Mathematics - 2017-10-22 (page 1 of 5)

well, the second case is not that easy. Why is $G$ abelian?

That's alot better Ted =P You have in your book the clearest things :D

I proved that mathei =P

Oh okay

all groups of order p^2 are abelian =p

12:22 AM

Good, Kasmir :)

Ted even has criminal examples on his book :D

criminal???!!!

anon

so, just to be clear, given a and b both have order p, you know G is isomorphic to Zp x Zp because?

calls lawyer

a drugdealer has 1 , 3 , 18 ons of drugs

12:22 AM

good ... answer @anon's question ...

or something like that

oh ted

can you check my proofs

I'd just apply the fundamental theorem for finitely generated modules over a PID :^)

jk

glares at Mathei

What proofs, Meow?

let me think a second @anon

12:23 AM

let me send

Too bad Karl didn't stick around, Mathei. He was an algebra jock here 3 years ago ... used to be just anon and he :)

Then Pedro turned into one. Then ... oh, never mind; it's depressing.

karl sounds like the name of an algebraist

I don't think it's his real name, but I'm not sure.

sent

If every element of $G$ has order $p$ and $G$ is abelian, then the thing is a vector space over $\Bbb F_p$, thus it is isomorphic to $\Bbb Z /(p) \times \Bbb Z /(p)$ by linear algebra and comparing orders

12:28 AM

is it because the intersection of <a> and <b> can only be 1

so it follows that <a> x <b> = G ?

@anon is that a good answer to your question ? :D

well i might add that both subgroups are normal

since G is abelian

@Meow: Answered.

Physics Guy

Hey Mr. Shifrin

@Kasmir: Have you proved some general result about when $H\times K$ must be all of $G$ if $H$ and $K$ are normal?

Hi @PhysicsGuy.

@TedShifrin Yes sir ! i used that argument here :D

Btw Ted hope you did not get offended when i said "criminal exercices "

Well, you only said "so it follows that ..."

12:41 AM

You had a funny example about that, to prove that relative prime numbers can generate the gcd =p

@TedShifrin I dont argue good when i type fast >< , had to keep watching the lecture :)

I'm not sure what you're talking about, but never mind.

I just came up with a stupid argument that groups of order $p^2$ are abelian

thanks

user76284

Would the negative of a distance or metric be considered a valid kernel? en.wikipedia.org/wiki/Kernel_method

I love the $Z(G)$ cyclic implies $G$ abelian tool, myself, @Mathei.

Physics Guy

12:45 AM

@TedShifrin I have a math question: Let $a\leq b\leq c$ with $abc=1$ Prove the following inequality: $(a+1)*(c+1)>3$ I've proven it for $a>\frac{1}{2}$ and for $c+1=2$ and $c+1=2$ but I'm not sure how to prove it with $a<\frac{1}{2}$ and $c+1>2$

user76284

Does abandoning semi-definiteness make sense? If not, what about using the kernel $k(x,y) = \exp(-d(x,y))$?

A druggist has five weights of 1,3,9,27 and 81 ounces and a two-pan balance, show that he can weight any integral amount up to and including 121 ounces @TedShifrin

I know nothing about machine learning, @user76284.

@TedShifrin that was the question in your book i found funny and criminal :D

because it is about selling drugs =p

that's just ternary

12:46 AM

That's just about thinking about numbers base $3$, @Kasmir. Druggists aren't dealing in illegal drugs.

ive never heard druggist

A druggist is a pharmacist :)

@TedShifrin yeah, it's way easier than the representation theory stuff I have in mind

i guess pharmacist has a less bad connotation

@TedShifrin oh sorry >< i missunderstood what druggist is :D

12:46 AM

Oy at representation theory. I like the class equation for the easiest proof, Mathei.

it could have been a drug dealer

breaking bad?

kind of stuff? :D

Druggist was a standard word in my younger days, as pharmacists actually mixed a lot of the potions they sold.

Yes I understand , we use that word here too =p

but now it is often understood to be the bad kind of drugs ><

Go back to studying, Kasmir.

12:47 AM

Yes sir !:)

Potions = lol

Hush, you.

user76284

Doh, nevermind. Of course you can use the metric (as in RBFs).

i deal with a lot of base 2

and base 16

lots of base 16

I remember when base 12 was supposedly important.

12:49 AM

i used to deal with a lot of base 2 but not anymore

base 12 is interesting in an alternative-universe kind of way

im past my cpu-building prime

you're composite now?

...now i find myself perversely curious about what pi looks like in base 12.

lol

12:51 AM

im compelled to be the smacker rather than the smackee now

@TedShifrin think thats the worst joke i have heard all day

Well, you haven't heard Balarka and Demonark, but OK.

i say 'perverse' mostly because there's really no reason why pi would look any more interesting in base 12 than base 10

@TedShifrin i have just not today

base 3 is interesting if only b/c of how it relates to the Cantor set.

