Mathematics

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

Adam

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

Mathphile

@Secret lol

Secret

The History of Mathematics and Its Applications

►

$0$

$n$

$1+1+\cdots=\omega$

$\omega n$

$(\omega^2 + 1)n$

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

Adam

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

Secret

$(\epsilon_0+\cdots + 1)n$

$\phi(n,0)$

$\phi \binom{1}{n}$

$f(n)$

$f(n)+1$

$f(f(n)+1)$

$f(\alpha)=\alpha$

$f(\alpha+1) > \alpha$

Semiclassical

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.)

Secret

hmm...

How likely is given a set of integers [-n,n] and a permutation $\pi \in S_{2n+1}$ that:

For all $a,b,c$: $\pi_i(a)+\pi_j(b)+\pi_k(c)=0$

Semiclassical

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

Secret

sorry typo

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

Semiclassical

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)

Secret

hmm...

0,1,2,3,4,5,6,7,...,n

$\pi ([0...7])$

Semiclassical

6:12 PM

and minus that, with zero only showing up once

Secret

$\pi_a(x)+\pi_b(x)+\pi_c(x)=0$ where $\pi_a(x) = x+a \mod (n+1)$

Thus:

$x+\pi_b(x)+\pi_c(x)=0$ gives:

$x+ (x+b) + (x+c) = 0 \mod (n+1)$

Adam

wow you guys make quwhite a team

Secret

$3x + b + c = 0 \mod (n+1)$

Since it has to hold for all $x \in [0,...,n]$ it means:

Adam

switches to semi classical monitor

damn it

Semiclassical

Okay, yeah, I'm convinced of this construction

I don't have a good name for this, but it works

Secret

6:18 PM

$3x$ has to be some additive inverse of $(b+c)$

What is $3x \mod 13$ for $x \in [0,...,6]$?

Semiclassical

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

Secret

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:

$x + \pi_a(x)+\pi_b(x) = 0 \mod (n+1)$

Semiclassical

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

Secret

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

PM 2Ring

This stuff is very reminiscent of magic squares and latin squares.

Or at least, latin rectangles.

Semiclassical

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}

Secret

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

Semiclassical

bah, still wrong

Secret

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

Semiclassical

okay, that shouold work

@PM2Ring it has that flavor, yeah

Secret

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

Semiclassical

6:28 PM

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

Secret

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)

Semiclassical

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

Secret

yeah

Semiclassical

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

Secret

hmm...

Semiclassical

6:35 PM

yeah, there's something a bit delicate going on.

Secret

0,1,2,3 will fail

Semiclassical

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

Secret

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

fail

Semiclassical

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

Dr. Nourian

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3

Secret

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

Semiclassical

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

Secret

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

Subset 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 { − 7 , − 3 , − 2 , 5 , 8 } {\displaystyle \{-7,-3,-2,5,8\}} , the answer is yes because the subset { − 3...

ÍgjøgnumMeg

7:16 PM

Evening chat

skullpetrol

welcome back

Balarka Sen

7:49 PM

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

happyEddie

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 $I_n$)

Semiclassical

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

Balarka Sen

Yikes

Ryan Unger

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.

ÍgjøgnumMeg

looks like a commutative square

Ryan Unger

@Semiclassical smbc-comics.com/comic/box

Balarka Sen

8:35 PM

@RyanUnger This is true, yes.

$f(\mathfrak{a}M) \subset \mathfrak{a}N$, so it passes to the quotients

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$Mathematics$ Mathematics