Mathematics - 2018-06-20 (page 2 of 4)

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1:42 PM

Hello :)

A strange thing happened during my PhD supervisor meeting yesterday. It was said that if we computed the order of a group to be zero, then that group is infinite. I think this can be said without any additional context.

I didn't understand it; I still don't.

Would anyone here like to enlighten me, please?

what's the order of a group ?

It's usually the number of elements in the group, right? So what would an order of zero mean?

@mercio

well you obviously don't have the same definitions as the other peoples

the empty set is not a group

I know.

so you should go back to the meeting and ask them to clarify their definitions

1:49 PM

Well, from what I remember from yesterday, apparently, if a group has order zero, then it is infinite. I don't understand it; it's just what was said.

there is nothing to understand if you don't have the definition

there is just no point

do you perhaps mean characteristic zero?

An infinite product was involved.

No, I don't.

I don't think an alternative definition was used.

In order to have an alternative definition of order, you'd actually have to

y'know

so yo uare seriously saying that if a group has 0 elements in it then it is infinite ?

1:51 PM

have one definition to start with

No, just that if it has order zero, then it is infinite.

that sentence is meaningless if you don't have the meaning of "order"

there is nothing to do with it

Plus, the usual definition of 'order' for groups is in terms of cardinality

and cardinalities aren't zero unless it's the empty set. so that definition will not deliver on that.

(the notion of 'characteristic' for rings, by contrast, is usually defined so that characteristic zero implies infinitely many elements)

I think we might have defined it that way, or that it's a definition itself, but I don't understand it. We were definitely talking about order. There's a formula for the order of a group and if that formula produced zero, the group was said to be infinite.

and what's that formula

1:58 PM

The product over all $n$th roots of unity $\theta$ of $f(\theta)$ for some polynomial $f$.

The group determines $n$ and $f$.

so some sort of group determinant?

Yes.

well, googling gives this right away:

Frobenius determinant theorem

In mathematics, the Frobenius determinant theorem was a conjecture made in 1896 by the mathematician Richard Dedekind, who wrote a letter to F. G. Frobenius about it (reproduced in (Dedekind 1968), with an English translation in (Curtis 2003, p. 51)). If one takes the multiplication table of a finite group G and replaces each entry g with the variable xg, and subsequently takes the determinant, then the determinant factors as a product of n irreducible polynomials, where n is the number of conjugacy classes. Moreover, each polynomial is raised to a power equal to its degree. Frobenius proved this...

That's not it.

books.google.co.uk/…

math.uconn.edu/~kconrad/articles/groupdet.pdf

seems related

2:05 PM

maybe they wanted to say that if the result of the formula was a positive number then the group was finite and that was its order, and if the result was 0 then the group was infinite

user image

lol

Ah yes, the usual convention $0 = \infty$. We've all been there.

5

It's a result from combinatorial group theory.

page 33 is also relevant

2:09 PM

Page 33 of what, @Semiclassical?

the book you linked

Simply Beautiful Art

<.<

excuse wut

Oh, I see now, @Semiclassical; thank you!

np

I think that answers my question.

Simply Beautiful Art

2:26 PM

>.>

Anyone wanna play a game of fast growing functions?

BB(x)

Simply Beautiful Art

:l

...sorry.

Simply Beautiful Art

Uh so rules:
Each turn write function(s) (perhaps infinitely many) such that at least one function grows faster than all the previous.
f grows faster than g if f(n) > g(n) for all sufficiently large n.

You are free to use other people's stuff too

f(n) = 0

2:29 PM

f(n)=1

Simply Beautiful Art

Also it is preferred you use N → N functions

All constant functions

:P

$f(n) = kn;\forall k\in \Bbb N$

f(n) = the sum of the first n digits of chaitin's constant

damnit i got killed

Simply Beautiful Art

kek

preferably k in N but whatevs

yeah fair my b

Simply Beautiful Art

2:31 PM

All polynomials

all exponentials with integer base > 1

Simply Beautiful Art

f(n) = n!

all iterated factorials

Simply Beautiful Art

Hey @Mr.Xcoder

@Fargle That exactly being?

