Mathematics - 2023-05-12

Ted Shifrin

00:25

Time to cite leslie: phbhfft

3

Semiclassical

@TedShifrin $\phi$bfft

leslie townes

00:51

@TedShifrin no, but it had a formula.

robjohn

$\phi\beta\varphi\varphi\tau$

冥王 Hades

01:08

I hate not being able to have lunch alone

D.C. the III

01:32

Well that is rather anti-social

shintuku

01:47

so we have that $z \in \mathbb Z$ gets mapped to $a-3b + \langle 3+i \rangle$, for some $a,b \in \mathbb Z$, but why are we getting surjectivity from this?

shintuku

02:02

can we argue that such a map has the form $\phi(z) = \underbrace{1+\cdots+1}_\textrm{z times}$, so since any $x \in \text{im}\ \phi$ has the form $\underbrace{1 + \cdots + 1}_\textrm{k times} \in \mathbb Z$, $\phi(k)$ goes to it?

Ted Shifrin

02:27

Absolute crap.

4

shintuku

right, the elements of $\text{im} \ \phi$ are not themselves in $\mathbb Z$

Ted Shifrin

First, what you wrote is not meaningful, as the image lives in the wring place. Second, you are arguing $\phi$ maps onto its image. Duh. I am going to say again that you need to back up a few months and learn basics.

冥王 Hades

I'm paying 0 attention in the History lecture and I'm growing the internet instead

robjohn

02:57

@D.C.theIII perhaps he meant having lunch without breakfast or dinner.

@TedShifrin |crap|

4

冥王 Hades

@robjohn Nah I had breakfast

That too alone

I also hate sharing food

shintuku

We argue instead as follows. Suppose $\phi$ is a ring homomorphism from $\mathbb Z$ to $\mathbb Z[i]\setminus \langle 3+i \rangle$.

Notice first that the identity $1_{R'}$ in the codomain is given by $1_{R'} =1 + \langle 3+i \rangle$, and therefore, by the properties of ring isomorphisms, $\phi(1) = 1 + \langle 3+i \rangle$.

Now, let $j \in \mathbb Z[i]\setminus \langle 3+i \rangle$. Then $j = a+bi + \langle 3+i \rangle$, with some $a,b \in \mathbb Z$. But notice that $i \equiv -3 \mod 3+i$, therefore, $a+bi + \langle 3+i \rangle = a-3b + \langle 3+i \rangle$. Let $a-3b = z$.

ted this must be it

but $1_{R'}(z + \langle 3+i \rangle) = \underbrace{1_{R'} + \cdots + 1_{R'}}_\textrm{z times}$ feels a bit weak

shintuku

03:21

i should have said instead: $z+I = (z \cdot 1) + I = (\underbrace{1+\cdots+1}_\textrm{z times}) + I = \underbrace{1_{R'} + \cdots + 1_{R'}}_\textrm{z times}$

shintuku

03:32

argh but $z$ might be negative

np just do the appropriate $(-1 + \cdots + -1) + I$ shenanigans

shintuku

04:50

what polynomial is denoted by that notation?

figured it out, it's $f(x_1, a_2, \cdots)$

Kabir Munjal

05:49

Hi Sir/Ma'am

I had aquestion to generalize this formula given in the link for more than 2 functions reader.elsevier.com/reader/sd/pii/…

Ajay

09:53

@PM2Ring Do you know how to use SAGE to determine if a function is even or odd?

PM 2Ring

12:54

@Ajay Here's a demo:

f(x) = sin(x)
print(bool(f(x) == -f(-x)))

Ajay

13:42

Thanks, it works.

Peter

13:55

Let $F_n$ denote the $n$ th Fibonacci-number and define $$f(n):=F_{F_n}$$ The numbers $f(23)$ and $f(29)$ are composite. What is the smallest prime factor in both cases ?

Semiclassical

@Peter it's probably obvious, but this is definitely not a problem that can be done by actually computing $f(23)$ and $f(29)$

if only b/c, for instance, $F_{23}=514429$ and so the approximation via the golden ratio $F_n\approx \phi^n/\sqrt{5}$ yields $f(23)\sim 10^{107467}$

shintuku

14:44

do you ever use fraktur notation on pen and paper and if you do, how do you write it?

