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Hw_Sol

# Hw_Sol - Solutions to selected exercises Use the Bisection...

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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 < 0 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 = 0, b 0 = 1, with f ( a 0 ) < 0, f ( b 0 ) > 0. Let p 0 = a 0 + b 0 a 0 2 = 0 . 5, and we have f ( p 0 ) = 0 . 6250 < 0 (the same sign as f ( a 0 ), therefore a 1 = p 0 = 0 . 5, b 1 = b 0 = 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 0.5 -0.6250000 1 0.75000000 +0.9843750 2 0.62500000 +0.2597656 3 0.56250000 -0.1618652 4 0.59375000 +0.0540466 5 0.57812500 -0.0526237 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 has a fixed point at p precisely when f ( p ) = 0, where f ( x ) = x 4 +2 x 2 x 3.

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