Three-term recurrence relation

Remarks

Example: the Chebyshev polynomials satisfy $T_{n+1}(x) - 2x\, T_n(x) + T_{n-1}(x) = 0$ in $\R[x]$, here read in the field $\R(x)$ of rational functions with $a_n = 1$, $b_n = -2x$, $c_n = 1$. Check for $n = 1$: $T_2(x) = 2x^2 - 1 = 2x \cdot x - 1$. Non-example: $u_{n+1} = u_n + u_{n-1} + u_{n-2}$ links four terms, so it is not a three-term recurrence. The condition $a_n \ne 0$ lets one solve for $u_{n+1}$, and $c_n \ne 0$ makes the order exactly two; so the solutions form a $K$-vector space of dimension $2$, fixed by $u_0$ and $u_1$. A common normalized form is $u_{n+1} = (x - \alpha_n) u_n - \beta_n u_{n-1}$, which is the three-term recurrence for orthogonal polynomials.