Mathematics

2:00 PM

@AkivaWeinberger Take a sphere, push the upper and lower hemisphere through each other to form a sphere with a toral lobe, emerging out of a great circle of double-points. This everts everything except the toral lobe, which is still "outside". But if you evert this torus, do you get a complete sphere eversion?

This is how I plan to evert the torus ^

Transport one of the toral lobes around the main torus to flip the position of the self-intersection loci (the two meridians), and then time-reverse the whole process to get a torus eversion.

Akiva Weinberger

I think you'd end up with a sphere with the torus on the inside

but maybe not, maybe the poles of the sphere get in the way

Balarka Sen

I don't think you can get that

Akiva Weinberger

All spheres are regularly homotopic to each other

jserv

Is there any area of stackexchange skilled with econometrics? I'm really having trouble interpreting a paper and determining where they're taking their numbers from for their formulae

Akiva Weinberger

That's the full version of Smale's theorem. The inside-out vs regular thing is just a surprising special case

Balarka Sen

2:03 PM

Oh of course I mean I don't think you can get it through my process.

As in my process always leaves the toral lobe outside the sphere it's sticking out of

Akiva Weinberger

Not me, sorry @jserv

@BalarkaSen Oh, I think you're right

Balarka Sen

So what's going on?

Akiva Weinberger

I guess the main question is what happens to the poles of the spherw

Balarka Sen

Yeah

@BalarkaSen I think if you go towards the pole by scanning it by the height function along the z-axis you get the homotopy of an arc which introduces/cancel a pair of loops.

Like that picture below

Like, there's a "lip" protruding out of the sphere, pole-to-pole, the upper lip outside the sphere, the lower lip inside.

Akiva Weinberger

Hm. But the homotopy will cancel on the front side and not the back side

Same picture of you do this all without the torus

Balarka Sen

2:07 PM

Hah, that's true. Hm.

Akiva Weinberger

The trick is to get it to switch where the homotopy happens

Balarka Sen

Yeah, there has to be a quadruple point somewhere.

That's where the switching happens

Akiva Weinberger

which I think is the idea behind Chéritat's eversion

Balarka Sen

(Any sphere eversion always has a quadruple point -- a theorem of Freedman, btw)

@AkivaWeinberger Oh, what's that?

Akiva Weinberger

Sphere Eversion : teaser

►

youtube.com/shorts/zOryEuVeTbE?feature=share

Balarka Sen

2:09 PM

Oh wow there's our torus eversion

It's happening along the blue part

Need to study this

Akiva Weinberger

youtube.com/?watch=zOryEuVeTbE

Hold on how did the first link expand and not the second one

Balarka Sen

It's a shorts

Akiva Weinberger

Sphere eversion : cut open version of the teaser

►

Balarka Sen

Aha OK so when you do the torus eversion you also have to pull the torus out from opposite sides.

Akiva Weinberger

In any case, the trick is to rotate the sphere partway through so you're not just dealing with horizontal slices

Balarka Sen

2:13 PM

There's always some 4-fold symmetry with these (of course, has to be, by Freedman)

Akiva Weinberger

Torus eversion

►

The "attachment line" to the sphere would end up on the opposite side I think

Balarka Sen

(I saw that edit) I agree

The attachment line is the line of double points. To get a quadruple point you must intersect this with itself.

It must go through itself.

Good to know Chéritat has thought this through. Truly nothing can be original these days!

Thanks a bunch @AkivaWeinberger

Akiva Weinberger

Look at this figure I found on Twitter from @HedronApp. Is it possible? @BalarkaSen

Balarka Sen

2:38 PM

@AkivaWeinberger Lol, this is possible, no?

Cheritat's paper is amazing, btw: arxiv.org/pdf/1410.4417.pdf

Akiva Weinberger

@BalarkaSen Yeah it is

Balarka Sen

What a troll

Akiva Weinberger

Just kinda feels similar in genre to impossible stuff but it's totally possible

Balarka Sen

Yeah

PM 2Ring

@AMDG For $2^a-1$, it's easy, you can just use binary long division, but that's O(n). Here's some example code:

sagecell.sagemath.org/…

AMDG

2:44 PM

O(n) is permissible if each n belongs to a unique term.

