Mathematics - 2016-04-29 (page 1 of 3)

anon

12:32 AM

every group has a unique subgroup of order 1

a subgroup of order 1 has exactly one element. what element must all groups have?

Benjamin R

1:03 AM

the identity, yep, as I said above I misspoke

what I mean to say is how is the <e> ... oh of course, it is a cyclic generator of itself, <e> = e

D'oh.

Sorry!

Benjamin R

1:20 AM

When determining the order of an element in a Direct Product, it's the lcm of the orders of the components as related to their parent groups, right?

e.g. $A \oplus B$ has the pair $(a,b)$

then the order of $(a,b)$ is going to be the lcm of $|a| \in A$ and $|b| \in B$, right?

anon

@BenjaminR yes

Benjamin R

1:38 AM

Thanks, sorry in Gallian it's not explicitly stated in such a way. It's the logical thing to infer, I just wanted to check

anon

It should be at least an exercise in Gallian.

Benjamin R

It probably is, I'm just reading the chapter on Direct Products now - the theorem as stated isn't explicit as I have said.

KKZiomek

2:18 AM

Can anyone tell if $\omega^{\omega^{\omega^{...}}}=\epsilon_0$ ?

anon

2:34 AM

@KKZiomek only looked up the definition of epsilons just now. define $\omega\uparrow\uparrow n=\omega^{\omega\uparrow\uparrow(n-1)}$ and $\omega\uparrow\uparrow 1=\omega$. let $\epsilon=\lim_n \omega\uparrow\uparrow n$. if $\omega^x$ is continuous in $x$, then by continuity $\epsilon$ is a fixed point of $\omega^x$, so $\epsilon$ is an epsilon number. finally, one may prove by induction that every epsilon number is $\ge\omega\uparrow\uparrow n$, hence every epsilon number is $\ge\epsilon$.

Therefore $\epsilon=\epsilon_0$.

Benjamin R

If $(a,b)$ is an element of direct product $A \oplus B$, the value of $(a,b)^2 = (a^2,b^2)$, right?

anon

$(a,b)\cdot(c,d)=(ac,bd)$, yes

nope, it doesn't matter

Benjamin R

thanks. Like I said, in Gallian there are zero explicit examples which make this clear. You would think you would just state that for the record but no.

anon

which edition are you using?

Benjamin R

9th = latest

I only had that intuition because we covered direct product groups in my discrete math paper last semester.

anon

2:40 AM

@BenjaminR surely Gallian defines the binary operation by $(a,b)\cdot(c,d)=(ac,bd)$?

then $(a,b)^2=(a^2,b^2)$ is a trivial consequence

Benjamin R

Nah, it doesn't even explicitly state the binary op.

Exactly, that's what I knew from my lecturer telling me last year and my faded memory, I just wanted to check

anon

I think you're lying.

Benjamin R

I mean, it probably does state it in the table in the back of the book somewhere or in the second chapter in groups

anon

The 8th edition has a chapter on external direct products which states the binary operation on the first page.

Benjamin R

Underneath the definition, using $g_1g'_1$ you mean?

anon

2:43 AM

yes

I bet you there is no introductory abstract algebra book in existence which introduces external direct products without explicitly saying what the binary operation looks like.

That would be like a calculus book not saying what a derivative is.

Not only that, but in the 8th edition the very first theorem in the chapter on external direct products states that the order of (a,b,c,...) is lcm(|a|,|b|,...), contrary to your earlier claim.

Benjamin R

Okay, I was exaggerating, but when I was (admittedly skimming) back over the chapter quickly I was looking for $(a,b)•(c,d) = (ac,bd)$ as you said which you must admit is a million times easier to read and parse.

Yes, but that statement is ambiguous (I think) about it meaning what I asked

that's why I had to ask the question

anon

Best would be $(a_1,a_2,\cdots,a_n)(b_1,b_2,\cdots,b_n)=(a_1b_1,\cdots,a_nb_n)$.

@BenjaminR If it's ambiguous, can you tell me another valid interpretation of what it says?

