DISCRETE FOURIER TRANSFORM
JOHN RANDALL
Contents 1. 2. 3. 4. 5. Representations of polynomials Comparison of the representations Roots of unity Discrete Fourier Transform Fast Fourier Transform 1 2 2 2 2
1. Representations of polynomials A polynomial f (x
Numerical Analysis/Fall 2009/Review #1 Solutions to Sample Questions. (1) (a) f (1) = 2 < 0, f (2) = 2 > 0, so by the Intermediate Value Theorem, f has a root in (1, 2). (b) f (x) = 3x2 3 = 3(x2 1) > 0 for x (1, 2), so f is increasing on [1, 2], and the r
Numerical Analysis/Exam #1/October 6, 2004
NAME:
Answer all questions. For full credit, you must show your work. Total points: 50. 1. [10 points] Let f (x) = x2 + 3x 1. (a) Show that f has a unique root in the interval [0, 2]. (b) Use the bisection method
Numerical Analysis/Exam #1/October 6, 2004 Solutions
1. (a) f (0) = 1 < 0, f (2) = 9 > 0, and f is continuous, so the Intermediate Value Theorem tells us that f has a root in (0, 2). (b) a b m e 0 2 1 1 0 1 0.5 0.5 0 0.5 0.25 0.25 0.25 0.5 0.375 0.125 0.2
Numerical Analysis/Exam #1/October 15, 2009
SOLUTIONS 1. (a) f (0) = 3 < 0, f (2) = 5 > 0, so there is a root in (0, 2). (b) f (x) = 3x2 0 on (0, 2), so f is increasing and root is unique. a b m 0 2 1 1 2 1.5 1 1.5 1.25 1.25 1.5 1.375 1.375 1.5 1.4375 ln
(1) (2)
(3) (4)
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Math 473/Fall 2009/Homework #1: Bisection Do 3 iterations of the bisection method for f (x) = x cos x on [0, 1]. Use the bisection method to nd solutions accurate to within 102 for f (x) = x3 7x2 + 14x 6 = 0 (a) on [0, 1] (b) on [3.2,
Math 473/Fall 2009/Homework #2: Newton Apply Newtons method to the following problems. In each case, you are given a function f and an initial estimate p0 to a root p of f . You need to nd p1 , p2 , p3 , . . . , using the iteration formula f ( pn ) pn+1 =
Math 473/Numerical Analysis/Fall 2009 Derivation of the Black-Scholes equation Terminology. Consider a European call option. t time T expiry S (t) value of underlying security V (S, t) value of option drift of underlying security volatility of underlying
Numerical Analysis/Quiz #1/September 14, 2009
SOLUTIONS 1. After n iterations, error is at most (b a)/2n . In this case, we need to nd n such that 1/2n < 103 , that is 2n > 1000. We could determine this by counting on our ngers: 29 = 512 < 1000 < 1024 = 2
Numerical Analysis/Fall 2009/Review #3/Solutions (1) (a) f (1) = 5 < 0, (3) = 5 > 0, so by the Intermediate Value Theorem, f has a root in [1, 3]. Since f (x) = 2x + 1 > 0 on [1, 3], f is increasing, and the root is unique. (b) 1.00000 3.00000 2.00000 2.0
Math 473/Numerical Analysis/Fall 2009 The Heat Equation The heat equation is the partial dierential equation ut = uxx . Solving this equation analytically is beyond the scope of the course. We instead solve it numerically. A solution of the heat equation