Mathematics

Akiva Weinberger

4:00 AM

Darude Sandstorm is pretty good tbh

MatheiBoulomenos

Honestly I have no idea how you prove $S_5 \cong \operatorname{PGL}(2,5)$

Balarka Sen

lol it's good elevator music after you slow it down x200 times

Secret

Either the dreams starting from the day I learn foundations of mathematics likes to pick ancient memories for some reason, or it is a sign I might be having alzhiemers

Balarka Sen

I think I like I'm blue (da ba dee da ba die da ba dee da ba die) unironically

MatheiBoulomenos

the literal translation of "being blue" in German means "to be drunk"

Balarka Sen

4:02 AM

I <3 Russian Sandstorm

Secret

Eiffel 65 - I'm Blue (Da Ba Dee Da Ba Die) + lyrics

►

Akiva Weinberger

Nyan Cat was only six years ago

Wow

Secret

Vitas - 7th Element (russian blblblblbl)

►

balulululululululul ba

Balarka Sen

vaporwave +9000

Akiva Weinberger

lol i remember when "What Does the Fox Say" was everywhere

Let's make a revival

Balarka Sen

4:04 AM

good idea

2017 is the greatest

Pink Season tops everything

Secret

Ylvis - The Fox (What Does The Fox Say?) [Official music video HD]

►

Akiva Weinberger

I have never heard of this thing @BalarkaSen

Balarka Sen

You have never heard of Filthy Frank?

Akiva Weinberger

So I guess Secret's about to link to it now probably

or not

Secret

PINK SEASON (FULL ALBUM)

►

Akiva Weinberger

4:06 AM

ah there we go

Balarka Sen

It's the greatest fucking album ever

Truly a hip hop masterpiece

Akiva Weinberger

holy shit is literally the whole thing like this

Balarka Sen

Yep!!!

Akiva Weinberger

I'm clicking to random parts in this and they're all like that

What the hell

This is like 70 minutes long

Balarka Sen

My favorite track is STFU

Google it and listen with music vid

Akiva Weinberger

4:09 AM

I am amazed at the power of the internet

Secret

@Semiclassical turner.faculty.swau.edu/mathematics/materialslibrary/pi/…

Akiva Weinberger

Hm do Collatz sequences look nice in base two

Oh uh Balarka did I ever show you the "O Fortuna" thing

O Fortuna Misheard Lyrics

►

frikkin enjoy

Secret

groups.google.com/forum/#!topic/sci.math/Cs_vx--GFyo

Balarka Sen

Checking

Secret

The binary is literally like a pine tree

Balarka Sen

4:12 AM

Oh right this is the thing you gave me when I told you about the Potato give me power thing

I love this

Akiva Weinberger

So I guess the only point is if you just look at the odd ones or something

re collatz

THIS OCTOPUS

LET'S GIVE HIM BOOTS

SEND HIM TO NORTH KOREA

frickin automod tellin me i'm posting too fast YOU CAN'T STOP ART

Balarka Sen

This is quality art

top 10 lyrics of all time

MatheiBoulomenos

@Kasmir do you want the solution? I don't think its the most effective use of your time to think that long about one exercise

given that you have an exam on monday

Akiva Weinberger

oh wow good luck

Secret

I cannot believe a misheard can be an entirely new song itself

Kasmir Khaan

4:15 AM

hmm well yeah it is true but i found it as a good challenge ><

thanks akivia :)

Akiva Weinberger

I don't know if my sleep-deprived brain is in the best position to wish you luck

but, uh, yeah, good luck

Kasmir Khaan

:D

well i had this 120 = 2^3 * 3* 5

( i dont know how we would use sylow theorem here )

but since you said we would use it so I did that

after that , i really could not figure how the pieaces are connected

MatheiBoulomenos

Big hint: how many 5-cycles are there in $S_5$ (that's not even group theory, it's combinatorics)?

