2Orthogonal polynomials

IB Numerical Analysis

2.2 Orthogonal polynomials

Definition

(Orthogonal polynomial)

.

Given a vector space

V

of polynomials

and inner product

h·, ·i

, we say

p

n

∈ P

n

[

x

] is the

n

th orthogonal polynomial if

hp

n

, qi = 0 for all q ∈ P

n−1

[x].

In particular, hp

n

, p

m

i = 0 for n 6= m.

We said the orthogonal polynomial, but we need to make sure such a polyno-

mial has to be unique. It is clearly not unique, since if

p

n

satisfies these relations,

then so does

λp

n

for all

λ 6

= 0. For uniqueness, we need to impose some scaling.

We usually do so by requiring the leading polynomial to be 1, i.e. it is monic.

Definition

(Monic polynomial)

.

A polynomial

p ∈ P

n

[

x

] is monic if the coeffi-

cient of x

n

is 1.

In practice, most famous traditional polynomials are not monic. They have

a different scaling imposed. Still, as long as we have some scaling, we will have

uniqueness.

We will not mess with other scalings, and stick to requiring them to be monic

since this is useful for proving things.

Theorem.

Given a vector space

V

of functions and an inner product

h·, ·i

,

there exists a unique monic orthogonal polynomial for each degree

n ≥

0. In

addition, {p

k

}

n

k=0

form a basis for P

n

[x].

Proof.

This is a big induction proof over both parts of the theorem. We induct

over

n

. For the base case, we pick

p

0

(

x

) = 1, which is the only degree-zero monic

polynomial.

Now suppose we already have {p

n

}

n

k=0

satisfying the induction hypothesis.

Now pick any monic

q

n+1

∈ P

n+1

[

x

], e.g.

x

n+1

. We now construct

p

n+1

from

q

n+1

by the Gram-Schmidt process. We define

p

n+1

= q

n+1

−

n

X

k=0

hq

n+1

, p

k

i

hp

k

, p

k

i

p

k

.

This is again monic since q

n+1

is, and we have

hp

n+1

, p

m

i = 0

for all m ≤ n, and hence hp

n+1

, pi = 0 for all p ∈ P

n

[x] = hp

0

, ··· , p

n

i.

To obtain uniqueness, assume both

p

n+1

, ˆp

n+1

∈ P

n+1

[

x

] are both monic

orthogonal polynomials. Then r = p

n+1

− ˆp

n+1

∈ P

n

[x]. So

hr, ri = hr, p

n+1

− ˆp

n+1

i = hr, p

n+1

i − hr, ˆp

n+1

i = 0 − 0 = 0.

So r = 0. So p

n+1

= ˆp

n−1

.

Finally, we have to show that

p

0

, ··· , p

n+1

form a basis for

P

n+1

[

x

]. Now

note that every p ∈ P

n+1

[x] can be written uniquely as

p = cp

n+1

+ q,

where

q ∈ P

n

[

x

]. But

{p

k

}

n

k=0

is a basis for

P

n

[

x

]. So

q

can be uniquely

decomposed as a linear combination of p

0

, ··· , p

n

.

Alternatively, this follows from the fact that any set of orthogonal vectors

must be linearly independent, and since there are

n

+ 2 of these vectors and

P

n+1

[x] has dimension n + 2, they must be a basis.

In practice, following the proof naively is not the best way of producing the

new

p

n+1

. Instead, we can reduce a lot of our work by making a clever choice of

q

n+1

.