Mathematics - 2019-03-25 (page 1 of 3)

Lewis

12:11 AM

Can anyone solve this inverse Z-transform problem? $\frac{1 - 2z^{-1} + z^-2}{1 - z^-2}

$

$\frac{1 - 2z^{-1} + z^-2}{1 - z^-2}$

Leaky Nun

12:25 AM

@Lewis I thought inverse Z-transform isn't well-defined since $x(n) = u[n]$ and $x(n) = -u[-n-1]$ give the same Z-transform?

Lewis

@LeakyNun yeah you have to specify the ROC but this problem doesn't have one

Leaky Nun

then maybe you should use $\frac{1}{1-z^2} = 1 + z^2 + z^4 + \cdots$

which has a ROC of 1

Lewis

@LeakyNun Hmm ok

Leaky Nun

I mean Z-transform is not injective so "inverse Z-transform" doesn't make sense

wait

yeah there's also a case where ROC is the complement of a disc

Wiki assumes a causal ROC for inverse Z-transform

@Lewis what do you think

Lewis

Well, the answer is x(n) = {1, -2, 2, -2, 2 ...}

I assume it was just long division but maybe I messed it up

Leaky Nun

12:37 AM

then that's anti-causal?

Lewis

Yeah

Leaky Nun

ok...

BAYMAX

2:54 AM

any help on this?

0

Q: Exactly one of the three points $L,M,N$ lies on the triangle $ABC$, where $AL, BM, CN$ are proper Cevian lines

$AL, BM, CN$ are proper Cevian lines and are concurrent at an ideal point. To prove - Exactly one of the three points $L,M,N$ lies on the triangle $ABC$. I was thinking that from Ceval's theorem we have, $\frac{BL}{LA}.\frac{BN}{NC}.\frac{BM}{MC} =1$. And $P$ is the only ideal point(I am t...

geometry

Leaky Nun

4:44 AM

> Orthogonal Polynomials: Monic orthogonal polynomials, three term recurrence relationship, best approximation by polynomials in inner product induced norms, Chebyshev and Legendre polynomials. Fourier series.
Polynomial Interpolation: Lagrange, Newton and canonical forms of interpolating polynomial, existence and uniqueness, divided differences, error analysis, Runge phenomenon.
Mini-Max Approximation: Best approximation by polynomials in the uniform
sense, Chebyshev equioscillation theorem, optimal interpolation points zeros

this course is very algorithmic

Silent

6:08 AM

Please provide an example of field extension that is not algebraic field extension.

Silent

6:36 AM

Can $F(\alpha)=\{a_0+\alpha a_1+\ldots+\alpha^{n-1} a_{n-1}:a_0,\ldots,a_{n-1}\in F\}$ hold if $\alpha$ is transcendental over $F$, for some $n$?

@AlessandroCodenotti

Akiva Weinberger

7:13 AM

Well $1/\alpha\in F(\alpha)$ @Silent

and if $1/\alpha=p(\alpha)$ for some polynomial $p$ then $\alpha$ is a root of $xp(x)-1$

which can't happen if $\alpha$ is transcendental

Silent

@AkivaWeinberger Why did you assume $1/\alpha=p(\alpha)$?

Akiva Weinberger

Isn't that the right-hand side of your equation?

$a_0+\alpha a_1+\dotsb+\alpha^{n-1}a_{n-1}$ is $p(\alpha)$ with $p(x)=a_0+\dotsb+a_{n-1}x^{n-1}$

(Also: did you mean $F(\alpha)$ or $F[\alpha]$)

(In the latter case you'd want to use $\alpha^n$ instead of $1/\alpha$ I guess)

Silent

oh! thank you. I thought that you were talking about some irreducible polynomial that one does in the beginning of field theory classes, while proving Kronecker Theorem, that $F[x]/(p(x))$ is field ext of F

Sorry for misunderstanding.

@AkivaWeinberger I meant smallest field containing $F$ and $\alpha$.

