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

TastyRomeo

00:00

Well, as it turns out, someone posted a thread "Why do people keep trying 2 Pi no scopes to end the game?" back in 2011

Balarka Sen

I am still pondering how to do a 4pi solid angle noscope

that would be some serious 2001 space odyssey shit

Meow Mix

BACK from burrito break

TastyRomeo

If you like burritos, you should play the new south park game :p

Semiclassical

the new south park game seems like "Fart Jokes: The Game"

Ted Shifrin

only fart jokes?

TastyRomeo

00:08

It's basically fart jokes and references to south park episodes, yeah

Balarka Sen

lol the new south park game

TastyRomeo

Still really fun, though

Balarka Sen

I mean the title of the game itself is a fart joke

Semiclassical

it has about the level of tastefulness you'd expect

Balarka Sen

"Fractured But Whole"

Ted Shifrin

00:09

I actually loved South Park when I used to watch it. I can't believe it's still on.

Meow Mix

does anyone here watch trailer park boys

Balarka Sen

@Ted !!

Ted Shifrin

I think Balarka just lost what little respect he had for me.

Meow Mix

jim lahey died :(

Balarka Sen

I was just really surprised haha

Kasmir Khaan

00:14

Hey Ted :)

I got a fast Q :)

Ted Shifrin

@Balarka: There are things you don't know about me :D

Yeah, Kasmir?

Kasmir Khaan

in the proof of all groups of order p^2

they are either iso to Z/ p^2 or Z/p x Z/P

Ted Shifrin

Yup

Kasmir Khaan

by cauchy theorem we have asubgroup of order p

let H= <a> has p elements in it

now let b be in G minus H

Ted Shifrin

Have you already proved the center of a $p$-group is nontrivial?

Kasmir Khaan

00:16

the group generated by " b" is either G or

Yes I did that :)

anon

(and G/Z(G) can't be cyclic, so G is abelian)

Ted Shifrin

OK

Kasmir Khaan

Am just wondering about a logical statement here

Yes that is good observation =p

Our prof said

this element b can generate whole G but it is not in H = <a>

so first we say that b is not a power of a

Ted Shifrin

Ugh

What's the order of $b$?

Kasmir Khaan

but then we say that b can be a power of a , since it generate all G

Ted Shifrin

00:18

cannot be

Kasmir Khaan

unkown right now

Ted Shifrin

Yes, but that's the right question to ask.

anon

<b>=G does not imply b is a power of a, it implies a is a power of b

MatheiBoulomenos

if b generates G, this means that a is power of b, not the other way around

sniped

Kasmir Khaan

ohhh :D

Ted Shifrin

00:18

It's fun being the algebra plebe here.

Kasmir Khaan

That clears things up :D

Thanks guys ! :D

Ted Shifrin

But I still think you should ask yourself, "Self: What is the order of $b$?"

Kasmir Khaan

well it can be 1 , p or p^2

anon

can't be 1

Ted Shifrin

I would do the proof differently. I would start with: "Is there an element of order $p^2$?"

Kasmir Khaan

00:19

right anon ><

it cant be 1

we took b from G - H

Hmm so now i see why

G is either Z/ p^2 or Z/p x Z/p

if this b has order p^2 then it is the first case , if not it has order p then the second case

Ted Shifrin

Yeah, I think it's clearer to start from my perspective. If there's an element of order $p^2$, it's cyclic and we're done. If not, every element has order $p$. Now do your thing.

You don't even need Cauchy.

MatheiBoulomenos

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

Kasmir Khaan

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

I proved that mathei =P

MatheiBoulomenos

Oh okay

Kasmir Khaan

all groups of order p^2 are abelian =p

Ted Shifrin

00:22

Good, Kasmir :)

Kasmir Khaan

Ted even has criminal examples on his book :D

Ted Shifrin

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?

Ted Shifrin

calls lawyer

Kasmir Khaan

a drugdealer has 1 , 3 , 18 ons of drugs

Ted Shifrin

00:22

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

Kasmir Khaan

or something like that

Meow Mix

oh ted

can you check my proofs

MatheiBoulomenos

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

jk

Ted Shifrin

glares at Mathei

What proofs, Meow?

Kasmir Khaan

let me think a second @anon

Meow Mix

00:23

let me send

Ted Shifrin

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.

Meow Mix

karl sounds like the name of an algebraist

Ted Shifrin

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

Meow Mix

sent

MatheiBoulomenos

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

Kasmir Khaan

00:28

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

Ted Shifrin

@Meow: Answered.

Physics Guy

Hey Mr. Shifrin

Ted 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.

Kasmir Khaan

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

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

Ted Shifrin

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

Kasmir Khaan

00:41

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 :)

Ted Shifrin

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

MatheiBoulomenos

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

Meow Mix

thanks

user76284

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

Ted Shifrin

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

Physics Guy

00:45

@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 $k(x,y) = \exp(-d(x,y))$?

Kasmir Khaan

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

Ted Shifrin

I know nothing about machine learning, @user76284.

Kasmir Khaan

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

because it is about selling drugs =p

Meow Mix

that's just ternary

Ted Shifrin

00:46

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

Meow Mix

ive never heard druggist

Ted Shifrin

A druggist is a pharmacist :)

MatheiBoulomenos

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

Meow Mix

i guess pharmacist has a less bad connotation

Kasmir Khaan

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

Ted Shifrin

00:46

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

Meow Mix

it could have been a drug dealer

Kasmir Khaan

breaking bad?

kind of stuff? :D

Ted Shifrin

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

Kasmir Khaan

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

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

Ted Shifrin

Go back to studying, Kasmir.

