Mathematics - 2019-12-04 (page 1 of 2)

Maximilian Janisch

00:00

@Ultradark $$\zeta(3) =\frac12\int_0^\infty \frac{t^2}{e^t-1}\,\mathrm dt$$

anakhro

I think so, Ted.

Maybe I need to take a break when I resort to asking about elementary things.

Maximilian Janisch

@Ultradark See for instance math.stackexchange.com/questions/2556986/…

Ultradark

Okay cool

I found another one

$$ \int_0^1 \frac{\ln(x)\ln(1-x)}{x}~dx=\zeta(3) $$

Maximilian Janisch

@Ultradark Ref math.stackexchange.com/questions/1581354/…

Ultradark

yeah I didn't look at that post

Maximilian Janisch

00:09

Did you just feel that this integral is Zeta(3)? :D

anakhro

No, he smelled the zeta on it.

Ultradark

each integral has a specific odor

Semiclassical

fwiw, that zeta integral (involving 1/(e^t-1)) shows up noticeably in statistical mechanics

namely, when doing certain computations re: photons

Ultradark

$$\int_0^1 \frac{\ln^3(x)\ln(1-x)}{x}~dx=6\zeta(5)$$

Semiclassical

that one, by contrast, I've never seen in physics :P

Ultradark

00:22

$$\int_0^1 \frac{\ln^5(x)\ln(1-x)}{x}~dx=120\zeta(7)$$

Exercise 2.65 show that the floor of this integral $$\int_0^1 \frac{\ln^7(x)\ln(1-x)}{x}~dx$$ equals the sum of the first $100$ natural numbers

I'm starting to notice a pattern

Leaky Nun

01:22

$\begin{array}{rcl} \displaystyle \int_0^1 \frac{ \ln^n(x) \ln(1-x) } x \ \mathrm dx &=& \displaystyle \int_{-\infty}^0 u^n \ln(1-e^u) \ \mathrm du \\ &=& \displaystyle \int_0^\infty -t^n \ln(1-e^{-t}) \\ &=& \displaystyle \sum_{k=1}^\infty \frac1k \int_0^\infty t^n e^{-kt} \\ &=& \displaystyle \sum_{k=1}^\infty \frac1{k^{n+2}} \Gamma(n+1) \\ &=& n! \zeta(n+2) \end{array}$

Akiva Weinberger

0

Q: Do North Koreans use Latin letters in their equations?

Do North Koreans use Latin (and Greek) letters in their equations? On the one hand, being such an isolationist country, I wouldn't be surprised if they used the Korean alphabet (조선글) in their equations. On the other hand, the similarly isolationist Soviet Union used Latin letters in its equation...

What are the odds of an answer

Leaky Nun

> Nearby China uses Latin as well (imagine using characters (漢字) as variables! …Actually that would be so cool. But never mind)

@AkivaWeinberger we actually do use characters as variables

Heavenly Stems

The ten Heavenly Stems or Celestial Stems (Chinese: 天干; pinyin: tiāngān) are a Chinese system of ordinals that first appear during the Shang dynasty, ca. 1250 BC, as the names of the ten days of the week. They were also used in Shang-period ritual as names for dead family members, who were offered sacrifices on the corresponding day of the Shang week. The Heavenly Stems were used in combination with the Earthly Branches, a similar cycle of twelve days, to produce a compound cycle of sixty days. Subsequently, the Heavenly Stems lost their original function as names for days of the week and dead...

we use 甲乙丙丁戊己庚辛壬癸

Akiva Weinberger

So 2甲+乙=6? That sort of thing? @LeakyNun

Leaky Nun

I think so

this was so long ago lol

Ultradark

01:44

I can speak nonsense

William Sun

Is there anyway one can "pack together" stuff when dealing with equivalence classes? I will explain what I mean by this:

In the construction of ring of fractions, completions of modules/rings/groups it is kind of annoying to verify the operations on them are "well defined" since they are defined through equivalence classes.

Is there like a general categorical theorem that includes all the cases?

This is kinda of a useless question since one can always go through these process but just curious...

