Mathematics - 2014-11-25 (page 1 of 5)

12:23 AM

@pedro Is it only possible to prove continuity of a function f on X, using the argument that whenever x_n -> x in X then f(x_n) -> f(x) then f is continuous, is this only valid if X is first-countable?

Twink

How can I constuct a series of meromorphic functions on $D_1(0)$ that converges locally uniformly to a meromorphic function with simple poles with residue $1$ at the points $1-1/k$, $k \in \mathbb N$?

anon

@JohnJack first-countable is sufficient but not necessary; en.wikipedia.org/wiki/Sequential_space

Twink

@PedroTamaroff Yes, in fact I was going to ask about how to prove the Cauchy criterion for products using the Weierstrass factorization theorem

or one of its corollaries

Pedro Tamaroff

@Twink Cauchy criterion for products?

You mean Cauchy sequences stuff?

Twink

yes

Pedro Tamaroff

12:36 AM

Eh, the Cauchy criterion is just saying "$\bf C$ is complete, so a sequence converges iff it is Cauchy".

It has little to do with Weiertrass products.

Twink

If $(z_n) \subset \mathbb C \backslash \{0\}$, then $\prod_{n=1}^\infty z_n$ converges iff for every $\epsilon>0$ there is $N \in \mathbb N$ such that for all $n, m \ge N$, $|1- \prod_{k=n}^m z_k | \le \epsilon$.

John Jack

@anon Don't you have that reversed? If f is continuous then it doesn't matter if it is first countable so x_n -> x oimplies f(x_n) -> f(x)...the reverse implication requires first-countability right?

anon

@JohnJack your previous comment was asking when sequential continuity implies continuity.

Twink

I found a proof of it but I think it might be shorter using that theorem since that's the topic we saw

John Jack

@anon okay I see I thought you were referring to the iff statement in the link. Do you have any experience working with Sobolev spaces?

anon

12:42 AM

nope

John Jack

@anon Damn...Can you use weak sequential continuity to prove continuity?

Mike Miller

what's weak sequential continuity

if it's weaker than sequential continuity, and you can't use sequential continuity, I'm going to bet no

John Jack

Can the weak topology be first countable?

Mike Miller

no clue

John Jack

Keeeel

Ted Shifrin

12:57 AM

@Pedro: Have you transmogrified to a canine?

Pedro Tamaroff

@TedShifrin I just used it to express my sadness.

Mike Miller

1:08 AM

What reason've you got to be sad, @Pedro?

Pedro Tamaroff

2 hours ago, by Studentmath

They all fleed from you Mr @Pedro

Twink

Question math.stackexchange.com/questions/1037454/…

@PedroTamaroff did you see the Cauchy criterion?

it's theorem 2.1 here math.ucdenver.edu/~spayne/prelim/inf.pdf

but I find this proof too long, and they don't use any important theorem

:(

hi @RandomVariable

Random Variable

Hello @Twink

Twink

did they delete the upvotes?

Random Variable

Yes

Mike Miller

1:39 AM

@Rafflesiaarnoldii Did you have a name, before Post No Bulls?

anon

1:54 AM

@MikeMiller thisismuchhealthier was I believe the first nym

Mike Miller

that was on his current account; post no bulls was a previous one, I'm quite sure

Ted Shifrin

The nym police?

Rafflesia arnoldii

@MikeMiller I could tell you but... you know... ☠

Mike Miller

Fair enough.

Mikhail

2:28 AM

so I got a PDE like this Ax=0, then I factor it into something like (B+) (B-) x=0, and I know how to solve one of those parts. If I know how to solve B+, did I solve the pde?

anon

3:10 AM

solve (B+)u=0 for u, then solve (B-)x=u for x

Jorge Fernández

3:31 AM

I need to solve this without sylow

let $N$ be a normal subgroup of $G$ so that $|N|$ and $|N/G|$ are relatively prime. prove $N$ is the only sub of $G$ of that size.

the question says to assume there is another sub of size $K$ and see what happends to $K$ under the natural projection from $G$ to $G/N$.

@anon

oops

the question says to assume there is another sub $K$ of size $|N|$ and ....

any ideas?

