Mathematics - 2024-05-03

John Zimmerman

00:28

2

Q: What is the volume of the largest surface of revolution that can be placed inside the unit cube?

Consider a surface of revolution $S$ and an embedding $e:S \hookrightarrow X^3$ for $X^3=(0,1)^3$ with points $p,q$ elements of $\partial X^3$ where $\partial X^3=X^3-(0,1)^3$ for $\mathrm {sup}~ \mathrm{dist}(p,q)=\sqrt{3}$. What is $\rho_{\mathrm{max}}=\mathrm{max} \lbrace \mathrm{vol}(S) \rbr...

Only thing I'm unsure whether i have defined well enough is the "ACCUMULATION" concept

Obliv

00:42

how does $$1+e^{-\beta \varepsilon}+e^{-2\beta \varepsilon}+...=\frac{1}{1-e^{-\beta \varepsilon}}$$

Thorgott

00:52

that's a geometric series

Obliv

oh thank u

Obliv

02:01

how does this approximation follow? is it a taylor expansion or something

I thought if $\beta \varepsilon \ll 1$ then $e^{\beta \varepsilon} \approx 1$..

oh wait

that makes sense

like intuitively

but I don't get why even include the $\beta\varepsilon$

EE18

02:24

I haven't studied Fourier analysis and the following is therefore coming only in the context of an engineering book, so no worries about a detailed explanation

At any rate, is it roughly correct to say that in the space of $T$-periodic functions (I think maybe this is called $L^2(T)$ or something like that?) one has that for any $f$, if we use $f_N$ to denote the approximation to $f$ with a trig fourier series of degree $N$ then we have $\lim_{N \to \infty} || f - f_N || = 0$ (where $|| f - f_N || = \sqrt{\langle f - f_N , f - f_N \rangle} = \sqrt{\int_T (f - f_n)^2}$)

(The integral denotes an integral over some interval of length $T$)

Also, I assume it's true that $\lim_{N \to \infty} || f - f_N || = 0 \iff \lim_{N \to \infty} || f - f_N ||^2 = 0$ but want to be sure

Obliv

is that the square root of the norm/inner product or something

the angle brackets

i wonder why they include \sqrt of that integral squared and not just leave it at the integral

EE18

@Obliv Looks like a physics text, maybe even Schroeder. We usually do arguments to first order (or the least order at which we can see interesting dependencies)

Obliv

yeah it was first order taylor series expansion of $e^x = 1+\frac{x}{1!}$ lol

i felt silly

EE18

square root of integral squared is not the same as the integral

Obliv

interesting :P

leslie townes

02:36

EE18: yes stuff like that is true

the L^2 norm was basically made for that

the gap between engineering and math widens when you want to talk about limits in senses other than that one (e.g. pointwise convergence)

no disrespect to engineers, the gap between "easy math" and "A+ undergrad student level math" and "grad student in functional analysis level math" also widens at pointwise convergence, which is among the hardest phenomena to study that i can think of

it is a cruel joke that this just happens to be a phenomenon that joseph fourier and a ton of other people thought about in the late 18th and early 19th centuries

pointwise convergence of fourier series had at least a little bit to do with cantor thinking about complicated sets of real numbers and hence set theory

which should all be very humbling indeed

Obliv

I think I confused you for copperhat. That makes a lot of sense that you aren't an engineer @leslietownes

I was thinking how is this person so proficient at math as an engineer lol

leslie townes

hahaha yes, because i didn't pop the Engineering Pills

not even once, kids

Obliv

I wonder if there are any ed wittens in engineering. I'm sure there are to some extent at least considering all of history

like heaviside i think was pretty dang good at math. he independently developed vector calculus for his work

EE18

Noted on all fronts Leslie, thank you for the historical commentary too, I always enjoy that

leslie townes

heaviside also had a chip on his shoulder about math, though

EE18: i hope "EE" doesn't stand for what i hope it doesn't stand for

EE18

02:46

I want to read Stillwell's history of math text someday soon, but want to understand undergrad math first

