Polynomials

Orthogonal sets

Two polynomials are orthogonal relative to a given functional if \(\langle\lambda | pq\rangle =0\). It is often convenient to create sets where there is one polynomial for every possible degree and polynomials of different degree are orthogonal. The further requirement that the coefficient of highest degree is 1, creates a set of “monic” polynomials. If the coefficient ring is an algebraic field, one can divide any polynomial by the value of its highest degree coefficient to give a monic polynomial. For convenience, we define \(\lambda_n = \langle \lambda | x^n \rangle\).

So let’s start. The monic definition requires that the initial zero-degree polynomial \(p_0(x)=1\). For \(n \gt 0\), \(\langle \lambda | p_0 p_n \rangle=\langle \lambda | p_n \rangle =0 \).

For \(p_1(x) = a + x\), this determines \(a\): \[\langle \lambda | a + x\rangle = \lambda_0 a + \lambda_1 =0 \implies a = -\lambda_1/\lambda_0\]

For this to make sense, we need \(\lambda_0 \neq 0\). We could redefine the functional so that \(\langle \lambda | 1\rangle = 1\).

We have \[p_n(x)=x^n +\sum\limits_{k=0}^{n-1} a_k p_k(x)\]

If \(m \gt n \), \[\langle \lambda | p_m x^n \rangle = \langle \lambda | p_m (p_n -\sum\limits_{k=0}^{n-1} a_k p_k ) \rangle=0,\] by orthogonality.

Since the leading terms of the \(p_n\) have been chosen to be \(x^n\), we also have \[\langle \lambda | p_n x^n \rangle = \langle \lambda | p_n x^m p_{n-m} \rangle\] and in particular, \[\langle \lambda | p_n x^n \rangle = \langle \lambda | p_n^2 \rangle\]

We can investigate \(p_{n+1}\) in stages. Now \(p_{n+1} - xp_n\) is a polynomial of degree n. If \(k\lt n-1\), \(\langle \lambda | p_k p_{n+1}\rangle =0\) and \(\langle \lambda | p_k xp_n \rangle\) is a sum of terms \(\langle \lambda | x^k p_n \rangle=0\) with \(k \lt n\).

One of the main uses of orthogonality is to simplify the expression of a general polynomial in terms of a given monic set \(\{p_n\}\). In this, we can certainly express, \(\deg(q)=M\) as: \[q=\sum\limits_{k=0}^M a_k p_k\]

Orthogonality then gives \(\langle \lambda | p_k q \rangle = a_k \langle \lambda | p_k^2\rangle\). So \(a_k = \langle \lambda | p_k q \rangle/\langle \lambda | p_k^2\rangle\) is well defined and unique, so long as \(\langle\lambda | p_k^2\rangle\neq 0\)

Applying this reasoning to \(p_{n+1} - xp_n\), we already have that there are no terms \(a_k p_k\) for \(k\lt n-1\). Hence \(p_{n+1} = (x+a_n)p_n + b_np_{n-1}\). This is a “three-term recurrence relation”.

Assuming we already know \(\lambda, p_n, p_{n-1}\), we can apply orthogonality to determine \(a_n, b_n\): \[0=\langle \lambda | p_{n+1}p_{n-1}\rangle = \langle \lambda | x p_n p_{n-1}\rangle + b_n \langle \lambda | p_{n-1}^2\rangle \implies b_n=-\langle \lambda | x p_n p_{n-1}\rangle/\langle \lambda | p_{n-1}^2\rangle = -\langle \lambda | p_n^2\rangle/\langle \lambda | p_{n-1}^2\rangle\] \[0=\langle \lambda | p_{n+1}p_n\rangle = \langle \lambda | x p_n^2\rangle + a_n \langle \lambda | p_n^2\rangle \implies a_n=-\langle \lambda | x p_n^2\rangle/\langle \lambda | p_n^2\rangle\]