91Ó°ÊÓ

The normalization factor for a Legendre polynomial (P_I(x)) in Legendre’s series is given by the integral-

∫ _-1¹[P_I(x)]² dx= 2/2I+1

Consider the graph:

The function for above graph is,

Substitute 0 for m and 1 for P₀ (x) in the above equation.

∫ _-1¹f(x) P₀ (x) dx=c₀ (2/2(0)+1)

∫ _-1⁰(1+x) dx + ∫ ₀¹(1-x) dx = 2c₀

(x+x²/2)_-1⁰ +(x-x²/2)₀¹ = 2c₀

1 = 2c₀

Solve further,

c₀ =1/2.

Substitute 1 for m and x for P₁(x) in equation (1) as follows:

∫ _-1¹f(x) P₁ (x) dx=c₁ (2/2(1)+1)

∫ _-1⁰(1+x) dx + ∫ ₀¹(1-x) dx = 2c₀

=2/3 c₁

c₁=0

Substitute 1 for m and (3x²-1)/2 for P₂(x) in equation (1) as follows:

∫ _-1¹f(x) P₁ (x) dx=c₂ (2/2(2)+1)

∫ _-1⁰(1+x) (3x²-1)/2 dx + ∫ ₀¹(1-x)(3x²-1)/2 dx = 2/5 c₂

- 1/4=2/5 c₂

c₂= - 5/8

From the above method the expression is evaluated and the values are obtained for the different coefficients and is given by,f(x) = 1/2 P₀(x) - 5/8 P₂(x) + ... .

Hence, the expanded form of the function in the Legendre series is f(x) = 1/2 P₀(x) - 5/8 P₂(x) + ... .