Rogue

02:04

In differential equations, the homogeneous and particular solution are both independently solutions, and I have a problem where I have three constants to solve for but only two initial conditions

It seems to be doable, as WolframAlpha solved it, but I'm wondering if I can use the solutions separately to solve for the constants instead of using just the general solution

as with the general solution alone, I'd only have two equations (one for x(0) and one for x'(0)), which is not enough to solve for 3 unknowns

The equation and initial conditions I have are x'' + 30.61x' = e^(-.2t); x'(0) = 0; x(0) = 0

Yourong Zang

@AkivaWeinberger I don't think they wrote equations. In ancient Chinese math books, equations were usually written in complete sentences. Also, they used graphs and charts to represent equations. Check this book for example Jade Mirror of the Four Unknowns.

Akiva Weinberger

@YourongZang Who said anything about ancient? I'm talking about today

Yourong Zang

@AkivaWeinberger Oh I saw @LeakyNun's comment.

@AkivaWeinberger If you are talking about today, then I believe everyone uses Latin characters, even in the nation that is the most reluctant to Western culture (since you list the USSR and China so I guess you are thinking of this problem politically). And @LeakyNun's is talking about ancient Chinese math.

Akiva Weinberger

Wait @LeakyNun when you said "this was so long ago lol" did you mean your childhood or your ancestors

Leaky Nun

my childhood

Akiva Weinberger

02:16

Yeah that's what I thought

Leaky Nun

20

Yourong Zang

@LeakyNun Wow that is unexpected lol. I thought using Chinese characters was like 50 years ago.

Leaky Nun

maybe I'm wrong

I really don't remember much

Semiclassical

02:43

@Rogue the homogeneous and particular solution are both solutions, but you can't add constant multiples of the particular solution. So I only see the two constants parametrizing the homogeneous solution.

Rogue

I'm not sure if I follow on that last statement

Ohhh my god I see now

Bless you

Semiclassical

02:58

np

Hawk

03:33

If a finite field F has characteristic p (prime), doesn't that mean p=0_F?

Shine On You Crazy Diamond

I think if that implies $F = \Bbb{Z}/p$ then yes

anakhro

@Hawk if you take p to mean 1 + 1 + ... + 1, p times, then yes. That's the definition.

Leaky Nun

yes, and I don't see what's wrong with that

Shine On You Crazy Diamond

Three answers! Spoiled

anakhro

@LeakyNun why would you think there is something wrong with that.

Leaky Nun

03:38

because he says "doesn't that mean"

anakhro

and?

Hawk

@anakhro, then consider the Frobenius endo where we take $a \to a^p$. This is supposed to be injective, but if $p = 0_F$ doesn't that mean $a \to 1_F$?

anakhro

Exponents are not in the field.

Exponents are integers.

Hawk

Isn't a^p = aaa....p times?

anakhro

Exactly.

Hawk

03:42

So the issue is idenfiying $p=0_F$ in the exponent? Because such law does not exist in a general field?

Leaky Nun

@anakhro "doesn't" is a rhetorical question

implying that it cannot be true

anakhro

Indeed. Exponents are all in integers.

Hawk

*fields with characteristic 0

i mean we are assuming $p$ is prime in a finite field, that's an integer

anakhro

No it is not. It's an element of the field.

It's 1+1+...+1, p times, where 1 is the identity element (and not necessarily "one" the integer).

Leaky Nun

let's just use $\uparrow p$ for the element of the field and $p$ for the integer

so $\uparrow p = \uparrow 0$

and $a \mapsto a^p$ is injective

Shine On You Crazy Diamond

03:44

^_^

Hawk

So what do we mean when we say charF is prime then?

Leaky Nun

char F is an integer

it is a prime number

Hawk

okay but that is the same number in the exponent

a^p

Leaky Nun

yes

anakhro

The notation p = 1+1+...+1 is not the same p as is in "char F = p".

Former is an element of the field. Latter is an integer.

Leaky Nun

03:47

$\uparrow p = \underbrace{1_F + 1_F + \cdots + 1_F}_{\text{$p$ times}} = 0_F$

$a^p = b^p \implies a = b$

$a^p = \underbrace{a \times a \times \cdots \times a}_{\text{$p$ times}}$

Balarka Sen

Weak Nullstellensatz still holds for $\Bbb C[\{x_i\}_{i \in I}]$, right? As in, the maximal ideals are of the form $(\{x_i - a_i\}_{i \in I})$.

