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7:00 PM
And I was wondering what transformations would preserve that structure.
 
I have no idea yet what you're talking about, @Semiclassic. Classically, incidence refers to points contained in lines, or lines contained in planes, etc. What are you talking about?
 
@iwriteonbananas I forget all this framing stuff. The point, informally, is that the orientation of the boundary of M x I is different on the two sides.
 
That's what I told him
 
Well, this is the picture that I showed earlier:
 
It's been months since Bananas has been here (or since I've seen him).
 
7:01 PM
Alright fuck I'll needa figure this out on my own
Ted I've been keeping a close eye on you
 
empty link @Semiclassic
As they say on main, I need context, @Semiclassic.
 
I installed some cameras in your house
 
should be right now
 
Sounds fun, bananas :P
 
And I refresh your FB page every 30 seconds
 
7:02 PM
Now, ignoring all of that except the lines (there's a lot of decoration in there that I don't care about)
 
But you're right, we havent talked in ages
Jeez
I've missed you
 
Where do these lines live, @Semiclassic? Are we talking points contained in lines? What's the ambient space?
Yeah, sure you have, bananas :D
 
Euclidean plane.
 
@TedShifrin ah right, i saw my flaw
 
So the obvious answer should be the affine linear group, which maps lines to lines, and will preserve the notion of points in lines.
 
7:03 PM
@Balarka Then why haven't you chosen a big goal to work towards and a bunch of small things on the way?
 
Good @Socrates :)
 
You're a fountain of wisdom, Ted.
 
That's the only way I can write. The big goal, obviously, is the paper. Slmetimes the small goals are "figure out what the norm map is", sometimes "write down the basic transversality argument", sometimes "get all that horrible notation into the page".
 
An old man.
 
7:04 PM
Sarcasm duly noted, bananas.
 
Young in spirit though
 
Yeah, I guess I shouldn't hope to go outside of that. I'd presumably preserve the crossing structure if I did a projective transformation, but they'd no longer be lines.
 
Bananas is never sarcastic.
 
@TedShifrin I think this can be shown nicely with a geometric representation
 
Hahaha
 
7:05 PM
In fact he has such a poor understanding of conversation and social mores he just says what he honestly thinks at every moment
 
Well, projective transformations won't stay in the Euclidean plane, @Semiclassic. If you enlarge my universe to $\Bbb P^2$, then, of course, projective transformations is what we want.
 
Mike knows what's up
 
@MikeMiller Ah, that's an interesting idea
 
Sure. But in that case lines are just circles as well, aren't they?
 
If he survived you for several days, he has more skills than most of us, @MikeM.
 
7:05 PM
I am a complete and utter social retard, I don't understand sarcasm
 
In the elliptic model of projective geometry, they're circles, @Semiclassic. Otherwise, they're lines.
 
hmm.
In any case, I don't have any reason to hope that what's being claimed in the paper to hold if I replace lines with circles. So affine linear group seems to be all that one can hope for.
 
@MikeMiller Oh god that looks horrible. Gotta check it out
 
But, as in the case of complex LFTs, some points (indeed a line) will map to infinity, so our universe isn't preserved if our universe is the Euclidean plane.
 
I still have no idea what I'm going to teach for topology.
 
7:08 PM
When is this? in an hour?
 
Two.
 
Yeah, you probably should decide soon.
Or you could do the lecture you did last year.
 
Like in two hours and ten minutes
 
I know you kept meticulous notes.
 
what did you teach the previous day
 
7:10 PM
He didn't @Balarka.
 
ok. then you obviously should start with casson handles
 
I feel unneeded :) None of Balarka, Alessandro, DogAteMy needs problems.
 
DogAteMy is working through your book, and Alessandro's going to do differential topology after 13th (I think?) so you'll have a lot on your plate soon
I don't know about me though.
I can bug you anytime you want
 
Bleh. I wish I felt awake enough right now to read this paper.
 
@TedShifrin can the equator be viewed as a open intervall [0,360[?
 
7:13 PM
@Ted What about me?
 
It's a circle, @Socrates.
 
@Balarka Pick your big goal.
 
I know nothing of any value for you, @MikeM.
I think it's healthy Balarka's taken a small break from math. When he misses it, he'll jump back in.
 
@MikeMiller Bring in a coffee mug and transform it into a doughnut for them
 
@MikeMiller I got one: learning foliations.
 
