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1:40 AM
@schn tbh, by looking at that graph and knowing that ln(2) is roughly 0.7
To actually compute it, I’d substitute $t=xs$ and then consider x->0
 
 
1 hour later…
3:01 AM
Is there a way to show that differentiable space filling curves cannot exist
 
 
1 hour later…
4:19 AM
Does the idea that "I suffered so you should too" have a name?
as opposed to the opposite idea, "I don't want you to suffer like I did"
 
Maybe German has a word for that
 
Other than, like, "bitterness"
 
Yeah, bitterness is too broad for that term. I don't know anything specific though.
Spite? That's not exactly the idea, but it's closer
 
Kind of having trouble w/ this combinatorics problem
2
Q: 7 distinguishable balls in 4 distinguishable boxes, probability that no box is left empty.

user372770I found a solution for this that uses multinomial however I tried to solve it in a different way and I don't know where I am going wrong. So I thought how about we fill each box with one ball first, and then put the remaining 3 in all possible boxes. This was my reasoning: First let's choose ...

 
Mooornin'
 
4:32 AM
Also: what's the opposite of "prefer"
 
ur up early
 
as in, "I prefer A to B" iff "I ___ B to A"
 
I don't understand how the double counting is taken care of
 
Commute takes an hour lol
 
Morning, Meg
 
4:33 AM
henlo
 
Maybe "I dislike B compared to A"
 
@ÍgjøgnumMeg wtf
 
@Ryan I live 20k outside of Heidelberg hahaha
 
my diagrams are slow to commute as well
 
4:36 AM
Anyone?
 
@Akiva en.wiktionary.org/wiki/disprefer how severely unsatisfactory
 
Sorry Krauser, but not me
 
Combinatorics is super difficult
 
ANyone better at probability haha?
 
@AkivaWeinberger in Cantonese slang we call that the fishball theory
and I can't find a (proper) English translation
 
4:50 AM
Why?
Like a ball of fish with the bones taken out?
 
it's a reference to a stand-up comedian who was using fishball as an example of how HK people have this "I suffered so you must to" mentality
 
oh great
yeah it's exactly that one
 
5:46 AM
Is the rubix cube group solvable?
 
@johnny133253 no
The Rubik’s Cube group is a group ( G , ⋅ ) {\displaystyle (G,\cdot )} that represents the structure of the Rubik's Cube mechanical puzzle. Each element of the set G {\displaystyle G} corresponds to a cube move, which is the effect of any sequence of rotations of the cube's faces. With this representation, not only can any cube move be represented, but also any position of the cube as well, by detailing the cube moves required to rotate the solved cube into that position. Indeed...
$G = (C_3^7 \times C_2^{11}) \rtimes ((A_8 \times A_{12}) \rtimes C_2)$
the $A_8$ and $A_{12}$ makes it not solvable
they correspond to permutations of the 8 corners and the 12 edges respectively
 
You know, once my high school CS teacher told me he didn't think there was a point to solving Rubik's cubes 'cause he heard there's an algorithm
That didn't make sense to me - it's a given that there's an algorithm, the puzzle is in discovering/constructing that algorithm
Incidentally, here's a puzzle: scramble the Rubik's cube using only the top and right faces, and solve it the same way
 
Yeah, like there are bots that automatically play video games. The act of playing a video game isn't invalidated by their existence
 
Even if you already know how to solve the Rubik's cube normally, it won't help you a lot, because now you can't use most of your algorithms
@Rithaniel By "algorithm" I don't really mean a bot, just something where "once you know it you can solve any position"
Incidentally, I doubt there's an algorithm for solving twisty puzzles in general, but I wonder if someone's made a machine learning program for it
Here's a question I don't know the answer to:
 
Well, a bot needs to have an algorithm to tell it how to act, but I suppose they're tangentially connected
 
5:58 AM
@LeakyNun I'm giving a talk today at Imperial.
You are expected to be there.
So long.
 
