@robjohn ORLY?

@PeterTamaroff I guess the day was not over yet.I thought it was a new day UTC already

@PeterTamaroff However, it still says 190. Interesting.

Perhaps today's rep isn't showing yet.

There.. now it shows. 3 seconds into today :-)

Hmm

@robjohn how are you, mister?

@Charlie pretty good. and you?

@robjohn I'm.alive

@Charlie No shit, Sherlock.

@PeterTamaroff It could be recorded keystrokes...

@PeterTamaroff no jokes for today, mister

@robjohn How dramatic. True.

@Charlie " Is everything a joke to you, Gordon?"

@PeterTamaroff remember the squirrels

@robjohn I should add Corgis. And spiders on coke.

@PeterTamaroff it's the real thing

@PeterTamaroff you male jokes at everything, Pedro

@Charlie Not everything. That would be impossible.

why you hate me?

@Charlie I don't hate you.

I don't hate, in general.

Except grammatical mistakes.

@PeterTamaroff so why don't you respect me?

And people that generalize for the sake of generalizing.

And inductivists.

@Charlie Respect?

Yes

Explain.

@PeterTamaroff you're always sarcastic, and mock everything I say

Even when I tell youu I'm not feeling okay

@Charlie I think you're taking me too seriously

You're a clown I'm so sorry

It is better if you just take me less seriously. Really.

And now I am serious. =)

@anon Hey. How's that popcorn?

huh

@anon bye, mr. anon

why bye?

@anon I'm leaving

:(

@Charlie Oh, Charles.

Let's see of this cheers you up, shall we?

@anon Don't be sad, Pedro is here

sigh

still hurt?

@anon I have a good memory, I don't forget things

wasn't talking about your memory

@anon it's a lame, because I'll never know your name, mr.

So, bye yo all

bye @Charlie

@Charlie Wait!

I am looking for something.

For you.

@PeterTamaroff what...want to mock me one last time?

@Charlie OK. When you are chill as a beer, come'a back. Si?

I will be waiting.

@PeterTamaroff don't worry, you know where else to find me :/

@Charlie Nope. This is the place.

@Charlie Exactly.

@Charlie Give me 5'

@PeterTamaroff Tchau, Pedro

@PeterTamaroff .....ok

I'm waiting.........

@Charlie I cannot find it.

@PeterTamaroff what are you looking for??

@Charlie I gave up.

Told you on the other side.

Okay...

@anon Dude.

huh?

@anon What proofs do you know that $C^1[a,b]$ is dense in $\mathcal L^1[a,b]$?

can one use dominated convergence to show every thing in the latter is a limit of things in the former?

I don't really remember

@anon I am showing that $C[a,b]$ is dense in $\mathscr R[a,b]$ now.

@anon One of my few linear algebra answers!

predual, cool word

@anon Heh, learnt it from Julien.

@anon If I give you a $2 \times 4$ matrix and if I know the determinants of all the $2 \times 2$ minors can I recover my original matrix?

how many minors are there? (hint)

6

and 8 entries in the matrix, so I wouldn't expect so

@anon You can I just recovered my matrix

@BenjaLim Burn the witch!

so there's a bijective polynomial map K^8 -> K^6?

hmm

why should there be one

the map from 2x4 matrices to the 6-tuple of their minors' determinants

if the 2x4 matrix can always be recovered from the 6-tuple, then the map is injective

(I suppose I should have said injective and not bijective)

and the polynomial map is linear in each coordinate, to boot

or wait, almost

Ok @anon I had a $2 \times 4$ matrix

with $m_{(1,3)} = 1, m_{(1,3)} = 2, m_{(1,4)} = 1,m_{(2,3)} = 1, m_{(2,4)} = 2, m_{(3,4)} = 3$

The matrix

$\left(\begin{array}{cccc} 0 & 2 & 4 & 2 \\ -1/2 & 1/2 & 3/2 & 3/2 \end{array}\right)$ works

how do you know you haven't "recovered" the wrong matrix?

if you apply a rotation matrix to every column of a matrix you preserve the minors' determinants, no?

