Numerical Diffrentiation

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
Recall
n
n
n
n
h
h
E
h
h
C
E
that
such
C
finite
real
h
O
E
h
O
h
x
f
h
x
f
h
x
f
x
f
h
x
f
h
x
f
h
x
f
dx
df
similar to
rate
at
zero
g
approachin
is
order
of
is
E
:
,
,
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(
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(
!
3
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2
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(
:
Theorem
Taylor
)
(
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(
lim
4
3
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3
(
2
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2
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0


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
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
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


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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
them?
judge
we
do
How
better?
is
method
Which
2
)
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:
Difference
Central
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(
:
Difference
Backward
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(
:
Difference
Forward
h
h
x
f
h
x
f
dx
x
df
h
h
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f
x
f
dx
x
df
h
x
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h
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dx
x
df











7. 7
The Three Formulas

8. 8
Forward/Backward Difference Formula
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Difference
Backward
_____
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__________
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__________
__________
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Difference
Forward
2
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

9. 9
Central Difference Formula
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Difference
Central
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h
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

10. 10
The Three Formula (Revisited)
answer.
better
a
give
to
expected
is
formula
difference
Central
accuracy.
in
comparable
are
formulas
difference
backward
and
Forward
)
(
2
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:
Difference
Central
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Difference
Backward
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Difference
Forward
2
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O
h
h
x
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h
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dx
x
df
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h
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dx
x
df
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h
x
f
h
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f
dx
x
df





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
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

11. 11
Higher Order Formulas
12
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h
f
Error
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

12. 12
Other Higher Order Formulas
.
order
error
the
obtain
and
them
prove
to
Theorem
Taylor
use
can
You
possible.
also
are
)...
(
),
(
for
formulas
Other
)
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with
Formulas
Central
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2
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4
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Error
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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