Mathematics - 2021-09-03

2:39 AM

Guys, I would appreciate any help with this. I’ve been thinking about this for about a day now, but I can’t seem to progress any further:

1

Q: Given a simple closed plane curve, find a relationship between the area of the original curve and the area of its parallel curve

For context, I have been introduced to simple plane curves from the perspective of single-variable calculus through the use of parameterization. Furthermore, my textbook only deals with plane curves that don't intersect themselves. The question from my textbook is stated as follows: Let $x = x(t...

I can’t seem to simplify the integral I obtain any further mostly due to the bounds of the integral

And frankly I’m a little burned out to flesh out my idea regarding using osculating circles and I don’t even know if it will work.

Ted Shifrin

2:56 AM

@DavidChoi Commented on main.

Clarinetist

Hi @Ted and everyone else here I can think of (@copper.hat): long time no see, just wanted to mention that I'll be putting in my two weeks soon. Leaving higher education entirely to join the corporate world in doing large-scale data work. I'm excited at the opportunity, because my colleagues and manager are all PhD statisticians, and there will be a lot for me to learn.

leslie townes

congrats!

David Choi

@TedShifrin thank you for your answer. I was wondering if you could elaborate on your method using Frenetic formulas or on how I could proceed with my method. Furthermore, I’m aware it’s a standard result, but I wasn’t able to find proofs online for case of simple plane curves. I’m completely stuck right now so I would really appreciate more detail if possible.

*Frenet formulas

Ted Shifrin

Congrats, @Clarinet!

David Choi

Furthermore, I wasn’t entirely sure on what you meant by

So you’re suggesting that we do not have double integrals bit do have the area formula from Green’s Theorem.

Ted Shifrin

3:06 AM

Do you know the Frenet equations? @David

David Choi

No, I am familiar with them.

I actually haven’t learned double integrals as well

Ted Shifrin

Then write the parametrization $\beta = \alpha - rN$, with $N$ the principal normal.

David Choi

From doing research online, i feel quite underequipped to tackle this problem without the knowledge of calc 3 (vector calculus etc.), but Courant seems to think I should be able to do it

Ted Shifrin

As you assumed, take $\alpha$ arclength-parametrized. Find $\beta’$.

You should also download my free diff geo text.

David Choi

I’ll definitely do that

Just out of curiosity, is it possible to simplify the integral I got further?

As in is my algebraic approach actually viable?

Or were both my attempts in vain

Ted Shifrin

3:14 AM

You need to recognize curvature and know $\int\kappa\,ds$ over a simple closed curve.

David Choi

Actually I was able to evaluate that using some reasoning

Since curvature is how “quickly” the tangent vector turns with regards to curve position

Ted Shifrin

Oh, you have curvature there. You can probably simplify your stuff using arclength parametrization, but I try to make everything geometricslly obvious by making the Frenet frame explicit .

David Choi

$\int\kappa ds$ over the entire length of the curve can be interpreted as the question how much does the tangent vector rotate as a point traces out the curve once

And for a simple curve that doesn’t intersect itself, I reasoned it should be 2*pi

Ted Shifrin

Right. A rigorous proof is complicated, but your intuition is correct.

David Choi

But the problem i have with my integral is the first term

Ted Shifrin

3:18 AM

Write out my way and then make things natch up.

David Choi

Ok, I’ll definitely try using frenet frame

Thanks so much for your input

Ted Shifrin

Your limits of integration and signs are not right, either.

David Choi

Oh really? Aren’t the limits meant to be $l_p$ and 0?

Oh wait

I assumed the original curve is arc length parameterized

But that doesn’t mean the parallel curve is

I think that’s where my error is

Ted Shifrin

Correct.

David Choi

Just to be clear, what did you mean by my signs are not correct?

Ted Shifrin

3:22 AM

$t$ is still the arclength parameter of $\alpha$.

The normal points out, not in.

David Choi

So you mean my equations for $x_p$ and $y_p$ are in correct?

*incorrect?

Ted Shifrin

Reread the problem. Do we go outward or inward?

David Choi

We go outwards

Ted Shifrin

That’s why I wrote $-rN$ in my formula. Go work on it.

David Choi

Will do! Thanks so much! Seems like I made some silly mistakes

Ted Shifrin

3:29 AM

Nah, it’s subtle.

