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Unformatted text preview: 151A HW 7 Solutions Will Feldman 1. This is not difficult just plug the values given in the problem into the second difference quotient formula f 00 ( x ) ≈ f ( x + h )- 2 f ( x ) + f ( x- h ) h 2 for h = 0 . 1 , . 01. The interesting part is that the approximation for h = 0 . 1 is better than the approximation for 0 . 01. This is due to rounding error. By using the data given in the table you are using approximations to f which are accurate up to = 5 × 10- 6 . Let ˜ f ( x ) be our approximation to the value of f at x which satisfies | ˜ f ( x )- f ( x ) | ≤ , i.e. it is an approximation of the actual value of f at x with error less than . Then the second difference quotient with ˜ f instead of f has the error, | ˜ f ( x + h )- 2 ˜ f ( x ) + ˜ f ( x- h ) h 2- f 00 ( x ) | ≤ | ˜ f ( x + h )- 2 ˜ f ( x ) + ˜ f ( x- h ) h 2- f ( x + h )- 2 f ( x ) + f ( x- h ) h 2 | + | f ( x + h )- 2 f ( x ) + f ( x- h ) h 2- f 00 ( x ) | ≤ 4 h 2 + h 2 M 12 where M is an upper bound for...
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