'Cuz ideals in a direct limit of rings is direct limit of ideals, I think.

Leaky Nun

well if that's true then maybe that's true

oh no

10

Q: Maximal ideals of polynomial rings in infinitely many variables

Let $k$ be an algebraically closed field. Nullstellensatz states that the maximal ideals of the polynomial ring $R=k[X_1,\dots,X_n]$ are precisely those of the form $\langle X_1-a_1,\dots,X_n-a_n\rangle$, with $(a_1,\dots,a_n)\in k^n$. What if we are working with infinitely many variables, sa...

abstract-algebra commutative-algebra ideals

Balarka Sen

Wow clever

Hawk

hang on so in the Froben map, the pth power a^p, the p in the exponent is not charF but the prime number p?

anakhro

Yes.

Hawk

03:52

wow that is confusing.

Leaky Nun

@BalarkaSen my example was wrong as I misunderstood

anakhro

The catch is that char F is the prime number p.

Leaky Nun

instead you take $I = \Bbb C(X)$

and project the polynomial ring onto it

Balarka Sen

So I need to take $I$ to have smaller cardinality than $\Bbb C$

That's all

anakhro

You are just confused over the notation "p" in the field, @Hawk

Leaky Nun

03:53

apparently so

@AkivaWeinberger hi

Akiva Weinberger

Hey

You are not wrong. But USSR was never as isolationist as North Korea. — Moishe Kohan 1 hour ago

Hawk

I am using Dummit and Foote and in prop 35 where he introduces the Frobenius Endo (548) he literally just calls everything p.

anakhro

@Hawk yes the fact that he abuses notation and says p is in the field (if that is what he does), that could be very confusing.

However, you should note in particular how exponents are defined.

Akiva Weinberger

@LeakyNun Hey you know the theorem that says a finite field of characteristic $p$ must have cardinality a power of $p$

anakhro

In particular, if a is in the field and k is a positive integer, we define a^k to be aaa...a, k times.

Leaky Nun

03:55

sure

Akiva Weinberger

and the proof is: note that the field is a vector space over $\Bbb F_p$

Balarka Sen

yep

Leaky Nun

I was about to say that

Akiva Weinberger

What other places can that trick be used

Leaky Nun

in Galois theory

Akiva Weinberger

03:56

where you want to show blah is a power of a prime, and it turns out blah counts a vector space

Leaky Nun

oh

in knot theory

anakhro

@Hawk note that a^k wouldn't necessarily make sense if k were in the field.

Hawk

@anakhro i mean as an example, if i have some field F where the elements are matrices and lets say charF = 2 (a prime) then 2I = 0_F. But charF = 2I right? it can't be charF = 2 as you say.

Akiva Weinberger

Ah right the number of n-colorings of a knot, yeah?

Leaky Nun

@AkivaWeinberger yes

Balarka Sen

03:57

@AkivaWeinberger You use a similar trick to prove Wedderburn's little theorem

Akiva Weinberger

Wait n doesn't need to be a prime does it

@BalarkaSen Which one's that again?

Leaky Nun

well usually n is a prime

Akiva Weinberger

What if it's not, do we still get a power of n?

anakhro

@Hawk Right about 2I = I + I = 0_F, but char F = 2, not 2I.

Leaky Nun

no idea

Balarka Sen

03:58

@Akiva A finite division ring is a field

anakhro

The characteristic of F is defined to be the number of times you must add the multiplicative identity in order to get 0.

So it will be an integer.

Balarka Sen

You view everything as a vector space over the center and use the class equation on the multiplicative groups

Hawk

@anakhro so in the same example, if i have a matrix A, and i take the pth power (2 in this case), A^2 \neq A^charF

Leaky Nun

you use H^2

Balarka Sen

And then some number theory trick

Leaky Nun

03:59

it's equivalent to H^2 of every finite field is trivial

anakhro

@Hawk no.

Balarka Sen

Also true

anakhro

Sorry, no, it's not true. They are the same.