Xander Henderson

15:18

@shintuku No. I'm an analyst. I don't use those dirty fraktur fonts.

That said, the fonts are based on German fonts which would normally be handwritten with a calligraphic pen. Not an easy thing to reproduce with a fixed-width pen or pencil.

Google suggests the following:

shintuku

oooooo nice, cursive fraktur

thanks

Jack Rod

15:41

@robjohn sir is there a mathematical technique named as "$$\TRILATERTION$$

shintuku

Trilateration

Trilateration is the use of distances (or "ranges") for determining the unknown position coordinates of a point of interest, often around Earth (geopositioning). When more than three distances are involved, it may be called multilateration, for emphasis. The distances or ranges might be ordinary Euclidean distances (slant ranges) or spherical distances (scaled central angles), as in true-range multilateration; or biased distances (pseudo-ranges), as in pseudo-range multilateration. Trilateration or multilateration should not be confused with triangulation, which uses angles for positioning; and...

PM 2Ring

@Peter How do you know they're composite? FWIW, F(514429) ~= 1.431439852411614618634259200E107467, its final digits are 109190851429. It has no prime factors <20000000. I did a brute-force check by calculating F(514429) mod p.

@shintuku ACM may have some suggestions.

in The h Bar, Dec 4, 2022 at 11:49, by ACuriousMind

@Feynman_00 I had an algebra prof who insisted on spending the better part of an entire lecture showing us how to draw the fraktur letters he used for ideals and algebras

shintuku

truly a man committed to his cause, committing others to the same

PM 2Ring

Teaching young German mathematicians to write in Fraktur is a reasonable cause. But that doesn't mean the rest of us need to learn it. ;)

There are actually several styles known as Fraktur. The WP article is kind of interesting. en.wikipedia.org/wiki/Fraktur

> In Germany, transition to more modern typefaces was controversial until 1941 when use of Fraktur typefaces was ended by (Nazi) government order.

OTOH, if someone protests against the use of Fraktur it's probably not a Good Idea to accuse them of being a Nazi. ;)

Koro

16:09

"If the reader wants to acquire facility with the Laplace transform or
to study the L2 convergence of the Fourier series of an L2 function she must look
elsewhere."

I'm seeing 'she' the first time in a preface. Usually it's a he or they.

Soumik Mukherjee

@shintuku I used them in commutative algebra course just by adding more loops in the English fonts :)

shintuku

16:30

looptyloops

K field. what do we use K-algebras for?

Thorgott

@Koro Körner?

冥王 Hades

💀

Koro

@Thorgott yup!

PM 2Ring

May 2 at 16:05, by Ted Shifrin

Well, let's ban all angle questions for two months.

shintuku

16:45

i approve

there are no angles only affine space

shintuku

16:56

@shintuku $\mathbb R[x_{1\cdots n}]$ is K-algebra

PM 2Ring

@冥王Hades Here's a 3D geometry question for you. What's the largest channel that can be cut through a cube such that another cube can pass through the channel? It looks like this:

2

Here's an animated version: i.sstatic.net/A9pzB.gif

user 858770

Perhaps such animations will bring back geometry to the high school math classroom.

And even more wishful thinking would be "proof assistances" causing a resurgence of proofs in the classroom.

Semiclassical

17:15

a source of madness when grading: students failing to distinguish between objects that are different, and failing to equate objects which are identical

shintuku

damn the students

they are the roots of all our torments

plato sucks

lmao dude didn't even brush his teeth

PM 2Ring

@user858770 Perhaps. Although modern systems can do some rather fancy graphics you generally need to understand what you're doing to make nice diagrams of stuff that isnt a simple shape or function. FWIW, I made those diagrams over a decade ago, using POV-Ray.

KZ-Spectra

Has anyone noticed the mathjax rendering when writing a post is very bad?

user 858770

To build the interest in understanding what you're doing could be motivated by such 3D animations @PM2Ring

Xander Henderson

@KZ-Spectra You are missing a $ somewhere.

