Now, since $f$ is monic, Gauß says that in fact $m_z \in \mathbb{Z}[X]$.

@DanielFischer is $m_z$ the minimal poly of $z$ over $\mathbb{Z}$ or $\mathbb{Q}$?

Over $\mathbb{Q}$.

okay

Sigh. I just got 4 downvotes. When does the bot sweep for those?

so, what is Gauß? this requires a result?

@Mike 3 a.m.

@DickSquizer Gauß' lemma.

the version of that i'm reading is that irreducible over $\mathbb{Z}$ implies irreducible over $\mathbb{Q}$

so $f$ is irreducible over $\mathbb{Q}$ so $m_z$ divides $f$ so $m_z$ is in $\mathbb{Z}[x]$

and that does it

Okay. Thanks @DanielFischer @PedroTamaroff

now time to move on to the actual question...fml

@Ted Clearly you taught him well.

@DickSquizer The product of primitive polynomials is primitive. If $f = q\cdot m_z$, then there are $a,b\in \mathbb{Z}$ such that $a\cdot q$ and $b\cdot m_z$ are primitive polynomials. Then $(a\cdot q)(b\cdot m_z) = ab\cdot f$ is a primitive polynomial, whence $a\cdot b = \pm 1$.

@Mike The bot might remove 3 but not all 4 downvotes.

okay so the coefficients of $(x-r)^2-q^2n$ are in $\mathbb{Z}$, we need to show that $r$ and $q$ are in $\mathbb{Z}$ to finish.

we have $-2r\in \mathbb{Z}$ and $r^2-q^2n\in\mathbb{Z}$

Is @Ted still here?

i don't actually see why we're done @Daniel @Pedro

No, @Mike, he isn't. (I must have taught him well. I taught him complex manifolds, and he corrected my answer on related material a few days ago.) :P

@Ted Do people have much interest in the topology of $\text{Homeo}(X), \text{Diffeo}(X), \text{Holomeo?}(X)$?

Holomeorphism is a word now BTW.

Lots of work has been done on such questions, @Mike, particularly in the 60's and 70's. I've never heard of the last. Why not biholomorphism?

Yeah, I realized I could name it that after the fact.

@DanielF @Pedro @Mike This one caught me a bit by surprise :)

Maybe it won't be interesting, @Ted... I woul be unsurprised if you can't deform a biholomorphism.

Depends on the symmetry of the manifold, @Mike. There are complex Lie groups going on in some cases. But, yes, for the generic compact complex manifold, there aren't many. Algebraic curves are a good place to start :)

Generalizations of Schwarz lemma for hyperbolic manifolds (noncompact) can be interesting, @Mike.

I guess that brings to mind a similar question: given a complex variety, can one deform a birational equivalence?

Well, there are a fair number of automorphisms of $\Bbb P^n$, @Mike :P

@Ted Sure... but are they close?

(I'm only talking compact here.)

Sure. $PGL(n+1,\Bbb C)$ is a nice Lie group. :P

Ahh... oops.

I even knew these things.

Yeah, you know one or two things.

The automorphisms of the disk, sphere, and plane are all rational (of degree 1!). So that makes me wonder... How nice are the automorphisms of a complex manifold that's also a variety?

IE: will biholomorphisms coincide with birational equivalences?

You need to look at Kobayashi's book on Transformation Groups. One of our grad students has absconded with my copy.

That devil!

@Ted Unfortunately there are more fascinating problems than minutes in the day.

Yeah, I even got mildly confused about one of my own homework problems today.

Oops.

Sometimes it's just easier to think matrices than linear transformations.

What was the questionV

sup guys

I bought "a copy of" AM today

Hi @Mike.

Hi @5space @Fernando

@Fernando But a hard copy you stupid nerd

>implying I have money

@Mike my country has stupid, ridiculous import taxes

@TedShifrin That's very Spivak.

@Fernando >implying you can't get books on the black market

@Mike the copy I bought cost me <4 USD

I like being cheap

@Mike, looks like I'm doing some topology over the summer!

@5space Tell me about it on gchat :)

@Fernando Is it an Ebook?

Okeydokey

@mike no

photocopies

well, actually they're not photocopies

they're just a printed copy of a .pdf

meh

that's at least more reasonable

the thing is now I can scribble on top of it

how cool is that?

and erase

anyone up for a localization question?

@Mike ?

@Fernando I am not up for a localization question, but I am up for talking about unrelated shit

:(

wut up ma hoes

Are you some sort of pimp daddy?

:D who enjoys math @Enjoys Math

can an inner product space be over a field other than the real or complex numbers?