7:15 PM
Doesn't DogAteMy ever go to class? :D
 
There's the oral part of the numerical analysis exam on the 20th @Balarka. But I'm starting to feel prepared enough, I was thinking about reading some G&P tonight. (Also because I finished doing all of the exams from previous years so I don't know what more to do now)
 
@Ted I might talk about the "homotopy principle" (not Gromov's) - if you can do something for an arbitrary choice, and the something you can do is well-defined up to homotopy, and your choices are connected by homotopy, you have an invariant.
 
While you're at it, learn more differential forms and learn the Frobenius Theorem, @Balarka.
 
@Ted do you have a big book of cool problems you keep by your desk :P?
 
Which is the entire point of G&P.
 
7:15 PM
@Alessandro Give Ted a hard time then
 
@TedShifrin I'm having lunch. (Well, had — it just finished)
 
No, @Alessandro. :)
It's almost lunchtime for me, DogAteMy. That makes it halfway to your dinnertime! :)
 
Maybe I'll talk about the "generalized degree" of a map between arbitrary closed smooth manifolds - a homology class in $H_{\dim M - \dim N}$ - and why it's worthless.
 
@TedShifrin Got it.
 
Fair enough @MikeM.
 
7:16 PM
On some days, lunch is ridiculously late in my school
 
Why is it worthless?
 
I couldn't survive your school, DogAteMy.
 
Without the information about framing it is
 
Have you done the error estimates for numerical integration (trapezoid, Simpson) yet, @Alessandro?
 
7:18 PM
I think those are cool.
I usually taught that in my Spivak Calc with Theory course.
 
My favorite version of that is when you do numerical integration of analytic periodic functions.
 
@Balarka: When you review forms and learn Frobenius, we can do cool stuff (including why $K=0$ gives a surface isometric to $\Bbb R^2$, assuming simple connectivity).
Actually, that's easy for surfaces, no Frobenius needed. But we can do higher-dimensional interesting stuff.
 
Since in that case the error from the trapezoid rule goes down geometrically not algebraically.
 
Hmm @Semiclassic
Oh yeah, when @Alessandro gets a bit further, we need to talk about what Morse means.
 
hellow Ted !
:D
 
7:20 PM
@Semiclassical It's usually zero.
 
I should have known @Kasmir would show up.
 
haha no worries sir
 
Done studying? :)
 
Iam here only to ask how you are :D
not asking things
no :)
 
Nah. Simple counterexample: $\int_0^{2\pi}\sqrt{1-m\sin^2 x}\,dx$ for $0<m<1$.
 
7:21 PM
taking break :)
 
Wow. Nothing to ask? I'm shocked :)
Ah, good idea.
 
the 2 big topics in the course were systems of linear equations and ODEs, but we also had a bit of numerical integration and polynomial interpolation
 
:D
Tomorrow i do more old exams and repeat theory questions
 
I wish more American math majors learned that stuff, @Alessandro.
 
@TedShifrin Interesting
 
7:21 PM
then next time is exam
 
What's a quick proof of that?
 
Huh? @Balarka To what do you refer?
You feeling more confident, @Kasmir?
 
simply connected $K = 0$ surfaces
 
That's certainly analytic in the neighborhood of the real line, but it has branch points and so isn't analytic everywhere.
 
Ohhh, that.
 
7:22 PM
@TedShifrin Yes sir ! thanks to you :)
 
$H^1=0$, @Balarka + learning moving frames finally
 
...which means I'm being silly. I said something stronger than I meant to.
 
Oh well
 
Probably more thanks to you, @Kasmir, but that's great. :)
 
@Balarka So start writing down your small goals.
 
7:23 PM
I meant periodic functions which are analytic in the neighborhood of the real line.
 
you alllways got the right things to say :D
 
He has one. Section 3.3 of my notes @Balarka :D
 
Those have an error that dies like $c^{-n}$ for some $c>1$ when you do the trapezoid rule with $n$ subdivisions.
 
@TedShifrin no i'll fight with you
 
Now that I know you're a math major, maybe you might enjoy the challenge of some of my videos, @Kasmir :)
 
7:24 PM
Also I need to be there when you talk to Alessandro about your favorite definition of Morse, @Ted. Not very comfortable with it.
 
Fight, huh?
 
(I know what it is)
 
OK, we can all talk about it @Balarka, but that shouldn't stall you from reading section 3.3.
 