Benjamin R

yep, that's fine, but I personally would start with 'product' of two pairs first, then show that follows.

anon

it doesn't follow, it generalizes

Benjamin R

There isn't, but it leaves it to 'interpretation' instead of saying i.e. etc etc

sorry, yes, generalises.

anon

2:47 AM

it gives three examples right off the bat before it gets to said theorem, illustrating the componentwise nature of the operation

Benjamin R

Bear in mind that you are looking at it from a mathematicians perspective, i am looking at it from a students perspective. I don't already KNOW it, and anything not made explicit is a sin in this context I think.

Oh for goodness sake.

It wasn't clear to me at all.

You can either dismiss me as a dummy or take me at my word.

anon

technically, those aren't mutually exclusive :P

Benjamin R

well, to be fair I did walk into that one.

anon

anyway, is there actually something in the text which in order to be understood requires the reader to already know the material it's explaining?

I doubt it.

Benjamin R

define 'already know'. In no paper with assignments and other papers with other assignments is there ever time to fully 'know' the topic completely.

anon

2:51 AM

students are raised from childhood to understand analysis and only the most basic algebra. learning abstract algebra is an entirely new way of thinking, and I think this is the reason why there is such a steep learning curve.

Benjamin R

You just do your best and rush to hand in the assignments in time.

anon

I was talking about the book. "already knows" meaning, the text's presentation of X cannot be understood except if one already knows X.

Benjamin R

And since I only give an answer when I know for SURE, that's why I ask what may seem inane questions.

Actually, that's frequently true for many math books. They are terrible for learning, excellent for reference. And sometimes the reverse holds too.

anon

It's true that many math textbooks require one understand A,B,C,...,Y in order to understand presentation of X. Not so sure about knowing X itself to understand X, though. The latter is characteristic of reference books, but the intro texts are not of that variety.

Benjamin R

Anyway, I shouldn't have got into this debate. I will just say this: I have masters in FINE ART as well as about to complete a degree in science and math. I doubt there is anyone else on the entirety of math.SE that has that combo. So, if I think differently from you, if it's easier for you than I, trust me there are many many many many many things that I can do which you have practically zero hope of doing by the standards it seems you are judging me by.

anon

2:57 AM

"Everyone you will ever meet knows something you don't." - Bill Nye

I have no illusions.

Or, at least you have not spotted any illusions under which I am laboring. :)

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Nobody is "judging" you :-)

Jessy Cat

I'm judging you.

Just kidding.

I have no idea what's going on.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

A bit of miscommunication

Jessy Cat

But I judge everyone. They call me Judge Judgy McJudge-Pants

Benjamin R

yep

I over-exaggerated about Gallian, true. But, there is no question that just bloody well stating the Binary Op in the way anon did at the top of the chapter would have been superior. And the thm about orders of direct product elements does require a conclusion on the reader's part which is unnecessary. Just make it explicit, don't leave anything to be concluded.

Forever Mozart

3:00 AM

"dont judge me, Dave"

That's what my friends in college said when they were stoned.

did I judge them...maybe

Benjamin R

Plus, I have a headache from not taking my crazy pills, so, y'know.

Jessy Cat

Gallian is a huge Beatles fan.

Benjamin R

(I forgot to take them this morning, they have a very short halflife)

So, in other words, he's entirely past it as a human being.

Forever Mozart

i like to watch Frasier, great show

Jessy Cat

No, the Beatles are cool.

Benjamin R

3:02 AM

Damn Boomers loving Frasier and The Beatles.

Jessy Cat

And not too weird.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Textbooks are what they are.

Jessy Cat

It would be weird if he was quoting from Kiss.

Benjamin R

The Beatles are overrated. Revolver is great, Sgt Peppers is great. The rest are meh.

Jessy Cat

And he showed up to AMS meetings wearing Kiss makeup.

They were extremely revolutionary for the time.

Benjamin R

3:04 AM

No, that would be random and awesome.

Jessy Cat

Nobody did what they did, and they changed pop music forever.

Benjamin R

damn boomers.

Jessy Cat

<-- not a boomer

my parents are boomers

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

So did prince.

Jessy Cat

yup

Benjamin R

3:04 AM

so are mine, that's why I hate The Beatles and love Prince.

Jessy Cat

The White Album is a great piece of art

Haha, Boehner called Ted Cruz Lucifer.

