ΑΡΙΘΜΗΤΙΚΗ ΑΝΑΛΥΣΗ
2. Αναπαράσταση αριθμών στον υπολογιστή
3. Σφάλματα
1
Scientific
computing is a
discipline
concerned with
the development
and
study of
numerical
algorithms for
solving
mathematical
problems that
arise in various
disciplines in
science and
engineering
2
Numerical methods are an essential part of an
engineer’s life
• They allow to solve a much wider range of physical/engineering
problems than analytical methods do.
• Numerical methods are however only one part of the solution: a
good physical understanding of the problem is essential.
• Numerical methods are the elementary pieces of more complex
codes and software used for scientific simulation.
• Studying numerical methods allows one to
- understand how more complex codes work
- be able to modify an existing code or create a new one
- understand the errors/limits introduced by the numerical simulation
3
2. ΑΝΑΠΑΡΑΣΤΑΣΗ ΑΡΙΘΜΩΝ ΣΤΟΝ
ΥΠΟΛΟΓΙΣΤΗ
Base 10
1
257.76  2 10  5 10  7 10  7 10  6 10
2
1
0
2
Base 2
 (1 23  0  2 2  1 21  1 20 )


(1011.0011) 2  
1
2
3
4 

(
0

2

0

2

1

2

1

2
) 10

 11.1875
4
• Numbers that have a finite expansion in one numbering
system may have an infinite expansion in another
numbering system:
(1.1)10  (1.000110011001100...) 2
• You can never represent 1.1 exactly in binary system.
5
Convert Base 10 Integer to binary representation
Converting a base-10 integer to binary representation.
Quotient
11/2
5
5/2
2
2/2
1
1/2
0
Remainder
1  a0
1  a1
0  a2
1  a3
(11)10  (a3 a 2 a1a0 ) 2
 (1011) 2
6
Fractional Decimal Number to Binary
Converting a base-10 fraction to binary representation.
0.1875  2
0.375 2
0.75  2
0.5  2
Number
Number after
decimal
0.375
0.75
1.5
1.0
0.375
0.75
0.5
0.0
Number before
decimal
0  a1
0  a2
1  a 3
1  a4
(0.1875)10  (a1a 2 a 3a 4 ) 2
 (0.0011) 2
7
Decimal Number to Binary
11.187510  
?.?
2
Since
(11)10  (1011) 2
and
(0.1875)10  (0.0011) 2
we have
(11.1875)10  (1011.0011) 2
8
All Fractional Decimal Numbers Cannot be
Represented Exactly
Converting a base-10 fraction to approximate binary
representation.
Number
0.3 2
0.6  2
0.2  2
0.4  2
0.8  2
0.6
1.2
0.4
0.8
1.6
Number
after
decimal
0.6
0.2
0.4
0.8
0.6
Number
before
Decimal
0  a1
1  a2
0  a 3
0  a4
1  a 5
(0.3)10  (a1a2 a3a4 a5 ) 2  (0.01001) 2  0.28125
Floating Point Representation (decimal)
• Floating Decimal Point : Scientific Form
256.78 is written as  2.5678 10
2
0.003678 is written as  3.678 10
3
 256.78 is written as  2.5678 10
2
10
Floating-Point Arithmetic (cont.)
• A computer number has three parts
– the sign (+ or -)
– the fraction part (called the mantissa)
– the exponent part
• There are three level of precision and these are the number of
bits used for mantissa and exponent.
Length
Sign
Mantissa
Exponent
Range
Single
32
1
23
8
10±38
Double
64
1
52
11
10±308
Extended
