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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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This note was uploaded on 01/30/2012 for the course MATH 131a taught by Professor Hitrik during the Spring '08 term at UCLA.
 Spring '08
 hitrik
 Math, Approximation

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