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Final_solutions

# Final_solutions - MAE 107 – Computational Methods Summer...

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Unformatted text preview: MAE 107 – Computational Methods Summer Session II 2009 Final Exam – Solutions Friday, September 4 Problem 1 – 12 points 1. (a) Newton’s formula can be applied directly to find the second order polynomial interpolating these three data points: p 2 ( x ) = f ( x 1 ) + f [ x 1 ,x 2 ]( x- x 1 ) + f [ x 1 ,x 2 ,x 3 ]( x- x 1 )( x- x 2 ) , (1) where the bracket notations f [ x 1 ,x 2 ] and f [ x 1 ,x 2 ,x 3 ] have the following meaning: f [ x 1 ,x 2 ] = f ( x 2 )- f ( x 1 ) x 2- x 1 , f [ x 1 ,x 2 ,x 3 ] = f [ x 2 ,x 3 ]- f [ x 1 ,x 2 ] x 3- x 1 = f ( x 3 )- f ( x 2 ) x 3- x 2- f ( x 2 )- f ( x 1 ) x 2- x 1 x 3- x 1 (b) Assuming h = x 3- x 2 = x 2- x 1 (equispaced data points), we have x 3- x 1 = 2 h and the previous brackets simplify into: f [ x 1 ,x 2 ] = f ( x 2 )- f ( x 1 ) h , f [ x 1 ,x 2 ,x 3 ] = f [ x 2 ,x 3 ]- f [ x 1 ,x 2 ] 2 h = f ( x 3 )- 2 f ( x 2 ) + f ( x 1 ) 2 h 2 Substitution in (1), leads to the final answer p 2 ( x ) = f ( x 1 ) + f ( x 2 )- f ( x 1 ) h ( x- x 1 ) + f ( x 3 )- 2 f ( x 2 ) + f ( x 1 ) 2 h 2 ( x- x 1 )( x- x 2 ) . 2. (a) Z x 3 x 1 ( x- x 1 )d x = ( x- x 1 ) 2 2 x 3 x 1 = ( x 3- x 1 ) 2 2 = 4 h 2 2 Therefore, Z x 3 x 1 ( x- x 1 )d x = 2 h 2 . (2) 1 (b) Using the hint and the result to the previous question, Z x 3 x 1 ( x- x 1 )( x- x 2 )d x = Z x 3 x 1 ( x- x 1 ) 2 d x + ( x 1- x 2 ) Z x 3 x 1 ( x- x 1 )d x = ( x- x 1 ) 3 3 x 3 x 1 + (- h ) × 2 h 2 = ( x 3- x 1 ) 3 3- 2 h 3 = 8 h 3 3- 2 h 3 and finally, Z x 3 x 1 ( x- x 1 )( x- x 2 )d x = 2 h 3 3 . (c) Using (2), the integral of p 2 on [ x 1 x 3 ] can be computed as I * = Z x 3 x 1 p 2 ( x )d x = f ( x 1 ) Z x 3 x 1 1d x + f ( x 2 )- f ( x 1 ) h Z x 3 x 1 ( x- x 1 )d x + f ( x 3 )- 2 f ( x 2 ) + f ( x 1 ) 2 h 2 Z x 3 x 1 ( x- x 1 )( x- x 2 )d x = f ( x 1 )( x 3- x 1 ) + f ( x 2 )- f ( x 1 ) h × 2 h 2 + f ( x 3 )- 2 f ( x 2 ) + f ( x 1 ) 2 h 2 × 2 h 3 3 = 2 hf ( x 1 ) + 2 h ( f ( x 2 )- f ( x 1 )) + h 3 [ f ( x 3 )- 2 f ( x 2 ) + f ( x 1 )] = h 3 [ f ( x 3 ) + 4 f ( x 2 ) + f ( x 1 )] The final result for the approximation of I by the integral I * of p 2 ( x ) is I * = h 3 [ f ( x 3 ) + 4 f ( x 2 ) + f ( x 1 )] . We recognize here Simpson’s 1/3 rule which is an approximation of I = R x 3 x 1 f ( x )d x by the integral of the second order polynomial interpolating f at the three points x 1 , x 2 and x 3 . 3. (a) First we recognize that d d x log(1 + x 2 ) = 2 x 1 + x 2 . Therefore we can compute the exact value of I as I = Z 1 x d x 1 + x 2 = 1 2 log(1 + x 2 ) 1 , and finally, I = log 2 2 ....
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Final_solutions - MAE 107 – Computational Methods Summer...

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