Mathematics - 2023-08-20

robjohn

00:30

The rain's a-comin'

Ted Shifrin

00:55

I wonder how bad it'll be. So far San Diego is a panic scene (sorta like people in Georgia when an inch or two of snow are predicted ... go buy all the milk and bread within miles).

Can your dogs feel something in the air? My cat seems indifferent, so far.

Xander Henderson

01:14

I just got a text from my daughter's middle school.

Which informed me of a whole notification system that my ex-wife has never disclosed to me.

This is going to be a fun email... :(

Ted Shifrin

01:39

I can imagine.

Xander Henderson

Moral of the story: never marry a shiksa.

Shaun

Hi :) I thought I'd advertise a bounty:

2

Q: Show, using presentations, that $\Bbb Z_m\times\Bbb Z_n\cong\Bbb Z_{{\rm lcm}(m,n)}\times \Bbb Z_{\gcd(m,n)} .$

Note: This is an alternative-proof question and so is not a duplicate. The Question: Show, using group presentations, that $$\Bbb Z_m\times\Bbb Z_n\cong\Bbb Z_{{\rm lcm}(m,n)}\times \Bbb Z_{\gcd(m,n)}.$$ Motivation: I was trying to answer this question in particular . . . $\Bbb Z_m \times \Bbb ...

Ted Shifrin

02:00

@XanderHenderson I personally draw other morals :)

Xander Henderson

I mean, there are others to be drawn. But it is still good advice, I think.

robjohn

02:14

@TedShifrin Neither our cats, nor our dogs, nor our bunny seem to be acting weird.

leslie townes

02:41

my daughter is acting weird

Ted Shifrin

03:14

@leslietownes Weirder than usual?

Thomas Finley

05:17

I think we will have no problem even if f(x)<=g(x)<=h(x) at x=c

Isn't it?

Koro

what does it mean to say that a hyperboloid of 1 sheet contains a line in it?

It is not clear by looking at its graph.

Thomas Finley

Hey @Koro! How r u? Long time no see...where have you been? On vacation? Ha ha....

leslie townes

thomas: indeed, no problem if you additionally assume that the functions are defined at c and satisfy that inequality there (or indeed that they are defined at c but do not satisfy that inequality there). the likely reason for excluding c from the hypothesis is just that it isn't necessary to consider what happens there to get the result

Thomas Finley

@leslietownes K. Got it. Thanks!

leslie townes

koro: to say that a subset of R^3 contains a line has only one meaning that i can think of (i.e., the standard one, where "line" means the usual thing and "contains" means contains as a subset). is this not in R^3? if so, where is it? is it at least embedded in some R^n?

Ted Shifrin

05:29

@Koro It in fact contains two lines through each point!.

leslie townes

i certainly join you in not finding it obvious that hyperboloids of one sheet are doubly ruled

Ted Shifrin

06:12

Look at a saddle surface. There it is obvious. But they’re projectively equivalent :)

leslie townes

06:58

i was going to say that you've proved my point, but having heard that just once, it's going to be hard to forget it. i will classify it as 'obvious, once you know why it is obvious.'

in the spirit of "there's no trick to this, it's just this little trick"

Jakobian

09:28

@XanderHenderson just absorb the energy from the sun like those next stage vegan people

one potato two potato

10:43

math.stackexchange.com/q/4755763/668308 I actually asked the exact same question to one of the senior graduate students, but he didn't answer it. Only saying that it would be quite difficult to draw an actual picture.

Koro

12:41

@TedShifrin I managed to show that the geodesics on a sphere are great circles. I just had to solve the following: $ f''(t)+ (f'(t). (N\circ f)') N\circ f(t)=0$ N is the unit normal vector. From here the details shape up quickly and I get the desired form of the geodesic.

I tried doing the same thing for a cylinder, i.e., showing that a geodesic on a cylinder ($x_1^2+x_2^2=r^2, r>0)$ is a helix. (vertical lines and points too I think. But let's not worry about that) But it doesn't work. The details are as under:

Suppose that f is a geodesic on the cylinder. $N(f(t))= \frac{(f_1(t), f_2(t),0)}{r}$. Then, I get $f''(t)+ \frac{\color{red}{f_1'(t)^2+f_2'(t)^2}}{r^2}(f_1(t), f_2(t),0)=0$. The problem is that the red colored term is not constant.

apart from this, I have $f_1^2(t)+f_2^2(t)+f_3^2(t)=r^2$ and $f'_{1}^2(t)+f'_2^2(t)+f'_3^2(t)=$ constant.

