the response to your comment is almost certainly dependent on thata

it means, nobody knows how many groups of order 2048 there are

@Mike to list all groups that have a given order, up to isomorphism

the OEIS entry stops at 2047

@LeakyNun A lot

@AlessandroCodenotti thanks

I didn't know

TIL

Fine, fine, I'll be precise just because of you: more than 7.

@ÍgjøgnumMeg Ok, run a computer program that enumerates all 2048 x 2048 tables and checks if they form valid groups, and then range over all $2048!$ permutations and identifies tables that permute to one another

I have classified them

I think usually 'classification' has to do with algorithmic procedures, and given the finitude here, there are algorithmic procedures available to you

I have a feeling the heat death of the universe will come before your program finishes running

Of course

I recognize I am being pedantic and generally worthless

it's all i can do

@Mike lol, I thought you were genuinely asking a question and gave a genuine answer

@AlessandroCodenotti Well, more than 52 billion also

Well, I was. I just also recognize that this is surely unsatisfying to everyone else

But if 'classification' means 'algorithmic procedure to list all groups of fixed order, only once up to isomorphism' we're already boned on time-complexity just because of how many of them there are, i think

How do you solve x^N+x-1=0 for N=2,3,4,5,...

Are there asymptotic estimates for how fast the number of groups (up to isomorphism) of order $2^n$ grows?

i shouldn't have wasted people's time on this lol

@TobiasKildetoft citation?

Real solutions only

@AlessandroCodenotti extrapolate from the 11 data points we have

@LeakyNun We get all the ones of order 1024 as well as easily enough new ones to go above 52 billion

@LeakyNun There's 49 billion of order 1024 and you get 3 more billions somewhere

lol

somewhere

I claim I can get 4 billion somewhere else

@LeakyNun At the dedicated groups store or even a big supermarket

@MikeMiller you see, we want to know how many groups of order 2048 there are

so writing a program that would run forever won't tell us this

Right

who said it would run forever?

you just need your immortal friend to check it every few hundred thousand years

eventually he will have your answer! :)

I don't really understand your pedantry.. classifying groups of order $n$ is listing groups of order $n$

Who calculated how many are there of order 1024? And how?

How does one solve the equation

@MikeMiller so we haven't classified them yet

because the program hasn't terminated yet

$x^n+x-1$

Equal to zero

that's very interesting

weird

@LeakyNun is it possible to solve $x^N+x-1=0$. Will it take a long time

@Ultradark depends on what sort of answer you are looking for

@LeakyNun ok, let me justify where I'm coming at this from. When talking about 3-manifolds, it is often said that 3-manifolds are classified. This is in the sense that given any two 3-manifolds, there is an algorithm with explicit runtime bounds checking if they are homeomorphic; and of course, we may produce every 3-manifold by enumerating larger and larger locally euclidean simplicial complexes

take a large enough splitting field, and it will still be decidable

Real solutions

this is about the most satisfying answer one can get. there are obviously infinitely many 3-manifolds, and one can also just as well say "Well, they have a unique geometric decomposition; hyperbolic manifolds are determined by their fundamental group; and the other geometric pieces are denumerable"

@MikeMiller how do you input the manifold?

but the hyperbolic pieces are nonetheless never going to be 'listed' on a piece of paper in any sense.

@LeakyNun as a simplicial complex

Also, I think that's the case with the classification of finite simple groups

For quadratic polynomials, one can appeal to the quadratic formula to get the solutions in terms of square roots

Similarly for cubic and quartic, though it's far more involved

we classified them up to isomorphism into 5 infinite families and 26 sporadic groups

@MikeMiller every 3-manifold has a triangulation?

@LeakyNun Now name those families :)

@TobiasKildetoft real solutions, x is between 0 and 1

@TobiasKildetoft A, B, C, D, E

For quintic and higher, though, there are polynomial equations for which there's no solutions of that kind.

Well, the infinite families are still easily denumerable, and (A,n), say, specifies a group you know how to state

@LeakyNun yes

@LeakyNun Those are some terribly uninspired names

@MikeMiller every manifold has a triangulation?

no

@TobiasKildetoft don't complain to me

The solutions exist as real numbers, but you will never be able to represent them in terms of stuff like square roots and the like

@MikeMiller what's the smallest dimension manifold that doesn't?

4

@LeakyNun to who then?

how do I know

ask the 60's guys who classified them

@Ultradark the simplest relevant case is n=5, i.e. x^5-x+1=0

@MikeMiller which manifold?

Freedman's E8 manifold

@LeakyNun What? Where did you see that they were called by those letters?

ugh, none of the lindlemann weistrass theorem proofs are comprehensible to me

i think that man has too much free time

Wil try again later

@TobiasKildetoft I must have, you know, misremembered :P

i am not so sure of that. you'll note that his 80s work was when he was a professor, so in fact, coming up with the E8 manifold was part of his job

perhaps he worked so hard he has too little free time

@LeakyNun 4 of those correspond to series of finite simple groups of Lie type (though one needs a bit more data apart from the letter and a number).

