Mathematics - 2018-08-29 (page 1 of 2)

Nick

00:08

Good morning!

rschwieb

00:45

evening all

Mason

01:28

@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.

Mason

02:23

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

rschwieb

02:59

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

Mason

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.

Secret

04:50

[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

user 2646

05:47

please continue here :-)

Secret

Wrong room

$Mathematics$ Rambles

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

user 2646

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

Secret

8

Q: Distributor? Distributive analog of commutator and associator?

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

Mithlesh Upadhyay

@MartinSleziak , I posted it on Meta.

0

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

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

Secret

06:02

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

user 2646

lol

Secret

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

Mithlesh Upadhyay

06:45

@MartinSleziak , can I ask a question?

Mithlesh Upadhyay

07:02

Thanks, already done.

Martin Sleziak

07:22

@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.)

user91500

08:03

@JasperLoy Any news on ANT?

Jasper Loy

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

user91500

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

Jasper Loy

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

user91500

@JasperLoy Just the question

Jasper Loy

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

user91500

08:09

@JasperLoy Thank you so much jasper!

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

Jasper Loy

@user91500 Are you the author?

user91500

@JasperLoy No.

Jasper Loy

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

user91500

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

Jasper Loy

@user91500 LOL

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

Mithlesh Upadhyay

08:38

@MartinSleziak , done.

0

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

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,...

Pherdindy

08:57

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?

Rudi_Birnbaum

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:

Tobias Kildetoft

@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.

Rudi_Birnbaum

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.

Pherdindy

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

Rudi_Birnbaum

@TobiasKildetoft any take on that?

Tobias Kildetoft

09:10

@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.

Rudi_Birnbaum

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

Tobias Kildetoft

@Rudi_Birnbaum Direction relative to what?

Rudi_Birnbaum

@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.

Tobias Kildetoft

What if there is not a unique such axis?

Rudi_Birnbaum

These are exactly the exceptions, the isometric groups.

Tobias Kildetoft

09:15

I see

Rudi_Birnbaum

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

Pherdindy

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

Tobias Kildetoft

@Pherdindy Chaos and fractals

Pherdindy

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?

Rudi_Birnbaum

@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

Tobias Kildetoft

09:25

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

Pherdindy

@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

Rudi_Birnbaum

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

Pherdindy

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

Rudi_Birnbaum

@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 ...

Pherdindy

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?

Rudi_Birnbaum

09:31

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

Pherdindy

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

Tobias Kildetoft

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

Rudi_Birnbaum

@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...

Pherdindy

@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

Tobias Kildetoft

@Pherdindy That was way more than 20 years ago.

Pherdindy

09:38

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

Rudi_Birnbaum

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

Pherdindy

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

Leaky Nun

@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

loch

what is STFGMPID

8

Leaky Nun

structure theorem of finitely generated module over principal ideal domain

loch

09:44

.... what kind of abbreviation is that

Leaky Nun

a kind that you don't understand

loch

too hard for me

Leaky Nun

so do you know such a proof?

loch

no

i dont even rmb how it's proved lmao

Leaky Nun

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 $\Bbb F_q$"

Leaky Nun

10:03

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

user 2646

\o @BalarkaSen

Rudi_Birnbaum

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

Poline Sandra

13:26

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

@LeakyNun hello

Leaky Nun

all linear transformations of finite dimensional vector spaces are continuous

Poline Sandra

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

13:43

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?

Poline Sandra

you have an idea@LeakyNun

Leaky Nun

@PolineSandra tell me what you need to prove

Poline Sandra

i need to prove that u is continuous

Leaky Nun

and for that what would you prove

Poline Sandra

13:54

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 $||u(x,y)||\leq c ||(x,y)||$

Leaky Nun

can you unfold that?

Poline Sandra

what is "unfold" ?

Leaky Nun

expand the definitions

Poline Sandra

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

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

rschwieb

@Mason What did you come across that involved modules?

Leaky Nun

14:03

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

Poline Sandra

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

Leaky Nun

expand u also

rschwieb

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

Leaky Nun

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

rschwieb

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

Leaky Nun

14:08

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

actually, linear + bounded -> lipschitz -> continuity

rschwieb

maximizing the function contributes to proving boundedness?