Secret

drew all the cycles on paper

Kasmir Khaan

there are 5 of them

MatheiBoulomenos

4:17 AM

you don't need to draw them all out

Akiva Weinberger

erm 24?

MatheiBoulomenos

@Kasmir, no that's wrong

Yes 24 is right

Kasmir Khaan

5 cycle?

there are 24 ? :O

Akiva Weinberger

'cause you can assume our cycle notation starts with 1, and then we can order the other four in 4! other ways

Kasmir Khaan

hmm neat._.

i never worked with more than S_4 =p

Anyway :D

Akiva Weinberger

4:19 AM

like (12345) and (12354) and (12435) and (12453) and (12534) and (12543) and

Secret

or box counting: first place there are 5 choice, then next has only 4 and so on thus 5!

Kasmir Khaan

we have 24 , 5 cycles

Yes i see the argument :)

MatheiBoulomenos

what does a subgroup of order 5 of $S_5$ like?

@Secret (12345) and (23451) are the same cycle

Kasmir Khaan

hmm

Secret

ah ok, so need to account for cyclic permutations

Kasmir Khaan

4:20 AM

look like you mean ?

MatheiBoulomenos

what elements does it contain?

in terms of cycle type

Kasmir Khaan

well i can contain 5 cycles or 1 cycle

Akiva Weinberger

@Secret Right so 5!/5

Kasmir Khaan

identity and 4 elements of 5 cycls

MatheiBoulomenos

right

Kasmir Khaan

4:21 AM

one is the inverse of the others between those 4

anyway ><

MatheiBoulomenos

so how many 5-Sylowsubgroups of $S_5$ are there?

Kasmir Khaan

well this is from sylow 3

MatheiBoulomenos

you don't need to use sylow 3 for that if you did the combinatorics

Kasmir Khaan

the number of sylow p-subgroups

MatheiBoulomenos

but you can use sylow 3

Kasmir Khaan

4:23 AM

oh okay ,cant do that since am not done with it yet ><

let me think 1 second

i give up ><

MatheiBoulomenos

can any 5-cycle be contained in two different subgroups of order 5?

Kasmir Khaan

well it cant

because if it is in one subgroup of order 5

then its inverse must be there

and to keep the group structure , the other element and its inverse cant be random

MatheiBoulomenos

I don't see how the inverse is related

Kasmir Khaan

i was thinking e, a ,a' , b, b'

MatheiBoulomenos

it's just due to the fact that any element of order 5 generates the whole group of order 5

Kasmir Khaan

4:29 AM

true

i want to say 24 such subgroups

but i dont feel it is right

MatheiBoulomenos

No!

Kasmir Khaan

:/

MatheiBoulomenos

each of these subgroups contains 4 5-cycles

Kasmir Khaan

ah

MatheiBoulomenos

and you know that every 5 cycle is contained in exactly one such subgroup

Kasmir Khaan

4:31 AM

6 then :D

MatheiBoulomenos

yes!

Kasmir Khaan

:DDD

MatheiBoulomenos

Can you finish the argument?

Kasmir Khaan

hmm thinking 1 second :D

we have 6 subgroups of order 5

grrrrrrrr

I dont see it

MatheiBoulomenos

can you think of any action of $G$ on the set of subgroups of order 5?

Kasmir Khaan

4:37 AM

i know what it is but i dont see how all the argument is built up to there

the action is conjugation

MatheiBoulomenos

yes

Kasmir Khaan

but i dont really see how it all fits

MatheiBoulomenos

now why is this action transitive?

Kasmir Khaan

we have a group S_5

well because conjugation action is a bijection

MatheiBoulomenos

no, that's not true

Kasmir Khaan

4:38 AM

oh

MatheiBoulomenos

what does it mean for this action to be transitive? write it out

Kasmir Khaan

action is transitive if it has 1 orbit

hmm

okay just if we back a little

we have a group S_5

we picked 6 subgroups of order 5

they are 6 unique subgroups of order 5

MatheiBoulomenos

yes

Kasmir Khaan

if we call one of those subgroup H

H acting on [1]

1 in the set 1-6

MatheiBoulomenos

what do you mean by "H acting on [1]"?