This is my $1000^\text{th}$ day on main :-)

@Alex sure, but it's usability may be limited.

my text has a true/false question on this, but it only states that "Except for section 6.8, we assume that all vector spaces are over the field $F$, where $F$ denotes either $\mathbb{R}$ or $\mathbb{C}$" when it talks about the fields for inner products... which isn't really exacting on yes or no to this true/false question....

@Alex Most I have seen are over $\mathbb{R}$ or $\mathbb{C}$, but the only thing that is specific to those is the conjugacy requirements.

$\langle x,y\rangle=\overline{\langle y,x\rangle}$

which is just symmetry when the field is $\mathbb{R}$

@robjohn Thank you. Yeah that and the other requirement that $\langle x,x \rangle \geq 0$ was trippin me up too, since this implies there is an ordering.

@Alex yes. The definitions seem to be geared for $\mathbb{R}$ or $\mathbb{C}$

Can someone tell me nontautological conditions are equivalent to the statement 'the square matrix A is invertible'

@rubito what do you mean by non-tautological but equivalent?

@rubito non-vanishing determinant?

@rubito independent rows

@rubito independent columns

Hello.

Hi.

Happy 1,000th day @robjohn :-)

@skullpatrol Thanks!

@user143125 hey there... what's up?

Anyone with a fairly good grasp of Walsh permutation permutations care to help a complete novice understand how to go from compression matrices to permutation matrices? (I'm extremely sorry for asking you to Charlie Eppes it)

Sorry, 3-bit walsh permutations

@user143125 sorry, don't know what those are.

No worries.

@user143125 someone might, but it seems there are some inactive people.

Oh I see. Well, it is late here (EDT anyway)

@user143125 I assume you've looked at this page.

Yeah, unfortunately I have no formal higher math background.. and I fully understand it. I understand Fano Plane -> Pemutation matrix. Compression Matrix -> Compression Vector.. but I have no idea how the compression matrices can actually be used to generate permutation matrices.

*I don't fully understand it rather

@user143125 I see that you've asked a question on the main site. If no one shows up here that can help, someone will probably answer your question.

Yeah, I hope so. Or perhaps I'll figure it out.

@robjohn @robjohn I'm not sure but I think it means sorry. The problem asks us to give 20 statements that are equivalent to the statement. Which I find very confusing

.

..

@Mike Can you give me examples of prosolvable group other than $\text{Gal}(\overline{\bf {Q}}_p/\bf{Q}{_p})$

LaTeX sucks.

@skullpatrol Hi

Heey! I'm having trouble with box plots. Anyone mind helping me out? math.stackexchange.com/questions/764654/…

Hi @icegirl how are you?

Greetings!

I just created a very very nice question.

$$\int_0^{\pi/2} \{\log(\tan(x)) \} \ dx$$ where $\{x\}$ - fractional part of $x$

@Chris'ssis $x \to \frac{\pi}{2} - x$ .. 0

@r9m It doesn't work ...

@r9m By the way, that is $0$ without the fractional part.

@Chris'ssis Wolphram says

@r9m W|A is not able to compute such a stuff. After a while you'll see (figure out) what I'm saying here.

@Chris'ssis that link is getting messed up ,, I tried 'integrate 0 to pi/2 frac(ln (tan x)) dx' as my search .. it gave the graph and the answer 0

@r9m This is because of the definition of frac(x) that W|A and Mathematica use.

@r9m mathworld.wolfram.com/FractionalPart.html

@r9m The mind blowing thing is that the answer is $\pi/4$.

@Chris'ssis so you mean $\{x\} = x - [x]$, for all x ?

@r9m yeah

@Chris'ssis it's either the fractional part or the fractional part + 1. It can't be greater than $\frac\pi2$.

@robjohn It's $\pi/4$.

@Chris'ssis that's the fractional part :-)

@Chris'ssis okay... I will work on it.

@robjohn hehe, glad to hear that. It's AWESOME! :-)

Meta looks awful at the moment. Several pages of posts bumped by Community♦. Couldn't they have converted the links without bumping the posts?

When viewing recently active questions, I have to go 10 pages back.

There are (as expected) some discussions about this on meta.stackexchange.

@MartinSleziak that is really annoying.

@MartinSleziak bots be crazy...

Hey, @Sawarnik

@BalarkaSen Hi.

@robjohn Back. Well-done! That's the way! :-)

@robjohn That's very cool.

Mmm

@Chris'ssis I am very familiar with the evenness of $\frac{e^u}{1+e^{2u}}=\frac12\mathrm{sech}(u)$ and that is pretty much the key.

