It is also known that $$\binom nk_q=\binom{n-1}{k-1}_q+q^k\binom{n-1}k_q$$

I want to prove that using the definition above.

I have a combinatorial proof of this using Ferrers diagrams and counting partitions and whatnot.

But I want that one! =D

Let $\pi \colon \mathbb{F}_q^n \to \mathbb{F}_q^{n-1}$ be the projection on the first $n-1$ coordinates. For a $k$-dimensional subspace $V$, $\pi(V)$ is $k-1$-dimensional if and only if $e_n \in V$, and $\pi$ induces a bijection between the $k-1$-dimensional subspaces of $\mathbb{F}_q^{n-1}$ and the $k$-dimensional subspaces of $\mathbb{F}_q^n$ containing $e_n$.

HELP @DanielF !! :)

That gives the first term on the right.

@TedShifrin You use all-caps? Atlanta must be burning.

LOL ... My analysis savior has failed me? :)

So, @Pedro, we need to see that every $k$-dimensional subspace has $q^k$ preimages under $\pi$.

@DanielFischer OK. Would you like to continue or would you rather I write things alone?

@TedShifrin This one? I guess it will turn out easy after one has found the right trick.

@TedShifrin I answered a uniform convergence question.

But finding the right trick isn't so easy.

Not that one though.

Yeah, @DanielF, that's the first uniform convergence proof I've gotten stuck on :) I'm usually good at these. But it's amazing the crap that people post, even after they've been warned ...

Yeah, @Pedro, it's sneaky.

Is it really acceptable to answer your own question? It feels so odd...

@Studentmath Yeah, it's cool.

It's good!

I have urged some people to do that after I hint them to figuring it out, @Studentmath.

Well, @Pedro, it's turning into summer ... I was dripping wet after tennis :(

Well, will do it, thanks! I managed the question last night\morning Prof @Ted!

@DanielFischer "Let's integrate over a grey elephant shaped countour..."

Yippee for you, @Studentmath!

@TedShifrin Well, sweating is good.

I prefer winter :P

@TedShifrin I'm currently - when I'm not otherwise occupied - trying to apply Dini, so I need to see that $$\limsup_{x\nearrow 1} \sum_{n=0}^\infty \frac{x^n}{(1+x^n)^n} \leqslant 2.$$

I prefer tempered weathers.

Is that like tempered distributions, @Pedro? :D

@Daniel: From my Mathematica graphing excursions, I would say that's definitely true, but I don't have a proof. I was trying to do my usual thing of turning the series into an integral, but haven't seen how to do so here.

@TedShifrin Well-tempered pianos.

@TedShifrin Haha, I meant "templado."

Ah, how Bach of you, @DanielF

"lukewarm"?

Nobody beats the Maestro. Not even Shostakovich.

Well, I was just listening to the Shostakovich Piano Quintet yesterday :)

I'm a huge fan of little-known piano quintets :)

@DanielF: If you're trying the Dini route, how do you know the limit is continuous?

@TedShifrin That is what needs to be shown. Since it is continuous on $[0,1)$ by standard arguments, we need only be concerned with continuity at $1$. $\liminf f(x) \geqslant 2$ is easy-peasy.

By Dini, the limit function is continuous if and only if the convergence is uniform.

Oh, you're starting your summation at $n=0$, and I was starting it at $1$. I see.

Constant difference, not essential.

Of course; but I was confuzled for a bit.

@DanielFischer LEL:

That was a good start.

Your algebra's not right, mr @Pedro.

@PedroTamaroff What is "LEL"?

It's the new way to say LOL.

I was wondering that.

Waaaaaaaay cooler.

Maybe he's embarrassed :P

Heya

@Chris'ssis I just came up with the series $$1+\sum_{k=1}^n(-1)^k\frac{\log(2k+1)}{k}$$ in a totally different way. I need to work on that sum.

but it's parktime now, then Mother's Day.

@robjohn Do think this has a (nice) closed form?

@Chris'ssis I don't really have any idea yet

OK

@PedroTamaroff I posted above a nice series you might enjoy.

Hi @Jas

@KarlKronenfeld (unprintable words here)

math is nice

@mito_562condrio meth is not, though

@PedroTamaroff Objection!

@PedroTamaroff right

font: angelustenebrae.livejournal.com/15908.html#

If you want to be the sort of user who answers book recommendations here are detailed instructions on how to succeed: First, do not read the question. Just skim the first few sentences and then post a list of some books for the topic asked for. Post at least 5 but more is better since more choice is a better recommendation (best recommendation would be to post the top 100 books on Amazon for that topic!).

Ok, I'll do that.

Excellent.

You should also post a long series of comments encouraging piracy and telling people how to pirate, too.

I'll get to that.

I guess I'm the only one disturbed by users posting differential geometry text recommendations when their primary tags in answers are precalc and elementary-set-theory and they have not a single answer (or question) in DG?

