Page 551 - Advanced_Engineering_Mathematics o'neil

15.2 Legendre Polynomials 531

First observe from Legendre’s differential equation that
2

(1 − x )P (x) − 2xP (x) + N(N + 1)P N (x) = 0.

N
N
If we put x = x k ,the kth zero of P N (x), then P N (x k ) = 0 and P (x k ) = 0, so

N
P (x k ) 2x k

N
= .
P (x k ) 1 − x 2

N k
The right side of this equation is the left side of the necessary and sufficient condition for equi-
librium. To complete the proof, we will show that the right side of this condition is also equal to

P (x)/P (x k ).

N N
Since P N (x) has all simple zeros, then for some number A,
P N (x) = A(x − x 1 )(x − x 2 )···(x − x N ).
In product notation,
P N (x) = A N (x − x i ).
i=1
Take the derivative of this product and use the fact that (d/dx)(x − x i ) = 1:
N

P (x) = A N (x − x m )
N m=1,m =i
i=1
so
N N

P (x) = A N (x − x m ).
N m=1,m =i, j
j=1 i=1,i = j
Then
N

P (x k ) = A N (x k − x m ).

N m=1,m =i
i=1
In this expression, all terms are zero except when m = k,so
N

P (x k ) = A m=1,m =k (x k − x m ).
N
Similarly compute
N
N
P (x k ) = 2A m=1,m = j,k (x k − x m ).

N
j=1, j =k

Since the zeros are simple, then P (x k ) = 0 and
N
P (x k ) 2

N
= .
P (x k ) j=1, j =k x k − x j

N
This is the right side of the necessary and sufficient condition for the beads to be in equilibrium.
Using j for the summation index on the right, we now have
N
P (x k ) 2

2x k N
= = .
1 − x 2 P (x k ) x k − x j

N
k
j=1, j =k
Therefore the zeros satisfy the necessary and sufficient condition for the beads to be in
equilibrium, completing the proof.
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October 14, 2010 15:20 THM/NEIL Page-531 27410_15_ch15_p505-562