Mathematics - 2017-05-24 (page 6 of 6)

Akiva Weinberger

20:03

Unrelated: I like how the union of $4$ circles of radius $1$, centered at the $(\pm1,\pm1)$ points, has the same area as a single circle of radius $2$ centered at $0$

It's a really basic fact, but a pleasing image I think

@TheGreatDuck Hint: Try factoring it in the complex plane first

TheGreatDuck

im not interested in the puzzle

Akiva Weinberger

Ah, fine

The answer is: Since $x^4+4=(x^4+4x^2+4)-(4x^2)$ is the difference of two squares,

you can factor it as $\big((x^2+2)+(2x)\big)\big((x^2+2)-(2x)\big)$

or, as it's usually written, $(x^2+2x+2)(x^2-2x+2)$.

@TheGreatDuck

Semiclassical

hi chat

TheGreatDuck

ok

Akiva Weinberger

Hi, Semi. I'm trying to prove a polynomial thing.

Apparently, if $f(x)=\prod(x-a_n)$ where the $a_n$ are distinct integers, then $f(x)-1$ is irreducible. I need to prove this somehow.

TheGreatDuck

20:09

@AkivaWeinberger would plugging in 10 into a polynomial be a test of its reducability?

Semiclassical

I see it.

I don't remember much about reducibility, unfortunately.

Akiva Weinberger

@TheGreatDuck Not necessarily. $x^2-2x-63$ is reducible, but when you plug in $10$ you get $17$

TheGreatDuck

oh ok

reducible into integer coefficient polynomials or real number polynomials?

Akiva Weinberger

The thing is, one of the factors evaluates to $1$ when $x=10$ in that case.

@TheGreatDuck Integer.

TheGreatDuck

ok

Akiva Weinberger

20:11

Well, rational

Semiclassical

Am I right that the hint amounts to "If you can't factor it over Z[x], you can't factor it over Q[x] either?"

Akiva Weinberger

but it can be proven that reducibility over the rationals is the same as reducibility over the integers

@Semiclassical Yeah

Semiclassical

Mmkay.

TheGreatDuck

@AkivaWeinberger if a polynomial has a leading coefficient of 1 and integer coefficients elsewhere then the only solutions are irrationals and integers.

Semiclassical

Problem is that it's not enough to prove that it doesn't have integer roots.

Akiva Weinberger

20:12

@TheGreatDuck Yes.

TheGreatDuck

that wasn't a question

i proved that

Semiclassical

I think it's easy enough that it won't have any linear factors.

Akiva Weinberger

Right

Semiclassical

If it did, that'd amount to having an integer root of $\prod_k (x-a_k)=1$. Clearly false.

Akiva Weinberger

Right.

Semiclassical

20:15

But, that's necessary but not sufficient.

Akiva Weinberger

Yeah, it still could have quadratic or higher factors

Semiclassical

Right.

Being simple-minded about it, a counterexample one would need $$\prod_k(x-a_k)-1=(\sum_j b_j x^j)(\sum_l c_l x^l)$$

Where the polynomials can be assumed to be monic, with integer coefficients, and of appropriate degree.

Akiva Weinberger

Hm. I guess, if we prove that ring generated by the roots of a monic polynomial contains no invertible elements, we'd be done, right?

@Semiclassical Yeah

Semiclassical

In which case when $x=a_k$ you'd have $-1=(\sum_j b_j a_k^j)(\sum_l c_l a_k^l)$

Which seems pretty unlikely.

Akiva Weinberger

You to use the fact that the $a_k$ are distinct, at some point.

Semiclassical

20:19

Presumably, yes.

But for now there's at least one thing we learn: We'd need one of those terms to be $+1$ and the other to be $-1$.

Akiva Weinberger

Ah, true.

Semiclassical

This is where the Gauss's law hint is handy, since otherwise we'd be stuck talking about rationals.

Akiva Weinberger

It might depend on which $a_k$ you use

Semiclassical

Yeah.