12:52 AM

No reason to stick to middle thirds, though.

not sure what you mean by that

You can do Cantor with things other than middle thirds.

well, sure

but the usual version

hmm, now I wonder

i should start tutoring

i need to become a better teacher

im often too all over the place

It takes practice (and a desire to be good). :)

12:55 AM

to be precise, what I have in mind is "In arithmetical terms, the Cantor [ternary] set consists of all real numbers on the unit interval that are expressible as a ternary (base 3) fraction using only the digits 0 and 2."

i should teach ted 6502 assembly

Yes, Semiclassic, we know that :)

lol sure ted we all know that...

Can one generalize that to a different base?

You don't know about the Cantor set, Faust?

12:58 AM

@Semiclassical yes, fat cantor set

say, real numbers on the unit interval that are expressible as a base-4 fraction using only the digits 0,2,3 ?

Semiclassic, I think 1/3 is the unique fraction that gives you measure 0.

@TedShifrin hmm, can't you remove the 2/4 from 1/4 + 2/4 + 1/4?

i have the base-b aspect more in mind than the Cantor construction

@Semiclassic: But there's a question of how many digits you allow or don't allow.

12:59 AM

@TedShifrin nope i have never taken topology

yeah.

or analysis @Faust

its a point set topology thats not dense or something?

hey im doing Anal sis right now

It's uncountable and every point is a limit point ... plus nowhere dense.

@Faust ...

1:00 AM

...

i have gotten 100% on every assinment and midterm thus far shooting for 100% in the class =)

@TedShifrin or a "perfect set" to be precise

I was avoiding jargon Faust doesn't know, Leaky.

It's no more precise than what I said. You're just showing off.

i mean, if you do decimal digits but limit yourself to the digits 0,2,4,6,8

is it actually a point set topology?

1:01 AM

That makes no sense, Faust, but it is a topological space, yes.

@TedShifrin I was introducing jargons that Faust doesn't know

@Faust "point-set topology" is a branch of topology

it isn't some specific topologies

then i'd think that the resulting set couldn't be measure zero

Why, Leaky? Think about good teaching.

ah

i am taking topology next semester im so excited

@Semiclassical why not?

1:03 AM

also taking a second Anal ysis class

Well, we need to do the computation. It looks like $\sum\limits_{n=1}^\infty \frac 5{10^n}$, right, Semiclassic?

should be?

yeah.

So that's $5/9$.

What's left has measure 5/9, so, no, it certainly hasn't measure 0. :)

lol

@TedShifrin what does the computation look like for the Cantor set?

1:05 AM

so you'd need to exclude exactly one specific digit to get a cantor set?

@Semiclassical I don't think that's right

No, as I said earlier, it's subtle. To get measure 0, the fraction 1/3 is crucial.

hmmm

I guess we have $\sum\limits_{n=1}^\infty \frac2{3^n} = 2\cdot \frac12 = 1$.

yeah. what i had in mind was $\sum_{n=1}^\infty \frac{b-1}{b^n}=\frac{b-1}{b-1}=1$ for any $b>1$.

1:08 AM

Ah, cool.

@TedShifrin I can't convince myself why it is $2$ not $2^n$

So you pull out only one segment each time, @Semiclassic.

no, @Semiclassical, I can't convince myself that it's right

Yeah, Leaky seems to have caught me in an error.

for 1/3, you removed $\displaystyle \sum_{n=1}^\infty \frac {(3-1)^{n-1}} {3^n} = \frac 1 {3 \left(1- \frac 2 3\right)} = 1$

1:09 AM

Yeah, that's right.

I was counting allowable digits and of course there are $2^n$ to go with $3^n$.

I flunk.

if you remove $r$ from $k$, you end up removing $\displaystyle \sum_{n=1}^\infty \frac {(k-r)^{n-1}} {k^n} = \frac 1 {k \left( \frac r k \right)} = \frac 1 r$

hmm, that coincides with @Semi's result

wait

my sum is wrong

yeah, that sum doesn't make sense for $r=0$

You're missing a $(k-r)/k$ in the numerator, @Leaky.

it isn't that simple: you need to specify where you are removing it from

$$\sum_{n=1}^\infty \frac{(k-r)^{n-1}}{k^n}=\frac{1}{k}\sum_{n=1}^\infty \left(1-\frac{r}{k}\right)^{n-1}=\frac{1/k}{1-(1-r/k)}=\frac{1}{r}$$

...wtf

1:14 AM

@Semiclassical my expression is wrong

The sum starts at $n=1$.

actually, your sum is fine. when $r=0$ the sum is divergent

This mistake I didn't make in my flawed claims earlier.

whether it represents the right thing, of course, is another matter

Oh, you shifted indices. Never mind.