@robjohn it seems that was easy to you. How would you rate my question as difficulty? Could it be considered for a Putnam contest?

Why is $$ \int_n^\infty e^{-t} t^{\sigma-1}\mathrm{d}t \leq C \int_n^\infty e^{-t/2} \mathrm{d}t $$, where C is some constant

$\sigma,n >0$

@N3buchadnezzar for all $\sigma,n\gt0$?

@Chris'ssis I hate Putnam problems.

@robjohn Yeah, my book says so. It is part of a proof that $\Gamma(s)$ is analytic for $\sigma>0$

@BalarkaSen :-)

@Chris'ssis They are more of a time-limit problems than the ones one have to sit over for days.

I like the latter ones, thank you.

@BalarkaSen This one you see above I created it this morning. It's so sweet!

(This is a personal view of things, though)

@Chris'ssis Why are you interested in fractional part-type integrals lately? =)

@BalarkaSen I have a lot of such questions! I mean I created a lot in the past too, and I love them.

But I do not see how this can help me - even though it seems related.

@N3buchadnezzar is $n$ an integer? that is do we have that $n\ge1$?

@Chris'ssis I hope you have collected all of them. Could make a very nice book.

@robjohn I guess so

@N3buchadnezzar $\Gamma$ can be easily proved to be analytic, I think. Prove that integral is uniformly convergent, then prove that integrands are analytic and voila.

@Sawarnik I'd like to write a book, but there is a problem: since I have no background in mathematics ... that would be an issue ...

This is just some thoughts.

@N3buchadnezzar if not, I think the inequality is false.

@Chris'ssis Why don't you write papers?

@BalarkaSen Yeah that is the path my book goes. But I do not understand the above inequality

@robjohn Want me to post the proof ?

papers are much well-read in mathematical community than books.

@N3buchadnezzar Uh, why not use De la Vallée Poussin's test for uniform convergence?

@BalarkaSen Well, I'd like to have something of big dimension that requires a book, like at least 1000 awesome calculus questions with multiple solutions.

@Chris'ssis That very less people would read.

@BalarkaSen Shooting birds with cannons

Try to publish papers on particular class of integrals.

@BalarkaSen If I had such a book, I'd read it day and night.

@N3buchadnezzar oh, I think I can prove it. I thought you were looking for a proof. I didn't know you had one.

@robjohn Yeah. But I do not get the above inequality. If you have a proof / an intuition please share it

@N3buchadnezzar wait... if we let $\sigma\to\infty$, the left hand side goes to $\infty$, does it not?

@N3buchadnezzar I like doing that. You know how I prove $\zeta$ is analytic? First, prove $\zeta$ is uniformly convergent for $\Re[s] \geq 1$. That is easy. Now dis : $$\oint_C \zeta(s) ds = \sum_{n \geq 1} \oint_{C} n^{-s} ds$$ and changing order is valid by uniform convergence, regardless of the Jordan contour $C$. Now $n^{-s}$ are all analytic, so by Cauchy $$\oint_C \zeta(s) ds = 0$$. By Morera, you have analyticity.

=p

@robjohn See here for the snippit i.sstatic.net/uGoYp.png

@robjohn sigma is a fixed constant < \infinity

@N3buchadnezzar Then $C$ depends on $\sigma$. Then it is simple.

@Chris'ssis Yes, it's just you. I know loads of people (students & mathematicians) who hate Gradshteyn & Ryzhik.

$$e^{-t/2}t^{\sigma-1}\le C$$

for all $t\ge1$

@Sawarnik

@BalarkaSen

Would you like to see the solution to the transformation problem I gave in the general room?

Yes.

Whenever I click the maximize button, the taskbar disappears. What is happening? :( @robjohn

So, the problem is to transform $x^3 + ax + b = 0$ into another cubic with no degree-1 term but a degree-2 term.

Right, this is possible. Note that we have $x_1 + x_2 + x_3 = 0$

Ok.

As well as $x_1x_2 + x_2x_3 + x_1x_3 = a$ and $x_1x_2x_3 = -b$

Fine, let's sub $x_1 = -x_2 - x_3$ in these.

@Sawarnik I don't see that, but I am not using Windows. It works okay in MacOS

@BalarkaSen Unfortunately, I didn't meet such books ... :-(. Moreover, it would be nice to have many such books with thousands of pages of integrals, series and limits, all nicely done, explained. Doesn't this world need such stuff?

