Transcript 4 significant figures

Matakuliah
Tahun
: S0262-Analisis Numerik
: 2010
Introduction and Analysis of Error
Pertemuan 1
Material Outline
• Introduction to Numerical Methods
• Approximation and Errors
• Propagation of Errors
Numerical Analysis: Methods for providing
numerical answer when analytic
procedures are either computationally
difficult or nonexistent.
Numerical Analysis: Always involving
complicated arithmetic computations 
Need the help of computing machines
(computers)
Numerical Analysis  Numerical Methods
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DR. Paston Sidauruk
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 Why Numerical methods are important:
 In real life many problems can not be solved
analytically.
 Numerical Analysis is a universal procedure (It can be
applied to many different engineering fields such as CE,
ME, EE, and IT)
 The fast growing development of computing machines
(computers)
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DR. Paston Sidauruk
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 INTRODUCTION
It is a fact that many mathematical or engineering
problems or or physics phenomena can not be solved
analytically (exact solution)  For this type of
problems or phenomena, numerical analysis can give
the solution.

Numerical Work in general will cover the following steps:
 Modeling : formulating a given work
(problem) into mathematical equations.
 Choosing appropriate Numerical methods.
 Developing a program
 Executing the program
 Analyzing the results
Note: Numerical Methods is always involving serious
arithmetical calculation  Need the help of computer.
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Some Concerns About Numerical
Analysis:
– Numerical Value: is an approximation of true
value that never known, hence we have to
concern about the errors that may occur.
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 ERRORS
In numerical solution, the results we get is the best
approximation to true value. Hence the numerical
solution is always associated to certain degree of
errors. For this purpose, a true value of any
parameter can be written as :
a  ã - Et
In which :
a = true value (exact)
ã = approximation (derived from measurements,
calculations etc)
Et = total error
Numerical Errors: arise from the use of
approximations to represent exact
mathematical operations and quantities.
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 RELATIVE ERRORS
It is sometimes desired to normalize the error with the true value
such as to account the magnitudes of the quantities being
evaluated, such error called relative error.
Relative error t=true error/true value
 Relative error of approximation:
a= approximation error/approximation
In the approximation of using iteration ( current approximation is
based on the previous approximation), then the relative error of
approximation is given below:
a= (current approximation – previous approximation)/ current
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approximation
In numerical computations or iteration process, it is sometimes
to pre specify the tolerance (s) such that the following eq is
satisfied,
| a|< tolerance (s) =(0,5 x 102-n) %
n= significant figure
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 Errors sources
1. Experimental errors (errors arise from
experiments, measurements etc)
2. Round off errors (errors because of
rounding)
3. Truncation errors (errors arise from a
process of simplification an algorithms,
computations, steps in algorithms)
4. Programming errors
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 Example:
A MacLaurin series of ex is given below:
2
3
x
x
ex  1 x 