Awesome.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

What about MJ?

Jessy Cat

Him too.

Benjamin R

well he sort of is

2

Jessy Cat

Lucifer with an extremely annoying voice.

Benjamin R

3:06 AM

MJ with Quincy Jones is very very good but definitely sub Prince.

Forever Mozart

Ted Cruz is finished

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Elvis?

Benjamin R

Prince is a polymath, maybe the closest to Mozart in the last 100 years

Jessy Cat

It would be nice if they all just went away and Tyrion Lannister became president.

Benjamin R

Good singer, okay

Elvis < CASH

Forever Mozart

3:06 AM

not Mozart! :)

by a mile

Johnny Cash?

yep

I hear that train a-comin'

It's rollin' round the bend

But I ain't seen that sunshine since I don't know when

Benjamin R

I shot a man in Reno

just to watch him die

Jessy Cat

3:08 AM

When I hear that whistle blowin' I hang my head and cry

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Elvis > Prince > MJ

Benjamin R

Pfffffffff

Jessy Cat

There's a video on YOutube of a 4 year old kid singing that song

Benjamin R

Prince > Cash > MJ > Elvis

Forever Mozart

Dylan > Neil Young > everybody else

Jessy Cat

3:08 AM

<3 Bob

Benjamin R

well that's true, but on average Neil is better

Jessy Cat

and don't forget Woody Guthrie

Benjamin R

peak Dylan better than Peak Young

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Pink Floyd >>> all

2

Benjamin R

Pfffffff

Jessy Cat

3:09 AM

That.

Forever Mozart

I actually like lots of Bob Dylan's newer stuff

Benjamin R

Radiohead >>>>>>>>>>>>>>>>>> Floyd

Jessy Cat

no...

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

No

Jessy Cat

But @TedShifrin>>>>>>>>>>>>>>>>>>>All of them and all y'alls.

Benjamin R

3:10 AM

Reactionaries, the lot of ya

Jessy Cat

The Runaways & Blondie >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

...

Benjamin R

Beefcake >>>>> Floyd

Jessy Cat

We heard you man, calm down yo

Benjamin R

haha

Beefheart I mean!

Shiz.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

:)

Jessy Cat

3:11 AM

No beef of any kind. This is all completely vegan.

Benjamin R

Captain Beefheart

Don Van Vliet (/væn ˈvliːt/, born Don Glen Vliet; January 15, 1941 – December 17, 2010) was an American singer, songwriter, musician and artist best known by the stage name Captain Beefheart. His musical work was conducted with a rotating ensemble of musicians called the Magic Band (1965–1982), with whom he recorded 13 studio albums. Noted for his powerful singing voice and his wide vocal range, Van Vliet also played the harmonica, saxophone and numerous other wind instruments. His music integrated blues, rock, psychedelia, and jazz with contemporary experimental composition and the avant-garde...

Jessy Cat

And don't use crack.

KKZiomek

@anon but you know that I'm talking about ordinal infinities right? Just to make sure

Jessy Cat

Sounds like some kind of failed McDonalds mascot.

Captain Beefheart says eat your meat, kids.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

lol

Benjamin R

3:13 AM

Jessy Cat

If you don't eat your meat, ye cannae have any pudding!

anon

@KKZiomek did I say anything to indicate I was talking about finite ordinals?

Jessy Cat

Tsk tsk, Jean-Luc.

Benjamin R

Well, I've sufficiently degraded this chat, time to run away

before the pitchforks come out

Jessy Cat

Too late.

sticks you

Benjamin R

3:15 AM

OWEEES

Jessy Cat

Yeah.

I know.

But it was necessary.

Anway, g'nite folks. I'm outtie.

::crickets::

cya

later

3:31 AM

ok the chat has gotten weird

Benjamin R

forgot to take mental pills.

is the inverse of any element of a direct product group e.g $(a,b) \in A \oplus B$ componentwise? It must be because $(a^{-1},b^{-1})•(a,b) = (e,e)$

Eric Stucky

Right answer, right reason :)

Benjamin R

thanks... having to rush on an assignment plus the afformentioned Efexor half-life.