80
1
64
15
10±4931
11
The form is
exponent
sign  mantissa 10
or
  m 10e
Example: For
 2.5678 10 2
  1
m  2.5678
e2
12
Floating Point Format for Binary Numbers
y    m  2e
  sign of number 0 for  ve, 1 for - ve 
m  mantissa 12  m  102 
1 is not stored as it is always given to be 1.
e  integer exponent
13
Example
9 bit-hypothetical word
the first bit is used for the sign of the number,
the second bit for the sign of the exponent,
the next four bits for the mantissa, and
the next three bits for the exponent
54.7510  110110.112  1.10110112  25
 1.10112  1012
We have the representation as
0
0
1
0
1
mantissa
Sign of the
number
1
1
0
1
exponent
Sign of the
exponent
14
Machine Epsilon
Defined as the measure of accuracy and found by difference
between 1 and the next number that can be represented
Ten bit word
Sign of number
Sign of exponent
Next four bits for exponent
Next four bits for mantissa
Next
number
0
0
0
0
0
0
0
0
0
0
 110
0
0
0
0
0
0
0
0
0
1
 1.00012  1.062510
mach  1.0625  1  2 4
15
Relative Error and Machine Epsilon
The absolute relative true error in representing a number
will be less then the machine epsilon
Example
0.0283210  1.11002  25
 1.11002  20110
2
10 bit word (sign, sign of exponent, 4 for exponent, 4 for mantissa)
0 1 0 1 1 0 1 1 0 0
Sign of the Sign of the
number
exponent
exponent
mantissa
1.11002  20110
2
 0.0274375
0.02832  0.0274375
a 
0.02832
 0.034472  2 4  0.0625
16
Subtracting two almost equal numbers
17
IEEE-754 Floating Point Standard
• Standardizes representation of floating
point numbers on different computers in
single and double precision.
• Standardizes representation of floating
point operations on different computers.
18
IEEE-754 Format Single & Double Precision
S
Exponent8
Fraction23
32 bits for single precision
00000000000000000000000000000000
Sign Biased
(s) Exponent (e’)
Mantissa (m)
.
Value  ( 1)  1 m 2  2
s
Double precision
S Exponent11
Fraction52
(continued)
e ' 127
19
Example#1
11010001010100000000000000000000
Sign Biased
(s) Exponent (e’)
Mantissa (m)
Value   1  1. m2  2
s
e ' 127
 1  1.101000002  2
  1  1.625  2162127
  1 1.625 235  5.5834 1010
1
(10100010) 2 127
20
Example#2
Represent -5.5834x1010 as a single precision
floating point number.
? ???????????????????????????????
Sign Biased
(s) Exponent (e’)
Mantissa (m)
 5.5834  10   1  1. ?   2
10
1
?
21
Exponent for 32 Bit IEEE-754
8 bits would represent
0  e  255
Bias is 127; so subtract 127 from
representation
127  e  128
22
Exponent for Special Cases
Actual range of
e
1  e  254
e  0 and e  255 are reserved for special numbers
Actual range of
e
 126  e  127
23
Special Exponents and Numbers
e  0
e  255
s
0
1
0
1
0 or 1
e
all zeros
all zeros
all ones
all ones
all ones
all zeros
all ones
m
all zeros
all zeros
all zeros
all zeros
non-zero
Represents
0
-0


