The document discusses numerical differentiation, which is used to approximate derivatives of functions when only tabulated data is available. It presents the forward, backward, and central difference formulas for approximating the first derivative. The central difference formula provides more accurate approximations, with errors on the order of h^2 rather than h. The document also shows an example applying the formulas to estimate the derivative of a sample function at different points using various step sizes h. Halving h reduces the error as expected based on the formulas.

1. 1
Numerical Differentiation

2. 2
Numerical Differentiation
 First order derivatives
 High order derivatives
Examples

3. 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)
0 30.1
5 48.2
10 50.0
15 40.2

4. 4
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5. For small values of h, the difference quotient [f (x0 + h) − f (x0)]/h
can be used to approximate f’(x0) with an error bounded by M|h|/2,
where M is a bound on |f’’(x)| for x between x0 and x0 +h.
CISE301_Topic6 5

6. 6
Three Formula
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7. 7
The Three Formulas

8. 8
Forward/Backward Difference Formula
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9. 9
Central Difference Formula
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10. 10
The Three Formula (Revisited)
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11. 11
Higher Order Formulas
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12. 12
Other Higher Order Formulas
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13. Example
 Use forward, backward and centered difference
approximations to estimate the first derivate of:
f(x) = –0.1x4 – 0.15x3 – 0.5x2 – 0.25x + 1.2
at x = 0.5 using step size h = 0.5 and h = 0.25
 Note that the derivate can be obtained directly:
f’(x) = –0.4x3 – 0.45x2 – 1.0x – 0.25
The true value of f’(0.5) = -0.9125
 In this example, the function and its derivate
are known. However, in general, only tabulated
data might be given.
13

14. 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 = -1.45
|t| = |(-0.9125+1.45)/-0.9125| = 58.9%
 Backward Divided Difference:
f’(0.5)  (0.925 – 1.2)/0.5 = -0.55
|t| = |(-0.9125+0.55)/-0.9125| = 39.7%
 Centered Divided Difference:
f’(0.5)  (0.2 – 1.2)/1.0 = -1.0
|t| = |(-0.9125+1.0)/-0.9125| = 9.6%
14

15. Solution with Step Size = 0.25
 f(0.5)=0.925, f(0.25)=1.1035, f(0.75)=0.6363
 Forward Divided Difference:
f’(0.5)  (0.6363 – 0.925)/0.25 = -1.155
|t| = |(-0.9125+1.155)/-0.9125| = 26.5%
 Backward Divided Difference:
f’(0.5)  (0.925 – 1.1035)/0.25 = -0.714
|t| = |(-0.9125+0.714)/-0.9125| = 21.7%
 Centered Divided Difference:
f’(0.5)  (0.6363 – 1.1035)/0.5 = -0.934
|t| = |(-0.9125+0.934)/-0.9125| = 2.4%
15

16. 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(h2)
 Halving the step size h approximately quarters
the error of the Centered difference.
16

17. 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

18. 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
= (0.64185389 − 0.58778667)/0.1 = 0.5406722.
Because f’’(x) = −1/x2and 1.8 < ξ < 1.9, a bound for this
approximation error is |hf’’(ξ )|/2
=|h|/2ξ2 < 0.1/2(1.8)2 = 0.0154321.
18

19. 19