Math 128A - HW5 Solutions.pdf

# Now we have p 1 0 43 f x 1 72 1 δ f x 1 1 72 64872 2

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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 . Exercise 3.3.6. Use the Newton backward-difference formula to construct interpolating polyno- mials 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. P n ( x ) = n k =0 ( - 1) k ( s k ) k f ( x 3 ), h = 0 . 25, s = (0 . 43 - 0 . 75) / 0 . 25 = - 1 . 28. 0 f ( x 3 ) = f ( x 3 ) = 4 . 48169 1 f ( x 3 ) = f ( x 3 ) - f ( x 2 ) = 4 . 48169 - 2 . 71828 = 1 . 76341 2 f ( x 3 ) = ( f ( x 3 ) - f ( x 2 )) - ( f ( x 2 ) - f ( x 1 )) = 1 . 76341 - (2 . 71828 - 1 . 64872) = 0 . 69385 3 f ( x 3 ) = 2 f ( x 3 ) - ( f ( x 2 ) - 2 f ( x 1 ) + f ( x 0 )) = 0 . 69385 - 0 . 42084 = 0 . 27301 Now we have: P 1 (0 . 43) = f ( x 3 ) - 1 . 28 1 f ( x 3 ) = 4 . 48169 - 1 . 28 · 1 . 76341 = 2 . 22453 P 2 (0 . 43) = P 1 (0 . 43) + 1 . 28 2 2 f ( x 3 ) = 2 . 22453 + 1 . 28 · (1 . 28 - 1) 2 · 1 · 0 . 69385 = 2 . 34887 . P 3 (0 . 43) = P 2 (0 . 43) - 1 . 28 3 3 f ( x 3 ) = 2 . 34887 - 1 . 28 · (1 . 28 - 1) · (1 . 28 - 2) 3! · 0 . 27301 = 2 . 36061 .

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Exercise 3.3.14(16). For a function f , the Newton divided-difference formula gives the interpo- lating polynomial P 3 ( x ) = 1 + 4 x + 4 x ( x - 0 . 25) + 16 3 x ( x - 0 . 25)( x - 0 . 5) , on the nodes x 0 = 0, x 1 = 0 . 25, x 2 = 0 . 5 and x 3 = 0 . 75. Find f (0 . 75). Solution. Since the interpolating polynomial agrees with f at the x i and we are asked to find f ( x 3 ), it suffices to compute P 3 ( x 3 ). We get P 3 (0 . 75) = 1 + 4(0 . 75) + 4 · 0 . 75 · (0 . 75 - 0 . 25) + 16 3 · 0 . 75(0 . 75 - 0 . 25)(0 . 75 - 0 . 5) = 6 . Exercise 3.3.19(10). Show that the polynomial interpolating the following data has degree 3. x - 2 - 1 0 1 2 3 f ( x ) 1 4 11 16 13 - 4 Solution. The divided difference table for the data is given below. Since the fourth and fifth divided differences are zero, P 5 ( x ) agrees with P 3 ( x ), a polynomial of degree 3. First Second Third Fourth Fifth Divided Divided Divided Divided Divided x f ( x ) Difference Difference Diference Difference Difference -2 1 3 -1 4 2 7 -1 0 11 -1 0 5 -1 0 1 16 -4 0 -3 -1 2 13 -7 -17 3 -4
Exercise 3.3.22(20). Show that f [ x 0 , x 1 , . . . , x n , x ] = f ( n +1) ( ξ ( x )) ( n +1)!
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• Spring '08
• Rieffel
• Polynomial interpolation, Divided differences, = F, Use Neville’s method

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