NaN
24
IEEE-754 Format
The largest number by magnitude
1.1........12  2
127
 3.40 10
The smallest number by magnitude
1.00......02  2
126
38
 2.18 10
38
Machine epsilon
 mach  2
23
 1.19  10
7
25
Significant Digits
•
Significant digits are those digits that can be
used with confidence.
•
Single-Precision: 7 Significant Digits
1.175494… × 10-38 to 3.402823… × 1038
•
Double-Precision: 15 Significant Digits
2.2250738… × 10-308 to 1.7976931… × 10308
26
Using Floating Point Numbers
• Beware of meaningless precision!
– In 1424, Jamshid Masud al-Kashi published
 = 3.141 592 653 589 793 25…
– …but noted that the error in computing the perimeter of a
circle with a radius 600’000 times that of earth would be less
than the thickness of a horse’s hair.
• Donald E. Knuth :
– Floating point arithmetic is by nature inexact, and it is not
difficult to misuse it so that the computed answers consist
almost entirely of “noise”. One of the principal problems of
numerical analysis is to determine how accurate the results
of certain numerical methods will be.
27
Using Floating Point Numbers
• Never use equality between two
floating point numbers !!!!!!!!
• Use a special method to compare
them!!!!! (define an acceptable error
for the specific problem)
28
3. Σφάλματα (Errors)
• Η λύση ενός προβλήματος με τη βοήθεια αριθμητικών μεθόδων
διαφέρει πάντοτε από την ακριβή λύση λόγω της παρουσίας
σφαλμάτων. Διάφορες πηγές μπορούν να προκαλέσουν
σφάλματα στα αριθμητικά αποτελέσματα
• Πηγές σφαλμάτων:
– Μετρήσεις φυσικών ή χημικών συσκευών και μηχανισμών (δεδομένα που
περιέχουν εγγενώς σφάλματα), ονομάζονται αρχικά σφάλματα.
– Σφάλμα του μαθηματικού προβλήματος ή σφάλμα της μαθηματικής
περιγραφής. Προέρχεται από τη μετατροπή του προβλήματος σε
μαθηματικό πρόβλημα, περιέχοντας απλοποιήσεις ή και παραλείψεις.
– Σφάλματα από προβλήματα κακής κατάστασης :αριθμητικά
προβλήματα που είναι πολύ ευαίσθητα σε μικρές μεταβολές των
δεδομένων.
29
Accuracy and Precision
• Accuracy refers to how closely a computed or measured value
agrees with the true value, while precision refers to how
closely individual computed or measured values agree with
each other (Μέτρο της ικανότητας διάκρισης μεταξύ σχεδόν
ίσων τιμών).
a) inaccurate and imprecise
b) accurate and imprecise
c) inaccurate and precise
d) accurate and precise
30
Four possible sources of errors
➡ Measurement error
➡ Modeling error
➡ Truncation error
➡ Round-off error
31
A. Measurement error
➡ Any instrument has a limit on its precision, and an
experimental result is always obtained with a tolerance:
e.g. 20.2 ± 0.1 cm
B. Modeling error
➡ Difference between the real system and the simplified
description used.
➡ Example of the bridge: representing the elements as
homogeneous or with a simplified geometry.
32
C. Truncation error
➡ This error arises due to the discrete or iterative nature of the
numerical methods used.
➡ For example, an iterative scheme can be developed to obtain the
physical quantity G.
- If the scheme is well-designed then G(n) approaches the true
value of G0 when n→∞.
- However, we always have to stop at a finite value of n=N.
The truncation error is the difference between G(N) and G(∞).
➡ A truncation error also arises when approximating a continuous
quantity by a discrete form or a derivative by a discrete limit:
33
D. Round-off error
• This error is intrinsic to the use of a computer.
• A computer does not use the real number but finiteprecision numbers (e.g. some decimals are discarded.)
• The propagation of rounding errors from one floating
point operation to the next is the most frequent source
of numerical instabilities.
34
Sources of Numerical Errors
1) Round off error (σφάλμα στρογγυλοποίησης)
– Προκύπτει από την ανάγκη αναπαράστασης των αριθμών με
πεπερασμένο πλήθος ψηφίων
2) Truncation error (σφάλμα αποκοπής)
– Δημιουργείται από τον αλγόριθμο που χρησιμοποιείται και
την προσέγγιση που επιλέγεται (κατά την αντικατάσταση
μιας ακριβούς διαδικασίας υπολογισμού με μία
προσεγγιστική)
35
Round off Error
• Caused by representing a number approximately
1
 0.333333
3
2  1.4142...
36
Rounding errors are random
37
Problems created by round off error
• 28 Americans were killed on February 25, 1991 by an
Iraqi Scud missile in Dhahran, Saudi Arabia.
• The patriot defense system failed to track and intercept
the Scud. Why?
38
Problem with Patriot missile
• Clock cycle of 1/10 seconds was represented in
24-bit fixed point register created an error of 9.5
x 10-8 seconds.
• The battery was on for 100 consecutive hours,
thus causing an inaccuracy of
 9.5  108
 0.342s
s
3600s
 100hr 
0.1s
1hr
• The shift calculated in the ranging system of the
missile was 687 meters.
• The target was considered to be out of range at a
distance greater than 137 meters.
39
Example: quadrature of a circle
40
41
42
A large collection of software bugs
• Careless numerical computing does occasionally lead to
disasters
http://wwwzenger.informatik.tumuenchen.de/persons/huckle/bugse.html
43
Άσκηση (σε MATLAB)
44
• Why measure errors?
1) To determine the accuracy of numerical results.
2) To develop stopping criteria for iterative algorithms.
45
True Error (απόλυτο σφάλμα)
• Defined as the difference between the true value in a
calculation and the approximate value found using a
numerical method etc.
True Error = True Value – Approximate Value
46
Example—True Error
The derivative, f (x) of a function f (x) can be
approximated by the equation,
f ' ( x) 
If
f ( x  h)  f ( x)
h
and h  0.3
a) Find the approximate value of f ' (2)
b) True value of f ' (2)
c) True error for part (a)
f ( x)  7e 0.5 x
47
Example (cont.)
Solution:
a) For x  2 and h  0.3
f ( 2  0.3)  f ( 2)
0.3
f (2.3)  f (2)

