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4:00 AM
but I mean do we need to rely on "$(x+1)^n - (a+1)^n = 0$" being a minimal polynomial?
 
@TedShifrin I want to say something like "I learned from the best" here
@Typhon It isn't.
 
Haha, DogAteMy. :)
 
ok
let me ask a clearer question
 
I should probably know this, but what's a current? The name Leray comes to mind but I can't place it.
 
Currents are essentially differential forms with distribution coefficients, @Semiclassic.
 
4:01 AM
are the conjugates of a also solutions in any polynomial to which a is a root?
 
Not $\vec J$ :)
 
Yes @Typhon
 
(Assuming rational coefficients)
 
You know the word distribution, Semiclassic?
 
4:01 AM
I can see why the interpretation of $dz/z$ as a current would make sense, then.
 
Generalized functions, if you like.
 
Sure.
 
@AkivaWeinberger so why bring up the minimal polynomial? We can use any polynomial with rational coefficients?
 
Thought so. Sorry. :)
 
(This needs proof, but as I said before, because it's such a fundamental property of conjugates, we can assume it for now)
 
4:01 AM
yeah fair enough
 
Psh, I"m a physicist. Dirac delta and all that :P
 
@Typhon The minimal polynomial business was just for defining "conjugates" in the first place.
 
@AkivaWeinberger ah ok.
 
In any case, we return?
 
yeah ok
 
4:02 AM
So is it something like $d(dz/z)=\delta(z)\,dz\wedge d\overline{z}$? I'm guessing, to be sure.
 
so we need to show $a$ is the only solution to $x^n - a^n = (x+1)^n - (a+1)^n$?
 
${}=0$.
 
(and I am not even trying to worry about whether it's normalized properly)
 
hm?
 
We want to show that $a$ is the only common root.
 
4:03 AM
Right, @Semiclassic. I just wasn't sure you knew the terminology. Most physicists call 'em generalized functions, I thought.
 
Sorry about giving you headaches lately @Semiclassical
 
so setting them equal does that, right?
 
@LasVegasRaiders No worries.
 
We want to show that it's the only thing that satisfies both $x^n-a^n=0$ and $(x+1)^n-(a+1)^n=0$. @Typhon
 
@ted I see both terminologies.
 
4:04 AM
Up to constants, yeah, @Semiclassic.
 
@AkivaWeinberger yeah and therefore setting them equal allows us to find that common root
 
Skull, you excel at headache-giving :)
 
Though tbh the only one we really ever use is Dirac delta and its various forms in different coordinate systems.
 
@Typhon But we also want them to equal $0$ so that we get a root in the first place rather than some other random value where they happen to be zero
 
Oh, I should send you my 2-page handout for the physics majors in my old multivariable course, @Semiclassic. One exercise was to figure out grad, div, curl in different coordinate systems using forms. The other was a dirac delta thing.
 
4:05 AM
We also had to deal with the Dirac comb a few times, I think
 
I don't know that.
 
"For every dirac epsilon, there exists a dirac delta such that..."
 
smacks Demonark
 
$\sum_{k\in\mathbb{Z}} \delta(x-2\pi k)$.
 
In any case, we'll actually use a slightly different approach @Typhon
Let's look at where the roots of $x^n-a^n$ are.
 
4:06 AM
@Ted sorry I tabbed out, that's weird... my intuition tells me it should be like a dirac mass orsomething
 
Oh, I remember that came up in the applied math course I taught in 1986, Semiclassic :P
 
Yeah, it's handy.
 
@Ted sorry I tabbed out, that's weird... my intuition tells me it should be like a dirac mass orsomething
 
Actually, first—Do you know about roots of unity, how they lie on a circle? @Typhon
 
Mostly because it's equivalent to something like $\sum_k e^{i k x}$.
 
4:07 AM
oh Semi beat me to
 
Right. But obviously as a differential form, it's closed. So currents see the difference.
 
Sniped
 
right that's cool
 
Yeah, that's neat.
Though, is it really $\delta(z)$? Somehow that seems wrong.
It feels asymmetric between $z$ and $\overline{z}$.
 
4:08 AM
I can email this if people want, but this is the handout to which I'm referring.
 
"tabbed out" sounds like UFC @EricSilva
 
Though I guess $z=0$ should mean $\overline{z}=0$ too? Hrm.
 
Neat.
 
yeah @LasVegasRaiders I was in a serious fight with uhhh
 
4:09 AM
Let's just say $\delta_0$ and be done with it, @Semiclassic :P
 
chrome
 
Lol, fair enough.
 
