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1. requires f(n) � For n = n + = 2h andfor a positive integer h (see (2. 1 . 5 n log2 n ops. The next result can be obtained based on a converse divide-and-conquer process [BP94, p. 1 2] or as a simple corollary of Theorems 2. 6, 2. 2. 1 . 2. For n = n + using n + 1 . 5n log n ops. 2 (see (2. 1 . 1)), the inverse DFTn problem can be solved by [Ch. 3 Fast Fourier transform (matrix version) A matrix version of FFf leads essentially to the same algorithm. Based on this version, the algorithm has been proved to be stable numerically.

L. :l. :l. A,B (M)M - I A , if A is a non-singular matrix. Proof. The claimed equations can be immediately verified by inspection. Let us also show a proof with derivation. 5 . 3 . Pre-multiply it by B - 1 and then interchange B and B - 1 throughout to obtain the second equation. Post-multiply it by A - I and then interchange A and A - I throughout to obtain the last equation of Theorem 1 . 5 . 3 . A , B . More specialized and more explicit expressions are known i n the important special case where M is a Toeplitz or Hankel matrix (see (2.

Lf = f The straightforward solution algorithm uses 9 (m + 1 ) (n + 1 ) - m - n - l ops. 2 For m = n, this is 9 (n + 1 ) - 2n - 1 . 4 . 1 ) and can b e applied over any field or even over any ring o f constants. Let us show a superfast algorithm that relies on the evaluation/interpolation tech­ niques of A. L. Toom, 1 963. The solution assumes computations in the field containing a primitive 2h root of 1 for h = rlog2 (m + n + 1 )1 , that is, h rlog2 (2n + 1 ) 1 for m = n, and then supports the complexity bound of M(n) ops for = M(n) Solution: Write h= = O (n log n ) .

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