Orthogonal polynomials/Proofs

From Wikipedia, the free encyclopedia

This mathematics article is devoted entirely to providing mathematical proofs and support for claims and statements made in the article orthogonal polynomials. This article is currently an experimental vehicle to see how well we can provide proofs and details for a math article without cluttering up the main article itself. See Wikipedia:WikiProject Mathematics/Proofs for some current discussion. This article is "experimental" in the sense that it is a test of one way we may be able to incorporate more detailed proofs in Wikipedia.

1 Proof of the recurrence relation
2 Proof of the existence of real roots
3 Proof of interlacing of roots

[edit] Proof of the recurrence relation

Any orthogonal series has a recurrence formula relating any three consecutive polynomials in the series.

$p_{n+1}\ =\ (a_nx+b_n)\ p_n\ -\ c_n\ p_{n-1}$

The coefficients a, b, and c depend on n. They also depend on the standardization, obviously.

Proof:

We will prove this for fixed n, and omit the subscripts on a, b, and c.

First, choose a so that the $x n + 1$ terms match, so we have

$a\ x\ p_n\ -\ p_{n+1}\ =\$ a polynomial of degree n.

Next, choose b so that the $x n$ terms match, so we have

$(ax+b)\ p_n\ -\ p_{n+1}\ =\$ a polynomial of degree n − 1

Expand the right-hand-side in terms of polynomials in the sequence

$(ax+b)\ p_n\ -\ p_{n+1}\ =\ \sum_{i=0}^{n-1} {\lambda}_i p_i$

Now if $j\le n-2$ , then

$\langle (ax+b)\ p_n,\ p_j \rangle\ -\ \langle p_{n+1},\ p_j \rangle\ =\ \sum_{i=0}^{n-1} {\lambda}_i \langle p_i,\ p_j \rangle\ =\ {\lambda}_j \langle p_j,\ p_j \rangle.$

But

$\langle p_n,\ p_j \rangle\ =\ 0$ and $\langle p_{n+1},\ p_j \rangle\ =\ 0,$

$a\ \langle x\ p_n,\ p_j \rangle\ =\ {\lambda}_j \langle p_j,\ p_j \rangle.$

Since the inner product is just an integral involving the product:

$\langle x\ p_n,\ p_j \rangle\ =\ \langle p_n,\ x\ p_j \rangle$

we have

$a\ \langle p_n,\ x\ p_j \rangle\ =\ {\lambda}_j \langle p_j,\ p_j \rangle$

If $\ j < n-1$ , then $x\ p_j$ has degree $\le n-1$ , so it is orthogonal to $\ p_n$ ; hence ${\lambda}_j \langle p_j,\ p_j \rangle\ =\ 0$ , which implies ${\lambda}_j\ =\ 0$ for $\ j < n-1$ .

Therefore, only $λ n - 1$ can be nonzero, so

$(ax+b)\ p_n\ -\ p_{n+1}\ ={\lambda}_{n-1}\ p_{n-1}\$

Letting $c\ =\ {\lambda}_{n-1}$ , we have

$p_{n+1}\ =\ (ax+b)\ p_n\ -\ c\ p_{n-1}.$

[edit] Proof of the existence of real roots

Each polynomial in an orthogonal sequence has all n of its roots real, distinct, and strictly inside the interval of orthogonality.

Proof:

Let m be the number of places where the sign of P_n changes inside the interval of orthogonality, and let $x_1 \dots x_m$ be those points. Each of those points is a root of P_n. By the fundamental theorem of algebra, m ≤ n. Now m might be strictly less than n if some roots of P_n are complex, or not inside the interval of orthogonality, or not distinct. We will show that m = n.

Let $S(x) = \prod_{j=1}^m (x - x_j)$

This is an m^th-degree polynomial that changes sign at each of the x_j, the same way that P_n(x) does. S(x)P_n(x) is therefore strictly positive, or strictly negative, everywhere except at the x_j. S(x)P_n(x)W(x) is also strictly positive or strictly negative except at the x_j and possibly the end points.

Therefore, $\langle S, P_n \rangle$ , the integral of this, is nonzero. But, by Lemma 2, P_n is orthogonal to any polynomial of lower degree, so the degree of S must be n.

[edit] Proof of interlacing of roots

The roots of each polynomial lie strictly between those of the next higher polynomial in the sequence.

Proof:

First, standardize all of the polynomials so that their leading terms are positive. This will not affect the roots.

Next, a lemma: For all n and all x,

$P_{n+1}'(x)\ P_n(x) > P_{n+1}(x)\ P_n'(x)\,$

Proof by induction. For n = 0, $P_1'(x) > 0\,$ , $P_0(x) > 0\,$ , and $P_0'(x) = 0\,$ .

Otherwise, the recurrence formula has

$P_{n+1}(x) = (ax + b) P_n(x) - c P_{n-1}(x)\,$

with $a = \frac{k_{n+1}}{k_n} > 0\,$ and $c = a\,\frac{k_{n-1}h_n}{k_{n}h_{n-1}} > 0\,$ .

$P_{n+1}' = a P_n + (ax + b) P_n' - c P_{n-1}'\,$ .

$P_{n+1}'\ P_n - P_{n+1}\ P_n' = [a P_n + (ax + b) P_n' - c P_{n-1}'] P_n - [(ax + b) P_n - c P_{n-1}] P_n'\,$

$= [a P_n - c P_{n-1}'] P_n + c P_{n-1}\ P_n'\,$

$= a P_n^{ 2} + c (P_n'\ P_{n-1} - P_n\ P_{n-1}')\,$

$\ge c (P_n'\ P_{n-1} - P_n\ P_{n-1}')\,$

But $P_n'\ P_{n-1} - P_n\ P_{n-1}' > 0\,$ by the induction step.

Now if x is a root of P_n+1, the lemma tells us that

$P_{n+1}'(x)\ P_n(x) > P_{n+1}(x)\ P_n'(x) = 0\,$

So $P_{n+1}'(x)\,$ and $P_n(x)\,$ have the same sign. But $P_{n+1}'(x)\,$ must change sign from any root of P_n+1 to the next. Therefore, P_n must change sign also, so P_n must have a root in that interval.