12:11 AM
Can anyone solve this inverse Z-transform problem? $\frac{1 - 2z^{-1} + z^-2}{1 - z^-2}$
$\frac{1 - 2z^{-1} + z^-2}{1 - z^-2}$

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?

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

then maybe you should use $\frac{1}{1-z^2} = 1 + z^2 + z^4 + \cdots$
which has a ROC of 1

@LeakyNun Hmm ok

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

Well, the answer is x(n) = {1, -2, 2, -2, 2 ...}
I assume it was just long division but maybe I messed it up

12:37 AM
then that's anti-causal?

Yeah

ok...

2 hours later…
2:54 AM
any help on this?
0

$AL, BM, CN$ 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...

2 hours later…
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

1 hour later…
6:08 AM
Please provide an example of field extension that is not algebraic field extension.

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

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

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

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)

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 $\alpha$.

OK
So that contains $1/\alpha$
If $\alpha$ is transcendental then $F[\alpha]$ and $F(\alpha)$ are infinite-dimensional vector spaces over $F$

i see

7:29 AM
4

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 ...

@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 $1/\alpha$

I have no idea how I did that
(re: digamma thing)
@Silent $\{\alpha^n:n=0,1,2,\dots\}$ isn't a basis for $F(\alpha)$

OH!!

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

wow

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?

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?

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

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

Alright, danke

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

7:37 AM
@LeakyNun Not if the denominator is a power of an irreducible, I'm pretty sure

what we said aren't mutually exclusive

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.

Yeah
Smallest ring containing $F$ and $\alpha$
$\Bbb R[x]$ is the polynomial ring

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

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?

7:47 AM
working

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

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

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

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

And you want to escape the class in O(ln n) time
What's the course?

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

All I know about that is Lagrange interpolation

yes, they spent 9 lectures talking about it

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

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

Yeah
Proof: vector spaces

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

@LeakyNun Oh, that's another proof

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

@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

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

No it's not
You also need $\sqrt[3]4$
${}=\sqrt[3]2^2$

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

@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?

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

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

oh!

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

8:05 AM
yes

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