      (t^2/y) * (1+t^3)   ?

Derso

(and people really think math is discovered, mother of Christ)

Soumik Mukherjee

@Nasser the former

Vladimir Lysikov

@Derso That is Dedekind's construction of reals

Nasser

@SoumikMukherjee Ok, but how did you decided? why it can't be the second one?

Derso

@VladimirLysikov Yup

Vladimir Lysikov

17:38

@Nasser $t^2/(y(1+t^3))$

But this is indeed terrible

leslie townes

if i had to give it an interpretation, i would give it the former interpretation, but if i could talk to the author i would strongly caution them against writing like this, and if i saw anything like that in a textbook i would recommend that nobody use that textbook

Nasser

The book is Differential equations. An introduction to modern methods and applications. James Brannan, William E. Boyce. Third edition. Wiley 2015 I think they use microsoft word to typeset math. This is from page 79

leslie townes

i recommend that nobody look at that image

Derso

But now I'm thinking of one more detail. How do we define a metric on the set of Dedekind cuts (in order to talk about Cauchy sequences), if we still don't have $\mathbb{R}$ to define what a distance function is... 🤔

You know, $d:X\times X \to ... \mathbb{R}$?

leslie townes

the one thing that any author of any math textbook should do, is not reinforce negative stereotypes about math, including "you need to know secret rules of inference to figure out what i'm saying"

@Nasser it wouldn't surprise me if they did, although if this is just one thing on one page it wouldn't surprise me if this escaped the editorial process

Xander Henderson

17:42

@Derso $q^2 < 2$.

Ben Steffan

you shouldn't use frac for inline fractions, but not parenthesizing properly is inexcusable

Soumik Mukherjee

@Nasser it's bad ofc but the later could be written as (1+t^3)t^2/y, so to choose one, I'll go with the former

Ben Steffan

my impression is that people use frac too much on average, in particular on this site

Xander Henderson

@Derso You don't need Cauchy sequences to get all the non-metric properties of $\mathbb{R}$ that you want. The Dedekind construction gives you "easy" access to the least upper bound property, which is what you really need to show that there are no "gaps".

leslie townes

i'm fine with it if it means people aren't constantly presenting riddles to the site

Derso

17:45

@XanderHenderson Oh, so that's how we approach "gaps".

Soumik Mukherjee

@BenSteffan petition to change the site name to math frac exchange

Ben Steffan

$\mathfrak{exchange}$

\mathfrak exchange :)

Soumik Mukherjee

haha

Xander Henderson

@Derso Anything that you can show with the Cauchy construction (about $\mathbb{R}$ being a complete ordered field, or whatever set of adjectives is needed to uniquely define $\mathbb{R}$) can be gotten from the Dedekind construction, and vice versa.

Derso

Yeah, but for completeness, we can't think of Cauchy sequences interpretation prior to having $\mathbb{R}$ already constructed lol

leslie townes

17:48

we can't?

Xander Henderson

@Derso Note that I am not using completeness here to mean metrically complete.

Derso

Even though completeness for me will always be the Cauchy sequences thing (because it applies for every metric space), never the supremum thing.

Xander Henderson

Rather, it is the property that every bounded, nonempty set has a least upper bound.

Derso

@XanderHenderson Yeah, yeah

Xander Henderson

Again, the least upper bound property.

Derso

17:50

@leslietownes We can't, because to talk about Cauchy sequences, we need distance functions, and to talk about distance functions, we need $\mathbb{R}$...

Ben Steffan

You can define $\mathbb{R}$ as the Cauchy completion of $\mathbb{Q}$, even if you have no a priori access to $\mathbb{R}$. You simply take the metric on $\mathbb{Q}$ to be $\mathbb{Q}$-valued.

You can.

leslie townes

@Derso what ben said

Nasser

@leslietownes the book is full of such ambiguities, Here is another example on page 44, problem 22 (image below).

Derso

Yeah, ok. A $\mathbb{Q}$-valued distance.

Nasser

Xander Henderson

17:51

@BenSteffan I take pains in my thesis to do this for the $p$-adic numbers. The absolute value is $\mathbb{Q}_+$-valued, not real valued.

Derso

@Nasser My eyes bleed.

leslie townes

nasser i am torn between validating your completely justified concerns and not wanting to validate the idea that every textbook is or should be flawless

Xander Henderson

@leslietownes Errors are there to keep you on your toes.

leslie townes

it's very possible for stuff like this to occupy like 5 cycles of brain time and then you just sigh and move on

i remember paying between $50 and $100 in 1997 for an overpriced differential equations book that boyce co-authored, and it was too much, and it is still too much for crap like that

Xander Henderson

An academic sibling of mine had some estimate in a presentation she gave of the form $$[\text{something or anther}] \le \frac{1}{\ln 1 + N}.$$

leslie townes

17:54

but at some point i have accepted that we live in a flawed world

Ben Steffan

@leslietownes can't trust boyce

books written by girls are much better

Xander Henderson

I kept asking "Isn't $\ln 1 = 0$?" It was five minutes before we figured out that she meant $$\frac{1}{\ln(1+N)}.$$

Ben Steffan

@XanderHenderson ahhhhhh

Xander Henderson

@BenSteffan Books written by girlce* are much better.

