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(i) b 11 a 11 =1 ,
b 11 = a 1
11
(ii) b 11 a 12 + b 12 a 22 =0
b 11 a 12 a 1
a 3
b 12 =
22 =
11 a 12 ,
(B.8)
(iii) b 11 a 13 + b 12 a 23 + b 13 a 33 =0
b 13 = ( b 11 a 13 + b 12 a 23 ) a 1
33 = a 1
11 ( a 12 a 1
22 a 23 − a 13 ) a 1
33 ,
(iv) b 11 a 14 + b 12 a 24 + b 13 a 34 + b 14 a 44 =0
b 14 = ( b 11 a 14 + b 12 a 24 + b 13 a 34 ) a 1
44 = a 1
11 [ a 12 a 1
22 ( a 24 − a 23 a 1
33 a 34 )
+ a 13 a 1
a 14 ] a 1
33 a 34
44 .
Consult Box B.5 for the general representation of b 1 n .
Notable for the GKS algorithm is the following. In the first step or the forward substitution,
a set of equations
x, x 2 , ..., x n− 1 ,x n
is constructed by substituting x ( y )intothepowers
x, x 2 , ...,x n− 1 ,x n , finally written into a matrix equation. The upper triangular matrix
{
}
A is
gained by a multinomial expansion as indicated. In contrast, the second step or the backward
substitution is based upon the upper triangular matrix
1
B :=
A , the inversion of
A .Its
first row contains the unknown coecients we are looking for:
{
b 11 ,b 12 ,..., b 1 n− 1 ,b 1 n }
The
1
construction of
A can be based on symbolic computer manipulation. The
algebraic manipulation becomes more concrete when we pay attention to Examples B.1 and B.2 .
The first example aims at the inversion of an univariate homogeneous polynomial of degree n =2,
namely y ( x )= a 11 x + a 12 x 2
A as well as
x ( y )= b 11 y + b 12 y 2 . The GKS algorithm determines the set of
coecients
from the two given coecients a 11 and a 12 . In contrast, the second example
focuses on the inversion of an univariate homogeneous polynomial of degree n =3,namely
y ( x )= a 11 x + a 12 x 2 + a 13 x 3
{
b 11 ,b 12
}
→ x ( y )= b 11 y + b 12 y 2 + b 13 y 3 . The GKS algorithm determines the
set of coecients {b 11 ,b 12 ,b 13 } from the three given coecients a 11 ,a 12 ,and a 13 .
Example B.1 (Inversion of an univariate homogeneous polynomial of degree n =2).
Assume the univariate homogeneous polynomial y ( x )= a 11 x + a 12 x 2 to be given and find the
inverse univariate homogeneous quadratic polynomial x ( y )= b 11 y + b 12 y 2 by the GKS algorithm.
1st step:
x ( y )= b 11 y + b 12 y 2 = b 11 a 11 x +( b 11 a 12 + b 12 a 11 ) x 2 + β 13 ,
(B.9)
x 2 ( y )= b 22 y 2 + β 23 = b 22 a 11 x 2 + β 23 .
 
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