Lecture Notes in Differential Equations - Bruce E. Shapiro

257
Changing the index of the second sum only to k = m − 1,
∞∑
∞∑
∞∑
0 = ka k t k − a k t k+1 − a k t k (28.10)
=
k=0
∞∑
ka k t k −
k=0
k=1
k=0
m=1
k=0
∞∑
∞∑
a m−1 t m − a k t k (28.11)
k=1
∞∑
∞∑
= (k − 1)a k t k − a k−1 t k (28.12)
=
k=1
∞∑
[(k − 1)a k − a k−1 ]t k (28.13)
k=1
In the last line we have combined the two sums into a single sum over k.
Equation (28.13) must hold for all t. But the functions { 1, t, t 2 , ... } are
linearly independent, and there is no non-trivial set of constants {C k } such
that ∑ ∞
k=0 C kt k = 0, i.e., C k = 0 for all k. Hence
a 0 = 0 (28.14)
(k − 1)a k − a k−1 = 0, k = 1, 2, ... (28.15)
Equation (28.15) is a recursion relation for a k ; it is more convenient to
write it as (k − 1)a k = a k−1 . For the first several values of k it gives:
k = 1 :0 · a 1 = a 0 (28.16)
k = 2 :1 · a 2 = a 1 ⇒ a 2 = a 1 (28.17)
k = 3 :2 · a 3 = a 2 ⇒ a 3 = 1 2 a 2 = 1 2 a 1 (28.18)
k = 4 :3 · a 4 = a 3 ⇒ a 4 = 1 3 a 3 = 1
3 · 2 a 1 (28.19)
k = 5 :4 · a 5 = a 4 ⇒ a 5 = 1 4 a 1
4 =
4 · 3 · 2 a 1 (28.20)
.
The general form appears to be a k = a 1 /(k − 1)!; this is easily proved by
induction, since the recursion relationship gives
a k+1 = a k /k = a 1 /[k · (k − 1)!] = a 1 /k! (28.21)