@BalarkaSen I've gotten this far: the order of $S_p$ is $p!$; the order of any particular Sylow p-subgroup of $S_p$ is $p$. Then the index of such a subgroup in $S_p$ is $(p-1)!$. I feel like I'm supposed to use the fact that the number of Sylow p-subgroups in $S_p$ is congruent to 1 mod p, but I don't know how.

You are on the right track. Can you prove that there are (p-2)! Sylow p-subgroups of S_p?

Then you'll be done.

Hello @bolbteppa

@Nick I think chemistry is all about memorization if you do it badly, for example organic chemistry seemed like memorization to me until I read Linus Pauling's General Chemistry and I seen how he 'derives' a ton of the functional groups in a kind of systematic way in the sense that all you had to do was think of a few basic principles and work your way down.

I'm not saying it's not memorzation, it just seems less like memorization and is more pedagogically like math if you do it carefully, which is very hard :(

Hey Balark

@Chris'ssis who is that quote about doing poorly in math in middle school by?

@bolbteppa Fields medalist Maryam Mirzakhani.

And they make even harder by going too fast.

@bolbteppa theguardian.com/science/2014/aug/13/…

Wow, cool. I very much felt the same, having had no interest in math until I was 21, I guess I will be a Fields medalist :|

It's actually interesting that I am doing a project on complex geometry this year :p

@BalarkaSen To do that, I would only have to prove that the normalizer of a Sylow p-subgroup has order p(p-1), right?

Or is that too complicated?

I think it gets complicated that way. But I guess you can try.

I'll try to find a simpler means. I'm pretty tired, so my brain's just being slow.

Hello,

@Fargle Hint : the sylow p-subgroups of S_p are cyclic

Let $f\in C^2(H,\mathbb{R})$ where $H$ is a Hilbert space, $u\in H$ is a critical point of $f$. $0\notin \sigma(f''(u))$ the spectral of $f''(u)$. Let $H_-$ and $H_+$ be the negative and positive spaces of the self-adjoint operator $f''(u):H\rightarrow H$ with spectral decomposition respectively. Then $H_-\bot H_+$ and $H= H_-\oplus H_+.$

what it means this text please

f is a continuously differentiable function from a Hilbert space H to R, whose 2nd order derivative is also continuous. u is a critical point, so that f'(u) = 0. Taking the second derivative at u, f''(u), the spectrum of f''(u) not having a zero eigenvalue is a fancy pants way for formalizing the basic calculus notion that the second derivative is non-zero so that we have curvature at that point.

@bolbteppa what about the spectral decomposition ?

Then you want to decompose the second derivative into areas where it's positive, i.e. curving upwards, and negative, i.e. curving downwards, and want the part of the domain where you're curving upwards, H_+, to be orthogonal to the part where you're curving downwards, H_-, so that the domain decomposes into two distinct parts, hence why you can write the direct sum.

we can allways do this decomposition ?

You're basically just using the signs of the eigenvalues to characterize whether the curvature is positive or negative, and decomposing the space into the set of eigenvectors with positive eigenvalue (relative to the operator) versus the set of negative ones, so geometrically you're just decomposing the domain of the operator into regions where the curvature is negative vs. positive.

Well, there are $(p-1)!$ elements of order $p$ (all possible arrangements of 5 letters, divided to eliminate duplications). $p-1$ elements of every Sylow group must be drawn from this pool (the other element being the group identity, the only other element which is $1$ when taken to the $p$ power), so that the number of Sylow groups is $(p-2)!$. Because $(p-2)! \equiv 1 \textrm{ mod } p$, $(p-1)! \equiv p-1 \textrm{ mod } p = -1 \textrm{ mod } p$.

@Fargle Exactly.

Well done.

So I really only had to prove that part, no mucking about with the index and whatnot. That's really cool, though, and weirdly elegant.

Why are the Sylow p-subgroups of $S_p$ cyclic?

Yeah, it is elegant. One of my favorite proof using finite grumber theory =P

@bolbteppa but why we have that $H_-\bot H_+$ ?

ugggh internet

Oh, duh, it's of prime order. Must be cyclic.

@Fargle well, what's the largest power of p that divides p!?

@bolbteppa ?

Brain fart of massive proportions. One, to answer your question

@Vrouvrou math.stackexchange.com/questions/82467/…

@bolbteppa this still true for an infinit dimentional space

?