Kasmir Khaan

00:47

Yes sir !:)

Faust

Potions = lol

Ted Shifrin

Hush, you.

user76284

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

Meow Mix

i deal with a lot of base 2

and base 16

lots of base 16

Ted Shifrin

I remember when base 12 was supposedly important.

Meow Mix

00:49

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

Semiclassical

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

Meow Mix

im past my cpu-building prime

Ted Shifrin

you're composite now?

Semiclassical

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

Faust

lol

Meow Mix

00:51

im compelled to be the smacker rather than the smackee now

Faust

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

Ted Shifrin

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

Semiclassical

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

Faust

@TedShifrin i have just not today

Semiclassical

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

Ted Shifrin

00:52

No reason to stick to middle thirds, though.

Semiclassical

not sure what you mean by that

Ted Shifrin

You can do Cantor with things other than middle thirds.

Semiclassical

well, sure

but the usual version

hmm, now I wonder

Meow Mix

i should start tutoring

i need to become a better teacher

im often too all over the place

Ted Shifrin

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

Semiclassical

00:55

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."

Meow Mix

i should teach ted 6502 assembly

Ted Shifrin

Yes, Semiclassic, we know that :)

Faust

lol sure ted we all know that...

Semiclassical

Can one generalize that to a different base?

Ted Shifrin

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

Leaky Nun

00:58

@Semiclassical yes, fat cantor set

Semiclassical

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

Ted Shifrin

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

Leaky Nun

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

Semiclassical

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

Ted Shifrin

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

Faust

00:59

@TedShifrin nope i have never taken topology

Semiclassical

yeah.

Ted Shifrin

or analysis @Faust

Faust

its a point set topology thats not dense or something?

hey im doing Anal sis right now

Ted Shifrin

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

Leaky Nun

@Faust ...

Semiclassical

01:00

...

Faust

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

Leaky Nun

@TedShifrin or a "perfect set" to be precise

Ted Shifrin

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

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

Semiclassical

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

Faust

is it actually a point set topology?

Ted Shifrin

01:01

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

Leaky Nun

@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

Semiclassical

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

Ted Shifrin

Why, Leaky? Think about good teaching.

Faust

ah

i am taking topology next semester im so excited

Leaky Nun

@Semiclassical why not?

Faust

01:03

also taking a second Anal ysis class

Ted Shifrin

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

Semiclassical

should be?

yeah.

Ted Shifrin

So that's $5/9$.

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

Semiclassical

lol

Leaky Nun

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

Semiclassical

01:05

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

Leaky Nun

@Semiclassical I don't think that's right

Ted Shifrin

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

Semiclassical

hmmm

Ted Shifrin

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

Semiclassical

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

Ted Shifrin

01:08

Ah, cool.

Leaky Nun

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

Ted Shifrin

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

Leaky Nun

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

Ted Shifrin

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

Leaky Nun

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

Ted Shifrin

01:09

Yeah, that's right.

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

I flunk.

Leaky Nun

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

Semiclassical

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

Ted Shifrin

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

Leaky Nun

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

Semiclassical

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

...wtf

Leaky Nun

01:14

@Semiclassical my expression is wrong

Ted Shifrin

The sum starts at $n=1$.

Semiclassical

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

Ted Shifrin

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

Semiclassical

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

Ted Shifrin

Oh, you shifted indices. Never mind.

Leaky Nun

01:16

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 $\dfrac {\frac12} {1-\frac52} = ???$

Semiclassical

it's a divergent sum in that case

Leaky Nun

it can't be divergent

Semiclassical

as you've written it, it certainly is

Leaky Nun

then what I've written is wrong, certainly

Semiclassical

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

Leaky Nun

01:22

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

Maneesh Narayanan

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

y?

Please help me. How to convert it automatically?

MatheiBoulomenos

01:38

@LeakyNun Hi

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

Leaky Nun

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

@MatheiBoulomenos which one...

MatheiBoulomenos

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

Leaky Nun

:o

its discriminant, lol

MatheiBoulomenos

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

Leaky Nun

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 @_@

Maneesh Narayanan

01:46

From where will I get solved problems of syllows theorem

?

MatheiBoulomenos

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

Kasmir Khaan

mathei :D

I got a fast Q :)

MatheiBoulomenos

sure

Kasmir Khaan

:D

Well the second sylow theorem

from what i understood

Leaky Nun

@ManeeshNarayanan out of the depths of the ocean

aus der Tiefen

rufe ich Herr zu dir

MatheiBoulomenos

01:48

*aus den Tiefen

Maneesh Narayanan

In examination, I am strugglig

Leaky Nun

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...

Kasmir Khaan

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

Maneesh Narayanan

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

MatheiBoulomenos

I see, that's archaic German I guess

Kasmir Khaan

01:49

what does this means in easiar terms ? :D

Maneesh Narayanan

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

how to master in that?

Leaky Nun

@MatheiBoulomenos is the der archaic or the Tiefen?

@ManeeshNarayanan practise makes master

Maneesh Narayanan

Can you suggest excersise book?

Leaky Nun

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

Maneesh Narayanan

@LeakyNun

Can you suggest excersise book?

Leaky Nun

01:52

@ManeeshNarayanan I have no idea

@MatheiBoulomenos fixed by conjugation, you know

Maneesh Narayanan

@LeakyNun then where shall I do exercises?

MatheiBoulomenos

@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

Kasmir Khaan

@MatheiBoulomenos Ithink that was meant for leaky :)

Leaky Nun

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

I think archaic dative is only for masculine nouns

Kasmir Khaan

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

Leaky Nun

01:54

@KasmirKhaan I'm an admin of en.wikt

I just don't speak German fluently

Kasmir Khaan

aha nice :D

did not knew you speak german =p

how many languages you know?

english french german and?

Transcript for

$Mathematics$ Mathematics