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5:12 AM
Hey I've got a question about Buddhabrots.
So wikipedia and a few other sources say that the way to calculate a Buddhabrot is to take points(c values) that aren't in the Mandlebrot set (at least up to a certain number of iterations through z^2 + c) and trace their paths to escape, since a point(c value) that escapes is by definition not in the Mandlebrot set. My question is this, how do people render Buddhabrots that show activity within the area associated with the Mandlebrot set since any value that is part of the Buddhabrot set is, again, by definition not part of the Mandlebrot set?
I'm curious because I've been trying to render a Buddhabrot from instructions on articles I find but when I try to render points that are explicitly not in the Mandlebrot set and trace their paths to escape, it produces a negative of the Mandlebrot set... which makes sense! So why do Buddhabrots produce renderings that are significantly different than the Mandlebrot set?
 
6:04 AM
@Michael Hardy : A small edit to your answered here: https://math.stackexchange.com/questions/148177/what-is-the-difference-between-⊢-and-⊨?rq=1
Change "proof-checking algorithm" to "provability-checking algorithm" . Since we are after an algorithm that checks the existence of a proof rather than checking whether a proof is correct or not.
Excellent answer by the way (plus easy to understand). I bookmarked it.
 
6:19 AM
If A is acute, how do I find which is greater- sin A or cos A? I think it should be $sin A>=cos A$ because sin's value increases from 0 to 90 while that of cos decreases. Research effort: 2 textbooks don't have any answer. Googled too but couldn't find the answer to my question. Used calculator and found that sin 35 is less than cos 35. Can someone account for that also?
I think I should ask that on Math.SE
 
@Abcd below 45, cos A is greater; above 45, sin A is greater
desmos is a great tool
 
@LeakyNun How?
Ok, let me check
@LeakyNun How to use it? desmos.com/calculator
 
@LeakyNun 1 for degrees please
Is it not possible to draw a graph with values in degrees?
 
just choose degrees in settings
 
6:27 AM
ok, thanks
 
Hey guys
 
6:44 AM
hi
 
hello!
 
Ah, you've finally rejoined us!
 
i exist occasionally
 
Wao reax only
 
oh hey araske
what did i miss in the chatlog
i can't keep up with the transcript [travolta.jpg]
 
6:50 AM
.jépg
 
JPИSKY
I am a Russian, not a French
 
Interesting. For my case, the actual dream changing is not necessary obvious to me because of how time travel can appear in my dreams and thus the narrative of the dream can mask that, but something like these do happen

Yeah, I heard people said the more you participate in dream recall, the better one remembers a dream

I also heard that the style of dreaming of each individual may be unique, thus expect that your dreams might be different form everyone else
 
@BalarkaSen Vodka for you, comrade
Hey @Alessandro!
 
I am pretty certain I found that top secret discord group, but for some reasonI cannot join
 
7:05 AM
That would suggest that it's pretty top secret
 
Meanwhile somewhere in the other side of the world there's this massive chat group
Erdős Math Group
 
Hi chat
 
Hai @Astyx
 
What's up ?
 
7:22 AM
Hey @Astyx
 
@Secret It's a rather private discussion room between a few of us, sort of like the number theory room (which is gallery; nobody other than the selected few who are involved in the study group can join), except mostly not math. I very much doubt that you found it.
 
@Astyx back: not too much, how about you?
 
Enjoying my freedom by not doind anything
So not much either
 
I should be doing... Riemannian geometry!
 
I should be rereading the beginning of Diff Top
 
7:26 AM
re-re-ading
jfc why are you doing stuff 3 times??
 
Anyway, one thing that suprised me the msot is that discord is only 2 years old
I thought it has existed for some time already
 
@Daminark I like to ad things
 
So it seems
 
7:56 AM
Well, to be REALLY picky:
yesterday, by Astyx
I've never been redundant nor have I ever said the same thing twice in a row
-> Hi chat
but anyway, it is not important
 
 
2 hours later…
9:37 AM
[Random]
It amazed me how we can have a countably infinite and mostly unique decimal expansion for any real number, but there are continuumly many real numbers
So, what about trying to extend the scope of our alphabet by having a continuum decimal expansion:
Reals:
$$s=\sum_{k=-\infty}^{\infty}\frac{a_k}{10^k}$$
??? number decimal expansion
$$s=\int_{-\infty}^{\infty}\frac{a(x)}{10^x}dx$$
 
@AkivaWeinberger I know that "$10=(3+i)(3-i)$, so it's a multiple of $3+i$", I did write that :) Thanks
 
hi ! do you guys know if this book is a good one for introduction or not?

