Transcript PPT

Differentiation-Continuous
Functions
Electric Engineering Majors
Authors: Autar Kaw, Sri Harsha Garapati
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
7/12/2016
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1
Differentiation – Continuous
Functions
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Forward Difference Approximation
lim f  x  Δx   f  x 
f  x  
Δx  0
Δx
For a finite ' Δx'
f  x  x   f  x 
f  x  
x
3
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Graphical Representation Of
Forward Difference
Approximation
f(x)
x
x+Δx
Figure 1 Graphical Representation of forward difference approximation of first derivative.
4
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Example 1
The velocity of a rocket is given by


14  104
 t   2000 ln 
  9.8t,0  t  30
4
14

10

2100
t


where 'ν' is given in m/s and 't ' is given in seconds.
a) Use forward difference approximation of the first derivative of νt  to
calculate the acceleration at t  16s . Use a step size of Δt  2s .
b) Find the exact value of the acceleration of the rocket.
c) Calculate the absolute relative true error for part (b).
5
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Example 1 Cont.
Solution
ati  
 ti 1   ti 
t
ti  16
Δt  2
ti 1  ti  t
 16  2
 18
a 16  
6
 18  16 
2
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Example 1 Cont.


14  104
 18  2000 ln 
 9.818

4
14  10  210018
 453.02m/s


14 104
 16  2000 ln 
  9.816
4
14 10  210016
 392.07m/s
Hence
a 16  
7
 18  16 
2
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Example 1 Cont.

453.02  392.07
2
 30.474m/s 2
b) The exact value of a16 can be calculated by differentiating


14 104
 t   2000 ln 
  9.8t
4
14 10  2100t 
as
d
at   νt 
dt
8
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Example 1 Cont.
Knowing that
d
ln t   1
dt
t
and
d 1
1


dt  t 
t2
 14  104  2100t  d 

14  104


a t   2000
4
 dt  14  104  2100t   9.8
14

10

 

4


 14 10 4  2100t 
14

10

 2100  9.8
 1
 2000
2 
4
4

14 10


 14 10  2100t  

9
 4040  29.4t
 200  3t
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Example 1 Cont.
a16 
 4040  29.416
 200  316
 29.674m/s 2
The absolute relative true error is
t 

True Value - Approximat e Value
x100
True Value
29.674  30.474
x100
29.674
 2.6967%
10
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Backward Difference Approximation of the
First Derivative
We know
f  x  
lim f  x  Δx   f  x 
Δx  0
Δx
For a finite ' Δx' ,
f  x  
f  x  x   f  x 
x
If ' Δx' is chosen as a negative number,
f  x  

11
f  x  x   f  x 
 x
f  x   f  x  Δx 
Δx
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Backward Difference Approximation of the
First Derivative Cont.
This is a backward difference approximation as you are taking a point
backward from x. To find the value of f x at x  xi, we may choose another
point ' Δx' behind as x  x . This gives
f  xi  
f xi   f xi 1 
x

f  xi   f  xi 1 
xi  xi 1
i 1
where
Δx  xi  xi 1
12
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Backward Difference Approximation of the
First Derivative Cont.
f(x)
x
x-Δx
x
Figure 2 Graphical Representation of backward difference
approximation of first derivative
13
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Example 2
The velocity of a rocket is given by


14  104
 t   2000 ln 
  9.8t,0  t  30
4
14

10

2100
t


where 'ν' is given in m/s and 't ' is given in seconds.
a) Use backward difference approximation of the first derivative of νt 
to calculate the acceleration at t  16 s . Use a step size of Δt  2s .
b) Find the absolute relative true error for part (a).
14
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Example 2 Cont.
Solution
at  
 ti   ti 1 
t
ti  16
Δt  2
ti 1  ti  t
 16  2
 14
a 16  
15
 16   14 
2
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Example 2 Cont.


14  104
 16  2000 ln 
 9.816

4
14  10  210016
 392.07m/s


14  104
 14  2000 ln 
 9.814

4
14  10  210014
 334.24m/s
a 16  

 16   14 
2
392.07  334.24
2
 28.915m/s 2
16
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Example 2 Cont.
The exact value of the acceleration at t  16 s from Example 1 is
a16  29.674m/s 2
The absolute relative true error is
t 
29.674  28.915
x100
29.674
 2.5584%
17
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Derive the forward difference approximation
from Taylor series
Taylor’s theorem says that if you know the value of a function ' f ' at a point
xi
and all its derivatives at that point, provided the derivatives are
continuous between
xi and xi 1 , then
f xi 1   f xi   f xi  xi 1  xi  
f  xi 
xi 1  xi 2  
2!
Substituting for convenience Δx  xi 1  xi
f  xi 1   f  xi   f  xi Δx 
f  xi 
Δx 2  
2!
f xi 1   f xi  f xi 
x   

