I have the following problem:
In our lecture, we discussed the Legendre polynomials twice.
At first we took the standard basis of $\mathbb{R^n}$ to do the following:
$q_{0}=1, q_{1}=x$ and for $k>1:$
$q_{k+1}(x)= x^{k+1} - \sum_{l=0}^k\gamma_{kl}q_{l}$, where
$\gamma_{kl} = \frac{\int_{-1}^{1} x^{k+1}q_{l}(x)dx}{\int_{-1}^{1} q_{l}^2(x)dx}$
Then, instead of taking $x^{k+1}$, we constructed the Legendre polynomials by taking $x\cdot p_{k}$ like this:
$p_{0}=1, p_{1}=x$ and for $k>1:$
$p_{k+1}(x)= x\cdot p_{k}(x) - \sum_{l=0}^k\beta_{kl}p_{l}$, where
$\beta_{kl} = \frac{\int_{-1}^{1} x\cdot p_{k}(x)p_{l}(x)dx}{\int_{-1}^{1} p_{l}^2(x)dx}$
I know they should give the same polynomials, but I don't see why.
What I have done so far:
In the second algorithm, I can see that
$\beta_{kl} = \frac{\int_{-1}^{1} x\cdot p_{k}(x)p_{l}(x)dx}{\int_{-1}^{1} p_{l}^2(x)dx}=0$ for $l<k-1$, because
$\int_{-1}^{1} x\cdot p_{k}(x)p_{l}(x)dx =\int_{-1}^{1} p_{k}(x)x\cdot p_{l}(x)dx$ and $x\cdot p_{l}(x)$
is a linear combination of the first $l+1$ constructed polynomials and thus orthogonal to $p_{k}(x)$.
Also $\beta_{kk} = \frac{\int_{-1}^{1} x\cdot p_{k}^2(x)dx}{\int_{-1}^{1} p_{k}^2(x)dx}=0$,
because I have shown earlier, that $p_{k}(x)$ either an even or an odd function, and so $x\cdot p_{k}^2(x)$ is an odd function and zero in zero and so the integral is zero.
With the requirement that $p_{k}(1)=1 \forall k$, I computed $\beta_{k,k-1}= \frac{2k-1}{2k+1}$ and with this I got a recursion formula for the polynomials of the second algorithm, wich is
$(2k+1)\cdot p_{k+1}(x)= (2k+1)\cdot x\cdot p_{k}(x) -(2k-1)\cdot p_{k-1}(x)$.
But this is not the standard recursion formula for the Legendre polynomials and none of my work does help me with my actual question..
Does anyone see how to prove that both algorithms give the same polynomials?
It would be really nice, if you could help me with this!
Suppose you have $2$ monic polynomials of degree $n$, $p_n(x)$ and $q_n(x)$ such that $$\int_{-1}^1p_n(x)r_{n-1}(x)dx=\int_{-1}^1q_n(x)r_{n-1}(x)dx=0$$ For any polynomial $r_{n-1}(x)$ of degree at most $n-1$. Then $p_n(x)-q_n(x)$ is a polynomial of degree at most $n-1$, the leading $x^n$ term canceling out in subtraction, so $$\int_{-1}^1p_n(x)\left(p_n(x)-q_n(x)\right)dx=\int_{-1}^1q_n(x)\left(p_n(x)-q_n(x)\right)dx=0$$ So $$\int_{-1}^1\left(p_n(x)-q_n(x)\right)\left(p_n(x)-q_n(x)\right)dx=0$$ Since $\left(p_n(x)-q_n(x)\right)^2\ge0$ it follows that $p_n(x)-q_n(x)=0$ identically.