It sure does :)

leslie townes

the problem with being too good of an engineer is that it is a supreme inoculant against writing papers about generalities

or the good thing, i guess

EE18

certainly no papers have been written by me, so i get to skate by that concern

I just get to read/learn from the masters (in math, physics, and EE). A pleasure :)

Obliv

I think math/physics provides the tools for engineers to actually put to use though. So the "mastery" is in the entire process. I guess if you consider experimental physics then physics/math are self contained

as in it'll keep progressing forward

leslie townes

physicists who are mostly physicists still frighten me, i had one on my phd committee and met another through my advisor, and i thought i was safe from tough questions from them because my stuff was too 'mathy'

... nope

Obliv

yeah depends on the physicist & area of study. I know some phd students at my school in cosmology that are pretty weak at math

EE18

02:53

were you an analyst leslie? Trying to think where the connection to physics would be and guess DEs broadly?

Obliv

it takes more coding than physics at this point lol

leslie townes

ee18 in some parts of operator theory there is basically no distinction

between math and physics

EE18

that's fair

leslie townes

as i learned

i remember wanting to tell my outside member "do you understand that, as my outside member, you sort of don't get to ask me real questions?"

EE18

Obliv: cosmology or astrophysics? I usually think of the former as more theoretical, requiring mastery of GR and QFT

LOL

Obliv

02:56

I think they're pretty much the same thing but maybe they're more on the experimental side of things idk

GR isn't as scary as QFT

and QFT is necessary for ST so that goes to show how it goes lol

but pure math is scarier because it's so vast. at least in physics the hardest theories are typically constrained to a few main branches or something

mathematicians just invent things for the heck of it

X4J

if $\sum ln(1+a_k)$ converge, so does $\sum a_k$?

I know it's true if the convergence is absolute

but I struggle to think about the general case

Semiclassical

@Obliv GR is hard but we have a pretty good handle on what GR actually is

QFT...

Obliv

@X4J i would guess no but idk

CowperKettle

04:17

user 70432

05:26

g without π is the 👿s number.

one potato two potato

05:52

@BalarkaSen It's quite a naive question but in general, if one says "a twisted $I$-bundle", then is it an $I$-bundle over an nonorientable manifold or a nontrivial $I$-bundle?

leslie townes

you didn't ask, but i'd think the latter

one potato two potato

nontrivial $I$-bundle?

It's also written in Hemple's 3-manifold textbook but as far as I know, it's not formally defined

leslie townes

yeah, i would not expect anything like that kind of use of language to be formally defined

but were i to give it a formal definition, it would be closer to the latter than the former

one potato two potato

Ok thank you

leslie townes

it's still worth asking balarka to be sure! i don't know if maybe there is some convention within people who actually know topology

i'm just giving my own instinct

EE18

13:48

Am following along with Enderton's construction of the reals via Dedekind cuts but am a bit confused about his argument here (but not the conclusion, as I would have used a different argument)

I've only shown half the proof above, and my only confusion is about the argument for injectivity of $E$

How does $r \in E(s)$ but $s \notin E(s)$ establish the claim?

I would have thought instead that $r \in E(s)$ but $r \notin E(r)$ establishes the claim?

ope never mind, i found an old errata list and i am right

Thorgott

I agree

EE18

:)

EE18

14:08

Does the Dedkind cut method require that the ordered field be archimedean?

Some other source I'm reading (Mendelson, 1973) seems to suggest as much but as I glance at my proof of the LUB property from enderton i don't see it slipping in

Thorgott

what precise statement is "the Dedeking cut method"

EE18

I am not sure I have such a precise statement, but the method where you define the reals as (lower) cuts and then show that they satisfy all the relevant axioms?

