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solhw4 - Solutions for Homework 4 Foundations of...

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Solutions for Homework 4 Foundations of Computational Math 2 Spring 2011 Problem 4.1 Suppose we want to approximate a function f ( x ) on the interval [ a,b ] with a piecewise quadratic interpolating polynomial with a constant spacing, h , of the interpolation points a = x 0 <x 1 ...<x n = b . That is, for any a x b , the value of f ( x ) is approximated by evaluating the quadratic polynomial that interpolates f at x i 1 , x i , and x i +1 for some i with x = x i + sh , x i 1 = x i h , x i +1 = x i + h and 1 s 1. (How i is chosen given a particular value of x is not important for this problem. All that is needed is the condition x i 1 x x i +1 .) Suppose we want to guarantee that the relative error of the approximation is less than 10 d , i.e., d digits of accuracy. Specifically, | f ( x ) p ( x ) | | f ( x ) | 10 d . (It is assumed that | f ( x ) | is sufficiently far from 0 on the interval [ a,b ] for relative accuracy to be a useful value.) Derive a bound on h that guarantees the desired accuracy and apply it to interpolating f ( x ) = e x sin x on the interval π 4 x 3 π 4 with relative accuracy of 10 4 . (The sin is bounded away from 0 on this interval.) Solution: On each interval we have f ( x ) p ( x ) = ( x x i 1 )( x x i )( x x i +1 ) 6 f ′′′ ( ξ ) = h 3 s ( s + 1)( s 1) 6 f ′′′ ( ξ ) = h 3 q ( s ) 6 f ′′′ ( ξ ) To bound the term with q ( s ) we note that q ( s ) = s 3 s q ( s ) = 3 s 2 1 extrema are s ± = ± radicalbig 1 / 3 q ( s ± ) 6 = ± 1 9 radicalbigg 1 3 0 . 06416 < 2 3 1 10 0 . 067 We must bound | f ′′′ ( x ) | from above and | f ( x ) | from below to set an upper bound on h 1
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that guarantees the desired accuracy. We have on π 4 x 3 π 4 f ′′′ ( x ) = 2 e x (cos x sin x ) π 4 x π 2 → | cos x sin x | ≤ 1 π 2 x 3 π 4 → | cos x sin x | ≤ 2 e x e 3 π/ 4 < 10 . 6 | f ′′′ ( x ) | < 43 | f ( x ) | ≥ f ( π/ 4) = e π/ 4 2 1 . 6 Therefore we can put all of these bounds together to get h 3 | q ( s ) | 6 | f ′′′ ( x ) | | f ( x ) | < 0 . 067 × 43 1 . 6 h 3 2 h 3 10 4 h 3 0 . 00005 0 . 04 Given that π/ 4 0 . 8 and 3 π/ 4 2 . 35 h = 0 . 04 implies n 38. Problem 4.2 4.2.a (i) Find the cubic polynomial p 3 ( x ) that interpolates a function f ( x ) at the values: f (0) = 0 , f (0) = 1 f (1) = 3 , f (1) = 6 (ii) Find the quartic polynomial p 4 ( x ) that interpolates a function f ( x ) at the values: f (0) = 0 , f (0) = 0 f (1) = 1 , f (1) = 1 f (2) = 1 Solutions: Consider the data: f (0) = 0 , f (0) = 1 f (1) = 3 , f (1) = 6 2
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We can use the Newton form to derive p 3 ( x ) via the table x 0 x 0 x 1 x 1 f 0 f 0 f 1 f 1 f 0 f [ x 0 ,x 1 ] f 1 f [ x 0 ,x 0 ,x 1 ] f [ x 0 ,x 1 ,x 1 ] f [ x 0 ,x 0 ,x 1 ,x 1 ] The values are 0 0 1 1 0 0 3 3 1 3 6 2 3 1
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