Mathematics - 2020-01-27

Thorgott

00:05

Konformist Liberal

00:16

Hi everyone! In this page https://en.wikipedia.org/wiki/Skolem_normal_form there is a part with
"... by the axiom of choice ..." - Is AC even required? Can't we just say that, there is an element in the model which satisfies this formula, let's say b, then we create the constant b function

Bob

Hi

vesii

Hi guys! I have two points A and B. I also have a point C so the $\angle ACB=120$. If I insert the triangle $ABC$ in a circle. How should it look like?

Konformist Liberal

@vesii, just choose a portion of the circle which is 240 degrees, choose A and B from the boundaries and C randomly from the smaller side

vesii

I'm trying to find the "geometric place" of all points of $C$ so the angle is $ACB=120$ using analytical geometry. I took three points on a circle. So If the angle $ACB$ is 120 degrees then $AOB$ is also 120 degrees. But now I'm stuck.

As I understand I need to set $C$ to be $(x_C,y_C)$ and somehow to get an equation using only those points.

I though of using the cos theorem, but I also have $A$ or $B$ in the final equation.

Any ideas?

I need to find the "Locus"

Konformist Liberal

00:35

you can select a 240 degrees portion by creating a 60 degrees and adding a 180 degrees and then the placement of C doesn't matter, it's 120 degrees in all cases (for the smaller side)

vesii

But how I find the Locus ?

Konformist Liberal

Inscribed angle

In geometry, an inscribed angle is the angle formed in the interior of a circle when two secant lines (or, in a degenerate case, when one secant line and one tangent line of that circle) intersect on the circle. It can also be defined as the angle subtended at a point on the circle by two given points on the circle. Equivalently, an inscribed angle is defined by two chords of the circle sharing an endpoint. The inscribed angle theorem relates the measure of an inscribed angle to that of the central angle subtending the same arc. == Theorem == === Statement === The inscribed angle theorem states...

vesii

Yes I know that I need to use the inscribed angle. Thats why I told $AOB$ is $120$.

Lucas Henrique

@Thorgott oh my god.

Thorgott

00:54

:)

Lucas Henrique

so easy lmao

What I don't understand is that I thought about this

it was one of the first things I thought but, for some unknown reason, $\frac{mn}{d^2} = l/d$ didn't look like a solution

hmm, what about the last implication?

ah, forget it

$\left((ab)^d\right)^{l/d} = (ab)^l$

Ted Shifrin

02:19

@LucasHenrique Yeah, you might be right. We just need to make sure we have $m/r$, $n/s$ relatively prime and their product has to be bigger than $n$.

And, yes, to create $r$ and $s$ I found it easiest to use prime factorizations.

Lucas Henrique

I usually want to avoid prime factorization, for no reason at all

Commodity maybe

Ted Shifrin

@vesii You have the answer. $C$ can be anywhere on that arc. You do still need to prove there are no other solutions.

@Lucas: Well, if you figure out a proof (recipe) without it, please tell me.

I think you need it to make sure that $m/r$ and $n/s$ are relatively prime. But maybe you can be clever.

Lucas Henrique

@Thorgott's proof, just above, is very clever and doesn't deal with primes

Ted Shifrin

Oh, is Thorgott being smart? :)

Oh, that looks cleverer than I was. Darn.

I worked with $rs=d$, by the way, so it would come out exactly right.

Lucas Henrique

02:42

but doesn't @Thorgott proof imply that $4+2 = 6$ has order $6$ over $\mathbb Z_{12}$?

the last implication is the problem. because, in fact, it says that among $(ab)^d, (ab)^{2d},\ldots, (ab)^{d\frac ld}$, the only term that equals $e$ is $(ab)^l$

that doesn't guarantee that there aren't other divisors of $l$, namely $r$, not being multiples of $d$, with $(ab)^r = e$

e.g.: in my example, $m = 3$, $n = 6$ so $d = 3$, $l = 6 \implies l/d = 2$ and $ab = 6$. It's correct that $(ab)^3 = (ab)^d \neq e$ but $(ab)^{6} = (ab)^l = e$. However, $(ab)^2 = e$, so the order is not $6$

@Thorgott: I'm happy I wasn't too picky when thinking about the last implication :P

Thorgott

03:05

looks like I was too clever to be true

Mike Miller

Here is what I think is a short argument, no explicit numbers or painful fiddling

Let A be a finite abelian group. Let x be an element of order m. Let y be an element of maximal order n. If (m,n) = m, then we are done, as this means m divides n. If m > (m,n) > 1, then replacing x by (m,n)x, we see that there is an element z of order k with (k, n) = 1. I claim this is impossible.