Koro

4:44 AM

Suppose that $G$ is a group with the property that given any subgroups $A,B$ of $G$, either $A\subseteq B$ or $B\subseteq A$ holds, then is $G$ cyclic?

For finite $G$, the result is true.

For finite $G$, it can also be shown that $G$ is cyclic and is of prime power order.

Is the result true for infinite $G$?

copper.hat

5:20 AM

@Clarinetist Best of luck!!!

Koro

hi @copper.hat

copper.hat

Hi @Koro!

I'm just heading to bed! I have a work meeting in the morning :-(

Koro

Good night, copper :)

if $G$ is finite and has the above stated properties then $G$ must be of prime power order because if not, then suppose on the contrary that order of $G$ is divisible by two primes $p$ and $q$ then by Cauchy's theorem $G$ has two elements $x,y$ of order $p$ and $q$ respectively.

The cyclic subgroups $\langle x\rangle$ and $\langle y\rangle$ then don't satisfy the stated property so a contradiction. Hence, $G$ must be a prime power order group.

leslie townes

well, there's only one infinite cyclic group, and it does not have that property. so are you wondering if there are infinite groups that do?

(necessarily non-cyclic)

Koro

Hi Leslie

yeah, I am looking for an infinite group which has that property and is non-cyclic.

that is a counterexample to the above theorem which is true for finite groups

leslie townes

5:30 AM

one of my favorite weird infinite abelian groups is G = {exp(i pi k/2^n): all nonnegative integers k and n}. it is not readily apparent, but all of its nontrivial subgroups are isomorphic to G (and hence to each other), and subgroup containment may be not too difficult to work out.

(removed)

Koro

Intuition says the counterexample exists. The intuition comes from trying to prove that $G$ is cyclic (it is already proven above that $G$ is of prime power order or what is the same- $G$ is a p-group)

@leslietownes fixing $n$ and varying $k$ over all integers will give a subgroup. Right?

leslie townes

or any other prime p instead of 2. the last time i brought this group up, someone told me it was some kind of algebraic limit of Z_{2^n}'s so maybe it is a natural extension to look for an analogue of the finite case. but i am weak on the details of that.

Koro

I can get two different subgroups by fixing different $n$'s say $n_1$ and $n_2$ and it's not apparent that one of these will contain the other

leslie townes

if n_2 = n_1 + a (without loss of generality) and k is given then k/2^{n_1} = (k*2^a)/(2^{n_1+a) = (k*2^a)/2^{n_2} = (integer)/2^{n_2}, suggesting that you do have containment.

in that example anyway. then there's the question of whether this example exhausts all possible subgroups of G.

Koro

Let $x$ be an element of order $p$ in $G$. Then if $\langle x\rangle=G$, we are done. If not, let $y\notin \langle x\rangle $ then clearly $\langle x\rangle \subset \langle y\rangle$ (by the hypothesis; the other containment is ruled out as $y\notin \langle y\rangle $ but $y\in \langle y \rangle$). Then if $\langle y\rangle=G$, I am done else I repeat the above process, which ends at some point as $G$ is finite.

6 mins ago, by Koro

Intuition says the counterexample exists. The intuition comes from trying to prove that $G$ is cyclic (it is already proven above that $G$ is of prime power order or what is the same- $G$ is a p-group)

@leslietownes thinking ...

leslie townes

5:40 AM

we need a proper algebraist. some of the europeans will be waking up soon.

Koro

"ruled out as $y\notin \langle x\rangle$ but $y\in \langle y \rangle$"

leslie townes

look at alex youcis's answer here. he handles the infinite case. math.stackexchange.com/questions/496254/…

if i'm not mistaken, the group up above is Z(2^oo) and yes i am using oo for \infty. but he reasons with it in a very abstract way. there must be some way of shaking it out of the above concrete realization.

Koro

Thanks. I'll take a look at that.

I liked how you wrote infinity. 2^oo

$\ddot \smile$

leslie townes

it drives some people crazy, but that's not why i do it.

i'm trying to extend the lifetime of my shift keys by not typing the dollar sign.