Akiva Weinberger

OK

So that contains $1/\alpha$

If $\alpha$ is transcendental then $F[\alpha]$ and $F(\alpha)$ are infinite-dimensional vector spaces over $F$

Silent

i see

Akiva Weinberger

7:29 AM

4

Q: Digamma equation identification

I was messing around with the digamma function the other day, and I discovered this identity: $$\psi\left(\frac ab\right)=\sum_{\substack{\large\rho^b=1\\\large\rho\ne1}}(\rho^a-1)\ln(1-\bar\rho)-\gamma$$ when $0<\dfrac ab\le1$. It's unusual in that it sums over the $b$-eth roots of unity (which ...

special-functions rational-numbers polygamma

Silent

@AkivaWeinberger But doesn't that mean that there is some linear combination of basis vectors that gives $1/\alpha$? While your argument above shows that no finite combination gives $1/\alpha$

Akiva Weinberger

I completely forgot about this

I have no idea how I did that

(re: digamma thing)

@Silent $\{\alpha^n:n=0,1,2,\dots\}$ isn't a basis for $F(\alpha)$

Silent

OH!!

Akiva Weinberger

You'd need $\{\alpha^n:n=\dots,-2,-1,0,1,2,\dots\}$

Silent

wow

Akiva Weinberger

7:31 AM

Oh wait

That's not even enough actually

because we need $1/(\alpha+1)$ in it as well

so I dunno what would be an easy basis

If I had to guess: $\alpha^n$ for nonnegative $n$, plus $\alpha^n/(p(\alpha)^m)$ with $n<\deg(p^m)$ for all irreducible $p$s

'cause you can do partial fractions

Hm. Can you write $\dfrac x{(x+1)^2}$ as a sum of "simpler" rational functions? You can't, right?

Rithaniel

So, $a$ is nilpotent if there exists $n\in\mathbb{N}$ such that $a^n=0$. This is in regards to the Ring's multiplication operation, correct?

Leaky Nun

@AkivaWeinberger hint: partial fractions (i.e. Chinese Remainder Theorem)

@Rithaniel yes

Akiva Weinberger

Whereas, like, $\dfrac1{x^2-1}=\frac12\dfrac1{x-1}-\frac12\dfrac1{x+1}$, and $\dfrac x{x^2-1}=\frac12\dfrac1{x-1}+\frac12\dfrac1{x+1}$

Rithaniel

Alright, danke

Leaky Nun

@AkivaWeinberger oops I just saw you said "partial fractions"

Akiva Weinberger

7:37 AM

@LeakyNun Not if the denominator is a power of an irreducible, I'm pretty sure

Leaky Nun

what we said aren't mutually exclusive

Silent

What does $F[\alpha]$ mean by the way? I think it means set of all elements which we get by plugging in $\alpha$ in each polynomial.

Akiva Weinberger

Yeah

Smallest ring containing $F$ and $\alpha$

$\Bbb R[x]$ is the polynomial ring

Silent

So, if $\alpha$ is algebraic then $F[\alpha]=F(\alpha)$?

Akiva Weinberger

Yes

(Prove this!)

How do you know that $\dfrac1{\sqrt[3]2+\sqrt[3]3}\in\Bbb R[\sqrt[3]2+\sqrt[3]3]$?

Hint: vector spaces

Hm actually that might not be the best example

How do you know that $\dfrac1{\sqrt[3]2+1}\in\Bbb R[\sqrt[3]2]$?

Or, in general, how do you know that $\dfrac1{\alpha+1}\in\Bbb R[\alpha]$ if $\alpha$ is algebraic?

Silent

7:47 AM

working

Akiva Weinberger

I have a second small hint with more words if you want

Silent

@AkivaWeinberger Shall i take cube of $\dfrac1{\sqrt[3]2+1}$ to begi?

Akiva Weinberger

That would give you $\dfrac1{2+3\cdot2^{2/3}+3\cdot2^{1/3}+1}$ or $\dfrac1{3\sqrt[3]4+3\sqrt[3]2+3}$ if I'm not mistaken

Leaky Nun

@AkivaWeinberger this compulsory course I'm taking is as algorithmic as possible ._. help

Akiva Weinberger

And you want to escape the class in O(ln n) time

What's the course?

Leaky Nun

7:52 AM

"numerical analysis"

they spent like 9 lectures (out of 30) talking about polynomial interpolation

it's probably actually interesting

but like I need to spend time to learn this stuff lol

Akiva Weinberger

All I know about that is Lagrange interpolation

Leaky Nun

yes, they spent 9 lectures talking about it

Akiva Weinberger

Why would you need 9 lectures for it? You can describe it in a page

Leaky Nun

(interpolation is unique btw, there's only one polynomial of degree n to interpolate n+1 points, so every interpolation is "Lagrange interpolation")

Akiva Weinberger

Yeah

Proof: vector spaces

Leaky Nun

7:55 AM

@AkivaWeinberger corollary: a polynomial of degree n has at most n roots? :P

@AkivaWeinberger they talk about how to compute the coefficients and how much the error is

Akiva Weinberger

@LeakyNun Oh, that's another proof

Silent

@AkivaWeinberger I can't think of any other way!

