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Chopping and rounding
For any real number x, let
x = ±1.a1a2· · · atat+1at+2· · · × 2m,
denote the normalized scientific binary representation of x.
1 chopping: simply discard the excess bits at+1, at+2, . . .to obtain
f l(x) = ±1.a1a2· · · at× 2m.
2 rounding: add 2−(t+1)× 2m to x and then chop the excess bits to obtain a number of the form
f l(x) = ±1.δ1δ2· · · δt× 2m.
In this method, if at+1= 1, we add 1 to atto obtain f l(x), and if at+1 = 0, we merely chop off all but the first t digits.
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Chopping and rounding
For any real number x, let
x = ±1.a1a2· · · atat+1at+2· · · × 2m,
denote the normalized scientific binary representation of x.
1 chopping: simply discard the excess bits at+1, at+2, . . .to obtain
f l(x) = ±1.a1a2· · · at× 2m.
2 rounding: add 2−(t+1)× 2m to x and then chop the excess bits to obtain a number of the form
f l(x) = ±1.δ1δ2· · · δt× 2m.
In this method, if at+1= 1, we add 1 to atto obtain f l(x), and if at+1 = 0, we merely chop off all but the first t digits.
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Definition 7 (Roundoff error)
The error results from replacing a number with its floating-point form is calledroundoff errororrounding error.
Definition 8 (Absolute Error and Relative Error)
If x is an approximation to the exact value x∗, theabsolute error is|x∗− x|and therelative erroris |x|x∗−x|∗| , provided that x∗ 6= 0.
Example 9
(a) If x∗ = 0.3000 × 10−3and x = 0.3100 × 10−3, then the absolute error is 0.1 × 10−4and the relative error is 0.3333 × 10−1.
(b) If x∗ = 0.3000 × 104and x = 0.3100 × 104, then the absolute error is 0.1 × 103 and the relative error is 0.3333 × 10−1.
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Definition 7 (Roundoff error)
The error results from replacing a number with its floating-point form is calledroundoff errororrounding error.
Definition 8 (Absolute Error and Relative Error)
If x is an approximation to the exact value x∗, theabsolute error is|x∗− x|and therelative erroris |x|x∗−x|∗| , provided that x∗ 6= 0.
Example 9
(a) If x∗ = 0.3000 × 10−3and x = 0.3100 × 10−3, then the absolute error is 0.1 × 10−4and the relative error is 0.3333 × 10−1.
(b) If x∗ = 0.3000 × 104and x = 0.3100 × 104, then the absolute error is 0.1 × 103 and the relative error is 0.3333 × 10−1.
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Definition 7 (Roundoff error)
The error results from replacing a number with its floating-point form is calledroundoff errororrounding error.
Definition 8 (Absolute Error and Relative Error)
If x is an approximation to the exact value x∗, theabsolute error is|x∗− x|and therelative erroris |x|x∗−x|∗| , provided that x∗ 6= 0.
Example 9
(a) If x∗ = 0.3000 × 10−3and x = 0.3100 × 10−3, then the absolute error is 0.1 × 10−4and the relative error is 0.3333 × 10−1.
(b) If x∗ = 0.3000 × 104and x = 0.3100 × 104, then the absolute error is 0.1 × 103 and the relative error is 0.3333 × 10−1.
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Definition 7 (Roundoff error)
The error results from replacing a number with its floating-point form is calledroundoff errororrounding error.
Definition 8 (Absolute Error and Relative Error)
If x is an approximation to the exact value x∗, theabsolute error is|x∗− x|and therelative erroris |x|x∗−x|∗| , provided that x∗ 6= 0.
Example 9
(a) If x∗ = 0.3000 × 10−3and x = 0.3100 × 10−3, then the absolute error is 0.1 × 10−4and the relative error is 0.3333 × 10−1.
(b) If x∗ = 0.3000 × 104and x = 0.3100 × 104, then the absolute error is 0.1 × 103 and the relative error is 0.3333 × 10−1.
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Remark 1
As a measure of accuracy, the absolute error may be misleading and the relative error more meaningful.