$f_1'^2(t)+f_1'^2(t)+f_3'^2(t)=$ constant.

I found a solution online but it doesn't make sense. 3rd line from below: I think that it should be instead-maximal geodesic is uniquely determined by initial position and initial velocity.
Otherwise it's circlular reasoning.

@ThomasFinley Hi, I'm good, thanks. How about you?

sunny

13:00

Consider $\sum_{n=0}^\infty (-1)^n x^{2^n}$, then $$a_n=\begin{cases}(-1)^k & \text{ if }n=2^k \\ 0 & \text{ if }n\neq2^k. \end{cases}$$ I'm interested in computing $|a_n|^{1/n}$, but I see a problem when $n=0$. Then $a_0=0$, but what is $|a_0|^{1/0}$? What if $a_0=1$, what would then $|a_0|^{1/0}$ equal?

Xander Henderson

@sunny Why are you "interested in computing $|a_n|^{1/n}$?

What is the end goal?

sunny

@XanderHenderson to compute $|a_n|^{1/n}$ to obtain the radius of convergence :) recall (Cauchy-)Hadamard's formula

Xander Henderson

@sunny Yes. This is basically the root test. Does the root test care what $a_0$ is?

Indeed, does it care what the first $n$ terms are for any finite $n$?

The goal is to take some kind of limit (the $\limsup$, to be precise)...

sunny

True, no, I guess $a_0$ is not so important then.

geocalc33

i don't get where this is valid in that complex plane: $$i s \frac{\partial ^2f(s,x)}{\partial s^2}=x \frac{\partial f(s,x)}{\partial x}$$

I wanna say it's valid for a particular solution $f(s,x)$ wherever that solution is valid

So I conclude that it's valid at least on some subset of $\Bbb C.$

Koro

13:17

@Koro nvm, it's done now.

geocalc33

chat.stackexchange.com/transcript/message/64227123#64227123 you have to look at the Schrodinger wave equation

sunny

13:34

@XanderHenderson what would be $\limsup_{n\to\infty} |a_n|^{1/n}$ of the sequence $a_n$ I stated above? I guess I could define $|a_n|^{1/n}$ to be whatever when $n=0$. So $$|a_n|^{1/n}=\begin{cases}1 & \text{ if }n=2^k \\ 0 & \text{ if }n\neq2^k \text{ and } n\neq 0 \\ C & \text{ if } n=0, \end{cases}$$ where $C$ is some arbitrary constant. Then $\limsup_{n\to\infty} |a_n|^{1/n}=1$ in any event, which gives the correct radius of convergence of the series.

Xander Henderson

@sunny I mean, you could. I'm not sure that you would want to.

A sort of analysis-y way of thinking about it would be to say that $a_n^{1/0}$ (which is nonsense) is $\lim_{x\to 0^+} a_n^{1/x}$. This is $1$ if $a_n = 1$, $\infty$ if $a_n > 1$, $0$ if $a_n \in (-1,1)$, and oscillatory otherwise.

But, like, why?

Who cares?

it is not relevant to the convergence or divergence of the series.

You should get used to the idea that it is always possible to ignore any finite number of terms when you take a limit (including a limit superior or limit inferior).

sunny

true, the expression $|a_n|^{1/n}$ for $n=0$ just put me off, but as you say, we can ignore it or just define it to be whatever. Either way, it doesn't matter!

Xander Henderson

13:51

Skip the "we can...just define it to be whatever". This isn't really correct (you can't just define things to be "whatever"). The point is that you don't need to define it in order to answer the question you are trying to answer. Ignore it. Don't worry about it. It doesn't matter.

PM 2Ring

14:11

6

one potato two potato

14:37

O'RLY?

Xander Henderson

ya rly.

Thomas Finley

15:31

Is this a standard theorem/lemma ?

If so, where can I find the proof ?

@PM2Ring With all due respect, is this a meme? Or a "true" book?

The way it's posted, ig the former is the case...

