Question

Find the best (in the least squares sense) second-degree polynomial approximation to each of the given functions over the interval -1<x<1.

|x|

Short answer

The best second-degree polynomial is 3/16 (5x²-1).

Step-by-step solution

Concept of the Legendre series:

An expansion in Legendre polynomials up to the second degree gives the best second-degree polynomial.

So, this corresponds to the usual Legendre polynomial fit with l=0 and l=2.

f(x)=c₀P₀(x)+c₁P₁(x)+c₂P₂(x)

The coefficients are given by the following integral:

c_l= 2l+1/2 ∫_-1¹ dx P_l(x) f(x)

c_l = 2l+1/2 [∫_-1⁰ dxP_l(x) (-x)+ ∫₀¹ dxP_l(x) (x)]

Calculate the value of different coefficients of the series:

Consider the following function:

Now doing calculation of coefficients as follows:

If l=0, calculation for the coefficient should be as:

c₀=2(0)+1/2 ∫_-1¹dxP_l(x) f(x)

c₀= 1/2 [∫_-1⁰dx (1) (-x) + ∫₀¹dx (1) (x)]

c₀= [(- x²/2)_-1⁰ + (x²/2)₀¹]

c₀= 1/4 (0+1+1-0)

c₀=1/2

If l=1, calculation for the coefficient should be as:

c₁= 2(1)+1/2 [∫_-1⁰ dx P_l(x)(-x)+∫₀¹ dx P_l(x)(x)]

=3/2 [∫_-1⁰ dx(x)(-x)+∫₀¹ dx(x) x]

=3/2 [-∫_-1⁰ x² dx+∫₀¹ dx (x) x]

=3/2 [-∫_-1⁰ x² dx+∫₀¹ (x)² dx]

=3/2 [(-x³/3) _-1⁰ + (x³/3)₀¹]

So,

c₁=1/2 [-(0+1)+(1-0)]

=0

If l=2, calculation for the coefficient should be as:

c₂= 2(2)+1/2 [∫_-1⁰ dx P₂(x) (-x) + ∫₀¹dxP₂(x) x]

=5/2 [∫_-1⁰ dxP₂(x) (-x)+∫₀¹ dxP₂(x) x]

=5/4 [∫_-1⁰(-3x³+x) dx+∫₀¹ (3x³-x) dx]

=5/4 [(-3 x⁴/4+x²/2)_-1⁰+(3 x⁴/4-x²/2)₀¹]

Simplify further as follows:

c₂ =5/4 [0-(-3/4+1/2)+(3/4-1/2)-0]

=5/4 [3/4-1/2+3/4-1/2]

=5/4 (1/2)

=5/8

Put the value of coefficients in the function and find the best fit polynomial:

The required best fit polynomial is,

f(x) = c₀P₀(x)+c₁P₁(x)+c₂P₂(x)

|x| = 1/2 P₀(x)+(0)P₁(x)+5/8 P₂(x)

|x| = 1/2 P₀(x)+5/8 P₂(x)

Replace the values of P₀ and P₂.

|x| = 1/2 +5/8 (3x²-1)/2

|x|=1/2 +15/16 x²-5/16

|x| = 15/16 x² - 3/16

|x| = 3/16 (5x²-1)

The best fit polynomial for the given function is 3/16 (5x²-1).