Mathematics

12:24 AM

I think I got them, N is conditioning on $X$, thus, $f_{X,N}=f_Xf_{N\,|\,X}$

12:51 AM

back to the first problem, Let $Y=N-X$, find the joint distribution of $Y$ and $X$. This should be $f_{Y,X}=P(Y=N-X=y,X=x)=P(N=x+y,X=x)=P(X=x\,|\,N=x+y)P(N=x+y)$

anakhro

@Thorgott quick question if you are still around: D&F define an algebraic closure of $F$ to be an algebraic extension $K$ such that every polynomial in $F[x]$ splits completely over $K$.

This seems to admit arbitrarily large closures?

I mean to ask, what is it in this definition that stops us from considering say $\mathbb C(x)$ as the algebraic closure of $\mathbb R$? Why is it that this definition somehow implies minimality of $K$?

Thorgott

$K$ being an algebraic extension of $F$ places a severe restriction on the size of the extension

anakhro

OH

I didn't even notice algebraic.

Thanks. That makes total sense now.

:)

So merely $x$ would not be algebraic over $\mathbb R$ and yadda yadda.

Oddly enough I had the same issue with the definition in D&F of normal extension.

"algebraic extension which is a splitting field for a family of polynomials".

I kept on wondering what the difference of being a splitting field for a family and a polynomial was, but I kept on skipping "algebraic".

Thorgott

The cool thing is that you can indeed have arbitrarily large algebraically closed extensions, but an algebraic closure embeds in any of them. So you can think of an algebraic closure as the "smallest" algebraically closed extension (and this smallness is captured by being itself algebraic over the base field).

anakhro

That is a neat way of putting it.

D&F kind of hides such subtleties.

Thorgott

1:10 AM

I prefer to define an algebraic closure as an algebraically closed algebraic extension for that reason, but D&F prove that this is the same in the next proposition.

Interestingly, even if $K/F$ is algebraic and every nonconstant polynomial over $F$ has a root in $K$, then $K$ is an algebraic closure of $F$.

This is a considerably weaker hypothesis than having the nonconstant polynomials split, so this fact is actually a bit more involved to prove.

A neat consequence of this is that in the construction of algebraic closures that goes back to Artin (also the one given in D&F), you can actually stop after the first step instead of iterating the construction.

anakhro

I like the Artin proof.

I think I read it first in one of Conrad's notes.

Thorgott

Yeah, it nicely generalizes the construction of adjoining a root

Those notes also contain the proof of the fact I mentioned above

Another cool variant is to attempt constructing a "maximal" algebraic extension like here: www2.im.uj.edu.pl/actamath/PDF/30-131-132.pdf

anakhro

Speedy

Does Zorn pose as just a helpful tool here or is there an unconstructive element being played off of?

Thorgott

1:26 AM

Another cool variant I learned from the user moderator of the Basic Mathematics chatroom is to well-order the irreducible polynomials and split them by transfinite recursion.

This last variant actually gives a concrete way to construct an algebraic closure without any choice for finite and countable fields, as we can always well-order these.

Lukas Heger

you can't prove existence of algebraic closures for all fields in ZF, so there is some nonconstructive element to it @anakhro

my favorite proof for the existence of algebraic closures uses the compactness theorem from logic

anakhro

Wild.

Thorgott

Uniqueness and existence of algebraic closures only requires the ultrafilter lemma and whether it implies the ultrafilter lemma is still open, according to MO

Jeff

Hi folks. Using the direct comparison test, what function do I compare $1/\sqrt{x+e^x}$ to?

Thorgott

direction comparison test for?

Jeff

1:32 AM

For the integral $\int_1^\infty 1/\sqrt{x+e^x} dx$

I need something whose denom is bigger. But the only benchmark we have is integral of $1/x^p$

Akiva Weinberger

$\text{I think}\le\text{I am}$

If we put $\text{I think}$ and $\text{I am}$ on two separate axes, we'd get a Cartesian plane

Thorgott

You want to find a function that's pointwise larger, so you want a smaller denominator. Taking roots is monotonic, so you want something that's smaller than $x+e^x$. Is there a likely candidate? Does it work?

Leaky Nun

@LukasHeger hei

anakhro

For the fact that $K/F$ finite means $K/F$ splitting field $\iff$ each irreducible $f\in F[x]$ with a root in $K$ splits completely, is finiteness only needed for $\Leftarrow$?

Lukas Heger

@LeakyNun hi

@anakhro if you allow families of polynomials, this holds without assuming finiteness

Leaky Nun

1:39 AM

@LukasHeger was hast du gelernt

Jeff

@Thorgott Oh right. I got it backwards (for a change).

anakhro

@LukasHeger splitting field for families, or families of such $f$?

