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5.7 Orthogonal Polynomials

We have already seen how polynomials can be used for data fitting and for approximating continuous functions. Since both of these problems are least squares problems, they can be simplified by selecting an orthogonal basis for the class of approximating polynomials. This leads us to the concept of orthogonal polynomials.

In this section, we study families of orthogonal polynomials associated with various inner products on C[a, b]. We will see that the polynomials in each of these classes satisfy a three-term recursion relation. This recursion relation is particularly useful in computer applications. Certain families of orthogonal polynomials have important applications in many areas of mathematics. We will refer to these polynomials as classical polynomials and examine them in more detail. In particular, the classical polynomials are solutions of certain classes of second-order linear differential equations that arise in the solution of many partial differential equations from mathematical physics.

Orthogonal Sequences

Since the proof of Theorem 5.6.1 was by induction, the Gram–Schmidt process is valid for a denumerable set. Thus, if $x_{1}, x_{2}, …$ $x_{1}, x_{2}, …$ is a sequence of vectors in an inner product space V and $x_{1}, x_{2}, …, x_{n}$ $x_{1}, x_{2}, …, x_{n}$ are linearly independent for each n, then the Gram–Schmidt process may be used to form a sequence $u_{1}, u_{2}, …,$ $u_{1}, u_{2}, …,$ , where ${u_{1}, u_{2}, …,}$ ${u_{1}, u_{2}, …,}$ is an orthonormal set and

Span (x_{1}, x_{2}, …, x_{n}) = Span (u_{1,} u_{2}, …, u_{n})

$Span (x_{1}, x_{2}, …, x_{n}) = Span (u_{1,} u_{2}, …, u_{n})$

for each n. In particular, from the sequence $x, x, x^{2}, …,$ $x, x, x^{2}, …,$ , it is possible to construct an orthonormal sequence $P_{0} (x), p_{1} (x), …$ $P_{0} (x), p_{1} (x), …$ .

Let P be the vector space of all polynomials and define the inner product $〈, 〉$ $〈, 〉$ on P by

〈 p, q 〉 = \int_{a}^{b} p (x) q (x) w (x) d x

$〈 p, q 〉 = \int_{a}^{b} p (x) q (x) w (x) d x$ (1)

where w(x) is a positive continuous function. The interval can be taken as either open or closed and may be finite or infinite. If, however,

\int_{a}^{b} p (x) w (x) d x

$\int_{a}^{b} p (x) w (x) d x$

is improper, we require that it converge for every $p \in P$ $p \in P$ .

Theorem 5.7.1

If $p_{0}, p_{1}, …$ $p_{0}, p_{1}, …$ is a sequence of orthogonal polynomials, then

$p_{0}, …, p_{n - 1}$ $p_{0}, …, p_{n - 1}$ form a basis for $P_{n}$ $P_{n}$ .
$P_{n} \in P_{n}^{⊥}$ $P_{n} \in P_{n}^{⊥}$ (i.e., $P_{n}$ $P_{n}$ is orthogonal to every polynomial of degree less than n).

Proof

It follows from Theorem 5.5.1 that $p_{0}, p_{1}, …, p_{n - 1}$ $p_{0}, p_{1}, …, p_{n - 1}$ are linearly independent in $P_{n}$ $P_{n}$ . Since dim $\dim P_{n} = n$ $\dim P_{n} = n$ , these n vectors must form a basis for $P_{n}$ $P_{n}$ . Let p(x) be any polynomial of degree less than n. Then

p (x) = \sum_{i = 0}^{n - 1} c_{i} p_{i} (x)

$p (x) = \sum_{i = 0}^{n - 1} c_{i} p_{i} (x)$

and hence

〈 p_{n}, p 〉 = 〈 p_{n}, \sum_{i = 0}^{n - 1} c_{i} p_{i} 〉 = \sum_{i = 0}^{n - 1} c_{i} 〈 p_{n}, p_{i} 〉 = 0

$〈 p_{n}, p 〉 = 〈 p_{n}, \sum_{i = 0}^{n - 1} c_{i} p_{i} 〉 = \sum_{i = 0}^{n - 1} c_{i} 〈 p_{n}, p_{i} 〉 = 0$

Therefore, $p_{n} \in P_{n}^{⊥}$ $p_{n} \in P_{n}^{⊥}$ .

