Mathematics - 2017-08-30 (page 4 of 8)

Secret

10:00 AM

but how I can get rid of the g s?

Alessandro Codenotti

@SteamyRoot why?

Leaky Nun

@AlessandroCodenotti how do you show that F(2,6) is infinite?

Alessandro Codenotti

I just learnt what the Fibonacci groups are, no idea

Leaky Nun

but you aren't new to presentations right

SteamyRoot

Let $\varphi$ be an automorphism of $F(2,6)$. Define an equivalence relation (twisted conjugacy) on $F(2,6)$ by $x \sim y \iff \exists z: x = zy\varphi(z)^{-1}$. Then there are either 2 or infinitely many equivalence classes

Leaky Nun

10:01 AM

@SteamyRoot :o

so fast

SteamyRoot

@AlessandroCodenotti It's isomorphic to the (3D) Hantzsche-Wendt group, which is one of the most nicely behaving crystallographic groups, which I do research on.

Leaky Nun

:o you do research

on something i just told u to look up in google

SteamyRoot

I'm a PhD student, so... yeah.

Leaky Nun

sorry :p

Alessandro Codenotti

@SteamyRoot Ah, I see, nice

SteamyRoot

10:03 AM

Well, I don't really know anything about the other Fibonacci groups, except which ones are finite

Leaky Nun

@SteamyRoot ugh, the latter is highly googlable

Secret

Showing gaagaagagaagaagagaagagaagaagaggaag=x^29 is some variation of the word problem, right?

Leaky Nun

is that mess the identity?

Secret

it is, since (fedcbag)a=a

and that mess is what happens when trying to unravel fedcbag into as many a s as possible

Leaky Nun

a note on symmetry might or might not be helpful

since a->b->...->a is an isomorphism

Secret

10:06 AM

a(bcdefg)=a also

so bcdefg is also a representation of the identity

Leaky Nun

hmm?

Secret

uh wait, mistake

Leaky Nun

summoning the phd @SteamyRoot :p

SteamyRoot

I don't have a PhD (yet) :P

Secret

a(bdfaceg)=a

(Can be easily seen by chasing that diagram above until you get back to a)

Leaky Nun

10:08 AM

nice pattern

Secret

so we know that bdfaceg=fedcbag=1

Leaky Nun

right

Secret

but without figuring out what the inverses are, we cannot go further.

hmm...

Using the diagram: g(aceg)=a

now checking aceg unravelled:

aceg=aabcdg=aagaabbcg=aagaagagaabg=aagaagagaagag

g (aagaagagaaga) agagaagagaagaagaggaag, that's off by a g, hmm...

fg=a
f=de=bccd=gaababbc=gaagaagagaab=gaagaagagaaga

gaagaagagaagaagagaagagaagaagaggaag=1
aaagaagagaagaagagaagagaagaagaggaag=f
aaagaagagaagaagagaagagaagaagaggaaa=gaagaagagaagaaagaagagaagag
aaagaagagaagaagagaagagaagaagaggaaaaagaagagaagag=gaagaagagaagaaagaagagaagaa
aaagaagagaagaagagaagagaagaagaggaaaaagaagagaagaa=gaagaagagaagaaagaagagaagaaaagaagagaagag
gaagaagagaagaaaagaagagaagaagagaagagaagaagaggaaaaagaagagaagaa=aaagaagagaagaaagaagagaagaaaagaagagaagag
gaagaagagaagaaaagaagagaagaagagaagagaagaagaggaaaaagaagagaagaaaagaagagaagag=aaagaagagaagaaagaagagaagaaaagaagagaagaa

user84215

10:23 AM

Questions about learning math are off-tpoic?

Secret

ga=b
b(cegbdf)=g
Therefore $a^-1=cegbdf$

ab=c
c(dfaceg)=a
Therefore $b^-1=dfaceg$

bc=d
d(egbdfa)=b
Therefore $d^-1=egbdfa$

There's a pattern of sorts among the inverses...