Oh wait I just realized you said binary long division and I was thinking of something else for some reason when you said that.

Long division is not a possibility here. It's for a division algorithm :)

What is m exactly?

PM 2Ring

@AMDG The long division algorithm uses integer subtraction.

AMDG

Yes, and each is dependent on the previous iteration which is what makes it slow in the first place.

PM 2Ring

@AMDG $m = (2^a - 1) / i$. The Python // operator is floor division.

Balarka Sen

@AkivaWeinberger: I forgot to mention, I was thinking about this because I wrote a short note on Smale's original proof of existence of the sphere eversion for a friend (and ended up getting sidetracked with this while drawing cartoons of a constructive eversion at the corner), you might appreciate it: drive.google.com/file/d/10F-8G1GBK3teDV9KN6cWyAhuJNPvv-dO/view

You only need to take some fibrations I have written on faith, and the homotopy long exact sequence.

AMDG

@PM2Ring Is there a way to compute $i$ here independently of $a$?

Balarka Sen

2:54 PM

There's a typo at the bottom of page 2, but you'll figure out what I meant.

Akiva Weinberger

@BalarkaSen I don't have access

Balarka Sen

Oh, cruds, let me see.

Akiva Weinberger

I hit the "request access" button

Balarka Sen

Gave access.

Akiva Weinberger

spheve

rhymes with Steve

Balarka Sen

2:56 PM

Lol

Spheve Smale

AMDG

@PM2Ring Or am I not understanding this correctly? I'm confused because you used $nc = 2^a - 1$ and here changed the variable names.

PM 2Ring

@AMDG Not really. The maximum loop length occurs with some primes, where the period = p-1. They're called cyclic primes. Some well-known examples in base 10 are 7, 17, 19, 43, 97. Eg, 1/7 = 142857 / 999999

That is, $7 | (10^6 - 1)$

AMDG

Well, I guess I'll just have to find a closed form of prime(n). Lemme just do that here really quick... /s :P

Well that's why I want to know if there's any way to make it so that we can easily compute $c$ and $a$ here if we allow more than just integers.

PM 2Ring

@AMDG Sorry. n turned into m, and c turned into i.

AMDG

Ah, ok.

Question: why did they not design sage cell server so that I can expand the text area? This is cringe.

There's an expand button that makes it fill the page, but I can't adjust the dimensions of the text area.

Like, bruh, UI 101. C'mon. Anyways...

PM 2Ring

3:02 PM

If you have the prime factors of m, then you use the Euler phi approach I posted earlier. That function is also known as the totient function. en.wikipedia.org/wiki/Euler's_totient_function

AMDG

Rip URLs with html hex entities

geocalc33

geometry is nothing without algebra

PM 2Ring

@AMDG Yeah, it's a bit annoying. Especially on the phone, since the CodeMirror editor is a bit flakey (although that could be partly the fault of this Samsung browser). I ended up writing my own code to embed it into a HTML page, and I use a plain textarea to do most of my editing. :)

AMDG

Well, thank you for your efforts, @PM2Ring, but this really isn't suitable for my needs here as-is. Clearly a different approach must be taken. What about $n = 2^a \pm b$? This form of reciprocal would work for $\frac{1}{n}$.

geocalc33

algebra is the essence of what the fundamental one utters

PM 2Ring

3:06 PM

@AMDG Oops. Euler's totient function

AMDG

No, wait, I take it back. I don't know why I said $n = 2^a \pm b$.

All I really need is literally just anything for c and any integer for a in $nc = 2^a \pm 1$.

If the constant term there is anything other than 1, then I end up requiring exponentiation using multiplies which is what I'm trying to avoid.

PM 2Ring

@AMDG Sorry, but I'm not completely clear on what your needs are. I only know it's got something to do with doing division / calculating reciprocals. I originally thought you were trying to implement a BigNum package. But then you said it's for some kind of 64 bit numbers.

robjohn

@PM2Ring that was the name I learned. The "phi function" came about because that was the main character used to denote the totient function.