Introduction to the theory of computation . By Michael Sipser
I checked Amazon's rating it had 4 stars :D
I'm not sure if the rating is trustworthy or not !
 
10:19 AM
One thing that is obvious is that the set $$\left\{s=\int_{-\infty}^{\infty}\frac{a(x)}{10^x}dx\right\} \subset \Bbb{R}^{\Bbb{R}}$$ since the latter has nonintegrable and even uncomputable functions
however the cardinality of this set is to be found out
 
@Secret in other words, $2^{\Bbb N} = \mathfrak c$
 
Did I assumed continuum hypothesis accidentally? (though I tend to assume GCH holds for most of my analysis as I am not ready to deal with cardinalities outside GCH yet)
 
user84215
How can I draw a diagonal arrow in a commutative diagram with AMScd in MathJax?
 
I have no idea. I think you need tikz for that?
 
user84215
What is tikz?
 
10:41 AM
A LaTeX package
26
A: How to draw a commutative diagram?

Zev ChonolesIt is possible to do (somewhat primitive) commutative diagrams using \array: $$\begin{array}{ccccccccc} 0 & \xrightarrow{i} & A & \xrightarrow{f} & B & \xrightarrow{q} & C & \xrightarrow{d} & 0\\ \downarrow & \searrow & \downarrow & \nearrow & \downarrow & \searrow & \downarrow & \nearrow & \do...

 
If you want to make commutative diagrams, I can recommend TikZ + TikZ-cd
Doubt it works in MathJax, though.
 
user84215
What is TikZ?
 
It's drawing stuff in TeX
 
user84215
I think AMScd is better than {array} for creating commutative diagrams
 
[Infinite set] A naive dedekind cut arguement:
Consider the subset of the reals [0,1]
 
user84215
10:54 AM
Infinity is an illusion.
 
We can perform dedekind cuts to obtain a pair of intervals of the form [0,a],(a,1]
Now $a \in [0,1]$ and we can use the bijective map $\frac{1}{\pi} \text{arctan}(x)+\frac{1}{2}$ to show that there are continuum many elements in [0,1]
We can also pick some bijective map to show that given some a, (a,1] also has continuumly many elements. Therefore the set obtained by taking dedekind cuts over [0,1] by having $a \in [0,1]$ (i.e. the set consists of all intervals of the form (a,1], [0,a] )has in total $2^{\mathfrak{c}}$ elements.
Now, because of the way the intervals are constructed, the set $\{[0,a], a \in [0,1]\}$ is one of the two continuum sized chains ($\{(a,1],a\in [0,1]\}$ being the other)
 
I don't understand how you get $2^{\frak c}$
I'm pretty sure you don't
Your elements are of the form $\langle(a,1],[0,a]\rangle$? Biject it to $a$
That forms a bijection between your set and $[0,1]$
 
For each $a$ we have the pair of sets $[0,a]$ and $(a,1]$. Now since $a \in [0,1]$ and $[0,1]$ is known to have continuumly many singletons then in total there are continuumly many $[0,a]$ and continuumly many $(a,1]$ so in total with have $\mathfrak{c}+\mathfrak{c}$, $=$ oopss... $2\mathfrak{c}=\mathfrak{c}$...
bah, I fail counting class again...
 
11:10 AM
Fun fact: In ZF, it is consistent that there exists no total order on the set $2^{\Bbb R}$
and thus on any set of cardinality $2^{\frak c}$
 
@AkivaWeinberger total???
 
(Total order = linear order = order)
 
how can it not have total order...
 
(The extra word is just to distinguish it from partial orders)
@LeakyNun It's very counterintuitive.
 
@AkivaWeinberger what about $\Bbb R$?
 