Balarka Sen

But that's not how Wedderburn's original proof goes

Leaky Nun

this chat is distracting, I have an assignment due in an hour that I still haven't started, bye

anakhro

04:00

Char F, the characteristic, and the power in the Frobenius endomorphism are ALL integers.

Akiva Weinberger

I'm guessing if n is not prime then the number of n-colorings need not be a power of n but at least it's a product of powers of the prime factors of n

anakhro

The thing which is NOT an integer, is p = 1 + 1 + ... + 1.

But that's merely an abuse of notation

Hawk

what's the abuse of notation? 1_F + 1_F = 2(1_F)

anakhro

The abuse of notation would be if you called that "1_F + 1_F" by "2".

As you wrote, it is more correctly, 2(1_F).

Akiva Weinberger

People sometimes will write $\bar p$ for the element of the finite field, and $p$ for the element of the integers

In any case, it's the image of of $p$ under the (noninjective) homomorphism $\Bbb Z\to{}$whatever you're working with

Hawk

04:07

But in every situation charF = p(1_F)= 0_F. right? But is there no division in the field to say p = 0_F?

in the context of beginning Galois theory

anakhro

No.

In every situation (excepting those that abuse notation) char F is an integer and NOT an element of the field.

Hawk

so you are saying my last statement is incorrect

the division part

the one that identifies the integer prime p with the 0_F

anakhro

You'd be abusing notation if you identified an integer with an element of the field.

(assuming the field is not built as a subset of the integers---it often isn't).

Hawk

So basically when we talk about A^p (pth power) we are NOT talking about A^(p*1_F). Because $p(1_F) = 0_F.$ i think this is my confusion

anakhro

@Hawk how would you even define A^k where k is an element of the field?

Hawk

04:16

i guess it isn't always defined

But is that my confusion?

or what you were trying to explain to me?

anakhro

I think there are multiple points of confusion.

That's one of them.

ÍgjøgnumMeg

p is just tellin' you how many times to do shit

waddup yo

Balarka Sen

sup

Hawk

@anakhro what are the others u think i might have? i think i cleared them right? Because I think i recalled I initially asked if a^p = a^0, but we just dismissed it as this would be asking a^(p*1_F)

ÍgjøgnumMeg

i don't understand how this has been going on for like an hour

anakhro

04:20

@ÍgjøgnumMeg some people need time to understand things.

@Hawk do you understand how writing "p=0_F" is an abuse of notation now?

Hawk

yeah i think my explanation clears that up right?

Balarka Sen

It's worse than abuse of notation because a^(p*1_F) is nonsense - it doesn't make sense

So it should be clear from the outset that the exponent there is just an integer, and like Ig said, it's telling you how many times to multiply

Hawk

I know because that's what i was trying to do

Balarka Sen

Plain and simple

Hawk

i m writing a^(p*1_F) as a demonstrative artifice. I know this thing has no meaning under certain fields.

Balarka Sen

04:23

OK

anakhro

@BalarkaSen for being so plain and simple I think you missed what that comment was saying. :P

Balarka Sen

I am not reading hours of rambling about how an exponent is something from the field

anakhro

I understand.

@BalarkaSen you recall that companion matrix commuting thing?

Balarka Sen

yes

anakhro

Is there an elementary way of showing that only those maps which are polynomials in the matrix commute with it?

Balarka Sen

04:28

Yeah, you can prove by hand that the space of matrices commuting with an $n \times n$ companion matrix has dimension $n$

And since powers of the companion matrices all commute with it, they have to span

anakhro

Ah, that's not a bad idea at all!

I was thinking some mess with rational canonical form would be the easiest.

Balarka Sen

Yeah in general it's a good exercise to count the dimension of matrices commuting with a given matrix $A$ in terms of it's rational canonical form

It's some Frobenius's theorem. Had that as an assignment problem

Wait, what was the original problem again?

anakhro

I don't remember.

Balarka Sen

Was it counting the dimension of $k[x]$-linear endomorphisms of $k[x]/(f^n)$ as a $k$-vector space?

$f$ being some irreducible polynomial

anakhro

Yeah. Something like that.

I think it was just finding all the k[x] linear ones.