x
2!
f xi 1   f xi 
f xi  
 0x 
x
f xi  
18
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Derive the forward difference approximation
from Taylor series Cont.
The 0x  term shows that the error in the approximation is of the order
of Δx  Can you now derive from Taylor series the formula for backward
divided difference approximation of the first derivative?
As shown above, both forward and backward divided difference
approximation of the first derivative are accurate on the order of 0x 
Can we get better approximations? Yes, another method to approximate
the first derivative is called the Central difference approximation of
the first derivative.
19
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Derive the forward difference approximation
from Taylor series Cont.
From Taylor series
f xi 1   f xi   f xi Δx 
 
f xi 
Δx2  f xi Δx 3  
2!
3!
f xi 1   f xi   f xi Δx 
 
f xi 
Δx2  f xi Δx3  
2!
3!
Subtracting equation (2) from equation (1)
f  xi 1   f xi 1   f  xi 2Δx  
f xi  
f xi  
20
2 f  xi 
Δx 3  
3!
f xi 1   f xi 1  f xi 
x 2  

2x
3!
f xi 1   f xi 1 
2
 0x 
2x
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Central Divided Difference
Hence showing that we have obtained a more accurate formula as the
error is of the order of 0Δx 2 .
f(x)
x
x-Δx
x
x+Δx
Figure 3 Graphical Representation of central difference approximation of first derivative
21
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Example 3
The velocity of a rocket is given by


14  104
 t   2000 ln 
  9.8t,0  t  30
4
14

10

2100
t


where 'ν' is given in m/s and 't ' is given in seconds.
(a) Use central divided difference approximation of the first derivative of νt 
to calculate the acceleration at t  16s . Use a step size of Δt  2s .
(b) Find the absolute relative true error for part (a).
22
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Example 3 cont.
Solution
ati  
 ti 1   ti 1 
ti  16
t  2
2t
ti 1  ti  t
 16  2
 18
ti 1  ti  t
 16  2
 14
a16 
23
 18  14
22
 18  14 

4
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Example 3 cont.


14  104
 18  2000 ln 
 9.818

4
14  10  210018
 453.02m/s


14  104
 14  2000 ln 
 9.814

4
14  10  210014
 334.24m/s
a 16  
 18  14 
4
453.02  334.24

4
 29.694m/s 2
24
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Example 3 cont.
The exact value of the acceleration at t  16 s from Example 1 is
a16  29.674m/s 2
The absolute relative true error is
t 
29.674  29.694
100
29.674
 0.069157%
25
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Comparision of FDD, BDD, CDD
The results from the three difference approximations are given in Table 1.
Table 1 Summary of a (16) using different divided difference approximations
Type of Difference
Approximation
Forward
Backward
Central
26
a16
m / s 
t %
30.475
28.915
29.695
2.6967
2.5584
0.069157
2
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Finding the value of the derivative
within a prespecified tolerance
In real life, one would not know the exact value of the derivative – so how
would one know how accurately they have found the value of the derivative.
A simple way would be to start with a step size and keep on halving the step
size and keep on halving the step size until the absolute relative approximate
error is within a pre-specified tolerance.
Take the example of finding
v t 
for


14  104


 t  2000 ln 
  9.8t
4
14

10

2100
t


at t  16 using the backward divided difference scheme.
27
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Finding the value of the derivative
within a prespecified tolerance Cont.
Given in Table 2 are the values obtained using the backward difference
approximation method and the corresponding absolute relative
approximate errors.
Table 2 First derivative approximations and relative errors for
different Δt values of backward difference scheme
t
2
1
0.5
0.25
0.125
28
v t 
28.915
29.289
29.480
29.577
29.625
a %
1.2792
0.64787
0.32604
0.16355
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Finding the value of the derivative
within a prespecified tolerance Cont.
From the above table, one can see that the absolute relative
approximate error decreases as the step size is reduced. At t  0.125
the absolute relative approximate error is 0.16355%, meaning that
at least 2 significant digits are correct in the answer.
29
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Finite Difference Approximation of
Higher Derivatives
One can use Taylor series to approximate a higher order derivative.
For example, to approximate f x  , the Taylor series for
f xi 2   f xi   f xi 2Δx  
 
f xi 
2Δx2  f xi 2Δx 3  
2!
3!
where
xi  2  xi  2Δx
f xi 1   f xi   f xi x  
 
f xi 
x2  f xi x 3 
2!
3!
where
xi 1  xi  Δx
30
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Finite Difference Approximation of
Higher Derivatives Cont.
Subtracting 2 times equation (4) from equation (3) gives
f xi 2   2 f xi 1    f xi   f xi Δx  f xi Δx 
2
f xi  
f xi  
31
f xi  2   2 f xi 1   f xi 
Δx 
2
 f xi Δx   
f xi  2   2 f xi 1   f xi 
Δx 
2
3
 0Δx 
(5)
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Example 4
The velocity of a rocket is given by