Hi, SBA

2:33 PM

f(n) = the maximum number of steps of a turing machine of size n for which there is a proof of size <= n in ZFC that the machine halts

@SimplyBeautifulArt $n!!!\dots!$

Simply Beautiful Art

>.>

2f(n), where f(n) is as mercio defined

I am looking at the following problem:
If you add one-third of Peter's savings money to one-fifth of that savings, then that sum is exactly x Euro more than half of his savings.

Form that we get $$\frac{1}{3}\cdot s+\frac{1}{5}\cdot s=x+\frac{1}{2}\cdot s\Rightarrow \frac{1}{3}\cdot s+\frac{1}{5}\cdot s-\frac{1}{2}\cdot s=x\Rightarrow \frac{1}{30}\cdot s=x$$ Or not?

Then it says the following:
This text task can somehow be helpful for problem solving associated with mappings / functions.

What exactly does this mean?

@MaryStar That does look to be the right answer, but I'm not sure what the second half menas.

Simply Beautiful Art

2:37 PM

Let $\operatorname{ZFC}(n)$ be the $f(n)$ mercio last defined. My functions are $f_0(n)=n$, and inductively, $f_{k+1}(n)=\operatorname{ZFC}(f_k(n))$, for each $k$.

let f(n) = the same procedure but with ZFC + a bunch of axioms

instead of ZFC

Simply Beautiful Art

kek

maybe a large cardinal axiom or two

but i don't really know one by heart

so lol

Simply Beautiful Art

(suddenly inconsistent theory)

<.<

what about ZFC + Con(ZFC)

Simply Beautiful Art

2:39 PM

Would prefer we stick to recursively defined functions

I dunno if that one improves much

Simply Beautiful Art

it isn't

lol

it isn't what ?

Simply Beautiful Art

much of an improvement

ah lol

Simply Beautiful Art

2:41 PM

<.< can we stick to recursively defined functions tho

Ok! I can write the whole exercise statement, maybe it will be clearer to you with that. Unfortunately I don;t have an idea about that :/

If you add one-third of Peter's savings money to one-fifth of that savings, then that sum is exactly x Euro more than half of his savings.

a) How much savings does Peter have if x is 7?
b) Which must be x so that Peter has € 300 savings?
c) How much savings does Peter have (term depending on x)?

At the last question we get the relation $s(x)=30x$, right? Do we maybe have to check if this is a function? And if it is injective, surjective, bijective?

@SimplyBeautifulArt in that case I'll bow out

Simply Beautiful Art

lol why tho

iterated factorials were my last trick

Simply Beautiful Art

x'D

am cri a bit inside

2:42 PM

I could keep doing "multiply the last thing said by natural numbers" but that seems lazy

Simply Beautiful Art

wouldn't work actually

for example, when I said "all polynomials"

@MaryStar It seems like it's saying that this looks a bit like a functional equation in general?

@SimplyBeautifulArt Was about to point that out :P

If someone else does the trick I've been hosed

f(n)!

Simply Beautiful Art

@Fargle lol, if anyone wants to continue, let's start over from here

That being $f_0(n)=n$ and $f_{k+1}(n)=f_k(n)!$ for each k.

Hello @AkivaWeinberger

one could then do $g_k(n)=\prod_{j=1}^k f_k(n)$

Simply Beautiful Art

2:47 PM

@Semiclassical No, you couldn't actually

According the rules I have laid out, you must say at least one function which grows faster than all the previous

fair enough

Simply Beautiful Art

Hm so I guess I'll use my new function iteration operator:
$$(I_0f)(n)=f^n(n)=\underbrace{f(f(\dots f(}_nn)\dots))$$

for any choice of $k$ that'll work, but not for arbitrary $k$

Simply Beautiful Art

yeah

(which makes the game more interesting $\ddot\smile$)

Let $s(n)=n+1$.

My function shall be $(I_0^3s)(n)=(I_0(I_0(I_0s)))(n)$

How do you visualize an infinite set?

Or is this something you should not try to do :P

2:51 PM

depends which set you have in mind

@90intuition Depends on what you mean. The circle is an infinite set.