Leaky Nun

eh, sure

rschwieb

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

mercio

o..o

Leaky Nun

interesting

mercio

14:11

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

rschwieb

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

mercio

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

Poline Sandra

@mercio i must use the norms

mercio

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

Poline Sandra

??? i must prove that u is continuous

rschwieb

14:16

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.

mercio

but $u$ is complicated

rschwieb

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.

Poline Sandra

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?

mercio

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?

Mason

14:29

@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.

rschwieb

hmmm

Semiclassical

@Evinda one way to approach this in a constructive way is to consider what the adjacency matrix of each case would be

Evinda

Isn't it like that, that each vertex of each side is connected to each of the same side? @Semiclassical

Semiclassical

what?

I mean, your logic may be right

I'm just giving an alternative approach

for $K_{6,10}$, the adjacency matrix would be of the form $$\begin{pmatrix} 0_{6\times 6} & 1_{6\times 10} \\ 1_{10\times 6}& 0_{10\times 10}\end{pmatrix}$$

Poline Sandra

i don't understand what you say @mercio

Semiclassical

14:35

where $0_{n\times n}$ is a matrix of 0's and $1_{m\times n}$ is a matrix of ones

So that contains 6*10=60 edges.

mercio

do the numbers I chose confuse you ?

but the larger they are, the easier it should be to prove

right ?

Semiclassical

By contrast, the complete graph on 10+6=16 vertices would contain 16*15/2 =120 edges

so the graph complement $\overline{K_{6,10}}$ should contain 120-60=60 edges as well

which, happily, matches your answer.

Evinda

I see, thanks a lot :) @Semiclassical

Semiclassical

the nice thing is that we can check this logic beyond the case of 6 and 10: the complete bipartite graph $K_{n,m}$ would by this logic contain $nm$ edges. the complete graph $K_{n+m}$ would contain $(n+m)(n+m-1)/2$ edges. So the graph complement should contain $(n+m)(n+m-1)/2-nm$ edges.

on the other hand, your logic would suggest that the number of edges in $\overline{K_{n,m}}$ should be $n(n-1)/2+m(m-1)/2$

so the two answers will agree if $(n+m)(n+m-1)/2-nm = n(n-1)/2+m(m-1)/2$

which, happily, is true

So your logic is just as sound as mine in all cases.

Evinda

I see @Semiclassical

Semiclassical

14:51

@Evinda the nice thing about your line of thinking is that it reflects the fact that $\overline{K_{6,10}}$ is nothing more than the disjoint union of $K_6$ and $K_{10}$

i.e. you draw it by drawing edges between the first 6 vertices and edges between the other 10 vertices

Poline Sandra

@mercio i don't understand how to apply your idea

mercio

my idea is divide and conquer

you prove a bunch of easier things

to prove the complicated thing

Poline Sandra

but how i can maximize |x|?

mercio

where have i ever said i wanted you to maximize |x|

also, finding the exact maximum of a set is often harder than showing that a set is bounded

you do not need to get the best possible $c$

Evinda

15:10

Yes, right @Semiclassical

Mary Star

Hello!!

At the ground floor of a building 6 men and 5 women get into the elevator. The elevator gets up and stops at each floor and stops at the floor 8. In how many ways can the people get out if all men are considered identical and all women are considered identical and it is possible that at one floor no one gets out.

We have 11 people. So are there $8^{11}$ ways?

Leaky Nun

all women are not identical

Secret

Geeeeeeeeeoooooommmmmeeettrryyyy

Braaaaaaaaasaaaaaains

Leaky Nun

are you ok

Mary Star

Is my idea wrong? @LeakyNun

Secret

15:13

sleepy...

Leaky Nun

i don't know

Evinda

@Semiclassical I have also an other question. If $A=\{ 0, \pm 1, \pm 2, \dots, \pm 10\}$, how can we calculate the number of even functions $f: A \to A$ ?

Semiclassical

well, how many options are there for $f(1)$?

Leaky Nun

what is an even function?

oh nvm

Semiclassical

(also, it may be easier to work with the more general case $f:A\to B$ where $B$ is some arbitrary codomain.)

