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Measuring Errors
Son Dao, PhD
Why measure errors?
1) To determine the accuracy of
numerical results.
2) To develop stopping criteria for
iterative algorithms.
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True Error
Defined as the difference between the true
value in a calculation and the approximate
value found using a numerical method etc.
True Error = True Value – Approximate Value
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Example—True Error
The derivative, f (x) of a function f (x) can be
approximated by the equation,
f ( x h) f ( x)
f ' ( x)
h
If f ( x) 7e and h 0.3
0.5 x
a) Find the approximate value of f ' ( 2)
b) True value of f ' ( 2)
c) True error for part (a)
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Example (cont.)
Solution:
a) For x 2 and h 0.3
f ( 2 0.3) f ( 2)
f ' (2)
0 .3
f (2.3) f (2)
0.3
7e 0.5( 2.3) 7e 0.5( 2)
0.3
22.107 19.028
10.263
0 .3
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Example (cont.)
Solution:
b) The exact value of f ' ( 2) can be found by using
our knowledge of differential calculus.
f ( x) 7e 0.5 x
f ' ( x ) 7 0.5 e 0.5 x
3.5e 0.5 x
So the true value of f ' ( 2) is
f ' ( 2) 3.5e 0.5( 2 )
9.5140
True error is calculated as
Et True Value – Approximate Value
9.5140 10 .263 0.722
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Relative True Error
Defined as the ratio between the true
error, and the true value.
True Error
Relative True Error ( t ) =
True Value
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Example—Relative True Error
Following from the previous example for true error,
find the relative true error for f ( x) 7e 0.5 x at f ' (2)
with h 0.3
From the previous example,
Et 0.722
Relative True Error is defined as
True Error
t
True Value
0.722
0.075888
9.5140
as a percentage,
t 0.075888 100% 7.5888 %
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Approximate Error
What can be done if true values are not
known or are very difficult to obtain?
Approximate error is defined as the
difference between the present
approximation and the previous
approximation.
Approximate Error ( E a ) = Present Approximation – Previous Approximation
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Example—Approximate Error
For f ( x) 7e 0.5 x at x 2 find the following,
a) f (2) using h 0.3
b) f (2) using h 0.15
c) approximate error for the value of f (2) for part b)
Solution:
a) For x 2 and h 0.3
f ( x h) f ( x)
f ' ( x)
h
f ( 2 0.3) f ( 2)
f ' ( 2)
0. 3
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Example (cont.)
Solution: (cont.)
f (2.3) f (2)
0.3
7e 0.5( 2.3) 7e 0.5( 2)
0.3
22.107 19.028
10 .263
0.3
b) For x 2 and h 0.15
f (2 0.15) f (2)
f ' ( 2)
0.15
f (2.15) f ( 2)
0.15
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Example (cont.)
Solution: (cont.)
7e 0.5( 2.15) 7e 0.5( 2 )
0.15
20.50 19.028
9.8800
0.15
c) So the approximate error, E a is
Ea Present Approximation – Previous Approximation
9.8800 10 .263
0.38300
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Relative Approximate Error
Defined as the ratio between the
approximate error and the present
approximation.
Approximate Error
Relative Approximate Error ( a) =
Present Approximation
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Example—Relative Approximate Error
For f ( x) 7e at x 2 , find the relative approximate
0.5 x
error using values from h 0.3 and h 0.15
Solution:
From Example 3, the approximate value of f (2) 10.263
using h 0.3 and f (2) 9.8800 using h 0.15
Ea Present Approximation – Previous Approximation
9.8800 10 .263
0.38300
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Example (cont.)
Solution: (cont.)
Approximate Error
a
Present Approximation
0.38300
0.038765
9.8800
as a percentage,
a 0 . 038765 100 % 3 . 8765 %
Absolute relative approximate errors may also need to
be calculated,
a | 0.038765 | 0.038765 or 3.8765%
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How is Absolute Relative Error used as a
stopping criterion?
If |a | s where s is a pre-specified tolerance, then
no further iterations are necessary and the process is
stopped.
If at least m significant digits are required to be
correct in the final answer, then
|a | 0.5 10 2 m %
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Table of Values
For f ( x) 7e at x 2 with varying step size, h
0.5 x
h f (2) a m
0.3 10.263 N/A 0
0.15 9.8800 3.877% 1
0.10 9.7558 1.273% 1
0.01 9.5378 2.285% 1
0.001 9.5164 0.2249% 2
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