Leaky Nun

@Secret as I said, transposing the symbols is an isomorphism

Secret

$\phi (abcdefg)=bcdefga$ is an isomorphism?

Leaky Nun

yes

because of how our group is defined

Kasmir Khaan

Hi @LeakyNun do you have double accounts? :D

Leaky Nun

10:35 AM

@KasmirKhaan I have not used the second one since April and I won't use it again

I never use both simultaneously

Kasmir Khaan

@LeakyNun hmm you are no the guy who helped me yesterday ? :o

he had the same name as urs

Secret

fedcbag=1
$\phi (fedcbag)=\phi(1) \implies edcbagf=1$
hmm, that will allow us to get all expressions of the inverses

Leaky Nun

I changed my profile picture ._.

Kasmir Khaan

Leaky Nun

33.5k 2 74 220

oh

Haha so you are the same guy :D

Leaky Nun

@Secret more like phi(a)=b, phi(b)=c, ... and then you are correct

@KasmirKhaan yes

Kasmir Khaan

10:37 AM

@LeakyNun Nice nice :D , I got some questions but need to get them orginazed =p

mostly about homomorphism and iso , epimorphism and stuff

Leaky Nun

@KasmirKhaan have you done the exercise yesterday?

googles epimorphism

Kasmir Khaan

Yes but did not find simpler way

Leaky Nun

ok, go on

Kasmir Khaan

I did it that way you told me a overkill =p

Leaky Nun

do you want the answer?

Kasmir Khaan

10:39 AM

but then I gave different solution u said there is better way

Hmm not sure really

xy=zx =e

i need to prove that y=z

Leaky Nun

@KasmirKhaan I'm still the same guy.

Kasmir Khaan

But the thing is that I think that xy=e implies that yx=e

Haha I know but you helped others for sure so =p

So i found the most easy way is that

yx=zx

and cancel from the right

Leaky Nun

My memory isn't that bad.

Kasmir Khaan

I know but I allways have to assume worst situation so its allways clear =p

Now I know I can just continue from where we left =p

Leaky Nun

continue then.

Kasmir Khaan

10:42 AM

Thats the first solution

xy=e ==> yx =e

so yx=e=zx

right cancellation law implies that y=z

Leaky Nun

I already said that's overkill

Kasmir Khaan

all the axioms of the group are obyed now =p

Yes I know

The other method was

hmm let me think a bit

Leaky Nun

do you recall the proof of xy=e => yx=e?

Kasmir Khaan

yes

Leaky Nun

xy=e; yxy=ye=y; yxyy'=yy'=e

Secret

10:45 AM

fedcbag
edcbagf
dcbagfe
cbagfed
bagfedc
agfedcb
gfedcba

hmm....

Kasmir Khaan

@LeakyNun hmm yes nice proof :D

Leaky Nun

@Secret heh

Secret

but that does not seemed to help simplify any of them into just one element powered

Leaky Nun

right

Kasmir Khaan

@LeakyNun hmm what is the other proof of xy=e=zx

am gonna post the nuke way to my teacher =p

because that was mine, i only curious what ur method is

Secret

10:51 AM

Btw, I have a new interpretation of homomorphisms: It is the set of symmetries of an algebraic structure such that it preserves its structure.

Just as the elements of the 3x3 rotation matrices acting on a cube rotate the cube around and can result it to remain unchanged, homorphisms are transformations that acts on the whole algebraic structure such that its structure remains unchanged

Mary Star

We have that A is not invertible, so there is a nonzero x such that Ax=0, where x has unit norm. How does it follow that I-A has an eigenvalue of 1?

Secret

(I-A)x=Ix-Ax=x-0=x, hence eigenvalue is one by definition

Kasmir Khaan

@LeakyNun there exist an element e in G such that xe=x for all x in G
for every x in G there exist an element z in G such that zx =e
show with example that (G,*) does not have to be a group .
show with example means I give them a group from the groups we leaned?