Akiva Weinberger

Hold on. Look at $\Bbb Q[x]/\langle x^n+a_{n-1}x^{n-1}+\dotsb+a_0\rangle$ (using Q for a second)

Semiclassical

20:22

Let's be more specific, though. Let's take the degrees of the proposed factorization to be $p,q$ and the total degree to be $n=p+q$.

Mmkay.

Akiva Weinberger

What's the inverse of $x$ in that?

Assuming it has one

Semiclassical

well, you'd need a polynomial such that $xP(x)=1+R(x)(x^n+a_{n-1}x^{n-1}+\cdots+a_0)$.

Akiva Weinberger

$-\frac1{a_0}(x^{n-1}+a_{n-1}x^{n-2}+\dotsb+a_1)$ I think

Right?

Semiclassical

That'd give $-\frac{1}{a_0}(x^n+a_{n-1}+\cdots+a_1 x)\equiv (-a_0)/(-a_0)=1$, yeah.

Akiva Weinberger

Hm, that's less helpful than I was thinking. Let me try to regain my train of thought for a second

8 mins ago, by Akiva Weinberger

Hm. I guess, if we prove that ring generated by the roots of a monic polynomial contains no invertible elements, we'd be done, right?

^That's obviously false

The roots of $x^2+3x+1$ are each other's inverses, for example

Semiclassical

20:27

Just following up on my previous line of thinking: $$\prod_{k=1}^{p+q}(x-a_k)=1+(\sum_{j=1}^p b_j x^j)(\sum_{l=1}^q c_l x^l)$$

I'm moving the 1 over because I want to.

And let me write those two factors as $B(x),C(x)$ .

Akiva Weinberger

Sure.

Semiclassical

What we need is $B(a_k)=-C(a_k)=\pm 1$ for all $k$.

Akiva Weinberger

16 mins ago, by Semiclassical

I think it's easy enough that it won't have any linear factors.

Semiclassical

Ooh, which means in particular $B(x)+C(x)$ would vanish at all the $a_k$

Akiva Weinberger

16 mins ago, by Semiclassical

If it did, that'd amount to having an integer root of $\prod_k (x-a_k)=1$. Clearly false.

Generalizing that earlier line of thought,

we want to show that it can't have higher factors,

which means that we'd need to show that we can't have a root of $\prod_k(x-a_k)$ that's the root of a monic polynomial of $<n$ degree.

And each of the $(x-a_k)$ terms is distinct.

Semiclassical

20:33

I think I can actually give an argument based on what I just said.

Akiva Weinberger

Can you?

Semiclassical

Yeah.

By Gauss's lemma, we'd have to be able to factorize $\prod_k(x-a_k)-1$ into monic integer polynomials $B(x),C(x)$ if we're going to factorize it at all.

But for that to work we'd need $B(a_k)C(a_k)=-1$ for all $a_k$; since the polynomials are integers, the only way for this to work is for $B(a_k)=-C(a_k)=\pm 1$.

In particular, we'd need $B(x)+C(x)$ to vanish at all the $a_k$.

Since all the $a_k$ are distinct, this would mean that $B(x)+C(x)$ would have the same degree as our original polynomial.

But the degrees of the polynomials should also add to the degree of our original polynomial, and the only way for this to work is if one of the polynomials is degree 0.

Akiva Weinberger

Or maybe $B=-C$

Semiclassical

?

Akiva Weinberger

If $B(x)=-C(x)$ everywhere, then $B(x)+C(x)$ is the zero polynomial, which vanishes at all the $a_k$s

Semiclassical

20:38

Hmm.

Akiva Weinberger

That would mean that $\prod(x-a_k)-1=-B(x)^2$

Or $\prod(x-a_k)=1-B^2=(1-B)(1+B)$?

Abdullah UYU

Sorry, i'm interrupting :D
A sequence have $4$ positive term. The first $3$ term makes an arithmetic sequence. The last three term makes an geometric sequence. The difference between the last and the first term is $48$. Find the sum of all terms.

