Mathematics - 2021-10-12 (page 2 of 2)

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Semiclassical

23:02

oh. here's something geometric i saw lately which is weird

consider two different rotations on the sphere. First one I'll call R1: rotate the sphere $\pi/4$ around the z-direction

second is R2: rotate the sphere $\pi/4$ around the x+z direction

claim: doing R1, R2, R1, R2 is equivalent to a rotation around some axis by an angle which is not a rational multiple of $\pi$

i guess one way to understand it it is to look at the eigenvalues of R1.R2, which seem to be $1,(1\pm 3i\sqrt{7})/8$

if R1.R2 was a rotation by a rational multiple of pi, then (R1.R2)^n would eventually be the identity matrix

so $(1\pm 3i\sqrt{7})/8$ would have to be a root of unity

i don't know how you prove that it isn't tho

Leaky Nun

23:19

@Semiclassical just use quaternions lol

this reminds me of the proof of the Banach--Tarski paradox using sphere rotations

@Semiclassical I think the easiest way to prove it is to use matrices

so R1 is $\begin{pmatrix} 0&1&0 \\ -1&0&0 \\ 0&0&1 \end{pmatrix}$?

I don't understand what "x+z direction" means

Joe Shmo

Ted is there a mean value inequality on integrals for vector valued functions?

Leaky Nun

@JoeShmo is Mean value theorem for vector-valued functions what you seek?

or does it have to be integrals

Joe Shmo

say $|f| \le M$, and I need to bound the following integral - $|\int f(X_n, Y_n) - f(X_n, c)|$ screams bound me by an MVT esque inequality

no I need integrals

$X_n, Y_n$ real RVs FWIW

Leaky Nun

try substituting $f$ in the statement by an integral and see what happens maybe

Ted Shifrin

There’s an obvious thing, yes, for integrating over a region of finite volume.

Joe Shmo

23:25

yeah mvt for integrals should follow from the regular mvt I think

Ted Shifrin

Not really.

Leaky Nun

@JoeShmo no, this screams bound me by triangle inequality

Ted Shifrin

This not a single-variable FTC setting.

Leaky Nun

idk if that's what you call it

or if it even has a name

Joe Shmo

I need an upper bound @LeakyNun not a lower bound

Leaky Nun

23:26

oh right nvm

i'll leave it to ted then

leslie townes

joe, do you want something like theorem 9.19 in rudin?

Ted Shifrin

It’s the usual $\sup$ norm integrand times volume of region.

Semiclassical

@LeakyNun was being sloppy---should be $(1,0,1)/\sqrt{2}$ direction

Leaky Nun

wait triangle inequality gives you an upper bound lol

@Semiclassical oh boy

can you just give me the matrix :P

Semiclassical

lol, sure

that's how i got the eigenvalues anyways

Joe Shmo

23:28

not the bound that I need @LeakyNun

you mean triangle inequality on the integral. not it..

Leaky Nun

what is the bound you need

Joe Shmo

what ted said is probably what I need

Semiclassical

$R_1=\begin{pmatrix} 1/\sqrt{2} & -1/\sqrt{2} & 0 \\1/\sqrt{2} & 1/\sqrt{2} & 0 \\ 0 & 0 & 1\end{pmatrix}$

Leaky Nun

oh boy

Semiclassical

for rotation around the z-axis

Leaky Nun

23:30

you mean R2

wait what

Joe Shmo

the other clever way to do this is probably to approximate $f$ (continuous) by lipschitz functions

Leaky Nun

oh i thought you said pi/2, nvm

Semiclassical

R2 is worse:

Joe Shmo

and I'm figuring unnecessary, and I should be able to get away with something similar, simpler.

Leaky Nun

this sounds like the "an engineer needed the Green's function for a circle so he asked for the Green's function for regular n-gons"

Semiclassical

23:33

$R_2=\begin{pmatrix} 1/2 + 1/(2 \sqrt{2}) & -1/2& 1/2 - 1/(2 \sqrt{2})\\ 1/2& 1/\sqrt{ 2}& -1/2\\ 1/2 - 1/(2 \sqrt{2}) & 1/2 & 1/2 + 1/(2 \sqrt{2})\end{pmatrix}$

yeeeah

Leaky Nun

and you found the eigenvalues of $R_1 R_2$?

