00:00 - 16:0016:00 - 00:00

4:04 PM
@Adam What if Ted, me, Semiclassical, Mathphile, Ultradark, Erico, Alessandro are the same person and thus this chat is just you and avatars ;P

Well it's none the bother to me, where else would I go for social interaction, facebook? of course I prefer stack exchange if you think therefore you are far more worth while than most social media streams

@Secret lol

$0$
$n$
$1+1+\cdots=\omega$
$\omega n$
$(\omega^2 + 1)n$
$(\omega^{\omega} + \cdots + 1) n$

4:26 PM
well see here we get to the belly of the beast, because if we let mathematics have it's say on physical matters then half of the pseudo science threads they pump to keep public interest alive in physics gets brought into question and eventually exposed as complete rubbish

$(\epsilon_0+\cdots + 1)n$
$\phi(n,0)$
$\phi \binom{1}{n}$
$f(n)$
$f(n)+1$
$f(f(n)+1)$
$f(\alpha)=\alpha$
$f(\alpha+1) > \alpha$

1 hour later…
5:57 PM
an observation that I don't know what to do with
-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6
3, 6, 4, -2, 5, -5, -1, -3, -6, 2, 0, 1, -4
3, -1, 0, 5, -3, 6, 1, 2, 4, -5, -4, -6, -2
Each column of that array sums to zero, and each row is a permutation of the numbers -6 through 6
I think that this phenomenon should be generic, i.e. for any positive integer n there should exist a 3-row array such that each row is a permutation of -n through n and every column sums to zero
But I don't really see a pattern in the above. (It's a solution that Mathematica generated via the FindInstance command, and therefore isn't expected to be unique.)

hmm...
How likely is given a set of integers [-n,n] and a permutation $\pi \in S_{2n+1}$ that:
For all $a,b,c$: $\pi_i(a)+\pi_j(b)+\pi_k(c)=0$

think I've found a construction that works:
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
6 4 2 0 -2 -4 -6 5 3 1 -1 -3 -5
0 1 2 3 4 5 6 -6 -5 -4 -3 -2 -1

sorry typo
$\pi_i(a)+\pi_j(a)+\pi_k(a)=0$

so first line is start from -6 and go up one at a time
second is to start from 6 and go down two at a time, wrapping around -8 = 5 mod 13
last is to start at zero and go up 1 at a time, wrapping around 7 = -6 mod 13
those each hit all integers from -6 to 6
and each column sum is shifted by 1-2+1 = 0 (with the 'wrapping' cases being taken mod 13)

hmm...
0,1,2,3,4,5,6,7,...,n
$\pi ([0...7])$

6:12 PM
and minus that, with zero only showing up once

$\pi_a(x)+\pi_b(x)+\pi_c(x)=0$ where $\pi_a(x) = x+a \mod (n+1)$
Thus:
$x+\pi_b(x)+\pi_c(x)=0$ gives:
$x+ (x+b) + (x+c) = 0 \mod (n+1)$

wow you guys make quwhite a team

$3x + b + c = 0 \mod (n+1)$
Since it has to hold for all $x \in [0,...,n]$ it means:

switches to semi classical monitor
damn it

Okay, yeah, I'm convinced of this construction
I don't have a good name for this, but it works

6:18 PM
$3x$ has to be some additive inverse of $(b+c)$
What is $3x \mod 13$ for $x \in [0,...,6]$?

0,3,6,9,12,2,5

o wait, my formulation does not work, because it forgot the permutations that are not shifts
Still, it is clear that this property has to satisfy this equation somehow:
$x + \pi_a(x)+\pi_b(x) = 0 \mod (n+1)$

my construction for -n...0...n :

-n, -n+1, -n+2, ... , 0, 1, 2, ... , n-1, n
n, n-2, n-4, ... , -n, n-1, n-3, ... , -n+3, -n+1
0, 1, 2, ... , n, -n, -n+1,..., -2, -1
hmm

somehow, one gets the additive inverse of $x$ by combining two permutations

This stuff is very reminiscent of magic squares and latin squares.
Or at least, latin rectangles.