Evinda

15:17

An even function satisfies $f(-x)=f(x)$ @LeakyNun

Leaky Nun

codomain @Semiclassical

Evinda

@Semiclassical We want f(1) to be equal to f(-1)

Semiclassical

right, careless of me

@Evinda sure. that constrains $f(-1)=f(1)$. but is there any further constraint on $f(1)$?

Evinda

It can take 21 possible values right? @Semiclassical

No 21

Right?

Semiclassical

yep, all 21 values are allowoed

of course, once you pick $f(1)$, then $f(-1)$ is determined

How about $f(2)$ through $f(10)$ ?

Poline Sandra

15:21

@mercio i don't know how to start?

mercio

what about writing what $||(x,y)||$ is ?

Evinda

f(1) has 21 possible values, f(2) has 20 possible values, .... , f(10) has 11 possible values, and so on, right? @Semiclassical

Leaky Nun

no

Semiclassical

Why does knowing $f(1)$ constrain $f(2)$?

All we're asking is that $f$ be a function. We're not asking that it be a bijection.

Evinda

Ah it doesn't

Semiclassical

15:23

Right.

(It can't be a bijection in any case, since $f(x)=f(-x)$.)

Evinda

So, there are $21^{11}$ even functions, because f takes 21 possible values at 0,..., 10

Semiclassical

Right.

Mary Star

Does someone of you have an idea about my question: math.stackexchange.com/questions/2898435/…

?

Semiclassical

More generally, if you had $f:A\to B$, all that would change is that you'd have $(\text{card }B)^{11}$ even functions

Poline Sandra

@mercio $||(x,y)||= \sqrt{\frac{x^2}{16}+\frac{y^2}{9}}$

mercio

15:26

so how about proving $|x| < 5000000 \sqrt {x^2/16 + y^2/9}$ ?

in fact, would you know any expression that only depends on $x$ and not on $y$, that is less than $\sqrt {x^2/16 + y^2/9}$ ?

Semiclassical

@Evinda for a variant of this, you could count the number of odd functions. to bump up the difficulty further, limit it to odd bijections.

Poline Sandra

$x/4$ ? @mercio

mercio

yes, for example, so now can you prove that $|x| < 5000000x/4$ ?

(or is there a small problem with this statement ?)

Poline Sandra

i really don 't know

mercio

what about $|x| < 5000000|x|/4$ would you be able to prove that ?

Evinda

15:37

Ok @Semiclassical

From a set of $12$ students $s_1, s_2, \dots, s_{12}$, we want to create a commitee of seven people that has a president and 6 participants at which either $s_1$ is president or $s_2$ is a participant or both. With how many ways can this happen?

So, $s_1$ is either president, or participant or nothing. So there are three possibilities for $s_1$. For $s_2$ there are two possibilities, that he is participant or nothing.

So are there $3 \cdot 2 \cdot \binom{10}{4}$ ways ?

Semiclassical

That way of counting would allow $s_1$ to be nothing and $s_2$ to be nothing.

But that's not an option, so this counting can't be correct.

Evinda

Ok, but how can we calculate it then? @Semiclassical

Semiclassical

Well, what's the complement of this case?

That is: Consider the the ways of creating a committee of seven people, including a president, from 12 people, without restriction. Which of these combinations are we not allowing?

Evinda

The number of ways of creating a committee of seven people, including a president, from 12 people, without restriction is $\binom{12}{7}$.

Mary Star

We have the bipartite graph $G=K_{5,9}$. We construct a new graph $G'$ by adding a new vertex u that is connected with each vertex of G. Then $G'$ has an euler circuit, because every vertex has an even degree and G' is connected, right?

quallenjäger

15:44

Is it true that the exponential of a finite dimensional Lie algebra is always a lie group?``

Semiclassical

@Evinda Right. So what is our criteria for rejecting one of these committee selections?

Evinda

It is not allowed that $s_1$ is not a president and $s_2$ is not a participant.

Perturbative

Say I have a group $G$ and a normal subgroup $H$, let $K = G/K$ and pick $x \in K$. Then $x = g+ H$ for some $g \in G$. Is the equivalence class of $x$, $[x] = \{y \in G \ | y+ H = g+H\}$?

Semiclassical

Right. So what we want for the moment is the selections for which $s_1$ isn't a president and $s_2$ is not a participant.