Leaky Nun

@anakhro splitting fields are always finite (if you allow a single polynomial)

Lukas Heger

@LeakyNun working on my undergrad thesis

Leaky Nun

@Jeff even if you wanted something bigger, there's $e^x + e^x$

Jeff

1:41 AM

@Thorgott Well, $\sqrt x$ is a smaller denominator (that is, $1\/sqrt x$ is larger). But $1\/sqrt x$ does not converge.

anakhro

I thought splitting fields are not necessarily finite....

Otherwise they are algebraic, and thus they are normal

Lukas Heger

a splitting field of a family of polynomials is not necessarily finite

Thorgott

Right, so $x+e^x>x$ is too weak of an inequality. What happens if you try $x+e^x>e^x$ (note that this is not a bad idea, because the $e^x$ term dominates the sum for large $x$)

Lukas Heger

a splitting field of a single polynomial is finite, though

Leaky Nun

@LukasHeger was schreibst du in das These daruber?

anakhro

1:43 AM

Does calling K a "splitting field" for F necessitate "for a single polynomial"?

Thorgott

A splitting field for a single polynomial is generated (as a field extension) by the roots of that polynomial, all of which are algebraic over the base field. Being generated by finitely many algebraic elements means being finite (this is an equivalence, in fact).

Leaky Nun

depends on who you're talking to, I guess @anakhro

Lukas Heger

@LeakyNun Modulformen, Elliptische Kurven, Galoisdarstellungen

anakhro

I don't even know anymore.

Leaky Nun

@LukasHeger and what's the unbutchered version of my question?

Jeff

1:44 AM

@Thorgott Yes, so $1/\sqrt{x+e^x} < 1/\sqrt{e^x}$, but I think (from the instructions) that I need to use a comparison to $1/x^p$.

Lukas Heger

@LeakyNun "Worüber schreibst du deine Arbeit?"

Leaky Nun

close enough, I guess

Lukas Heger

@anakhro I personally never say splitting field without saying splitting field of what. If you want to say splitting field (without qualification), you can just say normal

Leaky Nun

but now normal allows infinite

anakhro

Normal = splitting in either case for the finite case.

So for the question I guess it doesn't matter since finite is assumed?

Thorgott

1:46 AM

I don't see how to make that work @Jeff , the exponential comparison works though

Lukas Heger

from taylor expansion $e^x \geq \frac{x^2}{2}$ for $x \geq 0$

Jeff

Well, we (obviously) have the integral test which i can use to prove $\int_1^{oo} e^{-x/2} dx$ converges.

@Thorgott Maybe the hint was for the other direction, since the full problem is determine the convergence of $\int_0^{oo} 1/\sqrt{x+e^x} dx$.

Leaky Nun

@LukasHeger hast du Algebraische Topologie gestudiert?

Lukas Heger

*studiert

Leaky Nun

what why

Lukas Heger

1:51 AM

ja, ich habe algebraische topologie 1 und 2 gehört

Leaky Nun

denn was ist eine Kofibration

Lukas Heger

*Kofaserung

Leaky Nun

why on earth is it studeirt

Thorgott

Well, $e^x\ge x^2/2$ is too weak, but $e^x\ge x^3/6$ does the job, though on the other hand you can get the convergence of $\int_1^{\infty}e^{-x/2}dx$ directly from FTC, which is more straightforward imo

Leaky Nun

by e^x >= x^2/2 I believe he means e^(x/2) >= x^2/8

Jeff

1:53 AM

@Thorgott Agree with the last. Already typed that up.

Lukas Heger

@LeakyNun alle Verben, die auf -ieren enden, bilden das Partizip 2 ohne ge-

Leaky Nun

:o

ich hab niemals diese Regel gehort

Lukas Heger

Verben auf -ieren sind normalerweise Fremdwörter (bis auf wenige Ausnahmen wie verlieren), vielleicht kommt das daher

Hey @Amin

Leaky Nun

verlieren ist klar kein Fremdwort

Amin Idelhaj

Hey Lukas and everyone, how's it going?

Leaky Nun

1:55 AM

wait lieren isn't a verb?

Lukas Heger

pretty well, thanks. And yourself?

Thorgott

oh, I see, taylor after taking the square root

that works indeed

Lukas Heger

@LeakyNun no, but liieren exists (though it's unrelated to verlieren)

Amin Idelhaj

Doing alright, gonna finish up some AG and hopefully get quizzes graded

Thorgott

@Jeff the integral over $[0,1]$ is absolutely no issue, since there is no pole

Amin Idelhaj

1:57 AM

What are you up to?

Jeff

@Thorgott Well, there is that, too! :D

Lukas Heger

@Amin still writing my undergrad thesis

Leaky Nun

@LukasHeger if $U \subseteq V \subseteq \Bbb C$ are open sets, what does the cokernel of $\mathcal O(V) \to \mathcal O(U)$ look like?