 ..........
2!
3!
If the 1st term is considered as 1st estimate, the first 2
terms as the second estimate of ex, how many do
you have to include in the series such that the
relative errors | a|< s. In which a pre specified
tolerance conforming to 3 significant figures. Find
the true error and approximation errors.
Note: e0,5= 1.648721271
Solution: s =(0,5 x 102-3)%= 0,05 %
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•
Example (Cont.):
e0,5= 1,648721271
Solution: s =(0,5 x 102-3)%= 0,05 %
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Ith tern
Result
t (%)
a (%)
1.
1
39,3
2
1,5
9,02
33,3
3
1,625
1,44
7,69
4
1,645833333
0,175
1,27
5
1,648437500
0,0172
0,158
6
1,648697917
0,00142
0,0158
Therefore, the minimum of the first 6 terms have to be used to
estimate ex such that the error of approximation is less than
the pre-specified tolerance.
2nd column: Series value for x= 0.5,
3rd column= (2nd column)/1.648721271.
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
FINITE NUMBERS: can be written in two ways
1. fixed- point system
( is written according to the specified
number of decimal place)
Example: 62.358; 0.013; and 1.000 (3 decimal
place)
2. floating- point system is written
according to certain significant figures
Example:
0.6238 * 103; 1.7130 * 10-13; 2000 * 104
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 Significant Figures
The concept of significant figures designate the reliability
of a numerical value.
All digits that can be used with confidence.
Example:
4 digit significant figures
1.360 ; 1360 ; 0.001360
All zeros that are needed only to locate the decimal point
are not counted as significant figures:
Example: all of the following numbers are in 4 significant
figures 0.01845; 0.0001845; 0.001845
Also:
4.53 * 104 (3 significant figures)
4.530 * 104 (4 significant figures)
4.5300 * 104 (5 significant figures)
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
Rounding a number to certain significant figures
1.
2.
3.
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Those digits that are not significant will be omitted. The
last digit that is saved will rounded up if the first digit in
the omitted digits >5 and if the 1st in the omitted digits
=5 and the last digit in the saved digits is odd number.
The final results of summation or subtraction will be
rounded to the most significant figures of all the
numbers (quantities) that are being operated.
The final results of multiplication or division will be
rounded such that the number of significant figures will
be equivalent with the least number of significant figures
of all the numbers (quantities) that are being operated.
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
•
Rounding a number to certain significant figures
Examples:
–
–
Rounding
5.6723  5.67 (3 significant figures)
10.406  10.41 (4 significant figures)
7.3500  7.4 (2 significant figures)
88.21650  88.216 (5 significant figures)
1.25001  1.2 (2 significant figures)
Summation/Subtraction
Evaluate: 2.2 – 1.768
 2.2-1.768= 0.432  0.4
4.68 x 10-7+8.3x10-4-228x10-6= ….? …….
–
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(6.0x10-4)
Multiplication/Division
0.0642x 4,8= 0.30816 0.31
945/0.3185= 2967.0329672970
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

Error Propagation
The purpose is to study how errors in numbers
propagates through mathematical functions
•
Function of single Variable
y  f ( x)
f ( ~
x )  f ' (~
x ) ~
x
f ( ~
x )  estimate of error of f ( ~
x )  f ( x)  f ( ~
x)
~
x  approximat ion value
x  true value
~
x  estimate of error of ~
x  x~
x
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 Error Propagation
•
Function of single Variable
Example: Given a value of
~
x  2.5 with an error ~
x  0.01, estimate the resulting
error in the function f ( x)  x 3
Answer:
f ( x)  x 3  f ' ( x)  3x 2
f ( ~
x )  f ' (~
x ) ~
x  3(2.5) 2 (0.01)  0.1875
f (2.5)  15.625  0.1875
15.4375  f (2.5)  15.8125
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
Error Propagation
•
Function of more than one variable
y  f ( x1 , x2 ,, xn )
f ~ f ~
f ~
~
f ( x ) 
x1 
x2  
xn
x1
x2
xn
f ( ~
x )  estimate of error of f ( ~
x1 , ~
x2 ,  , ~
xn )
~
x ,~
x ,, ~
x  approximat ion values
1
2
n
~
x1 , ~
x2 ,, ~
xn  estimate of error of ~
x1 , ~
x2 ,  , ~
xn
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
Error Propagation
• Function of more than one variable
Example/Exercises:
xy4
y  f ( x, y , z , t ) 
8 zt
estimate the resulting error in y given that
~
x  50  ~
x  2 ~y  30  ~y  0.1
~z  1.5  ~z  0.01 ~t  0.6  ~t  0.006
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 Error Propagation
Estimated Error for common mathematical operations
Operation
Addition
Subtraction
Multiplication
Division
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Estimated Error
( ~
x~
y)
( ~
x~
y)
( ~
x~
y)
 ~x 
 ~ 
y
~
x  ~
y
~
x  ~
y
~
x ~
y ~
y ~
x
~
x ~
y ~
y ~
x
2
~
y
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