0.3
f ' ( 2) 
7e 0.5( 2.3)  7e 0.5( 2)

0.3
22.107  19.028
 10.263

0.3
48
Example (cont.)
Solution:
b) The exact value of f ' (2) can be found by using
our knowledge of differential calculus.
f ( x)  7e 0.5 x
f ' ( x)  7  0.5  e0.5 x
 3.5e 0.5 x
So the true value of
f ' (2)  3.5e 0.5( 2 )
 9.5140
f ' ( 2)
is
True error is calculated as
Et  True Value – Approximate Value
 9.5140 10.263  0.722
49
Relative True Error (σχετικό απόλυτο σφάλμα)
• Defined as the ratio between the true error, and the
true value.
Relative True Error ( t ) =
True Error
True Value
50
Example—Relative True Error
Following from the previous example for true error,
find the relative true error for f ( x)  7e 0.5 x at f ' (2)
with h  0.3
From the previous example,
Et  0.722
Relative True Error is defined as

t 
True Error
True Value
 0.722
 0.075888
9.5140
as a percentage,
t  0.075888  100%  7.5888%
51
Approximate Error (προσεγγιστικό σφάλμα)
• What can be done if true values are not known or are
very difficult to obtain?
• Approximate error is defined as the difference
between the present approximation and the previous
approximation.
Approximate Error (Ea ) = Present Approximation – Previous Approximation
52
Example—Approximate Error
For f ( x)  7e 0.5 x at x  2 find the following,
a) f (2) using h  0.3
b) f (2) using h  0.15
c) approximate error for the value of f (2) for part b)
Solution:
a) For x  2 and h  0.3
f ( x  h)  f ( x)
h
f ( 2  0.3)  f ( 2)
f ' ( 2) 
0.3
f ' ( x) 
53
Example (cont.)
Solution: (cont.)

f (2.3)  f (2)
0. 3
7e 0.5( 2.3)  7e 0.5( 2)

0.3

22.107  19.028
 10.263
0.3
b) For x  2 and h  0.15
f (2  0.15)  f (2)
0.15
f (2.15)  f (2)

0.15
f ' (2) 
54
Example (cont.)
Solution: (cont.)
7e 0.5( 2.15)  7e 0.5( 2)

0.15

20.50  19.028
 9.8800
0.15
c) So the approximate error,
Ea is
Ea  Present Approximation – Previous Approximation
 9.8800  10.263
 0.38300
55
Relative Approximate Error (σχετικό σφάλμα)
• Defined as the ratio between the approximate error
and the present approximation.
Relative Approximate Error ( a ) =
Approximate Error
Present Approximation
56
Example—Relative Approximate Error
For f ( x)  7e at x  2 , find the relative approximate
error using values from h  0.3 and h  0.15
Solution:
From Example 3, the approximate value of f (2)  10.263
using h  0.3 and f (2)  9.8800 using h  0.15
0.5 x
Ea  Present Approximation – Previous Approximation
 9.8800  10.263
 0.38300
57
Example (cont.)
Solution: (cont.)
a 

