First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 5. APPROXIMATION THEORY 203
Section 4.3 3 . They can be obtained from the following recursion
n =1, 2,..., and they satisfy
L n+1 (x) =
2n +1
n +1 xL n(x) −
n
n +1 L n−1(x),
〈L n ,L n 〉 = 2
2n +1 .
Exercise 5.3-1:
L 2 (x) are orthogonal.
Show, by direct integration, that the Legendre polynomials L 1 (x) and
Example 91 (Chebyshev polynomials). If we take w(x) =(1− x 2 ) −1/2 and [a, b] =[−1, 1],
and again drop the orthonormal requirement in Gram-Schmidt, we obtain the following
orthogonal polynomials:
T 0 (x) =1,T 1 (x) =x, T 2 (x) =2x 2 − 1,T 3 (x) =4x 3 − 3x, ...
These polynomials are called Chebyshev polynomials and satisfy a curious identity:
T n (x) =cos(n cos −1 x),n≥ 0.
Chebyshev polynomials also satisfy the following recursion:
T n+1 (x) =2xT n (x) − T n−1 (x)
for n =1, 2,...,and
⎧
0 if j ≠ k
⎪⎨
〈T j ,T k 〉 = π if j = k =0
⎪⎩ π/2 if j = k>0.
If we take the first n +1Legendre or Chebyshev polynomials, call them φ 0 , ..., φ n , then
these polynomials form a basis for the vector space P n . In other words, they form a linearly
independent set of functions, and any polynomial from P n can be written as a unique linear
combination of them. These statements follow from the following theorem, which we will
leave unproved.
3 The Legendre polynomials in Section 4.3 differ from these by a constant factor. For example, in Section
4.3 the third polynomial was L 2 (x) =x 2 − 1 3 , but here it is L 2(x) = 3 2 (x2 − 1 3
). Observe that multiplying these
polynomials by a constant does not change their roots (what we were interested in Gaussian quadrature),
or their orthogonality.