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Introduction to computing
Dr. Asaf Varol
[email protected]
1
Basic operation performed by a
computer
• Arithmetic Operations: Addition,
subtraction, multiplication and division
• Logical operations: the sign or the
comparative magnitude of two numbers
• Data transfer: Moving data from one
location to another in the memory.
• Input-output operations: Controlling the
reading/writing of information into or out of
the computer
2
On Digital Computers
Digital computers store numbers in an entity (or
unit) called a word which consists of a string of
binary digits, or bits. Various number systems
are used to represent mathematical numbers.
Some commonly used number systems are:
hexadecimal (base 16), decimal (base 10), octal
(base 8), and binary (base 2). For example, in a
decimal system the number 8,410 is
represented in powers of ten as:
8103 + 4102 +1101 + 0100 = 8000 + 400 +
10 + 0 = 8,410
3
On Digital Computers (Cont’d)
A method known as the doubling procedure is as follows. Given a
decimal number N, it can be decomposed as:
N = 2Q1 + R1
Q1 = 2Q2 + R2
.
.
.
Qk = 0 + Rk+1
(Q1 = N/2 - remainder)
(Q2 = Q1/2 - remainder)
.
.
.
etc.
The corresponding binary number is obtained by writing the
remainders Rk+1, Rk, ... , R1 in the reverse order as:
B = Rk+1RkRk-1 ... R1
4
Example
Convert the decimal number N = 8,410, to a binary number.
Solution:
Perform sequential division by 2 as follows:
8,410 = (24,205) + 0
65 = (232) + 1
4,205 = (22,102) + 1
32 = (216) + 0
2,102 = (21,051) + 0
16 = (28) + 0
1,051 = (2525) + 1
8 = (24) + 0
525 = (2262) + 1
4 = (22) + 0
262 = (2131) + 0
2 = (21) + 0
131 = (265)
+1
1 = (20) + 1
The binary equivalent of 8,410 is then given by collecting the remainder
digits from the last to the first:
10000011011010 = 11213 + 0212 + 0211 + 0210 + 029 + 028 + 127 + 126
+ 025 + 124 + 123 + 022 + 121 + 020
5
Representations of Numbers
Numbers are usually represented using the
normal form notation. That is,
x = m.10E for
10-1 < m < 1
where for x  0, m is called the mantissa and E
is the exponent. By convention, the number
zero has the normal notation, 0.100.
6
Significant Digits
If a number is written in standard decimal,
floating-point form, or in normal form such that:
x = 0.d1 d2 d3 ... dk10n
with d1  0 and dk  0, we say that this number
has k significant digits (or significant figures)
which indicates those digits that can be used
with a confidence relative to the true value of the
number.
7
Significant Digits (Cont’d)
Note that the zeros which are used only to shift the decimal point are
not counted as significant figures. The leading zeros may or may
not be significant. For example,
x = 0.0002815 has 4 significant figures!
x = 1,200 may have 4, 3, or 2 significant figures!
Some examples are:
46.45072800
significant digits)
-335.12
0.00517
0.74
= 0.46450728102
(with
8,
9,
or
10
= -0.33512103 (with 5 significant digits)
= 0.51710-3
(with 3 significant digits)
= 0.74100
(with 2 significant digits)
8
Computer Representation of Numbers
The decimal equivalent of the binary number represented in Figure
1.4.1 is given by:
-(026 + 025 + 024 + 123 + 022 + 121 + 120)
= -(0 + 0 + 0 + 8 + 0 + 2 + 1)
= -11
{sign}

