Construct the sequence $\{p_n\}_{n\in \mathbb{N}_0}$ by applying the Gram-Schmidt procedure on $\{x^n\}_{n\in \mathbb{N}_0} \in L^2([0,1])$ whitout normalizing so that they are monic ($p_0=1$ , $p_1=x-1/2$ , $p_3=x^2-x+1/6$ , ...).
Prove that there exists sequences $a_n$, $b_n$ and $c_n$ such that $\forall n\ge 1$ $$p_{n+1} = \big(a_nx+b_n \big)p_n - c_np_{n-1}$$
Observation: we would expect $a_n=1$ for all $n$ since $p_{n+1}$ must be monic (this means that the coefficient of the largest exponent term is $1$).
Could you please help with this?
I was trying to prove it by induction but was not getting anywhere...
I managed to come up with this argument
From the construction of our $p_i$ we can deduce that every polynomial of degree $m$ ($a_mx^m + \mbox{ ... } + a_1x + a_0$) is linear combination of the polynomials $p_0$, $p_1$, ... ,$p_m$, in fact $\forall k\in \{ 0$, $1$,... ,$m\}$ $$x^k= p_k(x) + \sum_{j=0}^{k-1}\frac{<x^k,p_j>}{<p_j,p_j>}p_j(x) ;$$ in particular $xp_m$ is linear combination of $p_0$, $p_1$, ... , $p_m$, $p_{m+1}$.
So, since $\langle p_j,p_i \rangle = 0$ when $i\not = j$, let $n$ e $m$ be two positive integers such that $m+1<n$, then $$\langle xp_n, p_m \rangle = \int_0^1 xp_n(x) \cdot p_m(x)dx = \int_0^1 p_n \cdot xp_m dx = \langle p_n,xp_m \rangle = 0 ,$$ which implies that $xp_n$ is linear combination only of $p_{n+1}$, $p_n$ e $p_{n-1}$. Let $A_n$, $B_n$ and $C_n$ be 3 constants (depending on a fixed $n$) such that $$xp_n = A_np_{n+1} + B_np_n + C_np_{n-1};$$ we can observe that $A_n \not = 0$, otherwise $xp_n$ would not have a term of degree $n+1$, so $$p_{n+1} = \Big( \frac{1}{A_n}x - \frac{B_n}{A_n} \Big)p_n - \frac{C_n}{A_n}p_{n-1}.$$ This is exactly what we wanted since we can choose $a_n= \frac{1}{A_n}$, $b_n=-\frac{B_n}{A_n}$ and $c_n=\frac{C_n}{A_n}$.