14  104
 t   2000 ln 
 9.8t,0  t  30

4
14  10  2100t 
Use forward difference approximation of the second derivative νt 
of to calculate the jerk at t  16s. Use a step size of Δt  2s .
32
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Example 4 Cont.
Solution
j ti  
 ti  2   2 ti 1   ti 
ti  16
t 2
t  2
ti 1  ti  t
 16  2
 18
ti  2  ti  2t 
 16  22
 20
j 16  
33
 20   2 18  16 
22
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Example 4 Cont.


14  104
 20  2000 ln 
 9.820

4
14  10  210020
 517.35m/s


14  104
 18  2000 ln 
 9.818

4
14  10  210018
 453.02m/ s


14  104
 16  2000 ln 
 9.816

4
14  10  210016
 392.07m/s
34
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Example 4 Cont.
j 16  
517.35  2453.02   392.07
4
 0.84515m/s 3
The exact value of j 16 can be calculated by differentiating


14  104
 t   2000 ln 
  9.8t
4
14

10

2100
t


twice as
d
νt 
a t  
dt
35
and j t  
d
at 
dt
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Example 4 Cont.
Knowing that
d
ln t   1
dt
t
and
d 1
1


dt  t 
t2
 14  104  2100t  d 

14  104


a t   2000
4
 dt  14  104  2100t   9.8
14

10

 

4


 14  10 4  2100t 
14

10

 2100  9.8
 1
 2000
2 
4
4

14  10


 14  10  2100t  

36
 4040  29.4t
 200  3t
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Example 4 Cont.
Similarly it can be shown that
d
at 
dt
18000

(200  3t ) 2
j t  
j 16 
18000
[200  3(16)]2
 0.77909m/s 3
The absolute relative true error is
t 
0.77909  0.84515
 100
0.77909
 8.4797 %
37
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Higher order accuracy of higher
order derivatives
The formula given by equation (5) is a forward difference approximation of
the second derivative and has the error of the order of Δx  . Can we get
a formula that has a better accuracy? We can get the central difference
approximation of the second derivative.
The Taylor series for
f xi 1   f xi   f xi x 
 
 
f xi 
x2  f xi x3  f xi x 4 
2!
3!
4!
(6)
where
xi 1  xi  Δx
38
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Higher order accuracy of higher
order derivatives Cont.
f xi 1   f xi   f xi x 
 
 
f xi 
x2  f xi x3  f xi x4 
2!
3!
4!
(7)
where
xi 1  xi  Δx
Adding equations (6) and (7), gives
f  xi 1   f  xi 1   2 f  xi   f  xi Δx  
2
f  xi  
f xi  
39
f  xi 1   2 f xi   f  xi 1 
Δx 2
f xi Δx 

12
f xi 1   2 f xi   f xi 1 
Δx 2
4

Δx 
f  x 
i
12
2
 0Δx 
2
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Example 5
The velocity of a rocket is given by


14  104
 t   2000 ln 
 9.8t,0  t  30

4
14  10  2100t 
Use central difference approximation of second derivative of νt  to
calculate the jerk at t  16s . Use a step size of Δt  2s.
40
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Example 5 Cont.
Solution
ati  
 ti 1   2 ti   ti 1 
t i  16
t 2
t  2
ti 1  ti  t
 16  2
 18
ti 1  ti  t
 16  2
 14
j 16 
41
 18  2 16  14
22
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Example 5 Cont.


14  104
 18  2000 ln 
 9.818

4
14  10  210018
 453.02m/s


14  104
 16  2000 ln 
  9.816
4


14

10

2100
16


 392.07m/s


14  104
 14  2000 ln 
 9.814

4
14  10  210014
 334.24m/s
42
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Example 5 Cont.
j 16 

 18  2 16  14
22
453.02  2392.07   334.24
4
 0.77969m/s 3
The absolute relative true error is
t 
0.77908  0.78
 100
0.77908
 0.077992%
43
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Additional Resources
For all resources on this topic such as digital audiovisual
lectures, primers, textbook chapters, multiple-choice tests,
worksheets in MATLAB, MATHEMATICA, MathCad and
MAPLE, blogs, related physical problems, please visit
http://numericalmethods.eng.usf.edu/topics/continuous_02
dif.html
THE END
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