Amin Idelhaj

Nice, good luck

What's it on exactly? I remember you were recently working on some computational things

Lukas Heger

it't not really computational

I'm writing on the Galois representation associated to Hecke eigenforms

the existence of these was conjectured by Serre and proved by Deligne

if you reinterpret modular forms as automorphic forms for $\mathrm{GL}_2$, that's a case of global Langlands

and I also do some kind of comparison with local Langlands (at least for the unramified primes)

@LeakyNun no idea in general

Leaky Nun

2:04 AM

@LukasHeger schreibst du deine Arbeit auf Englisch oder auf Deutsch oder auf Italianisch?

Lukas Heger

Englisch

lol

Leaky Nun

lol

Jack Ohara

Hey

Does anyone here know about the miller rabin test ?

@LukasHeger Hey

Thorgott

If $K$ is a field, is there an easy argument for $K(X)\not\cong K(X,Y)$ (rational function fields)? I can argue using transcendence degree, but there might be something more direct.

Lukas Heger

Purtroppo il mio italiano non è tanto buono, che posso scrivere la mia tesi in italiano. Un'altro problemo sarebbe che il mio supervisore non parla italiano. @Leaky

Jack Ohara

2:09 AM

@LukasHeger have you ever used the program macaulay2?

Leaky Nun

@LukasHeger qui e il tuo supervisore?

Lukas Heger

@Thorgott do you mean as extensions of $K$? Otherwise I think $K=k(X_1,X_2, \dots)$ is a counterexample

@LeakyNun Si chiama Böckle

@JackOhara I haven't

Leaky Nun

@Thorgott essentially using transcendence degree: if X maps to f(X) and Y maps to g(X) then g(f(X)) = f(g(X))

Jack Ohara

@LukasHeger it is a good one if you are intrested in AG

Leaky Nun

but g(X) != f(Y)

oops what am I thinking about

the relation isn't g and f but yeah you can find a relation

Lukas Heger

2:12 AM

@JackOhara probably, I don't know

Jeff

OK, folks, using direct compare on $\int_1^{oo} 1/\sqrt{\left(x^5+5x^2+6\right)^{5/16}}$. What do we compare it to? It's smaller than $1/x^{15/32}$, but that's no help, since that integral diverges.

Leaky Nun

then maybe it diverges!

the trick is to just say that eventually x^5 dominates

so you can replace the other two terms by x^5

and obtain a smaller divergent integral

Jeff

@LeakyNun I need to show why x^5 dominates, though. But hold on a sec.

Thorgott

yeah, as extensions of $K$, should've made that clear

Leaky Nun

I can't do everything for you

x^5 dominates because higher powers always dominate

one cannot focus on syntax and ignore semantics

Amin Idelhaj

2:18 AM

@LukasHeger yo that's pretty cool!

Jeff

@LeakyNun Huh? I'm working on it now, not asking you do anything. I don't understand this syntax vs. semantics comment at al.l

Leaky Nun

@LukasHeger do you know in maths there's a set that contains itself

Lukas Heger

@LeakyNun doesn't seem like a well-founded statement to me

Leaky Nun

@LukasHeger pt = {pt}

Lukas Heger

fair enough

Thorgott

2:48 AM

@LukasHeger I think the function field in countably many variables you gave is a "minimal" counterexample for isomorphism as fields. If $K$ has finite transcendence degree over its prime field $F$, then one can still argue by transcendence (over $F$) that $K(X)\not\cong K(X,Y)$.

CroCo

3:02 AM

Hi there,
any hints why the following limit is equal to one?

Thorgott

Can you simplify $\sum_{n=-N}^N1$?

CroCo

@Thorgott It seems to me equal to infinity?

Thorgott

not the limit, just the term $\sum_{n=-N}^N1$ for a fixed $N$

CroCo

If N=2, I see the summation as 5 or 2N+1?

I see it now.

thanks

Thorgott

indeed, it's $2N+1$, now ask yourself why

CroCo

3:19 AM

+1 for N=0, and from -N to 0 I have N ones and from 0 to N, I have N ones

or |-N| + N + 1

Thorgott

yup, you're summing $1$ $2N+1$ times :)

CroCo

got it thanks.

Lol Flo

11:22 AM

Are the even #s a subsequence of the positive integers ? I am thinking it lacks cofinality because there's no element comparable to 1

It is not cofinal right ? So it's not a subsequence ?

Can anyone confirm that thanks.

Lol Flo

11:54 AM

The way I am reading it there should be an element less than or equal to 1 in the subsequence, thru the relation......

Should I instead be reading that, 1 and 2 are the first in the sequences ????

@LeakyNunmy u available or are u busy????

@educ can you help ??

rustypaper

12:46 PM

hey chat

you think there is much work to be had as a mathematics typist?