@Vrouvrou exercise ;)

i really don't know how to do

halloooooooo

:)

Guys I got a quick question. How do you go about reading a math book?

It sounds trivial but I get the feeling I'm not doing it right

how are you reading it?

I read a page, make some notes

and try to figure out what's happening by writing it

@Sabಠ_ಠ Here

what book are you reading?

Apostol

Spivak

Then the theory matters less. Do the exercises

@skullpatrol thanks. Lemme read it

@BalarkaSen I do the exercises as well, but before moving to exercises, is reading the chapter enough?

not really

yes. enough for the first time.

@JasperLoy also try to watch this one youtube.com/watch?v=HiyYEVcU1tI It's great!

we disagree here, @skull

A math book requires a different type of reading than a novel or a short story. Every sentence in a math book is full of information and logically linked to the surrounding sentences. You should read the sentences carefully and think about their meaning. As you read, remember that math builds upon itself. Be sure to read with pencil and paper: Do calculations, draw sketches, and take notes.

That's exactly what I do. I guess I'm doing it right then. Maybe I'm slow cuz my mathematical maturity isn't high enough :)

Then you are fine :)

Math maturity takes patience and hard work to develop pal

I guess I should be patient. It's tedious

Ahh beat me to it :P

your job as a reader is to find the "logical links" between the sentences

yeah

but first, the information should be committed to memory

yeah

I don't mean memorize, but know where things are if you need them later

^Give that man a star

@Chris'ssis Starred

@bolbteppa we don't need to have that H_-is a sub space of H ?

@Vrouvrou this book says it's true, books.google.ie/… , do you believe them? Proof is the only way ;)

@Vrouvrou what are you studying from?

@BalarkaSen Glad you like it. In mathematics we need to have the gladiator's attitude, to fight the impossible if needed. :-)

@bolbteppa the book speak about finite dimensional

-_-

@Chris'ssis Great, now we have to do duel wearing flimsy skirts and 24feet long swords? Don't you think, er, that it's too much?

@BalarkaSen :-)))

@Sabಠ_ಠ Spivak and Apostol are hard if you're not ready for them, an easier book is Piskunov's Calculus (if you want to know calculus), and the problems in Demidovich problems in mathematical analysis. The only book I wish I'd read before the first times I picked up Spivak and Apostol, to prepare me, is two old books on Algebra by Chrystal.

Piskunov is awesome.

How did you come across it Balark?

@bolbteppa Were dumped in my father's old bookrack for years.

Russian binding. Original.

Wow cool, lucky you.

That's where I learned calculus, actually.

@Sabಠ_ಠ also constant DAILY review is essential

@bolbteppa ?

I think Landau and Lifshitz physics books are in the same spirit, if you ever study physics, you'll like those

Man, I leave for uni in 3 days, I hope my copy of Munkres gets here by then

@bolbteppa I have only read a little physics and the most fun book among the ones I have read is Perelman.

@BalarkaSen Madagascar 3 :-)

@r9m Seen that.

@BalarkaSen would you like to read the physics book Einstein read when he was your age?

@Vrouvrou I will say it's true, but that you should prove it, this might involve finding a good book and studying it. Also, H_- is a subspace of H. I would like to know what you are studying if you don't mind.

@skullpatrol which one?

@r9m mama-mia-santa santa-mia mama santa ferro

by Bernstein

=P

@skullpatrol Oh?

Perelman? That's two MIR books, they are so good and they are all over India as far as I can tell, so it doesn't surprise me :p

@bolbteppa Yeah. I have both of the copies.

you can find it on line for free

Can someone tell me why a biholomorphism from $\mathbb{C}$ to $\mathbb{C}$ must be a linear FLT?

@BalarkaSen I don't remember it that well .. somehow I remembered Kowalski's 'Nuculer' Reactor :P

I also have the four copies of Landua-Kitagorodsk lying somewhere around in the dust, @bolbteppa

@r9m Yeah, haha

@Anthony There is a linear version of fermats last theorem? News to me.

@BalarkaSen Wat?

@BalarkaSen I think he meant fractional linear transformation, as in a Moebius transformation

Unless you knew that, in which case, screw me, right?

I know.

@Fargle Yeah, sorry.

@BalarkaSen Ha....ha...

But i couldn't resist doing the pun

Sweet, I've only read a big chunk of L&L

@bolbteppa What's fun is that there is a mathematics book written by Perelman.

