Mathematics

12:45 AM

evening all

1:28 AM

@rschwieb. My most recent mathematical explorations have brought me up against algebraic object I figured I would never have to wrestle with...

you'll never guess.

2:23 AM

If you would have only said... this stuff about modules... it's going to be crucial.

2:59 AM

@Mason I did say that, you just made some "yap yap yap" motion with your hand and rolled your eyes.

That sounds like me.

Also I had been introduced to the defn of a group within like 1 month of our first convo on modules.

4:50 AM

[Random]

0q=1,q0=0,q0q=0q=1=q1=q X. q0q=0q=1=q1=1, q1x=qx abs

x0=0,1x=x,0q=1
x0q=0q=1,x0q=x1=1
x1y=xy,x1y=1y=y,xy=y
x1q=xq,x1q=1q=q,xq=q

. x0=0,x1=x,0q=1

x0q=0q=1,x0q=x1=x = ded

. x0=0,x1=1,q0=1
q10=10=0,q10=q0=1 ded

-> 0,1 same = ded

. x0=0,1x=x,0q=1
x0q=0q=1,x0q=x1->1
x1y=1y=y,x1y=xy => xy=y
x1q=xq=q
01q=1q=q,01q=0q=1 ded

. x0=0,1x=x,q0=1

q0q=1q=q,q0q=0q=1 ded

. x0=0,1x=x,0q=1

x0q=0q=1,x0q=x1->1
(theorem xy=y)
01q=1q=q,01q=0q=1 ded

=>

RRR,LLL=ded

RRL,LLR=ded

RLR,LRL=ded

RLL,LRR=ded

RLR,LRL=ded

Division by zero no-go theorem in magmas: Given any Magma $(M,\cdot)$, if there exists an one-sided absorber $z$ and there exists an one-sided identity $e$, then it is not associative

Proof: See above

(also note some of these are not flexible either)

Associator

In abstract algebra, the term associator is used in different ways as a measure of the nonassociativity of an algebraic structure. Associators are commonly studied as triple systems. == Ring theory == For a nonassociative ring or algebra R {\displaystyle R} , the associator is the multilinear map [ ⋅ , ⋅ , ⋅ ] : R × R × R → R {\displaystyle [\cdot ,\cdot ,\cdot ]:R\times R\times R\to R} given by ...

hmm...

(xy)z=x(yz)(x,y,z)

Classes: RRR,RRL,RLR,RLL

Ok this is taking too long, will deal with it later

update, maybe...:

(x,y,z)=1
(x,x,x)=1
(x,x,y)=1,(y,x,x)=1
(x,y,x)=1

Test: (0,1,0)

nah, still too long, need a more flexible way to work with magma associators

sciencedirect.com/science/article/pii/S0747717101905103

5:47 AM

please continue here :-)

Wrong room

$Mathematics$ Rambles

For all my maths rambles that are not even qualified to fit in...

I was going to use that but thought you may get offended.

Q: Distributor? Distributive analog of commutator and associator?

8

Motivation: "the commutator gives an indication of the extent to which a certain binary operation fails to be commutative" (http://en.wikipedia.org/wiki/Commutator). For example (courtesy of wikipedia), in a group, $G$, for each $g,h \in G$, the commutator of $g$ and $h$, $[g,h]$ is defined by, $...

That will be useful

Q: Is it OK to ask for reopen my post?

@MartinSleziak , I posted it on Meta.

0

I had asked How many rectangles can be observed in the grid? but there were already a similar post How many rectangles or triangles. . I want to tell you that my post has 12 times more views than linked post. People prefer my post to understand concept and I received Gold badge for more than 10,...

discussion support review exact-duplicates re-open

6:02 AM

Conjecture: Almost every division by zero magma are non associative, some of these are non flexible and possibly many more are non alternative

If all these nonassociativity rules are all broken, then we can confidently conclude that no nontrivial division by zero algebras exists

Always overkill your enemy when you are given the chance, so they will never ever be able to fight back

lol

PS we previously knew that power associativity is broken for division by zero algebras with the axiom 00=/=0

If you were to answer the question on "why we cannot divide by zero" then give the maximal answer so that nobody will be able to rebut it anymore and anyone who dared to do so will be met with extreme ridicule like those who said perpetual motion machines exists

💥(Answering a question) = Answering a question so well that the study field ceased to exists forever

and that is the pinnacle of perfectionism

Having said that, the completeness of the answer set to all possible questions is an undecidable problem in general

so nature is cunning as always

6:45 AM

@MartinSleziak , can I ask a question?

