Mathematics - 2022-03-21

robjohn

02:27

$$x!=\prod_{k=1}^\infty\frac{\left(1+\frac1k\right)^x}{\left(1+\frac xk\right)}$$

@AMDG This works for $x!=\Gamma(x+1)$ for all $x$

It may not be very fast, however.

Koro

05:58

C (the field of all complex numbers) is algebraically closed, which means that C has no proper algebraic extension, which is equivalent to -every polynomial in C[x] splits in C.
Is there a non-polynomial function f(x) in C[x], which has no zero in C but has a zero in some extension of C?

Have such non-polynomial functions been studied?

Koro

06:26

what should I know so that I can answer this: Is the ideal $I=(X+Y,X−Y)$ in the polynomial ring $C[X,Y]$ a prime ideal?
To answer this, I have the following two ideas, are these enough to answer the above question?:
1) One way to solve this of course would be to solve it by definition i.e., taking f(x,y) and g(x,y) from C[x,y] and then noting that there exist p(x,y) and q(x,y) in C[x,y] such that f(x,y)g(x,y)=p(x,y)(x+y)+q(x,y)(x-y) and then concluding from here that either p(x,y) or q(x,y) is in (x+y,x-y). But this seems very lengthy too me.

Martin Sleziak

07:36

A short discussion of the above question can be found in the Linear & Abstract algebra chatroom.

love_sodam

09:27

How can I show a group of order 1320 is not simple?

Rithaniel

12:21

@love_sodam One idea is to start by figuring out the necessary numbers of the Sylow subgroups for some prime and then consider the group action of conjugation acting on those subgroups. That'll act as a homomorphism to $S_n$, where $n$ is the number of subgroups, and you might be able to get traction from there

love_sodam

@Rithaniel I actually tried that. Let $G$ be a group of order $1320$. If $G$ is not simple then the number of Sylow $11$ subgroup should be $12$. So there is an induced monomorphism $G\to S_{12}$. Now, let $P_{11}$ be a Sylow $11$-subgroup. Then $|N_G(P_{11})| = 110$ and $|N_{S_{12}}(P_{11})| = 110$.

Now I want that the map $G\to S_{12}$ actually factor through $A_{12}$ so I can get a contradiction.

Mad Spaces

Hello. If i calculate the divergence of a vector Feld in Cartesian Coordinates and then transfer it to Cyllinderical coordinates, i still get the same result if i to transfer it first and then calculate the divergence? (just checking if i am doing something wrong or not=)

Jakobian

@AlessandroCodenotti Is Nadler a good book for continuum theory, or is there a better one? I'm still quite new to this all, but I licked it a little bit

Leaky Nun

@love_sodam @LukasHeger 😂😂

LearningCHelpMeV2

why is combinations and permutations such a hard topic :(

user193319

12:41

The veronese map $\nu_d : \Bbb{P}^n \to \Bbb{P}^N$ is defined by sending $$[x_0,...,x_n] \mapsto [...x^I...]$$, where $x^I$ ranges over all monomials of degree $d$ in $x_0,...,x_n$. I am trying to argue that this map is injective but I am having difficulty doing so...

I'm pretty certain that every entry of something in the image is of the form $x_0^{i_0} ... x_n^{i_n}$, where $i_k \in \Bbb{N}_0$ and $i_0 + ... + i_n = d$. So, we assume that $[....x^I ... ] = [...y^I ...]$, so there exists a non-zero scalar $\lambda$ such that $x^I = \lambda y^I$, so $x_0^{i_0} ... x_n^{i_n} = \lambda y_0^{j_0} ... y_n^{j_n}$.

But I don't see how that implies $[x_0,...,x_n] = [y_0,...,y_n]$...

Alessandro Codenotti

I really like that book! Also hyperspaces by Illanes and Nadler contains a lot of continuum theory (and
a lot of material on hyperspaces as well ofc)

In any case I wouldn't lick the books

user4539917

13:01

😋 licking it, in the sense of getting a little taste of what it's like, 🆚 sinking your teeth into it

Prithu biswas leftmse

13:18

Is there some kind of Rigorous Formalism for Classical Mechanics?

