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A.5. VANDERMONDE SYSTEMS 219
whereas we need to solve a system of the form
k 1 x 1 + k 2 x 2 + · · · + k n x n = w 1
k1x 2 1 + k2x 2 2 + · · · + knx 2 n = w 2
.
k1 n x 1 + k2 n x 2 + · · · + knx n n = w n .
(A.11)
By comparison with (A.10), we have transposed the matrix (which is not a
problem, since the inverse of the transpose is the traspose of the inverse), and
multiplied column i by an extra factor of k i . From Corollary 25, we can deduce
the criteria for this system to be soluble.
Corollary 26 If all the k i are distinct and non-zero, then the system (A.11) is
soluble.
The following variant of Algorithm 43 will solve the system.
Algorithm 44 (Vandermonde variant solver)
Input: Vandermonde style data (k 1 , . . . , k n ), right-hand side w.
Output: Solution x to (A.11)
x := 0
P := ∏ n
i=1 (z − k i) #O(n 2 )
for i := 1 to n
Q := P/(z − k i ) #nO(n)
D := k i P (k i ) #nO(n) by Horner’s rule
for j := 1 to n
coeff(Q,z
x i := x i + w j−1 )
j D