Definition 10
The number x is said to approximate x∗to tsignificant digitsif t is the largest nonnegative integer for which
|x − x∗|
|x∗| ≤ 5 × 10−t.
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Remark 1
As a measure of accuracy, the absolute error may be misleading and the relative error more meaningful.
Definition 10
The number x is said to approximate x∗to tsignificant digitsif t is the largest nonnegative integer for which
|x − x∗|
|x∗| ≤ 5 × 10−t.
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If the floating-point representation f l(x) for the number x is obtained by using t digits and chopping procedure, then the relative error is
|x − f l(x)|
|x| = |0.00 · · · 0at+1at+2· · · × 2m|
|1.a1a2· · · atat+1at+2· · · × 2m|
= |0.at+1at+2· · · |
|1.a1a2· · · atat+1at+2· · · | × 2−t. The minimal value of the denominator is 1. The numerator is bounded above by 1. As a consequence
x − f l(x) x
≤ 2−t.
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If the floating-point representation f l(x) for the number x is obtained by using t digits and chopping procedure, then the relative error is
|x − f l(x)|
|x| = |0.00 · · · 0at+1at+2· · · × 2m|
|1.a1a2· · · atat+1at+2· · · × 2m|
= |0.at+1at+2· · · |
|1.a1a2· · · atat+1at+2· · · | × 2−t. The minimal value of the denominator is 1. The numerator is bounded above by 1. As a consequence
x − f l(x) x
≤ 2−t.
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If the floating-point representation f l(x) for the number x is obtained by using t digits and chopping procedure, then the relative error is
|x − f l(x)|
|x| = |0.00 · · · 0at+1at+2· · · × 2m|
|1.a1a2· · · atat+1at+2· · · × 2m|
= |0.at+1at+2· · · |
|1.a1a2· · · atat+1at+2· · · | × 2−t. The minimal value of the denominator is 1. The numerator is bounded above by 1. As a consequence
x − f l(x) x
≤ 2−t.
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If the floating-point representation f l(x) for the number x is obtained by using t digits and chopping procedure, then the relative error is
|x − f l(x)|
|x| = |0.00 · · · 0at+1at+2· · · × 2m|
|1.a1a2· · · atat+1at+2· · · × 2m|
= |0.at+1at+2· · · |
|1.a1a2· · · atat+1at+2· · · | × 2−t. The minimal value of the denominator is 1. The numerator is bounded above by 1. As a consequence
x − f l(x) x
≤ 2−t.
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If t-digit rounding arithmetic is used and
at+1= 0, then f l(x) = ±1.a1a2· · · at× 2m.A bound for the relative error is
|x − f l(x)|
|x| = |0.at+1at+2· · · |
|1.a1a2· · · atat+1at+2· · · | × 2−t≤ 2−(t+1), since the numerator is bounded above by 12 due to at+1= 0.
at+1= 1, then f l(x) = ±(1.a1a2· · · at+ 2−t) × 2m.The upper bound for relative error becomes
|x − f l(x)|
|x| = |1 − 0.at+1at+2· · · |
|1.a1a2· · · atat+1at+2· · · | × 2−t≤ 2−(t+1), since the numerator is bounded by 12due to at+1= 1.
Therefore the relative error for rounding arithmetic is
x − f l(x) x
≤ 2−(t+1)= 1 2× 2−t.
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If t-digit rounding arithmetic is used and
at+1= 0, then f l(x) = ±1.a1a2· · · at× 2m. A bound for the relative error is
|x − f l(x)|
|x| = |0.at+1at+2· · · |
|1.a1a2· · · atat+1at+2· · · | × 2−t≤ 2−(t+1), since the numerator is bounded above by 12 due to at+1= 0.
at+1= 1, then f l(x) = ±(1.a1a2· · · at+ 2−t) × 2m. The upper bound for relative error becomes
|x − f l(x)|
|x| = |1 − 0.at+1at+2· · · |
|1.a1a2· · · atat+1at+2· · · | × 2−t≤ 2−(t+1), since the numerator is bounded by 12due to at+1= 1.