Jam

13:42

every subspace in Gr(n, k) can represented
by a unique matrix in row echelon form why is that true

every point of the grassmannian is k vector space

now why every subspace of G can represented in such a form

Means that given a subspace of the grassmanian $S$ exist a vecotr space in $R^m$ which includes all the subspaces which correspond to the points of the subspace $S$

anak

@Prithubiswasleftmse Yeah, sort of. It might not be as rigorous as some would like, but most of classical mechanics can be phrased fairly formally.

@Jam First you have to choose a basis. Only then is it unique.

Are you trying to show that Gr(n,k) is a manifold? This is the way of showing that it is a manifold, usually (finding its coordinates).

Alessandro Codenotti

14:15

The only problem with both continuum theory and hyperspaces is that paper copies are crazy expensive @Jakobian but of course one can easily find a pdf

Koro

@copper.hat math.stackexchange.com/questions/4409080/…

convex problem you might be interested in :-)

Koro

14:46

Let F be a field of char 0 and a and b are algebraic over F. Suppose that p(x) and q(x) are the minimal polynomials for a and b respectively over F. If $K_1$ is a splitting field of p(x) over F; and $K_2$ for q(x) over F then $p(x)$ and $q(x)$ both split in field $K:=K_1\cup K_2$. I want to know why $K$ is a field?

In general, union of fields should not be a field.

Jam

15:18

@anak no tryna see that every subspace of G is again represented by a matrix

copper.hat

@Koro Thanks :-), will take a look

anak

@Jam Same method anyway! Have you figured it out yet, or are you still in need of help?

Jam

thinkin of it

a subspace of G in order to be represented by a single matrix there must exist somehwere a vector space which contains all vector subspaces( points of G) right?

im tryna think the subspaces of G in terms of the vector spaces

i know each point in G represents a k-dimensional vector space

what does a subspace of G represents

anak

You are familiar with subspaces, right?

Jam

a subcollection of vector spaces

which must in turn be a vector space in itself

subspaces given what structure

anak

15:30

vector subspaces

Jam

yeap

G is not a vector space

just a parameter space of vector spaces

anak

It's a collection of k-dimensional subspaces of an n-dimensional vector space V.

You could just take it as V = R^n.

Jam

yeap

anak

So every element of G is a k-dimensional subspace of R^n. Fix a basis of R^n: which one do you want to use?

Jam

use the stndrd is fine

now a subspace of G would be?

AMDG

15:34

@robjohn Depends once again on the convergence of the series. Is there a general algorithm for computing the convergence of a series?

anak

Oh, I think there might be a misunderstanding, @Jam

Are you sure you want to look at subspaces of G, and not subspaces in G?

Jam

whats the difference?

anak

Every element of G is a subspace. The former only makes sense if you have a vector space structure on G. What is your vector space structure on G?

Jam

you cant have a vector structure

on G

but you can talk about its subspaces

anak

What subspaces? Elements you mean?

robjohn

15:37

@AMDG the terms for the log of this product are $\sim\frac{2s^2-s}{2k^2}$, so the error after $n$ terms is on the order of $\frac1n$

Jam

no subsets of G that are varieties also

anak

Should probably use "subvariety" not "subspace".

But I don't understand your original question then. Are you sure it is asking to associate to each subvariety of Gr(n,k) a matrix in row echelon form? Are you sure it's not to each element of Gr(n,k)?

Jam

apparently i was confused from the pdf i was reading

was saying we can represent every subspace of G as matrix

it meant points of G

not actual subspace

as in topological terms or varieties

anak

Yes, that's what I have been saying.

AMDG

@robjohn Ah, ok, so not necessarily the best, although log Gamma isn't necessarily what I'm looking for, just Gamma itself is what I want.

Jam

15:45

yeah my bad im interested in the actual subsets of G

why call em subspaces haha

anak

You wouldn't call them subspaces. The elements are "subspaces" (but of R^n, or another n-dimensional vector space).