Approximate Error
Present Approximation
 0.38300
 0.038765
9.8800
as a percentage,
a  0.038765 100%  3.8765%
Absolute relative approximate errors may also need to be
calculated,
a | 0.038765 |  0.038765 or 3.8765 %
58
How is Absolute Relative Error used as a stopping
criterion?
If |a |  s where s is a pre-specified tolerance, then
no further iterations are necessary and the process is
stopped.
If at least m significant digits are required to be correct
in the final answer, then
|a | 0.5  10 2m %
59
Table of Values
For
f ( x)  7e 0.5 x
at x  2 with varying step size, h
h
f ( 2)
a
m
0.3
10.263
N/A
0
0.15
9.8800
3.877%
1
0.10
9.7558
1.273%
1
0.01
9.5378
2.285%
1
0.001
9.5164
0.2249%
2
60
Θεώρημα
If a positive number x has n correct digits in the narrow
sense, the relative error ER of this number does not
exceed
1 𝑛−1
10
divided by the first significant digit of the
given number or 𝐸𝑅 ≤
1
1 𝑛−1
𝑎𝑚 10
,where 𝑎𝑚 is first
significant digit of number x.
61
Παράδειγμα
How many digits are to be taken in computing 20
so that the error does not exceed 0.1%?
62
Σφάλμα, Απόλυτο Σφάλμα και σημαντικά ψηφία
• Αν η τιμή 𝑥 είναι μια προσέγγιση της τιμής 𝑥 τότε :
– Σφάλμα : 𝐸𝑥 = 𝑥 − 𝑥
– Απόλυτο σφάλμα : 𝐸𝑥 = 𝑥 − 𝑥
– Σχετικό σφάλμα : 𝑅𝑥 =
𝑥− 𝑥
,
𝑥
𝑥≠0
• Ο αριθμός 𝑥 λέμε ότι προσεγγίζει την πραγματική τιμή
𝑥 με 𝑑 σημαντικά ψηφία αν 𝑑 είναι ο μεγαλύτερος
θετικός ακέραιος για τον οποίον ικανοποιείται η
ανισότητα:
𝑥− 𝑥
𝑥
<
1
10−𝑑
2
• Εφαρμογή για 𝑥 = 3,14 και 𝑥 = 3,141592
63
Problem conditioning and algorithm stability
• The problem is ill-conditioned if a small perturbation in
the data may produce a large difference in the result.
• The problem is well-conditioned otherwise.
• The algorithm is stable if its output is the exact result of
a slightly perturbed input.
64
An unstable algorithm
65
A stable algorithm
66
Effect of Carrying Significant
Digits in Calculations
67
Find the contraction in the diameter
Tc
D  D   (T )dT
Ta
Ta=80oF; Tc=-108oF; D=12.363”
α = a0+ a1T + a2
2
T
68
Thermal Expansion Coefficient vs
Temperature
D  D  T
T(oF)
α (μin/in/oF)
-340
-300
2.45
3.07
-220
-160
4.08
4.72
-80
5.43
0
40
6.00
6.24
80
6.47
69
Regressing Data in Excel
(general format)
α = -1E-05T2 + 0.0062T + 6.0234
70
Observed and Predicted Values
α = -1E-05T2 + 0.0062T + 6.0234
T(oF)
α (μin/in/oF)
Given
α (μin/in/oF)
Predicted
-340
2.45
2.76
-300
3.07
3.26
-220
4.08
4.18
-160
4.72
4.78
-80
5.43
5.46
0
6.00
6.02
40
6.24
6.26
80
6.47
6.46
71
Regressing Data in Excel
(scientific format)
α = -1.2360E-05T2 + 6.2714E-03T + 6.0234
72
Observed and Predicted Values
α = -1.2360E-05T2 + 6.2714E-03T + 6.0234
T(oF)
α (μin/in/oF)
Given
α (μin/in/oF)
Predicted
-340
2.45
2.46
-300
3.07
3.03
-220
4.08
4.05
-160
4.72
4.70
-80
5.43
5.44
0
6.00
6.02
40
6.24
6.25
80
6.47
6.45
73
Observed and Predicted Values
α = -1.2360E-05T2 + 6.2714E-03T + 6.0234
α = -1E-05T2 + 0.0062T + 6.0234
T(oF)
α (μin/in/oF)
Given
α (μin/in/oF)
Predicted
α (μin/in/oF)
Predicted
-340
2.45
2.46
2.76
-300
3.07
3.03
3.26
-220
4.08
4.05
4.18
-160
4.72
4.70
4.78
-80
5.43
5.44
5.46
0
6.00
6.02
6.02
40
6.24
6.25
6.26
80
6.47
6.45
6.46
74
Truncation error
• Error caused by truncating or approximating a
mathematical procedure.
75
Example of Truncation Error
Taking only a few terms of a Maclaurin series to
x
e
approximate
2
3
x
x
e x  1  x    ....................
2! 3!
If only 3 terms are used,
2


x
x
Truncation Error  e  1  x  
2! 

76
1 .2
e
Calculate the value of
with an absolute
relative approximate error of less than 1%.
e1.2
1.2 2 1.2 3
 1  1.2 

 ...................
2!
3!
n
e 1 .2
Ea
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.
How many are required to
get at least 1 significant
digit correct in your
answer?
77
Taylor Series (1)
78
Taylor Series (2)
To gain insight consider the mathematical formulation that is
used widely in numerical methods - TAYLOR SERIES.
A Taylor series :provides a means to predict a function value
at one point in terms of the function value at and its
derivatives at another point
Some examples of Taylor series which you must have seen
x2 x4 x6
cos( x)  1     
2! 4! 6!
x3 x5 x7
sin( x)  x     
3! 5! 7!
x2 x3
e  1 x 