7:02 AM

Thanks, already done.

Martin Sleziak

7:22 AM

@MithleshUpadhyay As suggested before, it might be better to post in the reopen request thread. (Surely you have noticed that your recent post on meta was closed as a duplicate of this one.)

8:03 AM

@JasperLoy Any news on ANT?

@user91500 News? No, I don't know any news, lol.

@JasperLoy Please downvote this post math.meta.stackexchange.com/questions/28842/…

@user91500 Am I supposed to downvote the question or one of the answers?

@JasperLoy Just the question

@user91500 Done, for your sake, and I also think the new theme doesn't add any value.

8:09 AM

@JasperLoy Thank you so much jasper!

@JasperLoy I strongly suggest you to read this book springer.com/us/book/9781493967933

@user91500 Are you the author?

@JasperLoy No.

@user91500 It does look like something I will enjoy, judging from the description.

@JasperLoy Yes, It's very amazing book, I am reading that currently and trying to solve all the problems!

@user91500 LOL

For reference, the book is Amazing and Aesthetic Aspects of Analysis by Paul Loya.

A: Requests for Reopen & Undeletion Votes (volume 07/2018 - today)

8:38 AM

@MartinSleziak , done.

0

I had asked How many rectangles can be observed in the grid? but there were already a similar post How many rectangles or triangles. . I want to tell you that my post has 12 times more views than linked post. People prefer my post to understand concept and I received Gold badge for more than 10,...

8:57 AM

What branches of mathematics do I need to learn so I can understand the poincare conjecture as shown in this link: ims.cuhk.edu.hk/~ajm/vol10/10_2.pdf

I can only see that linear algebra, geometry, and calculus is used but did I miss anything?

Hi all!

Still thinking about my old group theory problem. I have a new formulation which I hope is more concise, than what I had before. So here it goes:

@Pherdindy Well, "geometry" covers more than you probably think. You will need a decent amount of topology as well as a good knowledge of manifolds.

When you ortho-normalize the basis of a two-dimensional real representation (over $\Bbb R$) of some group $G$ to get say $\hat{v_1},\hat{v_2}$, is there a relation between the rotation transforming $\hat{v_1}$ to $\hat{v_2}$ and any of the group operations?

Empirically (for point groups) I find that for all non-isometric (isometric being tetrahedral, octahedral and icosahedral) groups the axis of this rotation is parallel to the axis belonging to the highest order of rotation in the group elements.

@TobiasKildetoft thanks but man I've never seen so much math in 1 paper in my entire life

@TobiasKildetoft any take on that?

9:10 AM

@Pherdindy Well, it is a very long and detailed paper

@Rudi_Birnbaum Well, there does not seem to be any reason for there to be any group element sending one vector to the other.

@TobiasKildetoft: Yes, the element does not send one to the other but the direction of the rotation axis might be related.

@Rudi_Birnbaum Direction relative to what?

@TobiasKildetoft Those group elements have at least order $3$ while that rotation has order $4$ for example

Relative to the roational axis of the highest order (rotation) group element.

What if there is not a unique such axis?

These are exactly the exceptions, the isometric groups.

9:15 AM

I see

i.e. the groups which cannot distinguish $x,y$ and $z$.

If I want to model the stock market's price action behavior what branch mathematics do you guys think I should study deeper?

@Pherdindy Chaos and fractals

Thanks i'll look into those

Most people say you cannot model the stock market because of its dynamic behavior but I really think there are universal principles behind it that can be mathematically modeled that is true all the time. What do you guys think?

@TobiasKildetoft do you think the question is now clear?

