B.1 Derivatives
613
x = a 1 y + a 2 y
2
+ a 3 y
3
+ a 4 y
4
+ · · ·
= a 1 [b 1 x + b 2 x
2
+ b 3 x
3
+ b 4 x
4
+ · · · ] + a 2 [b 1 x + b 2 x
2
+ b 3 x
3
+ · · · ]
2
+ a 3 [b 1 x + b 2 x
2
+ · · · ]
3
+ a 4 [b 1 x + · · · ]
4
+ · · ·
in which we have retained specifically only the number of terms in the assumed
expansion of y as a function of x needed to generate all terms having powers of x
up to and including x 4 on the right-hand side of the resulting equation.
If we now collect powers of x on the right-hand side of this equation, we arrive
at the identity
x ≡ a 1 b 1 x + [a 1 b 2 + a 2 b
2
1 ]x
2
+ [a 1 b 3 + 2a 2 b 1 b 2 + a 3 b
3
1 ]x
3
+ [a 1 b 4 + a 2 b
2
2 + 2a 2 b 1 b 3 + 3b
2
1 b 2 a 3 + a 4 b
4
1 ]x
4
+ O(x
5 ) ,
from which we obtain the set of relations
a 1 b 1 = 1
a 1 b 2 + a 2 b
2
1 = 0
a 1 b 3 + 2a 2 b 1 b 2 + a 3 b
3
1 = 0
a 1 b 4 + a 2 b
2
2 + 2a 2 b 1 b 3 + 3a 3 b
2
1 b 2 + a 4 b
4
1 = 0 ,
that must be satisfied by the coefficients b j . Note that we have expressed the
b j in terms of the coefficients a i and b k , with k < j in order to keep our
expressions relatively manageable. These equations can be solved sequentially for
the b j coefficients in terms of the a i coefficients and the b k coefficients for k-values
already obtained in the solution process. We thus obtain the first four coefficients b j
as
b 1 =
1
a 1
,
b 2 = − a 2 b
3
1 = −
a 2
a 3
1
,
b 3 = −b
2
1 (2a 2 b 2 + a 3 b
2
1 ) =
1
a 5
1
(2a
2
2 − a 3 a 1 ) ,
b 4 = −
1
a 5
1
(a 4 + 5a
3
2 b
2
1 − 5a 3 a 2 b 1 ) =
1
a 7
1
[5a 1 a 2 a 3 − 5a
3
2 − a
2
1 a 4 ] .