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A.5. VANDERMONDE SYSTEMS 217
Clearly c ≡ a
about (A.7)?
(mod M), so satisfying (A.8) means that (A.6) is satisfied. What
c = a + λM(b − a)
= a + (1 − µN)(b − a)
≡ a + (b − a) (mod N)
so, despite the apparent asymmetry of the construction, c ≡ b (mod N) as
well.
As in Algorithm 40, X, a and b may as well be polynomials in x whose
coefficients are polynomials in y (but M and N are still in y only).
Algorithm 42 (Chinese Remainder (Multivariate))
Input: Polynomials a = ∑ n
i=0 a ix i , b = ∑ n
i=0 b ix i ∈ K[y][x], and M and
N ∈ K[y] (with gcd(M, N) = 1).
Output: A polynomial = ∑ n
i=0 c ix i satisfying Theorem 36.
Compute λ, µ such that λM + µN = 1;
#As in Lemma 1 (page 45). Note µ isn’t needed in practice.
for i := 0 to n do
c i :=a i + λM(b i − a i );
It is even possible for x (and i) to represent numerous indeterminates, as we are
basically just doing coefficient-by-coefficient reconstruction.
Observation 14 We have explicitly considered K[y] (e.g. Q[y]), but in practice
we will often wish to consider Z[y]. Even if all the initial values are in Z[x],
λ and µ may not be, as in M = (y − 1) and N = (y + 1), when λ = −1
2
and
µ = 1 2
. This may not be an obstacle to such reconstruction from values (often
called interpolation, by analogy with interpolation over R). Interpolation over
Z[y] may still be possible: consider reconstructing X with X ≡ 1 (mod M)
and X ≡ −1 (mod N), which is 1 + −1
2
M(−1 − 1) = y. But interpolation over
Z[y] is not always possible: consider reconstructing X with X ≡ 1 (mod M)
and X ≡ 0 (mod N), which gives 1 2
(y − 1).
A.5 Vandermonde Systems
Definition 92 The Vandermonde matrix 4 generated by k 1 , . . . , k n is
⎛
1 k 1 k1 2 . . . k n−1 ⎞
1
1 k 2 k2 2 . . . k n−1
2
V (k 1 , . . . , k n ) = ⎜
⎟
⎝ . . . . ⎠ .
1 k n kn 2 . . . kn
n−1
4 This section is based on [Zip93, §13.1]. Our algorithm 43 is his SolveVanderMonde, and
our algorithm 44 is the one at the top of his p. 214.