∎

If ${p_{0}, p_{1}, …, p_{n - 1}}$ ${p_{0}, p_{1}, …, p_{n - 1}}$ is an orthogonal set in $P_{n}$ $P_{n}$ and

\begin{matrix} u_{i} = (\frac{1}{‖ p_{i} ‖}) p_{i} & \begin{matrix} for & i = 0, …, n - 1 \end{matrix} \end{matrix}

$\begin{matrix} u_{i} = (\frac{1}{‖ p_{i} ‖}) p_{i} & \begin{matrix} for & i = 0, …, n - 1 \end{matrix} \end{matrix}$

then ${u_{0}, …, u_{n - 1}}$ ${u_{0}, …, u_{n - 1}}$ is an orthonormal basis for $P_{n}$ $P_{n}$ . Hence, if $p \in P_{n}$ $p \in P_{n}$ , then

\begin{matrix} p & = & \sum_{i = 0}^{n - 1} 〈 p, u_{i} 〉 u_{i} \\ = & \sum_{i = 0}^{n - 1} 〈 p, (\frac{1}{‖ p_{i} ‖}) p_{i} 〉 (\frac{1}{‖ p_{i} ‖}) p_{i} \\ = & \sum_{i = 0}^{n - 1} \frac{〈 p, p_{i} 〉}{〈 p_{i}, p_{i} 〉} p_{i} \end{matrix}

$\begin{matrix} p & = & \sum_{i = 0}^{n - 1} 〈 p, u_{i} 〉 u_{i} \\ = & \sum_{i = 0}^{n - 1} 〈 p, (\frac{1}{‖ p_{i} ‖}) p_{i} 〉 (\frac{1}{‖ p_{i} ‖}) p_{i} \\ = & \sum_{i = 0}^{n - 1} \frac{〈 p, p_{i} 〉}{〈 p_{i}, p_{i} 〉} p_{i} \end{matrix}$

Similarly, if $f \in C [a, b]$ $f \in C [a, b]$ , then the best least squares approximation to f by the elements of $P_{n}$ $P_{n}$ is given by

p = \sum_{i = 0}^{n - 1} \frac{〈 f, p_{i} 〉}{〈 p_{i}, p_{i} 〉} p_{i}

$p = \sum_{i = 0}^{n - 1} \frac{〈 f, p_{i} 〉}{〈 p_{i}, p_{i} 〉} p_{i}$

where $p_{0}, p_{1}, …, p_{n - 1}$ $p_{0}, p_{1}, …, p_{n - 1}$ are orthogonal polynomials.

Another nice feature of sequences of orthogonal polynomials is that they satisfy a three-term recursion relation.

Theorem 5.7.2

Let $p_{0}, p_{1}, \dots$ $p_{0}, p_{1}, \dots$ be a sequence of orthogonal polynomials. Let $a_{i}$ $a_{i}$ denote the lead coefficient of $p_{i}$ $p_{i}$ for each i, and define $p_{- 1} (x)$ $p_{- 1} (x)$ to be the zero polynomial. Then

\begin{matrix} α_{n + 1} p_{n + 1} (x) = (x - β_{n + 1}) p_{n} (x) - α_{n} γ_{n} p_{n - 1} (x) & (n \geq 0) \end{matrix}

$\begin{matrix} α_{n + 1} p_{n + 1} (x) = (x - β_{n + 1}) p_{n} (x) - α_{n} γ_{n} p_{n - 1} (x) & (n \geq 0) \end{matrix}$

where $α_{0} = γ_{0} = 1$ $α_{0} = γ_{0} = 1$ and

\begin{matrix} α_{n} = \frac{α_{n - 1}}{α_{n}}, & β_{n} = \frac{〈 p_{n - 1}, x p_{n - 1} 〉}{〈 p_{n - 1}, p_{n - 1} 〉}, & γ_{n} = \frac{〈 p_{n}, p_{n} 〉}{〈 p_{n - 1}, p_{n - 1} 〉} & \begin{matrix} (n \geq 1) \end{matrix} \end{matrix}