KKZiomek

@anon no, but that's what I meant. $\omega$ is ordinal equivalent of $\aleph_0$, and I asked if omega exponentiated omega times to omega is epsilon0.

But I found the answer already.

anon

@KKZiomek while I am new to epsilons as I said, is what I gave not a proof?

EsX_Raptor

3:44 AM

hi

Eric Stucky

hey

how you

EsX_Raptor

preparing for finals. what are you guys up to

KKZiomek

I actually don't know @anon I'm new to all this as well.

floorcat

I'm watching the failarmy videos Raptor. Priorities and all.

KKZiomek

@anon Now I looked again, you're right :)

Benjamin R

3:49 AM

$U(7) = \mathbb{Z}_7 \setminus \{0\}$ under multiplication mod 7, right?

anon

\backslash or \setminus

also yes

Benjamin R

thank you very much anon.

It must therefore be that $U(p) = \mathbb{Z}_p \setminus \{0\}$ under $\times_p$ where $p$ is prime then

anon

mmhmm

Benjamin R

d'oh my damn TeX noobishness

EsX_Raptor

If $V\subseteq\mathbb A^n_\mathbb R$ is algebraic, I know there exists an $f\in\mathbb R[x_1,\dots,x_n]$ such that $V=\mathcal Z(f)$ because $\mathbb R$ is a PID.

I don't know if our professor is trying to trick us or what by asking if this also holds for $\mathbb C$.

I mean, $\mathbb C$ is also a PID.

ლ(　ﾟДﾟლ)

Forever Mozart

3:59 AM

LaTOUR - People Are Still Having Sex (Original 7'') (1991)

►

2

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

4:14 AM

Released^ at the peak of the AIDS epidemic.

Benjamin R

4:34 AM

Condoms FTW

How can I use the fact that all groups of order 4 must be isomorphic to $\mathbb{Z}_4$ or $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ to show that all proper subgroups of $D_4$ are Abelian?

I mean, I can see that the subgroups which are Abelian, but...

anon

what order does D4 have? what then can be the order of a proper subgroup?

Benjamin R

dihedral group of order 8

I have the Cayley table in front of me, it's pretty blatant which are the proper subgroups and that they are Abelian

anon

yep, D4 has order 8. so the order of a proper subgroup must be what?

Benjamin R

less than 8

anon

more specific

Benjamin R

4:39 AM

and dividing 8

anon

yes

1,2, or 4

is there any group of order 1,2 or 4 which is not abelian?

Benjamin R

pass

oh okay, I can see it in my notes, an abelian group (up to isomorphism) of order $n$ must be isomorphic to the (External) Direct Product of $\mathbb{Z}_p$ and $\mathbb{Z}_q$ where $p,q$ are the prime factors of $n$

(my generalisation)

anon

not really relevant to proving proper subgroups of D4 are abelian

also statement doesn't make sense if n has more than 2 prime factors (or bigger prime power factors)

Benjamin R

true.

okay... well from the Cayley Table of $D_4$ all the proper subgroups look abelian to me.

anon

cayley table of D4 is irrelevant as well

Benjamin R

4:46 AM

well, please can you give me the hint that clues me to the thing I need to know?

anon

7 mins ago, by anon

is there any group of order 1,2 or 4 which is not abelian?

Benjamin R

I don't know the answer.

I haven't studied that as far as I know thus far

anon

math isn't about regurgitating memorized facts. you have to think in order to determine answers. if you don't know an answer to a question immediately, the problem is not that you didn't learn it somewhere or that you forgot the answer. there isn't any problem with not knowing the answer. the next step is to figure out what it is.

Benjamin R

true, but I only have 20 mins left to answer this....

okay I am just going to take the time and figure it out and hand my assignment in late, because you are 100% right.

anon

a group of order 1 is the trivial group. a group of order 2 is the cyclic group Z/2Z. you know what the groups of order 4 are, you just said what they are. (of course I mean "up to isomorphism.")

Benjamin R

4:50 AM

Well, any finite group of order n is isomorphic to $\mathbb{Z}_n$ I do know that

anon

way, way wrong

Benjamin R

okay.

No worries, you are right, no point in shortcuts I will figure it out eventually.

anon

I've already told you the answer.