@MikeMiller No, wait, I know what you're going to do now. Just gimme a sec, I'll do it for you.

Textbook rec questions usually suck, and so do their answers.

I was agreeing with you.

: )

Usually when I complain about people on the internet I get the picture above^

Especially annoying because I think xkcd is not funny.

Holy hell, someone who shares my xkcd opinion

we're rare nowadays, @MattN.

Some are funny, some aren't.

@MikeMiller Yeah, actually I thought I was the only one. : D

What is xkcd

Heya

@DanielFischer I haven't come across a single funny one.

I sometimes find PhD comic amusing.

Anyone familiar with maple ?

I think PhD comics are also unfunny

They had like, five or six gags

And then just repeated them for ten years

I have L := {x=0} {x=1} and trying to find a command that gives me the number of elements of L.. =(

@N3buchadnezzar I was forced to use Maple for a year

What does L do

@MikeMiller Oh, I don't follow any comics. But I remember that I laughed at at least two of them. I'm not sure I saw more than that.

There's only one internet comic I've ever really liked.

In reality I define L := singularities(f(z),z) and it spits out L := {z=0} {z=1} ... and I am trying to figure out how many singularities it has.

@MikeMiller Oglaf?

@MikeMiller What's that?

Why the hell would you kill a lizard?

Yeah. Might put it into a cup and throw it out the window?

I find killing animals disturbing (unless they're insects or fish).

No.

But it doesn't mean I could kill the animals I eat.

Ok, except chickens maybe.

@JasperLoy Yes.

I once opened a window when I was a child and accidentally injured a lizard. It limped away and I was extremely disturbed by it.

I think humans are a pest.

Humans are the same wherever you go.

I doubt it.

Actually I don't.

You're right.

That's wonderful, congratulations on making the world a better place! : )

Hi @Jas

I'm not hiding anything, just couldn't make up my mind which I wanted to write.

That's great! I would think a PTSD is much more difficult to deal with than OCD.

@MattN. "Great!"

@MikeMiller Yes, great. Some real progress.

Great!

@MattN. Humans are the greatest pests ever on Earth.

Haha.

It gets worse with every page.

I will.

It's somewhat evil.

It's still evil.

But mosquitoes, just fine.

@JasperLoy It's not. Kill the ones that does real harm to you =)

But avoid killing otherwise.

Mmm

Why is $$ \frac{P'(z)}{P(z)} = \sum_{k=1}^n \frac{1}{z - a_k} $$

?

Try writing $P(z)$ in product form, @N3buchadnezzar

Mmm

@N3b Write out $P$ explicitly and use the product rule; you'll see it.

Alternatively take the derivative of $\log P$.

@MikeMiller That, I didn't think of.

Yep...

Thanks =)

@JasperLoy Smooth Criminal

@JasperLoy Well, try not to kill any animal from now on. You'll feel better.

@BalarkaSen do you see the link between this math.stackexchange.com/questions/790285/… and Galois theory ?

Let me see.

What do you mean by same complex root and coefficients?

I mean the roots are the coefficients

You mean that roots are all equal (so it is something like $(z - a)^n$) and coefficients are all equal?

@GabrielR. Ah.

$\prod_{i=1}^n (z-a_i)$ so that the elementary symmetric polynomials are $(-1)^k a_k$, correct?

I believe loads of them can be found by simply programming stuffs.

You know, they're only a few, and my guess is that there are none with real coefficients for $n\geq 5$

@JasperLoy OK if you can't see them.

@GabrielR. Oh, real coefficients?

That's pretty hard then.

@GabrielR. You already pointed out a family : $X^n(X-1)(X+2)$

@BalarkaSen it has $0$ as a root, so it's unoriginal

@GabrielR. So you want no $0$?

@BalarkaSen since if $P$ is one such polynomial then so is $XP$

@BalarkaSen and you can on with the family $X^nP$

Actually, this is a pretty good question.

but more of number theoretic interest than galois theoretic.

@BalarkaSen But consider the following mapping $T:(x_0,\ldots,x_{n-1}) \rightarrow (e_0(x_i),...,e_n(x_i)$ where $e_p(x_i)$ is the p-th elementary polynomial for (x_0,\ldots,x_n)

@GabrielR. OK.

OMG

The polynomial is over the purely transcendental field extension $\Bbb Q(x_1, \cdots, x_n)$

This question is looking for fixed points of $T$ (which may be defined alternatively as a mapping over R^n or C^n)

and the roots are there too.

so we are looking for polynomials over $\Bbb Q(x_1, \cdots, x_n)$ that splits over there.

but this somewhat loosens the conditions up, so I guess this is just a fact.

and if you look for the inverse of $T$ you solve polynomial equations (D'alembert) For $n\geq 5$, Galois tells that $T^{-1}$ has some properties.

So T is a pretty good mapping I guess :P

Yeah, you're right.