Semiclassical

yeah. more accurately, mathematica did

Leaky Nun

and what are they

Semiclassical

$1,(1\pm i3\sqrt{7})/8$

Leaky Nun

(you don't need to do $R_1 R_2$ the second time)

Semiclassical

23:34

yeah, i figured that much

the eigenvalue of 1 is easy: a rotation has an axis, after all

Leaky Nun

this sounds like those phenomena that 3B1B can tell you why but i can't

Semiclassical

evidently the story here is that no power of $(1\pm i 3\sqrt{7})/8$ is equal of $1$

Ted Shifrin

@Leaky What country are you in these days?

Leaky Nun

Joe Shmo

so Ted $|∫f(X_n,Y_n)−f(X_n,c)| \le 2 \|f\|_\infty \cdot \|(X_n, Y_n) - (X_n, c)\|_2 = 2M |Y_n - c|$?

Semiclassical

23:36

i.e., $(1+i3\sqrt{7})/8$ is not a root of unity

i wonder how you test this kind of thing more generally. say, take $(1+i\sqrt{n^2-1})/n$

Leaky Nun

well you can only get quadratics

Semiclassical

for what $n$, if any, is that a primitive root of unity?

Leaky Nun

and the only ones that are roots of unity are 1,2,3,4,6th roots of unity

Joe Shmo

what are you up to these days leaky?

Ted Shifrin

No, @JoeShmo. Where is the integral? You’re still mixing in standard MVT.

Semiclassical

23:38

@LeakyNun can you elaborate on that? including the quadratics bit

Leaky Nun

so the number you gave me is a root of $x^2 - \frac14 x + 1 = 0$

robjohn

@Semiclassical $n=2$

Leaky Nun

@Semiclassical you always get a quadratic because it's $\chi(A) / (x-1)$

where $\chi(A)$ is the characteristic polynomial of $A$

Ted Shifrin

@LeakyNun Cool. I wondered.

Leaky Nun

remember that you find eigenvalues by finding the roots of the characteristic polynomial

Semiclassical

23:39

right

Leaky Nun

oh no but you can't guarantee that the coefficients are in Q lol nvm

Semiclassical

@robjohn i guess that is the obvious case

is it evident that that's the only one?

Leaky Nun

I wonder how come the matrix you gave me has rational characteristic equation

@Semiclassical it isn't

Semiclassical

@LeakyNun that is rather wacky, yes

Leaky Nun

but basically numbers of the form $(1+i\sqrt{n^2-1})/n$ are solutions to quadratic equations with rational coefficients

Semiclassical

23:41

right

Leaky Nun

and there only roots of unity that are also solutions to quadratic equations with rational coefficients are the 1,2,3,4,6-th roots of unity

Semiclassical

$(nx-1)^2=(n^2-1)$

Leaky Nun

the only proof I know uses Galois theory

Semiclassical

yikes

Leaky Nun

basically, $\Bbb Q(\zeta_n)$ has degree $\varphi(n)$ over $\Bbb Q$

where $\varphi$ is Euler's totient function, i.e. the number of integers between $1$ and $n$ that are coprime to $n$

i.e. the minimum degree of a polynomial with Q coefficients that has $\zeta_n$ as a root is $\varphi(n)$

and you can imagine that if $n$ is not 1,2,3,4,6 then there would be more than 2 numbers coprime to $n$

Semiclassical

23:46

a slightly nicer version of this (derived from a related but different formulation) is $1/4(3+i\sqrt{7})$

i think that shouldn't be a root of unit either

Leaky Nun

yeah it wouldn't be

Semiclassical

but it sounds like showing it is not going to be obvious, at least not an elementary level

robjohn

@Semiclassical no, its just the one I saw. Niven's Theorem might be useful.

Joe Shmo

@leslietownes yeah that's what I'm getting at, but in the other direction (the function to be differentiated is the integral but I forget exactly how that works)

Leaky Nun

@Semiclassical yeah idk this feels like you're just multiplying two random roots of $A^8 = I$ together

Semiclassical

23:52

@robjohn hmm, I see what you mean

I think Niven’s theorem does it, in fact

Rational multiple of pi = rational number of degrees

And one of the components of our point on the unit circle is rational

So Nivens rules it out