So it seems on closer inspection the comment was not made when proving the LUB property

But rather to this (stronger?) related statement

My question regards the footnote given

LC is the "set of all lower cuts". This is from Appendix F of Mendelson's Number Systems and the Foundations of Analysis

Oh crap I am way off, this isn't about upper bounds in $R$ at all but rather in $Q$. Please ignore the above.

Jakobian

@EE18 obtaining Dedekind-completion of an ordered set - no

Obtaining Dedekind-complete field - I think so yeah

Which are subfields of R from what I recall. So its not that relevant

EE18

Really? How could that possibly be? The lub property doesn't interact with the field operations at all so seems if true for one then true for other no?

Like in Enderton he just proved that $R$ has lub property for $R$ considered as an ordered set

Jakobian

@EE18 but you need to prove that you can define operations that lead to a field

On the construction

EE18

14:21

agreed, but why would the archiemedean property slip in when defining the field operations? Just to be clear on what Enderton did as far as I can tell: show R has lub property as ordered set. Then show R is (can be made into) an ordered field. No interaction between the two as far as I can tell

Jakobian

12

Q: Dedekind completion of ordered fields

Let $\mathbb S$ be an ordered field of cardinality larger than $\mathbb R$. Let $\mathbb S^*$ be the completion of $\mathbb S$ via Dedekind cuts. Now it is well-known that $\mathbb R$ is the unique complete ordered field. So in what was does $\mathbb S^*$ fail to be a complete ordered field, a...

logic set-theory model-theory nonstandard-analysis

It looks like the obtained construction might not be a field because in general it lacks inverses.

So yeah. What I was saying

The construction doesn't lead to a field

EE18

Interesting, thanks Jakobian. So in doing the completion you get the lub property but some of the ordered field properties fail (unless it's R)

Jakobian

It works for any Archimedean field but not otherwise

Which in this case we can find injection into R, and then just take smallest Dedekind-complete subset containing this image

EE18

So if $\Bbb S$ in the above link were archiemedean then it would work? Hmm, interesting. Still curious as to where the archiemedean assumption slipped into some of the subsequent o.f. properties proved by enderton, will look again

super weird, he didn't even show $Q$ was archimedean. i guess the relevant facts were embedded in the construction though

Jakobian

14:41

@EE18 if it were Archemedean then it wouldn't have cardinality strictly greater than continuum

EE18

I certainly don't know enough set theory to prove that. Maybe that will be an exercise in the next chapter of Enderton :) anyway, i won't waste any more of your time, but this has been very interesting as always. thanks jakobian!

Jakobian

@EE18 its simply because all Archimedean fields embedd into R

EE18

Is that fact easy to show? It's not a priori obvious why that property would be enough to guarantee an injection into R

to me anyway

Jakobian

EE18

Cool, will read this now. Where is it from out of curiosity?

Jakobian

14:50

Rings of continuous functions by Gillman and Jerison

Chapter 0

Not really for you at your current education, in my opinion

EE18

That is for sure

Thorgott

ah, that's a cute proof

2

Pizza

15:34

hi

Soumik Mukherjee

Hi

John Zimmerman

hi

Soumik Mukherjee

15:51

Finite Simple Group (of Order Two)

►

Balarka Sen

16:57

@onepotatotwopotato I agree with @leslietownes: it means a nontrivial $I$-bundle

leslie townes

so when the kids say "don't get it twisted," they mean, "do not become confused in that you mistake a trivial bundle for a nontrivial one"

EE18

18:31

In my proof (for an exercise) that $Q$ is archimedean i basically reduced it to that $Z$ is archemedean. I'm happy to also post my solution in here if it helps for context but basically am curious if that's the "usual"/"best"/"most direct" strategy?

Xander Henderson

@EE18 Honestly, the "usual" / "best" / "most direct" proof is going to depend on what tools you have available to yourself.

leslie townes

EE18 all of this stuff about the archimedean property and dededkind cuts and whatever is "i'm stuck in this puddle of wet mud right outside my house" and the city is 10 miles away

EE18

That's fair and I guess (other than Leslie who has read Enderton) it will therefore be tough for me to explain, but basically it's the constructive approach

But i like puddle jumping!