Kasmir Khaan

4:43 AM

yeah exactly does not make any sense

do we see thos 6 subgroups as "elements " here?

MatheiBoulomenos

yes

Kasmir Khaan

are those 6 subgroups , do they form a group ?

MatheiBoulomenos

no

they just form a set

the set of all subgroups of order 5

Kasmir Khaan

hmm okay

how do we have a an action from a set to another set?

MatheiBoulomenos

But $S_5$ is a group!

Kasmir Khaan

4:45 AM

grrrrrrrrr

(

MatheiBoulomenos

It's the group that we want let to act on something

Kasmir Khaan

:(((

oh if we let S_5 act on that set

ahhh :D

I think i see it :D

the elements of the set i should see them as subgroups of order 5 not as {1,2 .. ,6 }

MatheiBoulomenos

exactly

Kasmir Khaan

So S_5 acting on that set of subgroups

hmm

this is very odd for me tbh

where did you find such question -.-

MatheiBoulomenos

you're almost there

Kasmir Khaan

4:47 AM

okay :D

MatheiBoulomenos

you asked me "what is Sylow 2 good for" or something like that, and this is what I came up with

Kasmir Khaan

haha nice =p

okay if we let sigma be in S_5

MatheiBoulomenos

So what does it mean for two subgroups to be in the same orbit under this action?

Kasmir Khaan

sigma H sigma inverse is also a subgroup of order 5

so now I know that elements of S_5 acting by conjugation on that set

they keep it in the same form, ie subgroup of order 5

MatheiBoulomenos

you need to invoke Sylow 2 at some point

Kasmir Khaan

4:51 AM

hmm

well all 5 -sylow subgrups are conjugate to each other

MatheiBoulomenos

yes, that's it! One way to phrase that part of Sylow 2 is that the action of a group on its p Sylow groups by conjugation is transitive

Kasmir Khaan

:D

nice nice :D

I definitly need now to do sylow 2 and 3 in a better way

and then try to solve this question without looking at our convo ><

MatheiBoulomenos

So now we have constructed a transitive action of $S_5$ on a $6$ element set, it's pretty nontrivial that such a thing exists

Kasmir Khaan

I hope you ull be here tomroow :D

yes it was =P

MatheiBoulomenos

I will, but not all day long

Kasmir Khaan

4:55 AM

I know ofc =p

But hmm ill be up untill 7 pm

maybe wake up 4 am monday

i have exam 9 am

If we meet for one hour on that intervall it would nice :D otherwise thanks alot for all your help :D

you are the best!

:D

MatheiBoulomenos

sorry for that exercise, it was a little bit too difficult, I guess

Kasmir Khaan

haha no i have to thank you for that ! it was difficult finding what set of 6 elements

it will act on

I mean was just focusing on the set {1,2..,6 } did not figure out that the set will come as subgroups

MatheiBoulomenos

in group theory, a lot of times you let a group act on subgroups or cosets. I used that so often that I forgot it's not the most obvious thing in the world

Kasmir Khaan

haha =p yeah same with my teacher, he keeps saying that stuff are obvious when we have no idea of what he is talking about

Secret

5:29 AM

[Random]

Leaky Nun

5:44 AM

♦ $Mathematics$ Logic

This room is meant for discussion about logic, including found...