$-(x_2+x_3)x_2 + x_2x_3 - (x_2+x_3)x_3 = a$

$x_2x_3 - (x_2+x_3)^2 = a$

And for the $-b$, we have $-(x_2+x_3)x_2x_3 = -b$

Ok.

Say $x_2x_3 = \alpha$ and $(x_2+x_3) = -\beta$

So we have $\alpha-\beta^2 = a$ and $\alpha \beta = -b$

Is this right?

Ok.

Do you recall these two?

Remember I mentioned about these resolvents before?

Yes!

OK, now let's do this.

$\alpha \beta = -b$ so $\alpha^2 \beta^2 = b^2$

$\beta^2 = b^2/\alpha^2$.

So $a = \alpha - \beta^2 = \alpha - b^2/\alpha^2$

Rearranging, $\alpha^3 - a \alpha^2 - b^2 = 0$

$\text{QED}$

Looks good. Just coming in 5 mins ...

Nevermind.

@robjohn I've just come up with a second solution to that question.

@Chris'ssis the fractional part question?

@robjohn sure!

@Sawarnik There is an easier (rather stupid) solution to my question, actually. Can you guess?

@BalarkaSen Back.

@BalarkaSen Maybe using the formulas here and getting an equation? en.wikipedia.org/wiki/… :/

@Sawarnik A depressed form is one with no $n-1$ term. But I ask for no $1$ term yet the $2$ term.

OK, here : $x^3 + ax + b = 0$. Thus $1 + ay^2 + by^3 = 0$ with $y = 1/x$

Hence $y^3 + a/b y^2 + 1/b = 0$

@BalarkaSen Yes, so can't we get equations?

My formula is essentially a variation of it, you see. $\alpha = x_2 x_3 = -b/x_1$

@Sawarnik What equations?

@BalarkaSen Ok, leave it.

@BalarkaSen I studied the definition of limits recently and they were so beautiful, how can you say its not!

@Sawarnik I just don't feel the need of them (up until now) in any of my works.

@BalarkaSen That is because you are implicitly using them, if not directly.

Quite true.

@Sawarnik Have you heard of Euclid's postulate?

@BalarkaSen Yes.

@Sawarnik Then you must also know how many people passed their whole life to prove the 5th through 1-4th?

@BalarkaSen Yes, I have read the history, but none of the actual works :)

@Sawarnik And must also know that they were not successful?

@BalarkaSen Yes.

@Sawarnik I learned the proof recently that 5th is not provable through all the 1-4th of Euclid's postulates.

I am curious whether you know about it, @Sawarnik?

@BalarkaSen I haven't read the proof :(

@Sawarnik I happened to think that we don't have any proof yet, up until recently Mahan Mj taught me the proof.

It amounts to construct a geometric model s.t. 5th postulate (or in another form, Playfair's axiom) does not hold.

Namely, hyperbolic geometry.

It was very beautiful.

Oh, wayy beyond me, as I thought. But nice :)

It's nothing advanced, don't let the name get you. It's simple calculus is what is needed.

Let's move to general discussions room.

Ok :)

I think MSE is one of the best things that has ever happened. It has allowed me to meet such great persons that I would have never met otherwise.

@Sawarnik no I think you are wrong...

@Sawarnik I will be back after an hour or 2

orange

Is meer2kat usually the only girl in chat? :)

@meer2kat Blue.

@Alex Oh dear! I was mentioned! :D

@WillHunting purple

@meer2kat Did you listen to the song?

@WillHunting crap. i'm terrible with follow through these days

Maybe you would be mentioned more if your pic was bigger. Or you could talk about maths...there's always that.

@Alex I'm mentioned often actually....

@Alex Nah, I'm at work. No time for math. I answer people's questions occassionally or hint them in the right direction.

@meer2kat oh okay, my mistake. So you aren't a student?

@meer2kat You should also watch Good Will Hunting. Then you will understand me, lol.

@Alex Oh I am. But I've already taken five calculus classes and a lot of the guys on here are young so I'm able to help out some I guees

@Alex I'm currently an engineering major and am switching to mathematics. I'm technically (almost) a senior in college. Off by like 5 credits.

@meer2kat Oh okay kwl. What area of maths are you interested in pursuing after your degree?

I don't believe this. One of the cats ate my lunch! The leftovers of the Chinese I cooked yesterday. No beef left. Next time I will make it at least twice as spicy.

Did they just eat the beef? Matt N.

Now I haz the hunger. But although there are also leftover beans the image of them being test-licked does not inspire temptation to eat them.

@Alex Yes. In the process scattered some peppers all over the floor.

@Alex I'll be getting my master's in middle school education

@MattN. quote of the day

I guess my lunch will be a sandwich.