Euler 2

$\ddot \frown$

leslie townes

5:49 AM

that's clever

TheReal__Mike

Could someone help me in this situation : academia.stackexchange.com/questions/175142/…

Koro

6:13 AM

@leslietownes As of now, I don't understand many things in the answer after "Edit". I think I'll understand them after I have covered torsion, rank of a group etc.

leslie townes

koro i wouldn't bother with the way he gets at the solution, which is pretty advanced, and maybe more useful for establishing that the Z(p^oo)s are the only examples.

with just the concrete construction above it ought to be possible to show in a more elementary way that the group G above does have the property you want

shintuku

1:03 PM

the academia stackexchange is horrifying

i will pretend it does not exist

Thorgott

coincidentally, I had briefly thought about which groups have totally ordered subgroup lattices just recently

but it seems my commentary is not needed anymore, Alex Youcis' answer is very good and resolves everything

TheReal__Mike

2:07 PM

@shintuku why ?

and what you think in my case

shintuku

most top posts convey a feeling of helplessness

@TheReal__Mike for instance, in your case, my usual technique would be to suck it up, which is the definition of helplessness. you're pretty much powerless as an undergrad if you don't have any prior business with your department which could provide you some leverage

but the workload as an undergrad makes the situation bearable, at least, and it is super temporary

TheReal__Mike

2:43 PM

@shintuku thanks for having the understanding. Actually this deal is made prior and I can challenge it with administration. I am going to talk to professor. Life is very short to do something we aren't interested in.

I have choice right now, atleast later I won't regret that I didn't asked. I seriously don't like doing thing even for hour i am not interested in.

shintuku

@TheReal__Mike yeah, first step is of course to talk to your supervisor very directly if you haven't done that yet

TheReal__Mike

yes just messaging it. And thanks for understanding. I really mean it

shintuku

there's actually absolutely no reason to escalate things if you haven't talked to your supervisor first hehe

TheReal__Mike

Not going to lie but I am in India and it's very common here to find helpless people working on things they don't like

Education system is very bad here compared to europe and usa

shintuku

oh also: you have to be very wary of isolating yourself by escalating, even if you have legitimate reasons to escalate. so the literal best way forward is to entertain a friendly relationship with the supervisor

TheReal__Mike

2:49 PM

I am worried so much because usually in lot of colleges they don't listen, but only difference in my case is that we made deal of topic before

and it's not ethical to chanfe it

Yes, I am messaging him with sincere request

Euler 2

yes

I can now pronounce xkcd

shintuku

@TheReal__Mike be sure to appear helpless and pitiful at first hehe

Euler 2

two hisses and a 'duh' sound

TheReal__Mike

@shintuku okay I will

shintuku

best to not even mention the deal/contract/administration, at first

TheReal__Mike

2:55 PM

Yeah I know. It would be bad

I will be respectful and sincere and professional as much as possible

shintuku

you should also attempt to inquire with fellow students or grad students about how common this is. it would be very unfortunate if you eventually escalate and you are principle right but in practice complaints are not enforced, or slight disregard of official policy is not typically punished

TheReal__Mike

Most student in India don't really care about these things. "Don't complain enjoy the pain, follow the herd, get degree or job and die" . I am going to talk to supervisor in good manner telling what is in my heart

He was student once and he is human, he should understand

shintuku

i think the steps you can safely undertake, if you are not well-connected and are unaware of typical practice in your department, is the pitiful/helpless request. anything more forceful or assertive than that risks isolating you if you don't have more information

TheReal__Mike

Man, I would want to be clear. I can't live my life more in fear. Notice that it's not impulsive decision, I have thought about it in last two days. I will do my best to convince him and change his mind

Ofcourse I will present myself as piful and helpless

*pitful

shintuku

good luck hehe

TheReal__Mike

3:07 PM

thanks

If anyone is professor here, How would you see such student requesting you ?

I want to see from shoes of professor also

So that I can present myself better

sent the message

:)

TheReal__Mike

3:26 PM

Thanks shintuku for discussion. see you

shintuku

good luck, see you around!

Derivative

4:15 PM

how do I logically justify this passage? I'm asked to prove that if A and B then C, and I want to prove instead that A and not C implies not B

I verified that the truth tables are the same but I wanted a more direct way of justifying it

as in if there's a way of saying "oh and then we just use the contrapositive and such and such..."

shintuku

@Derivative use the fact $A \implies B \iff \neg A \lor B$

Derivative

I get to not A or not B and not C and then?

I got it, nevermind

robjohn

4:38 PM

@Derivative $(A\land B)\implies C$ is the same as $A\implies(B\implies C)$ and $B\implies C$ is the contrapositive of $\lnot C\implies\lnot B$

so you get $A\implies(\lnot C\implies\lnot B)$ is the same as $(A\land\lnot C)\implies\lnot B$

Ted Shifrin

I vote for @robjohn's viewpoint. I was about to say it myself.