$\begin{matrix} α_{n} = \frac{α_{n - 1}}{α_{n}}, & β_{n} = \frac{〈 p_{n - 1}, x p_{n - 1} 〉}{〈 p_{n - 1}, p_{n - 1} 〉}, & γ_{n} = \frac{〈 p_{n}, p_{n} 〉}{〈 p_{n - 1}, p_{n - 1} 〉} & \begin{matrix} (n \geq 1) \end{matrix} \end{matrix}$

Proof

Since $p_{0}, p_{1}, \dots, p_{n + 1}$ $p_{0}, p_{1}, \dots, p_{n + 1}$ form a basis for $P_{n + 2}$ $P_{n + 2}$ , we can write

x p_{n} (x) = \sum_{k = 0}^{n + 1} c_{n k} p_{k} (x)

$x p_{n} (x) = \sum_{k = 0}^{n + 1} c_{n k} p_{k} (x)$ (2)

where

c_{n k} = \frac{〈 x p_{n}, p_{k} 〉}{〈 p_{k}, p_{k} 〉}

$c_{n k} = \frac{〈 x p_{n}, p_{k} 〉}{〈 p_{k}, p_{k} 〉}$ (3)

For any inner product defined by (1),

〈 x f, g 〉 = 〈 f, x g 〉

$〈 x f, g 〉 = 〈 f, x g 〉$

In particular,

〈 x p_{n}, p_{k} 〉 = 〈 p_{n}, x p_{k} 〉

$〈 x p_{n}, p_{k} 〉 = 〈 p_{n}, x p_{k} 〉$

It follows from Theorem 5.7.1 that if $k < n - 1$ $k < n - 1$ , then

c_{n k} = \frac{〈 x p_{n}, p_{k} 〉}{〈 p_{k}, p_{k} 〉} = \frac{〈 p_{n}, {x p}_{k} 〉}{〈 p_{k}, p_{k} 〉} = 0

$c_{n k} = \frac{〈 x p_{n}, p_{k} 〉}{〈 p_{k}, p_{k} 〉} = \frac{〈 p_{n}, {x p}_{k} 〉}{〈 p_{k}, p_{k} 〉} = 0$

Therefore, (2) simplifies to

x p_{n} (x) = c_{n, n - 1} p_{n - 1} (x) + c_{n, n} p_{n} (x) + c_{n, n + 1} p_{n + 1} (x)

$x p_{n} (x) = c_{n, n - 1} p_{n - 1} (x) + c_{n, n} p_{n} (x) + c_{n, n + 1} p_{n + 1} (x)$

This equation can be rewritten in the form

c_{n, n + 1} p_{n + 1} (x) = (x - c_{n, n}) p_{n} (x) - c_{n, n - 1} p_{n - 1} (x)

$c_{n, n + 1} p_{n + 1} (x) = (x - c_{n, n}) p_{n} (x) - c_{n, n - 1} p_{n - 1} (x)$ (4)

Comparing the lead coefficients of the polynomials on each side of (4), we see that

c_{n, n + 1} a_{n + 1} = a_{n}

$c_{n, n + 1} a_{n + 1} = a_{n}$

c_{n, n + 1} = \frac{a_{n}}{a_{n + 1}} = a_{n + 1}

$c_{n, n + 1} = \frac{a_{n}}{a_{n + 1}} = a_{n + 1}$ (5)

It follows from (4) that

\begin{matrix} c_{n, n + 1} 〈 p_{n}, p_{n + 1} 〉 & = & 〈 p_{n}, (x - c_{n, n}) p_{n} 〉 - c_{n, n - 1} 〈 p_{n,} p_{n - 1} 〉 \\ 0 & = & 〈 p_{n}, x p_{n} 〉 - c_{n n} 〈 p_{n,} p_{n} 〉 \end{matrix}

$\begin{matrix} c_{n, n + 1} 〈 p_{n}, p_{n + 1} 〉 & = & 〈 p_{n}, (x - c_{n, n}) p_{n} 〉 - c_{n, n - 1} 〈 p_{n,} p_{n - 1} 〉 \\ 0 & = & 〈 p_{n}, x p_{n} 〉 - c_{n n} 〈 p_{n,} p_{n} 〉 \end{matrix}$

