Expand the following functions in 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_lis obtained by the similar method to the one use in determining the formula for the coefficients in the Fourier series.

The value of c_lwill be as follows:

f(x) =∑_l=0^∞c_lP_l (x)

c_m 2/2m+1 = ∫_-1¹ f(x) p_m(x) dx

For m=0:

2c₀ = ∫₀¹ (log 1/x)² P₀(x) dx ..... (1)

To solve the equation further, let

log 1/x=u

1/x=e^u

x=e^-u

dx= -e^u du

Substituting value of x and dx in equation (1), you get-

2c₀ = ∫₀^∞e^-u u² du

2c₀ = 2

c₀ = 1

For m=1:

2/3 c₁ = ∫_-1¹f(x) P₁ (x) dx

2/3 c₁ = ∫₀¹x (log 1/x)² dx

c₁ = 3/8

For m=2:

2/5 c₂ = ∫_-1¹f(x) P₂ (x) dx

2/5 c₂ = 1/2 ∫₀¹(3x²-1) (log 1/x)² dx

c₂ = -20/9

Therefore, the given function can be expanded as f(x) = P₀(x)+3/8 P₁(x) -20/9 P₂(x).