For consider the element z+y. We have kn(z+y) = 0, so ord(z+y) | kn. If k' is a proper divisor of k, then k'n(z+y) = (k'n)z, which is not zero, as k'n is not zero in Z/k (here we use that (k,n) = 1 and that k' is a proper divisor). Similarly for any divisor n' of n. Thus ord(z+y) = kn > n; this contradicts the assumption that y had maximal order.

Lucas Henrique

03:45

okay, primitive element theorem is good to go for finite fields. :)

why is separability needed to prove the primitive element theorem for finite extensions (with infinitely many elements)?

the proof I know uses the polynomial $p(t) = \prod\limits_{i=1}^r (\alpha+t\beta - \sigma_i(\alpha)- t\sigma_i(\beta))$, where $\sigma_i$ is an embedding of the extension into is algebraic closure and argues that since this polynomial has finitely may roots, there's a $c \in E(\alpha, \beta)$ which is not fixed by any of the embeddings

and I honestrly can't remember why, but this implies that $E(\alpha, \beta) = E(c)$. (maybe something using polynomials? is it where separability is needed?)

Thorgott

04:27

I think what's going on is that the image of an element $\alpha$ under an embedding $\sigma$ is a zero of the minimal polynomial of $\alpha$ still. So if no $\sigma$ fixes $\alpha$, the extension $K(\alpha)/K$ has degree at least the number of $\sigma$s. For $\alpha$ to be primitive, you want this to be the degree of your initial extension, say $L/K$. There are precisely enough $\sigma$s for this to work (namely $[L\colon K]$ many) if the extension is separable.

Lucas Henrique

05:20

I think I get it. Why wouldn’t we have enough $\sigma$s, otherwise?

Leaky Nun

@LucasHenrique it's one of the characterisations of separability

but you can prove it for simple extensions and then use two tower laws

Lucas Henrique

Also, mainly @Ted and @Thorgott: I apologize for being so insistent and bragging about the same topics over and over. I’m really interested in the topics I’m studying but my book really does not help me, and I don’t know better resource.

Thanks to everyone helping me. :)

@LeakyNun ok, that makes sense.

Leaky Nun

@LucasHenrique that isn't what bragging means

o que queria dezir voce?

Lucas Henrique

Blabbering*

Leaky Nun

ah

ah, dizer not dezir

spanish is decir

Lucas Henrique

05:29

@LeakyNun hahaha. That would be “o que você queria dizer?”

Leaky Nun

oh you put the "voce" there?

Lucas Henrique

Mobile chat often sends duplicated messages

Leaky Nun

como voce pode ver, eu nao falo portugues bem

Lucas Henrique

That’s perfect

I don’t speak English very well so we’re even :P

@LeakyNun there’s a weird thing about verb conjugation in Portuguese. “I, you, he” and “they” have each different verb conjugations

To drink, for example, in the order I specified above: bebo, bebes, bebe, bebem.

Leaky Nun

em espanhol tambem

Lucas Henrique

05:35

The weird thing in portuguese is that we don’t use the “you” conjugation in informal language, because you $\mapsto$ tu, and we use “você” in informal language, opposed to Spanish which uses “usted”

That usually confuses people, even other members of Latin America

@LeakyNun asking questions is like claiming sentences but with an “?” in the ending

You like Ted $\mapsto$ Você gosta do Ted

Do you like Ted? $\mapsto$ Você gosta do Ted?

Leaky Nun

I see

@LucasHenrique voce usa "tu"?

Lucas Henrique

Rarely

I live in São Paulo, and that’s unusual. The other regions use “tu”, but each in its own way

In Rio they use “tu” a lot, but not “tu bebes”, but “tu bebe”

Leaky Nun

eu acho que o sotaque brasilero nao usa "tu" jamais

interesting

Lucas Henrique

Well, “tu” is never used “correctly”, but it’s used a lot

The pronoun that is really awkward to hear anywhere is “vós”, the plural of “tu”

No one knows how to conjugate the verbs for “vós”

It’s only used in very respectful and formal reunions, like in churches or in very specific, technical/bureaucratic contexts

Eu imagino que vós, leitores, estais deveras entretidos por um linguajar tão sofisticado e desnecessariamente pomposo! :p

Jacksoja

05:58

@LucasHenrique @LeakyNun Hello !