Thus,

c_{n n} = \frac{〈 p_{n}, x p_{n} 〉}{〈 p_{n,} p_{n} 〉} = β_{n + 1}

$c_{n n} = \frac{〈 p_{n}, x p_{n} 〉}{〈 p_{n,} p_{n} 〉} = β_{n + 1}$

It follows from (3) that

\begin{matrix} 〈 p_{n + 1}, p_{n - 1} 〉 c_{n, n - 1} & = & 〈 x p_{n}, p_{n - 1} 〉 \\ = & 〈 p_{n}, x p n - 1 〉 \\ = & 〈 p_{n}, p_{n} 〉 c_{n - 1}, n \end{matrix}

$\begin{matrix} 〈 p_{n + 1}, p_{n - 1} 〉 c_{n, n - 1} & = & 〈 x p_{n}, p_{n - 1} 〉 \\ = & 〈 p_{n}, x p n - 1 〉 \\ = & 〈 p_{n}, p_{n} 〉 c_{n - 1}, n \end{matrix}$

and hence, by (5), we have

c_{n, n - 1} = \frac{〈 p_{n}, p_{n} 〉}{〈 p_{n - 1}, p_{n - 1} 〉} α_{n} = γ_{n} α_{n}

$c_{n, n - 1} = \frac{〈 p_{n}, p_{n} 〉}{〈 p_{n - 1}, p_{n - 1} 〉} α_{n} = γ_{n} α_{n}$

∎

In generating a sequence of orthogonal polynomials by the recursion relation in Theorem 5.7.2, we are free to choose any nonzero lead coefficient $a_{n + 1}$ $a_{n + 1}$ that we want at each step. This is reasonable, since any nonzero multiple of a particular $p_{n + 1}$ $p_{n + 1}$ will also be orthogonal to $p_{0}, \dots, p_{n}$ $p_{0}, \dots, p_{n}$ . If we were to choose our $a_{i} ’s$ $a_{i} ’s$ to be 1, for example, then the recursion relation would simplify to

p_{n + 1} (x) = (x - β_{n + 1}) p_{n} (x) - γ_{n} p_{n - 1} (x)

$p_{n + 1} (x) = (x - β_{n + 1}) p_{n} (x) - γ_{n} p_{n - 1} (x)$

Classical Orthogonal Polynomials

Let us now look at some examples. Because of their importance, we will consider the classical polynomials beginning with the simplest, the Legendre polynomials.

Legendre Polynomials

The Legendre polynomials are orthogonal with respect to the inner product

〈 p, q 〉 = \int_{- 1}^{1} p (x) q (x) d x

$〈 p, q 〉 = \int_{- 1}^{1} p (x) q (x) d x$

Let $P_{n} (x)$ $P_{n} (x)$ denote the Legendre polynomial of degree n. If we choose the lead coefficients so that $P_{n} (1) = 1$ $P_{n} (1) = 1$ for each n, then the recursion formula for the Legendre polynomials is

(n + 1) P_{n + 1} (x) = (2 n + 1) x P_{n} (x) - n P_{n - 1} (x)

$(n + 1) P_{n + 1} (x) = (2 n + 1) x P_{n} (x) - n P_{n - 1} (x)$

By the use of this formula, the sequence of Legendre polynomials is easily generated. The first five polynomials of the sequence are

\begin{matrix} P_{0} (x) & = & 1 \\ P_{1} (x) & = & x \\ P_{2} (x) & = & \frac{1}{2} (3 x^{2} - 1) \\ P_{3} (x) & = & \frac{1}{2} (5 x^{3} - 3 x) \\ P_{4} (x) & = & \frac{1}{8} (35 x^{4} - 30 x^{2} + 3) \end{matrix}

$\begin{matrix} P_{0} (x) & = & 1 \\ P_{1} (x) & = & x \\ P_{2} (x) & = & \frac{1}{2} (3 x^{2} - 1) \\ P_{3} (x) & = & \frac{1}{2} (5 x^{3} - 3 x) \\ P_{4} (x) & = & \frac{1}{8} (35 x^{4} - 30 x^{2} + 3) \end{matrix}$