6:25 PM
\begin{pmatrix}
-n &-n+1 & -n+2 & \cdots &0 & 1 & 2 & \cdots & n-1 & n\\
n & n-2 & n-4 & \cdots & -n & n-1 & n-3 & \ldots & -n+3 & -n+1\\
0& 1& 2& \cdots & n & -n & -n+1 & \ldots & -2 & -1
\end{pmatrix}

$[-n,..,n]$ is a set that forms an additive group

bah, still wrong

$\pi_a(x)+\pi_b(x)$ forms the additive inverse of $x$

okay, that shouold work
@PM2Ring it has that flavor, yeah

Let $y+x=0$, it means that we have $y = \pi_a(x)+\pi_b(x)$, that is, for each $x$, there exists a unique decomposition of $y$ into pairs of numbers

6:28 PM
@Secret yeah. typically one would write the elements as 0 through 2n+1 but it's the same group

What's mysterious is why there always a $\pi_a,\pi_b$ such that $\pi_a(x)+\pi_b(x) = -x$ for all $x$. Its as if $\pi_a,\pi_b$ quotient the additive group somehow
(each column can be thought of as cosets of this additive group with the property they sum to the identity)

If I were to guess, I'd suspect that one should view this in terms of $[-n,n]\times [-n,n]\times [-n,n]$
which is itself a group
and now we're considering the orbit generated by repeatedly adding $(1,-2,1)$
Since 1 and -2 both generate $[-n,n]$, each element shows up exactly once for each component
and by construction one has that the sum of the three components remains zero

yeah

Only issue I could see is that we're only strictly guaranteed that the sum remains zero mod 2n+1

hmm...

6:35 PM
yeah, there's something a bit delicate going on.

0,1,2,3 will fail

yeah, it's definitely possible to have the sum of the three components be 13 if I pick a generic seed and orbit it
I don't think that the seed my construction above was using runs into this, tho

0,1,2,3
1,3,0,2
3,0,2,1
fail

well, for my construction with n=3, it'd be
-3 -2 -1 0 1 2 3
3 1 -1 -3 2 0 -2
0 1 2 3 -3 -2 -1
which looks legit
On the other hand, if I take the left-most column to be 0 0 0 then I get
0 1 2 3 -3 -2 -1
0 -2 3 1 -1 -3 2
0 1 2 3 -3 -2 -1
in which case the column sums are 0, 0, 7, 7, -7, 0
so zero mod 7 but not strictly zero

Hi guys, I started a StackExchange page about Graphing Calculator 3D here:
https://area51.stackexchange.com/proposals/120787/graphing-calculator-3d
It already has 316 members, it needs 75 more members with 200+ reputation before it can go public. If it doesn't reach that number this week it will shut down.
It'd be great if you support it by joining it.
3

6:46 PM
So it does seems that it boils down to whether given m rows, the underlying set forms an additive group and can be decomposed into a product of m groups and that the m possible shifts has a solution such that it sums to zero mod 2n+1

Sounds right? I'm not touching the case of m rows tho

hmm...
0,1,2,3
1,2,3,0
3,0,1,2
that still fails despite it is made by (0,1,3) which all the orbit actions should cancel
I think we need the additive group to have unique additive inverses except zero
which here it clearly fails because 2+2=0 mod 4
0,1,2,3,4,5
3 will fail
0,1,2,3,4,5,6,7
so at least for 3 rows, this will fail for all even n
since there always exists an element (the median) such that it is its own additive inverse
It will be useful though if our investigations above can be generalised to semigroups because then we have a way to tackle the sunset sum problem
In computer science, the subset sum problem is an important decision problem in complexity theory and cryptography. There are several equivalent formulations of the problem. One of them is: given a set (or multiset) of integers, is there a non-empty subset whose sum is zero? For example, given the set { âˆ’ 7 , âˆ’ 3 , âˆ’ 2 , 5 , 8 } {\displaystyle \{-7,-3,-2,5,8\}} , the answer is yes because the subset { âˆ’ 3...