NV-US

11:07 AM

hello

anyone herer?

here?

Leaky Nun

@KasmirKhaan so you want the answer?

@NV-US just ask

Kasmir Khaan

@LeakyNun No ! :D just what do they mean by "show by example"

Leaky Nun

@KasmirKhaan no... it means give them a non-group that satisfies the two criteria

Kasmir Khaan

Hmm

Leaky Nun

@KasmirKhaan no I mean for the xy=zx=e thing

NV-US

11:11 AM

Let V be a finite dimensional vector space and T : V -> V be a linear
transformation . If there exists a basis B of V consisting of eigenvectors
of T, then T is diagonalizable.

Kasmir Khaan

@LeakyNun Oh yes I proved it in two ways, but i wanted to see your proof =p

NV-US

how to prove this?

Leaky Nun

@KasmirKhaan y = ey = (zx)y = z(xy) = ze = z

NV-US

help please

Leaky Nun

@NV-US have you done any diagonalization at all?

NV-US

11:12 AM

yes

Leaky Nun

@NV-US we aren't entitled to answer your questions

NV-US

this was given as a theorom in my book

you said just ask

Leaky Nun

@NV-US then just use the same method as you diagonalize

i said just ask, if anyone want to answer then they will

Kasmir Khaan

@LeakyNun that is much better proofs =p uses exactly what we were given , I took alot from theorems on part a) of the question

Leaky Nun

@NV-US A[v1 v2 ... vn] = [L1v1 L2v2 ... Lnvn]

Kasmir Khaan

11:16 AM

@LeakyNun Il be looking for an example >< so brb =p

Leaky Nun

= [v1 v2 ... vn] diag(L1,L2,...,Ln)

@KasmirKhaan ok

Evinda

Hello

Leaky Nun

@Evinda hi

Evinda

How can I check if the natural cubic spline that is defined in respect to the partition points {0,1,2,3} and gets the values s(0)=1, s(1)=1, s(2)=0, s(3)=10 is identitical with the following function f?

$f(x)=\left\{\begin{matrix}
1+x-x^3 &, 0 \leq x\leq 1 \\
1-2(x-1)-3(x-1)^2 +4 (x-1)^3& , 1\leq x \leq2\\
4(x-2)+9(x-2)^2-3(x-1)^2 &, 2 \leq x \leq 3
\end{matrix}\right.$

Leaky Nun

@Evinda just check the endpoints and the second derivatives?

Evinda

11:25 AM

I have checked that f(0)=1, f(1)=1, f(2)=0, f(3)=10

@LeakyNun So now should I find the second derivative of f?

if so, why?

Leaky Nun

how is a spline defined?

Evinda

@LeakyNun Ah I saw the definition...So now it suffices to show that f''(0)=0 and f''(3)=0, right?

Leaky Nun

whatever the definition says :p

you just need to apply

user84215

The room Math Workshops has been created.

user84215

in Math Workshops, 7 hours ago, by MathematicsAminPhysics

This room has been created to hold math workshops. Please inform me of your suggestions to create events for them. Please note that your suggestions need not be very sophisticated; they can be elementary.

Leaky Nun

11:41 AM

@KasmirKhaan how goes the search?

Kasmir Khaan

@LeakyNun well so far icame to conculusion that , the operation should not be abelian and hmm

@LeakyNun the example can be done on matrcies

Leaky Nun

you can't search for non-groups if you keep seaching groups

Kasmir Khaan

@LeakyNun dont understant

Secret

\begin{align}
gaagaagagaagaagagaagagaagaagaggaag & =1\\
f & =gaagaagagaaga\\
gaagaagagaaga & =(agagaagagaagaagaggaag)^{-1}\\
gaagaagagaagag & =(agagaagagaagaagaggaag)^{-1}g\\
a & =(agagaaga gaagaagaggaag)^{-1}g\\
gaaga & =(agagaaga gaagaagag)^{-1}g\\
(aagag)^{-1}gaaga & =(agagaaga gaagaagag aagag)^{-1}g\\
(aagag)^{-1}gaaga & =(agagaagaa)^{-1}g\\
(gagaaga)^{-1}(aagag)^{-1}gaaga & =(aga a)^{-1}g\\
(aagaggagaaga)^{-1}gaaga & =(aga a)^{-1}g\\
(aagaggagaag)^{-1}gaaga & =(aga)^{-1}g\\
(aagaggagaagg)^{-1}gaaga & =(aga)^{-1}

Leaky Nun

@KasmirKhaan do you know what you want to find?