@AkivaWeinberger nope. Do you mean like graphing x and a in an XA-plane?
 
ding ding ding
 
Reminds me of a formula that has always bothered me a little
 
4:10 AM
Demonark: Did you notice that we have (had?) another Amin here?
 
aminliverpool?
 
right
 
Not sure I can find it easily, though. Hrm hrm hrm
 
weird that the current like "sees the residue"
 
That's just the beginning, Eric.
 
4:11 AM
Eh, it makes sense to me if only because I can see the analogy with magnetostatics.
 
@Daminark At first I had misread it as "Am I in Liverpool"
@Typhon I mean, in the complex plane
There are five complex numbers that satisfy $z^5=1$. $~1$ is one of them.
 
@AkivaWeinberger oh... I've never looked at the roots of a polynomial in that sense no
but hold up a sec
1 = x^2
 
Where taking the curl of a magnetic field should yield the volume current density, so taking the curl of the field produced by a line current should give that current back.
 
there's 2 solutions
-1 and 1
 
@AkivaWeinberger Kek
 
4:13 AM
@Typhon $(-1)^3\ne1$.
 
therefore... 1 is irrational?
 
(I wouldn't be shocked if there's a direct analogy. But I'm jumping far.)
 
@AkivaWeinberger course not
oooh
 
@EricSilva what u learn
 
1 = x^2 has two solutions
 
4:13 AM
If you were to write $x^2-1$, though, the solution is that $x^2-1$ isn't the minimal polynomial of $1$; $~x-1$ is.
And the only root of $x-1$ is $1$.
 
of course
duh
I was thinking that disproved your claim
 
Similarly, $(x^2-2)(x^2-3)$ has four roots, $\pm\sqrt2$ and $\pm\sqrt3$.
 
Today @Mike? I learned that reading federer is probably a bad idea
 
Hi Tern!
 
hiya
 
4:14 AM
in that it showed that the conjugates weren't always co-solutions
 
This doesn't mean that $\sqrt2$ and $\sqrt3$ are conjugates, though.
 
user84215
In order to choose math questions which we want to answer, it is better not to involve our personal emotions to each other.
 
i was thinking stupidly
anyways
 
(Cont'd) Because the minimal polynomial of $\sqrt2$ is $x^2-2$, which doesn't have $\sqrt3$ as a root.
 
@aminliverpool: It's fine for you to believe that. But human beings have emotions and you don't make all the rules.
 
4:15 AM
the final result comes from showing the intersection of the polynomials solutions sets has only one element?
 
thanks
the rest is too complex for me right now
 
$$\int_{\mathbb{C}^2}e^{-\overline{z}w z+\overline{u}z+\overline{z}v}\,dz\wedge d\overline{z}=\frac{2\pi i}{w}e^{\overline{u}v/w}$$
siiiigh
 
maybe when I...
wait I don't take number theory
 
The main idea is that the roots of one of them lies on a circle on the complex plane, and the roots of the other lie on a tangent circle on the complex plane
 
4:16 AM
@Eric Marianna told us that when she was in school in Hungary, they had an English language exam
 
and so they only have one intersection
 
damn
 
Someone used Federer because it barely had words
 
Oh good, only one error. I hope.
 
In hopes that it'd be easier because of that
 
4:16 AM
@Daminark you've told me this story before
 
I'll have to read up on number theory on my own. heheh
 
@Typhon Look up roots of unity. They're interesting, and hopefully the above will make more sense
 
@AkivaWeinberger yeah I shall.
 
@Semiclassic: I'm not thinking about your formula, but remember that you need $$\frac i2 \,dz\wedge d\bar z = dx\wedge dy.$$
Demonark, you told that one before.
 
@Typhon The main result is this: the $n$-th roots of unity form the vertices of a regular $n$-gon in the complex plane!
They all lie on the unit circle, as well.
 
4:17 AM
@AkivaWeinberger ooooh!
 
Damn, now DogAteMy truly is discussing Algebra: A Geometric Approach :D
 
all roots of a polynomial form a polygon?
that's high school math, man
 
@ted Heh. Want to know what's funny? The physics book I'm lifting this from uses the notation $\int d(z,\overline{z})=\int_{-\infty}^\infty\int_{-\infty}^\infty dx\,dy$
 
Roots of unity, Typhon
 
XD
@AkivaWeinberger which ones are not roots of unity?
 
4:18 AM
Yeah, they're wrong, @Semiclassic.
 
