Page 143 - Advanced engineering mathematics

4.1 Power Series Solutions 123

∞ ∞
n−1 n+1 1
na n x + 2a n x = . (4.1)
1 − x
n=1 n=0
We would like to combine these series and factor out a common power of x to solve for the a n ’s.
To do this, write 1/(1 − x) as a power series about 0 as
∞
1
= x n
1 − x
n=0
for −1 < x < 1. Substitute this into equation (4.1) to obtain
∞ ∞ ∞
n−1 n+1 n
na n x + 2a n x = x . (4.2)
n=1 n=0 n=0
n
Now rewrite the series so that they all contain powers x . This is like a change of variables in the
summation index. First,
∞ ∞
n−1 2 n
na n x = a 1 + 2a 2 x + 3a 3 x + ··· = (n + 1)a n+1 x ,
n=1 n=0
and next,
∞
n+1 2 3
2a n x = 2a 0 x + 2a 1 x + 2a 2 x + ···
n=0
∞

n
= 2a n−1 x .
n=1
Now equation (4.2) can be written as
∞ ∞ ∞

n n n
(n + 1)a n+1 x + 2a n−1 x − x = 0. (4.3)
n=0 n=1 n=0
These rearrangements allow us to combine these summations for n = 1,2,··· and to write the
n = 0 terms separately to obtain
∞
n
((n + 1)a n+1 + 2a n−1 − 1)x + a 1 − 1 = 0. (4.4)
n=1
Because the right side of equation (4.4) is zero for all x in (−1,1), the coefficient of each power
of x on the left, as well as the constant term a 1 − 1, must equal zero. This gives us
(n + 1)a n+1 + 2a n−1 − 1 = 0for n = 1,2,3,···
and
a 1 − 1 = 0.
Then a 1 = 1, and
1
a n+1 = (1 − 2a n−1 ) for n = 1,2,3,··· .
n + 1
This is a recurrence relation for the coefficients, giving a n+1 in terms of a preceding coefficient
a n−1 . Now solve for some of the coefficients using this recurrence relation:
1
(n = 1) a 2 = (1 − 2a 0 ),
2
1 1
(n = 2) a 3 = (1 − 2a 1 ) =− ,
3 3

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October 14, 2010 14:17 THM/NEIL Page-123 27410_04_ch04_p121-136