11:12 AM
@LeakyNun What do you mean?
 
is there total order on $\Bbb R$?
 
Of course there's a total order on $\Bbb R$; take the usual one
$<$
 
I mean, under ZF, of course.
 
What's the cardinality of $\omega_2$, is it $\aleph_2$?
 
Yeah @Secret
@LeakyNun Yeah
 
11:19 AM
@AkivaWeinberger could you expand on that?
 
I think it's consistent with ZFC, even
Not sure
Let me find out
Oh wait duh
 
user84215
To be Well-ordered needs the axiom of choice not for having a total order.
 
Of course it's total orderable with ZFC; it's even well-orderable
But yeah it's consistent with ZF that $\mathcal P(\Bbb R)$ has no total order
 
user84215
I mean for R not for 2^R.
 
It is consistent with ZF that $\Bbb R$ has no well-order, yes
but it definitely has a total order (the usual one)
 
11:27 AM
Ah, $\omega_1$ is different from $\Bbb{R}$. You can have subsets in $\Bbb{R}$ that has an infinitely decreasing sequence, but not in $\omega_1$
 
The statement "Every set can be well-ordered" is famously equivalent to the axiom of choice. The statement "Every set can be totally-ordered" is strictly weaker than the axiom of choice, but still stronger than ZF.
@Secret They could be equinumerous (that's the continuum hypothesis)
 
yup
But I do need to be careful of the ordering properties of infinite sets if I want to understand the higher cardinals
 
user84215
I have a problem with the following formula. I want the parentheses (left and right) to have the same size.
$$\begin{aligned} a= & \Biggl (1+2+3+ \cdots \\ & \left. \cdots + \infty -\frac{(x+2)^7}{\frac{\frac{\frac{x^2}{(x+3)^2}}{(x+2)^5}}{(x+y)^3}}+\infty -1+\infty \right ) \end{aligned}$$
 
If you split it, then you have to manually change the bracket sizes
 
@aminliverpool \vphantom might be helpful
 
user84215
11:31 AM
What is that?
 
\vphantom{stuff} inserts a zero-width invisible thing with the height of the "stuff"
 
It puts some invisible stuff there
 
Stands for "vertical phantom"
Without the v you get a space with the width of "stuff"
 
user84215
Thanks, good idea.
 
$\left(\vphantom{\frac{(x+2)^7}{\frac{\frac{\frac{x^2}{(x+3)^2}}{( x+2)^5}}{(x+y)^3}}}x\right)$
^Example
 
11:33 AM
why are we creating LaTeX Frankensteins
 
Why aren't you creating LaTeX Frankensteins
 
Hmmm...
$S_0 = \{0\} = \{|\}$
$S_1 = \{-1,0,1\} = \{|0\},\{|\},\{0|\}$
$S_2 = \{-2,-1,-0.5,0,0.5,1,2\} = \{|-1,0,1\},\{|,0,1\},\{-1|0,1\},\{-1|1\},\{-1,0|1\},\{-1,0|\},\{-1,0,1|\}$
$S_3 = \{-3,-2,-1.5,-1,-0.75,-0.5,-0.25,0,0.25,0.5,0.75,1,1.5,2,3\}$
...
$S_{\omega} =\Bbb{N} \cup \{\frac{1}{2^n},n\in \Bbb{N}\}$
ok this is getting too messy to follow...
 
user84215
What is LaTeX Frankensteins?
 
Mashing disgusting looking code together to get a working solution, I guess.
 
Another example of a latex frankenstein
$$\overset{+}{\left.\begin{matrix}
\\
\\
\\
\end{matrix} \right|} \begin{matrix} * & * & * \\0 & 0 & * \\0 & 0 & *\end{matrix} \overset{+}{\left.\begin{matrix}
\\
\\
\\
\end{matrix} \right|}=0$$
 
12:01 PM
Hi @MikeMiller
 
12:26 PM
How many ways are there to make 200 from 1,2,5,10,20,50,100,200 up to permutation?
@Secret obviously it isn't just $\frac1{2^n}$
 
Yeah, it in fact includes all reals and also the first generation infinite and infintesimal numbers
and STILL not bigger than $\Bbb{R}$
 
