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15.2 Legendre Polynomials 529

More than this can be shown. A zero x 0 of f (x) is simple if f (x 0 ) = 0but f (x 0 ) = 0. For

2
2
example, f (x)= x −4 has simple zeros at x =2 and x =−2, while g(x)=(1+ x) has a zero at
x =−1, but this zero is not simple because g(−1)= g (−1)=0. We call −1 a double or repeated

zero of g(x).
THEOREM 15.7 Zeros of Legendre Polynomials

For each positive integer n, P n (x) has n simple zeros, all lying between −1 and 1.

Proof Let n be a positive integer.
First we will show that, if P n (x) has a zero x 0 in (−1,1), then this zero must be simple.
Since P n (x) satisfies Legendre’s equation with λ = n(n + 1), then

2
(1 − x )P + n(n + 1)P n = 0.
n
If P n (x 0 ) = P (x 0 ) = 0, then P n (x) satisfies the initial value problem

n
2

[(1 − x )y ] + n(n + 1)y = 0; y(x 0 ) = y (x 0 ) = 0.
But y(x) = 0, the identically zero function, is also a solution of this problem. By uniqueness of
the solution to this problem, we would then have y(x) = P n (x) = 0 for all x, and this is false.
Therefore, any zero that P n (x) has must be simple.
Next we will show that P n (x) has at least one zero in (−1,1). Because P n (x) is orthogonal
to P 0 (x) = 1,
1 1

P 0 (x)P n (x)dx = P n (x)dx = 0.
−1 −1
If P n (x) were strictly positive, or strictly negative, on (−1,1), then this integral would be,
respectively, positive or negative. Therefore for some x 0 in (−1,1), P n (x 0 ) = 0.
We now know that P n (x) has at least one zero in (−1,1), and any zero must be simple.
Now we will show that P n (x) has n zeros between −1 and 1. Let x 1 ,··· , x m be all the zeros
of P n (x) in (−1,1). Then 1 ≤ m ≤ n. Order these zeros from left to right across the interval
−1 < x 1 < x 2 < ··· < x m < 1.
If m < n, then the polynomial

q(x) = (x − x 1 )(x − x 2 )···(x − x m )
has degree m < n and the same zeros as P n (x) in (−1,1). Then P n (x) and q(x) change sign at
exactly the same points in (−1,1). This means that P n (x) and q(x) are either of the same sign
on each interval (−1, x 1 ),(x 1 , x 2 ), ··· ,(x m−1 , x m ), (x m ,1), or of opposite sign on each of these
intervals. But then q(x)P n (x) is either strictly positive or strictly negative on (−1,1), except at
x 1 ,··· , x m . Then
1

q(x)P n (x)dx
−1
must be positive or negative. But this integral is zero if the degree of q(x) is less than the degree
of P n (x). We conclude that m = n, hence P n (x) has n simple zeros on (−1,1), as was to be
proved.

The zeros of Legendre polynomials have quite fascinating properties, one of which we will
now explore.

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October 14, 2010 15:20 THM/NEIL Page-529 27410_15_ch15_p505-562