Ted Shifrin

3:55 AM

@Jorge: Let $\phi$ be the projection. What can you say about the order of $\phi(K)$?

Twink

4:23 AM

How can I find a meromorphic function that has simple poles at the positive integers with residue $-k^2$ at $k$, $k \in \mathbb N$?

Jorge Fernández

It is the order of $K$ divided by $K\cap N$? @TedShifrin

Ted Shifrin

4:42 AM

But it has to divide $|G/N$, @Jorge!

@Twink: This is Mittag-Leffler. Write it down as a sum and then modify to makevsure you get convergence on compact sets.

Jorge Fernández

5:00 AM

@TedShifrin oh, I finished. thanks Ted

Twink

@TedShifrin I have some examples and I try to solve this the same way but I can't :(

usukidoll

6:37 AM

grrrrrrrrrrrrrrrrrrr I didn't downvote Genomeme's answer AND I AM BEING FALSELY ACCUSED OF DOING IT

Vincenzo Oliva

@usukidoll Kill 'em all

Nick

@usukidoll You were just in the wrong place at the wrong time, kid. You're not guilty till proven so in a court of law.

So, take a deep breath. Relax while you're at it.

Breath out knowing that there's no reason left to growl

Mike Miller

6:54 AM

He didn't accuse you of downvoting his answer. That comment didn't have an @usukidoll in it. He was saying that to whoever the downvoters were.

Nick

@VincenzoOliva "An eye for an eye will make the whole world blind" - Ben Kingsley .... no wait, that was Gandhi.

Twink

@Nick why are you so upset?

Nick

Upvoting and Downvoting, in the eyes of our new and growing community, are seen as equivalent to the Liking and Disliking.
I dislike bitter gourds but I know that they're right and they're good for me. I like dark chocolate and pizza cheese but I know that it's bad for me.
I feel downvotes are a necessary measure but I don't think they should be abused to such an extent that the OP of a Q/A feels offended because of a lack of feedback as to why their post was disliked.

Mike Miller

I don't understand how that relates to what I said, other than that we both said downvote.

Twink

what is wrong with you?

Nick

7:04 AM

@MikeMiller It's not related to what you said really but to why you had to say it.

@Twink Nick turns his head up and down, looking around. Are you talking to me?

Twink

4 mins ago, by Twink

@Nick why are you so upset?

Nick

Yes, I thought you mis-tagged.

Twink

no, you are the one

Nick

@Twink I'm not upset. Why do you think I'm upset?

Twink

no, you said I belong to you,so you know me

Nick

7:10 AM

I know you well enough to know that you can't know how I feel when I don't know how to end this sentance knowing not as to what I should be knowing inorder to know what I need to know.

See, that's one thing you didn't know about me. My native tongue is gibberish.

Vincenzo Oliva

@Nick I love Gandhi too, asd

Twink

are you sure?

Nick

@Twink Am I sure if I know gibberish? I'm pretty darn supercalifragilisticexpialidocious about it.

Twink

yes because I need to give the information to the resort

Nick

....

Am I going to the island?

Twink

7:15 AM

it's like you're very

very upset

don't be like that, be the nice guy

@Nick and what is your job?

Nick

What you are experiencing is what I like to call "Emotions-Contrary-to-Actual-Emotions-Received-Through-Text".

Twink

@Nick and what is your job?

Nick

@Twink I'm batman

Twink

it is somebody with the name of john kimbo

@Nick you're the one

you're the policeman?

Nick

@Twink John Kimble ... portrayed by Arnold Schwarzenegger in the movie Kindergarten Cop?

Twink

7:22 AM

you're the policeman?

what kind of cop are you?

Nick

... Good god man, I'm nothing. I'm an infant compared to @Ted and a middle ager compared to @Balarka ... So, I'm pretty much unemployed as I have always been for the past 17 years of my life.

Twink

I'm going to call the police

I need a bacation

I wanna make the reservation

Nick

@Twink Oh, this charade was in reference to that "court of law" line I told usukidoll earlier? I always pictured myself as Bruce Willis (when he was in the film Nancy Drew playing Bruce Willis playing a detective delivering the Miranda rights in a time when it didn't exist) when I say things like that. I can't pull of Schwarzenegger.

