lecture_21 - Polynomials & FFT Reading Cormen et al. :...

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Polynomials & FFT Reading Cormen et al. : Chapter 30 (recommended) Dasgupta et al. : 2.6 Lipson, Elements of Algebra & Algebraic Computing (available in the library)
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Fast Fourier transform: math & algorithm From Wikipedia , the free encyclopedia A fast Fourier transform ( FFT ) is an efficient algorithm to compute the discrete Fourier transform (DFT) and its inverse. FFTs are of great importance to a wide variety of applications , from digital signal processing and solving 2 partial differential equations to algorithms for quick multiplication of large integers . This article describes the algorithms, of which there are many; see discrete Fourier transform for properties and applications of the transform.
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Mathematics background Def. ω is said to be the n-th root of unity if ω is a root of the equation x n -1 = 0. That is, ω n = 1. Def. ω is said to be a primitive n-th root of unit if ω n = 1, and ω k ≠ 1 for all 0 < k < n. xample 1. n = 2. a 2 d ot of unity, and a rimitive 3 Example 1. 1 is a 2 nd root of unity, and -1 is a primitive 2 nd root of unity. For n > 2, the primitive n-th roots of unity are non-real complex numbers. Let i = √-1.
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Note that e i = 1, e i π/2 = i and e i π = -1 (polar representation). Fact . ω = e i 2π/n is a primitive n-th root of unity. Example 2. n = 4, e i 2π/n = i. n Complex numbers n = 8, e i 2π/n = √2/2 + i (√2/2)
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Trivial fact 1 Assume n is an even number. Lemma 1 . If ω is a primitive n-th root of unity, then ω 2 is a primitive n/2-th root of unity. 5 Proof. • (ω 2 ) n/2 = ω n = 1. • For any 0 < k < n/2, 2 ) k = ω 2k ≠ 1 because 0 < 2k < n.
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Trivial fact 2 Assume n is an even number. Lemma 2 . if ω is a primitive n-th root of unity, then ω n/2 = -1. Proof. Recall that ω n = (ω n/2 ) 2 = 1. 6 Thus, (ω n/2 ) 2 satisfies the equation x 2 = 1. The equation x 2 = 1 has two roots, namely, 1 and -1. Since ω n/2 ≠ 1, ω n/2 = -1.
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Polynomials A degree n-1 polynomial P(x) can be represented by • n coefficients a 0 , a 1 , a 2 , … a n-1 . I.e., P(x) = a 0 + a 1 x + a 2 x 2 + … + a n-1 x n-1 . • values at n points : (x 0 ,y 0 ), (x 1 ,y 1 ), . ., (x n-1 ,y n-1 ). .g., P(x) = x + 1 can be 7 Interpolation Theorem. Given n point-value pairs { (x 0 ,y 0 ), (x 1 ,y 1 ), . ., (x n-1 ,y n-1 ) }, there is a unique degree (n-1) polynomial P(x) such that P(x i ) = y i for all i. E.g., P(x) = x + 1 can be represented by two points (0,1) and (-1,0). In fact, any two points on the green line can represent p(x) = x+1.
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Evaluation of polynomials How to evaluate P(x) = a 0 + a 1 x + a 2 x 2 + … + a n-1 x n-1 at a certain value, say, x 1 . Horner’s rule: n-1 multiplications (plus n additions). (a x + a ) 8 n-1 1 n-2 (a n-1 x 1 + a n-2 ) x 1 + a n-3 ((a n-1 x 1 + a n-2 ) x 1 + a n-3 ) y + a n-4 . . (..((a n-1 x 1 + a n-2 ) x 1 + a n-3 ) x 1 + a n-4 ) x 1 + …) x 1 + a 0
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Polynomial multiplications (convolution) Given P(x) = a 0 + a 1 x + a 2 x 2 + … + a n-1 x n-1 and Q(x) = b 0 + b 1 x + b 2 x 2 + … + b n-1 x n-1 , computing P(x) × Q(x) requires O(n 2 ) multiplications.
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This note was uploaded on 03/01/2010 for the course CS 1234 taught by Professor Chan during the Spring '10 term at University of the Bío-Bío.

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lecture_21 - Polynomials & FFT Reading Cormen et al. :...

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