2! 3!
x
79
General Taylor Series
The general form of the Taylor series is given by
f  x  2 f  x  3
f  x  h   f  x   f  x h 
h 
h 
2!
3!
provided that all derivatives of f(x) are continuous and exist
in the interval [x, x+h] .
As Archimedes would have said,
“Give me the value of the function at a single point, and the
value of all (first, second, and so on) its derivatives at that single
point, and I can give you the value of the function at any other
point”
80
81
Example—Taylor Series
Find the value of f 6 given that f 4  125, f 4  74,
f 4  30, f 4  6 and all other higher order derivatives
of f x  at x  4 are zero.
Solution:
h2
h3
f x  h   f x   f x h  f x   f x   
2!
3!
x  4,
h  64  2
82
Example (cont.)
Since the higher order derivatives are zero,
22
23
f 4  2  f 4  f 42  f 4  f 4
2!
3!
 2 2   23 
f 6  125  742  30   6 
 2!   3! 
 125  148  60  8
 341
Note that to find f 6 exactly, we only need the value
of the function and all its derivatives at some other point, in
this case x  4
83
Derivation for Maclaurin Series for ex
Derive the Maclaurin series
x2 x3
e  1 x 


2! 3!
x
The Maclaurin series is simply the Taylor series about the
point x=0
h2
h3
h4
h5
f x  h   f x   f x h  f x   f x   f x   f x   
2!
3!
4
5
h2
h3
h4
h5
f 0  h   f 0  f 0h  f 0  f 0  f 0  f 0  
2!
3!
4
5
84
Derivation (cont.)
Since
f ( x)  e x , f ( x)  e x , f ( x)  e x , ... , f n ( x)  e x
and
f n (0)  e0  1
the Maclaurin series is then
(e 0 ) 2 (e 0 ) 3
f ( h )  (e )  (e ) h 
h 
h ...
2!
3!
1
1
 1  h  h 2  h 3 ...
2!
3!
0
0
So,
x 2 x3
f ( x)  1  x    ...
2! 3!
85
The exponentia l function is computed by
2
3
4
n
x
x
x
x
e x  1  x     ...   ...
2 3! 4!
n!
When x = 0.5
Terms Result
εt (True percentage relative εa (Approx. percentage
error)
relative error)
1
1
(1.6487-1)/1.6487 = 39.3%
2
1.5
(1.6487-1.5)/1.6487 = 9.02% (1.5-1)/1.5 = 33.3%
3
1.625
1.44%
4
1.645833333 0.175%
1.27%
5
1.648437500 0.0172%
0.158%
6
1.648697917 0.00142%
0.0158%
(1.625-1.5)/1.625 = 7.69%
86
How many terms should we use?
Terms
Result
εt
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%
• Computation stops when |εa|
εa
< εs
– εs = pre-determined acceptable percentage relative error
87
Error in Taylor Series
The Taylor polynomial of order n of a function f(x) with
(n+1) continuous derivatives in the domain [x, x+h] is
given by
n
h2
h
f x  h   f x   f x h  f ' ' x     f n  x   Rn x 
2!
n!
where the remainder is given by
Rn
n 1

x  h
x  
(n  1)!
f n 1 c 
where
x  c  xh
that is, c is some point in the domain [x, x+h]
88
Example—error in Taylor series
The Taylor series for
e x at point x  0 is given by
2
3
4
5
x
x
x
x
ex  1 x 
 
 
2! 3! 4! 5!
It can be seen that as the number of terms used
increases, the error bound decreases and hence a
better estimate of the function can be found.
How many terms would it require to get an
approximation of e1 within a magnitude of true error of
less than 10-6.
89
Example—(cont.)
Solution:
Using n  1 terms of Taylor series gives error bound of
n 1

x  h
x  0, h  1, f ( x)  e x
Rn x  
f n 1 c 
n  1!
n 1

0  1
Rn 0 
f n 1 c 
n  1!
n 1

 1

ec
n  1!
Since
x  c  xh
0  c  0 1
0  c 1
1
e
 Rn 0 
(n  1)!
(n  1)!
90
Example—(cont.)
Solution: (cont.)
So if we want to find out how many terms it would
1
require to get an approximation of e
within a
6
10
magnitude of true error of less than
,
e
 10 6
(n  1)!
(n  1)! 106 e
(n  1)! 106  3
(as we do not know the value of but it is less than 3)
n9
So 9 terms or more are needed to get a true error
6
10
less than
91
Exercise-1
Use Taylor series to find the value of function
f ( x)  sin( x)
for x  2
Compare this value with one get from a
calculator ( 0.909296723 )
92