Twink

just listen to this youtube.com/watch?v=W8m-8hcEz1A

and this youtube.com/watch?v=XZ3ydHuk6-o

Nick

:18766223 I'm sorry but you're most likely referencing things from our previous conversations like for example the fact that "I own you" ... To be completely honest, I don't remember saying these things. Infact, you can't even expect me to remember.

@Twink My internet connection is not so good as to give me the luxury of watching 20 minutes of youtube video.

I have to wait 20 minutes just for 1 minute of something to load.

Twink

7:38 AM

do you know where i am?

Nick

@Twink Unless you're in the Swiss Bank stealing bricks of gold or in the pentagon hacking security codes, I really couldn't care. But please do go on and tell me.

Twink

I am at work

Kaj Hansen

Flag/close/downvote: math.stackexchange.com/questions/1037832/…

Twink

Swapnil Tripathi

Hello everyone!

Please open the link: ''https://math.stackexchange.com/questions/1037071/show-that-gx-sqrtx-is-uniform‌ly-continuous-on-0-infty/1037084?noredirect=1#comment2114955_1037084''

Kaj Hansen

7:51 AM

Hey there @SwapnilTripathi

Swapnil Tripathi

Hey! @KajHansen Thank you so much for your help that day! :)

I don't know what the person who has commented on my answer talking about!!

Kaj Hansen

I'll take a look

Your first part is definitely true

Continuous on a compact set $\Rightarrow$ uniformly continuous on that set.

Can't say I'm all that great at analysis though :/

Swapnil Tripathi

Haha. :)

Ok. Yes. What would be the basis for cyclotomic extension? @KajHansen

I am quite confused on that one since a past few days.

Kaj Hansen

Hmmm. Let's see. It's been a while.

If you join an $n$th root of unity to $\mathbb{Q}$, the automorphism group will be isomorphic to the multiplicative group of integers $\pmod{n}$.

Oh wait, you're asking for a basis.

Swapnil Tripathi

Yes! I know that there would be $\phi(n)$ elements where $\phi$ denotes the Euler phi function.

Kaj Hansen

8:00 AM

$[K:\mathbb{Q}] = \phi(n)$

For simple extensions $K = F[a]$, you can always get a basis by simply taking powers of the adjoining element $a$.

Swapnil Tripathi

So, if 'a' is a primitive root. Then, $\{1,a,a^2,....,a^{phi(n)-1}\}$?

Kaj Hansen

Indeed

Swapnil Tripathi

Yes! Just like the proof of Kronecker's! My bad. :/

Friend request sent! @KajHansen :P

Kaj Hansen

Oh you're on facebook I see. Awesome!

You'll get some real-time updates into all the boring, tedious details of my life now :)

Swapnil Tripathi

@KajHansen: Haha. Mostly maths based!

Kaj Hansen

8:16 AM

:D

Nick

8:50 AM

Apparently all mediums of contact just boil down to facebook. Every other network seems to be a way for people to find friend requests. The number of friends one has is now equivalent to Mario coins.

►

Kaj Hansen

You mean other SE sites have some sort of "friend" functionality?

And I somewhat disagree, at least personally. I have relatively few friends on Facebook. I do try to add fellow mathematicians though.

Nick

@KajHansen No. I don't think so.

Kaj Hansen

Ah

Nick

@KajHansen Exactly my point. I like Google's concept of circles rather than "friend"

Kaj Hansen

I guess it doesn't really matter what you call them. I think "acquaintance" myself. I consider very few people actual friends.

2

Nick

8:58 AM

The terminology "friend" just boils everyone in a person's life to friends, strangers and foes.

Swapnil Tripathi

@Nick even me. That's why I have shared my google link in my profile. Still I don't like opening it very often.

Studentmath

Blah. I just do't get what I am doing in here

I.e. in the question

Not in the chat

Nick

@Studentmath don't donig nuthing, foo.

Swapnil Tripathi

@Nick sending a friend request just means that you'd like to hear from that person often. It has nothing to do with actual "friendship"

Kaj Hansen

The problem is that English doesn't have a 1-syllable word for "acquaintance". And so we default to "friend".