Derso

@BenSteffan So in that way $\mathbb{R}$ would just be "rational Cauchy sequences", right? Using a $\mathbb{Q}_+$-valued distance function...

Xander Henderson

17:55

(Which is why I never leave off the parenthesis when dealing with logarithms and trig functions. I hate it when people write things like $\sin x$.)

leslie townes

derso yes this is a standard construction of the reals

Ben Steffan

all cauchy sequences in $\mathbb{Q}$ are rational cauchy sequences

talking about "metrics" and "distance functions" is putting the cart before the horse anyways. at this stage you only have absolute values

Derso

Actually the equivalence classes of Cauchy sequences, I suppose? Two sequences are "equal" if they get indefinetly close

leslie townes

yes but in a standard construction of the reals the 'sequences' are rational valued

Derso

@leslie townes But it's a bit different from the Dedekind cuts.

Ben Steffan

17:57

$\mathbb{R} = \text{Ring of cauchy sequences in } \mathbb{Q} / \text{ideal of sequences converging to 0}$

Derso

Nice

leslie townes

yes, it's quite different from dedekind cuts. sequence constructions make managing the definitions of arithmetic easier, at the cost of introducing these really large sets of sequences and equivalence relations on them

Ben Steffan

if you have access to this much commutative algebra (which at this point of course you don't but) you immediately get out that $\mathbb{R}$ is a field :)

then of course you need to set up the order, which I believe is more painful

Xander Henderson

@leslietownes The Dedekind construction makes the least upper bound property a cinch, but arithmetic is a pain, and some of the metric properties are a little harder. The Cauchy construction gives you all the metric properties for free and doesn't hurt too much for arithmetic, but the lub is a bit of a pain.

Thus striketh the Law of Conservation of Difficulty.

Derso

@Ben Steffan Oooh, the inverse of a Cauchy sequence $a_n$ is essentially just $\frac{1}{a_n}$, right?

And also the completeness is kinda trivial

leslie townes

18:00

i like books that don't construct R from anything, or even suggest that this is a worthwhile project. just give me the axioms for a complete archimedean ordered field

it absolutely isn't a worthwhile project

Ben Steffan

@Derso Yes

Algebraically, you show that the ideal I wrote down is maximal

leslie townes

a huge portion of students will miss the point of literally everything, and the 'good' students will enter into a spiral of the universe of all possible constructions and what their favorite is

Ben Steffan

which boils down to an argument like that

Soumik Mukherjee

Sep 19, 2022 at 19:18, by Xander Henderson

sinx/cosx

leslie townes

#facts #busted

Derso

18:02

@leslie townes What about books that construct $\mathbb{N}$?

leslie townes

derso: not in my dictatorship :)

Sine of the Time

@SoumikMukherjee lol

Xander Henderson

18:14

@SoumikMukherjee Those aren't trig functions. That is the punchline of a joke.

Soumik Mukherjee

but how will you get the joke tanx if you don't see them as trig functions? :P

The Empty String Photographer

Most people agree that the three primary trig functions are $\sin, \cos, \tan$

However, $\cos$ is just co-$\sin$.

So $\sin, \sec, \tan$ makes more sense to me.

(don't forget the arcs and the hyperboles)

___

Sine of the Time

don't forget the versine :D

The Empty String Photographer

I have an issue with ChatJax on my computer

@SineoftheTime The versine died.

Sine of the Time

when was the funeral?

The Empty String Photographer

18:25

@SineoftheTime I don't know, $\sin$ just won't tell me!

Xander Henderson

@TheEmptyStringPhotographer $\cos$ and $\sin$ are primary---geometrically, they are the $x$- and $y$-coordinates of points on the unit circle.

Everything else is derived. :P

Derso

@XanderHenderson But $\cos$ is just $\sin (\cdot+\frac{\pi}{2})$, so...

The original $\sin$

Thorgott

the only trig function is $\exp$

2

Derso

18:42

@Thorgott You mean hyperbolic trig?

Thorgott

it's all the same to me

Derso

:0

The Empty String Photographer

@Thorgott If you use $\exp$ for a trig function, things get a little... complex...

When I use the ChatJax applet, and send a message, the $\LaTeX$ works fine.

However, as soon as someone else sends or edits a message, the $\LaTeX$ duplicates.

Sine of the Time

try ctrl+w

The Empty String Photographer

And it can happen again and again

Xander Henderson

18:52

or [ALT]+[F4]...

The Empty String Photographer

@SineoftheTime Nice try

does actually work though

Xander Henderson

19:19

@Derso At the end of the day, it comes down to what you believe are the things which fundamentally define the trig functions. In most of the classes I teach, the functions are defined in terms of the unit circle. $\cos(\theta)$ is the $x$-coordinate of the point where the angle $\theta$ intersects the unit circle, and $\sin(\theta)$ is the $y$-coordinate of that point.

