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Unformatted text preview: Solutions to selected exercises • Use the Bisection method to find solutions accurate to within 10 − 2 for x 3 − 7 x 2 + 14 x − 6 = 0 on [0,1]. Solution: Let f ( x ) = x 3 − 7 x 2 + 14 x − 6 = 0. Note that f (0) = − 6 < and f (1) = 2 > 0, therefore, based on the Intermediate Value Theorem, since f is continuous, there is p ∈ (0 , 1) such that f ( p ) = 0. Let a = 0, b = 1, with f ( a ) < 0, f ( b ) > 0. Let p = a + b − a 2 = 0 . 5, and we have f ( p ) = − . 6250 < 0 (the same sign as f ( a ), therefore a 1 = p = 0 . 5, b 1 = b = 1 and repeat: p 1 = 0 . 75, ... This yields the following results for p n and f ( p n ): n p n f ( p n ) 0.50.6250000 1 0.75000000 +0.9843750 2 0.62500000 +0.2597656 3 0.562500000.1618652 4 0.59375000 +0.0540466 5 0.578125000.0526237 6 0.58593750 +0.0010313 • Use the theorem from the course to find a bound for the number of iterations needed to achieve an approximation with accuracy 10 − 3 to the solution of x 3 − x − 1 = 0 lying in the interval [1 , 4]. Solution: Let’s first verify that f has a zero in the interval [1 , 4]: f (1) = − 2 < 0, f (4) = 64 > 0, therefore, since f is continuous, by the Intermediate Value Theorem, f has a zero in [1 , 4]. By the theorem from the course, we impose:  p n − p  ≤ b − a 2 n = 3 2 n ≤ 10 − 3 , then 3 · 10 3 ≤ 2 n ⇒ n ≥ log 10 (3 · 10 3 ) log 10 (2) ≈ 11 . 55 • Use algebraic manipulation to show that each of the following functions...
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This note was uploaded on 10/28/2010 for the course MATH 151a taught by Professor Staff during the Spring '08 term at UCLA.
 Spring '08
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