1. What is The Error ?
How did it happen ?
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2. Definition
• The Error in a computed quantity is defined
as:
Error = True Value – Approximate Value
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3. Examples:
a. True Value : phi = 3.14159265358979
Appr. Value : 22/7 = 3.14285714285714
Error = phi-22/7= -0.00126448926735
b. True Value : 12
Appr. Value: 11.78
Error = 12-11.78 = 0.22
c. True Value : 100
Appr. Value: 95.5
Error = 100-95.5 = 4.5
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4. Kind of Error
• The Absolute Error is measure the
magnitude of the error
Ea Error
• The Relative Error is a measure of the error
in relation to the size of the true value
Ea
Er
True Value
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5. Examples
• True value : 10 Ea = 10-9 = 1
Appr. Value : 9 Er = Ea/ 10 = 0.1
Ea 1, Er 0 .1
• True value : 1000 Ea = 1000-999= 1
Appr. Value : 999 Er =Ea/1000=0.001
Ea 1, Er 0 . 001
• True value : 250 Ea =250-240= 10
Appr. Value : 240 Er = Ea/250= 0.04
Ea 10 , Er 0 . 04 5

6. Sources of Error
a.Truncation Error
b.Rounding Error
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7. a. Truncation Error
• errors that result from using an approximation
in place of an exact mathematical procedure
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8. Example of Truncation Error
Taking only a few terms of a Maclaurin series to
approximate e
x
2 3
x x x
e 1 x .......... ..........
2! 3!
If only 3 terms are used,
2
x x
Truncation Error e 1 x
2!
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9. Example 1 —Maclaurin series
Calculate the value of e with an absolute
1 .2
relative approximate error of less than 1%.
2 3
1 .2 1 .2 1 .2
e 1 1 .2 .......... .........
2! 3!
n 1 .2
Ea %
e a
1 1 __ ___
2 2.2 1.2 54.545
3 2.92 0.72 24.658
4 3.208 0.288 8.9776
5 3.2944 0.0864 2.6226
6 3.3151 0.020736 0.62550
6 terms are required.
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10. Example 2 —Differentiation
f (x x) f ( x)
Find for using
2
f ( 3) f ( x) x f ( x)
x
and x 0 .2
' f (3 0 .2 ) f (3)
f (3)
0 .2
2 2
f (3 .2 ) f (3) 3 .2 3 10 . 24 9 1 . 24
6 .2
0 .2 0 .2 0 .2 0 .2
The actual value is
' '
f ( x) 2 x, f (3) 2 3 6
Truncation error is then, 6 6 .2 0 .2
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11. Example 3 — Integration
Use two rectangles of equal width to approximate
the area under the curve for
x over the interval [ 3,9 ]
2
f ( x)
y
90
9
y = x2 2
60
x dx
30 3
0 x
0 3 6 9 12
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12. Integration example (cont.)
Choosing a width of 3, we have
9
2 2 2
x dx (x ) (6 3) (x ) (9 6)
x 3 x 6
3
2 2
(3 )3 ( 6 )3
27 108 135
Actual value is given by
9 9
3 3 3
2 x 9 3
x dx 234
3 3 3
3
Truncation error is then
234 135 99
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13. b. Rounding Errors
• Round-off / Chopping Errors
• Recognize how floating point arithmetic operations
can introduce and amplify round-off errors
• What can be done to reduce the effect of round-off
errors
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14. There are discrete points on the
number lines that can be
represented by our computer.
How about the space between ?
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15. Implication of FP representations
• Only limited range of quantities may be
represented.
– Overflow and underflow
• Only a finite number of quantities within the range
may be represented.
– round-off errors or chopping errors
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16. Round-off / Chopping Errors
(Error Bounds Analysis)
Let
z be a real number we want to represent in a computer, and
fl(z) be the representation of z in the computer.
What is the largest possible value of ?
z fl ( z )
z
i.e., in the worst case, how much data are we losing due to round-off or chopping
errors?
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17. Chopping Errors (Error Bounds Analysis)
Suppose the mantissa can only support n digits.
e
z 0 .a1a 2  a n a n 1a n 2
 , a1 0
e
fl ( z ) 0 .a1a 2  a n
Thus the absolute and relative chopping errors are
e e n
z fl ( z ) ( 0 .00 ... 0 a n 1 a n
 
 2
) ( 0 .a n 1a n 2
)
n zeroes
e
z fl ( z ) ( 0 . 00 ... 0 a n 1 a n 2
)
e
z ( 0 .a 1a 2  a n a n 1a n 2
)
Suppose ß = 10 (base 10), what are the values of ai such that
the errors are the largest?
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18. Chopping Errors (Error Bounds Analysis)
Because 0 .a n 1a n 2 a n 3
 1
e n e n e n
z fl ( z ) 0 .a n 1a n 2
 z fl ( z )
e
z fl ( z ) 0 . 00 ... 0 a n 1 a n 2

e
z 0 .a 1a 2  a n a n 1a n 2

e n
e
0 .a1a 2  a n a n 1a n 2

e n
e
0 .100000 a n 1 a n
 2

n digits
e n e n
e n ( e 1) 1 n
z fl ( z ) 1 n
e 1 e
0 .1 z
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19. Round-off Errors (Error Bounds Analysis)
e
z 0 .a 1 a 2  a n a n 1
 , a1 0
1 ( sign ) base e exponent
e
( 0 .a 1 a 2  a n ) 0 an 1
2
fl ( z ) Round down
e
[( 0 .a 1 a 2  a n ) ( 0 ) ]
00 ...
. 01 an 1
n 2
Round up
fl(z) is the rounded value of z
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20. Round-off Errors (Error Bounds Analysis)
Absolute error of fl(z)
When rounding down
e
z fl ( z ) 0 . 00  0 a n 1 a n a
2 n 3

e n
0 .a n 1 a n a
2 n 3

e n
z fl ( z ) 0 .a n 1 a n a
2 n 3

1 1 e n
an 1
(. a n 1
) z fl ( z )
2 2 2
Similarly, when rounding up
1
i.e., when an 1 z fl ( z )
e n
2 2
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21. Round-off Errors (Error Bounds Analysis)
Relative error of fl(z)
1 e n
z fl ( z )
2
n
z fl ( z ) 1
e
z 2 z
n
1 e
because z (. a 1 a 2  )
2 (. a 1 a 2  )
n
1
because (. a 1 ) ( 0 . 1)
2 (. 1)
n
1
2
1 z fl ( z ) 1 1 n
z 2
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22. Summary of Error Bounds Analysis
Chopping Errors Round-off errors
Absolute e n 1 e n
z fl ( z ) z fl ( z )
2
Relative z fl ( z ) 1 n
z fl ( z ) 1 1 n
z z 2
β base
n # of significant digits or # of digits in the mantissa
Regardless of chopping or round-off is used to round the numbers,
the absolute errors may increase as the numbers grow in magnitude
but the relative errors are bounded by the same magnitude.
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