Akiva Weinberger

@Silent Hint: give me a basis for $\Bbb R[\sqrt[3]2]$

What does multiplication by $\sqrt[3]2+1$ do to that vector space

@LeakyNun This sounds like it could be interesting but it's not the sort of thing I know a lot about

Silent

one basis is $\{1,\sqrt[3]2\}$

Akiva Weinberger

No it's not

You also need $\sqrt[3]4$

${}=\sqrt[3]2^2$

Leaky Nun

8:01 AM

@AkivaWeinberger btw there is an alternative basis of the polynomials of degree <= n given n+1 complex numbers $z_0, \cdots, z_n$: $\prod_{k \ne j} \frac{z - z_j}{z_k - z_j}$

that's Lagrange interpolation

Silent

@AkivaWeinberger how did you figure that out? By multiplying by $\sqrt[3]2+1$? Also, how do we know there is no other element in basis?

Leaky Nun

@AkivaWeinberger replace $\Bbb R$ with $\Bbb Q$ everywhere

Akiva Weinberger

Oh crap you're right @LeakyNun

$\Bbb R[\sqrt[3]2]$ is just $\Bbb R$

lol

Right so $\Bbb Q[\sqrt[3]2]$

but the deal is

Silent

oh!

Akiva Weinberger

let $\alpha$ be $\sqrt[3]2$. We know that $\alpha^3-2=0$, yeah?

Silent

8:05 AM

yes

Akiva Weinberger

If $\alpha^2$ were a combination of $1$ and $\alpha$, then we'd have $\alpha^2=A+B\alpha$

which would mean $\alpha^2-B\alpha-A=0$

which would mean that $\alpha$ is the root of a quadratic polynomial

but $\sqrt[3]2$ isn't the root of any quadratic polynomial

So we get $\{1,\alpha,\alpha^2\}$ as a basis

or $\{1,\sqrt[3]2,\sqrt[3]4\}$

Well, we still need to show that those generate everything in $\Bbb Q[\sqrt[3]2]$

Silent

right

Akiva Weinberger

$\alpha^4\in\Bbb Q[\alpha]$, right?

When $\alpha=\sqrt[3]2$, what's $\alpha^4$?

Silent

yeah

Akiva Weinberger

It's $\alpha^3\cdot\alpha=2\sqrt[3]2$

Silent

8:09 AM

i was typing :)

Akiva Weinberger

Aw, sorry

so $\alpha^4$ is in the span of $\{1,\alpha,\alpha^2\}$

Silent

yes

Akiva Weinberger

We can show inductively that $\alpha^n$ is in the span of $\{1,\alpha,\alpha^2\}$ for all $n$

Silent

yes

Akiva Weinberger

and we know that the $\alpha^n$ span $\Bbb Q[\alpha]$

Silent

8:11 AM

wow

thank you!

Akiva Weinberger

OK so now we have our vector space and we have a basis for it. $\Bbb Q[\sqrt[3]2]=\{a+b\sqrt[3]2+c\sqrt[3]4:a,b,c\in\Bbb Q\}$

Silent

yes

Akiva Weinberger

What does multiplication by $\sqrt[3]2+1$ do to our basis vectors?

Silent

it gives elements like $(a+c)+\sqrt[3]2(a+b)+\sqrt[3]4(b+c)$

Zee

any idea what X^reg means? for a topological space X

Akiva Weinberger

8:19 AM

$a+2c$ at the start

because $\sqrt[3]4\cdot\sqrt[3]2=2$ @Silent

We can make it a matrix if we want to
$$\begin{bmatrix}
1&0&2\\
1&1&0\\
0&1&1\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}a+2c\\a+b\\b+c\end{bmatrix}$$

Silent

yes

Akiva Weinberger

That's one way to think about it

Let's go with this way first

Do you know how to compute the inverse of a matrix?

Silent

yes

computing

Akiva Weinberger

Well what would we want to solve

in order to write $\dfrac1{\sqrt[3]2+1}$ as $a+b\sqrt[3]2+\sqrt[3]4$

This is the same as solving $1=(\sqrt[3]2+1)(a+b\sqrt[3]2+\sqrt[3]4)$

Yeah?