thus, "the" matrix corresponding to the 6-tuple is actually not uniquely determined

yeah I know there are many choices

so sorry

When I said recover

My question was an existence question not a uniqueness question

ah, finding elts in preimages is a weaker activity than "recovering" things

Hmmm but this is funny because no matter what matrix you recover the kernels are all supposed to be the same

Basically for a subspace $V$ of codimension $d$ in say $\Bbb{C}^m$

we can write $V$ as the kernel of some linear map $A : \Bbb{C}^m \to \Bbb{C}^d$

yes

and then we have an induced map $\bigwedge^d A : \bigwedge^d(\Bbb{C}^m) \to \bigwedge^d(\Bbb{C}^d)$

The codomain is one dimensional

indeed

Thus a vector $e_{i_1}\wedge \ldots e_{i_d}$ is taken to some number $x_{i_1,\ldots,i_d}$

which is called the Plucker coordinate of $e_{i_1}\wedge \ldots \wedge e_{i_d}$ @anon

@anon My question was basically given points with Plucker coordinates blah

can I recover my subspace $V$

I see

perhaps seek to create a basis for it

@anon These sorts of things come up if you want to study flag varieties, etc

oh, by the way

yea

I have to find the question, hold on

type ahead I'm going to toilet

Introduction to Modern Number Theory states Kronecker-Wedderburn basically as: "every finite-order homo $G(\overline{\Bbb Q}/{\Bbb Q})\to{\Bbb C}^\times$ factors through $G({\Bbb Q}(\zeta_n)/{\Bbb Q})$ for some $n$," and says the usual version ("every abelian extension of Q is contained in a cyclotomic field") is a restatement of this. I can't see how to recover the usual version from the galois-rep version though. [I decided to retype out my question instead of find it in the transcript.]

I'll be back in a bit, have to go get some stuff from the store. Eventually I'll post on main with a screenshot if I don't understand.

@anon Mark this day! I "read" you say " I can't see how..."

You're human, after all.

@anon Can you help me count?

Lattice points. Count lattice points.

what about em?

@anon I want to count the number of solutions to $$|x_1|+\cdots+|x_p|\leq n$$

Now, one can use generating functions as in P&S. Namely, $$\left(1+2x+2x^2+2x^3+\cdots\right)^p=\sum_{k=0}^\infty a_kx^k$$ and the multiply by $$\frac{1}{1-x}$$ to collect $$\sum_{k=0}^n a_k$$

yay i answered a hard one for once math.stackexchange.com/a/464019/1284

How can one use a clear counting argument, @anon ?

dunno off the top of my head

I have to go to a party though

@anon Have fun.

pens never seem to work when you really need them to

@anon That's why you need either pencil or fountain pens.

@DanBrumleve Reads

had to write a program

had the idea and tested a few moduli of the form 2^n-1 and 3^n-1 until i found 4095

then wrote another prog to check the differences mod 4095

@DanBrumleve Maybe you can add that.

Else I look taken off a top hat.

ok

i linked the programs

i just found a bug in the second one but it doesn't break the proof

ok fixed that too

sup

If anyone is interested, I've been looking for an answer to this (original, I believe) question in elementary number theory.

I know neither how to approach a proof of the conjecture nor what the consequences would be if it were shown to be true.

Quite a difficult group-theoretical problem

Most of the questions here are way over my head. I'll get there one day.

Same here. :)

Did you take a look at my problem?

It's a really elementary statement, but I'm not sure the proof is easy.

BTW, that problem turns out to have an answer here

hi, can anyone pls tell me how to put consequently symbol?

I'm afraid I don't know

ok, np

do you mean the therefore symbol with three dots?

or the implies symbol that looks like a thick white arrow?

ya, 2nd one

implies symbol with thick white arrow

but it would be nice to know even therefore symbol

\implies and \therefore

nice, thanks

would either of you care to look at the problem I posted above?

$\therefore$

This symbol is rare in university textbooks.

$\because$

Hvorfor har du på deg hatt

cool tune

? =)

Indeed

:D

this is altså pretty cool.

I prefer this

Just remember that size matters

If a function (polynomial) is not reducible over the integers, is it possible that the function's range is only composite numbers (if the domain is the integers)?

