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
.