Silent

yes

Akiva Weinberger

8:25 AM

Which means it's the same as solving $1=(a+2c)+(a+b)\sqrt[3]2+(b+c)\sqrt[3]4$

Silent

yes

Akiva Weinberger

Which means it's the same as solving
$$\begin{bmatrix}
1&0&2\\
1&1&0\\
0&1&1\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}1\\0\\0\end{bmatrix}$$

Silent

Why 1 in place of a+2c only?

I mean, can we take e.g., (1/3,1/3,1/3)?

Akiva Weinberger

I need to go for a bit

Silent

ok ok. good bye

Akiva Weinberger

8:31 AM

6 mins ago, by Akiva Weinberger

Which means it's the same as solving $1=(a+2c)+(a+b)\sqrt[3]2+(b+c)\sqrt[3]4$

Think of the LHS as $1+0\sqrt[3]2+0\sqrt[3]4$

You want $a+2c=1$, $~a+b=0$, and $b+c=0$

Silent

@AkivaWeinberger so sorry. so solution is $(a,b,c)=(1/3,-1/3,1/3)$.

Rithaniel

So, I'm looking for a proof of a fairly basic fact. That $1+(-1)^nx^{n+1}=(1+x)\sum_{i=0}^n(-1)^ix^i$. I'm blanking on what exactly to search for, though.

Silent

@AkivaWeinberger, so sorry that i wasted so much of your time.

Akiva Weinberger

@Silent Not at all

So $\dfrac1{\sqrt[3]2+1}={}?$

Rithaniel

8:59 AM

(I have found the proof I was looking for. It's weird, I feel that I must have somehow skipped a class at some point where polynomial division was explored and verified.)

Akiva Weinberger

@Rithaniel $\begin{matrix}1&-x&+x^2&-x^3&\\&x&-x^2&+x^3&-x^4\end{matrix}$

Rithaniel

Ah, that's a pretty simple way to show it.

Okay, yeah, I knew it was a basic fact.

I'm gonna keep that arrangement of elements in my head for future reference.

Akiva Weinberger

This is a geometric progression @Rithaniel

$\sum(-x)^n$

Rithaniel

Yeah, related to the geometric series.

Leaky Nun

9:24 AM

well there's a better way to do it using Cayley Hamilton (@AkivaWeinberger @Silent)

Akiva Weinberger

He needs to answer $\dfrac1{\sqrt[3]2+1}={}?$ first

Leaky Nun

sure

@AkivaWeinberger Let $p$ interpolate $(z_0, f_0), \cdots, (z_n, f_n)$ and $q$ interpolate $(z_0, f_0), \cdots, (z_{n-1}, f_{n-1})$

then $p-q$ is zero at $z_0, \cdots, z_{n-1}$, so $p-q = C (z-z_0) (z-z_1) \cdots (z-z_{n-1})$

evaluate at $z_n$ to give $f_n - q(z_n) = C \prod (z_n - z_k)$

so $C = \dfrac{f_n - q(z_n)}{\prod (z_n - z_k)}$

which is also the $z^n$-coefficient of $p$

and is called the divided difference, notated $f[z_0, \cdots, z_n]$

it has a nice formula: $f[z_0, \cdots, z_n] = \dfrac{f[z_0, \cdots, z_{n-1}] - f[z_1, \cdots, z_n]}{z_0 - z_n}$

which should remind you of the first difference method

Silent

10:16 AM

@AkivaWeinberger It is $\frac{1}{3}(1+\sqrt[3]2+\sqrt[3]4)$

Sorry! $\frac{1}{3}(1-\sqrt[3]2+\sqrt[3]4)$. Can't edit now

ÍgjøgnumMeg

10:32 AM

errrm if $M$ is an f.g. free $\Bbb Z$-module, why is it also f.g. as an $\mathcal{O}_K$-module ? (ring of integers of number field $K$) I'm pretty sure this is obvious but I can't remember why! Is it literally just because $\Bbb Z \subset \mathcal{O}_K$

Silent

@LeakyNun oh. i think i spotted in artin algebra

Leaky Nun

@ÍgjøgnumMeg yes

ÍgjøgnumMeg

fair one

Akiva Weinberger

10:47 AM

@Silent Ya

Arright so I mentioned there were two ways to see this

This is the more computational way

The other is an existence proof

Silent

ok

Akiva Weinberger

Remember that our original question was just showing that $\dfrac1{\sqrt[3]2+1}$ is in $\Bbb Q[\sqrt[3]2]$

So the deal is that $\Bbb Q[\sqrt[3]2]$ is a finite-dimensional vector space (specifically, three-dimensional)

The important thing is, any injective linear map from a finite-dimensional vector space to itself is also a surjection.

