NUMERICAL ERROR
ENGR 351
Numerical Methods for Engineers
Southern Illinois University Carbondale
College of Engineering
Dr. L.R. Chevalier
Copyright© 2003 by Lizette R. Chevalier
Permission is granted to students at Southern Illinois University at Carbondale
to make one copy of this material for use in the class ENGR 351, Numerical
Methods for Engineers. No other permission is granted.
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Objectives
• To understand error terms
• Become familiar with notation and
techniques used in this course
Approximation and Errors
Significant Figures
• 4 significant figures
• 1.845
• 0.01845
• 0.0001845
•
•
•
•
43,500 ? confidence
4.35 x 104 3 significant figures
4.350 x 104 4 significant figures
4.3500 x 104 5 significant figures
Accuracy and Precision
• Accuracy - how closely a computed or
measured value agrees with the true value
• Precision - how closely individual computed or
measured values agree with each other
• number of significant figures
• spread in repeated measurements or
computations
Accuracy and Precision
increasing precision
increasing accuracy
Error Definitions
• Numerical error - use of approximations to
represent exact mathematical operations and
quantities
• true value = approximation + error
•
•
•
•
error, et=true value - approximation
subscript t represents the true error
shortcoming....gives no sense of magnitude
normalize by true value to get true relative error
Error definitions cont.
true error
et 
 100
true value
• True relative percent error
Example
Consider a problem where the true answer is
7.91712. If you report the value as 7.92, answer
the following questions.
1. How many significant figures did you use?
2. What is the true error?
3. What is the relative error?
Error definitions cont.
• May not know the true answer apriori
approximate error
ea 
100
approximation
Error definitions cont.
• May not know the true answer apriori
approximate error
ea 
100
approximation
• This leads us to develop an iterative
approach of numerical methods
Error definitions cont.
• May not know the true answer apriori
approximate error
ea 
 100
approximation
• This leads us to develop an iterative
approach of numerical methods
approximate error
ea 
100
approximation
present approx.  previous approx.

100
present approx.
Error definitions cont.
• Usually not concerned with sign, but
with tolerance
• Want to assure a result is correct to n
significant figures
Error definitions cont.
• Usually not concerned with sign, but
with tolerance
• Want to assure a result is correct to n
significant figures
ea  es
e s  0.5 10
2n
%
Example
Consider a series expansion to estimate
trigonometric functions
x3 x5 x7
sin x  x     .....   x  
3! 5! 7!
Estimate sinp/ 2 to three significant
figures
Error Definitions cont.
• Round off error - originate from the fact
that computers retain only a fixed
number of significant figures
• Truncation errors - errors that result
from using an approximation in place of
an exact mathematical procedure
Error Definitions cont.
• Round off error - originate from the fact
that computers retain only a fixed
number of significant figures
• Truncation errors - errors that result
from using an approximation in place of
an exact mathematical procedure
To gain insight consider the mathematical
formulation that is used widely in numerical
methods - TAYLOR SERIES
TAYLOR SERIES
• Provides a means to predict a function
value at one point in terms of the
function value at and its derivatives at
another point
• Zero order approximation
TAYLOR SERIES
• Provides a means to predict a function value
at one point in terms of the function value at
and its derivative at another point
• Zero order approximation
f  xi 1   f  xi 
This is good if the function is a constant.
Taylor Series Expansion
• First order approximation
{
f  xi 1   f  xi   f '  xi  xi 1  xi 
slope multiplied by distance
Taylor Series Expansion
• First order approximation
f  xi 1   f  xi   f '  xi  xi 1  xi 
slope multiplied by distance
Still a straight line but capable of predicting an
increase or decrease - LINEAR
Taylor Series Expansion
• Second order approximation - captures
some of the curvature
Taylor Series Expansion
• Second order approximation - captures
some of the curvature
f ' '  xi 
2
f  xi 1   f  xi   f '  xi  xi 1  xi  
 xi 1  xi 
2!
Taylor Series Expansion
f ' '  xi  2 f ' ' '  xi  3
f  xi 1   f  xi   f '  xi h 
h 
h 
2!
3!
f n  xi  n
...... 
h  Rn
n!
where h  step size  xi 1  xi
Taylor Series Expansion
f  xi 1   f  xi   f '  xi h 
f n  xi  n
...... 
h  Rn
n!
f ' '  xi  2 f ' ' '  xi  3
h 
h 
2!
3!
where h  step size  xi 1  xi
f  n 1   n 1
Rn 
h
xi    xi 1
n  1!
Example
Use zero through fourth order Taylor series
expansion to approximate f(1) given f(0) = 1.2
(i.e. h = 1)
f  x  01
. x 4  015
. x 3  0.5x 2  0.25x  1.2
1.4
1
f(x)
Note:
f(1) = 0.2
1.2
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
x
0.8
1
Solution
• n=0
• f(1) = 1.2
• et = abs [(0.2 - 1.2)/0.2] x 100 = 500%
• n=1
•
•
•
•
f '(x) = -0.4x3 - 0.45x2 -x -0.25
f '(0) = -0.25
f(1) = 1.2 - 0.25h = 0.95
et =375%
Solution
• n=2
• f "=-1.2 x2 - 0.9x -1
• f "(0) = -1
• f(1) = 0.45
• et = 125%
• n=3
• f "'=-2.4x - 0.9
• f "'(0)=-0.9
• f(1) = 0.3
• et =50%
Solution
• n=4
• f ""(0) = -2.4
• f(1) = 0.2 EXACT
• Why does the fourth term give us an exact
solution?
• The 5th derivative is zero
• In general, nth order polynomial, we get an
exact solution with an nth order Taylor series
Solution
1.4
1.2
f(x)
1
0.8
True Solution
Zero Order
0.6
1st Order
0.4
2nd Order
0.2
3rd Order
0
0
0.2
0.4
0.6
x
0.8
1
1.2
Exam Question
How many significant figures are in the following
numbers?
A. 3.215
B. 0.00083
C. 2.41 x 10-3
D. 23,000,000
E. 2.3 x 107
Taylor Series Problem
Use zero- through fourth-order Taylor series
expansions to predict f(4) for f(x) = ln x using a base
point at x = 2. Compute the percent relative error et
for each approximation.
1.6
1.4
1.2
y
1
0.8
0.6
0.4
0.2
0
0
1
2
3
x
4
5
f ' '  xi  2 f ' ' '  xi  3
f  x i  1   f  x i   f '  x i h 
h 
h
2!
3!
f n  xi  n
. . . . . .
h  Rn
n!
where h  step size  xi  1  xi

1. Determine the step size h = 4 - 2 = 2
2. Determine the analytical solution f(4) = ln(4) = 1.3863
3. Determine the derivatives for f(2)