Simply Beautiful Art

lol

the set of numbers between 0 and 1 is an infinite set, and that's just an interval

the cantor set is also infinite, but visualizing that is far less trivial

Simply Beautiful Art

For some comparisons:
$(I_0s)(n)=2n$
$(I_0^2s)(n)=2^nn$
if that helps

So yeah, it depends too much on the particular infinite set for there to be a generic answer

2:58 PM

If you draw a line with a pencil, I'm not saying that I can accurately, but I could try to visualise this as all the atoms that this line is made of. I imagine zooming in, until I reach the atoms, and then just following the molecules that make up the line. I can kind of imagine this in my mind.

But that is something different as I imagine the amount of atoms as being something finite in the end. If I draw it at my screen, I can imagine zooming in and visualising the individual pixels.

But if I try to visualise an infinite set even as simply as the natural numbers, I just get headaches.

Well I think for a set like the natural numbers, having a "local" picture is a sufficient visualization.

Like 1,2,3

Yeah. Knowing that between any two points, you have a (possibly empty) bunch of sequential points.

you obviously can't visualize the integers all at once, but you can visualize any finite portion of it

As far as the line, I'll draw a very rough analogy to pixels vs vector based graphics.

When you imagine zooming in, you imagine eventually having the image "resolve" into a bunch of atoms--a bunch of pixels. But instead imagine if every time you zoom in, the stuff that looked like it was about to resolve into atoms instead looks straight and continuous again.

3:20 PM

A question for myself: Let $G$ be a real symmetric matrix. If $e_i^\top G e_j>0$ for any $i,j$, does it follow that $x^\top G x>0$ for any vector $x$?

No, that condition shouldn't be sufficient.

Wait why not? Is it not possible to just distribute?

Oh no that doesn't make sense.

It could be that $x_ix_j < 0$

Counterexample: Take $x=e_2-e_1$. Then $x^\top G x = G_{22}+G_{11}-G_{12}-G_{21}$

Right.

If $G_{12}=G_{21}$ is too big, then that's negative and one has a counterexample.

@Fargle hmm nice analogy

In my case I furthermore have $G_{ii}=1$ so I can can be more precise and say that $G_{12}<1$ is necessary

similarly, if I do $x=e_2+e_1$ then I get the condition $G_{12}>-1$

@Semiclassical I would argue then that there are finite portions of it that you can not visualise with the atoms we have in this universe, but maybe the mind can :P

3:26 PM

lol

you can visualize any small portion of it, is perhaps the better statement

and since every small portion of it pretty much looks the same...

@90intuition Thanks. I think trying to "directly" visualize infinite sets is kind of futile (our brains are finite in capability, after all), but there are a number of ways to do it indirectly. For sets like the line and the circle, just thinking of them as being continuous, unbroken entities is pretty much enough; for something like the Cantor set, understanding the finite parts of its usual construction is a good first step.

From what I know of this stuff, though, the condition $-1<G_{12}<1$ is sufficient to claim that $x^\top G x>0$ for any $x\in\text{span}(e_1,e_2)$

Hi all!

Hello @Rudi

Hi @Fargle!

Simply Beautiful Art

3:28 PM

hi person I don't yet know :)

@SimplyBeautifulArt: me?

When do you know someone? @SimplyBeautifulArt

just for the case, I'm @Rudi_Birnbaum, nice to meet you!

Peter Sheldrick

what can we say about two operators on the function space f:[0,1]->R that share the same eigenfunctions and corresponding eigenvalues?

they're probably very nice functions

Peter Sheldrick

no what can we say about the two operators

they commute

?

just a guess

3:43 PM

if $f_1 v=\lambda v$ and $f_2 v=\lambda v$, then $(f_1-f_2)v=0$

yeah they commute :)

So $f_1-f_2$ has zero as an eigenvalue for every common eigenfunction of $f_1,f_2$

Peter Sheldrick

okay so the difference between the two operators has a pretty large kernel

If you can prove that the eigenfunctions form a complete basis, then you can claim that $f_1-f_2=0$

Peter Sheldrick

well they form a complete basis

3:49 PM

Are f surjective? is that implied in f:[0,1]->R?

Peter Sheldrick

yes

"yes, yes" or "yes, no"?

