Polynomials

Determinant

An alternative approach uses the determinant: \[P(x)=\left|\begin{array}{ccccc}\lambda_{0} & \lambda_{1} & \cdots & \lambda_{n-1} & \lambda_{n}\\ \lambda_{1} & \lambda_{2} & \cdots & \lambda_{n} & \lambda_{n+1}\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ \lambda_{n-1} & \lambda_{n} & \cdots & \lambda_{2n-2} & \lambda_{2n-1}\\ 1 & x & \cdots & x^{n-1} & x^{n} \end{array}\right| \]

We note that: \[\langle \lambda | x^k P\rangle = \left|\begin{array}{ccccc} \lambda_{0} & \lambda_{1} & \cdots & \lambda_{n-1} & \lambda_{n}\\ \lambda_{1} & \lambda_{2} & \cdots & \lambda_{n} & \lambda_{n+1}\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ \lambda_{n-1} & \lambda_{n} & \cdots & \lambda_{2n-2} & \lambda_{2n-1}\\ \lambda_k & \lambda_{k+1} & \cdots & \lambda_{k+n-1} & \lambda_{k+n} \end{array}\right| \]

The bottom row repeats one of those above if \(k \lt n \) and hence the determinant is equal to zero and thus \(P\) is orthogonal to \(x^k\). The polynomial is not monic, but this is simply corrected in: \[ p_n(x)=\dfrac{\left|\begin{array}{ccccc} \lambda_{0} & \lambda_{1} & \cdots & \lambda_{n-1} & \lambda_{n}\\ \lambda_{1} & \lambda_{2} & \cdots & \lambda_{n} & \lambda_{n+1}\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ \lambda_{n-1} & \lambda_{n} & \cdots & \lambda_{2n-2} & \lambda_{2n-1}\\ 1 & x & \cdots & x^{n-1} & x^{n} \end{array}\right| }{ \left|\begin{array}{ccccc} \lambda_{0} & \lambda_{1} & \cdots & \lambda_{n-2} & \lambda_{n-1}\\ \lambda_{1} & \lambda_{2} & \cdots & \lambda_{n-1} & \lambda_{n}\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ \lambda_{n-2} & \lambda_{n-1} & \cdots & \lambda_{2n-4} & \lambda_{2n-3}\\ \lambda_{n-1} & \lambda_{n} & \cdots & \lambda_{2n-3} & \lambda_{2n-2} \end{array}\right| } = \sum \limits_{k=0}^n \frac{\Delta^{(n)}_k}{\Delta^{(n)}_n}x^k=\sum \limits_{k=0}^n d^{(n)}_k x^k\]

The \(\Delta\)s are the appropriate minors needed to expand the determinant of \(P\). These can be used to express the three-term recurrence coefficients in terms of the moments of the linear functional defining orthogonality, \(\lambda_n\). We look at \[ p_{n+1}-x p_n = \sum \limits_{k=0}^n (d^{(n+1)}_k - d^{(n)}_{k-1})x^k\]

We define \(d^{(n)}_{-1}=0\) to enable the sum to make sense. The upper limit is \(n\) rather than \(n+1\) since \(d_{n+1}^{(n+1)} - d_n^{(n)}=1-1=0\). The first recurrence coefficient can be obtained from the leading power, \(a_n=d_n^{(n+1)} - d_{n-1}^{(n)}\).

Subtracting out the \(a_n p_n\) term to give \(b_n p_{n-1}\) and looking at the leading power: \[b_n = d^{(n+1)}_{n-1}-d^{(n)}_{n-2}-a_nd^{(n)}_{n-1}\]

We note that \(\langle \lambda | p_n^2 \rangle = \langle \lambda | x^n p_n \rangle = \Delta^{(n+1)}_{n+1}/\Delta^{(n)}_n\), giving the simpler formula \(b_n= -\langle \lambda | p_n^2\rangle/\langle \lambda | p_{n-1}^2\rangle = -\Delta^{(n+1)}_{n+1}\Delta^{(n-1)}_{n-1}/(\Delta^{(n)}_n)^2\)