Soumik Mukherjee

@ThomasFinley if $x$ is a non zero element of $W$ then think about the set ${\alpha x}$ where $\alpha \in F$

Thomas Finley

@Koro Yeah, I am good. Where have you been, recently? Earlier, you were active on MSE a lot! Hehe...

PM 2Ring

@ThomasFinley From en.wikipedia.org/wiki/O_RLY%3F#Parodies

> O’Reilly Media's book covers on programming and technology have been parodied online using the term O RLY?, first popularised by a meme generator by Ben Halpern.

one potato two potato

O'RLY?

Soumik Mukherjee

chat.stackexchange.com/transcript/message/64228101#64228101 $\{ \alpha x \}$

PM 2Ring

15:44

Another popular work by Leslie:

3

Sep 12, 2022 at 7:50, by PM 2Ring

Thomas Finley

@PM2Ring Oh! Really?

@PM2Ring where are u getting these pics?

@SoumikMukherjee I don't understand, what are you trying to imply?

The set alpha *x, may not be finite but also may not be infinite

Soumik Mukherjee

why not infinite?

The given field is infinite

Thomas Finley

@SoumikMukherjee but, it may happen, that c,d is a scalar and c * alpha=d * alpha... I mean why finite? Sorry, if I am missing something

Soumik Mukherjee

you mean $cx=dx$?

Thomas Finley

@SoumikMukherjee yes

Soumik Mukherjee

15:49

then that implies either $x=0$ or $c=d$

Thomas Finley

@SoumikMukherjee Will the inverse of x always exist? Otherwise, how is c=d ?

Soumik Mukherjee

what is inverse of $x$? $x$ is a vector

$(c-d)x=0$ implies either $c=d$ or $x=0$

Thomas Finley

@SoumikMukherjee I know I sound stupid! But recently I read about integral domains and zero divisisors in Ring Theory and my brain is goin haywire

But is that always the case in vector spaces ?

Soumik Mukherjee

@ThomasFinley what?

Thomas Finley

@SoumikMukherjee I mean, if ab=0 where a is a scalar and b is a vector or vice versa, then either a=0 or b=0 ?

PM 2Ring

15:55

@ThomasFinley There are several sites that generate them. You can see genuine O'Reilly covers and O'Rly covers on Google: google.com/search?q=O%27Reilly+Media+book+covers&tbm=isch

Soumik Mukherjee

@ThomasFinley yes, I think you also proved that in chat

Thomas Finley

@SoumikMukherjee when ?

I didn't get that...

@PM2Ring oh! Ha ha, funny

Soumik Mukherjee

@ThomasFinley Maybe it was a different question, like $0x=0$ using distributive property

Thomas Finley

@SoumikMukherjee Oh, yes. But that was, like if we multiply 0 by x or x by 0 we get 0. But here, isn't the thing we're discussing bout the reverse of it?

Soumik Mukherjee

yeah it is different

Thomas Finley

16:02

@SoumikMukherjee hmm...so I'm saying will that always hold?

I never came accross that strangely!

Soumik Mukherjee

but if $\alpha x=0$ and $\alpha$ is nonzero then you can multiply both sides by $\alpha^{-1}$ to get $x=0$

Xander Henderson

@ThomasFinley The scalars come from a field. Do fields contain zero divisors?

Thomas Finley

@SoumikMukherjee Oh! ok, so that proves it... But then returning back to the initial problem: We are considering that the set $\alpha'=c\alpha$ where c is a scalar and $\alpha\neq 0$ is a vector in a subspace S . Now, if all the $\alpha'$ ('s) are distinct we are done as then $S$ is infinite. But if, there exists $c,d$ such that, $c\alpha=d\alpha$ then, $c=d$ as $\alpha\neq 0$.

This means if, $c\neq d$ then, $c\alpha\neq d\alpha$ and since, the scalars come from an infinite field so, $S$ is infinite as well as each $\alpha'$ will be distinct for each distinct scalars. Did I get you?

@XanderHenderson Yeah, I got badly confused. Thanks! I think I get this.

@XanderHenderson A field never has a zero divisor.

Soumik Mukherjee

@ThomasFinley yes.

Xander Henderson

Oh, actually, it is even simpler than the non-existence of zero divisors. If $a\ne b$ are scalars and $u$ is a nonzero vector, then $$au - bu = (a-b) u \ne 0. $$ Hence $au \ne bu$. This implies that $|\{au : a\in \mathbb{F}\}| = |\mathbb{F}|$, where $\mathbb{F}$ is the underlying field.