I mean, if anything, it is $\frac{k}{2^n}$
 
good morning frands
 
12:54 PM
@GFauxPas good morning
 
1:37 PM
Consider an integral domain $A$ integrally closed in $K := K(A)$, $L|K$ a finite extension, and $B$ the integral closure of $A$ in $L$.
I can prove that every element of $L$ is $b/a$ for some $b \in B$ and $a \in A$, so $K \subset K(B) \subset L$.
Is $L = K(B)$?
(I'm probably just being dense)
s/is/isn't/
 
1:54 PM
Hi @Soham
 
Z is integrally closed in Q; take the extension Q(sqrt(5))/Q.
Hm
maybe that doesn't work
 
Hi, I think I have solve sat, how speak in french, here the solution :
NP=complexity of convexity optimisation
That's correct ?
 
what is "sat"?
 
En informatique théorique, le problème SAT ou problème de satisfaisabilité booléenne est le problème de décision, qui, étant donné une formule de logique propositionnelle, détermine s'il existe une assignation des variables propositionnelles qui rend la formule vraie. Ce problème est très important en théorie de la complexité. Il a été mis en lumière par le théorème de Cook,, qui est à la base de la théorie de la NP-complétude et du problème P = NP. Le problème SAT a aussi de nombreuses applications notamment en satisfaction de contraintes, planification classique, model checking, diagnostic, et...
 
2:03 PM
Oh
 
do you speak in french @GFauxPas ?
 
no but I have Google translate
what are the variables, and what are the subscripts? I need context
are they Boolean variables, 1 = TRUE, 0 = FALSE?
 
In computer science, the Boolean satisfiability problem (sometimes called Propositional Satisfiability Problem and abbreviated as SATISFIABILITY or SAT) is the problem of determining if there exists an interpretation that satisfies a given Boolean formula. In other words, it asks whether the variables of a given Boolean formula can be consistently replaced by the values TRUE or FALSE in such a way that the formula evaluates to TRUE. If this is the case, the formula is called satisfiable. On the other hand, if no such assignment exists, the function expressed by the formula is FALSE for all possible...
 
I'm looking at that, but that is a page about statements. Your variables seem to be numbers
 
yes in {0,1}
 
2:06 PM
@SohamChowdhury Ok, I'm a little confused. Your theorem says every element of $L$ is $b/a$ for some $b\in B, a \in A$. So that automatically says every element of $L$ is of the form $b_1/b_2$ for $b_i \in B$, hence $L \subset K(B)$ (because elements of the fraction field $K(B)$ are precisely of that form?)
 
right, I was calling those Booleans
 
Am I dumb?
 
but the trick is to work on real
 
and what is the relationship between $x_1$ and $x_{-1}$ Dattier?
 
with convex function
 
2:07 PM
what's the site's convention/policy for asking questions in another language, anyone know?
 
t
 
@BalarkaSen that makes sense, I think I have the containments wrong.
how do I know that $K(B) \subset L$?
 
Thanks it's difficult for me to be understanding in english, I go speak in french forum @GFauxPas
 
every element of $K(B)$ is $b_1/b_2$
ah, hm
 
Dattier ask your question in French and request a translation
 
2:09 PM
we need to use the finiteness/integrality conditions, ofc
 
Ok.
 
say at the beginning
 
J'aurais résolu le probléme sat
 
no, on the website
on the main website, say "I don't speak English, please translate my question" or similar, and then ask
 
@Soham Well, $b_i$ are elements of $L$ because $B \subset L$, right?
 
2:10 PM
j'ai donné un lien de ce que c'était, il permet de résoudre les problèmes de résolution de problémes de maths (par exemple et tout les problème de la classe NP)
 
fractions are closed in fields, so $b_1/b_2 \in L$
 
@BalarkaSen oh duh
 
right, so it was more of a symbol pushing thing
 
I have not done any math since we last spoke in chat / I went to MM, and I get a little dumb when I'm away
 
oh, haha, it's fine
i am dumb too
 
2:12 PM
check Hangouts if you can
yep, star-board confirms
 
je ne sais pas si NP=P, mais cette algo permet de résoudre sous la condition que l'on sache résoudre les problémes d'optimisation convexe
 
oh ****
 
ce qui semble être le cas
 
(checking)
 
2:24 PM
Hello!! We have that $\sigma$ is a reflection along a line and $\delta$ is a rotation. How can we check if $\sigma\circ\delta$ and $\delta\circ\sigma$ are reflections or glide reflections?
 