@TedShifrin they helped me on the first course this year about linear algebra =p but i will ofcouse watch them all when those things cames up or on summer just to learn more
 
@Balarka Is that what you care about (foliations), or is that just because I suggested it?
 
7:25 PM
As opposed to $n^{-2}$? Weird, @Semiclassic.
 
@MikeMiller I dunno. It sounds fun, and I don't have anything in particular that I care about for the moment.
 
If you're a math major, @Kasmir, thinking of the derivative as a linear map will be quite helpful. Somehow I thought you were more engineering, so I discouraged you.
 
I don't have any issues with learning it, besides my laziness
 
Proving simply connected + $K=0$ is isometric to a subset of the plane sounds like a good little project, @Balarka.
But if you want to be lazy for a while, I don't think that's bad.
3
 
there's transversality before (pretty soon actually) and you said you had stuff to tell me about it too if I remember correctly
 
7:27 PM
@TedShifrin am not major yet >< i will be that after 5 years of study , i did only 1 and half now =p
 
@Ted I agree, but my impression is he doesn't want to.
 
When I taught the course, @Alessandro, I stated all the transversality theorems and used them, and then proved them a month or so later. Just so students would see the beauty and power of using them first.
 
I'll keep it at the back of my head, @Ted. I agree being lazy is not bad, but the thing is I feel bad about it.
 
@Kasmir: In the US we refer to the subject you're studying seriously as the major. After a few years you'll earn a degree. :)
 
One way to understand it is via Euler-Maclaurin. In that case, you'd expect to get terms like $n^{-k}$. But the coefficients of those terms essentially measure to what extent the function (or its higher derivatives) must 'jump' in going from one endpoint to the other.
 
7:28 PM
@Balarka: Take a few weeks completely away from here and math. Read books, watch movies. It'll be good for you
 
Ah, it was Morse actually
 
So there ought to be some sort of Cauchy integral control, @Semiclassic. I've never pondered it.
Sure, I know, @Alessandro. And I haven't sent you the problem sets yet, either :P
 
@TedShifrin We are going to repeat that with more depth next years, we just hit the main points untill now but thats the plan as my teacher told me
 
@Ted You never say that to me!
 
You're not 15, @MikeM. At least I don't think you are.
 
7:29 PM
you sent them to me @Ted
 
so, to formalize the statement, f(x)=f(x+180) ?
 
Oh, OK. I get forgetful in my old age, @Alessandro.
 
But really, I agree breaks are good. But if you want to do something, you need to make an active effort to do it. And when you want to not do something, you need to make an active effort not to.
 
@Socrates, yes, for some $x$.
 
@Balarka Download vba and a ROM of Advance Wars.
I think you'll like it a lot.
 
7:30 PM
Yeah I am kinda stuck between doing something and not doing something
 
@MikeMiller Oh hey, I remember that game
 
The sequel is better but the first is a lot of fun.
 
I was not very good at it.
 
@BalarkaSen allways keep it simple and take small steps, that works for me
 
Anyhow, I'm gone for a while. You all have fun doing nothing!
 
7:31 PM
lol
 
haha thank you Ted ! see you :)
 
@MikeMiller Interesting
 
I think Ted should be nominated for greatest teacher :D
@semiclassical am still mad at you for downvoting and closing my questions -.-
 
Good for you.
 
meh that was a flop
 
7:33 PM
I had exam that week and you closed my Q's but thank god i passed
:(
 
@Balarka I laughed
 
Not going to apologize for voting to close questions that shouldn't remain open.
 
the joke's too old
 
yo what's everyone up to
 
Maths and stuff
 
7:34 PM
@KasmirKhaan Improve your question then. It gets added to the reopen queue the first edit after being closed.
You may find that the people closing questions are often the same people voting to reopen them.
 
I get that but i was stressed over exam
that is why i did not bother wasting an hour typing on latex my attempt
but eh ._. that is past now
we focus on future what we can change
 
@Balarka I still suggest you watch "The Lobster" by Lanthimos if you don't know what to do, it's a great movie
 
the lobster is WEIRD
it's Weird
 
is it tho
 
@AlessandroCodenotti Hmm ok.
 
7:43 PM
it's just straight uncomfortableness for its whole duration
 
8:08 PM
how is this not already answered? really that hard? math.stackexchange.com/questions/2025264/…
 
It was answered on MO and they linked the question there in the comments
 
two people, Alice and Bob move on a line. Alice starts at Y at t=1, and reaches X at t=2. Bob starts at X at t=0, and reaches Y at t=3. Now, how can we reason that they meet each other on the way?
 
