Transcript Example

What is The Error ?
How it is happened ?
1
Definition
• The Error in a computed quantity is defined
as:
Error = True Value – Approximate Value
• Example:
- True Value : phi = 3.14159265358979
- Appr. Value : 22/7 = 3.14285714285714
- Error = -0.00126448926735
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Kind of Error
• The Absolute Error is measure the
magnitude of the error
Ea  Error
• The Relatife Error is a measure of the error
in relation to the size of the true value
Ea
Er 
True Value
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Example
• True value : 10 m
Appr. Value : 9 m
Ea  1,
Er  0.1
• True value : 1000 m
Appr. Value : 999 m
Ea  1,
Er  0.001
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Source of Error
•
Truncation Error
–
•
Caused by approximation used in the
mathematical formula of the scheme
Round-off Error
–
–
Caused by the limited number of digits that
represent numbers in a computer and
The ways numbers are stored and additions and
substractions are performed in a computer
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Background of
The Truncation Error
• Numerical solutions are mostly approximations for
exact solution
• Most numerical methods are based on approximating
function by polynomials
• How accurately the polynomial is approximating the
true function ?
• Comparing the polynomial to the exact solution it
becames possible to evaluate the error, called
truncation error
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Taylor Series
• The most important polynomials used to derived
numerical schemes and analyze truncation errors
• With an infinite power series, it is exactly represents a
function within a certain radius about a given point
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Taylor’s Theorem
(See Introduction to Real Analysis
by Bartle and Sherbert for proves)
Let n  N , let I  [a, b], and let f : I   be such t hat
f and it s derivat ives f ' , f ' ' ,  , f ( n ) are cont inouson I
and t hat f ( n 1) exist son a, b . If x0  I , t henfor any x in I
t hereexist sa point c bet ween x and x0 such t hat
f ' ' ( x0 )
f ( x)  f ( x0 )  f ' ( x0 )(x  x0 ) 
( x  x0 ) 2  
2
( n 1)
f ( n ) ( x0 )
f
(c )
n

( x  x0 ) 
( x  x0 ) n 1
n!
(n  1)!
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Applications
• For practice we denote h  x  x0
x0  
• Find the Taylor expansion of sin(x) about
2
answer:
2
3
h
h
sin(x)  sin( )  h cos( )  sin( )  cos( )
2
2
4
2!
2
3!
2
5
h
h

 sin( )  cos( )  
2
2
4!
5!
where h  x  
2
9
• How about the Taylor expansion of Tan(x) at
x0  
2 ?
• The Taylor expansion of a function about
x0  0 is called the Maclaurin series. Thus,
Maclaurin series for sin(x) is,
3
4
7
x
x
x
sin(x)  x    
3! 5! 7!
• It is impossible in practical applications to include
an infinite number of terms. Therefore, the Taylor
series has to be truncated after a certain order term
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• If the Taylor series is truncated after the N th
order term, it is expressed as
h2
h3
f ( x)  f ( x0 )  hf ' ( x0 ) 
f ' ' ( x0 ) 
f ' ' ' ( x0 )
2!
3!
h 4 ( iv)
h n (n)