KZ-Spectra

17:34

This is a copy/paste of what I wrote, Xander: mathb.in/75291 It renders just fine here

PM 2Ring

@user858770 I agree. And if the 3D anim is interactive, it's even better. Eg, see the "interactive 3D view" link near the end of astronomy.stackexchange.com/a/43350/16685

Xander Henderson

@KZ-Spectra I maintain my previous comment. Unless you can link to the document which isn't working, I can't help you.

PM 2Ring

Here's another version of Prince Rupert's Cube

Semiclassical

test: Consider a two-sided real sequence $q = \{q_n\}_{n\in\mathbb{Z}}$. I want to obtain bounds for $\sum\limits_{n\in\mathbb{Z}}\sum\limits_{n\in\mathbb{Z}}\frac{|q_n - q_m|^2}{|n-m|^{1+\alpha}}$ where $\alpha \in \mathbb{R}^{+}$

PM 2Ring

Is that ok, Xander? Or should I hide it?

Semiclassical

17:43

renders fine here too

user 858770

Nice 👍

Xander Henderson

@PM2Ring I really appreciate you thinking about my pathological aversion to animated .gifs. Thank you. In this case, it does not bother me that much.

PM 2Ring

Oh, good. :)

Semiclassical

@PM2Ring this makes me think of the trammel of archimedes

but that's mostly b/c i have the trammel on the brain

user 858770

The "structure" is what makes it less optical illusionary.

PM 2Ring

17:55

@Semiclassical Nice. Linkages that convert between linear motion & smooth curves are fascinating, and somewhat hypnotic.

Peaucellier–Lipkin linkage

The Peaucellier–Lipkin linkage (or Peaucellier–Lipkin cell, or Peaucellier–Lipkin inversor), invented in 1864, was the first true planar straight line mechanism – the first planar linkage capable of transforming rotary motion into perfect straight-line motion, and vice versa. It is named after Charles-Nicolas Peaucellier (1832–1913), a French army officer, and Yom Tov Lipman Lipkin (1846–1876), a Lithuanian Jew and son of the famed Rabbi Israel Salanter.Until this invention, no planar method existed of converting exact straight-line motion to circular motion, without reference guideways. In 1864...

> The mathematics of the Peaucellier–Lipkin linkage is directly related to the inversion of a circle.

From futilitycloset.com/2014/12/05/straight-and-narrow-3 Sylvester writes that when he showed a model of the linkage to Lord Kelvin, he “nursed it as if it had been his own child, and when a motion was made to relieve him of it, replied ‘No! I have not had nearly enough of it — it is the most beautiful thing I have ever seen in my life.'”

Semiclassical

@PM2Ring yeah, linkages are fun

i like the trammel a bunch

i'd like to turn it into a way to teach the fact that straight-line motion is equivalent to a sum of left- and right-circular motions, and thus motivate why linear optical polarization is a superposition of left- and right-circular polarizations

i think there's a good way to do that

PM 2Ring

Sounds plausible...

user 858770

...and a lot of work.

shintuku

18:11

kin-groups of pre-industrial communities studied in anthropology have an algebraic group structure

neat

geocalc33

sounds right up one's alley...

PM 2Ring

18:34

in Python on Stack Overflow Chat, Sep 22, 2021 at 16:08, by PM 2Ring

@Kevin Australian Aboriginal society has (had) some interesting schemes that are quite good at reducing inbreeding in a smallish population. They used moiety systems which control who can marry whom. Systems with 4 moieties were the most common, and the mathematical relationship between the 4 moieties is given by the Klein four-group.

in Python on Stack Overflow Chat, Sep 22, 2021 at 17:20, by PM 2Ring

People who grow up in a moiety system can do the calculations almost instantly. If you give a 5 year old kid part of a family tree, and tell them the moiety of one person on the tree, they can tell you the moiety of anyone else on the tree with ease.

shintuku

very cool stuff

user 858770

5 year old :O

PM 2Ring

Klein 4 group arithmetic is pretty easy. :) But yeah, it's different to natural number arithmetic, and addition mod 4.

But I must admit that I was impressed when I saw a little kid doing it on a documentary.

Ted Shifrin

It’s also the mathematical basis of twelve-tone music.

Theme/retrograde/inversion/retrograde-inversion.

user 858770

18:49

If little kids can do it, there must be something there worthy of attention.