1
0

Value
0
0
1
0
1
1
Figure 1.4.1 Binary representation of an integer using an
8 bit word (or Byte)
9
Example
Determine the largest integer that can be represented by an 8 bit
machine.
Solution:
Imax = +(126 + 125 + 124 + 123 + 122 + 121 + 120)
= +( 64 + 32 + 16 + 8 + 4 + 2 + 1 )
= +(127)
= +(27 - 1)
In general:
Imax = +[2(n -1) - 1]; Imin = -[2(n -1) - 1]
For a binary computer utilizing 32 bit words,
Imax = 2,147,483,647
10
Floating-Point Representation
A floating-point number is written as:
x = (sign)m.b(sign)E
where m is the mantissa, b is the base (b = 2 for a binary
system), and E is the exponent.
Sign of Mantissa
d1
Digits of Mantissa
d2
d3
d4
Sign of Exponent Digits of Exponent
d5
d6
d7
d8
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Example
Determine the smallest, positive, nonzero, floating point
number that can be represented by an eight bit machine
using binary system with one bit spared for the sign of
the mantissa, one bit for the sign of the exponent, and
two bits for the digits of the exponent:
Sign of Mantissa
0
Digits of Mantissa
0
0
0
Sign of Exponent Digits of Exponent
1
1
1
1
Solution:
m = +(023 + 022 + 021 + 120)
m = +( 0
+0 + 0 + 1 )=1
E = -[(121) + (120)] = -(2+1) = - 3
Number = 12-3 (which is equal to 0.1250 in decimal system)
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Errors
the approximation of numbers, accuracy, and precision. Neither physical
measurements nor arithmetic calculations can be carried out exactly. The
engineer's motto should be:
“There is nothing which is absolutely correct or exact in
science”
Accuracy is the measure of how close an estimated value or
answer is to its true (or exact) value. Since in many situations this exact
value is not known, the accuracy of an answer is usually measured with
respect to the best-estimated value.
Precision implies how closely the repeatedly measured (or
calculated) values of a certain quantity agree with each other. Thus it
represents the number of significant figures in representing that quantity as
a single average number with a spread (variation) around its mean.
13
Errors
True absolute error
Et =(true value) - (approximate value)
Approximate absolute error
Ea =(best estimate) - (approximate value)
True relative error
et =(true value) - (approximate value)/true value
Approximate relative error
ea =(best estimate) - (approximate value)/best
estimate
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Errors (Cont’d)
For series, sequences, and iterations, approximate error can be defined as:
Approximate absolute error (for iterative calculations):
Ea =(current value) - (previous value)
Approximate relative error (for iterative calculation):
ea = ((current value) - (previous value))/current value
When the exact value of a quantity is not given (or known), it is not possible
to calculate a true error. However, the approximate error can be used to
calculate error bounds for an approximate answer. To this end, a theorem
called Scarborough criteria can be very useful.
Scarborough Criteria
If the approximate relative error, ea < 0.510-m, then the result (or the
answer) is correct to at least m significant digits.
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Example E1.5.1
Calculate the number of terms that is necessary to estimate the
value of  to two significant digits from the Taylor series expansion
for Arctan(x) about the base point x = 0. The number  =
3.141592653589793 (to 16 significant digits) is related to this
function by:
 = 4.0Arctan(1.0)
Solution:
The infinite series (convergent in the range –1.0 to 1.0) is given by:

x 3 x 5 x 7 x 9 x11
 1 x 2 n1
Arc tan x   x     
 ... 
 ...
3
5
7
9 11
2n  1
 n 1
where n = 1, 2, 3, ...
Check your answer by actually calculating  from the above series.
Plot the true relative error and the approximate relative error as a
function of the number of terms.
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Errors (Cont’d)
Note that for such series, the definition of ea reduces to:
ea =(current value) - (previous value)/current value
ea =last term used/current sum
If we assume that the current sum is accurate to two significant
digits ( = 3.1), then:
ea  4 (-1)(n+1) x2n-1/(2n-1)/(3.1) at x = 1
Using Scarborough criteria,
ea  (4/3.1)/(2n-1) < 0.510-2
Solving this equation for n (by trial and error) gives:
n  130
17
MatLab solution
18
Comparison of the true and approximate
relative error
19
Computer errors: Round off or Chop off errors
Most computers chop off (or simply ignore) the
digits beyond their capacity of representing
them. That is, when the number of significant
digits does not fit into the space allocated for the
mantissa, some computers round the number.
For example, a computer with a three digit
mantissa will represent 68.501 as 0.068E03
when chopping off is used, or as 0.069E03 when
rounding is used.
20
Computer errors: Round off or Chop off errors
Subtractive cancellation
This error occurs when subtracting two nearly equal numbers. Let us further
explain this with an example.
Example
If x = 40,000.01 and y = 40,000, what is x minus y using a 3-digit mantissa?
Solution:
0.4000001x105
-0.4000000x105
_______________________
0.000
x105 = 0.0
Smearing due to round-off errors
Significant errors can occur when adding a large and a small number. For
example, adding two temperatures, 0.4 K to 250 K, using a hypothetical decimal
computer with a mantissa of 3 digits, yields:
0.250x103
0.0004x103
_______________________
0.250 x103 = 250 K
21
Example E1.5.5
Consider the transient heat equation:
dT/dt = -(Ta - T)/h
where t is the time, h is a characteristic thermal relaxation time, and Ta is
the ambient temperature (the temperature of the surroundings).
Use a finite difference method to find the variation of the temperature T of
an object, initially at a temperature T0 = 950 K, after it is immersed in a fluid
having an uniform temperature Ta = 1000 K:
Tnew = Told + t(Ta - Told)/h
where t is the time increment between the old and new temperature and h
= 1000 sec. We can start at time t = 0. Set Told = T0 = 950 and march in time
to calculate T at subsequent times.
The results with t = 0.1 and 0.001 sec are depicted in figure along with the
exact solution :
T = Ta + (T0 - Ta)exp(-t/h)
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Figure E.1.5.5 Illustration of smearing due to
round off
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Truncation error
Truncation errors are those which usually arise from neglecting (cutting) a number of terms in a
series formulation.
The geometric series is given by:
in the interval -1 < x < 1 for n = 0, 1, 2, 3…
when x = 0.5, the exact value of (x) = 2. If we represent this function by only three terms (n = 2),
then:
(0.5)  1 + x + x2 = 1 + 0.5 + (0.5)2 = 1.75
Hence, the truncation error T.E. for n = 2, is given by:
T.E. = [(0.5)exact - (0.5)approx] = 2 - 1.75 = 0.25 = Et = true error
or the true relative error,
et = Et/2.0 = 0.125 (12.5%)
This truncation error is the sum of all the terms left out from the infinite series.
(x) = 1/(1 - x)  1 + x + x2 + (T.E.)n=2
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Taylor series expansion & approximation
•
•
•
•
•
•
•
•
Taylor series is used in Analysis to derive:
Integration formulas, functional approximations, finite
difference schemes and error analysis
A convergent series is a series whose sequence of
partial sums converges to a finite sum.
A divergent series is a series that does not converge.
The geometric series converges for -1< x < 1. As an
example let us calculate f(x=0.1):
n=0
S0=1
=1.000
n=1
S1=1 + x2
=1.100
n=2
S2=1 + x + x2
=1.110
n=∞
S∞=1.0/(1-0.1)
=1.111111……
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Determining of converges
• To determine whether a series is convergent or not, there are two
most commonly applied tests:
• The nth term test states that the series:
Sum=q1+q2+q3+…+qn
• Converges if qn approaches zero in the limit of the nth term must go to
zero as n approaches ∞.
• According to the ratio test, if a series with the sequence of the
summation of its term q1, q2, q3, ….qn has the property that.
• If R<1 or R=0, the series converges absolutely.
• If R>1, the series diverges.
• If R=1, the series may or may not converge.
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Example E.1.6.1
27
Approximation of Functions: Taylor Series
Taylor's Theorem (Taylor`s Formula)
Let (x) be a function defined and continuously differentiable in the closed
interval from a to x, and it has continuous derivatives of all orders in the
same interval. One can then express that function in terms of a power
series as follows:
(x) = (a) + (1)(a)(x-a) + (2)(a)(x-a)2/2! + (3)(a)(x-a)3/3! + …
where (n)(a) denotes the nth order derivative of  with respect to the
independent variable x, then, evaluated at the point x = a. The point x = a
about which the expansion is built is known as the base point.
If the Taylor series is truncated after a finite number of terms, then the sum
of all the terms truncated is called the remainder Rn(x). That is:
(x) = (a) + (1)(a)(x-a) + (2)(a)(x-a)2/2! + (3)(a)(x-a)3/3!
+ ... + (n)(a)(x-a)n/n! + Rn(x)
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Taylor Series
The remainder Rn(x) can be written (Stein, 1967) in the integral form as:
( x  t ) n f ( n 1) (t )dt
Rn ( x )  
n!
a
x
where t is a dummy variable used for the purpose of integration.
Using the first and second mean value theorems of calculus for integrals,
b
 f ( x)dx  (b  a) f ( ) ; a    b
a
b
b
a
a
 f ( x) g ( x)dx  f ( )  g(x) ; a    b
Eq. 1.6.5 can be written as:
( x  a) n1 f ( n1) ( )
Rn ( x) 
; a   x