The whole proof is here.

Benjamin R

The fact I can't see it proves I don't understand it, which means I need to think it through some more.

It's all good.

anon

The subgroups of D4 are all size 1,2,4 which means they are all isomorphic to one of (trivial group), Z/2Z, Z/2ZxZ/2Z or Z/4Z, all of which are abelian.

Benjamin R

4:53 AM

I can't give that as the answer unless I understand why that is so, is my point.

I am sure it is ultra blatant

Forever Mozart

would your math skills improve with cocaine?

Benjamin R

nah

anon

when you use the word "that" it makes it seem like I just gave a claim. I gave a sequence of claims, each following from the previous. specifically which claim in the chain, or which logical implication, would you like to understand?

Benjamin R

I admit I am dumb

or, at least, being dumb

anon

(1) the subgroups of D4 are all size 1,2,4
(2) they must each be isomorphic to one of (trivial group), Z2, Z2xZ2, Z/4Z
(3) all of said groups are abelian
(4) therefore, all of the subgroups of D4 are abelian

Forever Mozart

4:55 AM

dumb is relative

Benjamin R

okay I get that if there is an isomorphism from $G$ onto $H$ then $G$ is Abelian $\iff H$ is Abelian

anon

mmhmm

Benjamin R

sorry forgot the TeX for iff

anon

\iff, but "iff" is shorter :)

Benjamin R

d'oh

anon

5:00 AM

also \Leftrightarrow is a smaller symbol, but longer to type...

Benjamin R

Oh yep, that's the one I was forgetting

What though, is the isomorphism between the proper subgroups of $D_4$ and $\mathbb{Z}_4$, $\mathbb{Z}_2 \oplus \mathbb{Z}_2$

that's the blatatnly obvious bit I am missing

does that make it clearer

You say that it's because they are of the same order... but... I haven't seen the proof for that (or I have skimmed past it or forgotten it)

I 100% believe it is blatant and embarassing

rorty

Do you guys mind if I interject with a question of my own?

Benjamin R

of course, I am being dumb, ask something more worthy.

anon

Suppose $H$ is a subgroup of $D_4$. If $|H|=1$ then $H$ is the trivial subgroup, which is abelian. If $|H|=2$ then $H\cong\Bbb Z_2$, which is abelian. If $|H|=4$ then either $H\cong\Bbb Z_4$ (which is abelian) or $H\cong\Bbb Z_2\oplus\Bbb Z_2$ (which is abelian). That covers all possible cases, since $|H|$ is one of $1,2,4$ and if $|H|=4$ the $H$ must be isomorphic to one of $\Bbb Z_4,\Bbb Z_2\oplus\Bbb Z_2$.

rorty

I come to seek guidance on a proof. I want to show that if $A$ and $B$ are compact sets, then $A+B$ (that is, the set $\{a+b : a \in A , b \in B\}$) is compact. I know that $A+B$ is bounded, but am having trouble showing that it is closed. That is, I am having trouble showing that if $x = \lim (a_n + b_n)$, then $x \in A+B$.

My attempt: Bolzano-Weierstrass guarantees that $(a_n)$ has some subsequence $(a_{n_k})$ that converges to some $a \in A$. Likewise, there is some subsequence $(b_{n_j})$ that converges to some $b \in B$. Consider the indexing set $I = \{n_k\} \cap \{n_j\}$. If $I$ is …

Benjamin R

5:06 AM

I know that $Z_4$ etc are Abelian, but not why groups of order 4 are isomorphic to $\mathbb{Z}_4$

that's the bit – the "must be" that I am missing.

anon

31 mins ago, by Benjamin R

How can I use the fact that all groups of order 4 must be isomorphic to $\mathbb{Z}_4$ or $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ to show that all proper subgroups of $D_4$ are Abelian?

Benjamin R

EPIC FACEPALM

MASS EMBARASSMENT

sorry, maybe I am secretly on cocaine.