It happens frequently that you want to pull one of the hypotheses $A$ and $B$ out as an initial assumption, and then think about the remaining implication.

shintuku

yes, maybe that viewpoint is clearer and more helpful. yes, these are professional mathematicians suggesting you proceed along a more intuitive path. but life is easy enough as it is, why make it easier?

Ted Shifrin

Because that's an important strategy. I think I went through that recently, either with you or with Koro, I forget.

robjohn

@shintuku Ohh... you want to make it HARDER? Just give me a few minutes...

Ted Shifrin

Oy vey.

shintuku

4:48 PM

no please it was a joke

Ted Shifrin

We do not joke in this room.

8

Precisely.

5:07 PM

That clip made me laugh :-).

shintuku

is there a proof for $a \leq b \iff \frac 1 b \leq \frac 1 a$ that doesn't involve proof by cases when $a,b \in (-1,1)$ vs $a,b \notin (-1,1)$

Ted Shifrin

Again, you're making false statements.

shintuku

oh. the above only works with $a,b > 0$

Ted Shifrin

Ah.

With correct hypotheses, there are no longer cases to consider.

shintuku

good stuff thanks

edited out a statement, nvmmm

robjohn

5:37 PM

@shintuku it works if $ab\gt0$.

shintuku

@robjohn noted, thanks

anakhro

@TedShifrin I have been struggling with something that should probably be basic. I want to verify that if a geodesic of M starts with an initial point in a submanifold S and a vector tangent to S, that it stays in S. I have a geodesic equation for M and the general form of the elements and tangent vectors to S. I feel like I should find a vector field or something related to the geodesic, but I am unsure of how to go about this.

Ted Shifrin

You need to know the submanifold is totally geodesic.

You need its second fundamental form in $M$.

anakhro

Well this is towards proving that it is totally geodesic that I am doing this.

Ted Shifrin

You need the second fundamental form, not a geodesuc.

Unless this is highly unusual.

anakhro

5:51 PM

It might be unusual. A paper I was following along uses this method to prove that a submanifold is totally geodesic.

I was just trying to prove it for another case.

Ted Shifrin

What is the setting of the paper ?

anakhro

If I find that the second fundamental form is zero, this is equivalent to proving total geodesy, right?

The paper is about GL(n) as a submanifold of Diff R^n

Ted Shifrin

It allows you to see that the geodesic stays inside, by uniqueness.

What metric?

anakhro

Wasserstein L²

Ted Shifrin

I have no clue.

anakhro

5:55 PM

Yeah, it is awfully specific.

For the second fund form, I need to compute either the Gauss/shape map or the connection, right?

Ted Shifrin

I suspect, regardless, that they’re doing what I suggested. Unless the geodesics are themselves “obvious.”

anakhro

Diff R^n is flat.

With the metric in question

SAJW

If there is a big cargo (to big for one truck). Is it always the best solution to send n trucks with 1/nth of the cargo?

Ted Shifrin

you’re in an infinite dimensional setting.

anakhro

But I am a little confused on if that means we can take everything as linear interpolation for geodesics

Ted Shifrin

5:57 PM

oh, what are the totally geodesic submanifolds of a flat space?

linear interpolation?

anakhro

Open subsets?

Ted Shifrin

blah

anakhro

Yeah, like p(t) = (1-t)x+ty

as all the geodesica in some chart

Ted Shifrin

what does that mean?

anakhro

By your reactionI am guessing I am dead wrong.

Ted Shifrin

6:01 PM

Start with finite-dimensional examples, maybe. What are totally geodesic submanifolds of flat $\Bbb R^n$?

anakhro

Affine subsets

Span of something and a translation

Ted Shifrin

Namely, exponentiating vector subspaces of any fixed tangent space. Maybe prove that?

not true in general, but with flatness …

anakhro

Well is there a fast way of proving total geodesy there? Or would you still have to go the longer route of the second fundamental form?

Ted Shifrin

they’re equivalent.

anakhro

Geometrically it is obvious as they contain all the straight line segments.

What is equivalent?

Ted Shifrin

6:07 PM

You need to exponentiate. I’m talking about the general case.

you do need to work with the connection and use flatness. So, yes, you think about parallel transport.

anakhro

General case being that the only total geodesic submanifolds are exp(U) for subspaces U of the tangent spaces?