Yo I didn't know you used chat!
and he's gone
That felt like an imperial summons
 
and they're gone
 
In any case, the question I don't know the answer to:
Rotating the top face is notated "U", rotating the right face is notated "R"
(clockwise, I think - counterclockwise has an apostrophe, so U' and R')
(or other way around? Doesn't matter)
So I just asked about scrambling a cube using just U and R, and solving it the same way
which I do know how to do
 
Rotating the top two faces together is notated "u", lowercase
and rotating the right two faces together is notated "r"
Problem: Scramble a cube using just u and r, and solve it the same way
This must be possible but I have been unable to do it even once
and I kinda want someone to make a machine learning program figure it out
(I could build one myself, but this involves learning how to code neural networks and that seems hard)
I bet there are tricks you could use to make the training easier
@AkivaWeinberger Oh lol inadvertent pun
Didn't even realize
Rotating the top two *layers together, whoops
 
6:04 AM
puns are never inadvertent
 
I would like to attend that seminar. Seems like an interesting topic
(In related-ish news, I really need to get to a more formal college than I'm currently at. I've recently discovered that it's more of a party college than I realized. Only took me two years)
 
Someone from UC Berkeley transferred to Yale because of the former's party atmosphere
 
is that Hebrew grammar?
 
(Funnily enough she got put into "Berkeley College", one of the residential houses at Yale. Yeah, the residential houses at Yale are called "colleges", it's confusing)
@LeakyNun ?
Oh whoops wow
 
I've heard that Berkeley is a very political environment, actually
 
6:10 AM
like is that grammatical mistake influenced by Hebrew grammar subconsciously
not that I am a prescriptivist
 
I see what you mean, "ha-sefer ha-ze" = "the this the book"
 
yeah that's what I mean
 
but no it came from editing the comment before posting and accidentally leaving traces of the original edit
 
oh
 
I do that too, Akiva. Always bugs me when I do
 
6:12 AM
how do you say "the red book"?
 
Ha-séfer ha-adom, literally "the book the red"
(Whoops the first one should be "the book the this")
 
cool
 
I think Icelandic is similar
@ÍgjøgnumMeg Yo how do you say "this book" and "the red book" in Icelandic? Do you duplicate the article?
 
I'm not aware that he speaks Icelandic?
 
I think he learned a bunch of Germanic languages or something
By the way, "German" in German is Deutsche. Guess what "Germanic" is? @LeakyNun
 
6:16 AM
German in German is Deutsch, and Germanic is Germanisch
 
Damn you knew it already
 
I've discerned that a lot of people in here seek to become polyglots
 
I think the 'e' in Deutsch(e) depends on the noun vs the adjective
@Rithaniel Leaky and I are just both linguistics nerds
(and also yeah he speaks Cantonese and I speak Hebrew)
I think Secret also is from Hong Kong? but I forget
 
I think I'd like to become a polyglot too. I suppose you could say it's "on my bucket list"
Languages are difficult for me, though. They're not simple and strict, like numbers are
 
Tip: Learn the pronunciation first if you can, then the grammar, then the vocabulary
You can pick up a noun a lot easier than you can pick up grammar
 
6:19 AM
@Akiva I don't speak Icelandic but in Faroese you would have "Henda bókin" which is "This the book" and "Henda reyða bókin" which is "This red the book" I believe
 
and there's lots of pronunciation subtleties that you might not notice despite years of study if you're not training your ears for them
@ÍgjøgnumMeg Is that for "this red book" or "the red book"
 
Yeah, but the grammar is the biggest thing. Like, in Deutsch, every noun has an associated gender. It's a lot to keep track of
 
Yeah
I want to learn ASL but I don't know of any resources that let me learn it in that order
 
@Akiva "This red book" I guess, I think "the red book" is "tann reyða bókin" which is "the red the book" like you said
 
I don't even know what the "phonology" of ASL could be
 
6:20 AM
"the book" by itself is "bókin" though
so the article reduplication comes from adjectives
 
Like obviously you can describe each sign in terms of hand shape and stuff, and you can boil it down to five elements I think, but that applies to all sign languages, not just ASL
What does ASL have in particular
Fun fact: despite all speaking Spanish, Latin America has tons of different sign languages
'cause in colonization, Spain imported their spoken language but not their signed language
 
also "Deutsche" means "German female" or "Germans"
 
(which I dunno if they even had one at the time beyond home signs)
 
Deutscher is German male
 
Wait so what's the German language
 
6:22 AM
Deutsch
 
Oh
Whoops
 
:D
although if you are speaking with someone then you are speaking mit einem Deutschen oder mit einer Deutschen
 
BANZSL = British, Australia, New Zealand Sign Language
I think
 
-em for male and -er for female, right?
 