Can you see how to count that?

Perturbative

I'm just guessing that's the definition of an equivalence class in the quotient group since the textbook I'm using calls something an equivalence class without defining it

Evinda

15:49

$s_1$ is not a president if one of the rest $s_2, \dots, s_{12}$ is the president.

So there are 11 ways, right?

$s_2$ is not a participant if from the rest we pick 6 others, right? So there are again 11 ways, or not? @Semiclassical

Semiclassical

On the first, I agree. But I don't follow you on the second.

Did you mean 11 choose 6?

mercio

@Perturbative $x$ is already an equivalence class, it is the equivalence class of $g$

Evinda

$s_2$ is not a participant, if we pick 6 others from $\{ s_1, s_3, \dots, s_{12}\}$.

So I thought $\binom{11}{6}$, am I wrong? @Semiclassical

Semiclassical

I think so, but you said "So there are again 11 ways"

mercio

$x = [g] = \{g+h, h \in H\} = \{y \in G \mid y+H = g+H\}$

Evinda

15:53

So there are $\binom{12}{7}-11 \cdot \binom{11}{6}$ ways, right? @Semiclassical

Perturbative

Ohh okay thanks! @mercio

Semiclassical

I'm trying to confirm whether 11-choose-6 is right or if there's an error I'm making

I feel like there may be an issue there. The concern I'm seeing is that either s2 is the president or s2 isn't included.

If s2 is the president (1 way), then the six participants can be any of the 11 remaining people (11-choose-6 ways).

If s2 is not the president (and s1 can't be president either) then there are 10 ways to choose the president; call them s3. Then the six participants can be anyone except s2 or s3, so that's 10-choose-6

so that'll be $1\binom{11}{6}+10\binom{10}{6}\neq 11\binom{11}{6}$

So by that logic it'd be $\binom{12}{7}-\binom{11}{6}-10\binom{10}{6}$

There's a way to check this, in any case: One can do the initial counting instead by inclusion-exclusion

# of ways to have s1 preside OR s2 participate = (# of ways to have s1 preside)+(# of ways to have s2 participate)-(# of ways to have s1 president AND s2 participate)

there are $1\binom{11}{6}$ ways to have s1 preside. there are $(1)\binom{11}{5}(6)$ ways to have s2 participate. there are $(1)(1)\binom{10}{5}$ ways to have s1 preside AND s2 participate

so it should be $\binom{11}{6}+6\binom{11}{5}-\binom{10}{5}$

@Evinda oh. Big problem with this possible answer: $\binom{12}{7}=792,$ while $11\binom{11}{6}=11(462)=5082$

So if that was the answer you'd have 792-5082 ways, which is negative

and, uh, no way to have negative counts :S

Evinda

16:11

:/

Semiclassical

Ah, the fix there is just that there's 12 ways to pick the president, and then 11-choose-6 ways to pick the remaining 6 people. So that's $12\binom{11}{6}=12(462)=5544$ ways

Math geek

-1

Q: Doubt on the definition given in the textbook

In the second definition, I don't understand completely. I know that edge cut is the set of edges which disconnect the graph if we remove it from the edge set of the graph. How these two definitions are equivalent. Is the definition complete?

graph-theory

how does these two definitions are equivalent

Semiclassical

ahah, and it's true that $\binom{11}{6}+6\binom{11}{5}-\binom{10}{5} = 12\binom{11}{6}-\binom{11}{6}-10\binom{10}{6}=2982$

(as in, I plugged those numbers in and verified it)

Drew Brady

Anyone familiar with Frank Almgren's work, like projecteuclid.org/download/pdf_1/euclid.bams/1183530286

Math geek

16:32

-2

Q: Equivalence of the definition

Definition in the textbook definition-edge cut is the set of edges which disconnect the graph if we remove it from the edge set of the graph. How does these two definitions are equivalent?

graph-theory

i have corrected the errors in the question. i made it precise. please help me.

KeJie

16:46

Hello everyone.

Could someone take a look at my newest question? Only need 3 more reputation for the next privilege.