PM 2Ring

From thatsmaths.com/2015/02/12/the-klein-4-group This tone row can then be transformed using reflection (left-right flip), inversion (up-down flip) or a combination of these (rotation through 180º). These transformations are completely equivalent to the symmetries of a rectangle, embodied in the group K_4

Ted Shifrin

When I was taking algebra in college and had just learned about the Klein group, it so happened that a pompous musician wrote a longggg article in Perspectives of New Music about the math of twelve-tone music. He went on and on and on. I explained the symmetries to my dad — who was a composer and music prof. He was livid that his colleague had done his best to obfuscate and condescend to his fellow musicians. I remember his saying, “Is that all?”

That sticks with me, 52 years later.

user 858770

He said "Is that all?" to your explanation?

PM 2Ring

Let's face it, strict 12 tone serial compositions sound pretty boring. I guess it was a worthwhile thing to explore, but it didn't lead to music that was interesting and engaging. People had been doing musical stuff with permutation groups for centuries, the purest examples being schemes for bell ringing. But various kinds of permutations are used in contrapuntal music. Guys like Bach & Mozart wrote numerous pieces using permutations that are both clever and interesting to listen to.

Ted Shifrin

19:06

@user858770 He was furious that his colleague had obfuscated to make his “mere” music colleagues who didn’t know the math feel stupid, yes.

I was of course a supreme teacher even then :)

user 858770

👍

Ted Shifrin

I don’t disagree, PM2. The standard sonata form is a basic mathematical structure, of course, but more global and hence not boring :)

leslie townes

and in counterpoint to what pm said, i find listening to a lot of bach as enjoyable as watching someone row reduce a 10x10 matrix of integers by hand.

"oh, he's on row 2 now. oh, he's on row 3 now. ooh, a pivot!"

Ted Shifrin

Definitely unfair to Bach!

You can have second-rate Tchaikovsky :)

PM 2Ring

I love Bach, but in small-ish doses. :) And preferably not with vocals.

leslie townes

19:14

mathematician noam elkies has a party trick where he improvises stuff that sounds like bach. it's impressive until you realize, he's chosen the one composer where you could probably use a computer from 1990 and a minor refinement of a gaussian elimination algorithm to generate compositions that "sound like bach."

i'm a music critic today. this is unexpected.

PM 2Ring

Lots of bluegrass players are Bach fans.

Sierra Hull - Sonata No. 3 in C Major (Allegro) | BACH

►

The Punch Brothers & friends doing some Bach:

Brandenburg Concerto No. 3 in G major: Allegro - 12/16/2017

►

geocalc33

$$I(k)=\int_0^k\pi(x)\pi(k-x)dx$$

$I(11)$ and $I(13)$ are the first odd numbers in the sequence. Pairing these together and going forward like this with the next pairing of odd numbers, does this pattern always hold? That is these pairs are separated by exactly 2 units

冥王 Hades

@PM2Ring you don't expect me to solve a problem that remained unsolved for nearly 100 years after it was proposed, right?

geocalc33

wolframalpha.com/…

PM 2Ring

@冥王Hades Yes, I do. :D My diagrams are a huge hint. At least try to find the ratios of the cubes' sides.

user 858770

19:29

"unsolved for 100 years" is a bit too much imo

冥王 Hades

@user858770 look at the Wikipedia page

user 858770

Will do.

冥王 Hades

@PM2Ring okay, I'll try to derive the ratios without looking at any online solutions

user 858770

Good luck.

PM 2Ring

@冥王Hades Excellent. You shouldn't have much of a problem figuring out what's going on in my diagrams. The hard part is in proving that it's the best solution, and I don't expect you to do that.

user 858770

19:36

We need @copper.hat for optimization!

PM 2Ring

Although pure 12-tone serial composition was a musical dead end, those experiments did influence later composers who were interested in "mathematical" composition techniques, both in "art" music and in jazz. It didn't have much of a direct effect on popular music, but there was a little bit of indirect influence, via jazz.

冥王 Hades

@PM2Ring found a solution

user 858770

So fast!