Mike Miller

weird choice of drug to do group yheory on

anon

@rorty if you want a secret method, you can use the fact + as continuous

rorty

I wish not to, because I haven't learned continuity yet... I'm towards the beginning of a very intro analysis text.

anon

5:11 AM

@rorty I think you want to write $(a_{n_k})$ and $(b_{m_k})$, not what you wrote

rorty

yes, ur right

anon

notice how both are functions of $k$

Benjamin R

sorry anon, that must have been very very painful for you.

anon

it's fine

Forever Mozart

this one's for you @BenjaminR youtube.com/watch?v=rLs_0h9_wXM

Benjamin R

5:14 AM

I prefer this one: youtu.be/5bJB51NZRh8

Forever Mozart

petarded? I dont get it

Benjamin R

Try this one: youtu.be/aCU1JUUSz2s

It's me recognising my own retardation

Actually, anon it is this: youtu.be/ZrewRRy5sP4

"Firetruck... firetrucks... what's the colour of those red firetrucks..."

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

:D

Benjamin R

"This plan's so good it's retarded"

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

here's a classic

Benjamin R

5:28 AM

That's the one I would have thought of re: Coke.

Valery Saharov

6:25 AM

6

Q: Approximate spectral decomposition

See attempt below I am interested in effective and constructive computations for finding approximate spectral decompositions in some suitable format. Namely, let $A: H \rightarrow H$ be a Hermitian operator on an $n-$dimensional Hilbert space $H$ with the spectrum $\{\lambda_1, ... \lambda_m\},...

Forever Mozart

@ValerySaharov oh. yeah.

Kaustabha Ray

6:50 AM

Consider the polynomial x5+ax4+bx3+cx2+dx+4x5+ax4+bx3+cx2+dx+4 where a,b,c,d are real numbers. If (1+2i) and (3−2i) are two roots of this polynomial then what is the value of a?

I see the complex conjugates as the other two roots of the polynomial

So where does the fifth root come from and how do I solve this?

Martin Sleziak

@KaustabhaRay You know also product of the five roots: en.wikipedia.org/wiki/Vieta%27s_formulas

Kaustabha Ray

@MartinSleziak So Sum is -a and Product is -4

Martin Sleziak

Yes.

Forever Mozart

7:15 AM

hi everybody

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

hi

Forever Mozart

this problem is going to kill me

I have noticed what some graduate students do in research - they create their own problems and constructions, and produce a lot. Meanwhile I am torturing myself with an unsolved problem by Paul Erdos

its so sick

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

that would give you a tortured-Erdos number of 1, since it is unsolved :)

user173049

let $ f $ is analytic in the disk $B(a;R)$ and $z\inB(a;R)$.The what can we say about $f(z)$? It is real, imaginary or power series representation.

Forever Mozart

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ the Supreme Fascist is winning

anon

7:23 AM

@KaustabhaRay why do you have so many terms in the polynomial written out twice?

Forever Mozart

The purpose of life - Paul Erdös

►

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

SF =?= God

Forever Mozart

or maybe just someone who tortures us with math problems

I

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

he believed in "The Book" of proofs too

Forever Mozart

yes, I have a book called "Proofs from 'the book'"

which is a collection of beautiful proofs of mostly combinatorial and geometric theorems

Kaustabha Ray

7:42 AM

@anon That was a copying error sorry

How to find a solution to $x = 15n + 2$ and $x = 21m+4$ ? Equating the two don't help as we have two unknows

Forever Mozart

8:19 AM

@DanielFischer you are good topologist

should I continue to torture myself with problem by Erdos?

@BalarkaSen how do you do

Balarka Sen

alright. what's up with you?

Forever Mozart

well, I constructed an interesting explosion point space

Balarka Sen

ah?

Forever Mozart

but it's not good enough for Erdos

very depressing

Balarka Sen

maybe there's no counterexample after all.

Forever Mozart

8:24 AM

for several days I thought it was good

Balarka Sen

you'd need to prove something at some point of time.

Forever Mozart

what do you mean?

Balarka Sen

I mean you're looking for a counterexample, no? Maybe there's none.

Forever Mozart

yes possibly

Balarka Sen

So you'd need to prove the conjecture.

Forever Mozart

8:26 AM

but first I have to see why there is no counterexample of the type I'm looking for

cause I am looking in a small area

A proof of the conjecture would have to eliminate a counterexample in the Cantor fan

and I don't even see how to do that

but I am getting closer

Good to hear.

what do you work on?