(and thus a bijection)

Silent

yes

Akiva Weinberger

Now, $x\mapsto(\sqrt[3]2+1)x$ is a linear map from $\Bbb Q[\sqrt[3]2]$ to itself

and it's injective

Therefore, it's surjective

Therefore $1$ is in its image

which means there's some $x\in\Bbb Q[\sqrt[3]2]$ such that $(\sqrt[3]2+1)x=1$

Leyla Alkan

Hi. Am I wrong about this solution?

Akiva Weinberger

10:52 AM

meaning that $\dfrac1{\sqrt[3]2+1}\in\Bbb Q[\sqrt[3]2]$

QED

Isn't that nice?

And the nice thing is that it generalizes

If $\alpha$ is algebraic, then $\Bbb Q[\alpha]$ is finite-dimensional (why? what's its dimension?)

and so for any $A\in\Bbb Q[\alpha]$, we have that $x\mapsto Ax$ is an injective linear map from $\Bbb Q[\alpha]$ to itself

which means it's surjective

which means there's an $x\in\Bbb Q[\alpha]$ solving $Ax=1$

which means $1/A\in\Bbb Q[\alpha]$

(Why doesn't this work when $\alpha$ is transcendental? For example, $1/\pi\notin\Bbb Q[\pi]$. Where does this proof go wrong in that case?)

Silent

its really nice. thinking about your last question

Akiva Weinberger

@LeylaAlkan If it's symmetric, wouldn't $t_{[321]}$ be zero because they'd all cancel out?

Leaky Nun

@AkivaWeinberger $S_4/V_4 = ?$

ok $S_4/V_4 = S_3$

why?

Leyla Alkan

@AkivaWeinberger yes

Akiva Weinberger

@LeakyNun 'cause you can write everything in $S_4/V_4$ as $[(1xyz)]$

Leaky Nun

11:01 AM

ah because $V_4$ is transitive

it's quite hard to imagine a surjection $S_4 \to S_3$

Akiva Weinberger

Or, rather, $\left[\begin{pmatrix}1&x\\y&z\end{pmatrix}\right]$

Leaky Nun

is there a geometric meaning to all this

Akiva Weinberger

because I view $V_4$ as symmetries of a rectangle

(acting on the corners of the rectangle)

Leyla Alkan

Isn't $(3,0)$ tensor symmetric? @AkivaWeinberger

Rithaniel

Question: What does the degree of an algebraic number refer to again?

Akiva Weinberger

11:03 AM

@LeylaAlkan I don't actually know much about tensors

Sorry

@Rithaniel Degree of its minimal polynomial

Leaky Nun

@Rithaniel the degree of its minimal.... ok

Akiva Weinberger

(the smallest polynomial of which it is a root)

Hah, snipe

@LeakyNun I also kinda sniped your "geometric meaning" question with the rectangle thing

but it's clear with $S_4$ being the symmetries of a cube as well

Silent

@AkivaWeinberger In that case it is not finite dimensional as you told me earlier today. Hence injective does not imply surjective.

Akiva Weinberger

Ya @Silent

Color the faces of a cube in three colors so that opposite sides get colored the same @LeakyNun

Rithaniel

Danke

Akiva Weinberger

11:04 AM

Symmetries of the cube permute the colors

Leaky Nun

@AkivaWeinberger nice

Silent

Thank you sooo much for all this help akiva

Akiva Weinberger

And the nontrivial elements of $V_4$ corresponds to 180 degree rotations about some axis @LeakyNun

@Silent No problem

@LeakyNun Do you know how to do a Rubik's cube?

Rithaniel

So, $[\mathbb{Q}[\alpha ]:\mathbb{Q}]=n$ where $n$ is the degree of $\alpha$. Is this correct?

