First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 5. APPROXIMATION THEORY 199
which gives the (n +1)normal equations for the continuous least squares problem:
n∑
∫ b
a j x j+k dx =
j=0
a
∫ b
a
f(x)x k dx (5.11)
for k =0, 1, ..., n. Note that the only unknowns in these equations are the a j ’s, hence this is
a linear system of equations. It is instructive to compare these normal equations with those
of the discrete least squares problem:
(
n∑ m
)
∑
m∑
a j = y i x k i .
j=0
i=1
x k+j
i
Example 86. Find the least squares polynomial approximation of degree 2 to f(x) =e x on
(0, 2).
Solution. The normal equations are:
2∑
∫ 2
a j x j+k dx =
j=0
k =0, 1, 2. Here are the three equations:
Computing the integrals we get
0
∫ 2
0
i=1
e x x k dx
∫ 2 ∫ 2 ∫ 2
a 0 dx + a 1 xdx + a 2 x 2 dx =
0
0
∫ 2 ∫ 2 ∫ 2
a 0 xdx + a 1 x 2 dx + a 2 x 3 dx =
0
0
∫ 2 ∫ 2 ∫ 2
a 0 x 2 dx + a 1 x 3 dx + a 2 x 4 dx =
0
0
0
0
0
∫ 2
0
∫ 2
0
∫ 2
0
e x dx
e x xdx
e x x 2 dx
2a 0 +2a 1 + 8 3 a 2 = e 2 − 1
2a 0 + 8 3 a 1 +4a 2 = e 2 +1
8
3 a 0 +4a 1 + 32
5 a 2 =2e 2 − 2
whose solution is a 0 =3(−7+e 2 ) ≈ 1.17, a 1 = − 3 2 (−37+5e2 ) ≈ 0.08,a 2 = 15
4 (−7+e2 ) ≈ 1.46.
Then
P 2 (x) =1.17+0.08x +1.46x 2 .
The solution method we have discussed for the least squares problem by solving the