Therefore the relative error for rounding arithmetic is
x − f l(x) x
≤ 2−(t+1)= 1 2× 2−t.
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If t-digit rounding arithmetic is used and
at+1= 0, then f l(x) = ±1.a1a2· · · at× 2m. A bound for the relative error is
|x − f l(x)|
|x| = |0.at+1at+2· · · |
|1.a1a2· · · atat+1at+2· · · | × 2−t≤ 2−(t+1), since the numerator is bounded above by 12 due to at+1= 0.
at+1= 1, then f l(x) = ±(1.a1a2· · · at+ 2−t) × 2m.The upper bound for relative error becomes
|x − f l(x)|
|x| = |1 − 0.at+1at+2· · · |
|1.a1a2· · · atat+1at+2· · · | × 2−t≤ 2−(t+1), since the numerator is bounded by 12due to at+1= 1.
Therefore the relative error for rounding arithmetic is
x − f l(x) x
≤ 2−(t+1)= 1 2× 2−t.
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If t-digit rounding arithmetic is used and
at+1= 0, then f l(x) = ±1.a1a2· · · at× 2m. A bound for the relative error is
|x − f l(x)|
|x| = |0.at+1at+2· · · |
|1.a1a2· · · atat+1at+2· · · | × 2−t≤ 2−(t+1), since the numerator is bounded above by 12 due to at+1= 0.
at+1= 1, then f l(x) = ±(1.a1a2· · · at+ 2−t) × 2m. The upper bound for relative error becomes
|x − f l(x)|
|x| = |1 − 0.at+1at+2· · · |
|1.a1a2· · · atat+1at+2· · · | × 2−t≤ 2−(t+1), since the numerator is bounded by 12due to at+1= 1.
Therefore the relative error for rounding arithmetic is
x − f l(x) x
≤ 2−(t+1)= 1 2× 2−t.
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If t-digit rounding arithmetic is used and
at+1= 0, then f l(x) = ±1.a1a2· · · at× 2m. A bound for the relative error is
|x − f l(x)|
|x| = |0.at+1at+2· · · |
|1.a1a2· · · atat+1at+2· · · | × 2−t≤ 2−(t+1), since the numerator is bounded above by 12 due to at+1= 0.
at+1= 1, then f l(x) = ±(1.a1a2· · · at+ 2−t) × 2m. The upper bound for relative error becomes
|x − f l(x)|
|x| = |1 − 0.at+1at+2· · · |
|1.a1a2· · · atat+1at+2· · · | × 2−t≤ 2−(t+1), since the numerator is bounded by 12due to at+1= 1.
Therefore the relative error for rounding arithmetic is
x − f l(x) x
≤ 2−(t+1)= 1 2× 2−t.
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Definition 11 (Machine epsilon)
The floating-point representation, f l(x), of x can be expressed as
f l(x) = x(1 + δ), |δ| ≤ εM, (1) whereεM ≡ 2−t is referred to as theunit roundoff erroror machine epsilon.
Single precision IEEE standard floating-point format The mantissa f corresponds to 23 binary digits (i.e., t = 23), the machine epsilon is
εM = 2−23≈ 1.192 × 10−7.
This approximately corresponds to7accurate decimal digits
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Definition 11 (Machine epsilon)
The floating-point representation, f l(x), of x can be expressed as
f l(x) = x(1 + δ), |δ| ≤ εM, (1) whereεM ≡ 2−t is referred to as theunit roundoff erroror machine epsilon.
Single precision IEEE standard floating-point format The mantissa f corresponds to 23 binary digits (i.e., t = 23), the machine epsilon is
εM = 2−23≈ 1.192 × 10−7.
This approximately corresponds to7accurate decimal digits
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Definition 11 (Machine epsilon)
The floating-point representation, f l(x), of x can be expressed as
f l(x) = x(1 + δ), |δ| ≤ εM, (1) whereεM ≡ 2−t is referred to as theunit roundoff erroror machine epsilon.
