Show that the recurrence relation

$$\begin{aligned} p_{0}(x) &=1 \\ p_{n+1}(x) &=q_{n+1}(x)-\sum_{k=0}^{n} \frac{\left\langle q_{n+1}, p_{k}\right\rangle}{\left\langle p_{k}, p_{k}\right\rangle} p_{k}(x) \end{aligned}$$

where $\langle\cdot, \cdot\rangle$ is an inner product on real polynomials, produces a sequence of orthogonal, monic, real polynomials $p_{n}(x)$ of degree exactly $n$ of the real variable $x$, provided that $q_{n}$ is a monic, real polynomial of degree exactly $n$.

Show that the choice $q_{n+1}(x)=x p_{n}(x)$ leads to a three-term recurrence relation of the form

$$\begin{aligned} p_{0}(x) &=1 \\ p_{1}(x) &=x-\alpha_{0} \\ p_{n+1}(x) &=\left(x-\alpha_{n}\right) p_{n}(x)-\beta_{n} p_{n-1}(x) \end{aligned}$$

where $\alpha_{n}$ and $\beta_{n}$ are constants that should be determined in terms of the inner products $\left\langle p_{n}, p_{n}\right\rangle,\left\langle p_{n-1}, p_{n-1}\right\rangle$ and $\left\langle p_{n}, x p_{n}\right\rangle$.

Use this recurrence relation to find the first four monic Legendre polynomials, which correspond to the inner product defined by

$$\langle p, q\rangle \equiv \int_{-1}^{1} p(x) q(x) d x$$