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128ahw5sum10

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

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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: Newton’s forward-difference formula gives P n ( x ) = n k =0 ( s k ) Δ k f ( x 0 ). In our case, we have x 0 = 0 and x k = x 0 + 0 . 25 · k for each k , so that h = 0 . 25; and solving x = x 0 + s · h for s given x = 0 . 43, x 0 = 0, and h = 0 . 25 yields s = 1 . 72. To get expressions for the polynomials, we need to compute Δ 0 f ( x 0 ), Δ 1 f ( x 0 ), Δ 2 f ( x 0 ), and Δ 3 f ( x 0 ). Δ 0 f ( x 0 ) = f ( x 0 ) = 1 Δ 1 f ( x 0 ) = f ( x 1 ) - f ( x 0 ) = 1 . 64872 - 1 = 0 . 64872. Δ 2 f ( x 0 ) = ( f ( x 2 ) - f ( x 1 )) - ( f ( x 1 ) - f ( x 0 )) = (2 . 71828 - 1 . 64872) - 0 . 64872 = 0 . 42084. Δ 3 f ( x 0 ) = ( f ( x 3 ) - 2 f ( x 2 )+ f ( x 1 )) - 0 . 42084 = 4 . 48169 - 2 · 2 . 71828+1 . 64872 - 0 . 42084 = 0 . 27301. Now we have: P 1 (0 . 43) = f ( x 0 ) + ( 1 . 72 1 ) Δ f ( x 0 ) = 1 + 1 . 72 · 0 . 64872 = 2 . 1158. P 2 (0 . 43) = P 1 (0 . 43) + ( 1 . 72 2 ) Δ 2 f ( x 0 ) = 2 . 1158 + 1 . 72 · (1 . 72 - 1) 2 · 1 · 0 . 42084 = 2 . 37638. P 3 (0 . 43) = P 2 (0 . 43) + ( 1 . 72 3 ) Δ 3 f ( x 0 ) = 2 . 37638 + 1 . 72 · (1 . 72 - 1) · (1 . 72 - 2) 3! · 0 . 27301 = 2 . 3606. 9. a. Approximate f (0 . 05) using the following data and the Newton forward divided-difference formula: x 0 . 0 0 . 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 = 0 and x k = x 0 + 0 . 2 · k for each k = 1 , . . . , 4. So in the formulas we have h = 0 . 2 and x = 0 . 05 = x 0 + 0 . 25 · h implying s = 0 . 25. The Newton forward divided-difference formula gives: Date : July 12, 2010. 1

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2 BENJAMIN JOHNSON P 4 (0 . 05) = 4 X k =0 0 . 25 k Δ 4 f ( x 0 ) = 0 . 25 0 Δ 0 f ( x 0 ) + 0 . 25 1 Δ 1 f ( x 0 ) + 0 . 25 2 Δ 2 f ( x 0 ) + 0 . 25 3 Δ 3 f ( x 0 ) + 0 . 25 4 Δ 4 f ( x 0 ) = 1 · 1 + 0 . 25 · 0 . 22140 + - 0 . 09375 · 0 . 04902 + 0 . 0546875 · 0 . 01086 + - 0 . 0375977 · 0 . 00238 = 1 . 05126 For this problem I computed both the choose functions and the Δ k ’s using Mathemat- ica. I have included the relevant session transcript below for reference. Even though Mathematica isn’t covered in this class, it may be useful to see what it can do easily. It
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