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Chapter 12: Q6P (page 584)

Substitute the P1(x)you found in Problems 4.3 or 5.3 into equation (10.6)to find, Plm(x); then let x=cos θto evaluate:

P32(cosθ)

Short Answer

Expert verified

The value of P32(cosθ) is 15 cosθsin2θ.

Step by step solution

01

Concept:

An equation similar to Legendre’s equation is-

(1-x2) y"-2xy'+[l(l+1)-m2/1-x2] y=0 (∀ m2≤ l2)

The solution to this equation is of the form

Plm (x)=(1-x2)m/2 dm/dxm (Pl(x))

These solutions are called Associated Legendre Functions.

It is given that,

Plm(x) =(1-x2)m/2 dm/dxm Pl(x) …… (1)

02

Put l=3 and m=2 in equation (1):

Putting l=3 and m=2:

P32(x) = (1-x2)2/2 d2/dx2 P3(x)

P32(x) = (1-x2)1 d2/dx2 (1/2 (5x2-3x))

P32(x) = (1-x2) (15x)

Substitute cosθ for x as,

P32(cosθ) = (1-cos2θ) 15 cosθ

=15 cosθ sin2θ

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Most popular questions from this chapter

Find the best (in the least squares sense) second-degree polynomial approximation to each of the given functions over the interval -1<x<1.

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To show J3/2(x)=x-1J1/2(x).

Substitute the Pl(x), you found in Problems 4.3 or 5.3into equation (10.6)to find Plm(x) then let x=cosθ to evaluate:

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4x2(x+1)y"-4x2y'+(3x+1)y=0
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