Chebyshev Polynomials

The Chebyshev polynomials are orthogonal with respect to the inner product

〈 p, q 〉 = \int_{- 1}^{1} p (x) q (x) (1 - x^{2})^{- 1 / 2} d x

$〈 p, q 〉 = \int_{- 1}^{1} p (x) q (x) (1 - x^{2})^{- 1 / 2} d x$

It is customary to normalize the lead coefficients so that $a_{0} = 1$ $a_{0} = 1$ and $a_{k} = 2^{k - 1}$ $a_{k} = 2^{k - 1}$ for $k = 1, 2, \dots$ $k = 1, 2, \dots$ . The Chebyshev polynomials are denoted by $T_{n} (x)$ $T_{n} (x)$ and have the interesting property that

_{} T_{n} (cos θ) = cos n θ

$_{} T_{n} (cos θ) = cos n θ$

This property, together with the trigonometric identity

cos (n + 1) θ = 2 cos θ cos n θ - cos (n - 1) θ

$cos (n + 1) θ = 2 cos θ cos n θ - cos (n - 1) θ$

can be used to derive the recursion relations

\begin{matrix} T_{1} (x) & = & x T_{0} (x) \\ T_{n + 1} (x) & = & 2 x T_{n} (x) - T_{n - 1} (x) & for n \geq 1 \end{matrix}

$\begin{matrix} T_{1} (x) & = & x T_{0} (x) \\ T_{n + 1} (x) & = & 2 x T_{n} (x) - T_{n - 1} (x) & for n \geq 1 \end{matrix}$

Jacobi Polynomials

The Legendre and Chebyshev polynomials are both special cases of the Jacobi polynomials. The Jacobi polynomials $P_{n}^{(λ, μ)}$ $P_{n}^{(λ, μ)}$ are orthogonal with respect to the inner

product

〈 p, q 〉 = \int_{- 1}^{1} p (x) q (x) (1 - x)^{λ} (1 + x)^{μ} d x

$〈 p, q 〉 = \int_{- 1}^{1} p (x) q (x) (1 - x)^{λ} (1 + x)^{μ} d x$

where $λ, μ > - 1$ $λ, μ > - 1$ .

Hermite Polynomials

The Hermite polynomials are defined on the interval $(- \infty, \infty)$ $(- \infty, \infty)$ . They are orthogonal with respect to the inner product

〈 p, q 〉 = \int_{- \infty}^{\infty} p (x) q (x) e^{- x^{2}} d x

$〈 p, q 〉 = \int_{- \infty}^{\infty} p (x) q (x) e^{- x^{2}} d x$

The recursion relation for Hermite polynomials is given by

H_{n + 1} (x) = 2 x H_{n} (x) - 2 n H_{n - 1} (x)

$H_{n + 1} (x) = 2 x H_{n} (x) - 2 n H_{n - 1} (x)$

Laguerre Polynomials

The Laguerre polynomials are defined on the interval (0, ∞) and are orthogonal with respect to the inner product

〈 p, q 〉 = \int_{0}^{\infty} p (x) q (x) x^{λ} e^{- x} d x

$〈 p, q 〉 = \int_{0}^{\infty} p (x) q (x) x^{λ} e^{- x} d x$

where $λ > - 1$ $λ > - 1$ . The recursion relation for the Laguerre polynomials is given by

(n + 1) L_{n + 1}^{(λ)} (x) = (2 n + λ + 1 - x) L_{n}^{(λ)} (x) - (n + λ) L_{n - 1}^{(λ)} (x)

$(n + 1) L_{n + 1}^{(λ)} (x) = (2 n + λ + 1 - x) L_{n}^{(λ)} (x) - (n + λ) L_{n - 1}^{(λ)} (x)$

The Chebyshev, Hermite, and Laguerre polynomials are compared in Table 5.7.1.