Lucas Henrique

Hey there, @Jacksoja

Jacksoja

Did you solve that question ? haha

Lucas Henrique

Yes, after a long time

Jacksoja

@LucasHenrique I did not think of it much since yesterday,have other courses

but ill give it a try today or tomorrow

we can compare proofs if you want

Lucas Henrique

Sure.

sevdaicmis

10:01

i think i've finally understood the diagonalization.

Lucas Henrique

11:01

Linear operator diagonalization?

Andrews

Where do people often use symbols $\imath, \jmath$?

Lucas Henrique

$\imath$ is usual to denote immersion from a ring into another one

Embedding*

Andrews

No result for this...

Some results about this

sevdaicmis

11:17

@LucasHenrique yeah

Lukas Heger

11:37

@Andrews $\iota$ looks very similar to $\imath$ and is probably more well-known, so I guess people just use that for embeddings

math.stackexchange.com/search?q=%5Ciota+embedding

Edward Evans

Afternoon @Lukas

Lukas Heger

Buon pomeriggio!

Edward Evans

Ah yes

Góðan dag

Lukas Heger

that looks like Icelandic

Edward Evans

I think it's probably the same in Icelandic but it's Faroese lol

Lukas Heger

11:45

Loquamur Latine! Non difficile est.

Quid hodie fecisti?

Edward Evans

Nie moge mowic po lacinski

lol

Lukas Heger

Pōrando-go desu-ka?

Edward Evans

si

Sagt man da als Antwort eigentlich einfach "desu" oder so?

lol

Lukas Heger

sō desu

oder

hai

oder

un

Edward Evans

hai kannte ich

ah nice

Lukas Heger

11:50

tiel falis la babilona turo

Edward Evans

Ich will den Vorlesungsangebot fürs Sommersemester sehen

I don't think the trick you told me to use works anymore lol

Lukas Heger

it does work

lsf.uni-heidelberg.de/qisserver/…

Edward Evans

oh damn

I'm doing it wrong hte

then*

Lukas Heger

Ĉu vi rekonas ĉi tiun lingvon?

Edward Evans

Looks romance but also not

Or maybe latvian or smth

lol

Lukas Heger

11:53

je suis esperantiste

Edward Evans

Ah that was Esperanto?

Lukas Heger

yeah

"tiel falis la babilona turo", too

Edward Evans

Ah nice

Quickly: what things do I need to change in the URL on the Vorlesungsverzeichniss?

with 1 s

lol

Lukas Heger

delete "&veranstaltung.semester=20192"

but it only seems to work on searches

so you can search for a particular lecture or seminar

and then deleting the above will give you the results for all semesters, including the next one

Edward Evans

yeah that's cool

thanks hahaha

Ah Algebra 2 is with Böckle

Lukas Heger

11:58

Böckle is great :)

there's also a Böckle seminar on noncommutative algebra

they also do Brauer groups in there

Edward Evans

Ah nice, might do that too

Lukas Heger

so it's relevant for NT

and since it's a seminar and not a proseminar you can get masters credit, most likely

Edward Evans

woohoooo

it doesn't look like AZT2 is there :(

Lukas Heger

yeah I think that was mentioned

Vogel has other teaching duties or something

but there'll be a CFT seminar at least

Edward Evans

yeah I spoke to Rustam in the tutorial and he said it probably won't be offered

Oh there's a Seminar called Algebraische Zahlentheorie

so I guess that'll be the CFT seminar lol

Und Analytische Zahlentheorie!

Lukas Heger

12:13

who's offering analytic NT?

oh, I see, it's Kohnen

then it'll be good :)

Kohnen is an internationally recognized expert on modular forms

there's even a book by Shimura which cites one of his papers

Edward Evans

wtf

hahaha

That's nice then

Lukas Heger

there's even a Kohnen space of modular forms

Edward Evans

and the Dozent for the AZT seminar is Leohardt

also woah

Quantenfeldtheorie, Teilchenphysik und Statistische Physik

lol

ah das ist ein Oberseminar

Nice ich glaub ich weiß jetzt was ich nehme hehe

Lukas Heger

12:32

was denn?

Edward Evans

uhm

Alg 2, AZT Seminar, Analytische Zahlentheorie, Funktionalanalysis

I can get credit for Alg 2 and Functional analysis

so that's fine

and I might take one more thing

Like Russian

or the non-commutative algebra seminar

or both lawl

Lukas Heger

@EdwardEvans Se esperanto intrigas vin, lernu.net vera bonas.