These two functions are "primitive" in that setting, and the remaining trigonometric functions can be defined in terms of those two primitive functions. Yes, they are horizontal translations of each other (i.e. $\cos(x) = \sin(x+\pi/2)$, or whatever), but that isn't really the primitive notion that I have in my head.

Indeed, that relation is most easily understood with respect to the unit circle by way of a rotation.

Of course, $\tan(\theta)$ is also relatively easily understood as the slope of the angle $\theta$, so there is a nice geometric interpretation there, as well.

leslie townes

atan2 is the only primitive trig function

you can tell because they call it atan2

Xander Henderson

@leslietownes I am going to come to your home and stuff that atan2 down your throat. X(

Derso

@XanderHenderson You can define $\cos (\theta)$ as the $x$-coordinate of the point as you said and then prove the $y$-coordinate is the $\cos$ (or "$x$-coordinate") of the figure rotated $-\frac{\pi}{2}$ lol

Xander Henderson

3 mins ago, by Xander Henderson

Indeed, that relation is most easily understood with respect to the unit circle by way of a rotation.

Yes, I said that.

Derso

;)

So

Xander Henderson

19:25

But, right before I said that, I said "...but that isn't really the primitive notion that I have in my head."

Derso

There's only one trig function

Xander Henderson

There are at least 12 trig functions, but I think that the best pedagogical approach is to present cosine and sine as the two most primitive trig functions, and define everything else in terms of those two.

Derso

Haha I would say the primitive one that I have in my head is actually that of a right triangle (there's not even a circle)

Xander Henderson

@Derso Many students have this in their heads when they get to my class. I think that this is the "wrong" primitive notion, however, as it only defines cosine and sine on $(0,\pi/2)$.

Derso

Yes, but that can be easily "solved" by imagining a circle with radius given by the hypotenuse and then extending the notion using the $x$ and $y$ coordinates as you do

Xander Henderson

19:29

@Derso Sure, but the pedagogy should not create problems which have to be "solved".

Derso

But then you need to introduce a less "primitive" notion in order to not have that "problem" lol

I mean, "coordinates", "functions"

Xander Henderson

@Derso I don't understand what you mean by "less" primitive...?

Derso

Shouldn't they come after triangles?

Jakobian

There is this weird phenomenon occurring, that it seems that to create topological spaces with zero-dimensional compactifications, one just removes the assumptions that $X$ is locally compact by $\beta X\setminus X$ being locally compact, then by $\overline{\beta X\setminus X}\setminus (\beta X\setminus X)$ being locally compact. I wonder how much deeper can I go and does it go infinitely deep.

Xander Henderson

@Derso Why should I care about triangles?

I teach mostly pre-calculus and calculus. I don't really care about triangles.

Derso

19:31

They're pedagogical

Xander Henderson

Though I do point out that you can recover right-triangle trigonometry from the unit circle definitions.

Derso

Oh, ok. We're not talking about kindergarten then

Xander Henderson

@Derso No. I am talking about the classes I teach.

Jakobian

I suppose I can replace this condition by, let $r(A) = \overline{A}\setminus A$, then I guess it suffices to assume that $r^n(A)$ is locally compact for some $n$

Or maybe equivalently, that $r^n(A) = \emptyset$ for some $n$

I wonder how to call this property which generalizes compactness and local compactness

(here $A$ is a subset of compact Hausdorff space)

Thorgott

triangles are super-important, everything is made up of triangles

(triangles = simplices)

Jakobian

19:36

Other things I am curious about, is that if $r^n(A) = \emptyset$ in a compact Hausdorff space $X$, and $A\to Y$ is an embedding into a compact Hausdorff space, does $r^n(A) = \emptyset$ in $Y$

Xander Henderson

@Thorgott Topology is dumb. :P

Thorgott

what if I say this is category theory

(nevermind, I know the answer)

Jakobian

this would make sense as local compactness and compactness have this property (n = 1 and n = 2)

I propose the name $n$-compactness. What do you think?

does this property even exist in literature. Why did no one tell me about it

damn I need some answers

wait I am confusing myself

No I'm not confusing myself. If $n = 2$ then $\overline{A}\setminus A$ is closed, so that compact, so that $A$ is open and dense in $\overline{A}$, so $A$ is locally compact

and conversely, if $A$ is locally compact then $A$ is open in $\overline{A}$

0

Q: Generalization of compactness and local compactness

Consider the operation $r_X(A) = \overline{A}\setminus A$ in a space $X$. Here, suppose that $X$ is a compact Hausdorff space. Call a space $A$ to be $n$-compact in $X$, if $r_X^n(A) = \emptyset$ where $r_X^n$ means the operation $r_X$ iterated $n$ times. Being $1$-compact is equivalent to compac...

general-topology compactness

$Mathematics$ Math Mods' Office

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$Mathematics$ Mathematics