An explicit bijection is given by $a \mapsto au$ (where $u\ne 0$).

Thomas Finley

16:17

@SoumikMukherjee thanks.

Xander Henderson

If $S$ is a subspace of $V$ and $u \ne 0$ is an element of $S$, then $S$ contains $\{au : a\in\mathbb{F}\}$. Thus $|S| \ge |\{au : a\in\mathbb{F}\}| = |\mathbb{F}|$.

Thomas Finley

@XanderHenderson yeah, it's indeed a good way to demonstrate the result. I have noted it. Thank you for showing the explicit proof! I learnt a good strategy.

geocalc33

16:34

6

Q: Involutive fourier transform

The writer here states I am introducing a viewpoint (the involutive convention) which makes the Fourier transform its own inverse (i.e., the Fourier transform so defined is an involution). If I am reading the notation correctly, the definition given is: $$F(f)(s) = \int_{-\infty}^{\inf...

calculus functional-analysis reference-request fourier-analysis integral-transforms

Koro

17:47

Let $S$ be an $n-$ surface in $\mathbb R^{n+1}$. Let $v\in S_p, p\in \mathbb R^{n+1}$ and $\alpha: I\to S$ be the maximal geodesic in $S$ with initial velocity. Show that the maximal geodesic $\beta$ with initial velocity $cv,c\in \mathbb R$ is given by $\beta(t)=\alpha(ct)$.

How to check maximality of $\beta$?

$\beta$ is a geodesic as $\ddot \beta(t)= c^2\alpha(ct)\perp S_{\alpha(ct)}$ as $\alpha$ is a geodesic.

not sure how to show that it is maximal too.

copper.hat

@geocalc33 was that a question?

Koro

here is a solution found online but it makes no sense.

nvm, I think I got it now.

geocalc33

18:10

@copper.hat I am just mulling over how the Fourier transform changes with the choice of metric. Trying to understand some things wrt to that.

Xander Henderson

@geocalc33 The Fourier transform is defined on an algebraic group with a natural Haar measure. I'm not sure that I see how a metric is involved, as the Fourier transform is not defined in terms of a metric.

geocalc33

@XanderHenderson Of course you are right.

Xander Henderson

I always am.

Except when I'm not.

Ted Shifrin

Riemannian metric leads to volume form (but is not required).

Xander Henderson

In Soviet Russia, volume forms YOU!

geocalc33

18:17

@TedShifrin thanks ted

@XanderHenderson ah okay this now makes a lot of sense. Working with Fourier over $(0,1)$ instead of $\Bbb R$ makes things complicated because you lose the algebraic group structure.

but when one door closes another one opens

copper.hat

Loosely I view Fourier as a change of basis.

Ted Shifrin

But you go to a different space!

Xander Henderson

@geocalc33 No. You generally regard $(0,1)$ as the torus, i.e. $\mathbb{R}/\mathbb{Z}$. The group operation is addition modulo $1$. The Fourier transform of a function on $(0,1)$ is a Fourier series (very roughly speaking).

Again, you should have a look at the concept of Pontryagin duality.

copper.hat

I am interpreting the word loosely very loosely.

geocalc33

18:32

@XanderHenderson okay I will look now and be happy

Ted Shifrin

That explains it.

Xander Henderson

@copper.hat Even loosey-goosey?

copper.hat

Only when plucked.

Ted Shifrin

Where are @leslie and Munchkin’s ducks/geese?

copper.hat

Apparently there is a storm in SoCal. My daughter drove to LA on Friday I think, hopefully the weather does not have too much impact.

Ted Shifrin

18:34

The impact is yet to arrive.

copper.hat

Sorry a storm warning.

So much news about it I presumed it much have arrived already.

Ted Shifrin

It seems to be shifting some , so it may be worse inland. Dunno yet.

But the strength has diminished.

copper.hat

The news is always a breathless doomsday when it comes to weather.

Ted Shifrin

Yeah. But consider the alternative in Hawaii .

copper.hat

Can't imagine how a Californian weather reported would survive in Ireland.

True.

Ted Shifrin

18:37

There is, however, a boy cried wolf effect eventually.

copper.hat

Apparently there is a systematic bias towards predicting worse weather from commercial weather reports than from the nws.

apparently people like when the weather is better than predicted. really. i am ashamed of the human race.