Nick

9:08 AM

@SwapnilTripathi True, but the popularity of social networks over mediums of communication is a pure monopoly in my opinion. I think social profiles are the reason for it. They provide a user the means to satisfy his/her narcissistic needs.

Kaj Hansen

Do you include text messaging in your mediums of communication? I suspect it is more pervasive than social networking.

Swapnil Tripathi

@Nick Agreed, that might be the case with most people. But for some, it isn't.

You don't generalize an example, you know! ;)

Nick

@KajHansen It's a nuance when used by a specific age group but I find WhatsApp (which has replaced SMS in the first world) to be more centered on contact. We can't deny the benefits of it.

Swapnil Tripathi

Thinks: If it had latex support too.....

Nick

@SwapnilTripathi MathJax extension to WhatsApp. We should totally patent that idea.

2

Sawarnik

9:22 AM

@Nick Hey!

Nick

School Children all over the world would cry out "Hooray!"

Sawarnik

@Swap @Kaj Hi.

Swapnil Tripathi

Hahaha, totally.

@Sawarnik hi!

Nick

@Sawarnik Hello Operator. Get me number nine.

Mike Miller

I'd never even heard of WhatsApp, so it's at least not a universal replacement...

Sawarnik

9:25 AM

Error... buzz.

@Nick I don't think so :/

Nick

@SwapnilTripathi I'm just saying narcissism is fairly common trait in a socializing community. And it's almost a necessity for introverts if they don't want to be depressed.

Swapnil Tripathi

I agreed to it! :)

Nick

@MikeMiller Do you still use SMS?

Mike Miller

Yes, as does everyone I know.

Nick

@MikeMiller You use SMS on a smartphone though, right?

Sawarnik

9:27 AM

@MikeMiller WhatsApp may be a cheaper and faster approach.

Mike Miller

Yes, @Nick... as does everyone I know :)

Sawarnik

Nah.

Nick

@MikeMiller Ah well, then I have my next patent idea.

MathJax for text mesage reader.

Sawarnik

lol

ok..bye.

Nick

@Sawarnik toodles. take are buddy :D

Swapnil Tripathi

9:29 AM

@Sawarnik see you!

Nick

@Sawarnik *care ... dang, I hate typos.

@SwapnilTripathi You're right. I overgeneralize too much :)

Jasper Loy

11:05 AM

@KajHansen The word friend is overused. A facebook friend is not really a friend.

Chris's sis

Greetings

@JasperLoy One of the worst punches I've ever received from life was in the moment when I learned the reality about that word, friend.

Jasper Loy

@Chris'ssis A girl I called friend actually ruined my life, lol. She is actually enemy.

Chris's sis

Facebook friend? What is that? At most a joke. :-)

@JasperLoy :-)))

Jasper Loy

I have no facebook, twitter or blog. Those are dangerous things when you think people could be spying on you.

Chris's sis

@JasperLoy On the other hand, there is another way of saying: "Hello, that's me!". I mean all depends on what I plan to share I think. It's very good not to talk about yourself, but you might like to post things about a hobby of your or something like that. You can find other people with passions like yours (in case you wanna have talks with such people).

skullpatrol

11:21 AM

Hello friends :D

Chris's sis

Hi

@JasperLoy almost anything in this world can be used in a good or bad way. It's up to us to decide how to use the things around us.

skullpatrol

Agreed, there are some things in this world that you have to ignore or you will drive yourself crazy.

Chris's sis

A mathematician told me once, some years ago, when I was very young and I was trying to understand everything, things just happen, try to get used to them.

Those words saved a lot from my time since that moment. Words change lives, that's true.

@JasperLoy I was spied too (even here). Things happen ... and now I'm downvoted daily by someone obssesed with me. :-)

robjohn

@Chris'ssis here?

Chris's sis

@robjohn Yeah. Actually I'm spied once in a while here, but not by the users here.

robjohn

11:33 AM

@Chris'ssis do you mean people looking at what you post in chat or what you write on main?

Chris's sis

@robjohn Both. I refer to someone that knows me in the real life.

robjohn

@Chris'ssis I figured that when you said that it was not users.

