If $$p_n(x) = A_n (x-x_1)\cdots(x-x_n) = A_nx^n + B_nx^{n-1} + \dots$$ is an orthogonal polynomial for some scalar product $\langle \cdot, \cdot \rangle$, and we note $\gamma_n = \langle p_n, p_n \rangle$. Then the sequence of orthogonal polynomials $(p_n)_{n \in \mathbb N}$ satisfies three term reccursion $$p_{n+1}(x)=(a_nx + b_n)p_n(x) - c_np_{n-1}(x)$$ where $a_n = \frac{A_{n+1}}{A_n},\quad b_n = a_n\left(\frac{B_{n+1}}{A_{n+1}} - \frac{B_n}{A_n}\right), \quad c_n = \frac{A_{n+1}A_{n-1}}{A_n^2}\frac{\gamma_n}{\gamma_{n-1}}$
My question is, since this reccursion depends on both $A_{n+1}$ and $B_{n+1}$, which were not calculated before, how do we calculate $p_{n+1}$. Do we just fix some arbitrary $A_{n+1}$ and $B_{n+1}$ and then calculate it or what?
For example, if $\langle f,g \rangle = \int_0^1 g(x)f(x) dx$ and we set $p_0(x) = 1, p_1(x) = x - \frac{1}{2}$, if we set both $A_2 = 1$ and $B_2 = 0$, with some calculation, we get $p_2(x) = x^2 - \frac{1}{3}$, which is orthogonal to $p_0$, but not $p_1$.
The construction is most easily understood, if we consider monic orthogonal polynomials $\pi_n,$ i.e. with highest coefficient $1.$ Then, the three-term relation is $$\pi_{n+1}(x)=(x-\alpha_n)\,\pi_n(x)-\beta_n\,\pi_{n-1}(x).$$ The skalar product with $\pi_n$ gives (using orthogonality) $$\langle(x-\alpha_n)\,\pi_n(x), \pi_n(x)\rangle=0,$$ i.e. $$\alpha_n=\frac{\langle x\,\pi_n(x), \pi_n(x)\rangle}{\langle\pi_n(x), \pi_n(x)\rangle},$$ and the skalar product with $\pi_{n-1},$ using $$\langle(x-\alpha_n)\pi_n(x),\pi_{n-1}(x)\rangle=\langle(x-\alpha_n)\pi_{n-1}(x),\pi_n(x)\rangle$$ and orthogonality, since $$(x-\alpha_n)\pi_{n-1}(x)=\pi_n(x)-(\alpha_n-\alpha_{n-1})\,\pi_{n-1}+\beta_{n-1}\,\pi_{n-2}(x), $$
gives the equation $$\beta_n=\frac{\langle\pi_n,\pi_n\rangle}{\langle\pi_{n-1},\pi_{n-1}\rangle}=\frac{\gamma_n}{\gamma_{n-1}}.$$ You see that you can construct $\alpha_n$ and $\beta_n$ recursively from skalar products, and with $p_n(x)=A_n\,\pi_n(x),$ $B_n$ is not arbitrary, but $$\frac{B_{n+1}}{A_{n+1}}=\frac{B_n}{A_n}-\alpha_n.$$