Kasmir Khaan

11:46 AM

Yes

xe=x for all x in G

that has to be ex=x for all x in G also

so the non -group wont fail that one

Leaky Nun

I take that as a no

Kasmir Khaan

I do

I have to find a non-group

Leaky Nun

what kind of non-group?

Kasmir Khaan

where xe=x for all x in the set G

Secret

More accurately, what kind of magma?

Kasmir Khaan

11:48 AM

and zx=e for all x in G

Leaky Nun

that isn't right

Kasmir Khaan

Hmm

I dont get it then

Secret

Are
xe=x
zx=e
the only eqautions?

Kasmir Khaan

they want an example of a set

Leaky Nun

nvm you edited

Kasmir Khaan

11:49 AM

okay =p

they want an example of a set, where those hold but it is not a group

so it fails somehow

Leaky Nun

no asso?

Kasmir Khaan

my point of attack was

the left and right inverse are not the same

hmm

if I work on matrices such that the determinant is =0

Leaky Nun

you can't say that yet

Evinda

@LeakyNun Ok, thanks

Kasmir Khaan

Yes I know just brain-storming

Secret

11:51 AM

@KasmirKhaan Can you list all the axioms and equations you have for you structure?

Evinda

I have also an other question @LeakyNun

Suppose that $\frac{1}{2} \leq q \leq 1, G(x)=\frac{3x-qx^2}{2}$.

I have found that a nonzero fixed point of $G$ is $x^{\ast}=\frac{1}{q}$. How can we find an interval for which the fixed point method will converge certainly to $x^{\ast}$ ? @LeakyNun

Kasmir Khaan

@LeakyNun there exist an element e in G such that xe=x for all x in G
for every x in G there exist an element z in G such that zx =e
this was given. so we have right identity and left inverse

Leaky Nun

yes

Kasmir Khaan

we are supposed to prove this by example not a real proof

Secret

Is the magma a semigroup (i.e. is it associative?)

Kasmir Khaan

11:53 AM

we cant take a abelian operation

Leaky Nun

@Evinda you want to compare G(x) and x

Evinda

So we take the difference? And then? @LeakyNun :/

Leaky Nun

@KasmirKhaan "find"ing a group might not be the right path

I mean, a non-group

Secret

abelian is not the same as associativity

Leaky Nun

@Evinda work out where it is positive

Secret

11:54 AM

We need to know whether this magma is associative

Kasmir Khaan

@LeakyNun hmm yes I know , I meant we take some elements , few matrices that have left inverse but not right inverse on that set

Leaky Nun

you don't need to do this

do you know the course you are studying?

Kasmir Khaan

okay , yes abstract algebra =p

Secret

Ok, so it's a semigroup with right identity and left inverse

Leaky Nun

matrices are not abstract enough

Kasmir Khaan

11:56 AM

okay i get it ><

abstract example =p

hmm

if I do a table

i need to have e, a,b,c say

ab=e but ba=c

but there is some structure I cant destroy =p

Evinda

@LeakyNun It is positive in one of the intervals (0, 1/q) or (-oo, 0) U (1/q,+oo), right?

Kasmir Khaan

like I cant just make up an example by putting info there

Evinda

What do we get from that?

Leaky Nun

@KasmirKhaan what does the axioms you have tell you?

