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6.7 The Function Space C[a,b] 183

EXAMPLE 6.22
Let n and m be positive integers, and let S n (x) = sin(nx) and C m (x) = cos(mx). These functions
are in C[−π,π].Let p(x) = 1 to use the dot product
π
f · g = f (x)g(x)dx
−π
in C[−π,π]. With respect to this dot product, S n (x) and C m (x) are orthogonal, because their dot
product is
π

S n · C m = sin(nx)cos(mx)dx = 0,
−π
by a routine integration. This type of orthogonality of functions will form the basis for Fourier
series in Chapter 13, and for more general eigenfunction expansions in Chapter 15.

Theorems 6.6, 6.7, and 6.8 and their proofs, while stated for vectors in R , depend only on
n
the vector space structure in which they were stated, and are valid in C[a,b] as well. Here is an
application of Theorem 6.8.

EXAMPLE 6.23
Suppose we want to approximate f (x) = x(π − x) on [0,π], using a sum of the form
c 1 sin(x) + c 2 sin(2x) + c 3 sin(3x) + c 4 sin(4x).
The term “approximate” has meaning only in the context of some measure of distance, since we
generally call one object a good approximation to another when the objects are close together in
some sense. The necessary structure is available to us if we work in the function space C[0,π],
which contains f (x) and the functions sin(nx). Using the integral dot product with p(x)=1, the
distance between two functions in C[0,π] is

π

2
F − G = (F − G) · (F − G) = (F(x) − G(x)) dx.
0
To make use of Theorem 6.8, let S be the four-dimensional subspace of C[0,π] spanned by
sin(x),sin(2x),sin(3x) and sin(4x). Then S consists of exactly the linear combinations
c 1 sin(x) + c 2 sin(2x) + c 3 sin(3x) + c 4 sin(4x)
that we want to use to approximate f (x). f is not in S. By Theorem 6.8, the object in S closest
to f is the orthogonal projection f S of f onto S.Thisis
f · sin(x) f · sin(2x)
f S = sin(x) + sin(2x)
sin(x) 2 sin(2x) 2
f · sin(3x) f · sin(4x)
+ sin(3x) + sin(4x).
sin(3x) 2 sin(4x) 2
All that remains is to compute these coefficients. First, for n = 1,2,3,4,
π π
2 2
sin(nx) = sin (nx)dx = .
2
0
Furthermore,
2(1 − (−1) )
π n
f · sin(nx) = x(π − x)sin(nx)dx = .
n 3
0

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October 14, 2010 14:21 THM/NEIL Page-183 27410_06_ch06_p145-186