$T = T^{-1}$ is identity in the case of these polynomials

yeah

math.stackexchange.com/questions/789890/… answered my own good question! So happy

Heya guys

Oh Linear Algebra...

so does it imply that such polynomials must necessarily be solvable over $\Bbb Q$?

i have to think about it now.

dang it, @GabrielR., i was revising my group theory.

read this @BalarkaSen books.google.fr/…

"Can one show that for n => 6 the transformations Tn^-1 are obtained by composition of Tm^-1 of lower degree m operating on suitable subspaces of E^n?"

Provably no.

@BalarkaSen Do you understand the definition of S ? I mean, why is that interesting ?

@GabrielR. S? What 'S'?

@BalarkaSen first read what H is. He defines then $S=H^{-1}TH$

and he considers the inverse of $S$ in the scope of Galois theory

why ?

I don't really see an algebraic significance of this.

What does he mean by "homeomorphism of $\Bbb E^n$"? Perhaps homeomorphism from $\Bbb E^n$ to itself?

yes of course

@Sawarnik I don't know .. but I don't think there is a mistake anymore :P

@GabrielR. I think I am starting to get what he is saying.

$H^{-1} T H$ really shifts the original $e_k$ of the roots by continuously mapping it to somewhere else.

Let's take an example.

Darn that is not a homeomorphism.

Give me a good example of a nontrivial homeomorphism, @GabrielR.

in which space ?

$\Bbb E^n \to \Bbb E^n$

I thought of exponentiation termwise.

But logarithm behaves weird at the negatives.

linear transforms are trivial ?

@GabrielR. Of course.

But let's do it this way.

$f : (\Bbb R^+)^n \to (\Bbb R^+)^n$ as $f : (x_1, \cdots, x_n) \mapsto (e^{x_1}, \cdots, e^{x_n})$

A subset consideration is almost nearly the same.

yeah yeah

Now, we are concerned about $f^{-1}Tf$, which acts like this

$(x_1, \cdots, x_n) \mapsto (\log(e_1(\exp(x_i))), \cdots, \log(e_n(\exp(x_n))))$

Now they are asking if the inversion of this map is can be done by radicals.

you may run in trouble with finding roots but let's suppose they exist

@GabrielR. good observation, but we are using this as an example here so ok.

the technicalities can be addressed later.

@GabrielR. Actually, this problem essentially reduces to the question of whether a polynomial of degree $\geq 5$ is solvable by radicals and expoenentials/logarithms.

hmmm I see now ! thank you @BalarkaSen

@GabrielR. The answer is negative assuming Schanuel's conjecture.

this

his final question is actually interesting

Gotta eat, be back later

ok, let me see what he asks.

@GabrielR. Well, it seems too general to me.

Hahah, the previous notation for Tate-Shaferevich group was TS.

@GabrielR. Actually, my explanation was wrong.

It actually asks whether a quintic over the field generated by rationals and exponential/logarithm values is solvable through radicals and exponential/logarithms.

So in general, this is like asking whether a polynomial of degree $\geq 5$ in the function field $\Bbb Q(f)$ is solvable over a radical extension of that field. For example, is $x^5 - \log(3) + \exp(2)$ solvable through radicals and exponential/logarithms?

@BalarkaSen Can you prove that $\sqrt{19}^{\sqrt{19}^{\sqrt{19}}}$ is not an integer? :P

@Sawarnik I believe no.

But I also believe that this is an open problem.

Orly? Not an integer?

@Sawarnik I can't guarantee though.

Oh, integer?

@BalarkaSen Yes!

It becomes much more interesting.

@Sawarnik I think one might have to do explicit bounding of $\sqrt{19}$ to derive that.

And it seems utterly non interesting and completely tedious job.

@BalarkaSen Ah! Don't you have a NTist way?

I hate trick theory problems. But let me think on it.

@r9m Hmm, how did you discover that mistake?

@Sawarnik i tried to apply a similar inequality elsewhere and got devastrating results :P

@Sawarnik I give up. Not interested. You tell.

@r9m So you should try to apply the new inequality elsewhere and try to get devastating results :P :P

@BalarkaSen Uh oh! It was just my curiosity. I don't know the answer!

OK, then I'll think about it!

:D

But I have to go now, bye.

@Sawarnik imp :P

@BalarkaSen Ok. Byes.

@r9m Hmm. Let me put up some courage to go through that!

@r9m How does sin(B-C/2) equal cos(B-C/2)?

@Sawarnik nope ..

@r9m You write $\frac{\sin A + \sin B}{2\cos \frac{C}{2}}=\cos(\frac{B-C}2)$? :/ .... :/..... :/

@Sawarnik the numerator $\sin A + \sin B = 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})=2\sin(\frac{\pi - C}{2})\cos(\frac{A-B}{2}) = 2\cos(\frac{C}{2})\cos(\frac{A-B}{2})$

@r9m Hmm, so the culprit lies in the fact that I remember the formulas wrong.