@Evinda you basically listed out the entirety of the reals

Evinda

What do you mean? @LeakyNun

Kasmir Khaan

11:59 AM

@LeakyNun any element * indentity = that element back and for each element , if we multiply it on the left it gives the identity

Leaky Nun

someone hasn't learnt the tabular representation of a grpup

Kasmir Khaan

><

Ermm the tabular representation makes sure that the table is "latin square"

Leaky Nun

@Evinda which one is it?

Kasmir Khaan

if it is not latin square structure then it is not a group

Leaky Nun

you don't need a group

Kasmir Khaan

12:01 PM

I know

so I make an example such that 2 elemets appear in same row or colum

while ofc obeying the 2 criteria given

x*e=x for all x and zx=e for all x

user84215

I think all inactive rooms have become active because of my posts.

Evinda

@LeakyNun Do we also check in which intervals G(x)-x<0 and the desired interval is the one that we didn't get at the two cases?

Secret

Do not post irrelevant stuff in the gamedev chat room, else insta ban

Leaky Nun

@KasmirKhaan so you can already fill in some spaces of the table

Kasmir Khaan

@LeakyNun I made one on papper but ill put in on paint and post it here =p

user84215

12:06 PM

@Secret They are not irrelevant. They are about Math Workshops .

Leaky Nun

@Evinda sorry I'm not too familiar

Secret

gamdev chat room is one of the most strict chat room. Even the slightest off topic-ness will attract mods and result in ban and delete posts

Leaky Nun

@KasmirKhaan or you can just type it out

you don't need to use LaTeX

Secret

That's what happened to me. but perhaps, because of how quaternions are related to 3D rotations, perhaps they might foudn that not so off topic and it stayed

user84215

Thanks for your advice.

Kasmir Khaan

12:11 PM

Daminark

@MathematicsAminPhysics they're game developers and your math workshops are at best tangentially related to what they do

Kasmir Khaan

@LeakyNun hopes you can see it ><

Daminark

Thus, they'll most likely take it as off-topic

Leaky Nun

@KasmirKhaan why isn't it a group?

Kasmir Khaan

@LeakyNun because for example ab= e and ba =c

if it was a group it would be equal , ab =ba =e

Leaky Nun

12:13 PM

alright.

Kasmir Khaan

:D

I wonder if this can be done using 3 elements only

user84215

@Daminark I hope this does not happen for my posts.

Leaky Nun

go ahead

Kasmir Khaan

Ill work on it now =p

user84215

They are very useful.

Kasmir Khaan

12:14 PM

but with 2 elements it cant

@LeakyNun It cant be done on 3 elements =p

Daminark

You gotta be careful about relying on these self-perceptions too much. People typically see their own suggestions as being far more useful than other people see them. And more often than not, the latter is right

Kasmir Khaan

ab= e, ba cant be equal to a , b nor e

Leaky Nun

why must ab=e?

Kasmir Khaan

because if a*b=a for example , means that b =e

xe=x for all x

Daminark

But the real point is that the context of your workshop is such that in a math room, it's relevant, elsewhere it borders on spam

Leaky Nun

12:17 PM

why would ab=a imply b=e?

@secret I'm asking him

Kasmir Khaan

a*b=a , can only be done if b were the identity

Leaky Nun

why?

Kasmir Khaan

from x*e=x for all x in G right?

well hmm

Secret

Leaky: I will let you to deal with it, I am heading back to chemistry

Leaky Nun

does xe=x for all x exclude ab=a? @KasmirKhaan

from the former you get ae=a.

does ae=a and ab=a imply e=b?

Kasmir Khaan

12:20 PM

@LeakyNun hmm let me think

Yes

we have the left inverse

Leaky Nun

when working with axiomatic disciplines, one must dispose of previous knowledge.

@KasmirKhaan continue

Kasmir Khaan

@LeakyNun Yes that what I need to train myself on =p

I need to base my proof on the axioms given =p

Leaky Nun

indeed.