Table 5.7.1 Chebyshev, Hermite, and Laguerre Polynomials

Chebyshev	Hermite	Laguerre $(λ = 0)$ $(λ = 0)$
$\begin{matrix} T_{n + 1} & = & 2 x T_{n} - T_{n - 1}, n \geq 1 \\ T_{0} & = & 1 \\ T_{1} & = & x \\ T_{2} & = & 2 x^{2} - 1 \\ T_{3} & = & 4 x^{3} - 3 x \end{matrix}$ $\begin{matrix} T_{n + 1} & = & 2 x T_{n} - T_{n - 1}, n \geq 1 \\ T_{0} & = & 1 \\ T_{1} & = & x \\ T_{2} & = & 2 x^{2} - 1 \\ T_{3} & = & 4 x^{3} - 3 x \end{matrix}$	$\begin{matrix} H_{n + 1} & = & 2 x H_{n} - 2 n H_{n - 1} \\ H_{0} & = & 1 \\ H_{1} & = & 2 x \\ H_{2} & = & 4 x^{2} - 2 \\ H_{3} & = & 8 x^{3} - 12 x \end{matrix}$ $\begin{matrix} H_{n + 1} & = & 2 x H_{n} - 2 n H_{n - 1} \\ H_{0} & = & 1 \\ H_{1} & = & 2 x \\ H_{2} & = & 4 x^{2} - 2 \\ H_{3} & = & 8 x^{3} - 12 x \end{matrix}$	$\begin{matrix} (n + 1) L_{n + 1}^{(0)} & = & (2 n + 1 - x) L_{n}^{(0)} - n L_{n - 1}^{(0)} \\ L_{0}^{(0)} & = & 1 \\ L_{1}^{(0)} & = & 1 - x \\ L_{2}^{(0)} & = & \frac{1}{2} x^{2} - x + 2 \\ L_{3}^{(0)} & = & \frac{1}{6} x^{3} + 9 x^{2} - 18 x + 6 \end{matrix}$ $\begin{matrix} (n + 1) L_{n + 1}^{(0)} & = & (2 n + 1 - x) L_{n}^{(0)} - n L_{n - 1}^{(0)} \\ L_{0}^{(0)} & = & 1 \\ L_{1}^{(0)} & = & 1 - x \\ L_{2}^{(0)} & = & \frac{1}{2} x^{2} - x + 2 \\ L_{3}^{(0)} & = & \frac{1}{6} x^{3} + 9 x^{2} - 18 x + 6 \end{matrix}$

Application 1

Numerical Integration

One important application of orthogonal polynomials occurs in numerical integration. To approximate

\int_{a}^{b} f (x) w (x) d x

$\int_{a}^{b} f (x) w (x) d x$ (6)

we first approximate $f (x)$ $f (x)$ by an interpolating polynomial. Using Lagrange’s interpolation formula,

P (x) = \sum_{i = 1}^{n} f (x_{i}) L_{i} (x)

$P (x) = \sum_{i = 1}^{n} f (x_{i}) L_{i} (x)$

where the Lagrange functions $L_{i}$ $L_{i}$ are defined by

L_{i} (x) = \frac{\begin{matrix} Π_{\underset{j \neq i}{j = 1}}^{n} & (x - x_{j}) \end{matrix}}{\begin{matrix} Π_{\underset{j \neq i}{j = 1}}^{n} & (x_{i} - x_{j}) \end{matrix}}

$L_{i} (x) = \frac{\begin{matrix} Π_{\underset{j \neq i}{j = 1}}^{n} & (x - x_{j}) \end{matrix}}{\begin{matrix} Π_{\underset{j \neq i}{j = 1}}^{n} & (x_{i} - x_{j}) \end{matrix}}$

we can determine a polynomial $P (x)$ $P (x)$ that agrees with $f (x)$ $f (x)$ at n points $x_{1}, \dots, x_{n}$ $x_{1}, \dots, x_{n}$ in $[a, b]$ $[a, b]$ . The integral (6) is then approximated by

\int_{a}^{b} P (x) w (x) d x = \sum_{i = 1}^{n} A_{i} f (x_{i})

$\int_{a}^{b} P (x) w (x) d x = \sum_{i = 1}^{n} A_{i} f (x_{i})$ (7)

where

\begin{matrix} {A_{i} = \int}_{a}^{b} L_{i} (x) w (x) d x & i = 1, \dots, n \end{matrix}

$\begin{matrix} {A_{i} = \int}_{a}^{b} L_{i} (x) w (x) d x & i = 1, \dots, n \end{matrix}$

It can be shown that (7) will give the exact value of the integral whenever $f (x)$ $f (x)$ is a polynomial of degree less than n. If the points $x_{1}, \dots, x_{n}$ $x_{1}, \dots, x_{n}$ are chosen properly, formula (7) will be exact for higher degree polynomials. Indeed, it can be shown that if $p_{0}, p_{1}, p_{2}, \dots$ $p_{0}, p_{1}, p_{2}, \dots$ is a sequence of orthogonal polynomials with respect to the inner product (1) and $x_{1}, \dots, x_{n}$ $x_{1}, \dots, x_{n}$ are the zeros of $p_{n} (x)$ $p_{n} (x)$ , then formula (7) will be exact for all polynomials of degree less than 2n. The following theorem guarantees that the roots of $p_{n}$ $p_{n}$ are all real and lie in the open interval (a, b).