Balarka Sen

04:36

Alright. There should be a direct way to argue that the vector space $\text{End}_{k[x]}(k[x]/(f^n))$ is $k\langle 1, x, x^2, \cdots, x^{n-1}\rangle$, where by $x^i$ I mean the multiplication by $x^i$ map.

Let's see

Lol, by $f^n$ I mean $f$ everywhere. Some degree $n$ polynomial $f$

anakhro

OH

I think I am onto something

(i.e. if you get it don't spoil it just yet)

Balarka Sen

Alright

It's actually exactly similar to the matrix argument

Ultradark

anyone wanna see something with many colors

anakhro

@BalarkaSen do you have it?

Balarka Sen

Ya

anakhro

04:49

Do you use an evaluation map to compare dimensions?

Balarka Sen

Hm. I am just proving that $\text{End}_{k[x]}(k[x]/(f))$ is isomorphic to $k[x]/(f)$ as $k$-vector spaces in a natural way.

Just send $x^i$ to $x^i$

Oh yeah, I see what you mean

Yes, send an endomorphism $\phi$ to $\phi(1)$

Exactly right

anakhro

Great.

Balarka Sen

It's still fun to count the dimension of matrices commuting with $C(f)$ by hand :)

anakhro

is sitting there picking glass out of his eyes.

looks DIRECTLY at Balarka Sen.

Balarka Sen

so anime of you

anakhro

04:56

No, that's what I'd rather do than watch anime.

Ultradark

cool colorful math art

Balarka Sen

$C(f) A = A C(f)$ will essentially tell you that the cyclic shift of $k$-th column of $A$ is equal to the $k$-th column of $A$ for all $1 \leq k \leq n-1$, so all entries of the columns through column $1$ to column $n-1$ are identical, contributing to $n-1$ dimensions

Leaky Nun

@BalarkaSen I'm back from my suffering

Balarka Sen

or something

Leaky Nun

what are you talking about?

Balarka Sen

05:01

@LeakyNun Go back

Leaky Nun

@BalarkaSen :c

Balarka Sen

Just matrix theory

Leaky Nun

what's the question?

Balarka Sen

Count $\dim_k \text{End}_{k[x]} (k[x]/(f))$ where $f$ is some irreducible polynomial of degree $n$

Leaky Nun

hmm

Akiva Weinberger

05:06

Leaky Nun

in general $\operatorname{Hom}_A(A/I, A/I) = \{ f \in \operatorname{Hom}_A(A, A/I) \mid f(I) = 0 \} = \{ x \in A/I \mid Ix = 0 \} = A/I$ @BalarkaSen

Akiva Weinberger

The worst part is Autocomplete was absolutely right, that's exactly what I wanted to Google

O.O

Ultradark

how aerodynamic is a cow HAHAHA

Leaky Nun

@AkivaWeinberger maybe you were reading about cows before

Balarka Sen

@LeakyNun True.

Leaky Nun

05:09

@BalarkaSen play?

Akiva Weinberger

Cow aerodynamics at 150 km/h

►

Balarka Sen

No

Akiva Weinberger

Now I'm no uh air scientist but I think from that video the answer is "not very"

Leaky Nun

why on earth would... nvm

Ultradark

Akiva Weinberger

05:10

The ships hung in the air in much the same way that cows don't

Answer to #3 is nothing, and uh for #4 when a mommy cow and a daddy ~~cow~~ bull

Ultradark

do cows moo at night????

Akiva Weinberger

Depends on if they're awake or not?

I mean I dunno maybe the one awake one doesn't want to wake up the others

Balarka Sen

Go do something productive for Christ's sake

Akiva Weinberger

Balarka I'm trying to figure out the efficiency of launching cows this is a matter of importance

Ultradark

I just made an artwork

Balarka Sen

05:13

Jesus Christ this discussion about cows is pissing the hell out of me

Akiva Weinberger

And if I do it at night they won't make a noise while I'm doing it, apparently

What is a cow scared of though

Ultradark

Currently reading: Why VR headsets on Russian cows are nothing to moo about

Akiva Weinberger

insert political joke here

@Ultradark baffled nonverbal response

> Cows are way less spooky than horses

I think they meant to write "spookable" or something (re: what are they scared of)

but I like the idea of spooky cows

Balarka Sen

Artin's proof of existence of algebraic closure is sneaky.