Argentine Sign Language and Kazakh Sign Language exist so I guess the map just has missing data
 
6:25 AM
depends on the position in the sentence, adjectives in the dative and genitive take -en after an article but standing alone they follow the same inflection rules for articles
so "in deutscher Sprache" for instance, but "in der deutschen Sprache" with an article
(because Sprache is feminine"
=
)
 
@AkivaWeinberger also they don't all speak Spanish
 
Details
Brazil is Span-"ish"
Suriname is weird
 
Pedantry is a skill for mathematicians
 
Also apparently there isn't a unified Kazakh Sign Language
>Zhasanova shared she never considered learning sign language herself because she always believed that one day her deaf child would start speaking.

“I realised I had to change my way of thinking a few years ago when my younger son thought I didn’t love him because I didn’t want to learn gestures. All my life I thought my child would be able to speak, but it’s been 30 years and it is clear that he will not,” she said.
Lol whut??
 
lol plot twist
 
6:29 AM
Yo if you don't speak your son's native language then what are you doing
 
@AkivaWeinberger Eric will disagree
 
I was thinking the kid would be like 6 or smth and she's like aaaiight I'll learn now
 
"Eric discordará" @LeakyNun
 
well played
 
@ÍgjøgnumMeg Yeah
 
6:35 AM
@AkivaWeinberger nice meme 10/10
 
I'm constructing $L$-functions today!
 
@ÍgjøgnumMeg cool
you're gonna prove dirichlet's theorem?
 
Idk what we're doing, it#s the first lecture of a course called Construction of L-functions
 
lol
 
so we'll probably just define a bunch of stuff
hahaha
 
6:38 AM
guess you might wanna bring plasticine
 
@Leaky I attended the first lecture in the ANT course and the first definition was "Def: A prime number is...."
turns out we are doing 2 weeks of elementary number theory first
 
and what is "..."?
 
You tell me
xo
 
a positive generator of a prime ideal in Z
 
hahaha
aye
nah but we are doing quadratic reciprocity and representations as sums of squares for 2 weeks before we do any ant
and in modular forms we just talked about möbius transformations for 2 hours
although I'm happy for the German maths practice before we get into anything hard
 
6:43 AM
 
Euclid or what
 
yeah
 
mad man
 
How do I prove that given two events, $X$ and $Y$, the min.{Prob[X], Prob[Y]) $\geq$ Prob [X $\cap$ Y] ?
 
6:45 AM
show that Prob[X] $\ge$ Prob [X ∩ Y] and Prob[Y] $\ge$ Prob [X ∩ Y]
 
@Leaky is $\Bbb Z/\Bbb Z$ an integral domain tho
 
I know that Prob [X $\cap$ Y] can be re-expressed as Prob[X] + Prob[Y] - Prob[X$\cup$ Y]
 
@ÍgjøgnumMeg clearly not
 
But I dont know if that will help
If I re-express the RHS then proving Prob [X] $\geq$ Prob[X] + Prob[Y] - Prob[X $\cup$ Y] gives 0 $\geq$ Prob[Y] - Prob[X $\cup$ Y].
Oh nevermind I think I get it
 
7:09 AM
0
Q: Why change of transformation matrix is unique

maths studentLet $V$ be an $n$ -dimensional vector space and let $B, B^{\prime}$ be two ordered bases for $V .$ Prove $P_{B B^{\prime}}$ is the only matrix with the following property: $$ [v]_{B^{\prime}}=P_{B B^{\prime}}[v]_{B} $$ So somehow I have to show if $P_{B B^{\prime}}[v]_{B}$ = $Q_{B B^{\prime}}[v]...

@LeakyNun any simple way to prove above problem.
 
 
2 hours later…
9:02 AM
This comforts me: $\textbf{The set of shit you need to know for any particular problem satisfies the descending chain condition.}$
just read this on reddit hehe
 
 
3 hours later…
11:57 AM
can somebody help me with setting up limits of a spherical coordinates....I am struggling to understand it
0
Q: Triple integral in different coordinate systems.

