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HW5solution

# HW5solution - Homework 5 Solution 1 Consider the following...

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Homework 5 Solution 1. Consider the following function: f ( x ) = 1 + 4 x - 2 x 2 + cos x (a) [2pt] Find an maximum using three iterations of the Golden Section search with initial guesses x l = - 5 and x u = 5. (b) [2pt] Find an maximum using three iterations of quadratic interpolation with initial guesses x 0 = - 5 , x 1 = - 2, and x 2 = 3. (c) [2pt] Find an maximum using three iterations of the Newton’s- method with an initial value x 0 = 5. Solution. (a) Golden Section Method Iteration 1:Given initial guesses x l = - 5 , x u = 5 d = 5 - 1 2 ( x u - x l ) = 6 . 18 x 1 = x l + d = - 5 + 6 . 18 = 1 . 18 , x 2 = x u - d = - 1 . 18 f ( x 1 ) = 3 . 3161 , f ( x 2 ) = - 6 . 1239 f ( x 2 )is smaller,let x l = x 2 = - 1 . 18 , x 2 = x 1 = 1 . 18 Iteration 2: d = 5 - 1 2 ( x u - x l ) = 3 . 8195 x 1 = x l + d = 2 . 6395 , f ( x 1 ) = - 3 . 2521 f ( x 1 )is smaller,let x u = x 1 = 2 . 6395 , x 1 = x 2 = 1 . 18 Iteration 3: d = 5 - 1 2 ( x u - x l ) = 2 . 3606 x 2 = x u - d = 0 . 2789 f ( x 2 ) = f (0 . 2789) = 2 . 9214 , f ( x 2 )is smaller,let x l = x 2 = 0 . 2789. Now the maxima is limited within the range[0.2789,2.6395]. (b) Quadratic interpolation Iteration 1: Given x 0 = - 5 , x 1 = - 2 , x 2 = 3 with f ( x 1 ) = - 15 . 4161 x 3 = 1 2 f ( x 0 )( x 2 1 - x 2 2 ) + f ( x 1 )( x 2 2 - x 2 0 ) + f ( x 2 )( x 2 0 - x 2 1 ) f ( x 0 )( x 1 - x 2 ) + f ( x 1 )( x

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HW5solution - Homework 5 Solution 1 Consider the following...

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