Let $P_{0}, P_{1}$ ... be the orthogonal polynomial for weight $w_{1}$ on $[0,1]$ and $Q_{0}, Q_{1}$...be the orthogonal polynomial for weight $w_{2}$ on $[1,2]$.
The inner product is define as usual $
The question is: suppose that $2w_{1}=w_{2}(2-x)$, find the relation between $P_{j}$ and $Q_{j}$.
Note: $P_j$ is a polynomial with degree j.
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I am asking this question for two reasons:
I am really not sure if it was actually asking us for the relation between $P_{j}$ and $Q_{j}$. Here is the original handwriting of my professor (I knew I should have asked him earlier)
Normally that little subscript that looks like a reserved c means s in his handwriting. What do you guys think is the something that he would be reasonably asking according to his notation?
Assuming that it is asking for the relation between $P_j$ and $Q_j$, then I thought the solution should be like this.
Since they are both orthogonal sets, we have
$\int_{0}^{1}P_{i}(x)P_{j}(x)\omega_{1}(x)dx=0$
$\int_{1}^{2}Q_{i}(x)Q_{j}(x)\omega_{2}(x)dx=0$
where $i\neq j.$
Using the relation $2\omega_{1}(x)=\omega_{2}(2-x)$
$\implies\omega_{1}(x)=\frac{1}{2}\omega(2-x)$ we have
$\int_{0}^{1}P_{i}(x)P_{j}(x)\frac{1}{2}\omega_{2}(2-x)dx=0$
Let $u=2-x$, then we have $x=2-u$, $dx=-du$
$\int_{x=0}^{x=1}P_{i}(2-u)P_{j}(2-u)\omega_{2}(u)dx=0$
$\int_{2}^{1}P_{i}(2-u)P_{j}(2-u)\omega_{2}(u)(-du)=0$
$\int_{1}^{2}P_{i}(2-u)P_{j}(2-u)\omega_{2}(u)du=0$
Since $\int_{1}^{2}Q_{i}(u)Q_{j}(u)\omega_{2}(x)dx=0$
we have $Q_{j}(u)=P_{j}(2-u)$
That last line of logic is where I felt skeptical. It seems that it requires that:
On a given domain and a fixed weight, the set of orthogonal polynomial on it is unique.
Thinking about this in detail, I realized that seems to be only true if we require that:
In this set of polynomials , the polynomial of each degree must appear and that for each degree, there is only one polynomial.
According to Wikipedia's definition of Orthogonal Polynomial, this seems to be implied, but not directly stated. Is this condition really true? Thanks.