AMDG

@PM2Ring If there exists a quick way to find either $nc = 2^a \pm 1$ or $n/c = 2^a \pm 1$ for $c \in \mathbb{R}$ and $\{n,a\}\in\mathbb{Z}$, then this satisfies my needs.

geocalc33

AMDG exponentiating things is bad

AMDG

3:12 PM

Yes, geo.

geocalc33

it causes confusion - don't exponentiate kids

robjohn

exponentiation leads to "hard math" (just as marijuana leads to "hard drugs")

AMDG

waitwhat

PM 2Ring

@robjohn Same here. Sage calls it euler_phi, which I guess is reasonable. Naming things is hard, especially when you have a lot of things to name. :)

AMDG

Call it euler_totient_function /shrug

user193319

3:15 PM

If $a,z$ are complex numbers in the unit disc, how does one show that $|z-a| < |1-\overline{a}z|$?

PM 2Ring

FWIW, the discrete Fourier transform is essentially using the properties of a geometric progression of complex numbers, specifically the complex nth roots of unity, for some n. The Fast Fourier transform typically use a power of 2 for n.

AMDG

Alright, I'm going to make some lunch. I'll be back.

geocalc33

exponentiation is where the devil lives - yes it's why we can't detect hell - because it's not embedded in Euclidean space. Also imagine a physical 3D world based on exponential space. Circles would not be circles and if you saw a circle that would mean another space was present

of course physics would work consistently but everything would be weird

@I'manalienImaneaglealien this is for you ^

robjohn

Differentitation reveals the finer structure of things. $\frac{\mathrm{d}}{\mathrm{d}x}e^x=e^x$. What are they trying to hide??

geocalc33

3:30 PM

@robjohn they differentiate differently. they do: $\frac{d^*}{dx}e^x=e$

Thorgott

$|z-a|=z\overline{z}-(a\overline{z}+\overline{a}z)+a\overline{a}$, where the three summands are real, $|1-\overline{a}z|=1-(a\overline{z}+\overline{a}z)+a\overline{a}z\overline{z}$, where the three summands are real, and then use $x+y<1+xy$ for any real numbers $x,y\in[0,1)$

PM 2Ring

3:47 PM

A few weeks ago, I was playing around with drawing a closed loop through a set of points, using cubic Bézier curves. I learned a nice algorithm that ensures that both the 1st & 2nd derivatives of the Bézier functions match at their endpoints. But for n points, it requires solving a system of n equations in n unknowns, which gets expensive for large n.

Fortunately, these equations can be expressed as circulant matrices, so the problem can be efficiently solved using Fast Fourier transforms. en.wikipedia.org/wiki/Circular_convolution

Here's a 3D example, which puts a Bézier curve through the points on a Hamiltonian path on a regular dodecahedron.

sagecell.sagemath.org/…

AMDG

4:16 PM

How does one represent any square wave function using only powers of two?

To clarify, I'm referring to square waves and their periodicity.

If I wish to make a function that oscillates between zero and one and alternates after three of either, then for a square wave function $f(t)$ with frequency 1, the corresponding wave is $f(\frac{t}{3})$.

So can I compute $f(\frac{t}{n})$ for all $n$ using $f(\frac{t}{2^a})$ and other simpler wave functions?

I mean the issue is that of creating a wave that describes a convolution between the two essentially.

robjohn

@geocalc33 who does that? that is the derivative evaluated at $1$.

schn

@robjohn , thanks for the reply. Could you clarify what you mean by "...carry that constant along..."?

AMDG

This is the kind of stuff I'm talking about: desmos.com/calculator/u0yqixnwq2 . You wouldn't happen to have any ideas, would you @PM2Ring ?

robjohn

@schn Once you have a constant, you know that $O\!\left(x^m\right)\le C|x|^m$. The decay is quantified, and not left unspecified as with little-o where all we have is $\lim\limits_{x\to0}\frac{o\left(x^m\right)}{x^m}=0$.

schn

But if the constant is unspecified, is the decay really quantified?

robjohn

4:31 PM

you have a constant. It doesn't matter that it is not given ahead of time. What do you have with little-o? a limit, with not even a function to specify how fast it vanishes.

If you would try some examples instead of trying to create some calculus with little-o and Big-O, I bet you would have a better understanding.

Transcript for

$Mathematics$ Mathematics