At least, it's highly misleading.
 
The $n$-th roots of unity are the things that, when raised to the $n$-th power, equal $1$.
 
I agree.
 
The solutions to $z^n=1$.
 
4:19 AM
I don't like that notation at all.
 
that notation makes me sweat
 
you lost me
 
It turns out that there are $n$ of these, and they lie on a regular (!!!) polygon in the plane.
 
But in their version of it it was only $\pi$ on the right; I added in $2i$ by hand.
(I may have gotten the sign wrong, though).
 
@Typhon Go to your calculator of choice and calculate $(-\frac12+i\frac{\sqrt3}2)^3$.
 
4:20 AM
Semiclassic: I gave you the correct formula. You can check it easily enough.
 
if I have a polynomial $c_nx^n + c_{n-1}x^{n-1} + \cdots$ ALL the solutions form a polygon right?
 
In defense of them, though (and this is what bothers me) the only way I know how to verify that Gaussian integral is to go back to $x,y$.
 
That's (.5+i*0.866)^3.
 
Oh kek
 
@Typhon No, they can be scattered essentially randomly
 
4:20 AM
@EricSilva ok fair enough
 
oh
ok
 
DogAteMy: You presume a certain ordering on the roots of unity to get the right polygon.
 
@AkivaWeinberger I'm pretty sure that is either 1 or -1
just from my knowledge of cosine and sine and polar coordinates
 
So whenever I saw that formula it always seemed like it was just rearrangement for rearrangement's sake.
 
Neves also gave me an overview lecture on a bunch of results relating to plateau's problem which was kind of cool @Mike
 
4:21 AM
and the fact that powers are rotations in the complex plane
 
@TedShifrin I don't understand
 
@EricSilva cool
 
Well, if I join the $n$ points in a random order, I don't get the regular $n$-gon!!
 
Sure, it's neat to write it over the complex plane...but if the only way I can verify it is by doing real integration, it seems superfluous.
 
@TedShifrin Oh, that's what you mean
 
4:22 AM
i dunno why i sent a blank message
 
Whatever, convex hull
There exists a regular $n$-gon such that these are the vertices of it. Happy? @TedShifrin
 
@AkivaWeinberger it is the point $(1/2,\sqrt{3}/2)$ which corresponds to 120 degrees. Therefore, it cubed it 1.
im well aware of how cubes form a triangle
 
Which I guess means I'm wondering if there's a way to do that kind of Gaussian integral without having to go back to $x,y$. Otherwise it just seems unnecessary.
 
and 4th roots etc
i assumed it was true for all roots
 
I was just partly defending typhon.
 
4:23 AM
Right. And 5th roots of unity (read: "unity" equals "one") form a regular pentagon
 
@TedShifrin hmm? I was responding to akiva.
 
@Typhon OK, yeah, it is
 
This conversation is becoming a directed graph, it seems
 
So you do know about roots of unity, then.
 
LOL
 
4:24 AM
@AkivaWeinberger sooo... how do the roots of 1 or -1 or any other unit in the complex plane help with the other thing?
 
@Semiclassic: I don't think I know how to do a Gaussian integral without real stuff ...
 
Me either.
 
Let's look at $x^n-a^n$ again. @Typhon
 
@AkivaWeinberger not by that name, but if you said roots of units and how they form rotations and equally spaced points, then yes.
actually
 
Which makes it seem like the complex version is just a redressing of the real version.
 
4:25 AM
give me five minutes to snatch my phone and login
I have to migrate to the toilet
 
And that just seems lame :/
 
well, Semiclassic, if you can write $|z|^2$ ... then ... :P
 
Just like how the roots of $x^n-1$ (which would be the $n$-th roots of unity) all lie on the unit circle, I claim that the roots of $x^n-a^n$ all lie on a circle of radius $|a|$ centered at the origin. @Typhon
 
lol
Anyways.
The reason it came to mind is that I can get a Dirac mass as the limit of such a Gaussian.
 
we've had that discussion before, Semiclassic.
 
4:27 AM
Have we? I suppose so.
(one obviously has to normalize the Gaussian etc.)
 
approximations to the identity ... mollifiers ...
 
I more was wondering if it links up to what you were saying about currents.
I'm not sure it does, though.
 
@Ted i feel like the definition of "$d$" of a form that you're thinking of as a current should be inducing the usual exterior derivative on forms, so like if you call $C_{\omega}$ the current associated to $\omega$ then we should have $dC_{\omega} = C_{d\omega}$ (i think a formula like this holds for distributions iirc, so it's the same thing) so maybe it "sees" the residue in the sense that this formula fails
 
well, it's just talking about distributions ... which are one level of currents.
 