Theorem 5.7.3

If $p_{0}, p_{1}, p_{2}, . . .$ $p_{0}, p_{1}, p_{2}, . . .$ is a sequence of orthogonal polynomials with respect to the inner product (1), then the zeros of $p_{n} (x)$ $p_{n} (x)$ are all real and distinct and lie in the interval $(a, b)$ $(a, b)$ .

Proof

Let $x_{1}, . . ., x_{m}$ $x_{1}, . . ., x_{m}$ be the zeros of $p_{n} (x)$ $p_{n} (x)$ that lie in $(a, b)$ $(a, b)$ and for which $p_{n} (x)$ $p_{n} (x)$ changes sign. Thus, $p_{n} (x)$ $p_{n} (x)$ must have a factor of $(x - x_{1})^{k_{i}}$ $(x - x_{1})^{k_{i}}$ , where $k_{i}$ $k_{i}$ is odd, for $i = 1, . . .$ $i = 1, . . .$ , m. We may write

p_{n} (x) = (x - x_{1})^{k_{1}} (x - x_{2})^{k_{2}} \cdot \cdot \cdot (x - x_{m})^{k_{m}} q (x)

$p_{n} (x) = (x - x_{1})^{k_{1}} (x - x_{2})^{k_{2}} \cdot \cdot \cdot (x - x_{m})^{k_{m}} q (x)$

where $q (x)$ $q (x)$ does not change sign on $(a, b)$ $(a, b)$ and $q (x_{i}) \neq 0$ $q (x_{i}) \neq 0$ for $i = 1, . . ., m$ $i = 1, . . ., m$ . Clearly, $m \leq n$ $m \leq n$ . We will show that $m = n$ $m = n$ . Let

r (x) = (x - x_{1}) (x - x_{2}) \cdot \cdot \cdot (x - x_{m})

$r (x) = (x - x_{1}) (x - x_{2}) \cdot \cdot \cdot (x - x_{m})$

The product

p_{n} (x) r (x) = (x - x_{1})^{k_{1} + 1} + (x - x_{2})^{k_{2} + 1} \cdot \cdot \cdot (x - x_{m})^{k_{m} + 1} q (x)

$p_{n} (x) r (x) = (x - x_{1})^{k_{1} + 1} + (x - x_{2})^{k_{2} + 1} \cdot \cdot \cdot (x - x_{m})^{k_{m} + 1} q (x)$

will involve only even powers of $(x - x_{i})$ $(x - x_{i})$ for each i and hence will not change sign on $(a, b)$ $(a, b)$ . Therefore,

〈 p_{n}, r 〉 = \int_{a}^{b} p_{n} (x) r (x) w (x) d x \neq 0

$〈 p_{n}, r 〉 = \int_{a}^{b} p_{n} (x) r (x) w (x) d x \neq 0$

Since $p_{n}$ $p_{n}$ is orthogonal to all polynomials of degree less than n, it follows that $\deg (r (x)) = m \geq n$ $\deg (r (x)) = m \geq n$ .

∎

Numerical integration formulas of the form (7), where the $x_{i} ’s$ $x_{i} ’s$ are roots of orthogonal polynomials, are called Gaussian quadrature formulas. The proof of exactness for polynomials of degree less than 2n can be found in most undergraduate numerical analysis textbooks.

Actually, it is not necessary to perform n integrations to calculate the quadrature coefficients $A_{1}, . . ., A_{n}$ $A_{1}, . . ., A_{n}$ . They can be determined by solving an $n \times n$ $n \times n$ linear system. Exercise 16 illustrates how this is done when the roots of the Legendre polynomial $P_{n}$ $P_{n}$ are used in a quadrature rule for approximating $\int_{- 1}^{1} f (x) d x$ $\int_{- 1}^{1} f (x) d x$

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Table of Contents for 5.7 Orthogonal Polynomials

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Table of Contents for
5.7 Orthogonal Polynomials