Mariana

1:54 PM

Hi to everyone! Do anyone ever heard of this version of Spectral theorem?

(Spectral Decomposition). Let E be a complex Banach space and T ∈ L(E).
Assume that σ(T) = σ1 ∪σ2, where σ1, σ2 are compact disjoint sets. Then there are closed spaces E1, E2 with:
1. E = E1 ⊕ E2;
2. T(E1) ⊂ E1 and T(E2) ⊂ E2;
3. σ(T|E1) = σ1 and σ(T|E2) = σ2.

If yes, can you please tell me where to find.

Jack Ohara

3:03 PM

@MatsGranvik Hi do you know how to code in mathematica?

Mats Granvik

3:20 PM

@JackOhara Yes short programs.

Jack Ohara

@MatsGranvik I would like to ask you if you have written or can write Miller Rabin test for primility testing, i did not find on the web such thing

I can provide you with the details on the test ofcourse if you do not know about it

Mats Granvik

@JackOhara I have heard of the Miller Rabin test. Most of my programs are things I have thought out myself, I have never really programmed other peoples work.

Jack Ohara

@MatsGranvik it is not a work , just something useful for me haha

I do not work, am a student

I have everything written down, just have no experience in using codes in mathematica

Mats Granvik

I am not an expert on the Miller Rabin rest, but looking at the Wikipedia page I judge that you could program this in Microsoft Office Excel.

Input #1: n > 3, an odd integer to be tested for primality
Input #2: k, the number of rounds of testing to perform
Output: “composite” if n is found to be composite, “probably prime” otherwise

write n as 2r·d + 1 with d odd (by factoring out powers of 2 from n − 1)
WitnessLoop: repeat k times:
pick a random integer a in the range [2, n − 2]
x ← ad mod n
if x = 1 or x = n − 1 then
continue WitnessLoop
repeat r − 1 times:
x ← x2 mod n
if x = n − 1 then
continue WitnessLoop
return “composite”
return “probably prime”

What I meant with other peoples work, is other peoples creative ideas, not work-work like a job or task.

Jack Ohara

@MatsGranvik I see thanks !

@AlessandroCodenotti Hi

Alessandro Codenotti

3:30 PM

Hi

Jack Ohara

where in Italy do you live?

I just heard the news about the virus spread there, you have any idea?

Here in Canada is also spreading and it is not looking good

Alessandro Codenotti

I live in one of the most affected regions, but there are no cases in the province I live in so far. Apart from a few villages the situation is looking pretty good I'd say, as long as it doesn't start spreading in a big city we should be fine

Jack Ohara

Hope so ! stay safe

wait what

Alessandro Codenotti

(I actually live in Germany since I'm doing my Masters there, but I happen to be back in Italy now, not the best time to go back home)

Jack Ohara

Oh never mind, I missunderstood what you wrote first time

Mats Granvik

3:35 PM

@JackOhara rosettacode.org/wiki/…

Jack Ohara

I see, but it is eveywhere i think

@MatsGranvik Super ! thanks! , I can try to modify it a bit for my use

Semiclassical

3:47 PM

@genescuba first: yes, those three outcomes are the only ones possible. Second: the only cases I could see replacement being an issue are the last two. But you specified that, upon drawing the mint the first time, you’d put it back in the bag. So you do replace it.

That is, I see nothing in the problem statement which prevents you from drawing the Dane mint twice. It’s certainly unlikely, but not forbidden.

geocalc33

4:07 PM

I'm so pleased I was able to make progress over the past year and write a pretty concise and coherent question :) math.stackexchange.com/questions/3560611/…

my new year's resolution of being more mathematically precise is looking do-able

thanks adderall

this one too even though it's kind of a moot point math.stackexchange.com/questions/3559324/…

RScrlli

4:53 PM

Hey guys a simple question, is there any difference between "polynomial chaos" and "homogeneous chaos" in the context of Wiener chaos?

geocalc33

Generally, how does one use a (non-linear) decomposition operator to split a smooth manifold into a ring-like structure?

I guess I should first ask if it's possible to do that for a linear operator

and maybe it's better to consider the smooth manifold to actually be a variety in the context of algebraic geometry

If $X$ is an affine algebraic set defined by an ideal $I$, then the quotient ring ${\displaystyle R=k[x_{1},\ldots ,x_{n}]/I}{\displaystyle~~~~ R=k[x_{1},\ldots ,x_{n}]/I}$ is called the coordinate ring of $X$. If $X$ is an affine variety, then $I$ is prime, so the coordinate ring is an integral domain. The elements of the coordinate ring R are also called the regular functions or the polynomial functions on the variety

Royi

5:41 PM

Could a moderator open this - math.stackexchange.com/questions/3558240 ?

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$Mathematics$ Mathematics