Chris's sis

Maybe I should not use that verb anymore. Someone might simply like to see what I post, say.

robjohn

@Chris'ssis I don't think many people I know IRL would care what I do here.

Chris's sis

@robjohn Neither do I. But I'm sure that 2 of them would do their best to know some more.

@robjohn were you ever subject to cyberbullying? I was, and it's not that nice. Also on internet all kind of unpleasant stuff can happen.

robjohn

11:38 AM

@Chris'ssis No. I've never had much presence on the net except on sci.math, and I don't think I was bullied there.

Chris's sis

I mean you're very nice, do your best to help others, and on one day someone decides to write things about you. That's bad because even if you report those blogs no one does anything to delete that content.

Things should definitely change in this area, but this is just a personal opinion.

Of course, there is no logical reason to believe those downvotes come from a rational thought, but from some hate or so.

(I deleted the message above not to give ideas ...)

Even for this answer I received a downvote (lol)

2

A: Value of $\psi\left(\frac{1}{2}\right)$

Note the simple fact that $$\lim_{N\to \infty} \sum_{k=1+\lfloor{\frac{N-1}{2}}\rfloor}^{N} \frac{2}{2k+1} =\lim_{n\to\infty} \int_{n}^{2n} \frac{2}{2x+1} \ dx= \log(2)$$ Q.E.D.

It's not because this is my answer, but that is a very natural way of doing things. No need for pen and paper.

I also received a downvote here

1

A: Limit $\lim_{n \to \infty }\sqrt[n]{ \frac{\left | \sin1 \right |}{1}\cdot\cdot\cdot\frac{\left | \sin n \right |}{n}} $

$$\lim_{n\to\infty}\sqrt[n]{ \frac{\left | \sin1 \right |}{1}\cdot\frac{\left | \sin2 \right |}{2}\cdot\cdot\cdot\frac{\left | \sin n \right |}{n}} \leq \lim_{n\to\infty}{\dfrac1{\sqrt[n]{n!}}}=\lim_{n\to\infty}{\dfrac n{\sqrt[n]{n!}}}\cdot\frac{1}{n}=\lim_{n\to\infty} \frac{e}{n}=0.$$ Q.E.D. (...

Again, that is the natural way to go, all done at high school level, very easy to understand.

Anyway!

robjohn

@Chris'ssis There is little legal precedent or power that anyone has over postings, especially since they can come from other countries and there is little power one country has to bear on another over things like this.

Chris's sis

11:54 AM

@robjohn Yeah, I know, I'm aware of it.

@robjohn btw, have you seen this one? $$\displaystyle \int_{0}^1\frac{\log(x)}{x^2-x-1}\text{d}x$$

@DanielFischer do you see a straightforward way to compute it by methods of real analysis, without using series?

Actually, the question is addressed to all that like this stuff.

Problems and awesome solutions to such questions will turn my book into a very nice book.

bbl - I need to finish some proofs and then bring to life new questions

Chris's sis

12:18 PM

@r9m news here ... :-)

I computed $$\displaystyle \int_{0}^1\frac{\log(1-x)\log(x)\log(1+x)}{x^2-x-1}\text{d}x$$

@r9m ^^^

bbl

robjohn

12:48 PM

@Chris'ssis Can we use properties of $\mathrm{Li}_2$?

Chris's sis

@robjohn Yeah :-)

robjohn

@Chris'ssis Okay... let me write it up :-) or do you already know it?

UserX

What group is U(1)?

Chris's sis

@robjohn Nice .... :-))))

@robjohn Of course I know it. Better don't write it up now. I think also r9m works on it. :-)

@robjohn that also means that you can do the second integral, right? Because one of the important keys to the first one is also a key to the second one.

Or this version $$\displaystyle \int_{0}^1\frac{\log(1-x)\log(1+x)}{x^2-x-1}\text{d}x$$

robjohn

@Chris'ssis All I did in the first one was partial fractions then use a formula I proved for $\int_0^1\frac{\log(x)}{x-a}\mathrm{d}x$

Chris's sis

1:00 PM

@robjohn Well, yeah, that works nice.

bbl - some work to do

robjohn

@Chris'ssis Are we allowed to use series here?

