The document defines error as the difference between the true value and approximate value of a computed quantity. It provides examples of absolute error, which is the magnitude of the error, and relative error, which measures the error relative to the true value. There are two main sources of error - truncation error from using approximations, and rounding error from limitations of floating point representations. Truncation error is analyzed using examples of Taylor series approximations and numerical integration. Rounding error bounds are derived, showing the absolute error is bounded by 1/2 the least significant digit, while relative error is bounded by 1/2 times the number of significant digits.

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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