👏👏👏

冥王 Hades

Assuming a unit square, the side length of that little triangular prism, let's say $x$, can be found with the equation $\sqrt{2x^2+1}=\sqrt{2}(1-x)$. This gives $x=\frac{1}{4}$. Now the side length of the square is just the hypotenuses of that "fat" triangular prism at the top and bottom, that has a side length of $1-x$. This gives $\frac{3\sqrt{2}}{4}$

Ted Shifrin

hypotenuses? hypoteneese?

leslie townes

19:48

honk

Ted Shifrin

(I have not even looked at the question.)

leslie has ducks/geese on his brain. Where's munchkin?

Semiclassical

is there a shorter way to say: need to pick basis vectors such that they're orthogonal to each other and give cross product in +z direction

user 858770

For beginners more words are better

PM 2Ring

@冥王Hades Well done. Note that $\frac{3\sqrt2}4=\sqrt{\frac{18}{16}}>1$

冥王 Hades

@PM2Ring yeah, the new cube seems to be larger than the unit cube. That's why I was double checking my answer to see if I screwed up

Turns out its correct

leslie townes

19:54

semi: probably not. i'd use this amount of words (or more) to emphasize that the basis sought is an ordered list, not just a list or set

PM 2Ring

Indeed! But it's still a bit mind-blowing that the red cube is bigger than the yellow & blue cube in chat.stackexchange.com/transcript/message/63559770#63559770

Ted Shifrin

@Semiclassic Your sentence is unnecessarily unclear. You want an orthogonal basis for the $xy$-plane?

Ah, you want a positive ordered basis.

Semiclassical

right. the orientation matters

shintuku

hypoteni

user 858770

Then tell them that.

Semiclassical

19:56

(it's in the context of plane waves, so need to have $\vec{E}\times \vec{B}\propto \vec{S}\propto +\hat{z}$

leslie townes

semi: even some symbols might help here, e.g. "choose an orthogonal basis u_1, u_2 for the plane such that u_1 x u_2 has positive z-coordinate"

user 858770

^

Semiclassical

ehhhh

leslie townes

i might even be more specific about 'the plane,' i mean there is more than one plane in R^3, unless this is somehow clear from other context

冥王 Hades

@PM2Ring the animation helped me realize that the sides of the red cube are perpendicular to the "fat" triangular prisms. That's what motivated my solution by the Pythagorean theorem

leslie townes

19:57

"when is something in R^2 also in R^3" is a bit of a stumbling block for some people, particularly if you use e.g. i, j, k notation where something with zero z coordinate might not be written with any express indication that it has a z coordinate

user 858770

As all good geometry reduces to.

Semiclassical

the problem was posed in a kinda annoying way tbh

they were in essence told that $\vec{E}\times \vec{B} = A \sin(kz-\omega t)\hat{k}$

leslie townes

if you think it's annoying, imagine what it's like going through life left handed (or maybe you don't have to)

user 858770

Ok lefty.

robjohn

@PM2Ring To check a uniform point-in-sphere generator, I made an animated GIF of two views of random points in a rotating sphere. Crossing eyes to get the 3-D view it looks great. Unfortunately, the image is 11 MB.

leslie townes

20:01

at least i can use the right hand rule without dropping my pen

Semiclassical

with a given value of $A$. from plane wave theory you know that you need to have $|\vec{E}|=E_m \sin(kz-\omega t)$ and $|\vec{B}|=B_m\sin(kz-\omega t)$ with $E_m/B_m=c$

user 858770

@leslietownes Coolio

Semiclassical

that's enough information to pin down the values of $E_m,B_m$

but the actual orientations of the fields aren't prescribed, aside from them being orthogonal and reproducing the right cross product

the problem put it as "find possible expressions for the E and B fields"

which...bleh

user 858770

Let's not get into a left-brain, right-brain argument.

robjohn

@leslietownes are you left handed?

user 858770

20:04

Munchkin is deciding.

leslie townes

rob: yes

whenever i make sign errors, that's why, btw. it's really the world that is making a sign error

user 858770

You put the sign on the right?

3- not -3

tough life

;-)

PM 2Ring

@robjohn Oh well. You could probably make a much smaller version using three.js, and it'd be interactive. But you'd need a site to host it on. The three.js library has a stereo camera, but it's wide-eyed, not cross-eyed. But I guess it wouldn't be too hard to hack it to swap the views.