7:16 PM
Evening chat

welcome back

7:49 PM
@user170039 Ah no but I have pictures of the lecture notes that I can share.

Hello, if $a_n$ are real numbers and $I_n$ are half open intervals of type $[a,b)$ and $\sum_{n=0}^\infty |a_n|\mu(I_n)<\infty$, then is it true that $\sum_{n=0}^\infty a_n\chi_{I_n}$ converges almost everywhere? ($\mu$ is the Lebesgue measure so that if $I_n=[a,b)$ where $b>a$ then $\mu(I_n)=b-a$, $\chi_{I_n}$ is the characteristic function of $I_n$)

This recent SMBC was uncomfortably on point: smbc-comics.com/comic/stress

Yikes

8:18 PM
@BalarkaSen If $f:M\to N$ is an $A$-module hom and $\mathfrak a$ is an ideal, do we always get an induced hom $M/\mathfrak aM\to N/\mathfrak aN$ or is there some condition
seems like it should be fine.

looks like a commutative square

8:35 PM
@RyanUnger This is true, yes.
$f(\mathfrak{a}M) \subset \mathfrak{a}N$, so it passes to the quotients

9:06 PM
hi chat

user280247
9:36 PM
Hi guys

user280247
One question:

user280247
is there any nice book to read about triangles and its properties, or maybe one covering all geometry, but from a very low level

user280247
?

what do you mean by low level ?

user280247
Well, showing where sin, cos, and so one come from

user280247
9:38 PM
and the different types of triangles

user280247
and so on

user280247
not one starting with complicated theorems

There are plenty of resources on the subject on diverse websites
If you want a book specifically I don't know
But any basic introduction to trigonometry should cover these things

user280247
yes you're right, I'm diving on the web to see if I found something good

For instance, googling "introduction to trigonometry" gives you mathsisfun.com/algebra/trigonometry.html

9:50 PM
Hey guys, anybody know if there's a "standard" $\sigma$-algebra for $\mathbb{N}_0 \cup \{\infty\}$?

@RyanUnger yeah, that was on point too. Reminded me of an old onion video: youtu.be/NYSxkqL9l_8

I'm reading some probability theory stuff and the book claims that a certain class of $\mathbb{N}_0 \cup \{\infty\}$-valued functions $N \to \mathbb{N}_0 \cup \{\infty\}$ are measurable for a certain $\sigma$-algebra $\mathcal{N}$ on $N$. But I'm wondering what the $\sigma$-algebra on the $\mathbb{N}_0 \cup \{\infty\}$ the measurability is referring to.

user280247
@Astyx that's cool, thanks

No reference to it in the book. So I assume that it's somehow a standard one.

user280247
I'm trying to learn mathematics not through many formulas but by ideas and visual thinking, but I can't find very good books

user280247
9:56 PM
Some from walter warwick sawyer are nice

If $k$ is a perfect field and $P\in k[X]$ then the Galois extension $k[X]/(P)$ has the symmetric group of order $\deg P$ for its Galois group iff $P$ is irreducible right ?

is $N$ the natural numbers?

10:11 PM
Depends on context @happyEddie

sorry, it was meant for zxmkn. what does he mean by $N$?

ah ok, no problem
btw what I wrote above about Galois groups is wrong and I know why (in case someone wants to answer it)

It's not even necessarily Galois

Is it not ?

e.g. $\Bbb Q[X]/(X^3 - 2)$

10:18 PM
Oh right, I meant $k[x_i]$ where the $x_i$ are the roots of a polynomial $P$
Didn't realise that wasn't the same thing

You mean a splitting field of $P$ I think

Yeah, probably

@zxmkn: if $N$ is a topological space then there's the Borel sigma-algebra which is generated by all the open sets

Are there any non trivial cases where the Galois group of a finite Galois extension of a perfect field is the symmetric group ?