Jam

yeah but the confusion is you have projective subspaces in general

so i got confused

i would call em points in G(k,n) not subspaces thats all

Is there a complete list of the subvarieties of a given Grassmannian or we just consider special cases like the schubert subvarieties?

anak

Probably something you can ascertain by papers off of google

Jam

ok thanks

i wont be bothered ill just read about the special cases

i just thought there was somethin like the proposition i wrote and was like hm.. was it so simple?

given a collection of points in G it should correspond to subspaces of a certain space or intersections of given subspaces tohave the structure of a variety

Ted Shifrin

@Jam What is your definition of subspace of $G$?

There are zillions and zillions of topological subspaces.

Jam

15:55

didnt have one. I thought of it a projectivation of some vector space

as a*

anak

@TedShifrin he was talking about points here but didn't realize it at the time.

Jam

if you thing G as a subset of the projective space

Ted Shifrin

You thought of what as projectivization of some vector space?

The Grassmannian is itself not a projective space.

Jam

in harris you can talk about subspaces of P^n

Ted Shifrin

Yes.

These are the projectivization of subspaces (of one dimension higher) of the vector space.

Jam

15:57

a linear subspace of Pn is a closed subspace defined by linear homogeneous equations.

and then i though G is a subset of Pn

Ted Shifrin

There is a thing called the Plücker embedding, but you don't want to think of the Grassmannian that way, usually.

Jam

yeap i know of it

and the whole construction of G

but we can talk about subvarieties of G

Ted Shifrin

Yes, and most of these have nothing to do with "linear subspaces."

Jam

we cant list em all only if we talk about certain ones

Ted Shifrin

In some sense the most important ones are called Schubert varieties.

AMDG

16:01

@robjohn I'm looking to approximate to (at least) 64-bit precision for Re and Im btw.

Jam

well i thought consider a subset of G . now each point of that set corresponds to a k-dimensional linear space. Maybe the whole set had a corresponding vector space which has all the corresponding linear spaces as subspaces

Ted Shifrin

No, a Grassmannian (other than $G(1,n)$ or $G(n-1,n)$) has completely different topological structure than a projective space.

Jam

yeye you are right

also even if existed such a space would have subspaces not of dimension k. where would those subspaces correspond to G

Ted Shifrin

I do not understand.

Jam

yes sorry never mind

i think i got it.

consider 2 points in G(2,4) so these points correspond to 2 planes in R^4

Ted Shifrin

16:05

(Probably $\Bbb C^4$ if you're reading Harris.)

Jam

i could find a space in R^4 where these 2 planes are subspaces of that space

and i thought that space corresponds to the subset of G somehow

Ted Shifrin

Almost surely that space is all of $\Bbb R^4$.

Jam

yeap

and would not correspond to the 2 element subset of G

thats what i mean even if such space existed.... would have alot more points

in my example has all the points

Ted Shifrin

Probably time to move on to something else.

I have to leave for now, anyhow. Good luck!

Jam

yeah sorry for the confusin talk!!

cheers

robjohn

16:59

@AMDG well, the formula I gave you was for Gamma. Alternatively, once you have the log of Gamma, it seems pretty easy to compute Gamma.

AMDG

19:03

@robjohn Sure, through the exponential function, but what are the error characteristics of the original formula you gave? For the $\frac{1}{n}$ proportional error, I would need something on the order of $2^{66}$ terms if taken literally.

user76284

21:30

Looking for an $f \in \mathrm{Diff}(\mathbb{R}^n, \triangle_n)$ such that $g = \sqrt{\det \langle f', f' \rangle}$ is rotationally symmetric, i.e. $g \circ Q = g$ for all $Q \in O(n)$, and $g$ has a simple closed form. Any ideas?