Here's a Bucky ball I raytraced in POV-Ray a decade ago, in cross-eyed stereo.

PM 2Ring

20:25

sagecell.sagemath.org/…

^ Interactive 3D Moon anim, wide eyed

D Left Adjoint to U

@PM2Ring that's amazing

that it took only 50 lines of code to render that

on the user code side, minusing all the libs

leslie townes

dippin dots truly are the ice cream of the future

D Left Adjoint to U

The app code I mean

@PM2Ring want to do a screenshare / share our projects with one another?

PM 2Ring

Well, there's a lot more code hiding in the three.js module. But yeah, it is pretty amazing. :) And it's pretty amazing that a modern browser can do those transforms in real-time, in JavaScript.

D Left Adjoint to U

That's just a few matrix multiplies, but I thought GL would do those

Thanks for bringing these 3D moons to our attention

LOL

PM 2Ring

20:37

It has to do the non-linear transform from the
equirectangular texture map to the sphere, as well as the rotations. But yeah, GL is doing the heavy lifting. I expect it'd be a bit slower if all the arithmetic was done by the JS engine.

D Left Adjoint to U

What else you working on?

Is this 3D stuff part of some bigger project?

shintuku

it is part of the unrelenting march of progress

D Left Adjoint to U

@shintuku hey mon

are you studying advanced algebra stuff ?

PM 2Ring

@DLeftAdjointtoU I only know a tiny bit of three.js. I use it a lot, but mostly via Sage. Sage handles the messy details, but it only exposes a tiny fraction of the full power of three.js threejs.org

shintuku

yes

D Left Adjoint to U

20:40

@shintuku what are you studying?

shintuku

localization of rings

D Left Adjoint to U

Nice. I know about that a little

I studied it from a CA book

It then relates to everything - you to prove what categorical operations it commutes with

shintuku

i need to prove that the kernel of a homomorphism to an integral domain is a prime ideal

leslie townes

well, it's an ideal because it's the kernel of something, and if f(pq) = 0 then . . .

shintuku

i think i can do it by showing the image of that homomorphism is isomorphic to $R\setminus \ker \phi$

Ted Shifrin

20:49

Oh boy.

D Left Adjoint to U

So you take the kernel P. Note that P is prime if and only if R/P is an integral domain. Define f* : R/P -> S by taking f(x + P) = f(x). Then since if x + P = y + P we have x - y in P or f(x - y) = 0 by definition of P the kernel of f. So f(x) = f(y). Now f(P) is a subring of S and also an integral domain. Done

@shintuku

Ted Shifrin

You mean quotient or setminus?

PM 2Ring

@DLeftAdjointtoU I've been fascinated by stereo images since I was a kid. I've done a lot of stereo ray-tracings, but not many of them are online. Here's a non-interactive ray-traced stereo anim that's too big for Imgur.

See hosting.photobucket.com/albums/y43/PM2Ring/…

shintuku

@TedShifrin quotient ring

@DLeftAdjointtoU woah let me read through that slowly

D Left Adjoint to U

@shintuku can I teach you my proof?

It's actually there, just not explained well

shintuku

20:50

sure just give me a sec to read things properly

D Left Adjoint to U

@PM2Ring can I come to you later when we wan to add 3D stuff to your Python site?

*our

@shintuku the third sentence starts the proof of well-definedness

we showed that R/P

$\approx S'$ the image $S \supset S' = f(P)$ by first isomorphism theorem for rings

Since the image $f(P)$ is an integral domain we have that $R/P$ is by isomorphism

I think when you take a subring, it automatically inherits integral domainness b/c the definitino of it is by elements

Note that f(P) need not be an ideal, which is why we took the subring in the first place

it is an ideal if f is onto

Any time you take a quotient, you've got to prove well-definedness unless the quotient is output by some already proven theorem

That is like a math law

got to do it

@shintuku make sense yet? If not, let's start a room :)

robjohn

@PM2Ring I swapped the images to make it cross-eyed

shintuku

@DLeftAdjointtoU why are you giving an explicit $f$ at all if you're citing the first isomorphism theorem?

nvm, that's the same mapping as the one in that theorem

yeah i get it

but you can skip the proof of well-definedness if you're citing the first isomorphism theorem no?

