Transcript PPT

Differentiation-Discrete
Functions
Computer 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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Differentiation – Discrete
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
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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.
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Example 1
The upward velocity of a rocket is given as a function of time in Table 1.
Table 1 Velocity as a function of time
t
s
0
10
15
20
22.5
30
v(t)
m/s
0
227.04
362.78
517.35
602.97
901.67
Using forward divided difference, find the acceleration of the rocket at t  16 s .
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Example 1 Cont.
Solution
To find the acceleration at t  16s, we need to choose the two values
closest to t  16s, that also bracket t  16s to evaluate it. The two
points are t  15s and t  20s.
 ti 1   ti 
ati  
t
ti  15
ti 1  20
t  ti 1  ti
 20  15
5
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Example 1 Cont.
 20  15
5
517.35  362.78

5
 30.914 m/s 2
a16 
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Direct Fit Polynomials
In this method, given ' n  1' data points x0 , y0 , x1 , y1 , x2 , y 2 ,, xn , y n 
one can fit a
n th order polynomial given by
Pn x   a0  a1 x    an 1 x n 1  an x n
To find the first derivative,
Pn x  
dPn ( x )
 a1  2a 2 x    n  1a n 1 x n 2  na n x n 1
dx
Similarly other derivatives can be found.
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Example 2-Direct Fit Polynomials
The upward velocity of a rocket is given as a function of time in Table 2.
Table 2 Velocity as a function of time
t
s
0
10
15
20
22.5
30
v(t)
m/s
0
227.04
362.78
517.35
602.97
901.67
Using the third order polynomial interpolant for velocity, find the
acceleration of the rocket at t  16 s .
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Example 2-Direct Fit Polynomials cont.
Solution
For the third order polynomial (also called cubic interpolation), we choose the velocity given by
vt   a 0  a1t  a 2 t 2  a3 t 3
Since we want to find the velocity at t  16 s , and we are using third order polynomial, we need
to choose the four points closest to t  16 s and that also bracket t  16 s to evaluate it.
The four points are
to  10, t1  15, t2  20, and t3  22.5.
to  10, vto   227.04
t1  15, vt1   362.78
t2  20, vt2   517.35
t3  22.5, vt3   602.97
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Example 2-Direct Fit Polynomials cont.
such that
v10  227.04  a0  a1 10  a2 10  a3 10
2
3
v15  362.78  a0  a1 15  a2 15  a3 15
2
3
v20  517.35  a0  a1 20  a2 20  a3 20
2
3
v22.5  602.97  a0  a1 22.5  a2 22.5  a3 22.5
2
3
Writing the four equations in matrix form, we have
100
1000  a0  227.04
1 10
1 15
  a  362.78
225
3375

 1   

1 20
400
8000  a 2  517.35

  

a
1
22
.
5
506
.
25
11391
602
.
97

 3  

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Example 2-Direct Fit Polynomials cont.
Solving the above four equations gives
a0  4.3810
a1  21.289
a2  0.13065
a3  0.0054606
Hence
vt   a0  a1t  a2t 2  a3t 3
 4.3810  21.289t  0.13065t 2  0.0054606t 3 , 10  t  22.5
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Example 2-Direct Fit Polynomials cont.
Figure 1 Graph of upward velocity of the rocket vs. time.
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,
Example 2-Direct Fit Polynomials cont.
The acceleration at t=16 is given by
a16 
Given that
d
vt  t 16
dt
 t   4.3810  21.289t  0.13065t 2  0.0054606t 3 ,10  t  22.5
d
a t   vt 
dt

d
 4.3810  21.289t  0.13065t 2  0.0054606t 3
dt
 21.289  0.26130t  0.016382t 2 , 10  t  22.5


a16  21.289  0.2613016  0.01638216
2
 29.664m/s 2
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Lagrange Polynomial
In this method, given x1 , y1 ,, xn , yn  , one can fit a n  1th order Lagrangian polynomial
given by
f n ( x) 
where ‘ n ’ in
n
 L ( x) f ( x )
i 0
i
i
f n (x) stands for the n th order polynomial that approximates the function
y  f (x) given at (n  1) data points as x0 , y0 , x1 , y1 ,......, xn1 , y n1 , xn , y n  , and
n
Li ( x)  
j 0
j i
x  xj
xi  x j
Li (x) a weighting function that includes a product of (n  1) terms with terms of
ji
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omitted.
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Lagrange Polynomial Cont.
Then to find the first derivative, one can differentiate f n x  once, and so on
for other derivatives.
For example, the second order Lagrange polynomial passing through
x0 , y0 , x1, y1 , x2 , y2 
f 2 x  
is
x  x1 x  x2  f x   x  x0 x  x2  f x   x  x0 x  x1  f x 
x0  x1 x0  x2  0 x1  x0 x1  x2  1 x2  x0 x2  x1  2
Differentiating equation (2) gives
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Lagrange Polynomial Cont.
2 x   x0  x2 
2 x   x0  x1 
2 x  x1  x2 

f 2 x  
f  x0  
f  x1  
f x 
x0  x1 x0  x2 
x1  x0 x1  x2 
x2  x0 x2  x1  2
Differentiating again would give the second derivative as
f 2x  
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2
x0  x1 x0  x2 
f x0  
2
x1  x0 x1  x2 
f  x1  
2
x2  x0 x2  x1 
f x2 
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Example 3
The upward velocity of a rocket is given as a function of time in Table 3.
Table 3 Velocity as a function of time
t
s
0
10
15
20
22.5
30
v(t)
m/s
0
227.04
362.78
517.35
602.97
901.67
Determine the value of the acceleration at t  16 s using the second
order Lagrangian polynomial interpolation for velocity.
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Example 3 Cont.
Solution
 t  t1  t  t 2 
 t  t 0  t  t 2 
 t  t 0  t  t1 









v(t 2 )
v(t )  
v(t 0 )  
v(t1 )  






 t 0  t1  t 0  t 2 
 t1  t 0  t1  t 2 
 t 2  t 0  t 2  t1 
at  
2t  t1  t 2 
2t  t 0  t 2 
 t 0  
 t1   2t  t0  t1  νt2 
t 0  t1 t 0  t 2 
t1  t 0 t1  t 2 
t2  t0 t2  t1 
216  15  20
227.04  216  10  20 362.78  216  10  15 517.35
a16 
20  10 20  15
15  1015  20
10  1510  20
 0.06227.04  0.08362.78  0.14517.35
 29.784m/s 2
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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/discrete_02
dif.html
THE END
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