Single precision IEEE standard floating-point format The mantissa f corresponds to 23 binary digits (i.e., t = 23), the machine epsilon is
εM = 2−23≈ 1.192 × 10−7.
This approximately corresponds to7accurate decimal digits
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Double precision IEEE standard floating-point format The mantissa f corresponds to 52 binary digits (i.e., t = 52), the machine epsilon is
εM = 2−52≈ 2.220 × 10−16.
which provides between15and16decimal digits of accuracy.
Summary of IEEE standard floating-point format
single precision double precision
εM 1.192 × 10−7 2.220 × 10−16
smallest positive number 1.175 × 10−38 2.225 × 10−308 largest number 3.403 × 1038 1.798 × 10308
decimal precision 7 16
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Double precision IEEE standard floating-point format The mantissa f corresponds to 52 binary digits (i.e., t = 52), the machine epsilon is
εM = 2−52≈ 2.220 × 10−16.
which provides between15and16decimal digits of accuracy.
Summary of IEEE standard floating-point format
single precision double precision
εM 1.192 × 10−7 2.220 × 10−16
smallest positive number 1.175 × 10−38 2.225 × 10−308 largest number 3.403 × 1038 1.798 × 10308
decimal precision 7 16
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Double precision IEEE standard floating-point format The mantissa f corresponds to 52 binary digits (i.e., t = 52), the machine epsilon is
εM = 2−52≈ 2.220 × 10−16.
which provides between15and16decimal digits of accuracy.
Summary of IEEE standard floating-point format
single precision double precision
εM 1.192 × 10−7 2.220 × 10−16
smallest positive number 1.175 × 10−38 2.225 × 10−308 largest number 3.403 × 1038 1.798 × 10308
decimal precision 7 16
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Let stand for any one of the four basic arithmetic operators +, −, ?, ÷.
Whenever twomachine numbersxand y are to be combined arithmetically, the computer will produce f l(x y)instead of x y.
Under (1), the relative error of f l(x y) satisfies
f l(x y) = (x y)(1 + δ), δ ≤ εM, (2)
where εM is the unit roundoff.
But if x, y arenotmachine numbers, then they must first rounded to floating-point format before the arithmetic operation and the resulting relative error becomes
f l(f l(x) f l(y)) = (x(1 + δ1) y(1 + δ2))(1 + δ3),
where δi ≤ εM, i = 1, 2, 3.
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Let stand for any one of the four basic arithmetic operators +, −, ?, ÷.
Whenever twomachine numbersxand y are to be combined arithmetically, the computer will produce f l(x y)instead of x y.
Under (1), the relative error of f l(x y) satisfies
f l(x y) = (x y)(1 + δ), δ ≤ εM, (2)
where εM is the unit roundoff.
But if x, y arenotmachine numbers, then they must first rounded to floating-point format before the arithmetic operation and the resulting relative error becomes
f l(f l(x) f l(y)) = (x(1 + δ1) y(1 + δ2))(1 + δ3),
where δi ≤ εM, i = 1, 2, 3.
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Let stand for any one of the four basic arithmetic operators +, −, ?, ÷.
Whenever twomachine numbersxand y are to be combined arithmetically, the computer will produce f l(x y)instead of x y.
Under (1), the relative error of f l(x y) satisfies
f l(x y) = (x y)(1 + δ), δ ≤ εM, (2)
where εM is the unit roundoff.
But if x, y arenotmachine numbers, then they must first rounded to floating-point format before the arithmetic operation and the resulting relative error becomes
f l(f l(x) f l(y)) = (x(1 + δ1) y(1 + δ2))(1 + δ3),
where δi ≤ εM, i = 1, 2, 3.
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Let stand for any one of the four basic arithmetic operators +, −, ?, ÷.
Whenever twomachine numbersxand y are to be combined arithmetically, the computer will produce f l(x y)instead of x y.
Under (1), the relative error of f l(x y) satisfies
f l(x y) = (x y)(1 + δ), δ ≤ εM, (2)
where εM is the unit roundoff.