And that is precisely where I learned algebra

@bolbteppa H_- is not a subspace of H

@BalarkaSen I don't think it's that Perelman ;)

I was gonna say. That dude gets around if so.

@bolbteppa I mean Yakov Perelman.

Not Gregori, of course not.

=)

@bolbteppa $H_-=\lbrace h\in H, \langle f''(u)h,h\rangle \leq 0 \rbrace $ right ?

@Vrouvrou how can you write H as a direct sum of H_+ and H_- if they are not subspaces?

Anyone here familiar with a hypergeometric expression for gamma function?

because this i ask this question if $H_-=\lbrace h\in H, \langle f''(u)h,h\rangle \leq 0 \rbrace $ then $H_-$ is not a sub spaces so we can't do the decomposition

@bolbteppa

@BalarkaSen maybe johndcook.com/special_function_diagram.html

Looks good but unfortunately that's the incomplete gamma.

In the limit you get gamma

It says on the page

yes, but it will get me a wrong kind of expression. i want a suitable one for performing the CF expansions. thanks for the link though

@bolbteppa ?

@Anthony Any bijective entire map $\Bbb C \rightarrow \Bbb C$ can be extended to a self-map of the sphere. Convince yourself of that.

Now convince yourself that $f(\infty)=\infty$.

By restricting to the north hemisphere d the sphere instead of the south hemisphere you have a meromorphic function with a pole only at 0.

Convince yourself that this pole is neither essential nor removable.

The only kind of singularity left is a pole. (I should have been saying singularity instead of pole. oops)

So $f$ must have been a polynomial. Because $f$ was bijective it must have been linear.

I wonder if we can do this without pen and paper $$\lim_{n\to\infty} \frac{1}{n^2} \int_0^1(x+2^2 x^2+3^2 x^3+\cdots+n^2 x^n) \log(1+x) \ dx$$

eeeeeeew that integral is icky.

bad integral bad

turns away

What I said was poorly phrased but you should get the gist @Anthony

please what it means the negative space of a self adjoint operator ?

What does 'vrouvou' means woman woman? @Vrouvrou or wifewife

@Vrouvrou I am trying to find you something that will explain it, patience grasshopper

@MikeMiller Yeah, thanks. Why can't it be removable... though?

What would the argument be? Then you could extend it to a bounded function, so it'd have to be constant? Something like that? I can't think of a good way to formalize that.

Oh I guess removable iff bounded.

@Vrouvrou I do not see how to use that definition of H_- to prove it is a subspace, because you get some terms like <T(x),y> + <T(y),x>, which don't make sense to me. However I can only think of defining H_- by following the path you take in the spectral theorem, where you decompose a vector space into a direct sum of subspaces corresponding to distinct eigenvalues, then calling the part of the direct sum with negative eigenvalues H_- etc... Thoughts?

@Yup, you got it. That it's not essential is harder.

@Anthony I mean, not @Yup. Essential is, if you're willing to use it, big picard. If you're not, use Casaroti-Weierstrass, or however it's spelled

guys can somebody tell me what is push forward of a vector from one tangent space to another

The keyword is "differential"

is this regarding push forward @Mike

Yes

what i dont understand is when defining a lie derivative why is it that after a pull back or push forward the second vector field doesnt come back to its value at the pulled/pushed to point and rather have a different value

ofcourse you would say if it is same value then we say lie derivative is zero

here i mean lie derivative of one vector field wrt another

@mike , are u online

@MikeMiller Yeah I had essential, I forgot to do removable, for some reason.

Thanks.

@everyone, somebody explain the lie derivative pls as i hv stated above

i think im having a brain aneurism, can someone help me understand the part where he says 'we conclude that either'.... math.stackexchange.com/a/364800/42808

Does anybody what's original proof of Weierstrass Approximation theorem? Baby rudin has a proof which he claims is the original and this pdf claims this one is original.

@ccorn This answer of yours and the comment about the validity of the product form (an analogue of the Weierstrass product for trig functions, apparently) made me think

about an analogy essentially coming up from this paper linked in the MO-version of this question. Does the Riemann surface $w = j(z)$ worth thinking about? What's the monodromy around it's fundamental zero $z = \zeta_3$? Should be $\mathbf{SL}_2(\Bbb Z)$ or some discrete subgroup of it but I don't see it.

For a rough version of this with the logarithms, see my comment here. [Warning : What I did was just an analogy. If I understand it correctly, the correct dictionary of this analogy is what Qiaochu did]