Akiva Weinberger

I mean only 11 months to Halloween right?

Leaky Nun

05:16

@BalarkaSen and actually the first step is already algebraically closed

but it's a whole nother story

Balarka Sen

Interesting, how so?

Akiva Weinberger

Are you talking about fields

'cause those might have cows in it

Balarka Sen

JESUS

Leaky Nun

@BalarkaSen it's long and I don't know how much time you have

I hope I still remember the proof

Balarka Sen

I have all the time in the world lol

Leaky Nun

05:18

so the claim is that if $L/K$ is a field extension such that every monic irreducible polynomial has a root in $L$ then every polynomial in $K$ splits in $L$

Akiva Weinberger

This picture exists

Leaky Nun

the first step is to eliminate separability issues by showing that the perfect closure of $K$ is contained in $L$

Akiva Weinberger

despite the previous conclusive evidence that cows are not aerodynamic

Leaky Nun

and I'll explain the two words if you don't know what they mean

Akiva Weinberger

Unrealistic

Leaky Nun

05:19

@BalarkaSen maybe we should go elsewhere

Balarka Sen

Already there

Akiva Weinberger

No it's OK I'll leave

user76284

06:04

Any tips for solving the functional equation $f(t)/f(at) = g(t)$ for $f$ in terms of $g$?

anakhro

@BalarkaSen does invertible $k[x]$-linear homomorphisms add any meaningful conditions to pick a special subgroup of polynomials in the companion matrix?

Or is that a much harder distinction to make than the $k[x]$ linear part.

I think it might be the conjugacy class of the companion matrix, then?

Leaky Nun

@user76284 I would think about evolution, i.e. suppose $f$ is defined on $[a^n, a^{n+1}]$, then the functional equation lets you extend the domain to $[a^{n-1}, a^n]$, etc

user76284

@LeakyNun I'm asking for stats.stackexchange.com/questions/439063/…, just to give some context.

So $f$ and $g$ here are characteristic functions of probability distributions, where I want to express $f$ in terms of $g$.

Balarka Sen

06:46

@anakhro Hm, I don't know, that seems like an interesting question. You want $k[x]$-linear automorphisms of $k[x]/(f)$? That corresponds to units of $k[x]/(f)$ I believe. That should depend on the polynomial $f$ heavily.

If you have like $k[x]/(x^n)$ everything outside of $(x)$ is a unit. If you have $k[x]/(x^n - 1)$, all the powers of $x$ are units.

In your case $f$ was some power of an irreducible polynomial, right? Then if $f = p^n$, the units of $k[x]/(p^n)$ are the ones outside the ideal $(p)$, which is the unique maximal ideal with quotient field $k[x]/(p)$.

It's what is known in literature as a local ring.

Leaky Nun

07:09

and by CRT you can decompose $k[x]/(f)$ into product of local rings, and the units decompose similarly

Adam

08:27

I understand things in the wrong order and it's disgusting

skullpatrol

08:52

the only other choice you have is to not move on until you have fully understood what is being presented, but that can make learning a painfully slow process in my opinion

dondeman

09:32

Is there anyone here who knows why this \sum_{j=p}^{r-1}\frac{1}{2} is equal to \frac{r-p}{2} ?

Rithaniel

Well, what is $\sum_{j=0}^{r-1}\frac{1}{2}$ equal to?

Or, maybe, what is $\sum_{j=0}^3\frac{1}{2}$? Or what is $\sum_{j=0}^7\frac{1}{10}$. Compute a few sums and see if you can tell what is happening when you sum a constant from 0 to r-1

(and then, when you have a good grasp on that, think about what happens if you start counting from p instead of 0)

famesyasd

10:34

what is wrong with some of the people who denote $\mathbb{R}^n$ as $\mathbb{R}^m$

skullpatrol

nothing, why?

provided we're talking about positive integers in both cases :P

Rithaniel

Hehe: $\mathbb{R}^z$ where $z\in\mathbb{C}$

ÍgjøgnumMeg

that's just $e^{z \log \Bbb R}$ of course

famesyasd