@MikeMiller I was making a pun

boo

(A question I don't actually know the answer to: How can one tell for which $a$ the equation $x^5-x+a=0$ is solvable by radicals?)

@rschwieb It's not too weird. The classification of finite simple groups + Jordan-Holder says that every group has a finite composition series, made up of finite simple groups. But it doesn't say how those pieces fit together - the group is not determined by its factors. In the case of p-groups, the simple groups are the `simplest' possible, $\Bbb Z/p$.

I have in mind rational $a$, though one could probably be a bit more generic than that

@TobiasKildetoft so what are the names?

@Semiclassical it's always solvable by Bring radicals

So as you increase the order, you need (1) the number of possible composition series with the right cardinality of their product (2) the ways they fit together

@LeakyNun I am not actually sure which ones are usually grouped together, since I mainly know about the Lie type ones (and mostly the Chevalley groups).

as in, just bring me the radical

@LeakyNun that's not 'solvable by bring radicals' so much as it's basically the definition of bring radicals

So $2^n$ is likely to have the most groups. Though you don't have many ways to cut it up into simple groups - they are always the group $\Bbb Z/2$ - you do have many ways to fit them together. $2^n$ is where you can fit the maximal number of factors

Actually, hmm

bring radical is the solution of x^5+x+a=0

I imagine the solution to x^5-x+a=0 should be related

@Semiclassical I think it never is unless $a = 0$ actually

@Semiclassical I just wanted to make a pun

lol

@TedShifrin: @TobiasKildetoft suggested to consider the projection map from the sequence to its even elements and see if it's continuous as the closure is preserved under continuous maps. I'm unsure about the details.

also boo

@OskarTegby Not preserved. But the set in question will be the preimage of a singleton.

@TobiasKildetoft that can't be right. when a=30, you get x^5-x-30=0 which has x=2 as an integer solution

@Semiclassical Ahh, right

@TobiasKildetoft: Mea culpa.

@LeakyNun every smooth manifold is triangulable (in such a way that the maps from simplices are all smooth embeddings, even). the 4-manifold I just mentioned is not smoothable. I believe that in 4D smoothable is equivalent to triangulable, but there is a piece of the argument I have in mind I was never comfortable with and I should ask an expert about.

Okay. So, it's closed because it's a point?

In dimension above 4, it was a longstanding open question whether or not every manifold was triangulable, that my advisor disproved in 2013, following work of Galewski-Stern and independently Matsumoto who reduced this question to a question in 3-manifold topology

@Secret Stop using this chat as your personal notebook.

I guess here's a question: For which $n$ is $x^n-x+1=0$ solvable by radicals?

(with $n>4$ being the nontrivial cases)

I miss algebra.

I do not say that often

Why?

@Semiclassical I'm trying to solve $logx=log(1-x)=2logx=2log(1-x)$

uh

do you have an extra equals sign

I Might be wasting my time

because as it is you've got log(x) = 2 log(x)

which is false unless x=1, and in that case log(1-x) isn't well-defined

A weaker version of my question above: For which $n$ is $x^n-x+1=0$ reducible?

I guess I should specify "reducible over Q"

@Semiclassical So, what is the discriminant of that?

@Leaky: Do you have any clue?

good question. I don't actually know how to compute the discriminant of an arbitrary polynomial off the top of my head

@OskarTegby question?

@Semiclassical I wrote it weird bc I'm on my phone but it's a system of four equations in which 1-x is always paired up with x

Me neither. I just am not good enough with symmetric polynomials for that.

@Ultradark Then I don't know what you mean

what you wrote was of the form A=B=2A=2B

which is only true when A=B=0

and that's never true when A=log x, B=log(1-x)

There you go, @LeakyNun.

@OskarTegby and where are you stuck at?

I don't really know how to show that they are closed. Tobias suggested looking at the projection map and showing that it's continuous. I'm a bit lost on the details.

what's l1 again?

absolute convergent series?

$\ell^p=\{x:\|x\|_p<\infty\}$

hint: arbitrary intersection of closed sets is closed

@Semiclassical so the system of equations is: 1) $\ln(x)=\ln(1-x)$ 2) $2\ln(x)=2\ln(1-x)$ 3) $\ln(x)=2\ln(1-x)$ 4) $\ln(1-x)=2\ln(x)$

hmm, looks to be that x^n-x+1=0 is reducible when n=2 mod 6. (i'm not saying that's the only such instance, but so far it's worked for n=8,14,20,26...

I'm on a computer now yay

@Semiclassical 3 is a prime, 5 is a prime, 7 is a prime, 9 is an insignificant error, 11 is a prime, 13 is a prime, ...

Okay. Hm... Would we consider sets with elements of the sequence?