Hi @Ted.

Hi Balarka.

Hi, not Ted

What have you been up to?

Balarka Sen

6:13 PM

Just kidding, hi.

Nothing particular

anakhro

It is okay, I am used to your ambivalence towards me. ;)

Grand news is I think that I understand why this paper can prove total geodesy the way they do.

Balarka Sen

Cool

anakhro

@TedShifrin sorry for bugging tou with this again. I will work on it some more. Thanks for your suggestions!

Ted Shifrin

Hi, a @Balarka.

Balarka Sen

Suppose $E \subset M \times \Bbb R^n$ is a rank $k$ vector subbundle of a trivial bundle; there's an associated Gauss map $M \to G_{k, n}$. The derivative of this Gauss map gives a bundle homomorphism $TM \to \mathrm{Hom}(E, E^\perp)$, which is the same as a fiberwise bilinear map $TM \times E \to E^\perp$. This is the "normal connection" on $E$.

That is if $D$ is the Euclidean connection on $M \times \Bbb R^n$ then this is $D_X s - (D_X s)^{\|}$ for $X \in \mathscr{X}(M)$, $s \in \Gamma(E)$.

Ted Shifrin

6:26 PM

Sounds like second fundamental forms ….

Balarka Sen

Yeah that's what it is I believe

But what if you just have a formal derivative of $M \to G_{k, n}$, namely, a bundle homomorphism $TM \to TG_{k, n}$ covering the Gauss map $M \to G_{k, n}$? Is it clear what this corresponds to?

Ted Shifrin

well, you’re talking about ruled submanifolds …

Balarka Sen

It's not literally the second fundamental form on $E$, though. It's like the second fundamental form story but bundle-valued.

Ted Shifrin

second fundamental form is normal bundle valued

Balarka Sen

Oh, sure, but I am saying, any bundle embedded in a trivial bundle gives you a "second fundamental form" taking in sections of the bundle (along with a tangent vector on the base), outputting sections in the normal bundle.

Specifying to $E = TM$ gives you back second fundamental forms of embedded submanifolds.

Ted Shifrin

6:30 PM

well, the geometry is the ruled submanifold of $\Bbb R^n$ parametrized by $M$.

I’m thinking projectively, though

Balarka Sen

But if you treated $E$ as a submanifold of $M \times \Bbb R^n$ the second fundamental form would be a bilinear map $TE \times TE \to NE$.

Which is more complicated than what I have. Your thing restricted to the zero section of $TE$ (aka $E$) is probably my thing.

Not quite, but something like that

Ted Shifrin

I’m just thinking of a map to the (projective) Grassmannian as specifying a submanifold.

ruled by $k$-planes

Balarka Sen

I follow that

Ted Shifrin

so how does that geometry relate to yours?

could start with a curve and a line to get a ruled surface …

of course you have a distinguished section of the $\Bbb P^1$-bundle, I guess

Balarka Sen

I don't understand your setup precisely enough, but if I understand it correctly you'd get a bilinear map $TE \times TE \to NE$ where $E \subset M \times \Bbb R^n$ is your ruled submanifold; $TE$ is the tangent bundle of $E$, $NE$ is the normal bundle. I reckon if you restrict to the subset $TM \subset TE$ (tangents to $M \subset E$, zero section) on the first coordinate, $E \subset TE$ (zero section) on the second coordinate, this maps into $E^\perp \subset NE$.

And that's my bilinear map $TM \times E \to E^\perp$.

Ted Shifrin

6:40 PM

well, maybe my viewpoint is hopeless … I’m thinking projectively to get an affine grassmannian or projective grassmannian. I’m thinking if $M$ as the directrix of the ruled submanifold.

this aligns with the Griffiths/Harris paper I’ve always loved on degenerate Gauss maps and linear systems of second fundamental forms

Aha, interesting.

Long time no see

Heya Krijn

Heya Ted

I have a nice mathematical thingy I'm struggling with, wanna think about it?

Balarka Sen

@Ted: But have you ever encountered these "formal 2nd fundamental forms" I was speaking of? Namely, a bundle homomorphism $TM \to TG_{k, n}$ covering a Gauss map $M \to G_{k, n}$. Heck, even for the hypersurface case, $k = 1$.

