128ahw5sum10 - MATH 128A, SUMMER 2010, HOMEWORK 5 SOLUTION...

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Unformatted text preview: MATH 128A, SUMMER 2010, HOMEWORK 5 SOLUTION BENJAMIN JOHNSON Homework 5: Due Monday, July 12 3.2; 4a, 9a, 16 3.3; 2a, 4a 3.4; 4b, 6b, 19 section 3.2 4. Use the Newton forward-difference formula to construct interpolating polynomials of degree one, two, and three for the following data. Approximate the specified value using each of the polynomials. a. f (0 . 43) if f (0) = 1, f (0 . 25) = 1 . 64872, f (0 . 5) = 2 . 71828, f (0 . 75) = 4 . 48169 solution: Newtons forward-difference formula gives P n ( x ) = n k =0 ( s k ) k f ( x ). In our case, we have x = 0 and x k = x + 0 . 25 k for each k , so that h = 0 . 25; and solving x = x + s h for s given x = 0 . 43, x = 0, and h = 0 . 25 yields s = 1 . 72. To get expressions for the polynomials, we need to compute f ( x ), 1 f ( x ), 2 f ( x ), and 3 f ( x ). f ( x ) = f ( x ) = 1 1 f ( x ) = f ( x 1 )- f ( x ) = 1 . 64872- 1 = 0 . 64872. 2 f ( x ) = ( f ( x 2 )- f ( x 1 ))- ( f ( x 1 )- f ( x )) = (2 . 71828- 1 . 64872)- . 64872 = 0 . 42084. 3 f ( x ) = ( f ( x 3 )- 2 f ( x 2 )+ f ( x 1 ))- . 42084 = 4 . 48169- 2 2 . 71828+1 . 64872- . 42084 = . 27301. Now we have: P 1 (0 . 43) = f ( x ) + ( 1 . 72 1 ) f ( x ) = 1 + 1 . 72 . 64872 = 2 . 1158. P 2 (0 . 43) = P 1 (0 . 43) + ( 1 . 72 2 ) 2 f ( x ) = 2 . 1158 + 1 . 72 (1 . 72- 1) 2 1 . 42084 = 2 . 37638. P 3 (0 . 43) = P 2 (0 . 43) + ( 1 . 72 3 ) 3 f ( x ) = 2 . 37638 + 1 . 72 (1 . 72- 1) (1 . 72- 2) 3! . 27301 = 2 . 3606. 9. a. Approximate f (0 . 05) using the following data and the Newton forward divided-difference formula: x . . 2 0.4 0.6 0.8 f ( x ) 1.00000 1.22140 1.49182 1.82212 2.22554 solution: Here we have x = 0 and x k = x + 0 . 2 k for each k = 1 ,..., 4. So in the formulas we have h = 0 . 2 and x = 0 . 05 = x + 0 . 25 h implying s = 0 . 25. The Newton forward divided-difference formula gives: Date : July 12, 2010....
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128ahw5sum10 - MATH 128A, SUMMER 2010, HOMEWORK 5 SOLUTION...

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