Semiclassical

I should also have included a statement at the start that the factorization is into monic integer polynomials that aren't constant. That's the assumption one would want in order to make a proof by contradiction.

Akiva Weinberger

Right, yeah

Semiclassical

So we've effectively reduced it to showing that $1-\prod_k(x-a_k)$ can't be a perfect square.

Hmm. Suppose we pick some integer $x$ other than an $a_k$.

If that polynomial were to be a perfect square, then the result at this $x$ would have to be nonnegative.

Actually, better yet: Let me pick the smallest integer that's bigger than all the $a_k$.

Then $\prod_k(x-a_k)$ would necessarily be positive.

Dodsy

20:44

@TheGreatDuck what's the bug

what does it do.

TheGreatDuck

@Dodsy what browser are you using?

Dodsy

Chrome of course.

TheGreatDuck

1

Q: Glitch in chat that allows me to produce a backspace character literal

I was editing a post over in the Math Stack Exchange chat. I clicked in chat once (like to click off the box) and hit backspace to go back a page. Weirdly enough, a weird symbol appeared. I could replicate this arbitrarily to produce blank messages. I searched the ascii character code and apparen...

oh well then you cannot do it

Akiva Weinberger

@AbdullahUYU I can do it if one of them is allowed to equal 0

TheGreatDuck

afaik

Semiclassical

20:45

@AkivaWeinberger I think this is almost a proof.

Akiva Weinberger

Oh. Just let $x\to\infty$

Semiclassical

...huh, yeah.

Akiva Weinberger

$1-\prod_k(x-a_k)\to-\infty$ then and thus can't be a square

Semiclassical

Right.

TheGreatDuck

Semiclassical

20:46

So unless we're missing something, that seems to close the loophole and force a contradiction.

Danu

o/

Akiva Weinberger

So the proof goes through two cases, but that's a QED now, right?

Abdullah UYU

@AkivaWeinberger why? what is special when one of them is 0?

TheGreatDuck

would someone please star my empty comment? I'm experimenting and I wish to see what that bug will do in the side bar.

Akiva Weinberger

If the first one is $0$, then I can do $(0,x,2x,4x)$

TheGreatDuck

20:47

thanks

Semiclassical

Let the two middle terms be $x,y$ in that order.

TheGreatDuck

:/

nothing special happened

that's good

Akiva Weinberger

Oh, wait, never mind, you don't need one of them to be 0

TheGreatDuck

the bug must just be an error in the code for the editing box then

Semiclassical

The difference between those would be $y-x$, so the first term would have to be $x-(y-x)=2x-y$.

Akiva Weinberger

20:48

2,4,6,9 is also a valid sequence. So I don't need 0. @AbdullahUYU

Semiclassical

To have a geometric sequence, we'd need last term / $y$ = $y/x$. So the last term would have to be $y^2/x$.

Abdullah UYU

Hmm, interesting. But question says that all of them is positive.

Semiclassical

So we'd need $y^2/x-(2x-y)=48\implies y^2+x y-2x^2-48x=0$

:/

I guess you also need $2x-y>0$, but eh

This doesn't feel constrained enough.

Abdullah UYU

i missed everything, how do you call first term?

Semiclassical

Look in the transcript.

If the second term is $x$ and the third is $y$, then the difference between them is $y-x$.

So the first term would have to be $x-(y-x)=2x-y$.

Akiva Weinberger

20:53

This has too many variables.

Semiclassical

Well, maybe the problem is undetermined?

Oh, but I'm being silly. It only asks for the sum of the terms, not what they all actually are.

user314159

hi chat

Semiclassical

And the sum of the terms would be $(2x-y)+(x)+(y)+(y^2/x)=3x+y^2/x$.

hrm. this still doesn't seem fully determined.

Abdullah UYU

i've read history. what a quick response.

Akiva Weinberger

Maybe it would be easier to see that if you rewrote it as $x(2-r)$, $x$, $xr$, $xr^2$

Semiclassical

20:57

Not a bad idea.