Xander Henderson

@leslietownes Oh, is there context I'm missing?

(I refuse to read back through the chat!)

leslie townes

EE18 that seems like a logical reduction to me

xander: sure, but everything is [two thumbs up]

Xander Henderson

18:36

@leslietownes GENE SISKEL?! IS THAT YOU?

Thorgott

yes, the Archimedean property of $\mathbb{Q}$ reduces immediately to that of $\mathbb{Z}$

EE18

19:23

I completed most of the exercises up through the end of Ch 5 of Enderton this morning, and most of them I was able to connect (in terms of "what they're really saying", or what properties of e.g. the reals they're talking about) to my previously less formalized knowledge from elsewhere, except for this one:

Is there anything interesting or general to be said about this one, or is it just an exercise? Exercises near it had me prove e.g. that the reals are archimedean etc so I wanted to check in on this one

Thorgott

19:51

this seems like a pretty meaningless observation, all things considered

Semiclassical

20:07

@XanderHenderson the ghost of thumbs departed

Obliv

20:34

trying to remember how to do partial sums for geometric series. Suppose $$s_{n-1}=1+e^{-\beta\varepsilon}+e^{-2\beta\varepsilon}+...+e^{-\beta\varepsilon(n-1)}$$ then $$e^{-\beta\varepsilon}s_{n-1}=e^{-\beta\varepsilon}+e^{-2\beta\varepsilon}+...+e^{-n\beta\varepsilon}$$ $$e^{-\beta\varepsilon}s_{n-1}-s_{n-1}=1+e^{-n\beta\varepsilon}$$ $$s_{n-1}=\frac{1+e^{-n\beta\varepsilon}}{e^{-\beta\varepsilon}-1}$$ what am I doing wrong

do I just approximate $e^{-n\beta\varepsilon}\approx 0$ as $n$ gets larger

ya apparently u do

also messed up sign

should be $-1-e^{-n\beta\varepsilon}$ on the 3rd line

ZaWarudo

21:26

When we say that a sequence of functions $\{f_n\}$ defined on $A \subseteq \mathbb{R}$ is bounded, we mean that for each $n \in \mathbb{N}$ there exists $M_n \ge 0$ such that $|f_n(x)| \le M_n$ for each $x \in A$? That is, we fix $n$ and we want the boundedness as a function of $x$ even if the sequence $M_n$ that bounds $|f_n(x)|$ is unbounded in $n$?

If yes, this means that $f_n(x)=\frac{n}{x^2+1}$ is bounded on $\mathbb{R}$ because $|f_n(x)| \le n$, right?

Thorgott

I would assume it means there is a uniform bound, but check however your source defines it

ZaWarudo

That's the problem: I searched on more than one textbook and no one defines rigorously with quantifiers the boundedness of a sequence of functions

But I have to say that later in the textbook they define the uniform boundedness as "there exists $M \ge 0$ such that $|f_n(x)| \le M$ for each $n\in\mathbb{N}$ and for each $x \in A$"

Knowing this, you think that the boundedness definition I gave above is probably what is meant when talking about boundedness?

Thorgott

21:45

that is impossible for me to tell

Obliv

23:50

anyone have a hint on how I can analyze this function $$\frac{Nk_B(\varepsilon\beta)^2 \exp(\beta\varepsilon)}{(\exp(\beta\varepsilon)-1)^2}$$ where I'm looking at both limits $(\beta\varepsilon) \gg 1$ and $(\beta\varepsilon)\ll 1$

$N$,$k_B$ for the purposes of this are constants

I wanted to try L'hopitals rule but I don't think it'll do much because there are exponentials so the degree doesn't change

can i write it as a geometric series again

i found on wiki for $\lim_{T\to \infty}$ we have $3Nk_B$

oh $\beta = \frac{1}{k_BT}$

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$Mathematics$ Mathematics