@meer2kat Haha : )

Determining the culprit will boil down to seeing which of them throws up Chinese. But since this must have happened hours ago and no one has thrown up yet I might never know.

I think it's best you just let it go. Matt N.

I agree.

To be honest I feel like eating Chinese now.

Me too.

@Alex Food or people?

lol

I have very boring maths to do now.

I'll be making my sandwich. Catch you later!

c yer

meer2kat do you want to do a house swop?

@Alex huh?

agh who else here is mechanically minded?

@DanielFischer @robjohn Let $x_1<-1$ and $x_{n+1}=\dfrac{x_n}{1+x_n} \forall n\ge 1$. Does $\{x_n\}$ converge? If yes, to which value? I was getting it to be bounded but not converging...please help on this one.

@Sawarnik I am back here.

@meer2kat you stay in my house and I stay in yours.

@Hawk it converges to $0$

@robjohn Yes, that is what the answer is, but why?

@Hawk I think I wrote an answer to that... let me look.

@Alex lol you wouldn't want to. i'm currently on co-op as an engineer at a powder metallurgy company.

@robjohn ok...

@robjohn sounds like my bank account

@meer2kat I stay in...wait for it...Africa...so what do you say now?

@Alex you live there?

@meer2kat sure do...in the heart of it...maybe not the heart, maybe the feet...

@Hawk consider the two cases $x_1<x_2$ or $x_2< x_1$ .. I think you will see a common pattern .. then multiply all $x_{n+1}=\dfrac{x_n}{1+x_n}$ from $n=1$ to $n=N$

@r9m I do not see any pattern :(

@Hawk monotone ??

@Hawk okay, I can't find it. However, note that $x_n\gt0$ for $n\gt1$ and $x_n$ is a decreasing sequence. Therefore, $x_n$ has a limit and it is $\ge0$.

$$\lim_{n\to\infty}x_n=\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}\frac{x_n}{x_n+‌​1}=\frac{\lim\limits_{n\to\infty}x_n}{\lim\limits_{n\to\infty}x_n+1}$$

@Alex Which country?

@robjohn Yes, I got that now. Thanks! But, I have a few more.

@meer2kat South Africa

@r9m Yes, I could find that...and I got the solution too.

@Alex Cool deal :)

@r9m Thanks!

@meer2kat where are you from?

@Alex southern United States. Not nearly as exciting

@Alex I was asking cuz I have friends from Zambia :)

@robjohn The sum of the first $n$ terms of an AP whose first term is an integer(not necessarily positive) and the common difference is $2$, is known to be $153$. If $n>1$, then the number of possible values of $n$ is?

@meer2kat I'm sure I don't know them :) Where in southern United States? I was in america for a while when I was younger.

@Alex LOL probably not. Whiles away haha. KY/VA

@r9m The sum of the first n terms of an AP whose first term is an integer(not necessarily positive) and the common difference is 2, is known to be 153. If n>1, then the number of possible values of n is?

@meer2kat you stay in Kentucky?

@Hawk do you know the formula for the sum of an AP?

@robjohn yes, I do, and I used that in my approach to the problem, and apparently, I find the number of possible values of $n$ to be $2$ but the answer given is $5$

@Hawk what's the final expression with the first term and n ?

@meer2kat Oh okay kwl. I drove through Kentucky when I was there. We stayed in Indiana. Have to go do some studying now. C yer.

@Alex bye

@r9m $153=\dfrac{n}{2}[2a+2n-2]$...so we have a quadratic in $n$

so we get the sum to be $n^2+(b+1)n=153$

@Hawk factor 153 :)

@robjohn yes, we do...

@robjohn now, which way?

@Hawk n must be a factor of 153

@robjohn yes, surely...

@robjohn 153 is a multiple of 9.

@Hawk so try the factors in computing $b+1$: $b+1=\frac{153-n^2}{n}$

@robjohn that seems scary...

@r9m I do not understand yet

@robjohn I do not understand yet...and this step says me that I am lost!

@Hawk Plug in any factor of $153$ for $n$ and get $b+1$ ($b+2$ is the starting point of the AP)

Hahahah.... i.sstatic.net/ELYWi.png

@Hawk Hiiee.

@Hawk There are 6 factors of 153.

@robjohn But, we cannot include 1, so we are done...right?

@Hawk does it say that an AP of length 1 is not allowed?

@robjohn There is given condition, $n>1$

@Hawk then 5 it is.

@robjohn Yes, I understand that now.

My brain is so fuzzed up it seems...

@robjohn Ok, I have another one for you.