1. $\text{im} \ \phi \backsimeq R\setminus \ker \phi$ (First Isomorphism Theorem)
2. $R\setminus \ker \phi$ is an integral domain if and only if $\ker \phi$ is a prime ideal.
3. Therefore $\text{im} \ \phi$ integral if and only if $\ker \phi$ prime.

D Left Adjoint to U

21:10

Not sure! I'd have to read and understand your proof lol

PM 2Ring

@DLeftAdjointtoU Sure, but depending on what you need I may not be able to offer much assistance.

D Left Adjoint to U

I got to do something though

@PM2Ring I will remember you!

PM 2Ring

@robjohn Cool!

D Left Adjoint to U

Well, definedness is needed

because the first isom assumes you already are given a morphism

it's not a morphism if it's not well-defined

@shintuku

In answer to your query :)

shintuku

hm there are pairs of rings for which no homomorphisms exist (edit: *unital homomorphisms)

PM 2Ring

21:13

Sage lets you read the three.js file it creates, so it's possible to modify that file, but I haven't experimented much with that.

shintuku

21:30

@DLeftAdjointtoU i think you only need the existence of one unital homomorphism for the rings in order to use the first isomorphism theorem, and if they're abstract rings you can't prove that existence anyhow so you can just proceed without an explicit homomorphism

PM 2Ring

@robjohn Here's a "random points in a sphere" thing I did late last year. It uses Poisson Disc Sampling "a technique for randomly picking tightly-packed points such that they maintain a minimum user-specified distance"

sagecell.sagemath.org/…

robjohn

@PM2Ring so it seems to place points so that there are no clumps or voids.

The random points in a sphere is interesting to see where the clumps form

PM 2Ring

By default, it projects the points to the surface of the sphere. Toggle the hollow checkbox to off to see the points distributed through the ball.

@robjohn Exactly.

robjohn

I'm going to see if I can make a smaller version of the points in a sphere that will be under 2 MB. It may not be as nice.

shintuku

ted, suppose there exists at least one ring homomorphism $\phi: R \to R'$. does the following ring true?

25 mins ago, by shintuku

1. $\text{im} \ \phi \backsimeq R\setminus \ker \phi$ (First Isomorphism Theorem)
2. $R\setminus \ker \phi$ is an integral domain if and only if $\ker \phi$ is a prime ideal.
3. Therefore $\text{im} \ \phi$ integral if and only if $\ker \phi$ prime.

PM 2Ring

21:35

I figured it would be a nice distribution for pseudo-random starfields. Now I need a vaguely realistic distribution for the stars' luminosity, and colour...

@robjohn Several decades ago, Clifford Pickover was generating 3D points using a uniform distribution from a well-known PRNG (pseudo-random number generator) and saw that the points were clumping in a way that showed that the numbers were a lot more correlated than they should be.

Back in those days, there were a lot of crummy PRNGs in standard libraries. ;) A famous example is RANDU en.wikipedia.org/wiki/RANDU IIRC, that's the one that Cliff Pickover was using. But he's a hard guy to search online because he's so prolific, with a very broad range of interests.

robjohn

22:30

@PM2Ring I hope this is random

leslie townes

i've seen randomer

robjohn

I generated 3-D uniform points to test an answer that I can't post because the question is a PSQ. The question is the mean reciprocal distance of uniformly random points inside a sphere of radius $R$.

I've tried to get the OP to improve the question, but no luck.

冥王 Hades

I think I'm up early at 7:40 AM on a Saturday....not in Japan, everyone is up and about at least an hour before me.

There's a huge waiting list at my favorite breakfast place

robjohn

I'm at the park with my dog at 7:30 AM on Saturday

冥王 Hades

I'm done waiting.

I'd rather eat rocks

robjohn

23:09

@冥王Hades 8:09 there?

冥王 Hades

@robjohn 8:11AM yes

D Left Adjoint to U

23:42

@shintuku on your previous comment nope

The homomorphism f is given abstractly

but with respect to any such f

you can define the map from the quotient as $f^*(x + P) := f(x)$

Is it well defined? Yes because P contains ker f

it is ker f

Or I mean it's contained in ker f (the other dir)

HTH

Transcript for

$Mathematics$ Mathematics