Question

Expand the following functions in Legendre series.

Hint: Solve the recursion relation (5.8e)for P_l(x)and show that ∫_a¹ P_l(x) dx=1/2l+1 [P_l-1 (a)-P_l+1 (a)].

Short answer

The expanded form of the function in the Legendre series is∫_a ¹ P_l(x) dx=1/2l+1 [P_l-1 (a)-P_l+1 (a)].

Step-by-step solution

Concept of Legendre series:

All the Legendre polynomials are orthogonal functions in the range (-1,1) and thus they satisfy the relation-

∫_-1¹ P_l(x)P_m(x) dx = 0

(∀I≠ m)

Calculation to show that from recursion relation of the function ∫a1 Pl(x) dx=1/2l+1 [Pl-1 (a)-Pl+1 (a)]:

From the recursion relation:

(2l+1) P_l(x) = P'_l+1 (x)-P'_l-1 (x)

P_l(x) = [P_l+1(x) - P_l-1(x) ]/2l+1

∫_a ¹ P_l(x) dx= ∫_a ¹ [P'_l+1 (a)-P'_l-1 (x)]/2l+1 dx.

∫_a ¹ P_l(x) dx=1/2l+1 {P_l+1 (1)-P_l-1 (1) -P_l+1 (a) + P_l-1 (a)}

∫_a ¹ P_l(x) dx=1/2l+1 {P_l-1 (a)-P_l+1 (a)}

Hence, the given relation is proved.