1

a discussion on incompleteness along with a walkthrough of its proof will take place here after 5 hours

anyone is welcome

Secret

8:58 AM

in Mathworks, 5 mins ago, by Secret

n < n+1 ordering criteria: More 1 s under +
n < n2 ordering criteria: More n s under +
n < n^2 ordering criteria: More n s under */more n lots of 1 s under +
n < n^n ordering criteria: More n s under ^/ more n lots of n s under */ more n^2 lots of 1 s under +
n < epsilon_n ordering criteria: More levels of n under ^ / (skip)
n < zeta_n ordering criteria: More layers of n under epsilon_
n < phi(,) ordering criteria: More nesting of n under phi()
n < Ferfermann_n ordering criteria: Limit of phi nestings

It appears there is no explicit ordering criteria to tell the set on how to well order countable well ordered and computable strings

Maneesh Narayanan

9:26 AM

How can I write statement of purpose by understanding the subject. I roughly check the area of research. it founded interesting. How do I get deep insigt in the subject. where will I get current papers in that subject?

@LeakyNun

@Secret

@KasmirKhaan

Secret

what research area you are reading on?

Maneesh Narayanan

Mathematics for Quantum Physics

where will I get the current papers?

@Secret

Albas

@ManeeshNarayanan Isnt that quite a broad topic and by the way you are stating it, it sounds vague.

Secret

That would be a better question for h bar, since quantum papers are all over the place. But as always, start with a textbook first

Maneesh Narayanan

I have done a course in quantum mechanincs

@Secret

I have done hilbert sapce theory also

I wanted to know, where will I get a current research papers in that subject?

I wish to join the phd course on that topic

what shall I do before that?

Secret

9:34 AM

Here's the problem: Quantum mechanics and quantum computing researches are published in a variety of journals thus I am unable to refer you to a single source

Have you tried physics review D and nature?

Maneesh Narayanan

best journal for Quantum Physics

Mathematical Physics journal

no

Secret

in The h Bar, 20 secs ago, by Blue

There's one for quantum information. Don't know about QM

in The h Bar, 13 secs ago, by Blue

International Journal of Quantum Information

The International Journal of Quantum Information was established in 2003 and is published by World Scientific. It covers the field of quantum information science, with topics on areas such as quantum metrology, quantum cryptography, quantum computation, and quantum mechanics. == Abstracting and indexing == The journal is abstracted and indexed in Zentralblatt MATH, Science Citation Index Expanded, CompuMath Citation Index, Current Contents/Physical, Chemical & Earth Sciences, Inspec, and Scopus. == References == == External links == Official website...

Maneesh Narayanan

Not interested in quantum information. i am interested in Mathematical physics

thank for your help :)

Secret

The h Bar

General chat for Physics SE (physics.stackexchange.com). For M...

6

7

Felix.C

10:47 AM

hey does someone know a good text with a proper definition of the Dirichlet condition for Fourier series? Everything I found so far states a bit something else than the other...I saw weak and strong variants...I'd like to see a definition that is as generell as possible :)

Yashas

I can come up with many answers to get 51/2010 but how do I prove that best answer I got is really the best answer?

Typhon

11:04 AM

is it me or are people quite rude here?

-3

Q: Is there any existing precedent that states that divergent limits return ranges of values rather than single numbers?

Now I know that infinitly diverging limits always do not exist. Infinity is not a number and assuming it exists leads to contradictions. Now consider something like $sin(x)$ as $x$ goes to infinity. We know the limit returns something between $1$ and $-1$. I also know that range algebra exists a...

Anonymous

11:28 AM

@Yashas It says "least"

Yashas

@Blue yea I am not able to show that the solution I got uses the least possible number of integers

I just have a solution

Anonymous

It' pretty straightforward isn't it? $a_1>1$ for sure. So $51/2010=(1-1/a_1)(1-1/a_2)...(1-1/a_n)\geq(1-1/2)(1-1/3)....(1-1/(n+1))$ [wlog $a_1<a_2<...<a_n$]. You get $n+1 \geq 2010/51$. It's easy to get the lowest integral bound for $n$ from there.