IVT
 
My attempt is to give them distance functions
 
On the distance between them
Well not exactly
More like $P_{Alice}-P_{Bob}$
 
8:20 PM
distance from the point X
 
Yeah for instance
 
and then, construct a distance function between both, and show that it is positive at the beginning and "negative" at the end, but that makes no sense i guess
I want to avoid assuming HOW they move
 
You can't call it a distance because it's negative, but you got the idea
Any segment is isomorphic to $[0,1]$
 
altho, does there exist some n-dimensional way, where they don't meet? maybe möbius strip? :-D
 
Meh :p
 
8:25 PM
if the point X is on the equator, and both start at antipodal points, then one could travel in the opposite direction and never meet each other
Y too obv
 
But that's not a line
 
well, the original excercise was a bycicle way, ah well, tough to ride over the seas^^
it's fun to lead it ad absurdum.
they drive in a roundabout for 3 hours, i suppose
 
Hello, i have an excersice for my school about showing a problem belongs to NP-Complete class. Would it be possible to write the problem and give me some hints if you can, at least to say to me if i am at the right direction? Thanks.
 
Read guidelines on the top right
But sure
 
@wdika I think it could be good if you question it on the mainsite. but shoot it
 
8:32 PM
Okay here is the problem. Given a undirected graph G(V,E) and a subset of the edges, named F (F subset of E), is there a simple cycle at G that comes from every edge of F? Show that this problem belongs to NP Complete
 
Checking an answer can be done in polynomial time
 
The reason i am not asking it as a question at the mainsite is because i am sure i would be downvoted many times. My knowdledges are verry little at computation theory and my professor doesn't explain anything. I am hardly trying to understand them, i just searching all the time and i am not sure where i should start. I am sure it would be something close to CLIQUE problem and HAMILTONIAN cycle.
 
Why do you think you would get downvoted ?
 
Because i am actually looking for a big part of the solution, as i said i don't know even where to start, so i can't provide what i have done so far. That's why. Maybe i am wrong.
 
You can say that you don't know where to start, write what you told us (About Clique and Hamiltonian), and you should be fine
Just be sure not to throw the exercise with no other form of context
Write where this exercise comes from too
 
8:38 PM
I would give a try if i don't find any other solution, i am trying to understand it so far. But anyway thanks for the hints.
 
And getting downvoted isn't such a big deal after all
 
if i have a 2d space filled with interactionless dust moving randomly from infinity, and i put a polygon in the space to catch the dust, the amount of dust i catch should be equal to the surface area of its convex hull, right?
 
Why would that be ? @zounds
 
i dunno that's my intuition and im trying to see if it's so
 
Why not the whole surface ?
 
8:43 PM
if it have pacman in there, anything that goes into its mouth would be caught eventually, so i can just wall of its mouth
 
I'm not sure I follow
 
8:58 PM
pro tip for philosophics: add to any question "true". Like "is this a true cheeseburger?"
 
Is this a true chatroom?
 
Is it true that i don't know the solution to my problem?
 
Hi @Akiva
 
Hi. I actually need to leave in 0.2 seconds.
Bye!
 
well that was quick! Bye
 
9:01 PM
Statements like that are fishy.
Because the timeframe of typing the message is around the timeframe of the time stated; it becomes very important at what moment those $0.2$ seconds started
 
"very important"
 
ah, simultaneity
 
If something can be proved wrong if it is false, can it necessarily be demonstrated if it is true ?
 
makes these things all the more interesting :)
Depends on the something, perhaps. For instance, if I were to claim that 27 is prime, I can easily be disproven by noting that 3|27. But no such counterexample is possible for 29, so 29 is prime.
But I'm not sure that's quite the same as what you said.
 
Let's say I have a property $P$ I claim true for all natural. If my statement is wrong, you can prove it's wrong by giving $n\in \Bbb N$ such that we don't have $P(n)$ after trying a few. However does that mean you can demonstrate the property if it is true ?
 
9:13 PM
if it is never true to begin with, then yes, it can be shown if it is true
 
For a concrete example: Based on heuristics, it's plausible that Goldbach's conjecture is true. Suppose it is. Must there be some proof of that fact?
 
because ex falso quod libet
 
It's been a while since I played tourist in logic classes.
 