I'm having trouble wrapping my head around these objects.

Ted Shifrin

6:51 PM

How is this a Gauss map, though?

ugh, safari does not handle chat well at all

is your $M$ a $k$-dimensional sub of $\Bbb R^n$?

Balarka Sen

That's what I call the map $M \to G_{k, n}$ parametrizing the $k$-planes in $\Bbb R^n$ whenever you have a rank $k$ bundle $E \subset M \times \Bbb R^k$.

In the case when $X \subset \Bbb R^n$ is an immersed codimension 1 submanifold, $E = NX$ is embedded in $X \times \Bbb R^n$, so in this case $X \to G_{1, n} = \Bbb P^n$ is actually the Gauss map. Lifts to $S^{n-1}$ for orientation reasons.

Ted Shifrin

I don’t see the geometry unless this fits my viewpoint.

safari is driving me nuts. Back later.

Balarka Sen

OK, let's make a compromise and ditch the bundle-talk, and use your setup. $X^k \subset \Bbb R^n$ be an immersed submanifold. There's now a genuine tangential Gauss map $X \to G_{k, n}$.

The derivative of this map is the second fundamental form

Question: What does simply a bundle homomorphism $F : TX \to TG_{k, n}$ covering the tangential Gauss map $f : X \to G_{k, n}$ mean? In case $F = df$, it's the second fundamental form.

Krijn

Well, Balarka, seems like you're stuck with me until Ted gets back

Balarka Sen

Indeed so

Krijn

7:03 PM

soooooo what's up

Balarka Sen

Nothing particular really

anakhro

What is particular?

shintuku

the unbearable responsability to choose and make decisions, which cannot be delegated to any other person

Krijn

what is up ?

anakhro

Now we are getting deep.

Ted Shifrin

8:29 PM

@Balarka if you have a non-symmetric bundle map $E\otimes E\to E^\perp$, I have no idea what it means. Even an arbitrary symmetric one …

robjohn

8:39 PM

@TedShifrin mobile or desktop?

@Krijn a preposition

@Krijn a movie

Ed Asner died recently. sad

August 29, 2021

SAJW

@shintuku well you could always just wait and have your futereself handle it

Krijn

@robjohn thanks

Ted Shifrin

Yup. I posted twice on FB that Lou Grant was the curmudgeon I always aspired to be. … mobile

robjohn

@TedShifrin Yes, mobile site is horrid

Ted Shifrin

Half the time I type and can’t post. Arrow disappears and return just returns.

robjohn

8:49 PM

I haven't had that problem, but then, I avoid the mobile browser as much as I can.

Ted Shifrin

9:03 PM

@Balarka In the complex projective case, there must be a Chern class interpretation of this stuff. Some of these computations in a couple of our papers. Aren’t these independent of the actual bundle maps, homotopy and all? Hmm …

SAJW

Can math change when switching between universes? Is one side than basically "wrong"?

or are proven theorems of universe a also true in universe b

shintuku

@SAJW i'm not sure what you mean by universes, but there are theorems that can be proven in ZFC and can't be proven in ZF

@SAJW for some examples, there's a nice diagram here mathoverflow.net/questions/114395/…

SAJW

9:24 PM

Many-worlds interpretation

The many-worlds interpretation (MWI) is an interpretation of quantum mechanics that asserts that the universal wavefunction is objectively real, and that there is no wavefunction collapse. This implies that all possible outcomes of quantum measurements are physically realized in some "world" or universe. In contrast to some other interpretations, such as the Copenhagen interpretation, the evolution of reality as a whole in MWI is rigidly deterministic. Many-worlds is also called the relative state formulation or the Everett interpretation, after physicist Hugh Everett, who first proposed it in...

shintuku

9:35 PM

hm, well if you have zfc you can prove all theorems of zfc in whatever universe you are

i don't think the provability of theorems depends on the characteristics of a physical universe, but rather on the logical models in use

SAJW

so math is not only a universal language, it's interuniversal? :D

shintuku

we should probably give more definite meanings to those terms hehe

what you can prove depends on your models

SAJW

But if there is an entity which one can call "God" or whichever name you want for an omnipotent entity: couldn't he create a universe where some or all of our proven theorems are false?

shintuku

no need to suppose God, we can prove that if we remove some axioms of zfc, some theorems become undecidable

SAJW

yeah but they are not necessarily false

or?