0

Q: Show that $10^n \gt 6n^2+n$

Show for all $n \in \mathbb{N}$ $(n\geq1):$ $10^n \gt 6n^2+n$ My solution: Base case: For $n=1$ $10^1 \gt 6 \cdot 1^2+1$ Inductive hypothesis: $10^n \gt 6n^2+n \Rightarrow 10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$ Inductive step: $10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$ $\Rightarrow$ $10^{n+1} \gt 6(n...

analysis inequality induction

Abdullah UYU

Hi, I have a sequence $x_{n+2}=4x_{n+1}-5x_n$ with conditions $x_1=8,\,x_2=0$. I know not how to find general solutions to these. I tried a method explained in en.wikipedia.org/wiki/… and found $x_n=\frac{-20}{3i-1}(2-i)^n+\frac{20}{3i-1}(2+i)^n$.

Obviously (I suppose) this is not the solution. I don't know any of this classifications and I think I failed to select the appropriate one. @Semiclassical do you know how to solve these?

mercio

how do you know it is not the solution ?

Abdullah UYU

17:06

$x_1$ yields $12-4i$ which isn't $8$; if I didn't make a mistake.

mercio

maybe one of the $3i-1$ should be a $3i+1$

didn't you choose the two coefficients so that $x_1$ would be $8$ and $x_2$ would be $0$ ?

if $x_1$ is not $8$ that means you made a mistake while solving for the coefficients

Abdullah UYU

Probably, I'm going to redo it.

Evinda

17:54

@Semiclassical I haven't understood... What do we change? :/

Jasper Loy

@Evinda Hi Evinda. Aha!

Evinda

Hi! Long time no see @JasperLoy

Jasper Loy

Wow, your avatar is gold! That is so rare...

Evinda

:)

Jasper Loy

Mine is completely blue. That is rare too...

Evinda

17:57

Yes, that's true :) @JasperLoy Tell me news about you ;)

Jasper Loy

@Evinda My news is the same. I am still trying to get well from my mental illness. I don't know how long more I will take. But I am inspired by John Nash.

Evinda

Aha @JasperLoy

Jasper Loy

@Evinda Tell me news about you now.

Evinda

I have finished with my studies... I have no other news @JasperLoy

Jasper Loy

Good.

I am still looking for a nice book on axiomatic geometry. Not that many books in this topic. Of course, there is John Lee's Axiomatic Geometry. But I also wanna find one that covers the three-dimensional case in some depth, because it is not so trivial to generalise everything from the two-dimensional case.

Another very good book I have seen is Edwin Moise's Elementary Geometry from an Advanced Standpoint. Of course, Moise also has another good book called Geometric Topology in 2 and 3 dimensions which includes a proof of the triangulation of surfaces theorem, something extremely rare in the literature.

Semiclassical

18:12

@Evinda yeah, my statements above were a bit confusing

first: If you count the number of committees without restrictions, you need to pick the president from the 12 people and then pick the six participants from the 11 people remaining.

Rudi_Birnbaum

Why did they vote for the nasty summer time? I want my winter time back!!!

Semiclassical

@JasperLoy my personal frustration in that regard is people not realizing how long recovery can take, and what it even means to recover.

Jasper Loy

@Semiclassical Yes. Indeed, unless we have thought about it for our own case, we wouldn't even know what that means for ourselves.

Semiclassical

Yeah.

Jasper Loy

Anyway, about academia, I don't know whether one will really be happy writing papers and teaching classes for the rest of their lives, with the danger of publishing or perishing.

Semiclassical

18:17

Right.

Jasper Loy

But you know, at least you are a doctor now!

Semiclassical

The frustrating thing for me is that I really do enjoy academic activities, but that's very different from liking academia institutions/bureaucracy

@JasperLoy Heh. PhD, heal thyself?

The PhD next to my name doesn't feel real yet

Jasper Loy

Dr. Semiclassical, next Bond movie, 2019 (First Bond movie was Dr. No, 1962)

Semiclassical

lol

The big mental roadblock for me is my resume/CV

Jasper Loy

I don't like how some journalists write of Bond as being sexist. I think the word sexism is overused. Bond doesn't discriminate against women.

@Semiclassical You just need to write the truth in a positive manner, that is all.

If you need help I will proofread it for you, lol.

$Mathematics$ Rambles

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$Mathematics$ Mathematics