Differential forms.

oh I have worked with those

several years ago I used Spivak's first two books to help with general relativity

Balarka Sen

Ah, I see. Nice.

Forever Mozart

8:30 AM

I was very interested in Godel's model of the universe which satisfied Einstein's equations

but I really never got around to it

Balarka Sen

I didn't know there was such a thing. But I no physics.

Forever Mozart

it was a really wild model, in which time is circular

if Godel invented it, you know it has to be crazy

Balarka Sen

Fair enough.

Daniel Fischer

@ForeverMozart Is it interesting and inspiring? Then definitely continue.

Forever Mozart

@DanielFischer I make small amounts of progress, bit by bit. On the surface the problem is very simple and would be of interest to any topologist.

There are several related problems and I switch between them every now and then

but they all seem very fundamental

Daniel Fischer

8:36 AM

If you're making progress, that's good, isn't it? You're learning from it.

Forever Mozart

yes

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

8:47 AM

try not to think of it as torturing yourself

perhaps, challenging yourself would be more positive

Evinda

Hello @DanielFischer
Suppose that we are over $\mathbb{F}_q$ and consider vectors of length $n$.
Then if we pick a primitive n-th root of unity, does it imply that $a^n=1$ and $a^i \neq a^j$ for $i \neq j$ and $1 \leq i,j \leq n$ ?

anon

yes, but not sure what you mean by "vectors of length n"

Evinda

@anon I am learning coding theory and there we have a code with words of length n.

Tobias Kildetoft

in fact none of that has anything to do with being over $\mathbb{F}_q$

anon

@Evinda but nowhere in the question you just asked is there any vectors

why tell us to consider vectors when your question has nothing to do with vectors?

Evinda

8:58 AM

The codewords are vectors of length n @anon

anon

there are no codewords in your question

Balarka Sen

"suppose I start with a topological quantum field theory. now an important question would be: what's a fish?"

Evinda

@anon Yes, I wrote vectors of length n because it doesn't make a difference

anon

you said something irrelevant because it was irrelevant? hrm..

:P

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

may i ask why you chose Erdos? @ForeverMozart

Evinda

9:02 AM

Why? Since we have vectors of length n , we take n-th root of unity, don't we? :D @anon

anon

not sure if serious

Evinda

I am

Forever Mozart

I found the problem first, then I discovered he asked it

but I was attracted to the problem first

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Interesting.

Tobias Kildetoft

@Evinda We can't read your mind. The actual question you asked was completely devoid of vectors or codes or anything like that. It was simply a question about what a primitive $n$'th root of unity is

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

9:04 AM

have you discussed this your adviser @ForeverMozart

Evinda

Ok, sorry... The first proposition I wrote is related to coding theory... @TobiasKildetoft @anon

Forever Mozart

yes, when I think I have something

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

I mean the whole difficulty of it.

After all, advise is his job :P

Mambo

10:08 AM

mean?

tatan

Anyone seen the movie "The man who knew Infinity"?

Mambo

when did it release?

tatan

@Mambo 8th April in UK and today in India...

Mambo

okay

tatan

@Mambo youtube.com/watch?v=oXGm9Vlfx4w

Balarka Sen

10:14 AM

@MikeMiller You're up early.

Mike Miller

A little.

Balarka Sen

Good morning, then. :)

tatan

@MikeMiller Quite a little...:-)

Mike Miller

Yeah well you're up a little early

pal

Valery Saharov

10:32 AM

Can somebody help please?

7

Q: Approximate spectral decomposition

See attempt below I am interested in effective and constructive computations for finding approximate spectral decompositions in some suitable format. Namely, let $A: H \rightarrow H$ be a Hermitian operator on an $n-$dimensional Hilbert space $H$ with the spectrum $\{\lambda_1, ... \lambda_m\},...

linear-algebra hilbert-spaces approximation computability constructive-mathematics

I got a bounty for this one that expires soon

Mike Miller

note that this is the same crowd that wasn't able to answer your question the last few times you've asked :)

Tobias Kildetoft

The chat... The chat never changes...

skill patrol

10:50 AM

it is what it is

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$Mathematics$ Mathematics