Question

Expand the following functions in Legendre series.

Short answer

The expression of the given function in Legendre series is, f(x) = 1/3 P₀(x) + 2/5 P₁(x) -11/42 P₂(x) ... .

Step-by-step solution

Concept of Legendre series:

The general expression for the function according to Legendre polynomial is given as follows:

f(x) = ∑_l=0^∞c_lP_l (x)

In the above equation the value of coefficient c_l is obtained by the similar method to the one use in determining the formula for the coefficients in the Fourier series.

The both side of both the equation is multiplied by P_m (x)and integrated from -1to 1 as the Legendre polynomial are orthogonal. All the integrals on the right are zero except the one containing coefficient c_m, the function is evaluated as follows:

∫_-1¹ f(x)P_m (x) dx= ∑_l=0^∞P_l (x)P_m (x) dx

∫_-1¹ f(x)P_m (x) dx= c_m∫_-1¹ [P_m (x)]² dx

∫_-1¹ f(x)P_m (x) dx= c_m 2/2m+1

The value of c_l will be:

f(x) = ∑_l=0^∞ c_lP_l (x)

c_m 2/2m+1 = ∫_-1¹ f(x)P_m (x) dx

Obtain the value of different coefficients of the series:

For m=0;

2c₀ = ∫_-1¹ f(x)P_m (x) dx

2c₀ = ∫₀¹ √(1-x) dx

c₀ = 1/3

For m=1:

2/3 c₁ = ∫_-1¹ f(x)P₁ (x) dx

2/3 c₁ = ∫₀¹ x√(1-x) dx

2/3 c₁ = ∫₀¹ x^2-1(1-x)^3/2-1 dx

2/3 c₁ = ∫₀¹ x^2-1(1-x)^3/2-1 dx

2/3 c₁ = β (1/2, 3/2)

So,

c₁ = 2/5

For m=2:

2/5 c₂= ∫_-1¹ f(x)P₂ (x) dx

2/5 c₂ = 1/2 ∫₀¹ (1-x)^1/2 (3x²-1) dx

2/5 c₂ = ∫₀¹ 3 (1-x)^3/2-1x^3-1 dx - 1/2 ∫₀¹ x^1-1 (1-x)^3/2-1x^3-1 dx

2/5 c₂ = 3/2 β (3, 3/2) - 1/2 × 2/3

So,

c₂ = -11/42

Therefore, f(x) = 1/3 P₀(x) + 2/5 P₁(x) -11/42 P₂(x).