Chris's sis

@robjohn Yeah.

Mary Star

1:20 PM

@TedShifrin Hello!!!

To show that the field extension $K \leq F$ with $[F:K]=2$ is normal, do I have to find a specific algebraic element $a$ such that $Irr(a,K)$ is of degree $2$??

How can I show then that the splitting field is $F$ without knowing $F$ ??

Ted Shifrin

1:31 PM

That's what I was saying is soecial about 2, @MaryStar

Hi @robjohn @Chris'ssis

robjohn

@TedShifrin hey

Chris's sis

@TedShifrin Hello Ted :-)

Nick

1:52 PM

Anyone know how to show that $L^{t}MLv \cdot v = MLv \cdot Lv$?

Or is it true? I'm rusty on my matrix rules.

Mike Miller

morning

Integrator

@robjohn Is there any badge for accepted answer with negative score?

robjohn

@Integrator There is tenacious and Unsung Hero, but I don't see any for negative score.

Ted Shifrin

what in the world are you doing awake, @Mike?

Integrator

@robjohn :(

Ted Shifrin

2:06 PM

I don't even know what that means, @Nick. You're multiplying two vectors?

robjohn

@Nick what does the dot mean? I think I know what you meant...

@TedShifrin I think he wrote \dot instead of \cdot

Mike Miller

@TedShifrin I teach at 8...

Ted Shifrin

ohhhh .... That makes more sense ... I thought this was physics with a time parameter.

Yes, @Nick, I call that Ted's favorite formula :) $Ax\cdot y = x\cdot A^\top y$ !

I guess you need 2 hours to be awake, @Mike :D

Mike Miller

The first hour is for waking up, but I usually get to my office at around ~7:15. Gives me time prepare more, or do the crossword.

robjohn

@Nick write in terms of matrices: $L^TMLv\cdot v=v^TL^TMLv=MLv\cdot Lv$ since $u\cdot v=v^Tu$

Ted Shifrin

2:11 PM

I've gotten in trouble this semester ... Usually I prepare several weeks ahead so that I can make up homework sets competently. But then I've not reviewed my notes adequately, thinking that — as I can in most classes — I can wing it ... :)

Mike Miller

ah... I just prepare the night before, because all it amounts to is picking problems to do in class and doing them myself, or doing their homework to see what will be tricky

The upper div will be more time consuming but I think I'm willing to trade that for more interested students.

Ted Shifrin

Sure, no question, @Mike. Plus, you might learn something — if not content, pedagogy. :)

Ugh, I'm simultaneously working on five rec letters for grad schools ... and more to come.

Mike Miller

Yikes... at least my letter requests didn't come as much of a burden when I applied, because nobody else from my school was applying

user2179021

hi

Ted Shifrin

Imagine what the faculty at places like MIT have to do :P

user2179021

2:26 PM

is it appropriate to ask a math.SE question to explain a mathoverflow answer?

Mike Miller

I would say so, yes.

user2179021

@MikeMiller great! Maybe I should try here first and someone can just tell me it is easy :)

Mike Miller

@TedShifrin Yeah... and it's gotta be rough differentiating the many applicants in your letters when many are applying to the same schools....

Mary Star

@TedShifrin Since we suppose that $a \in F$ is an algebraic element over $K$ we have that $K \leq K(a) \leq F$, right??

Ted Shifrin

Sure, @MarySTar :)

Mike Miller

2:35 PM

Kobayashi is slow-going... the individual details are easy to check since he's very careful, but it's usually difficult to see the big picture, and most of my toy examples are of the form $G \to G/H$ (or $M \times G \to M$).

Mary Star

@TedShifrin Since $[F:K]=2$ the degree of $Irr(a,K)$ should divide $2$. So, the degree is $1$ or $2$.

If the degree is $1$, $Irr(a,K)=x-a$. The solution is $a \in F$

If the degree is $2$, $Irr(a,K)=x^2+Ax+B, A,B \in K$. The one solution is $a \in F$ and the other one is $-a-A \in F$ ($a \in F$ and $A \in K \Rightarrow A \in F$)

That means that all the solutions are in $F$.