Besides, the world will end tomorrow anyway.

Xander Henderson

Aw... man.

The world's ending?

But that's where I keep all my stuff. :(

copper.hat

Just think, if it doesn't then you will feel sooo much better.

amWhy

@XanderHenderson There's a children's tale like that! It's been around a long time.

Ted Shifrin

Hi, amWhy!

Xander Henderson

18:50

It involves Chicken Licken, right?

Ted Shifrin

Not Henny Penny!

amWhy

@XanderHenderson Indeed!

@TedShifrin Who are you, anyhow? ;D

Xander Henderson

Chicken Licken is finger lickin' good!

For that chicken, the world really was ending. With breading and a deep fryer.

Ted Shifrin

Did they have deep fryers back then?

Xander Henderson

(with apologies to my vegetarian and vegan brethren and sistren)

amWhy

18:55

@XanderHenderson Reminds me of a bumper sticker of a chicken "speaking": "My worth can't me measured it how many chicken nuggets I make!"

Xander Henderson

@amWhy The bumper sticker is right. You also have to consider the wings, breasts, thighs, etc.

amWhy

@XanderHenderson boo!!!

Xander Henderson

Gotta go feed my hummingbirds again.

Ted Shifrin

Don’t they use cartilage to make “nuggets”?

amWhy

@TedShifrin The old way was heating oil in a very large dutch oven, over an open pit fire

Xander Henderson

18:59

For the record, I believe I said that I had gone through almost five pounds of sugar this summer. That was incorrect. The big ol' bag o' sugar I bought was a 25 lb bag.

For the last two weeks, they've been going through sugar water at a rate of about 1 lb of sugar every 36 hours.

Ted Shifrin

I’ve never bought larger than 10 lb. bags — when I lived in a house with plenty of kitchen storage.

Wow. The heat.

geocalc33

@XanderHenderson but isn't Fourier series a decomposoition of a periodic function as a linear combinatino of sines and cosines or complex exponentials?

Ted Shifrin

@amWhy What did they use for oil? That’s a lot of suet or tallow.

PM 2Ring

geocalc33

@XanderHenderson oh nevermind - I got it you are right

amWhy

19:03

@TedShifrin What they had: Whale oil, lard, etc. I'd like to see Gordon Ramsey in practice, back in the days of yore!

PM 2Ring

@copper.hat Regarding your angle convention question math.stackexchange.com/q/1504483 I suspect that your guess is correct, that it comes from astronomy. If you look "down" on the Solar System from the north, most motions are anti-clockwise. And when we draw planet orbits, we almost always draw the semi-major axis as horizontal, generally with the Sun at the origin and the perihelion on the +X axis.

Xander Henderson

@geocalc33 A function on the torus $T \simeq (0,1)$ is a periodic function. :P

copper.hat

@PM2Ring Navigation always fascinates me. Hence my curiosity re the difference in conventions.

Xander Henderson

$\DeclareMathOperator{\Hom}{Hom}$ I am very rusty at this, but if I recall correctly, the basic idea is that you first define the characters of a locally compact abelian group to be elements of $\Hom(G,T)$, where $T$ is the torus. That is, a character of a group is a homomorphism from that group to the torus. The collection of all characters ($\Hom(G,T)$) is itself a group. This group is the Pontryagin dual of $G$, usually denoted by $\hat{G}$.

The Fourier transform of a function on $G$ is a function on $\hat{G}$. So, for example, if $f$ is a periodic function on $\mathbb{R}$ (i.e. a function on the circle group $T$), then $\hat{f}$ is a function on the Pontryagin dual of $T$, which is the integers. So $\hat{f} : \mathbb{Z} \to \mathbb{C}$ eats an integer $n$, and spits out the coefficient of $\mathrm{e}^{Cinx}$ in the Fourier series expansion of $f$.

(Where $C$ is some constant which depends on the period of the function---I think that it is usually something like $2\pi / L$, where $L$ is the period).