1:16 AM

if your first removal leaves $s$ parts and you remove proportion $r$ of each piece every time: $\displaystyle \sum_{n=1}^\infty s^{n-1} r^n = \frac r {1-rs}$

For Cantor set, $s=2$ and $r=\dfrac13$

as I said, you need to know how many pieces you have left

if you remove the middle piece only, i.e. $s=2$, then you removed $\dfrac r {1-2r}$, which is only $1$ when $r=1-2r$, i.e. $r=\dfrac13$

if you leave $3$ pieces in the first time, then you removed $\dfrac r {1-3r}$ in total, which is only $1$ when $r=\dfrac14$

for your $n=0,2,4,6,8$ case, you have $s=5$ and $r=\dfrac12$, so you removed $\dfrac {\frac12} {1-\frac52} = ???$

it's a divergent sum in that case

it can't be divergent

as you've written it, it certainly is

then what I've written is wrong, certainly

the geometric factor you've written is $rs$

and the geometric sum converges when $0<rs<1$ i.e. $s<1/r$

so i dunno what to tell you

1:22 AM

I know the sum diverges, so I need to reconsider my expression

if I focus to this case, it is actually $\dfrac12 + \dfrac14 + \cdots = 1$

this is madness

the sum is obviously $\displaystyle \sum_{n=0}^\infty r(1-r)^n = 1$

yes, whatever $r$ is, you eventually remove the whole set

Step 1: remove $r$ from the whole set.
Step 2: remove $r$ of each remaining pieces.
...

This always removes the whole set

@Semiclassical fat cantor set is constructed by removing an exponentially smaller proportion

can't believe I got stuck in this for so long

@TedShifrin

I am getting only tex commands. not converting to equation in this chat

y?

Please help me. How to convert it automatically?

1:38 AM

@LeakyNun Hi

@LeakyNun did you solve the Galois theory exercise I gave you?

@ManeeshNarayanan you need to enable latex from a link in the room description

@MatheiBoulomenos which one...

@LeakyNun Show that the unique quadratic subfield of $\Bbb Q (\zeta_p)$ is contained in $\Bbb R$ iff $p \equiv 1 \pmod 4$

:o

its discriminant, lol

yeah, you can use discriminants, but it's possible with Galois correspondence only

it has to do with quadratic residues

no it doesn't

yes it does

The Galois group is $(\Bbb Z/p\Bbb Z)^*$

I need to find a subgroup of index $2$

I need to square every element

@MatheiBoulomenos @_@

1:46 AM

From where will I get solved problems of syllows theorem

?

How can you characterize an element being real in terms of automorphisms? @LeakyNun

mathei :D

I got a fast Q :)

sure

:D

Well the second sylow theorem

from what i understood

@ManeeshNarayanan out of the depths of the ocean

aus der Tiefen

rufe ich Herr zu dir

1:48 AM

*aus den Tiefen

In examination, I am strugglig

Aus der Tiefen rufe ich, Herr, zu dir, BWV 131

Aus der Tiefen rufe ich, Herr, zu dir (Out of the depths I call, Lord, to You), BWV 131, is a church cantata by the German composer Johann Sebastian Bach. It was composed in either 1707 or 1708, which makes it one of Bach's earliest cantatas. Some sources suggest that it could be his earliest surviving work in this form, but current thinking is that there are one or two earlier examples. The cantata was commissioned by the minister of one of the churches in Mühlhausen, the city where Bach worked at the time. It was possibly written for a special occasion. The text is based on Luther's German version...

we have K subgroup of G, whose order is divisible by p. let Hbe a p sylow subgroup of G ,
then there is a conjugate subgroup H' = gHg' of G st K intersect H' is a p-sylow subgroup of K

problems like checking given order cyclic or not, abelian or not etc

I see, that's archaic German I guess

1:49 AM

what does this means in easiar terms ? :D

@LeakyNun problems like checking given order cyclic or not, abelian or not etc

how to master in that?

@MatheiBoulomenos is the der archaic or the Tiefen?

@ManeeshNarayanan practise makes master

Can you suggest excersise book?

@MatheiBoulomenos oh it's neither: it's just the genitive

@LeakyNun

Can you suggest excersise book?

1:52 AM

@ManeeshNarayanan I have no idea

@MatheiBoulomenos fixed by conjugation, you know

@LeakyNun then where shall I do exercises?

@LeakyNun hmm, "der Tiefen" is archaic, I'd say "der Tiefe" (singular) or "den Tiefen" (plural). It's the ending "-en" for the dative case that is archaic in this case

@MatheiBoulomenos Ithink that was meant for leaky :)

@MatheiBoulomenos no, I don't think the archaic dative ends in -en here

I think archaic dative is only for masculine nouns

Leaky how can you argue with a german guy about german stuff :D

1:54 AM

@KasmirKhaan I'm an admin of en.wikt

I just don't speak German fluently