128a_su09_hw3_sol

128a_su09_hw3_sol - 2-1-f [1]-f [0] 1-2 = (8-1)-(1-0) 2 =...

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MATH 128A, SUMMER 2009: HOMEWORK 3 SOLUTIONS 3.1.5a. To minimize the error bound, we should take as our x i the ( n + 1) points nearest x = 8 . 4. When n = 2 there’s a tie; both reasonable answers are shown below. octave:1> format long octave:2> neville(8.4,[8.3,8.6], [17.56492, 18.50515]) %n=1 ans = 17.8783300000000 octave:3> neville(8.4,[8.3,8.6,8.1], [17.56492, 18.50515, 16.94410]) %n=2 ans = 17.8771300000000 octave:4> neville(8.4,[8.3,8.6,8.7], [17.56492, 18.50515, 18.82091]) %n=2 ans = 17.8771550000000 octave:5> neville(8.4,[8.3,8.6,8.1,8.7], [17.56492, 18.50515, 16.94410, 18.82091]) %n=3 ans = 17.8771425000000 3.1.19c. There are many ways to find the polynomial. A correct answer, when expanded with a calculator, should come out to P ( x ) = 0 . 1970056667 x 3 - 1 . 06259055 x 2 + 2 . 532453189 x - 1 . 666868305. The error, for x [1 , 1 . 4], is of the form ln (4) ( ξ ( x )) 4! ( x - 1)( x - 1 . 1)( x - 1 . 3)( x - 1 . 4). Since ln (4) ( ξ ) = ( - 3)( - 2)( - 1) x - 4 = - 6 ξ - 4 , it’s at worst - 6(1) - 4 = - 6. | ( x - 1)( x - 1 . 1)( x - 1 . 3)( x - 1 . 4) | is at most 4 · 10 - 4 , attained when x = 1 . 2. So P ( x ) approximates ln( x ) to within 10 - 4 on [1 , 1 . 4]. Extra. f [0 , 1 , 2] = f [1 , 2] - f [0 , 1] 2 - 0 = f [2] - f [1]
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Unformatted text preview: 2-1-f [1]-f [0] 1-2 = (8-1)-(1-0) 2 = 3. 3.2.11. (a) Calculation shows P ( x ) = f ( x ) and Q ( x ) = f ( x ) for x =-2 ,-1 , , 1 , 2. (b) Both polynomials are equal (to x 3-3 x + 1), just written in a dierent form. 3.3.1a. Construct a divided-dierence table as follows. x = 8 . 3 f [ x ] = f (8 . 3) = 17 . 56492 f [ x ,z ] = f (8 . 3) = 3 . 116256 z = 8 . 3 f [ z ] = f (8 . 3) = 17 . 56492 . 0594800 f [ z ,x 1 ] = = 3 . 134100-. 002022222 x 1 = 8 . 6 f [ x 1 ] = f (8 . 6) = 18 . 50515 . 0588733 f [ x 1 ,z 1 ] = f (8 . 3) = 3 . 151762 z 1 = 8 . 6 f [ z 1 ] = f (8 . 6) = 18 . 50515 (See, e.g., Table 3.13 in the text.) The coecients for the Newton form of the approximating polynomial are on the top edge of this table: H ( x ) = 17 . 56492+3 . 116256( x-8 . 3)+0 . 0594800( x-8 . 3) 2-. 002022222( x-8 . 3) 2 ( x-8 . 6). Date : Due Thursday 7/09. 1...
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