1 Numerical Differentiation

2  First order derivatives  High order derivatives  Examples

3 Motivation  How do you evaluate the derivative of a tabulated function.  How do we determine the velocity and acceleration from tabulated measurements. Time (second) Displacement (meters)

4 Recall

For small values of h, the difference quotient [f (x 0 + h) − f (x 0 )]/h can be used to approximate f’(x 0 ) with an error bounded by M|h|/2, where M is a bound on |f’’(x)| for x between x0 and x0 +h. CISE301_Topic65

6 Three Formula

7 The Three Formulas

8 Forward/Backward Difference Formula

9 Central Difference Formula

10 The Three Formula (Revisited)

11 Higher Order Formulas

12 Other Higher Order Formulas

Example  Use forward, backward and centered difference approximations to estimate the first derivate of: f(x) = –0.1x 4 – 0.15x 3 – 0.5x 2 – 0.25x at x = 0.5 using step size h = 0.5 and h = 0.25  Note that the derivate can be obtained directly: f’(x) = –0.4x 3 – 0.45x 2 – 1.0x – 0.25 The true value of f’(0.5) =  In this example, the function and its derivate are known. However, in general, only tabulated data might be given. 13

Solution with Step Size = 0.5  f(0.5) = 0.925, f(0) = 1.2, f(1.0) = 0.2  Forward Divided Difference: f’(0.5)  (0.2 – 0.925)/0.5 = | t | = |( )/ | = 58.9%  Backward Divided Difference: f’(0.5)  (0.925 – 1.2)/0.5 = | t | = |( )/ | = 39.7%  Centered Divided Difference: f’(0.5)  (0.2 – 1.2)/1.0 = -1.0 | t | = |( )/ | = 9.6% 14

Solution with Step Size = 0.25  f(0.5)=0.925, f(0.25)=1.1035, f(0.75)=  Forward Divided Difference: f’(0.5)  ( – 0.925)/0.25 = | t | = |( )/ | = 26.5%  Backward Divided Difference: f’(0.5)  (0.925 – )/0.25 = | t | = |( )/ | = 21.7%  Centered Divided Difference: f’(0.5)  ( – )/0.5 = | t | = |( )/ | = 2.4% 15

Discussion  For both the Forward and Backward difference, the error is O(h)  Halving the step size h approximately halves the error of the Forward and Backward differences  The Centered difference approximation is more accurate than the Forward and Backward differences because the error is O(h 2 )  Halving the step size h approximately quarters the error of the Centered difference. 16

Use the forward-difference formula to approximate the derivative of f (x) = ln x at x0 = 1.8 using h = 0.1, h = 0.05, and h = 0.01, and determine bounds for the approximation errors. 17 Example 2

Solution The forward-difference formula f (1.8 + h) − f (1.8) h with h = 0.1 gives (ln 1.9 − ln 1.8)/0.1 = ( − )/0.1 = Because f’’(x) = −1/x 2 and 1.8 < ξ < 1.9, a bound for this approximation error is |hf’’(ξ )|/2 =|h|/2ξ 2 < 0.1/2(1.8) 2 =

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