But if x, y arenotmachine numbers, then they must first rounded to floating-point format before the arithmetic operation and the resulting relative error becomes
f l(f l(x) f l(y)) = (x(1 + δ1) y(1 + δ2))(1 + δ3),
where δi ≤ εM, i = 1, 2, 3.
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Example
Let x = 0.54617 and y = 0.54601. Using rounding and four-digit arithmetic, then
x∗= f l(x) = 0.5462is accurate tofoursignificant digits since
|x − x∗|
|x| = 0.00003
0.54617 = 5.5 × 10−5 ≤ 5 × 10−4.
y∗ = f l(y) = 0.5460is accurate tofivesignificant digits since
|y − y∗|
|y| = 0.00001
0.54601 = 1.8 × 10−5≤ 5 × 10−5.
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Example
Let x = 0.54617 and y = 0.54601. Using rounding and four-digit arithmetic, then
x∗= f l(x) = 0.5462is accurate tofoursignificant digits since
|x − x∗|
|x| = 0.00003
0.54617 = 5.5 × 10−5 ≤ 5 × 10−4.
y∗ = f l(y) = 0.5460is accurate tofivesignificant digits since
|y − y∗|
|y| = 0.00001
0.54601 = 1.8 × 10−5≤ 5 × 10−5.
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Example
Let x = 0.54617 and y = 0.54601. Using rounding and four-digit arithmetic, then
x∗= f l(x) = 0.5462is accurate tofoursignificant digits since
|x − x∗|
|x| = 0.00003
0.54617 = 5.5 × 10−5 ≤ 5 × 10−4.
y∗ = f l(y) = 0.5460is accurate tofivesignificant digits since
|y − y∗|
|y| = 0.00001
0.54601 = 1.8 × 10−5≤ 5 × 10−5.
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The exact value of subtraction is r = x − y = 0.00016.
But
r∗≡ x y = f l(f l(x) − f l(y)) = 0.0002.
Since
|r − r∗|
|r| = 0.25 ≤ 5 × 10−1 the result has onlyonesignificant digit.
Loss of accuracy
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The exact value of subtraction is r = x − y = 0.00016.
But
r∗≡ x y = f l(f l(x) − f l(y)) = 0.0002.
Since
|r − r∗|
|r| = 0.25 ≤ 5 × 10−1 the result has onlyonesignificant digit.
Loss of accuracy
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Loss of Significance
One of the most common error-producing calculations involves the cancellation of significant digits due to the subtraction of nearly equal numbersor theaddition of one very large number and one very small number.
Sometimes, loss of significance can be avoided by rewriting the mathematical formula.
Example 12
The quadratic formulas for computing the roots of ax2+ bx + c = 0, when a 6= 0, are
x1= −b +√
b2− 4ac
2a and x2= −b −√
b2− 4ac
2a .
Consider the quadratic equationx2+ 62.10x + 1 = 0and discuss the numerical results.
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Loss of Significance
One of the most common error-producing calculations involves the cancellation of significant digits due to the subtraction of nearly equal numbersor theaddition of one very large number and one very small number.
Sometimes, loss of significance can be avoided by rewriting the mathematical formula.
Example 12
The quadratic formulas for computing the roots of ax2+ bx + c = 0, when a 6= 0, are
x1= −b +√
b2− 4ac
2a and x2= −b −√
b2− 4ac
2a .
Consider the quadratic equationx2+ 62.10x + 1 = 0and discuss the numerical results.
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Loss of Significance
One of the most common error-producing calculations involves the cancellation of significant digits due to the subtraction of nearly equal numbersor theaddition of one very large number and one very small number.
Sometimes, loss of significance can be avoided by rewriting the mathematical formula.
Example 12
The quadratic formulas for computing the roots of ax2+ bx + c = 0, when a 6= 0, are
x1= −b +√
b2− 4ac
2a and x2= −b −√
b2− 4ac
2a .
Consider the quadratic equationx2+ 62.10x + 1 = 0and discuss the numerical results.
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