True.
 
4:28 AM
Nope, Eric, $d$ is defined on currents by forcing Stokes's Theorem.
If you have a global smooth form, then you're right, of course.
Remember that submanifolds (chains) are also currents, because you can integrate over them. So you want $d$ to include $\partial$ in the appropriate way.
 
"There's a net circulation of the magnetic field, so there must be a current enclosed."
 
right @Semiclassic
 
hmmm lemme think about this
 
I maintain my suspicion that there's a rather exact analogy there.
 
ok i think i see how it comes from stokes
 
4:31 AM
though I guess maybe the analogy only works if I go to x,y.
 
I'm not disputing that, Semiclassic. My example earlier extends to point masses/charges in 3-space.
 
@Semi there's definitely a closed analogy
 
Ah, true.
 
It's like "distributional derivatives," Eric. They're defined by integration by parts. Same difference.
 
yeah
 
4:32 AM
@TedShifrin Hmm, I suppose this should also allow for point dipoles. That should still count as a distribution.
 
Of course, Semiclassic. I leave the physics to you :)
 
(And then the entire hierarchy of multipole moments ay-yie-yie)
 
@AkivaWeinberger and what about x+1?
 
Hi @BalarkaSen
 
Well, it's not just physics. For instance, it's stuff like $\nabla\delta(\vec{x})$ (3D guassian).
 
4:34 AM
hi
 
@AkivaWeinberger but a could be any number? How do you know the norm of all the a's are equal?
 
oh, no; it's a Balarka'.
 
run run run
 
It's him again
lock and load boys
 
::runs::
 
4:35 AM
i do feel strongly that for a form like the $dz/z$ that's locally integrable we should have a formula like $dC_{\omega} - C_{d\omega} = Residue$ or something
 
heeeere's johnny
 
It's about time that we 'differentiate' @BalarkaSen
 
Well, sure, EricS.
 
ill look at distributions again
we did them for like five minutes in my analysis class last fall
 
differentiates @BalarkaSen 32 times
 
4:37 AM
@Typhon You know polar coordinates. Let's say, for example, that $a$ is $2$. So, let's look at the roots of $x^n-2^n$.
So these would satisfy $x^n-2^n=0$, or $x^n=2^n$. Yeah?
 
Ok
but you miss my point
how do you know |a| is always one value?
 
Complex multiplication multiplies their lengths, yeah?
@Typhon What? It depends on $a$
 
But you said the solutions for mm a ring of radius |a|
 
so... which |a|?
 
4:39 AM
@Ted I hve just acquired a pdf of griffiths/harris
this looks cool
 
I didn't hear that, Eric.
 
What do you mean? The absolute value of $a$. @Typhon
 
Which a?
 
There's lots of mathematics for you to learn there. They make occasional errors, but it's still by far my favorite source for complex algebraic geometry and related topics.
 
ph I'm sorry
 
4:39 AM
Why is there more than one $a$?
 
i was thinking you said x
forgot that a was a real number
that was predetermined
carey on
 
@Ted I will probably use it on the side if I take algebraic geometry this fall! this look like something i'd enjoy
 
I was doing a specific example a few moments ago, where it's 2. So, let's look at the roots of $x^n-2^n$.
So that's $x^n-2^n=0$, or $x^n=2^n$, right?
 
ok?
 
4:41 AM
Complex multiplication multiples the magnitude, I think you mentioned before
 
i know... a ring of radius 2
 
Right, yeah
$|x|^n=2^n$, so $|x|=2$
 
@TedShifrin well, it's better to illegally acquire a copy of griffths-harris than legally acquire a copy of peter may's concise course, though
 
In general, the roots of $x^n-a^n$ lie on a circle of radius $|a|$. That makes sense?
 
I'm still not listening.
 
4:42 AM
:D
 
Although as certain people know, I've been known to share my own book(s).
 
@AkivaWeinberger I wrote a large section on complex numbers in the context of geometry for about 5 pages of my 50 geometry paper I wrote last fall. I know the algebra rules. I just don't know the fancy terms or "theory x"s.
 
i never said the pdf was illegally acquired
 
That was tacit, Eric.
 