PM 2Ring

19:27

@copper.hat I have this book, from 1964. It's not a big book, but it contains a lot of info. It doesn't go deeply into theory, and all the calculations use logarithms. :)

copper.hat

:-)

PM 2Ring

Despite its age, and shortcomings, it's still a great little reference book. The author assumes the reader knows basic trig, and is smart (and diligent) enough to be a navigator, but he doesn't expect that you have an advanced maths degree. He explains spherical stuff well, but is a bit sketchy on elliptical stuff. In those days, you'd use special tables to apply elliptical corrections to spherical calculations.

David Raveh

20:20

I find that the time I spend on MSE goes down when school starts (resumes tomorrow), since I can't spare the time. Does the traffic usually go up during schooltime, since people ask HW questions?

Xander Henderson

@DavidRaveh Traffic tends to peak around midterms and finals.

sunny

Let $f(x)=\frac{x}{e^x-1}=\sum_{n=0}^\infty B_n \frac{x^n}{n!}$, where $B_n$ are the Bernoulli numbers ($f(0)=1$). I have a hard time understanding why $B_n=0$ for odd $n\geq 3$. The argument goes like this; since $\frac{x}{e^x-1}+\frac12 x$ is even, the result follows. I can see why $\frac{x}{e^x-1}+\frac12 x$ is even, but how does this relate to $\sum_{n=0}^\infty B_n \frac{x^n}{n!}$?

e.g. see here for why $\frac{x}{e^x-1}+\frac12 x$ is even.

sunny

20:47

I think I might have found the answer. It follows probably from the fact that $B_1=-\frac12$. Then $\frac{x}{e^x-1}+\frac12 x=\sum_{n=0}^\infty C_n \frac{x^n}{n!}$, where $C_n=B_n$ for $n\neq 1$ and $C_1=0$.

And since the power series is even, all odd numbered coefficients vanish...

copper.hat

@PM2Ring A long time ago I had hoped to do a little project for a 3rd grade class I used to volunteer in, basically a small sundial with various angles marked out. The idea was to estimate latitude & longitude using a few measurements and a clock. I decided that it was to far beyond their attention span on we settled on a make-it-yourself slide rule.

Xander Henderson

21:06

My sister just sent this to me with the caption "Jesus math":

copper.hat

21:38

How Rood

I just assumed they meant BornBorn

Xander Henderson

Or maybe BBoorrnn?

Is this operation being defined on a commutative ring?

shintuku

which algebraic structure is the most divine?

copper.hat

Maybe its a rhetorical question as in Born to...

Not a fan of anyone telling me what to do or believe, even if implicit.

Going down a rat hole in the comments math.stackexchange.com/q/4756046/27978

I am one of Zorn's lemmings

David Raveh

Zorn's Lemma gives me nightmares sometimes

copper.hat

21:55

I just lose myself in a chain looking for a maximal element.

Xander Henderson

@copper.hat Just know that one exists. You don't need to see it.

copper.hat

I am a doubting Thomas

David Raveh

I believe AoC much more than ZL

Xander Henderson

Well, I mean, the Axiom of Choice is obvious. And the Well-Ordering Theorem is clearly false. As for Zorn's Lemma... who the hell knows?

David Raveh

Love that quote

Xander Henderson

22:04

4

Q: What's the point of maximal atlases?

If $M$ is topological space (lets say connected Hausdorff and second countable, because I don't think it matters particularly for the question) then we can give $M$ a smooth structure by specifying an open cover $\mathcal U = \{U_i : i \in I \}$ and homeomorphisms $\psi_i\colon U_i \to V_i \subs...

category-theory manifolds differential-topology smooth-manifolds

O_o

TL;DR

copper.hat

Yeah, I started reading Lee's manifold book and early on he gets into maximal atlases. I mean really?

Xander Henderson

My reaction to the question is to ask "What's the point of real numbers? Whenever we want to actually do any computations, we work with rational numbers, or algebraic numbers. Why do we need to know that the rationals have a metric completion if we never actually do anything with it?!"

geocalc33

I'm looking to define a function in the following fashion. Fix 9 "evenly spaced" green points on the real unit interval $I=(0,1)$. Let the points (where the y-coordinate is zero for all the points so I won't write it) be $S=[.1,.2,.3,...,.9].$ Now, randomly select 9 real numbers red in $I.$ Now this is how I want to develop the function. I want the function to count the number of red less than a given green. This will give a step function in $I^2$.