@Pherdindy You will be able to say stuff like it will always be some positive(?) rationals. But I am not sure about any quantitative modeling. I guess also others might have thought about it already

9:25 AM

@Rudi_Birnbaum It is clear in a sense, but perhaps very broad now (what sort of relation would be a satisfactory answer?)

@Rudi_Birnbaum I am so interested to find someone who can predict price action very accurately with models. It's a really controversial issue because plenty of the models were probably created around the wrong idea and gives the reputation that the market cannot be modeled due to its uncertainty

Surely it'll be in probabilities but if someone can hit majority of the calls correctly then it would be cool

@TobiasKildetoft I'd be happy about things like: "In (exactly) non-isometric point groups they are parallel".

I'm also wondering whether the new AI's can come up with some strategies on the stock market

Anyway thanks for the help i'm out have a nice day everyone :D

@Pherdindy If you consider there are single individuals who can by free will deliberately change the stock market value at any instance that is as good as a proof that there can be no model which considers only past and current input ...

Yes I get what you mean but if you have a portfolio or diversity of stocks

Then you can minimize the risk that cannot be avoided such as those I guess and will generally follow the general behavior?

9:31 AM

@TobiasKildetoft "parallel" $\to$ "parallel/antiparallel"

Anyway it's really something i'll just have to work on and see lol

Thanks

I just feel that back in the time there was no computers

And it's really speculative but since we have real time data now it's a whole different story

@Pherdindy What do you mean by "now"? Nothing major seems to have changed in this context in the past 10 or even 20 years

@Pherdindy It somehow will boil down on the ability to model mass psychology and stuff. But maybe there are interdependencies which average out those, who knows...

@Tobias During my parent's times stock positions were made through the phone

The only way they can make money in a stock was to know about the fundamentals behind the market and invest in a good company

@Pherdindy That was way more than 20 years ago.

9:38 AM

Yes that's a long time :P but the stock brokers in our country just give us charts

And indicators

But nothing holistically

Based on all the combined data of all stocks

Technical analysis then and now is still being used

@Pherdindy game theory could help as well, just a thought. I'd maybe search for a baby model and study this ...

Yes I might look into the chaos and fractals theories as well

As well as machine learning/deep learning

Since I can connect an API to my python application to try model things

I'm sure big institutions are already using lots of good models but maybe I can get a slice of the pie

@TobiasKildetoft do you know a proof of STFGMPID that isn't long?

I mean, I'd rather learn a lot more tools than to learn a long proof that wouldn't tell me much

what is STFGMPID

8

structure theorem of finitely generated module over principal ideal domain

9:44 AM

.... what kind of abbreviation is that

a kind that you don't understand

too hard for me

so do you know such a proof?

no

i dont even rmb how it's proved lmao

that's exactly the problem

the proof is long and doesn't give us much insight so you forget the proof

whereas the proof of classification of finite fields is "oh the field of order $q^n$ must be the splitting field of $X^{q^n}-X$ over $\Bbb F_q$"

10:03 AM

$0 = 9 - 9 = \pi^2 - 9 = g - 9 = 10 - 9 = 1$

\o @BalarkaSen

I give it a new try math.stackexchange.com/questions/2898191/…

1:26 PM

hello, someone help me on this: math.stackexchange.com/questions/2897675/…

@LeakyNun hello

all linear transformations of finite dimensional vector spaces are continuous

i ha

i have to mrove it, using norms , one suggest to me to find the tangent but i don't know how to continue

i found $\sqrt{2} x +y =\pm\frac{123}{11}$ is tangent to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$

how to find that $|u(x,y)|\leq c ||(x,y)||$

@LeakyNun

user131753

1:43 PM

in Homotopy Theory, 57 secs ago, by user 170039

Let $\mathbf{A},\mathbf{B}$ be two categories and $\mathscr{F}:\mathbf{A}\to\mathbf{B}$ be a non-full embedding such that if $X$ and $Y$ be two $\mathbf{A}$-objects and $f:\mathscr{F}(X)\to \mathscr{F}(Y)$ be a $\mathbf{B}$-isomorphism between them there there exists an $\mathbf{A}$-object $Y^\ast$ and an $\mathbf{A}$-isomorphism $g:X\to Y^\ast$ such that $\mathscr{F}(g)=f$. Does this type of embedding has a name?

