91Ó°ÊÓ

Legendre series of the function f(x) is,

f(x) = ∑_l=0^∞c_lP_l(x) ..... (1)

On Solving:

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

c_m[2/2m+1] =∫_-1¹f(x)p_m(x) dx ..... (2)

Given function is

f(x) = ³¦´Ç²õÏ€³æ

Now calculating for m=0:

2c₀ = ∫_-1¹f(x)P₀(x) dx

2c₀ = ∫_-1¹ ³¦´Ç²õÏ€³æ P₀(x) dx

c₀ = ∫₀¹ ³¦´Ç²õÏ€³æ P₀(x) dx

c₀ =0

Now calculating for m=1:

2/3 c₁ = ∫_-1¹f(x)P₁(x) dx

2/3 c₁ = ∫_-1¹ x ³¦´Ç²õÏ€³æ P₁(x) dx

c₁ =0

Now calculating for m=2:

2/5 c₂ = ∫_-1¹f(x)P₂(x) dx

2/5 c₂ = 1/2 ∫_-1¹ ³¦´Ç²õÏ€³æ (3x²-1) dx

2/5 c₂ = ∫₀¹ 3x² ³¦´Ç²õÏ€³æ - ∫₀¹ 1 dx

2/5 c₂ = 6/Ï€²cosÏ€ -1

Therefore,

c₂ = -5/2 (6/Ï€²+1)

On putting the coefficients in function, f(x) = -5/2 (6/Ï€²+1)P₂(x).

The best second-degree polynomial approximation is, -5/4 (6/Ï€²+1) (3x²-`1).