Anonymous

I get $n>38.41..$

Anonymous

So lowest value of $n$ is $39$. @Yashas

Yashas

there needn't be a solution with $n$ numbers

you showed that the absolute minimum is 39

at least 39 or more numbers are needed

but there needn't be a solution with 39 numbers

Anonymous

11:42 AM

It's easy to check. Try with 2,3,4,5,....,40. Adjust the last number

Anonymous

I think 67 will do

Anonymous

So the set is {2,3,4,....,40,67}

Lozansky

1:26 PM

How can I see that $\phi(t) = 0$ is an unstable solution to the equation $\dfrac{d \phi}{dt}(t) = [1+e^{\phi(t)}]sin(\phi(t))$?

Kasmir Khaan

@LeakyNun here?

@MatheiBoulomenos Morning mathei :D

Leaky Nun

there

Balarka Sen

@Lozansky I think stability means that solutions starting off with nearby initial points to $0$ stay close to $\phi = 0$?

Kasmir Khaan

:D

I have a basic question about rings

ÍgjøgnumMeg

@KasmirKhaan Go on

Balarka Sen

1:38 PM

I think that's equivalent to requiring $f(x) = (1 + e^x) \sin(x)$ having derivative $< 0$ at $x = 0$. (Which it doesn't)

So that should be unstable

Leaky Nun

@BalarkaSen it's strange to see those words coming from your mouth

Balarka Sen

Huh?

Leaky Nun

what

Balarka Sen

I don't understand what's strange

Kasmir Khaan

@ÍgjøgnumMeg how do we view the polynmial rings? R[x], in my notes they are not viewd the same as the function

@ÍgjøgnumMeg I know its not very well written question but I hope you understand what i mean

ÍgjøgnumMeg

1:40 PM

@KasmirKhaan I don't think your question makes sense

Kasmir Khaan

i give example =P

R = Z/2

there are 4 functions from R--> Rr

but infinitly many polynomials in R[x]

what does this suppose to mean ?

Leaky Nun

@BalarkaSen you just usually don't talk about those things lol

ÍgjøgnumMeg

Polynomials over $R$ just mean the polynomials have coefficients in $R$

Alessandro Codenotti

@Balarka I have an $A$-module $M$ and two $A$-submodules $N,P\subseteq M$. I want to show that $S^{-1}(N+P)=S^{-1}N+S^{-1}P$ and that $S^{-1}(N\cap P)=S^{-1}N\cap S^{-1}P$, this is quite straightforward to see just by writing down the elements in those sets

But the professor said it can be derived from the exactness of $0\longrightarrow N\cap P \stackrel{f}{\longrightarrow} N\oplus P\stackrel{g}{\longrightarrow} N+P\longrightarrow 0$, where $f(a)=(a,-a)$ and $g((a,b))=a+b$ and the fact that localization preserves the exactness of sequences, but I don't see how that'd work

Leaky Nun

@KasmirKhaan they are formal polynomials

Kasmir Khaan

1:41 PM

so we do view polynomials as a different thing here?

Balarka Sen

@LeakyNun ?? This is ODE theory, which is everything I do these days (I admittedly do not know about stability yet)

Kasmir Khaan

yes leaky exactly

what does that mean ?

Leaky Nun

view them as a function $\Bbb N \to R$

it's just a list of their coefficients

that's all there is to it

it's just a sequence of elements of $R$

Kasmir Khaan

so a typical element in R [Z/2] is 1+x

is this statement correct?

Leaky Nun

"typical"?

Kasmir Khaan

1:43 PM

i mean ><

Leaky Nun

no, that's wrong notation

Kasmir Khaan

example of that

Leaky Nun

it's Z/2[X]

here R = Z/2

Kasmir Khaan

Oh yeah ><

Sorry about that

Balarka Sen

@Alessandro Hm. I think $S^{-1}(N \oplus P) = S^{-1} N \oplus S^{-1} P$ is quite easy to see.

Lozansky

1:43 PM

@BalarkaSen What theorem is that?