shintuku

9:46 PM

exactly, they are proven to be impossible to determine to be right or wrong (undecidable)

SAJW

and there will always be some true statements which are not provable (can't remember who said it)

shintuku

there's Gödel's incompleteness theorems that might interest you

SAJW

are there contradicting systems in that they have one or more statements that are true in one and wrong in the other? which are used, not some bogus

shintuku

10:04 PM

@SAJW there's en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox, which happens in ZFC but not in ZF

it's not, however, false in zfc. just a sometimes unwanted consequence

SAJW

I even knew that one

it's making two balls out of one right?

shintuku

yeah

SAJW

that I call "growing some balls" haha

Given a huge cargo at Point A. All vehicles you use to transport the cargo to point B should be after doing so, again be at Point A. The cargo is too big for one truck. So you can either let 1 truck drive forth and back until all cargo is at point B. Or let multiple drive. Is the distance travelled back always the same?

(assuming no "useless" trucks and all trucks you use are fully loaded)

My feeling is: you are trading time vs used trucks

Avra

10:25 PM

Given the following series, where $m,n$ are integers such that $n<m$:

\begin{align}
\frac{1}{n}\sum_{i=0}^{n-1}{\frac{m}{m-i}}\\
=\frac{m}{n}\left( H_n-H_{m-n} \right)
\end{align}

We know that harmonic number $H_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + + \frac{1}{n}$. So,

\begin{align}
\frac{1}{n}\sum_{i=0}^{n-1}{\frac{m}{m-i}}\\
=\frac{m}{n}\sum_{i=0}^{n-1}{\frac{1}{m-i}}\\
=\frac{m}{n}(\frac{1}{m-0}+\frac{1}{m-1}+\frac{1}{m-2}+\cdots \frac{1}{(m-n)+1})\\
\end{align}

So,
\begin{align}
H_n=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots \frac{1}{n}\\

Anyone who have time, can you look this question please:

https://math.stackexchange.com/questions/4241036/writing-frac1n-sum-i-0n-1-fracmm-i-fracmn-left-h-n-h-m

robjohn

10:44 PM

@Avra Looks more like $\frac mn(H_m-H_{m-n})$

So the formula you stated is false.

Avra

@robjohn. Wow! You are very sharp

9

Now it's very straightforward. Thank you very much

$$
\frac{1}{1}+\cdots +\frac{1}{m}
$$
$$
-
$$
$$
\frac{1}{1}+\cdots +\frac{1}{m-n}+\frac{1}{m-\left( n-1 \right)}+\frac{1}{m-\left( n-1 \right)}+\cdots +\frac{1}{m-1}
$$

Avra

11:13 PM

@robjohn. Can I multiply the inner terms of logarithm $ ln \frac{m}{n}$ as follows please:

$ ln \frac{m}{n} = ln \frac{m \times \frac{1}{m}}{n \times \frac{1}{m}}$

robjohn

@Avra Nevermind... yes, you can since $\frac mn=\frac{m\cdot\frac1m}{n\cdot\frac1m}$

Avra

@robjohn. I got it.

I see it's allowed ;) Thanks again

robjohn

I thought you were taking ratios of logs

Avra

I did not remember we did it before! So I though I might be wrong

no not rations it's $ln (\frac{m}{n})$

Sorry, I should have added paranthesis

Balarka Sen

@TedShifrin You get a bundle map TM o E -> E^perp. Symmetry doesn't quite make sense I think.

Similar to how symmetry isn't a thing for bundle connections

Ah OK but we specialized to the manifold aka tangent bundle case. Yes, good point, it's mysterious

shintuku

11:49 PM

so, for the point-set topology i've been doing, i've only used a sort of $\mathbb{R} \times \mathbb{R}$ real space in order to picture my proofs and used blobs as sets and circles as neighborhoods, but I'm seeing that if I compressed this 2D real space into the 1D real line $\mathbb{R}$, the pictures would still work

is there any precise statement for this?

I'm trying to say there's a map from my 2D blobs in $\mathbb{R} \times \mathbb{R}$ onto the real line $\mathbb{R}$ that would preserve topological properties

PM 2Ring

@SAJW Greg Egan wrote a couple of stories on that topic. Luminous and its sequel Dark Integers. There's some info on kasmana.people.cofc.edu/MATHFICT/mfview.php?callnumber=mf19

Transcript for

$Mathematics$ Mathematics