mws_com_dif_txt_continuous_examples

# mws_com_dif_txt_continuous_examples - Chapter 02.02...

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Chapter 02.02 Differentiation of Continuous Functions-More Examples Computer Engineering Example 1 There is strong evidence that the first level of processing what we see is done in the retina. It involves detecting something called edges or positions of transitions from dark to bright or bright to dark points in images. These points usually coincide with boundaries of objects. To model the edges, derivatives of functions such as 0 , 1 0 , 1 ) ( x e x e x f ax ax need to be found. a) Use the forward divided difference approximation of the first derivative of   x f to calculate the functions derivative at 1 . 0 x for 24 . 0 a . Use a step size of 05 . 0 . Also, calculate the absolute relative true error. x b) Repeat the procedure from part (a) with the same data except choose 12 . 0 a . Does the estimate of the derivative increase or decrease? Also, calculate the absolute relative true error. Solution a)   x x f x f x f i i i 1 24 . 0 a 1 . 0 i x 05 . 0 x x x x i 1 05 . 0 1 . 0 15 . 0 ) 1 . 0 24 . 0 ( 1 ) 1 . 0 ( e f 023714 . 0 24 . 0 ( ) 15 . 0 1 ) 15 . 0 ( e f 035360 . 0

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02.02.2 Chapter 02.02   05 . 0 1 . 0 15 . 0 1 . 0 f f f 05 . 0 023714 . 0 0.035360 0.23291 The exact value of can be calculated by differentiating 1 . 0 f  0 , 1 x e x f ax as    x f dx d x f Knowing that  ax ax ae e dx d we find  ) 1 ( ax e dx d x f x ax e ae 24 . 0 24 . 0     ) 1 . 0 24 . 0 ( 24 . 0 1 . 0 e f 0.23431 The absolute relative true error is 100 Value True Value e Approximat Value True t 100 0.23431 0.23291 0.23431 % 0.59761 b) 0.12 a  ) 1 . 0 12 . 0 ( 1 1 . 0
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## This note was uploaded on 06/12/2011 for the course EML 3041 taught by Professor Kaw,a during the Spring '08 term at University of South Florida.

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mws_com_dif_txt_continuous_examples - Chapter 02.02...

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