Kasmir Khaan

okay thanks :)

Lozansky

That requires the derivative to be negative

Or that says that positive derivative implies unstable solution

Leaky Nun

@KasmirKhaan x^2+x+1 is also an element of R[X]

so is x^100+x^49+x^27+1

Balarka Sen

Just use the projection maps $N \oplus P \to N$ and $N \oplus P \to P$ to obtain a map $S^{-1}(N \oplus P) \to S^{-1} N$ and $S^{-1}(N \oplus P) \to S^{-1} P$, and take the direct sum to obtain the map $S^{-1}(N \oplus P) \to S^{-1} N \oplus S^{-1}P$

This is the potential candidate for the isomorphism

Alessandro Codenotti

@BalarkaSen everything is done component-wise so I agree

Balarka Sen

1:45 PM

Right.

Kasmir Khaan

@LeakyNun so we reduce only the coefficents mod 2 ?

Leaky Nun

@KasmirKhaan yes. As I said, it is a sequence of elements of R

1+x+x^2 means nothing more than (1,1,1)

Kasmir Khaan

ahh so we can use only 0 and 1

Balarka Sen

@Lozansky I am not sure. I'm 70% confident that you'd need to have that but I have to ponder if that's correct.

Kasmir Khaan

but we can have as many eleents as we want

of the form a_i x^i

Leaky Nun

1:46 PM

1+x+x^3+x^5+x^6 means nothing more than (1,1,0,1,0,1,1)

Kasmir Khaan

a_i is 1 or 0 here

I got it :D

thanks for claryfying that leaky :D

Lozansky

@BalarkaSen Okay I'll take your word for it

How can I see that there is an open interval $I$, containing $0$, such that the IVP $$\dfrac{d\phi}{dt}(t) = \phi^{2/3}(t) \\ \phi(0)= 0$$ has a solution on $I$?

ÍgjøgnumMeg

@KasmirKhaan Just think of regular old polynomials, but the coefficients are in your ring $R$

@KasmirKhaan That's all you need :P

Alessandro Codenotti

@BalarkaSen Ok, that makes sense. So I get a short exact sequence $0\longrightarrow S^{-1}(N\cap P) \stackrel{S^{-1}f}{\longrightarrow} S^{-1}N\oplus S^{-1}P\stackrel{S^{-1}g}{\longrightarrow} S^{-1}(N+P)\longrightarrow 0$

Kasmir Khaan

@ÍgjøgnumMeg my teacher confused me when he said "we wont view them as standard polys , like how we used to "

ÍgjøgnumMeg

1:50 PM

@KasmirKhaan What did he mean then? o.O

Kasmir Khaan

@ÍgjøgnumMeg er du norsk btw? :D

I dont know yet << but I feel it is just what u said

ÍgjøgnumMeg

@KasmirKhaan Nei er eg ikkje hahaha

Kasmir Khaan

dansk ?

><

Lozansky

I see that $f(t) = \phi^{2/3}(t)$ is not Lipschitz in an interval containing $0$, so I know there isn't a unique solution

ÍgjøgnumMeg

@KasmirKhaan Neeei engelsk

hahaha

Kasmir Khaan

1:51 PM

haha

Balarka Sen

@Alessandro I mean, I see why proving $S^{-1}(N \cap P) \cong S^{-1}N \cap S^{-1}P$ and then that $S^{-1} f$ becomes $S^{-1} f_N \oplus S^{-1} f_P$ under that isomorphism would prove $S^{-1}(N + P) \cong S^{-1} N + S^{-1} P$

But per se, I don't see how the exact sequence proves both isomorphisms

Kasmir Khaan

Okay kasmir back on thinking mode

got exam tomorrow and I know nothing about rings -.-

Balarka Sen

@Lozansky Try to write down a solution.

Kasmir Khaan

but at least I got the group part right :D

Balarka Sen

The Picard-Lipschitz indeed does not apply here

ÍgjøgnumMeg

1:52 PM