So, $F$ is the splitting field.

Ted Shifrin

Yes, @Mike, I don't advocate it as an ideal book from which to learn.

Mary Star

Is this correct??

Ted Shifrin

Yes, @MaryStar.

Mike Miller

Where else is there to learn the geometry of principal bundles/connections? One of the reasons I'm using it is... I don't know what else to learn from!

Hippalectryon

2:37 PM

Hello @TedShifrin @MikeMiller @MaryStar

Mike Miller

hi

Mary Star

@Hippalectryon Hello!!

Ted Shifrin

Salut, @Hippa.

user2179021

does anyone have a moment to look at the answer of KFabian at mathoverflow.net/questions/168474/… ?

Ted Shifrin

Spivak is very verbose, but does stuff somewhat less generally. It's actually all in Dieudonné's Treatise on Analysis, too :P

user2179021

2:42 PM

I (like the commenter) don't understand why "If a solution exists then all $v_k\in\{-1,0,1\}$" is true?

Ted Shifrin

There are all sorts of new diff geo books, including some French ones. I just don't know them, @Mike.

user2179021

any help very much appreciated?

Mike Miller

Yeah, well I don't either.

I'm not sure if you're actually recommending I read Dieudonne...

I suppose not.

Ted Shifrin

I sorta like Dieudonné :P

user2179021

I understand if everyone here is too busy of course

Mike Miller

2:45 PM

I can't comment on it, @user2179021.

Ted Shifrin

I don't know how to answer that, @user2179021. Not ignoring you :)

user2179021

@MikeMiller @TedShifrin ah.. ok thanks

Hippalectryon

@TedShifrin Do you know what the ARQS is in French ? (Physics)

Ted Shifrin

I thought about circulant matrices about 20 years ago when I taught applied math, but nothing jumps out at me.

nope @Hippa

user2179021

@TedShifrin I mean the answer not the question, in case it wasn't clear

Hippalectryon

2:47 PM

@TedShifrin Ah nvm then

user2179021

@TedShifrin the answer of KFabian

it's all clear until that magic line

I don't understand where it comes frmo

Ted Shifrin

I know, @user2179021. Of course, any scalar multiple of a solution is a solution, so he means "can be chosen ..."

user2179021

I mean it's elementary until then

Ted Shifrin

I thought about similar issues in applications of linear algebra to graph theory. Even for some problems in my own textbook. :P

user2179021

@TedShifrin ok so can you see why the solutions have to be scalar multiples of {-1, 0,1} vectors?

@TedShifrin basically.. do you believe the answer :)

?

Ted Shifrin

2:49 PM

LOL, no. In the graph theory applications, we did this by induction.

Mike Miller

The only other thing I see from a cursory google is Morita's "Geometry of differential forms"... which sounds up your alley.

user2179021

@MikeMiller who is this aimed at?

Mike Miller

Ted.

user2179021

@TedShifrin ok well at least you don't find the answer easy to understand either :)

I wonder if it's correct

Ted Shifrin

I think I once looked at that, @Mike, but I've forgotten.

Mike Miller

2:56 PM

Ah well.

Mary Star

@TedShifrin Let $f(x) \in \mathbb{Q}[x]$ is a polynomial of odd degree $\geq 3$ that has exactly one real solution $a \notin \mathbb{Q}$. I have to show that the extension $\mathbb{Q} \leq \mathbb{Q}(a)$ is not normal.

$f(x)$ is of odd degree $\geq 3$ that has exactly one real solution $a \notin \mathbb{Q}$. That means that it has at least two other solutions, which are complex. Let $z, \overline{z}$ be the two solutions. Then the splitting field is $\mathbb{Q}(a, z, \overline{z})=\mathbb{Q}(a,z) \neq \mathbb{Q}$.

Ted Shifrin

It's fine, but you need to correct your last statement. $\Bbb Q(a,z)\ne ??$

Mary Star

@TedShifrin It should be $\mathbb{Q}(a,z) \neq \mathbb{Q}(a)$, right??

Transcript for

$Mathematics$ Mathematics