Akiva Weinberger

There's a simpler version where you only allow yourself to make 180 turns of the faces

and try to solve it with those as well

Simpler 'cause there's far fewer combinations it can end up in

Every face can have at most two colors on it

@Rithaniel Yeah

Rithaniel

11:06 AM

Excellent

Akiva Weinberger

Also, here's a puzzle on the Rubik's cube that feels like it should be simpler than the full thing but isn't

Only allow yourself to turn the top and right faces

Mix it up with those and solve it with those

There's this 2x2x3 section of the cube that stays constant when you do it like that

and it feels like it should be easier, but it isn't

in my experience at least

Leaky Nun

11:48 AM

@AkivaWeinberger yeah

@AkivaWeinberger why would Tchebyshev polynomials be related to minimizing $\sup \prod |(x-x_j)|$

Akiva Weinberger

Those are $\cos(n\arccos x)$, yeah?

Leaky Nun

given $n$ find $x_0, \cdots, x_n \in [-1, 1]$ minimizing $\displaystyle \sup_{-1 \le x \le 1} \prod_{j=0}^n |x - x_j|$

Akiva Weinberger

Or something similar

Leaky Nun

yes

Akiva Weinberger

Hm so the log derivative is $\sum\frac1{x-x_j}$, yeah?

'cause the derivative of $\ln|x|$ is $\frac1x$, not $\frac1{|x|}$

Leaky Nun

11:57 AM

right

so we're looking at the.. harmonic mean of $x_j$?

maybe not

Akiva Weinberger

No, it's not translationally invariant

I dunno if that's the right term but it would make sense

Leaky Nun

anyway so $\sum \frac1{x^\ast-x_j} = 0$

(or $x^\ast = \pm 1$)

Akiva Weinberger

So you gave a hint with the Chebyshev there

Problem is I don't know what to do with that

What's $(x-\rho)(x-\bar\rho)$ where $\rho$ is a root of unity

$(x^2+1-2\cos\theta)$

Leaky Nun

right

brb digging up notes for Legendre transform

Semiclassical

Chebyshev of first or second kind?

Rithaniel

12:07 PM

Question: Are there any finite rings with a right identity but no left identity?

Semiclassical

Probably first kind if you’re doing minimax

Leaky Nun

@Semiclassical you probably know how to do this :P

13 mins ago, by Leaky Nun

given $n$ find $x_0, \cdots, x_n \in [-1, 1]$ minimizing $\displaystyle \sup_{-1 \le x \le 1} \prod_{j=0}^n |x - x_j|$

Semiclassical

Probably

It presumably hinges on Chebyshev polynomials as orthogonal

Akiva Weinberger

If it were the sum of the squares we'd want $x$ to be at their mean I think

Semiclassical

I don’t remember what the inner product for T_n is tho

Leaky Nun

12:08 PM

same as L^2

Akiva Weinberger

@Rithaniel Are there any infinite rings like that?

Semiclassical

I guess one should be able to deduce it from $\int_0^\pi \cos(m t)\cos(n t)\,dt$ for non-negative integers n,m

Leaky Nun

how does orthogonality help anyway?

@AkivaWeinberger there must be a 3b1b-style answer to this question lol

Semiclassical

good question

Leaky Nun

unfortunately I am not 3b1b

Akiva Weinberger

12:12 PM

@LeakyNun The gold standard

Rithaniel

Well I hope so, because one of my assignments asks for an example of a ring with a right identity but no left identity.

Semiclassical

My initial impulse is to look at the log of the objective function

Akiva Weinberger

Is $\{a,b\}$ with $a^2=a$, $b^2=b$, $ab=a$, and $ba=b$ associative?

In this case, both $a$ and $b$ are right identities

Semiclassical

Looking at the comments on this question: math.stackexchange.com/q/2710542/137524

Akiva Weinberger

And then we can make our ring $\{0,a,b,a+b\}$ (characteristic $2$) @Rithaniel

That's finite

Semiclassical

12:18 PM

Seems like the idea of looking for rings with more than one right identity is correct

Rithaniel

Indeed, I saw some stuff about a unique right identity automatically being a left identity.

Akiva Weinberger

Ah that's cool

If $e$ is a unique right identity then $b(ea-a)=bea-ba=0$

Wait no

$b(ea-a+e)=bea-ba+b=b$

Semiclassical

@LeakyNun my brain is not awake yet but I’ll note that you can instead consider the objective function to be a product of squares

Akiva Weinberger

so $ea-a+e$ is another right identity so $ea-a+e=e$ so $ea=a$ so $e$ is a left identity

This is just the proof from the link

Rithaniel

I think that's how it goes, yeah. The basic of ring theory are wild.

