Lecture Notes in Differential Equations - Bruce E. Shapiro

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into (28.114) and collect terms,
∞∑
∞∑
∞∑
0 = (1 − t 2 ) k(k − 1)a k t k−2 − 2t ka k t k−1 + n(n + 1) a k t k
k=0
k=0
k=0
(28.118)
Let j = k − 2 in the first sum and observe that the first two terms in that
sum are zero.
∞∑
∞∑
0 = (j + 1)(j + 2)a j+2 t j [
+ −k − k 2 + n(n + 1) ] a k t k (28.119)
j=0
k=0
We can combine this into a single sum in t k ; by linear independence all of
the coefficients must be zero,
(k + 1)(k + 2)a k+2 − [k(k + 1) − n(n + 1)]a k = 0 (28.120)
for all k = 0, 1, 2, ..., and therefore
a k+2 =
k(k + 1) − n(n + 1)
a k , k = 0, 1, 2, ... (28.121)
(k + 1)(k + 2)
The first two coefficients, a 0 and a 1 are arbitrary. The remaining ones are
determined by (28.121), and generates two sequences of coefficients
a 1 , a 3 , a 5 , a 7 , ...
a 0 , a 2 , a 4 , a 6 , ...
(28.122)
so that we can write
y = ∑
a k t k + ∑
a k t k (28.123)
k even k odd
These two series are linearly independent. In particular, the right hand side
of (28.121) vanishes when
k(k + 1) = n(n + 1) (28.124)
so that for k = n one of these two series will be a finite polynomial of order
n. Normalized versions of the solutions are called the Legendre Polynomials,
and the first few are given in the (28.125) through (28.135).