@Alyosha See this one

@FrankScience thanks, the link was useful. Is it possible to come up with a polynomial (degree $\ge1$) that spits out only prime numbers?

@Alyosha What? You mean the polynomial only takes non-prime values?

Sorry, I meant only prime values.

It seems impossible.

Is it provable, though?

There's a neat proof.

So it's possible, but improbable.

What's possible?

"It is not even known whether there exists a univariate polynomial of degree at least 2 that assumes an infinite number of values that are prime"

That's infinite number of, not all of.

Mea culpa

What?

My mistake, sorry.

In the proof, I don't see why it's necessary that $P(1+kp)\equiv 0 \mod p$.

Must go for a few minutes

It's obvious.

$(1+kp)^n\equiv1\pmod p$

It is obvious, yes. Silly me.

Can anybody help with my simple question

@Alyosha Never mind. It's becoming obvious as your experience grows.

@FrankScience thanks for the time. Now I've seen it, it looks similar to the proof of Lucas' theorem.

@Alyosha It's much easier than Lucas' theorem.

Hi @amWhy how are you?

Hi, @skull. I'm fine ;-)

@amWhy Good, good :D

@skullpatrol And are you "fine" as well? :^) (I hope)

@amWhy Yes I am, thanks.

@amWhy are you enjoying your time away?

@skullpatrol Actually, yes, it's been a good thing.

goodgoodgoodgoodgood...... ;-)

@amWhy Even Old John has not returned.

Helo.

What is up ?

@MarianoSuárez-Alvarez

I have to tell you something.

zzz

@anon

@HenningMakholm

Nevermind, sire!

@PeterTamaroff Hi

@HenningMakholm Yello.

Green

@N3buchadnezzar Maroon.

Jello

@HenningMakholm What is your field, Henning?

CS

@PeterTamaroff Um? I tend to do $\mathbb R$ most, but $\mathbb F_{65536}$ has a special place in my heart too.

@HenningMakholm HEHEHE, I mean field of expertise, or whatever it is called.

Ah yes, that would be software engineering, theoretical CS, formal logic ...

I have capped. For the first time since early April. Very nostalgic feeling.

@HenningMakholm Yay!

I posted a nice answer today, a rare but good feeling.

How does one prove that ${\sin(mx),\sin(nmx)}$ for $n,m \in R$ is linearly independent?

@Alyosha Orthogonality.

@PeterTamaroff I'm aware of the proof of their orthogonality (via integrals), but is there a more elementary way without using orthogonality?

@Alyosha Orthogonality is pretty elemental. Things would get more tortuous, but I guess one can find a proof.

In fact, you're simply using the fact that $$\int_0^{2\pi}\sin(mx)\sin(nx)dx=0$$ when $m\neq n$. That is all.

That is really elementary.

@Alyosha It may be even more elementary to observe that one of them has zeroes that the other doesn't have. (Of course if $n\in\{-1,0,1\}$, then you're in trouble).

@HenningMakholm Yes, that's the sort of thing I was looking for. I'm doing this as a motivator for the inner product/orthogonality of functions, which is why I was so pedantic in avoiding it, if arbitrarily. Thanks.

@HenningMakholm hey

@BenjaLim Hey.

do you know why any mobius transformation that sends the disk to itself has to send the unit circle to itself? @HenningMakholm

@BenjaLim By continuity of $z \mapsto |f(z)|$?

Hmmm I don't really see why that's what everybody's been saying to me even in class but I'm not convinced @HenningMakholm

@BenjaLim Maybe this is slightly related.

@BenjaLim Since the transformation is bijective and conformal, any point that's the image of a point on they unit circle must have neighbors that are in the image of the disk and neighbors that are in the image of the outside of the disk. You know where those images are, so the image of a point on the circle must itself be on the circle. Now repeat for the inverse transform.

ah ok

right @HenningMakholm the image of $f(z)$ for $z$ in the ball cannot be outside of the unit circle

otherwise it will have no neighbors $f(w)$ with $w$ in the unit disk

thanks @HenningMakholm

@BenjaLim In more fancy language, a Möbius transform is an auto-homeomorphism of the Riemann sphere, and the unit circle is the boundary of the unit disk -- so if it preserves the disk it also has to preserve the boundary.