@AkivaWeinberger yes. Was thinking a was the varisble
Got mixed up
 
4:43 AM
No, you politely emailed the authors and asked if they could possibly send you a copy @EricSilva
@Typhon OK, sure. Similarly, for $(x+1)^n-(a+1)^n$, it's essentially the same situation but shifted
 
ok i gotta run now
 
To where?
 
i will say nothing further to incriminate myself
 
Hi/bye, Balarka'.
 
A little puzzled
 
4:45 AM
@Typhon Like, one unit to the left in the complex plane
 
Ok
 
We have $(x+1)^n-(a+1)^n=0$, so $(x+1)^n=(a+1)^n$
 
Alrighty and that helps how?
 
Magnitude'ing, $|x+1|^n=|a+1|^n$
so $|x+1|=|a+1|$
or $|x-(-1)|=|a+1|$
So the roots of $(x+1)^n-(a+1)^n$ lie on a ring of radius $|a+1|$, centered on $-1$.
 
Ok
once again
how does this help?
 
4:46 AM
And the roots of $x^n-a^n$ lie on a ring of radius $|a|$, centered on the origin.
If you draw these, you'll see they only intersect at one point (they're tangent)
 
Oooh
 
which means, $x^n-a^n$ and $(x+1)^n-(a+1)^n$ have at most one root in common
 
making an appeal to geometry are you?
 
and we already know a root they have in common, $a$
 
i know that they intersect at a
 
4:48 AM
so the important thing is, they don't have any other roots in common
 
can you prove or at least justify saying that they don't intersect elsewhere?
 
In particular, we know the conjugates of $a$ are also roots of both polynomials.
 
i believe you
 
@Typhon Sure.
 
But does geometry?
wait..
 
4:49 AM
I think the MSE answer I stole this from used a bit of stuff using complex conjugation
 
if a is real, wouldn't the conjugate be real?
 
AGGGGGGHHHHH.
 
what?
 
@Typhon That stuff wasn't complex conjugation, remember
 
I know that
im assuming that the conjugate of a real number is real...
 
4:50 AM
Oh. Then, no, the conjugates of real stuff need not be real. I provided an example earlier
 
Which was? :/
 
There are three roots of $x^3-2$
One is $\sqrt[3]2$.
 
Oh of course
thinking of second order again
 
The other two are 120 and 240 degrees away, forming an equilateral triangle.
 
@AkivaWeinberger no, the conjugate of a real number is itself
unless you mean $x^3-2$ this kind of conjugate
how do you call that? algebraic conjugate?
 
4:51 AM
I do.
Uh, I dunno, probably?
 
no clue
 
I leave you to it, DogAteMy. Night, all.
 
I've never pondered upon that
 
If I were to guess, I would call it that, yeah
 
@TedShifrin how do you call it?
 
4:51 AM
Cya
 
ok well then I'll let geometry speak for itself
not as rigorous as I prefer but sufficient
 
Google seems to suggest "Galois conjugate"
 
Hmmm
wait
 
@Typhon I can make it rigorous, if you want, using complex conjugates
 
is this Galois theory?
 
4:53 AM
I fail to find the original post
 
Only a little.
We know $|x|^2=x\bar x$, right?
So, uh, see if I can remember how to do this
 
Yup
 
@Typhon what is the original question? @AkivaWeinberger que es la cuestion original?
@AkivaWeinberger now that's the complex conjugate
 
I'm getting off the toilet. Be back in a minute or two.
 
4:54 AM
@AkivaWeinberger I see
@Zacharý the last two digits go $02 \mapsto 04 \mapsto 13 \mapsto 31 \mapsto 02 \mapsto \cdots$
 
@LeakyNun Suppose $a$ is real, and let $n$ be a positive integer. Prove that, if $a^n$ and $(a+1)^n$ are rational, then $a$ is rational as well.
 
@AkivaWeinberger well that's easy
$a = \frac{(a+1)^n}{a^n} \in \Bbb Q$
because $\Bbb Q$ is a field i.e. closed under division
well, for non-zero elements
 
…Uh, you seem to have confused base and exponent
We have $\frac{a^{n+1}}{a^n}=a$.
 
yes, indeed I have.
 
So, in any case, $|x|=|a|$ becomes, after squaring, $x\bar x=a\bar a$ (complex conjugates now!) @Typhon
 
4:58 AM
i see
 
and $|x+1|=|a+1|$, similarly becomes $(x+1)(\bar x+1)=(a+1)(\bar a+1)$
 
ok
but if they are real numbers
 
so $x\bar x+x+\bar x+1=a\bar a+a+\bar a+1$
 
x = \bar x
 

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