Question

Expand the following functions in Legendre series.

f(x) = P'_n (x).

Hint: For I≥ n, ∫_-1¹ P'_n(x)P_l(x) dx=0 (Why?); for l<n, integrate by parts.

Short answer

The expanded form of the function in the Legendre series is f(x)=P₀(x)+ 5 P₂(x)+..... + (2n-1) P_n-1(x).

Step-by-step solution

Concept of Legendre series:

The general expression for the function according to Legendre polynomial is given by,

f(x) = Σ_l=0^∞c_l P_l(x) …… (1)

To find the coefficients c_l you multiply with P_m(x)and integrate. Because Legendre polynomials are orthogonal, all the integrals on right are 0 except the one contains c_m and you can evaluate it,

∫_-1¹ [P_m(x)]² dx = 2/2m+1

So you get:

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

∫_-1¹ f(x) P_m(x) dx = ∑_l=0^∞c_m ∫_-1¹ [P_m(x)]² dx

∫_-1¹ f(x) P_m(x) dx = c_m 2/2m+1

∫_-1¹ P'_n(x) P_m(x) dx = c_m 2/2m+1

Integrate the given function f(x) = P'n (x)

It is given that,

f(x) = P'_n(x) …… (2)

By integration:

∫_-1¹ P'_n(x) P_m (x) dx

m>n, m=n' m<n

You know that, ∫_-1¹ P_m(x).

(Any polynomial of degree< m) dx=0 …… (3)

Here, P_n(x) is a polynomial of degree n, therefore P'_n(x) is a polynomial of degree n-1.

∫_-1¹P'_n(x) P_m(x) dx

m≥n ∫_-1¹P'_n(x) P_m(x) dx =0

m<n

Simplify further as follows:

∫_-1¹P'_n(x) P_m(x) dx = [(P_m(x) ∫P'_n(x) dx)]_-1¹ - ∫_-1¹P'_m(x) (∫P'_n(x) dx) dx

=[(P_m(x) P'_n(x) dx)]_-1¹ - ∫_-1¹P'_m(x) P_n(x) dx

=[(P_m(x) P'_n(x) dx)]_-1¹ -0

Simplify further as follows:

∫_-1¹P'_n(x) P_m(x) dx = [(P_m(x) P'_n(x) dx)]_-1¹

= P_m (1) P_n(1) -P_m(-1) P_n (-1)

=1×1 - (-1)^m (-1)ⁿ P_l (1)

=1

So,

P_l (-1) = (-1)^l

P_l (-1) = 1- (-1)^m+n ..... (5)

Obtain the value of different coefficients in the series and evaluate the expression:

To obtain the value of differential coefficients, evaluate as follows:

∫_-1¹P'_n(x) P_m(x) dx = c_m 2/2m+1

0 = c_m 2/2m+1

c_m =0

c_n = c_n+1=c_n+2=...=0

So, m<n.

Evaluate further as follows:

∫_-1¹P'_n(x) P_m(x) dx = c_m 2/2m+1

1-(-1)^m+n=c_m 2/2m+1

c_m =2m+1/2 (1-(-1)^m+n)

Evaluate further for m=0:

c₀ = (2×0)+1/2 {1-(-1)⁰⁺ⁿ}

=1/2 (1-(-1)ⁿ)

Evaluate for m=1,

c₁=(2×1)+1/2 (1-(-1)¹⁺ⁿ)

=3/2 (1-(-1)ⁿ⁺¹)

=3/2 (1+(-1)ⁿ)

Evaluate further for m=2,

c₂ = (2×2+1)/2 (1-(-1)²⁺ⁿ)

=5/2 (1-(-1)²⁺ⁿ)

=5/2 (1-(-1)ⁿ)

Evaluate further for m=n-1:

c_n-1=2 (n-1)+1/2 (1-(-1)^n-1+n)

=2n-1/2 (1-(-1)^2n-1)

=2n-1/2 (1+(-1)²ⁿ)

=2n-1/2 (1+1)

=2n-1

Evaluate further as follows:

f(x) = Σ_l=0^∞c_lP_l(x)

f(x) = c₀P₀(x)+c₁P₁(x)+c₂P₂(x)+.....+c_n-1P_n-1(x)+0

f(x)=1/2 (1-(-1)ⁿ) P₀(x)+3/2(1-(-1)ⁿ) P₁(x)+ 5/2(1-(-1)ⁿ) P₂(x)+..... + (2n-1) P_n-1(x)

f(x)=1/2 (2) P₀(x)+3/2 (0) P₁(x)+ 5/2 (2) P₂(x)+..... + (2n-1) P_n-1(x)

Therefore,

f(x)=P₀(x)+ 5 P₂(x)+..... + (2n-1) P_n-1(x)