@Astyx sure, to take my previous example, the splitting field of $X^3 - 2$ has $S_3$ as its Galois group

10:22 PM
@zxmkn: but maybe try to find if $\mathcal{N}$ is defined somewhere else in the book

Hmm true

@happyEddie No, N here is actually a particular set of $\mathbb{N}_0 \cup \{\infty\}$-valued measures. I was thinking about the Borel sigma-algebra for $\mathbb{N}_0 \cup \{\infty\}$, but I don't know if that has some obvious topology.

That's because $3$ is prime though
It can't happen with $S_n$ where $n$ isn't prime, can it ?

@Astyx You can construct a splitting field for a polynomial by successively factoring your polynomial in each extension into irreducibles, picking one, and then quotienting
so that the degree of the splitting field is bounded above by $\deg f!$

@happyEddie The definition for the measure space $(N, \mathcal{N})$ is clear. But not the sigma-algebra on $\mathbb{N}_0 \cup \{\infty\}$ that the $\mathcal{N}/?$-measurability refers to. Thanks for the response, though!

10:26 PM
@zxmkn: sorry, i misread, you were looking for the sigma algebra on the codomain, not the domain

Right
I think I'm asking whether that bound can always be reached

@zxmkn: would it make sense if the sigma-algebra were the whole power set of $\mathbb{N}_0 \cup \{\infty\}$

In fact according to some other post, most polynomials will have $S_{\deg f}$ as their Galois group

@happyEddie Oh, right. I should check if that works out. If so, then it doesn't matter which sigma-algebra was actually meant, since measurability would then work for any sigma-alg. on the codomain. Thanks!

Hmmm that seems a bit counterintuitive to me for some reason

10:37 PM
No worries :)

@zxmkn: yeah then it would actually make sense that the sigma-algebra on the codomain was not specified

What a generator of the splitting field of $X^3-2$ over $\Bbb Q$ ?

Meh I'll give this some sleep
Thank you again for your help
Bye chat

no dont leave
anyways, i have a question
regarding math, of course

10:49 PM
@zxmkn: or another way to interpret that: measurable with respect to whichever sigma-algebra on the codomain, hopefully it works out

@Astyx given that its Galois group is $S_3$ it doesn't have just one generator

hello?
Is there anyone active in this chat?

Aren't all finite Galois extensions over perfect fields monogeneous ? I was thinking of $i\sqrt{3}\over{2^{1/3}}$

This is called the "Primitive element theorem"

10:55 PM
Right
So if I find any element of order $6$ (as for instance the one I gave above) I know it generates the extension right ?
Anyway II'm gone for good now
Good day/night

how come my silkroad room got deleted that's hate speech

@Astyx if that were the case then you could write down a $\Bbb Q$-linear map $T : \Bbb Q(\sqrt[3]{2}, j) \to \Bbb Q\left( \frac{i\sqrt{3}}{\sqrt[3]{2}}\right)$
where $j^3 = 1$ and $j \notin \Bbb R$

@Astyx actually that's not quite enough because they can be isomorphic as $\Bbb Q$-vector spaces
in fact I think you're right so ignore me
lol

11:35 PM
is it possible to find the solution to:
$x^3+y^3=1$ and $x^2+y^2=1$?
for $x,y \in \Bbb R(0,1)$

Uh..., that's just circles in l2,l3 missing out the axes intersections?

@skullpetrol well it's a website I don't know if it will ever get THAT liberal

@Secret yes i want find the point where they intersect

That's just (0,1), (1,0)
set x=0 or y=0 and the value of the remaining one becomes obvious

$x,y \in (0,1)$
the interval is open

11:42 PM
In that case, you have no intersections

I meant $(1-x)^3+y^3=1$
intersecting $x^2+y^2=1$
the natural density is $\approx .457$