 
2 hours later…
4:10 PM
@MaryStar get an object, and do it yourself
 
@MaryStar I'm thinking, for the reflection first and then rotation
We can assume the reflection is along the $x$-axis for a moment
and then we can always write the rotation as a composition of a rotation about the origin and then a translation.
So then if you rewrite that as (reflect across x-axis + rotate about origin) + translation
the first two combine into like a reflection about some other line $\ell$
and if the translation is purely perpendicular to $\ell$, it's a reflection
and otherwise it's a glide reflection
and, looking back, there was absolutely no reason to require the first reflection to be along the $x$-axis.
But whatever, what's done is done
 
Hi guys! I wonder if anyone had some knowledge in the percolation theory and could answer to some of my questions?
 
I have no idea what the word percolation means, sorry
 
Isn't that thing about trees ?
 
the proof is trivial site contains the three lines and nothing more?
 
4:22 PM
But post the question just in case someone who knows sees it later
 
And heaps
 
@Astyx Dictionary says it's passing slowly through lots of holes
 
(which are specific heaps)
 
Like coffee
 
4:45 PM
hi chat
 
hı hi hï
…hī
 
@Hippalectryon o/
 
@Waiting \o How are you ?
 
@AkivaWeinberger I calculated a rook polynomial can you please check my work ?
 
I prefer bishop polynomials
 
4:52 PM
the board is a 3x3 with the entry 3,3 deleted
 
Oh it's not even a typo
 
Hi @Astyx
is there really bishop poly ?
 
What's a rook polynomial ?
I have no idea
 
Yeah, what's a rook polynomial
 
4:53 PM
its the number of ways to place n rooks such that they cant take each other
 
@Hippalectryon I managed to do some outstanding discoveries lately, trying to investagate all the details related to them. How about you?
 
Oh ...
 
To be more precise: It's the generating function for those numbers.
I presume?
 
Then I'm pretty sure there is a bishop polynomial
 
4:54 PM
Hi @Semiclassical
Yes you are correct
 
So I guess the constant term is $1$
 
yes allways =p
 
and the linear term is $8x$
 
yes :D
 
4:54 PM
What ?
 
hey,I've a silly doubt, is a compact subset of an open set(in subspace topo) in the complex plane compact as a subset of the plane ?
 
those the easy cases =p
 
Wait, why 1 ?
 
@LucyferZedd Yes, compactness is an intrinsic property
 
zero rooks.
 
4:55 PM
1 way to place 0 rooks on any board
 
It doesn't depend on the space it's in @LucyferZedd
 
Though I'm not seeing why it'd be 8 ways if there's one rook.
 
We're deleting a corner apparently
 
@Aki
 
@Waiting That's great :D nothing special on my end :(
 
4:56 PM
1 rook is alone
 
Oh and you're dealing with a $3\times 3$ board where one (bottom right) is missing
 
cant take anything =p
 
@AkivaWeinberger How to show that it is closed in the complex plane ?
 
Oh, 3-by-3 minus a corner.
 
Yes @Astyx
Yes!
 
4:56 PM
Yeah, okay.
 
Scheisse
 
@LucyferZedd Do you know the "finite open cover" definition of compactness?
 
1+8x+14x^2+4x^3
 
I just put molten chocolate all over my bed
 
was my answer but I need some conformation i done it right =p
 
4:57 PM
...
 
@AkivaWeinberger and also, what do you mean by that it is intinsic?
 
I just counted 14 for two rooks
 
counted?
 
@AkivaWeinberger yea, okay I will try using that
 
@LucyferZedd It doesn't depend on the surrounding space. (Closedness is not intrinsic)
(but compactness is)
@KasmirKhaan Just went through all the cases
 
4:58 PM
@Hippalectryon I go to bed for 15-30 min, very tired here. Some research takes away all the energy you have. I need to rest a bit.
 