Akiva Weinberger

12:23 PM

I thought I could simplify it

I could not

Semiclassical

The x which maximizes that (for a particular choice of xi’s) will also maximize the original product

At which point I can maybe imagine how Chebyshev will enter? Not sure until I have a chance to write something down tho

Akiva Weinberger

@Rithaniel

Multiplication table

Semiclassical

Something along the lines of expressing $\prod_i (x-x_i)$ as a sum of Chebyshev polys

Leaky Nun

@Semiclassical wat

Akiva Weinberger

(Too lazy to figure out how to typeset that)

Thing on the left times thing on the top

Dunno if that's the usual convention but whatever

Notice that $a+b$ is an annihilator

Semiclassical

12:27 PM

Hrm. Not sure what I said actually helps tho

Akiva Weinberger

If you have two right identities then their difference must be an annihilator

Leaky Nun

@AkivaWeinberger just tell him it's the monoid ring

Akiva Weinberger

In fact I think that "unique right identity" is equivalent to "no right annihilators"

Leaky Nun

@Semiclassical how on earth is prod (x-x_i) a sum of Tchebyshev polynomials...

Akiva Weinberger

and then $ea-a$ is a right annihilator

Leaky Nun

12:28 PM

also they're only orthogonal in L^2 not L^infty

Akiva Weinberger

which means $ea-a=0$ so $ea=a$

Semiclassical

because it’s a polynomial

Akiva Weinberger

And that's a better proof that a unique right identity is a left identity

Add a concept but lose the annoying $ea-a+e$ thing

@Semiclassical Can we say anything about $\sup (f+g)$ given $\sup f$ and $\sup g$

Semiclassical

I doubt it

Akiva Weinberger

Well $\sup(f+g)\le\sup f+\sup g$ isn't it

But that's the only thing we can say

Semiclassical

12:33 PM

Yeah

ÍgjøgnumMeg

math.stackexchange.com/questions/1718702/… Brilliant

Semiclassical

The issue with my line of thinking is not “can you expand $\prod_{i=1}^n (x-x_i)$ in Chebyshev polynomials.” That’s trivial, since you can express the monomials 1,x,x^2,...x^n in terms of Chebyshevs

The problem is why it’s worth doing in the first place :/

Leaky Nun

@Semiclassical so what's your experience with minimax?

Akiva Weinberger

Are the Chebyshev polynomials monic?

Semiclassical

Very thin tbh

I’m aware of it but have not dealt with it

Leaky Nun

12:36 PM

@AkivaWeinberger no, the leading coefficient is n

Semiclassical

2^n

No, that’s not right either is it

Akiva Weinberger

How does $\int fg$ relate to $\sup f$ and $\sup g$

Semiclassical

For n>0, the leading coefficient is 2^(n-1)

Akiva Weinberger

$\frac12\int_{-1}^1fg~dx\le\sup|f|\sup|g|$, yeah?

Semiclassical

I don’t think that’s going to help much here

I guess my thought right now would be to take a few examples of small n

Akiva Weinberger

12:42 PM

No I think this is good

When is that an equality

Oh never mind

Also we need some way to keep the zeroes in $[-1,1]$

Semiclassical

Lol, one simple way

Akiva Weinberger

'cause otherwise we could just do $x^2-c$ for higher and higher $c$ or something and make $\sup_{[-1,1]}$ go to minus infinity

Semiclassical

Let $x=\cos t$ and $x_i=\cos t_i$ @AkivaWeinberger

b_jonas

@Semiclassical And there's in fact a funny combinatorial interpretation for the coefficient matrix when you do that.

For expressing the monomials that is.

Semiclassical

Sounds right

b_jonas

12:53 PM

The Chebishev polynomial U_n is defined by U_n(cos theta) = sin((n+1)theta) / sin(theta).

Semiclassical

That’s chebyshev of the second kind tho

b_jonas

Yes, that's how the funny path counting stuff works well.

Semiclassical

Also, it’s the quotient of those not the product

b_jonas

oh yeah

fixed

Now if x and y are positive integers, then let c_{y+x, y-x} be the number of paths from (0,0) to (x,y) where each edge of the path does a step of either (0,1) or (1,0), and the path never goes to a point (x',y') above the diagonal where x'<y'.

And if k and l have different parity then c_{k,l}=0.

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