HedgehogMy task is to write volume integral in 3 coordinate systems : Cartesian, cylindrical and spherical. This integral shows volume of intersection of 2 spheres, first with center at $(0, 0,-3)$ and radius $5$ and second with center at $(0,0, 3)$ and radius $\sqrt{13}$.I am able to do it for Cartesian...

i tried plotting graph or this and i got:
But how do i set up limits here in spherical coordinates....
not able to visualise thetha,phi here
from what i understood
ok what i understood:
all y values belong to x^2+y^2<=9
all x values same :
Z values if i am correct varies from the lower portion of the upper circle to z=point of intersection
how do i convert them to thetha,phi and z form
 
 
2 hours later…
2:05 PM
0
Q: Trying to better understand conformally compactified Euclidean space into the unit ball

UltradarkGoal: To gain a better understanding of Euclidean space conformally compactified into a unit ball. Question: How can I visualise and mathematically describe Euclidean space conformally compactified into a unit ball? My attempt: Start with the conformally compactified disk model of Eucl...

Thoughts?
 
 
2 hours later…
4:12 PM
If F is a linear transformation that has k eigenvectors belonging to k distinct eigenvalues, then the eigenvectors are linearly independent. Suppose that F only has k-1 distinct eigenvalues, but the eigenspace associated with one of these eigenvalues is spanned by two linearly independent vectors. Does F then still have k linearly independent vectors?
 
@schn Yes
 
4:34 PM
@TobiasKildetoft Cool. Thanks.
 
4:56 PM
$(a+b)(c+d)=(a-b)(c-d)$ iff $ad+bc=0$
This also means $(a+b)(c+d)=(a-b)(c-d)$ iff $(a+c)(b+d)=(a-c)(b-d)$
Hm. So if $ab=xy$ then$$((a+x)^n+(a-x)^n)((b+y)^n+(b-y)^n)=\\ ((a+x)^n-(a-x)^n)((b+y)^n-(b-y)^n)$$?
Er no
Some sign errors
 
5:48 PM
@Aladdin Answered. You get the award for one of the yuckiest problems I've ever seen!
 
help
 
6:12 PM
hi
 
6:32 PM
Evening
 
evening @ÍgjøgnumMeg
 
what's up algebra-people (and real people too)
 
hi @Tobias
wait does that mean that algebra-people aren't real?
 
I leave that as an exercise for the reader
 
@ÍgjøgnumMeg what do you think of the L-functions lecture?
 
6:53 PM
$$
f(x,y) = \left\{
\begin{array}{ll}
\dfrac{xy}{\sqrt{x^2+y^2}} & \quad \text{if $x^2+y^2 \ne $ 0} \\
0 & \quad \text{if $x^2+y^2$ = 0}
\end{array}
\right.
$$
does its partial derivative exists at (0,0)

If I use the definition of partial derivatives, then I get answer as 0.
But if I first partially differentiate and then substitute values then it does not exist.
 
@Mathein it was alright but i‘m Not sure if I’ll survive an exam in it haha
we will see, Rösner is veeeery fast
it’s really interesting tho
i hope he does more algebraic stuff than analytic though
Statement of Langlands correspondence sounds way too close to research hahahaha
 
well, I don't think we will state the Langlands correspondence in utmost generality
 
Fair, i liked the sound of Artin L-Functions and dirichlet dedekind hecke usw
 
7:08 PM
but L-functions are analytic objects, so there's going to be a certain amount of analysis
there's still some algebra involved of course
for example, the proof that Artin L-functions admit meromorphic continuation uses nontrivial representation theory
 
Yeah of course but I mean.. for instance Rösner mentioned that Riemann zeta function is a super analytic example, while there are like more algebraic examples idk
ofc I don’t mean I wanna avoid analysis completely hahaha
 
I just hope you stay in the lecture so that I can use your notes (as I mentioned I can't attend the lectures starting next week), lol
 
Hahaha sure, I TeX all my notes after the lectures so I can share them
 
oh wow
that's great
 
@MatheinBoulomenos algebra people are usually complex
 
7:14 PM
that’s how I learn hahah, write an outline of notes and then fill the gaps and if I don’t understand smth while I’m writing them up then I can do some calculations nebenbei and write my own explanations in and stuff
 
7:36 PM
Thought
The middle-halves Cantor set has Hausdorff dimension 1/2
That means its Cartesian product with itself has Hausdorff dimension 1
Does that mean it has a length?
Hm: we can Buffon it maybe
Throw it at some parallel lines
 
8:00 PM
@AkivaWeinberger are you sure the dimension is 1/2?
 
log(2)/log(4), no?
You scale it up by a factor of 4 and it doubles in size
 
oh
 
(aka log_4(2))
 
can you describe the product as a self-similar set?
 