@AkivaWeinberger hmm that would be hard when dealing with bigger board =p
 
well how many ways of placing 3 rooks by your method? =p
 
@Waiting Alright, rest well :-)
 
Oh, they all need to be in a diagonal with wraparound
so that'd be 6, but two contain the deleted corner so it's 4
 
4:59 PM
I also have 14
 
:D
so I did it right:)
thanks all ! :)
 
@Hippalectryon Thank you for that, I'm releaved !
 
@AkivaWeinberger, oh ,yeah, silly me ... hey can you help me with convergence of harmonic functions?
 
And 4 for three rooks
0 above
 
@LucyferZedd Probably not
Maybe someone else here could, though
 
5:03 PM
@Akiva is $\aleph_\alpha=\beth_\alpha$ independent of ZFC for all $\alpha\neq 0$?
 
@AlessandroCodenotti Isn't that the generalized continuum hypothesis?
Oh, the conjunction of all of them would be
 
GCH says that this holds for all $\alpha$, I'm wondering whether each instance is independent
 
I don't know if there's a specific $\alpha$ for which it's provable. I doubt it
 
for $\alpha=1$ that's just CH
 
<-- Not a set theorist
I have no idea how forcing works
@KasmirKhaan My guess for how to compute it for larger boards
So you'd want an $n\times n$ board to depend on the $(n-1)\times(n-1)$ version
 
5:08 PM
[Chemistry] The horror: Realised that 10 of the structures I calculated are not the lowest possibel energy, meaning that a total of 30 calculations need to be redone!
 
If it's the bottom-right corner that's dropped,
either none of the rooks are on the top or right edges,
or one of them is,
or two of them are.
 
Speaking of set theory I don't remember exactly who was interested in the set theory study group, but I'll create a room for that on Thursday or Friday @Mike @Dami @Perturbative @Astyx @Balarka
 
For each of those cases, you could do it in terms of smaller boards
looking at what other squares are available
 
Count me in (I might not be here on thursday and friday though, but then on should be fine)
 
@Alessandro I probably won't be available for 10 or so days, but definitely do it!
 
5:10 PM
If there were a set called $\in$
we'd have ${\in}\notin{\in}$
 
$\ni$
 
@AkivaWeinberger being able to write $\in\in\in$ is a good enough argument to reject the axiom of foundation to me
 
$\notin \ni$
$\niton$
Sad
 
@AkivaWeinberger Thanks but we have 2 formulas so far, one by recursion and other by product , I have to keep reading and do some examples =p
 
$\notni$
 
5:11 PM
or regularity maybe, I don't remember what's the usual name
 
$\in$_$\in$
 
$\in\!\ni$
$\in\!\equiv\!\ni$
 
I was lied to
 
Oh what
Pitchforks! Get your pitchforks here! $-\!-\!\!\!\in$
 
$\in\!\equiv\!\equiv\!\equiv\!\equiv\!\equiv\!\equiv\!\equiv\!\equiv\!\equiv\! \ni$
 
5:13 PM
@Akiva $\pitchfork$
I got mine.
 
Wait that's a thing??
 
$\psi$
 
Yeah it's used to denote "transverse intersection"
 
$\madeupbullshit$
 
$\newcommand{\madeupbullshit}{END OF WORLD}$
@Akiva edit that message
but don't change anything
 
5:16 PM
I see through your tricks
 
$\madeupbullshit$
 
the spacing in that "END OF THE WORLD" is super wonky on my screen
 
mine too
 
.endofline
 
@Astyx check discord
 
5:30 PM
$\newcommand{\madeupbullshit}{\text{BRACE THE INFINITE VOID}^{\text{BRACE THE INFINITE VOID}^{\text{BRACE THE INFINITE VOID}^{\text{BRACE THE INFINITE VOID}}}}}$
$$\Huge{\madeupbullshit}$$
$$\infty\infty\infty\infty\infty\infty\infty\infty\infty\infty\infty\infty\infty‌​\infty\infty\infty\infty\infty\infty\infty\infty\infty\infty \infty\infty\infty\infty\infty\infty\infty\infty \infty \infty\infty$$
 
5:52 PM
@Secret for the love of God
 

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