1) and 4) together imply $\ln x = \ln(1-x)=2\ln x$, which is only true when $\ln x=0\implies x=1$

@Oskar: Why not show the complement is open? That'll be easier for you to see.

Which intersection would we be doing?

Okay. I think that my intuition fails when thinking about sequence spaces.

Take a point outside $X$ and show that there's an $\ell^1$ neighborhood around it not meeting $X$.

but then you're back to ln(1-x) being ill-defined

Like, what's even a set there?

So no, these equations have no solution.

> what's even a set

but I'm looking at $0<x<1$

Then you won't satisfy the equations.

There is no 0<x<1 for which all four equalities can be valid.

hm

@TedShifrin do you like my approach?

i mean, if 0<x<1, then ln(x) and ln(1-x) are negative real numbers

and no real number is twice itself except zero.

It's fine, too, as long as you are careful to check closure given the $\ell^1$ topology.

i'm just talking about checking that they are closed

so even just 1) and 4) are enough to ensure that there's not going to be a solution with 0<x<1

So am I.

What would a point be in the sequence space? What does it mean to consider a $\ell^1$ neighborhood?

Well, you need to answer those questions.

What does it mean to say $\|(x_n)-(y_n)\|_1<\epsilon$?

I guess you want to show that the complement is open by considering neighborhoods of points outside $X$ as close to $X$ as possible in order to study the boundary.

I don't want to worry about boundary. What's a point in the complement of $X$?

That the limits of the sequences don't differ by more than epsilon.

@Semiclassical ln(x)=ln(1-x) has a solution for x in 0,1 it's x equals 1/2

A point that isn't in the sequence?

and? I've no objection to 1), 2), 3), 4) having solutions when considered by themselves

oh okay

but the claim that 1) and 4) can be true at the same time is plainly false

yeah I understand

I wasn't meaning to say that

I was on my phone

ah

well, two of those equations have easy solutions: equation 2) is just equation 1) with both sides multiplied by 2

so the solution is again just x=1/2

@loch are you going to tea

@Oskar: First of all, every element of $\ell^1$ is a sequence that converges to $0$ (why?).

additionally, equation (4) is equation (3) with x replaced by 1-x

It's just annoying because there's so many equations to solve. There's got to be an easier way...there is symmetry

But did you answer my question: What is a point (sequence) in the complement of $X$?

@Ultradark ...again, what do you mean there's 'so many equations to solve'

If you consider more than one of those at a time, then there's no solution.

So speaking about there being 'so many equations to solve' makes little sense

@Semiclassical I have to solve $ln(1-x)=ln(x)$, then I have to solve ln(1-x)=2ln(x)$, then $ln(1-x)=3ln(x)$

and it's just so many

I don't really know. I only know that $$\ell^1:=\{x:\|x\|_1<\infty\}$$ and $$\|x\|_1=\sum_{k=1}^\infty|x_k|.$$

Where's the question about $\ell^1$? I missed it

So, if $\sum |x_k|$ is finite, what does that tell you about the sequence $(x_k)$?

oh. so this is just $x^n-x+1=0$ again

Scroll up, @Alessandro. It's displayed.

$x^n+x-1=0$

fine

like I said earlier, you should not expect to be able to solve that by radicals when n>4

yeah its very un nice

it's possible that this will have a nice solution for some particular n>4. but in general it will not

maybe I'll just approximate numerically

that'll be fine

@Ted: It converges?

To what?

that's what I'd suggest, yes

@OskarTegby found it! I love how you can tell it's from Brezis because the space is called $E$

@Alessandro: Lots of people call Banach spaces $E$. :)

No one uses $V$. Blah, boring.

There are ways you can write the solution with 0<x<1 as an infinite series, fyi

I'd use X and Y

For some reason, $E$ is most common.

but you should not hope for any nice closed form when n is sufficiently large

@AlessandroCodenotti: Haha! Yeah! @TedShifrin: I don't know. (I know that you mentioned convergence to zero, but I don't know why that's the case.)

okay

what are the ways?

@Oskar: This is the first thing you learned in your calculus course when you learned about convergence of series.

taylor series?

@AlessandroCodenotti why does Italian speak other languages with a nasal accents even though there's no nasal vowels in Italian?

I have no idea

I feared that was the case. I should become a teacher assistant in calculus. I've forgotten way too much of it.

If the $n$th partial sum of the series is $s_n$ and $s_n\to\ell$, then what is the limit of $s_{n+1}$?

@TedShifrin 🅱anach s🅱ace

@Ultradark more or less, yeah

refers Leaky to the second item on the starboard

Is it $\ell$?

OK, so what is $\lim (s_{n+1}-s_n)$?

Sounds like no~

0

So what is $\lim x_n$?

0

There's the proof.

So any sequence in $\ell^1$ must in fact converge to $0$. Now I'm back to my original question. What's a point (sequence) in the complement of $X$?

Remember $X$ had a specific description.