Scale it up by a factor of four and you get four things
 
8:04 PM
it would be spooky if it had a length
 
How do you define length for a pointset?
 
but it's Spooktober so it's fine
 
Well OK if you superimpose a grid of resolution 4^3 it hits 4^3 squares of your grid
Yeah I think it has length 1
 
that's very spooky
 
Meaning if you do the middle-fourths construction on a set of length $1\rm cm$, the result has size $1\sqrt{\rm cm}$
2
 
8:08 PM
ok it would be extra spooky if it weren't 0 or 1 or infty
 
${\rm cm}^{1/2}$
Yeah if it had length pi that would really oodle your noodle
Right so what about Burton's needle problem with this shape
What was the answer for the normal needle problem? Like $2/\pi$ I think?
 
 
2 hours later…
10:09 PM
Hi all
0
Q: $ a_n = \frac{a_{n-1}(a_{n-1} + 1)}{a_{n-2}}.$ and $ T = 3.73205080..$?

mickConsider the following sequence : Let $a_1 = a_2 = 1.$ For integer $ n > 2 : $ $$a_n = \frac{a_{n-1}(a_{n-1} + 1)}{a_{n-2}}.$$ $$ T = \lim_{k \to \infty} \frac{a_k}{ a_{k - 1}}.$$ $$T = ??$$ What is the value of $T$ ? Is there a closed form or integral for $T$? I get $$ T = 3.73205080..$$...

Any ideas ?
 
10:39 PM
@Mathein when Rösner mentioned Plancherel or Parseval, he meant Poisson yeah?
hahaha
 
Sup
 
@ÍgjøgnumMeg yeah
 
11:18 PM
Hello
Guys
let $y_1(x)$ $=$ $x^2$ if $x\geq 0$ and $0$ if $x<0$
and $y_2(x)=0$ if $x\geq 0$
and $x^2$ if $x<0$
how do I show that they are linearly independent?
 
11:31 PM
@MatheinBoulomenos hej
@topologicalmagician draw the graph of $ay_1+by_2$
 
@LeakyNun hi
 
@LeakyNun I solved it, $A,B$ are supposed to be constants so I break it into cases, when $x\geq 0$ and when $x<0$.
 
@MatheinBoulomenos show that k[x]/(x) = k
@topologicalmagician cool
 
@LeakyNun obvious qed
 
do you have a nuke proof
 
11:36 PM
Hello guys, I was just wondering if the following statement is mathematically true: "For every x in the reals, there exists a y in the reals, such that x*y=1"?
 
@Abwatts no
 
is it because x can equal 0 and there can't exist a number that would make it a 1?
 
precisely
 
got it, thanks a lot!
 
@Abwatts For a more complicated problem, it might be of help to know the negation rules of quantified statements
@TedShifrin hey Ted!
 
11:40 PM
@LeakyNun by universal properties of quotients and polynomial rings, for any $k$-algebra $A$, we have a natural bijection $\mathrm{Hom}_{k\mathrm{-Alg}}(k[x]/(x),A) \cong \{a \in A \mid a=0\}$, this is of course always a one-object set, so $k[x]/(x)$ is an initial object in the category of $k$-algebras. So is $k$, so we have $k \cong k[x]/(x)$
 
yeah that's also my proof
 
I don't really consider this a nuke proof
 
perhaps we can reduce to showing that Z[x]/(x) = Z because of tensor product shenanigans
there must be a categorical interpretation of k[x]
I think I once did one
Dec 30 '18 at 20:44, by Leaky Nun
in fact let's form a category $\mathscr C$ whose objects are $k$-vector space $V$ with an endomorphism $T:V \to V$, and let a morphism $(V,T) \to (W,U)$ be a linear map $\varphi : V \to W$ such that $\varphi \circ T = U \circ \varphi$
 
one can show that for any ring $R$, the center $Z(R)$ is isomorphic to the endomorphism ring of the identity functor of $R$-Mod. This implies that Morita equivalent rings have isomorphic centers. In particular Morita equivalent commutative rings are isomorphic. So if we can show that $k$ and $k[x]/(x)$ are Morita-equivalent, then we are done.
 
Dec 30 '18 at 20:44, by Leaky Nun
I claim that $\mathscr C$ is equivalent to the category of $k[X]$-modules
 
11:46 PM
yes, that's what I was thinking of
 
yay
 

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