For example the first point in the function would be (.1, checks how many red are less than .1)

copper.hat

22:20

Yeah, but I think the maximal atlas stuff could be relegated to an appendix.

geocalc33

Hope that made sense^^

冥王 Hades

I’ve beaten COVID @TedShifrin

another victory for me!

geocalc33

And then I plan on making a fractal out of that method by letting $S$ go to 99 elements to 999 elements etc.

fractal step function.

Xander Henderson

@copper.hat *shrugs*

I "learned" out of Lee. I like the presentation

copper.hat

I needed something a bit more concrete first.

Xander Henderson

22:25

But I like to get my hands dirty with things like constructions of the real numbers, or of topologies from a basis. I guess I recognize that you don't need to do that nonsense.

But I like to start with that nonsense.

@geocalc33 Define "fractal". Though the kind of function you seem to be defining appears to be monotonically increasing, and monotonically increasing functions have Hausdorff dimension 1. Not fractal, so far as I am concerned.

copper.hat

i like to think of a fractal as a fixed point of some set valued mapping.

geocalc33

@XanderHenderson you're right...again. Not fractal

But yeah monotonically increasing

Xander Henderson

@copper.hat Okay. But most sets can be expressed as the fixed point of some set valued mapping...

copper.hat

matrix, i knew it.

Xander Henderson

E.g. the unit interval is the fixed point of the map $\Phi$, where $\Phi : \mathscr{P}(\mathbb{R}) \to \mathscr{P}(\mathbb{R})$ is defined to be $\varphi_1(A) \cup \varphi_2(A)$, with $\varphi_{1,2} : \mathbb{R} \to \mathbb{R}$ defined by $\varphi_1(x) = \frac{1}{2}x$ and $\varphi_2(x) = \frac{1}{2}x + \frac{1}{2}$.

copper.hat

22:33

restrict the set valued map to one induced by an ordinary function :-)

Xander Henderson

@copper.hat If that rules out the map I gave above, how do you deal with the Cantor set?

That's certainly a fractal, right?

copper.hat

you are right, i should stop here.

Xander Henderson

Just redefine $\varphi_1(x) = \frac{1}{3}x$ and $\varphi_2(x) = \frac{1}{3}x + \frac{2}{3}$. :P

copper.hat

perhaps the smallest fixed pint

Xander Henderson

@copper.hat That would be the emptyset.

copper.hat

22:35

i meant point, but now that i am thinking of pints

Xander Henderson

Or, if you insist on non-empty, then the Banach fixed point theorem tells us that there is a unique compact fixed "point" (the Cantor set is an example).

But that only applies to iterated function systems which are contractive.

copper.hat

i'm trying hard to be quiet

Xander Henderson

Heh.

Thanks to this bad answer, I found a typo in my thesis. :(

The second $0_{\mathbb{o}}$ should read $0_{\mathbb{k}}$. :'(

What is written there is NONSENSE!!!

冥王 Hades

Fried rice @XanderHenderson

Xander Henderson

Oh, joy. I'm being memed.

Thank goodness I have absolutely no idea what it means.

冥王 Hades

22:47

You aren’t. I’m just showing you fried rice

MathCrackExchange

23:06

Homology of functions summing over square-free numbers maybe?

0

Q: A family of polynomials $f_n \in \Bbb{Z}[X]$ whose solutions $x$ lying in $\{p_n + 2, \dots, p_{n+1}^2 -2\}$ are precisely the twin prime averages.

Derivation $$ x \in I_n = [p_{n} + 2, p_{n+1}^2 -2] $$ is not a solution to $X^2 - 1 \pmod {p_i}$ for all $i = 1..n$ where $p_n =$ the $n$th prime, for any prime $p_1, \dots, p_n \iff x$ is a twin prime average in $I_n$. That means we have a system of logically AND'd equations: $$ g_i(X) = X^2(X...

abstract-algebra homology-cohomology exact-sequence prime-twins monoid-ring

I have a chain complex at the end

copper.hat

when i see twin primes i think room temperature superconductors.

Xander Henderson

@copper.hat Weird. I think cold fusion.

Like, how you gonna fuse a pair of twin primes when it's cold out?

That looks wrong...

Hrm... that looks wrong, too...

Google says it's correct, now. Imma just hafta trust the Google.

冥王 Hades

Funny how I’m acting all nice and calm in this place while engaged in a fierce debate in another forum

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