Can there exist true statements about the natural numbers which are nevertheless unprovable?
 
That's my question
 
9:14 PM
But this all goes back to Gödel's incompleteness theorems
 
I think the answer is yes, because Godel.
 
Gödel greets ya @Astyx, i guess??
 
True but unprovable results exists.
 
@Semiclassical >:|
 
(well, probably)
 
9:14 PM
Among naturals ?
 
actually, not just probably
what do you mean with "among naturals"?
 
Yeah that's Gödel's theorem
 
there exists an integer, such that $\frac{1}{sin(n)}$ is undefined.(wrong)
 
Hmm, this is an interesting answer: math.stackexchange.com/a/1052384/137524
 
Like, can the result concern a countable set ?
 
9:16 PM
No, @Socrates: Nonzero multiples of $\pi$ can never be rational, let alone integers.
 
^ Beat me to it.
 
Sorry ..
 
np. On the other hand, you can certainly have integer n for which sin(n) is arbitrarily close to zero.
 
0 is a rational and an integer
 
Third line of wikipedia's article on Gödel's incompleteness theorem states that "The first incompleteness theorem states that no consistent system of axioms whose theorems can be listed by an effective procedure (i.e., an algorithm) is capable of proving all truths about the arithmetic of the natural numbers."
 
9:17 PM
bah
 
@Semi Thanks, this answers my question
 
BTW, the answer I linked above goes to the point re: true but unprovable statements
 
Is there any beautiful justification possible, saying that $\sum_{k=2}^{\infty} \frac{z^{k-1}}{k!} \to 0$ for $z \to 0$, or is that obvious? Thanks.
 
Oh, @Socrates, so $n=0$ makes your statement correct. facepalm
 
@TedShifrin but your answer is still correct, because 0 was not in my mind
like, "correct" as in helpful :d
 
9:19 PM
@Kirill This converges uniformely on $[-1, 1]$ if I'm not mistaken, so it's continuous
 
@Kirill: Note that the sum starts at $k=2$ so the value at $z=0$ is $0$. And the power series defines a continuous function (in fact, infinitely differentiable function).
 
@Kirill can't you rewrite that sum using the exponential?
 
Thus it goes to 0
 
It converges uniformly on any compact set in $\Bbb R$, @Astyx. sticks out tongue
 
Yeah, but that's irrelevant
:p
 
9:19 PM
Though I think I like this short answer from there best:
"There is no natural number, $n$, such that $n$'s interpretation as an ASCII string is a proof of this statement."
 
@SteamyRoot I have just made this from the exponential :)
 
Nice indeed
 
Not that I entirely know why that works, but as a short statement it's nice :P
 
@Astyx yes, but I am not sure that $\to 0$.
 
I see. Don't understand why you'd want to do that, but okay.
 
9:21 PM
Is there a chatroom here where a combinatorics question would be on topic?
 
Here is fine.
 
You just have to compute it at 0, since it's continuous it answers your question @Kirill
 
No guarantee anyone will respond, of course, but that's always true.
 
Okay, thanks. I'm looking for a reference/monograph on vector partitions
the results of my internet searches turned out to be rather sparse
 
9:23 PM
Close enough.
 
On a more serious note, I personnaly have no idea
 
@TedShifrin Hello. So should I show that this one is differentable or is this also obvious?
 
@Kirill Not sure if one can easily make a proof out of this, but that sum is a concave-up function for all $x$.
 
I don't know what you're trying to do, @Kirill. But it's a well-known fact about power series: They are infinitely differentiable on any open interval where they make sense (i.e., converge).
 
Hrmf. The "doctoral working group" at my campus is offering to review posters before the poster session in february.
 
9:26 PM
That sum is $$e^z - 1\over z$$ unless I'm mistaken
 
But nobody on that working group has a strong mathematical background, so I'm wondering if I should send in a poster filled with Mathsgen nonsense to troll them...
 
That would have a constant term, @astyx. The one above doesn't.
 
What do you mean ?
 
No, @Astyx. The sum vanishes at $0$. Yours does not.
 
9:28 PM
Oops, @Semiclassic beat me. I guess we're even :P
 
true, but you said it simpler
It should be $\dfrac{e^z-z-1}{z}$.
 
Yeah, I'm just bad
 
@SteamyRoot I just found about about mathgen from your comment. It's magnificent!
 
