Page 205 - Advanced engineering mathematics

6.7 The Function Space C[a,b] 185

These polynomials are orthogonal in C[−1,1], using the integral dot product
1
f · g = f (x)g(x)dx.
−1
This means that
1
P n (x)P m (x)dx = 0if n = m.
−1
Let S be the subspace of C[−1,1] spanned by P 0 (x), P 1 (x), P 2 (x). The orthogonal projection of
f onto S is

f S (x) = a 0 P 0 (x) + a 1 P 1 (x) + a 2 P 2 (x)
1
2
= a 0 + a 1 x + a 2 (3x − 1),
2
where
f (x) · P n (x)
a n =
P n (x) · P n (x)
1
x
e P n (x)dx
−1
1
=
P (x)dx
2
−1 n
for n = 0,1,2. These integrals are easily done using MAPLE and we find that
1 −1 −1 35 −1 5
a 0 = (e − e ),a 1 = 3e ,a 2 =− e + e.
2 2 2
Using these coefficients, f s (x) is the closest approximation (in the distance defined by this dot
product) to exp(x) on [−1,1]. Figure 6.17 shows graphs of f (x) and f S (x) on this interval.
We can improve the accuracy of this polynomial approximation by including more terms.
Suppose S is the subspace of C[−1,1] generated by the orthogonal basis consisting of the first
∗
four Legendre polynomials. These are the three given previously, together with
1 3
P 3 (x) = (5x − 3x).
2
S differs from S by the inclusion of P 3 (x) in the basis. Compute the orthogonal projection of
∗
f (x) onto S to obtain
∗
3

f S ∗(x) = a n P n (x).
n=0
where a 0 ,a 1 and a 2 are as before, and
1 x
e P 3 (x)dx
−1
1 2
a 3 =
P (x)dx
−1 3
259 35
= e −1 − e.
2 2
Figure 6.18 shows graphs of f (x) and f S ∗(x) on [−1,1]. These graphs are nearly indistinguish-
able in the scale of the drawing.
With a little more computation we can quantify the distance between f and f S and between
f and f S ∗. The squares of these distances are
1
2 2
f − f S = ( f (x) − f S (x)) dx ≈ 0.00144058
−1

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October 14, 2010 14:21 THM/NEIL Page-185 27410_06_ch06_p145-186