Peter Sheldrick

yes i only consider functions

for example polynomials

ok

So youre actual question is "What is sufficient that they will be identical?"?

Peter Sheldrick

they are not identical

there are plenty of v with f1v!=f2v

3:58 PM

Then those v can't be in the span of the eigenfunctions of f1,f2

and therefore the eigenfunctions aren't a basis

also my guess

Peter Sheldrick

is it possible to represent any monomial v=x^n as a series of a_mexp(2*piimx) terms?

any such series is a $1$-periodic function of $x$.

Whats the dimension of the basis?

Peter Sheldrick

infinite

4:02 PM

countable?

and no monomial is periodic, sooo...

Peter Sheldrick

yeah but on [0,1] there should be such a representation right?

(at least if you mean a series with integer m)

So some wicked convergence issue?

ah. then yes, sure

Peter Sheldrick

4:03 PM

countably infinite

this is just fourier analysis tho

Peter Sheldrick

well i'm getting f1v!=f2v for v=x^n for n=0,1,2,3,4...

okay? you haven't actually said what your f1,f2 are

Peter Sheldrick

okay not 0

f1*v=f2*v for v=x^0

(f1v)(x)=\int_0^1 (t-1/2)*v(x-t) dt

(f2v)(x)=\int_0^1 sgn(t-x)*(|t-x|-1/2)*v(t) dt

eigenfunctions are exp(2 * pi * i * m * x), m=...-2,-1,0,+1,+2

should that be *m rather than /m ?

Peter Sheldrick

4:09 PM

and corresponding eigenvalues are i/(2*pi*m)

if it's /m, then those aren't 1-periodic functions

also, I think you're missing an i

okay

Peter Sheldrick

auto formatting is messing it up

makes sense

the problem is that those two claims are not compatible

Peter Sheldrick

what claims?

if you've got functions for which f1 v != f2 v, then v is not in the kernel of f1-f2

4:11 PM

@Gérard : je passe en coup de vent pour te dire..............5

But if the common eigenfunctions are as you've written them, then all those eigenfunctions are in the kernel of f1-f2

Peter Sheldrick

yeah for example v=x then f1v=-1/12 but f2v=-x^2/2+x/2-1/12

So the trick is that they do not have all their eigenfunctions mutually in common? But isnt't that more an language issue?

Peter Sheldrick

i think they do have all eigenfunctions in common

That's what I'm saying isn't compatible

If there are functions which aren't in the kernel of f1-f2, then those functions can't lie in the span of the common eigenfunctions

4:14 PM

I dont get it. x is an eigenfunction of f1, no?

If f1*x = -1/12, it's certainly not an eigenfunction

Peter Sheldrick

no

oh sorry

misread that!

right!

Peter Sheldrick

f1x=-1/12

its constant

yes

Peter Sheldrick

4:17 PM

oh its
(f2v)(x)=-\int_0^1 sgn(t-x)*(|t-x|-1/2)*v(t) dt

or

(f2v)(x)=+\int_0^1 sgn(x-t)*(|t-x|-1/2)*v(t) dt

But is x surjective on your function space?

Peter Sheldrick

(f2v)(x)=+\int_0^1 sgn(x-t)*(|x-t|-1/2)*v(t) dt

of course x:[0,1]->R is surjective?

no the image is [0,1] I think so its not surjective because [0,1]!=R.

Here's the issue. If you do Fourier analysis, you can write $v=x=\sum_m a_m e^{2\pi i m x}$

But if $(f_1-f_2)x \neq 0$, then it can't be the case that $(f_1-f_2)e^{2\pi i m x}=0$ for all $m$

Mohammad Areeb Siddiqui

0

Q: Why does this integral vanish while doing integration by parts?

