Differentiation-Discrete
Functions
Industrial Engineering Majors
Authors: Autar Kaw, Sri Harsha Garapati
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
4/13/2015
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1
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 failure rate ht  of a direct methanol fuel cell (DMFC) is given by
the formula
R' t 
ht  
Rt 
Where Rt  is the reliability at a certain time t , and the values of the
reliability are given in Table 1.
Table 1 Reliability of DMFC system.
t hrs 0 1
Rt 
10
100
1000 2000 3000 4000 5000
1 0.9999 0.9998 0.9980 0.9802 0.9609 0.9419 0.9233 0.9050
Using the forward divided difference method, find the failure rate of the
DMFC system at t  50 hours.
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Example 1 Cont.
Solution
Rti 1   Rti 
t
ti  10
R ' ti  
ti 1  100
t  ti 1  ti
 100 10
 90
R100  R10
90
0.9980 0.9998

90
 2.0000105
R' 50 
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Example 1 Cont.
The reliability Rt  at hours t  50 is
R100  R10
50  10  R10
100 10
  2.0000105 40  0.9998
 0.999
R50 


The failure rate ht  at t  50 hours is then
R' 50
h50  
R50
  2.0000105 

 
0.999


 2.0020105
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Direct Fit Polynomials
In this method, given ' n  1' data points x0 , y0 , x1 , y1 , x2 , y2 ,, xn , yn 
one can fit a
n th order polynomial given by
Pn x  a0  a1x    an 1xn 1  an xn
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 failure rate ht  of a direct methanol fuel cell (DMFC) is given by
the formula
R' t 
ht  
Rt 
Where Rt  is the reliability at a certain time t , and the values of the
reliability are given in Table 2.
Table 2 Reliability of DMFC system.
t hrs 0 1
Rt 
10
100
1000 2000 3000 4000 5000
1 0.9999 0.9998 0.9980 0.9802 0.9609 0.9419 0.9233 0.9050
Using a third order polynomial interpolant for reliability Rt  , find the
failure rate of the DMFC system at t  50 hours.
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Example 2-Direct Fit Polynomials cont.
Solution
For the third order polynomial (also called cubic interpolation), we choose the reliability
given by
2
3
Rt   a0  a1t  a2t  a3t
Since we want to find the reliability at t  50 , and we are using third order polynomial,
we need to choose the four points closest to t  50 and that also bracket t  50 to
evaluate it.
The four points are t0  1, t1  10 , t2  100 and t3  1000 hours.
to  1, Rto   0.9999
t1  10, Rt1   0.9998
t 2  100, Rt 2   0.9980
t3  1000, Rt3   0.9802
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Example 2-Direct Fit Polynomials cont.
such that
R1  0.9999 a0  a1 1  a2 1  a3 1
2
3
R10  0.9998 a0  a1 10  a2 10 2  a3 10
3
R100  0.9980 a0  a1 100  a2 100  a3 100
2
3
R1000  0.9802 a0  a1 1000  a2 1000  a3 1000
2
3
Writing the four equations in matrix form, we have
1
1
1  a0  0.9999
1
1 10
  a  0.9998
100
1000

 1   

6
1 100 10000 110  a2  0.9980
 


6
9 
a
1
1000
1

10
1

10
0
.
9802

 3  

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Example 2-Direct Fit Polynomials cont.
Solving the above four equations gives
a0  0.9991
a1  1.0023 105
a2  9.9788 108
a3  9.01011011
Hence
Rt   a0  a1t  a2t 2  a3t 3
 0.99991 1.0023105 t  9.9788108 t 2  9.01011011 t 3 , 1  t  1000
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Example 2-Direct Fit Polynomials cont.
Figure 2 Graph of reliability as a function of time.
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,
Example 2-Direct Fit Polynomials cont.
The reliability at t  50 is given by,
R ' 50  
d
Rt  t 50
dt
Given that
Rt   0.999911.0023105 t  9.9788108 t 2  9.01011011 t 3 , 1  t  1000
R' t  
d
Rt 
dt
d

0.99991 1.0023105 t  9.9788108 t 2  9.01011011 t 3
dt
 1.0023105  1.9958107 t  2.70301010 t 2 , 1  t  1000

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
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Example 2-Direct Fit Polynomials cont.
R' 50  1.0023105  1.9958107 50  2.70301010 50
2
 1.9326105
Using the same function, we can also calculate the value of Rt  at t  50 .
Rt   0.99991 1.0023105 t  9.9788108 t 2  9.01011011 t 3 , 1  t  1000
R50  0.99991 1.0023105 50  9.9788108 50  9.01011011 50
 0.99917
2
The failure rate is then
ht   
3
R' t 
Rt 
 1.932610 

5
0.99917
 1.9343 105
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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 , yn1 , xn , yn  , 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 failure rate ht  of a direct methanol fuel cell (DMFC) is given by
the formula
R' t 
ht  
Rt 
Where Rt  is the reliability at a certain time t , and the values of the
reliability are given in Table 3.
Table 3 Reliability of DMFC system.
t hrs 0 1
Rt 
10
100
1000 2000 3000 4000 5000
1 0.9999 0.9998 0.9980 0.9802 0.9609 0.9419 0.9233 0.9050
Determine the value of the failure rate at t  50 hours using the second
order Lagrangian polynomial interpolation for reliability.
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Example 3 Cont.
Solution
For second order Lagrangian polynomial interpolation, we choose the
reliability given by
 t  t1  t  t2 
 t  t0  t  t2 
 t  t0  t  t1 

 R(t0 )  


 R(t1 )  
 R(t2 )
R(t )  
 t0  t1  t0  t2 
 t1  t0  t1  t2 
 t2  t0  t2  t1 
Since we want to find the reliability at t  50 , and we are using a second
order Lagrangian polynomial, we need to choose the three points closest
to t  50 that also bracket t  50 to evaluate it.
The three points are t0  1 , t1  10 , and t2  100.
Differentiation the above equation gives.
2t  t0  t2 
2t  t0  t1 
2t  t1  t2 
R' t  
Rt0  
Rt1  
Rt2 
t0  t1 t0  t2 
t1  t0 t1  t2 
t2  t0 t2  t1 
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Example 3 Cont.
Hence
R' 50 
250  10  100
0.9999  250  1  100 0.9998  250  1  10 0.9980
1  101  100
10  110  100
100 1100 10
 1.9102105
We must also find the value of Rt  at t  50 .
 50  10  50  100
 50  1  50  100
R50  

0.9999  

0.9998
 1  10  1  100 
 10  1  10  100 
 50  1  50  10 


0.9980
 100 1  100 10 
 0.99918
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Example 3 Cont.
The failure rate is then
ht   
R' t 
Rt 

 1.910210 

5
0.99918
 1.9118105
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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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