This is an over my level assignmet i got:
Let $\mu$ denote the Lebesgue (of arc-length) measure on the unit circle $\mathbb{T}:=\{|z|=1\}$. Prove that for the measere
$dv(z)=|1-z|^{2}d\mu$
the monic polynomial of degree n orthogonal with respect to $v$ is
$P_{n}(z;v)=\frac{1}{n+1}\sum_{j=0}^{n}(j+1)z^{j}.$
Analyze the location and the behavior of the zeros of $P_{n}$.
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Here's what I've gotten so far:
$d\mu(z)=|z|=ds$ on $\mathbb{T}$, so:
$dz=ie^{i\theta}d\theta=izd\theta$ $\implies$ $ds=\frac{dz}{iz}$, thus
$dv(z)=|1-z|^{2}d\mu$ $\implies$ $ds=\frac{|1-z|^{2}dz}{iz}$.
Now, the entries for the Gram matrix are (having in consideration we are working on the unit circle):
$<z^{j},z^{k}>=\int_{\mathbb{T}}\overline{z^{j}}z^{k}dv(z)=\int_{\mathbb{T}}|1-z|^{2}\overline{z^{j}}z^{k}\frac{dz}{iz}=\frac{1}{i}\int_{\mathbb{T}}(1-z)\overline{(1-z)}z^{k-j-1}dz=$
$=-\frac{1}{i}\int_{\mathbb{T}}(2+z+\overline{z})z^{k-j-1}dz=-\frac{1}{i}\left(2\int_{\mathbb{T}}z^{k-j-1}dz+\int_{\mathbb{T}}z^{k-j}dz+\int_{\mathbb{T}}z^{k-j-2}dz\right)$.
So the integral equals:
- $4\pi$ if $k=j$,
- $2\pi$ if $k=j+1$,
- $2\pi$ if $k=j-1$,
- $0$ in other case
but this is not leading me to the answear. Its going towars something like:
$P_{n}(z;v)=\sqrt{\frac{\Delta_{n-1}}{\Delta_{n}}}z^{n}$, where $\Delta_{n}$ is the determinant of the $n\times n$ Gramm sub-matrix.
which is not the desired.
......
Here are some updates: the Gram matrix this way has $4\pi$ on the diagonal and $2\pi$ in the upper and lower diagonal, the determinant can be calculated by some very serious double induction, but its not leading to the solution.
After this I tried just prooving that the solution proposed is orthogonal in $\mathbb{C}$ with $dv$, by calculating $, which comes down to this integral:
$<p_{a}(z),<p_{b}(z)>=\frac{N}{K}\int_{\mathbb{T}}\sum_{}z^{terms}dz=\frac{N}{K}\sum_{}\int_{\mathbb{T}}z^{terms}dz=\frac{N}{K}\sum_{}\int_{\mathbb{T}}z^{-1}dz$ = I.
if $p_{a}(z),p_{b}(z)$ are indeed orthogonal, then I must be null, but its not clear, so I proceeded to calculate $<p_{1}(z),<p_{2}(z)>$:
$p_{1}(z)=\frac{1}{2}(1+2z)$
$p_{2}(z)=\frac{1}{3}(1+2z+3z^{2})$
$<p_{1}(z),<p_{2}(z)>=\int_{\mathbb{T}}\overline{p_{1}(z)}p_{2}(z)|1-z|^{2}dz=$
$=\frac{1}{6}\int_{\mathbb{T}}\overline{(1+2z)}(1+2z+3z^{2})\overline{(1-z)}(1-z)dz=\frac{-1}{6}\int_{\mathbb{T}}z^{-1}dz=-\frac{\pi}{3}\neq0$
This means the exercise was not correct. Of course, this is never a good sign, so if you find any mistake in my work, please let me know.