@Astyx yes, it is. I split 2 and $z$ from it and have $1 + \texttt{the series above}$. I have to show that $\frac{e^z -1}{z}$ converges to 0 for $z \to 0$. Tha is why I am asking myself - is that ok, if I say that the series converges to 0, or should I prove it additionally?
 
9:30 PM
if someone wants to debunk a counterexample, go ahead math.stackexchange.com/questions/2092433/…
 
@Kirill It does not, it goes to 1, as Ted and Semi pointed out
 
@Socrates I don't feel like there's a counterexample to debunk when you interpret the assignment in a clearly unintended fashion.
 
$\dfrac{e^z-1}{z}=\sum_{k=1}^\infty z^{k-1}/k!$, not $\sum_{k=2}^\infty$.
 
You have to substract 1
 
It remains true, though (as it must---all we're talking about is just an extra $1$ overall)
 
9:33 PM
$$\lim_{z\to 0} = {e^z-1\over z}$$ is the derivative of $exp$ at 0, ie $e^0 = 1$
 
@SteamyRoot lol! But I must say this is really well done. I guess for someone with no maths background this is indistinguishable from a real maths paper. That you can actually get published with it is another story...
 
I'm trying to see if there's an obvious squeeze theorem approach to this. I presume there is, but I'm not seeing one :/
 
@Semiclassical $\displaystyle \frac{e^z - 1}{z} = \frac{z \cdot \sum_{k=1}^{\infty} \frac{z^{k-1}}{k!}}{z} = 1 + \sum_{k=2}^{\infty} \frac{z^{k-1}}{k!}$
 
@SteamyRoot omg, this stuff is hilarious (i mean the random false ones)
 
@Kirill Yes.
 
9:39 PM
One of the funnier things I've seen in mathsgen is probably a multi-line inequality going like $\lambda \leq a \geq b \leq c$.
 
eugh. that still something meaningful, but it just makes my eyes hurt.
 
$\lim$ from this one were $\lim 1 + \lim \texttt{series}$
 
Yeah. So proving the limit for one proves it for the other. Hence either form is fine.
 
@Semiclassical I have not understand You.
@Semiclassical should I prove the limit for 1?
 
I'm saying that if prove the limit as $z\to 0$ for $\sum_{k=1}^\infty \frac{z^{k-1}}{k!}=\frac{e^z-1}{z}$, then you also know it for $\sum_{k=2}^\infty \frac{z^{k-1}}{k!}$ (it only differs by 1). So if it proves simpler to do the series starting from k=1, then you can do that.
 
9:46 PM
@Semiclassical I have to show that it is 1, as @Astyx wrote before. I decided to split 1 and am trying to prove that the series converges to 0. My original question was if I should?
 
You can
Wether you should is another question
 
^
I am not staking an opinion as to whether that version is simpler to handle.
 
can you devide a circle's circomference into equal parts, which are alternatingly painted black/white, such that there exist antipodal points that share the same color?
 
@Astyx :) that is an exercise for the correction and should be done perfectly. So, if I've understood it right, I could better show it.
 
Basing on your knowledge of $z\mapsto e^z$ seems more "clever" to me because you don't have do the demonstration again
But it's a good exercise to still do it
@Socrates No
 
9:50 PM
White-Black-White-Black in four equal lengths?
 
thank you for the feedback!
 
Oh that share the same color
Then yes, always
 
and that share not the same color?
 
Antipodal points always share the same color with your configuration
So no
@Kirill My pleasure
 
@Astyx What about half white, half black?
In that case all antipodes have opposite colors.
 
9:52 PM
True
It's late
 
(Silly excuse, I know)
 
I don't really get the point of that question, for that reason. You can get either situation rather simply.
 
in how many parts do we have to devide it? a power of 2?
 
No.
Doing it in 12 parts would again give a same-color pairing.
 
9:54 PM
$2(2n+1)$ or $2(2n)$ gives you the different solutions
 
i mean, otherwise black might hit black at the start
or vice versa
 
$2(2n+1)$ is same colors, $4n$ different ones, unless I'm saying complete nonsense
 
That's what I get too.
Wait.
No, 4n gives the same colors. It's 4n+2 that gives opposite colors.
4(0)+2=2 being the operative example in the latter.
 
Indeed
I should stop now before saying anything stupid again
 
a function about the temperature around a circle has the domain [0,360[?
in degrees
or how is a "cyclic" domain expressed?
 
9:59 PM
For instance
You could use $\Bbb R/360\Bbb Z$
 

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