Ciao @Leaky

Leaky Nun

@LukasHeger hej

Edward Evans

@Lukas learn this en.wikipedia.org/wiki/Ithkuil

Leaky Nun

@EdwardEvans what's special about Ithkuil?

Lukas Heger

12:38

@Edward lol, how many speakers does Ithkuil have?

Edward Evans

It's an engineered language so probably 1

hahaha

It's a cool experiment though

96 grammatical cases

Lukas Heger

I like that for Esperanto it's not impossible to actually find other people to talk to, despite the fact that it's engineered

Edward Evans

yeah indeed

although aren't there some native speakers of it?

Like a couple of thousand

in the sense that they grew up speaking it as their first language

Lukas Heger

yes, there are some native speakers

feynhat

13:37

Hello

Suppose a function f is holomorphic in some disc, then if I write the power series of f about the center of this disc, will the radius of convergence of this power series be atleast as much as the radius of the disc?

Lukas Heger

@feynhat yes

Semiclassical

pretty sure you can strengthen that as such: the radius of convergence will be the radius of the largest disk, centered at the point of interest, on which f is holomorphic

feynhat

@LukasHeger How do I show this?

Semiclassical

(e.g., if it's holomorphic for |z|<1 and its power series about z=0 has radius of convergence 2, then it's also holomorphic for |z|<2)

feynhat

Is there any reference for this? I have checked Conway and Papa Rudin, and couldn't find anything.

I tried using identity theorem

Lukas Heger

13:47

the identity theorem won't help

you have to use the Cauchy integral formula

supppose wlog that the center of the disk is $0$ to make notation less cumbersome. Let $C$ be circle around $0$ traversed once and of radius $<r$, where $f$ is holomorphic on the disk with radius $r$. Then we have the Cauchy integral formula $f(z)=\frac{1}{2\pi i}\int_C\frac{f(w)}{z-w}\mathrm{d}w$

Now the idea is the following: use that $\frac{1}{w-z}=\frac{1}{w} \frac{1}{1-\frac{z}{w}}=\frac{1}{w}(1+\frac{z}{w}+\frac{z^2}{w^2} + \dots+(\frac{z}{w})^{n-1}+\frac{(\frac{z}{w})^n}{1-\frac{z}{w}})$ (this is just a partial geometric series)

Semiclassical

@LukasHeger denominator should be $w-z$

Lukas Heger

oops

thanks

too late to edit

Semiclassical

happens

feynhat

$|z| > r$, right?

Lukas Heger

no, $|z|<r$

so we obtain $\frac{1}{w-z}=\frac{1}{w}+\frac{z}{w^2}+\dots+\frac{z^{n-1}}{w^n}+\frac{z^n}{w^n} \frac{1}{w-z}$

now we plug that into the integral:
$f(z)=\frac{1}{2\pi i} \int_C \frac{f(w)}{w-z} \mathrm{d}w=\frac{1}{2\pi i} \int_C (\frac{f(w)}{w}+\dots+\frac{f(w)z^{n-1}}{w^n})\mathrm{d}w+\frac{1}{2\pi i} \int_C \frac{z^n}{w^n}\frac{f(w)}{w-z}\mathrm{d}w$

Now recall the generalized Cauchy integral formula: $f^{(n)}(z)=\frac{n!}{2\pi i}\int_C\frac{f(w)}{(w-z)^{n+1}}\mathrm{d}w$

(which is obtained from just differentiating the usual Cauchy integral formula)

apply that with $z=0$ and compare with the terms we have in our integral, then we see that $f(z)=\sum_{k=0}^n \frac{f^{(k)}(0)z^k}{k!}+R$, where $R$ is the remainder term $\frac{1}{2\pi i} \int_C \frac{z^n}{w^n} \frac{f(w)}{w-z} \mathrm{d}w$

so all that is left to show is that $R \to 0$ as $n \to \infty$

I'll leave that to you :P

it's just some standard inequalities you have to deal with a lot in complex analysis

use that continuous functions attain a maximum on a compactum etc.

but that's the idea, at least

hope this helps @feynhat

feynhat

14:08

We can bound the integral by the perimeter of C times the supremum of the integrand. But the integrand vanishes as n tends to infinity because |z| < |w| = r... right?

Lukas Heger

yeah exactly

feynhat

Right so this shows that f equal to the power series for any |z| < r.

So, the radius of convergence is atleast r.

gustaffIR

14:56

Can someone please help with the following: Is [max(0,x^2 sin(1/x))]^2 differentiable at all points?

sevdaicmis