Akiva Weinberger

Given $x(r^2+r-2)=48$, you'd want to find $x(r^2+3)$.

See? Very undetermined.

Semiclassical

Yeah.

There is at least one cute example I found: x=12,r=2

oh, wait.

Akiva Weinberger

Yeah, I mentioned it above

but the first term is 0 so it doesn't count

Semiclassical

Yeah.

Hmm, though, one still has to make the inequalities work.

Akiva Weinberger

Besides for 0,12,24,48, it seems impossible if you want them all to be integers.

Semiclassical

21:02

I guess an interesting question is: What's the smallest case with integers, if it can be done at all?

I don't -think- it's impossible.

Mike Miller

can someone remind me how to secretly name my theorems in tex so I can cite them later by number?

Akiva Weinberger

Google it

Semiclassical

lol

Akiva Weinberger

My TeX-fu is limited to MathJax.

@AbdullahUYU Are you sure you have the question right?

Mike Miller

\label

Abdullah UYU

21:04

answer is 333 according to book :D

Semiclassical

wtf

well.

Akiva Weinberger

I think you misread the problem probably. @AbdullahUYU

Semiclassical

that does give enough info to determine $r$ uniquely.

you'd need $x(r^2+r-2)=48$ and $x(r^2+3)=333$.

So that'd require $48(r^2+3)=333(r^2+r-2)$.

Abdullah UYU

yep

Semiclassical

...which has $r=6/5$ as a positive root, and from that we get $x=75$.

So $60,75,90,108.$

stare

Akiva Weinberger

21:07

…Oh

Abdullah UYU

so

Semiclassical

yeeeah

I have no earthly idea why they picked that one.

Danu

@MikeMiller Right, \label. I recommend using \label{thm:name} for clarity, especially if you will be working on this document after you've forgotten the names of all your theorems.

Abdullah UYU

do you think there are not enough info

Balarka Sen

Hey, how does one integrate stuff like $\dfrac{1}{(x+c)\sqrt{x^2+ax+b}}$? Wasn't there a super clever trick or something?

Semiclassical

21:09

I'm not convinced that's a unique answer, no.

Danu

and similarly \label{lem:name}, \label{prop:name} etc.

Mike Miller

What is name?

Oh, sorry, I see what you mean

Danu

The custom name you want to assign

Mike Miller

Yes, that's what I'm doing

Danu

Just distinguishing the types is all I meant ^^

Mike Miller

21:10

oh, I see. I don't care very much about that.

Semiclassical

To get an integer answer, we'd need to an integer $x$ such that $9x^2+192x$ is a perfect square.

$x=75$ works for that, apparently.

Danu

@MikeMiller You might later on. And it's very little effort.

Mike Miller

I think maybe I'll care about the subtle differences between those classes when I have a paper to start with...

right now I label everything as theorem :)

Balarka Sen

maybe i should dig up the infamous integral kokeboken

Mike Miller

not sure what should be a lemma, say

Semiclassical

21:11

But I don't know how that could be determined. :/

user284891

Is stackexchange useless for people with math Ph.Ds?

Danu

@MikeMiller Hah, regarding that issue... I'm taking this seminar on the Atiyah-Singer index theorem and the professor is a real stickler for doing that stuff "correctly".

Semiclassical

Do you mean: StackExchange the main site, or StackExchange sites more generally?

Mike Miller

yeah, I don't care about that...

Danu

He made one guy erase "Proposition" to put in "Theorem", twice.

Abdullah UYU

21:12

thank you for your effort, how can you write these so quickly @Semiclassical

Semiclassical

Experience, mostly.

Though I like the way @akiva wrote it better.

user284891

When does someone finding asking questions on stackexchange math helpful? Is that when they get their Ph.D in math?

Akiva Weinberger

@AbdullahUYU So, in short: Your problem required much more cleverness than either of expected, and I completely messed up the math and mistakenly thought it was impossible

Danu

@brittany Huh?