f
( x0 )   
f
( x0 )  ( h n 1 )
4!
n!
where h  x  x0
( h
n 1
) f
( N 1)
h N 1
( x0   h)
, 0  1
( N  1)!
• Since  cannot be found exactly, the error term is
often approximated by
N 1
h
(h n 1 )  f ( N 1) ( x0 )
( N  1)!
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Example 1:
x
Find the Taylor expansion of e about x0  0
which use the first two, three, four and five and
evaluate for x= 0.5, respectively.
Answer:
x
2
h  x0
e  1  x  ( x )
2
x
x
x
3
e  1  x   ( x )
2!
2
3
x
x
x
e  1  x    ( x 4 )
2! 3!
2
3
4
x
x
x
x
e  1  x     ( x 5 )
2! 3! 4!
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e0.5  1.648721..
e0.5  1.5  ( x 2 )
e
( x 2 )  0.14872..
( x )  0.023721..
0.5
 1.625 ( x )
0.5
 1.64583..  ( x )
( x )  0.002887..
0.5
 1.64843..  ( x )
( x )  0.0002837..
e
e
3
3
4
5
4
5
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Example 2:
x
Find the Taylor expansion of e about x0  0.25
which use the first two, three, four and five and
evaluate for x= 0.5, respectively.
Answer:
h  x  0.25
e x  e0.25  he0.25  (h2 )
2
 0.25
h 0.25
x
0.25
0.25
3
e  e  he  e  (h )
2!
2
3
h 0.25 h 0.25
x
0.25
0.25
e  e  he  e  e  (h 4 )
2!
3!
2
3
4
h 0.25 h 0.25 h 0.25
x
0.25
0.25
5
e  e  he 
e  e 
e 
(
h
)
14
2!
3!
4!
e
0.5
 1.60503.. (h )
2
(h )  0.04368...
2
e0.5  1.64515.. (h3 )
e
0.5
(h3 )  0.00356..
 1.64850.. (h )
4
(h4 )  2.1988..e- 004
e
0.5
 1.64871..  (h )
5
(h5 )  1.0900..e- 005
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Summary from example 1 and 2:
Order of
Trunc. Er.
x0  0, h  0.5 x0  0.25, h  0.25
Ratio
(h2 )
0.14872..
0.04368..
3.4
(h )
0.023721..
0.00356..
6.66
(h )
0.002887..
2.1988..e-04
13.13
(h5 )
0.0002837..
1.090..e-0.5
26.03
3
4
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Numbers on Computers
• Computers do not use the decimal system in
computations and memory but use the binary
system
• It caused by computer memory consists of a
huge number of electronic and magnetic
recording devices, of which each element has
only “on” and “off” statuses
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• Example: In a single precision, 4 bytes, or
equivalently 32 bits, are used to store one real
number. If a decimal number is given by input, it
first converted to the closest binary in the
normalized format:
 0.abbbbb...bbb2 x 2
z
where a is always 1 and b’s are binary digit that are
0 or 1; z is an exponent which is also expressed in
binary.
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Numbers Store in Computer’s Memory for
Single Precision (IBM 370)
1 1111111

 
11111111
11111111
11111111


s e (7 bits)
m (24 bits)
where s is as sign (+ or -), e is an exponent, and m
is a mantissa including the a and b’s.
Example:
(1.5)10  (0.1100 0000 0000 0000 0000 0000)2 x 21
(0.00001)10  (0.1010 01111100 0101 1010 1100)2 x 2-16
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Causes Round-off Error
The summation of these two numbers becomes
(1)10  (0.00001)10
 (0.1100 0000 0000 0000 0101 0011 1110 0010 1101 0110 0 ) 2 x 21
Because the mantissa has 24 bits, therefore, the
result of this calculation is stored in memory as
(1)10  (0.00001)10  (0.1100 0000 0000 0000 0101 0011)2 x 21
 (1.5000100
136)10
Thus, whenever 0.00001 is added to 1.5, the result
gains 0.0000000136 as an error which called round20
off error
The effects of round-off error can be minimized by
changing the computational algorithm although it
must be devised case by case. Some useful strategies
include:
o
o
o
o
o
Double precision
Grouping
Taylor expansions
Changing definition of variables
Rewriting the equation to avoid substraction
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• Double Precision (IBM 370)
In a double precision, 8 bytes, or equivalently 64
bits, are used to store one real number. In this format
1 bit is used for sign, 7 bits are used for exponent,
and 56 bits are used for mantissa.
• Grouping
when the small numbers are computed, e.g. addition,
substraction, etc., grouping them helps to reduce
round-off errors. For example, to add 0.00001 to
unity ten thousand times one can grouped into 100
groups and each group consists of 100 small values.
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• Taylor Expansions
as  approaches 0, accuracy of a numerical
evaluation for
sin(1   )  sin(1)
f ( ) 

becomes very poor because round-off errors. By
using Taylor expansion, we can rewrite the equation
so that the accuracy for  is improved as
f ( )  cos(1)  0.5 sin(1)
2
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• Rewriting the equation to avoid substraction
consider the equation of
f ( x)  x  x  1  x 
for an increasing of values x, the calculation of the
equation above has a loss-of-significance error.
To avoid this error one can reformulate it to get
f ( x) 
x
x 1  x
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