Sam

I see a lot of discussion in text books of using interaction terms in linear models... does it make sense to trial interaction terms of inputs in a non linear model?

you have an idea@LeakyNun

@PolineSandra tell me what you need to prove

i need to prove that u is continuous

and for that what would you prove

1:54 PM

with the norm ||(x,y)||=\sqrt{\frac{x^2}{16}+\frac{y^2}{9}}$

i must prove that there exists a posive constante c such that $||u(x,y)||\leq c ||(x,y)||$

can you unfold that?

what is "unfold" ?

expand the definitions

i must find that $||u||= \sup_{||(x,y)||=1} ||u(x,y)||\leq c$

where $ u(x,y)=\sqrt{2} x+y$

@Mason What did you come across that involved modules?

2:03 PM

7 mins ago, by Poline Sandra

i must prove that there exists a posive constante c such that $||u(x,y)||\leq c ||(x,y)||$

@PolineSandra no, start from here, expand the norms

$|u(x,y)|=|\sqrt{2} x+y| \leq c ||(x,y)||=c\sqrt{\frac{x^2}{16}+\frac{y^2}{9}}$

expand u also

@LeakyNun Do you follow the hint in the solution on Poline's question? math.stackexchange.com/a/2897693/29335 (Because I don't :( )

oh right that's useful

8 mins ago, by Poline Sandra

i must find that $||u||= \sup_{||(x,y)||=1} ||u(x,y)||\leq c$

yeah this exists because the domain is compact

@LeakyNun How does maximizing that function contribute to proving continuity?

2:08 PM

it's well-known that linear + bounded -> conitnuity

actually, linear + bounded -> lipschitz -> continuity

maximizing the function contributes to proving boundedness?

eh, sure

OK, I guess I see the connection to boundedness through the Lipschitz condition. Thanks

I forgot the user already proved linearity and that it was in play

It just struck me that the ideas were related to this stuff I've been reading lately in the calculus of variations leading up to the baby version of Noether's theorem on invariances

o..o

interesting

2:11 PM

maybe you can try to prove that $(x,y) \mapsto x$ is continuous

But looks like the connection was just illusory. That's what you get for thinking about something a lot.

then do the same for $y$ and then argue with composition of continuous functions

@mercio i must use the norms

yes, use the norms to show that $(x,y) \mapsto x$ is continuous

??? i must prove that u is continuous

2:16 PM

I miss being able to take college courses. I can't believe how luxurious it was, in retrospect, to think about new stuff all day.

but $u$ is complicated

Now it's a struggle to find the time :)

I guess technically there is an upside to making more than a pittance a year, though.

one suggest me to find c such that $\sqrt{2} x+y=c$ such that it is tangent to the ellipse but how to continue?

why struggle to find the best possible $c$ instead of showing that $|u(x,y)| < 100000000 ||(x,y)||$ ?

with very rough approximations

for example maybe you could show that $|x| < 500000 ||(x,y)||$ ?

(which is the same as showing that $(x,y) \mapsto x$ is continuous at the origin)

say you managed to show that $|x| < 500000 ||(x,y)||$ and $|y| < 800000 ||(x,y)||$, can you then deduce that $|\sqrt 2 x + y| < 100000000000 ||(x,y)||$ ?

Evinda

Hello @LeakyNun!!! We consider the complete bipartite graph $K_{6,10}$ and we symbolize with $\overline{K_{6,10}}$ its complement. I want to calculate the number of edges of $\overline{K_{6,10}}$.
I have thought the following:

In $\overline{K_{6,10}}$ each vertex of each side is connected to each other of this side but with none of the other side. So the number of edges is $\frac{6 \cdot 5}{2}+\frac{10 \cdot 9}{2}=60$. Am I right?

2:29 PM

@rschwieb. The most recent question I asked here is more algebra-y but it's related to the question I asked before that. In my notation $A_s$ is module (but I am not sure it's not just a vector space). $A_s$ has scalars of the form $p/q^s$ so for $s=1$ that's just the rationals so it's a vector space.