robjohn

22:41

@AMDG You can rearrange that formula to give

$$
\begin{align}
\log(x!)
&=\sum_{k=1}^\infty\left(x\log\left(1+\frac1k\right)-\log\left(1+\frac xk\right)\right)\\
&=\sum_{k=1}^\infty\sum_{j=1}^\infty\frac{(-1)^{j-1}x}{jk^j}-\sum_{k=1}^\infty\sum_{j=1}^\infty\frac{(-1)^{j-1}x^j}{jk^j}\\
&=\sum_{j=2}^\infty\frac{(-1)^{j-1}}j\left(x-x^j\right)\zeta(j)\\
&=x\log(2)-\log(1+x)+\sum_{j=2}^\infty\frac{(-1)^{j-1}}j\left(x-x^j\right)(\zeta(j)-1)
\end{align}
$$

This converges much faster

but only for $|x|\lt2$

So it limits the imaginary part of the argument

leslie townes

tempted to do some kind of pun on 'imaginary part of the argument,' but not coming up with much.

robjohn

or part of an imaginary argument

Ted Shifrin

Hard to imagine.

robjohn

the kind I have with my imaginary friends

Ted Shifrin

Tigger and Roo …

robjohn

22:49

Do you know them, too?

amazing

Ted Shifrin

Of course. I’m Eeyore.

robjohn

$e$ or $\pi$

it's a transcendent dish

leslie townes

the first thing my daughter did this morning was pick up the cat. as in, she took two steps out of bed, grabbed the cat, walked around the room with it, and then said "here you go, dad" before depositing olivia in my lap.

she's so obsessed with this right now.

livvy just keeps putting up with it. i don't understand this at all. i have a feeling that it won't stop until paws and claws come out.

Ted Shifrin

23:23

Olivia can start avoiding evil munchkin if she finds it necessary.

leslie townes

could it be that some part of her enjoys being the center of attention? i think so.

slate: i use your name as an opening guess in wordle. just thought i'd mention that.

it's actually been a really good opener, lately. two twos in the last few weeks.

four today :(

Ted Shifrin

Two for me today!

leslie townes

sorcery.

Ted Shifrin

A bit.

leslie townes

i used to think that twos were impossible, but now that i have two of them they are remotely possible.

you may still be a sorceror.

Slate

23:40

@leslietownes It's a pretty clean opening, if I do say so myself.

leslie townes

it's better for letter placement than STALE, which used to be my opener. but slightly less likely to give rise to abstract poetry.

Slate

I don't play Wordle as much anymore, but I still open with LATEN.

leslie townes

interesting. maybe i'll try that tomorrow.

Slate

I did some number crunching to get that word, to be fair. Number crunching that fits the strategy I enjoy most.

leslie townes

my switch to SLATE was the result of number crunching. there's no shame in it.

Slate

23:46

:)

Ted Shifrin

I use 3 vowels in my first word.

leslie townes

wow. i would have thought of ted as so foreign that it would be ADIEU.

amWhy

@TedShifrin E.g. HOUSE?

Slate

I tried for a while to come up with a set of five words that uniquely determined the sixth word, and I came to the tentative conclusion that I made a mistake in my code but that it was probably not possible to do.

I made that assessment on the basis of nothing, given that I think there's a mistake in the code, but hey, what can you do. At least the words it spat out were pretty good.

leslie townes

slate: that sounds right, at least if the five words are required to be universal and not dependent on information revealed during the words.

Slate

23:56

Yeah, that was the idea

There's just barely not enough information

leslie townes

i have pretty good luck with slate/round/picky as my 3. i can usually get it in 4 from that, but usually there's still a whole world of words left. i'm making guesses about what is more likely.

today for example, it didn't eliminate M, which made my fourth guess a genuine guess.

Ted Shifrin

@amWhy Not quite.

amWhy

@Slate My average wordle is 3.1. I don't seek guarantees; I use the feedback, from the first word I post, and onward, and given known letters, test a missing letter or two, or three.

leslie townes

ted jealously protects his trade secrets.

Slate

@amWhy See, I'd do that if I were smart. Sadly, I am a bird.

amWhy

23:58

@TedShifrin I meant only to exemplify a word with three vowels :P

leslie townes

blue jays are pretty smart birds.

Slate

that's true

Ted Shifrin

Quite is too :)

amWhy

@leslietownes I hear they can be pretty mean, as well? Or perhaps I'm thinking of another species.

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