Consider $$\int_{-\infty}^{+\infty} x \dfrac{\partial}{\partial x}(\Psi^*\dfrac{\partial \Psi}{\partial x} - \Psi\dfrac{\partial \Psi^*}{\partial x}) dx $$ If I apply integraton by parts here by bringing in the x inside the derivative then its $$\int_{-\infty}^{+\infty} \dfrac{\partial}{\partial...

integration definite-integrals physics quantum-mechanics

Peter Sheldrick

4:23 PM

wow i think the image on the wikipedia page is confusing

my analysis 1 course was a while ago

its sometimes called "onto" in English

none of Y should have an empty fiber.

one thing for sure in here is that $f_1 (1)=f_2(1)=0$

Peter Sheldrick

@Semiclassical yeah i have that as well

yeah, thought it was missing earlier for some reason

Peter Sheldrick

i overlooked it but then corrected myself

4:29 PM

gotcha

Peter Sheldrick

f1v!=f2v for v=x^n, for n=1,2,3,4 but not n=0

@Rudi_Birnbaum so what is the problem with non-surjective functions in this case?

I think I see the issue. The eigenfunctions are the same, but I don't think the eigenvalues are.

...maybe not

Peter Sheldrick

they are the same

at least if the only eigenvalues are i/(2*pi*m)

maybe there is more than the point spectrum

yeah, I'm perplexed

It doesn't look like $v=x$ can possibly be a common eigenfunction. But $f_1-f_2$ apparently annihilates all 1-periodic functions

Peter Sheldrick

no v=x isnt an eigenfunction

its just a function that i tried

the eigenfunctions are exp(2 * pi * i * m * x)

4:35 PM

(I really mean the 1-periodic extension of $v=x$ on $[0,1]$)

@PeterSheldrick I was just interested to know when you stated the problem. If all eigenfunctions images are restricted to a subset of R, then you of course could run into trouble with functions who are not.

So you've seemingly got an element which lies in the span of the eigenfunctions in the kernel...but which isn't itself in the kernel

that don't make sense

Peter Sheldrick

@Rudi_Birnbaum yeah the eigenfunctions exp(2 * pi * i * m * x) are restricted to a subset of C

Dumb question: if $T : V \to W$ is a linear operator, $V$ is infinite dimensional, and $W$ finite dimensional, then $T$ can never be injective, right?

Peter Sheldrick

i think it could be

oh wait

no

i havent used injective/surjective in ages

@Rudi_Birnbaum isnt exp(2 * pi * i * m *x): [0,1] in [-1,+1] + i*[-1,+1]?

4:44 PM

Here's a precise claim: for $v=x$ on $[0,1]$, one has $\displaystyle v=\frac12 - \sum_{m=1}^\infty \frac{\sin(2\pi m x)}{\pi m}$

one should then be able to apply $f_1-f_2$ term-by-term to both sides

well I am actually not quite sure if is an issue.

Yeah I am confuzzled

the series I wrote seems to indeed be annihilated term-by-term by $f_1-f_2$

no I think its not a problem. I just about even monomials and odd and there it cooured to me, but thats a different reason then

but $(f_1-f_2)v\neq 0$

@Semiclassical if i would be superhero, I'd be confusion-man!

2

4:52 PM

lol

yeah thats where I am really good at ;-)

" This is often
the content of a second course in algebraic geometry, and in an ideal world, people
would learn this material over many years, after having background courses
in commutative algebra, algebraic topology, differential geometry, complex analysis,
homological algebra, number theory, and French literature."

lol

heh

the inclusion of intro to french literature is interesting

I mean, I’m not crazy to be confused right

4:54 PM

hope so

It seems absurd that a Fourier series could be an operator term-by-term and yet not have the function itself vanish

Peter Sheldrick

i can even produce a third operator f3 again with the same eigenvalues and eigenfunctions except for m=0 but f1v!=f2v!=f3v for v=x^n

Wait...hmmmmm

Peter Sheldrick

maybe that isnt so exciting

I think I see the issue.

4:55 PM

let us know

If you go to the definition of f1

Peter Sheldrick

is it some convergence issue like rudi suggested earlier?

No, it’s sillier than that I think

What enters into the integrand is v(x-t), and ive been naively replacing that with x-t

Peter Sheldrick

dinner, ill check back in a bit

But x-t takes values outside of [0,1] as t ranges from 0 to 1

4:58 PM

!

As such, this substitution is only valid when 0<x-t<1

Ie from t=x-1 to x. Since x is assumed to be between 0 and 1, this amounts to t=0 to x

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