Kasmir Khaan

Ehm well if we multily by a' from the left

we know the left inverse exists

we do get e=b

Leaky Nun

@KasmirKhaan for the last time: do not skip any step.

Kasmir Khaan

12:23 PM

(a'a)b=a'a --> b=e

Yes sorry was writing it :D

Leaky Nun

a'ab isn't well-defined

Kasmir Khaan

Now it is :D

Leaky Nun

what justifies that bracket?

Kasmir Khaan

associative law ( given )

Leaky Nun

when you multiply ab on the left by a', you only get a'(ab).

when was it given?

Kasmir Khaan

12:24 PM

its part c) of the question

we are given associative binary operation

Leaky Nun

We (I and Secret) asked you three times whether associativity is given

Kasmir Khaan

Omg sorry :(

Did not see that

Leaky Nun

if it is then your 4-element example is wrong: a=ae=a(bc)=(ab)c=ec=c

Kasmir Khaan

Was to many texts and i was working on paint, looked only when it pings

hmm

Leaky Nun

back to your three-element example

so ab can't be a

Kasmir Khaan

12:26 PM

Okay

Leaky Nun

can it be b?

Kasmir Khaan

hmm nothing wrong with that =p

Leaky Nun

continue then

Kasmir Khaan

Okay working on it

Leaky Nun

hi fisherman @MatsGranvik

Secret

12:32 PM

blue=given

Mats Granvik

@LeakyNun hi

Leaky Nun

@Secret who gave you aa=ab=e?

Secret

o wait, nvm

Btw, Kasmir want ab=e I recall

Leaky Nun

all roads lead to rome

Kasmir Khaan

I nned 5 mins break

ill come back shortly

with a solution =p

Leaky Nun

12:35 PM

@Secret have you ever constructed the field with 4 elements?

Secret

One thing is clear: For 3 elements, You cannot have the ba=e (i.e., the es not forming a column), else you will find you are making $\Bbb{Z}/3$

nope, I have not done any field constructions in the past, because of how most of the things I interested in relies on breaking most of the nice things

Leaky Nun

@Secret you might be interested in building $a^2 b^2 a^{-2} b^{-2}$ using words of the form $gaba^{-1}b^{-1}g^{-1}$ where $g$ is a word with $a$ and $b$

Secret

Semigroup with left identity, right inverse

Leaky Nun

bonus: right identity :p

Secret

Another possibility (to be checked by having instead eb=e)

Leaky Nun

12:44 PM

how did you check associativity? I find it to be a logistic nightmare

(ab)b = eb = b; a(bb) = ae = a; back to the drawing board @Secret

Secret

ooops, forgot to check associativity

(I have yet to figure out how to do it for infinite groups, but for finite groups, if you multiply a row of the cayley table with some element on the left, it should return another row of the cayley table)

Leaky Nun

why?

Royi

Hello, Any of the moderators is here?

Leaky Nun

@Secret right, composition of left-map

Secret

My guess is that the morphism "multiplication on the left" or "multiplication on the right" on the cosets is a homomophism of some sort, but I have not checked that in detail yet

Leaky Nun

12:47 PM

your theorem is right but I don't think it applies

for all a in G: aG=G=Ga

wait, it might not be a homomorphism

never mind

Secret

but the way baS=cS suggest there is some kind of pattern

where b,a,c are some elements in the semigroup S

Leaky Nun

a = de = bce = eace = cdace = abdace => bdace=1; a = de = dcd = dcbc = dcbab = dcbaea => dcbae=1

@Secret right

@Secret maybe you would like this one: aa=bb=cc=(ab)^3=(bc)^3=(ac)^2=1

Secret

O btw, Kasmir is right, the 3 element semigroup with left inverse and right identity with ab=e is impossible since the triple abe will always fail the associative law

Leaky Nun

hmm

Secret

This is because $be\neq e$ by definition, and we already checked that $bb\neq e$

and $ba=e$ will turn it into $\Bbb{Z}/3$

Transcript for

$Mathematics$ Mathematics