When they run into a problem they would like to have solved for them, presumably?

Mike Miller

*stop finding, you mean?

Danu

21:13

ah

Even then, that question sounds very naive :P

Mike Miller

I don't see any reason to believe that just because someone has a PhD doesn't mean they have questions that are appropriate for MSE

Akiva Weinberger

@Semiclassical You know what would have been a better thing to start from?

Semiclassical

The answer? :P

Akiva Weinberger

$x$, $x+d$, $x+2d$, $x+48$.

Semiclassical

hmm.

yeah, subject to $(x+d)(x+48)=(x+2d)^2$.

Akiva Weinberger

21:15

Then we just need to solve $(x+48)(x+d)=(x+2d)^2$ for integer solutions.

Semiclassical

bah, that's much smarter. especially since $x^2$ cancels

user314159

@brittany most phds are on mathoverflow

Akiva Weinberger

So $x^2+(48+d)x+48d=x^2+4dx+4d^2$

Semiclassical

so you're left with $48x+48d=3xd+4d^2$

Abdullah UYU

ohh, impressive

Semiclassical

21:16

$48x-3xd=3x(16-d)=4d^2-48d=4d(d-12)$.

Akiva Weinberger

$3x(16-d)=4d(d-12)$?

Semiclassical

Yeah.

$3$ divides both sides, so need $d$ a multiple of 3.

But $d$ can't be 12 and shouldn't exceed 16, so $d=15$.

Akiva Weinberger

Oh.

Wow.

Semiclassical

in which case you need $3x=4(15)(3)=180\implies x=60$.

huh.

Akiva Weinberger

@AbdullahUYU You caught all that?

Semiclassical

21:18

I see it, but I'm not sure I believe it.

Akiva Weinberger

OK, Cantor

Semiclassical

lool

Abdullah UYU

yes,

Semiclassical

But yeah, I think that works.

I had not expected it to, but yeah.

That's kinda bizarre.

@AkivaWeinberger In any case, though, that last representation of the sequence is definitely the best one.

Akiva Weinberger

Yes.

Abdullah UYU

21:24

i have a test that have lots of questions like this. requiring to be considerate

user314159

then you've got a lot of work to do :-)

Semiclassical

What bothers me is how unexpected a solution that was

Akiva Weinberger

I just want to point out that you never explicitly said that they're integers

Semiclassical

I think the lesson is to focus on writing terms in an additive way and focus on the multiplicative step last

Akiva Weinberger

I am definitely unpracticed when it comes to Diophantine equations.

Whoa, my band-aid left a tan line

Semiclassical

21:32

Going back to my first attempt, I could just as easily written it as $2x-y,x,y,2x-y+48$ and required $x(2x-y+48)=y^2$

Akiva Weinberger

Stuff doesn't cancel then

Semiclassical

yeah

It's probably still tractable, but not anywhere as obviously as the other

Abdullah UYU

oh sorry, i looked it up the transcript and noticed that i didn't state they are integer as you said.

Zee

@Semiclassical do you use much algebra in your work?

Semiclassical

Sure

Abdullah UYU

21:37

they are positive $\textbf{integers}$

i mean there are a condition in question that states they are integers

by the way, i love here, thanks to all of you

it's 00:43 here, good night :D

Akiva Weinberger

Good night!

double-checks the time where Balarka is

Fargle

lol, I just woke up.

Balarka Sen

3:18

AM

user314159

why aren't you asleep? :P

Akiva Weinberger

@Semiclassical The next problem is showing that $f^2+1$ is irreducible

under the same circumstances

Hm, when is that false with repeated roots?

Something *Germain-y probably

Semiclassical

21:52

You think Germain is germane, then?

Akiva Weinberger

Fixed typo. And yes

Specifically, the problem is showing that $\left(\prod(x-a_k)\right)^2+1$ is irreducible when the $a_k$ are distinct integers.

To be formal.

Semiclassical

Well, you'd still have polynomials B,C such that BC=1 for a set of integer a_k

So $B=C=\pm 1$

Akiva Weinberger

(Sorry, had to go, back now)

Semiclassical

And therefore B-C is either identically $\pm 1$ or vanishes at the integer roots

Akiva Weinberger

Yeah, so consider $B-C$ now?

Semiclassical

21:56

Yeah.

Akiva Weinberger

"Either identically $\pm1$"?

Semiclassical

derp

Identically zero

Akiva Weinberger

In any case, since they're both of smaller degree, and since the roots are distinct, we must get $B=C$

And so $f^2+1=B^2$

or $(f-B)(f+B)=1$, I guess,

Semiclassical

Right. It again reduces to proving that's not a complete square.

Akiva Weinberger

which is impossible since, when you take $x\to\infty$, the LHS becomes $(\pm\infty)(\pm\infty)$.

QED.

Semiclassical

22:00

Eh?

In this case both sides go to positive infinity

Akiva Weinberger

Oh, right, $B$ has smaller degree so it's negligible

Fine

So the LHS becomes $\infty\cdot\infty\ne1$.

If you want, those were numbers 25 and 26 from this link: http://people.cs.uchicago.edu/~laci/REU12/puzzles.pdf

Daminark

22:19

Whoa @Akiva you are blasting through this stuff

Also hai everyone

Akiva Weinberger

Wait, $B$ has the same degree, what am I talking about @Semiclassical

Still, at least one of the terms goes to infinity

Balarka Sen

Hi @Daminark

I'll let Akiva do all this before I get my hands into it... I would need serious help :P

Akiva Weinberger

The first few were more puzzle-y. Right now it's all ring theory-y (polynomials and irreducibility), so it's gotten easier

And 25 and 26 were mostly Semi

Daminark

I see

Zee

@Daminark hey frenemy

Daminark

22:32

How's it going?

Zee

@Daminark good, am thinking about studying abstract algebra even though I don't like it that much

Any fav books?

Daminark

I haven't yet learned much of it, but I have heard good things about Jacobson

I'd say don't go for DF, that book drags too much

Zee

lol this is so funny

I was reading DF and stopped for the same reason you mentioned

Daminark

Artin may be your style

Zee

And I just ordered Jacobson like 10 mins ago

Daminark

22:35

(Assuming introductory)

Because that book situates algebra in a broader context of math

tinbananavrekriari

every underline fuction of monomorphism in a concrete category is injective??

Daminark

So if you're not into algebra for its own sake quite as much it may be wise

Zee

I had an undergrad course in algebra but I don't remember much of it

Daminark

Also lol that's hilarious

Zee

thats true but am preparing myself for gradute course so it needs to be mainstream

Prob gonna try Jacobson

Daminark

22:38

Probably a good plan

Zee

@Daminark I wanna quit smoking but I can't

Daminark

Well, get people around to help you and whatnot, quitting smoking is the probably one of the best choices you can make right about now

Zee

@Daminark I don't like it, am forcing myself to do it

To smoke that is

Balarka Sen

so you don't like smoking, but you can't quit it, you're forcing yourself to do it, but you don't want to?

is this a confusing existentialist choice or what

Zee

Am forcing myself since it will make me a better mathematician

Daminark

22:45

I don't think it will

Balarka Sen

lol

I am going back to work

Daminark

And I don't know if being a better mathematician is worth the eventual lung cancer

Zee

Well , research shows it has cognitive benefits

And being a better mathematcian is def worth dying 10 years younger

Mike Miller

I look forward to an early death but of all the ways to improve my math ability I think there are better uses of time than cigarettes

Daminark

Whoa @MikeM, that's... dark

Zee

22:49

Am listening mike

Akiva Weinberger

You know what else makes you a better mathematician? Regular exercise

Zee

Yes , it should balance the smoking habit too

Zee

23:25

Alright , my nocturnal brain is up and running now, its math time!

Transcript for

$Mathematics$ Mathematics