(n 1)!
29
Taylor Series (Cont’d)
Different forms of the Taylor series
In Taylor`s Theorem, let a = xi (meaning the ith point) and let x = xi+1 = a+h = xi+h
denote the next point. Then it takes the form:
(xi+1) = (xi) + (1)(xi)h + (2)(xi)h2/2! + (3)(xi)h3/3! + … + (n)(xi)hn/n!
This form is particularly convenient for developing finite difference formulae. This
equation can be further simplified with a change of notation:
i = (xi); i+1 = (xi+1); i(n) = (n)(xi)
Hence, we can write:
i+1 = i + i(1)h + i(2)h2/2! + i(3)h3/3! + … + i(n)hn/n! + ...
This form is convenient in deriving formulae for the integration of ordinary differential
equations (see Chapter 6).
Some simple formulae for computing the derivative of a function numerically are
given here as an introduction to numerical differentiation.
30
Numerical differentiation
We truncate the Taylor series after the second term and solve for (1) =  to obtain:
 = (i+1-i)/h
or
f ( xi 1 )  f ( xi ) f (a  h)  f (a)
 df 
Sf  


; Forward Difference
xi 1  xi
h
 dx  x  x
i
where h = xi+1-xi = x. This is called a first order, forward difference approximation for the
first derivative at the point x = xi = a. Note that the fundamental definition of the derivative of
a function is:
df
f x  h   f x 
 lim
dx h  0
h
Hence, Sf should become more and more accurate as we make the step size h smaller and
smaller. However, one must be careful with round off errors
Other formulae can also be derived with the help of Taylor series expansion. (see fig. below)
f ( xi )  f ( xi 1 )
 df 
Sb   

xi  xi 1
 dx  x  xi
; Backward difference
f ( xi 1 )  f ( xi 1 )
 df 
Sc   

xi 1  xi 1
 dx  x  xi
; Central difference
31
Graphical interpretation of finite difference
Sb
fi
y
fi 1
Sc
Sf
fi 1
y  f (x)
h
xi1
h
xi
xi 1
x
32
Application of Taylor Series
Example:
Using Taylor series, expand the function (x) = ln(1+x) about the base point a = 0. Also,
using Taylor's formula, approximate this function as a first, second, and third degree
polynomial. Compute ln(1.5) from the Taylor series with increasing an number of terms and
make a table of the true and approximate errors as a function of the number of terms used in
the summation. Further, estimate the truncation error.
Solution:
(x)
= ln(1+x)
(1)(x)
= 1/(1+x)
 (2)(x)
= -1/(1+x)2
 (3)(x)
= (2)(1)/(1+x)3
 (4)(x)
= -(3)(2)(1)/(1+x)4
.
.
(n)(x)
= (-1)n+1(n-1)!/(1+x)n
a = 0; (0) = 0,  (1)(0) = 1,  (2)(0) = -1,  (3)(0) = 2,  (4)(0) = -6
Substituting all of this into Eq. 1.6.4 yields:
(x) = ln(1+x) = 0 + x - x2/2 + x3/3 - x4/4 + ... + (-1)n-1 xn/n + ..
To determine for what values of x this series is convergent, apply the ratio test:
x n 1
R  lim n n 1  x
n  x
n
Hence, the series is convergent in the interval -1 < x < 1.
33
Application of Taylor Series (Cont’d)
Polynomial approximation (truncate Taylor series with finite terms)
One term
Two terms
Three terms
ln(1+x)  x
ln(1+x)  x - x2/2
ln(1+x)  x - x2/2 + x3/3
(straight line)
(parabola)
(cubic polynomial)
These approximate functions are plotted in the figure below in comparison with the
original function, (x) = ln(1+x). It is seen that as we take more and more terms, the
approximate function (polynomial) represents the original function better and better in
the neighborhood of the base point x = a = 0.
34
Application of Taylor Series (Cont’d)
Figure: Taylor series approximation to (x) = ln(1+x).
35
Case Study
Case Study C1.7.1 Numerical Evaluation of Derivatives
Using the approximate forward difference formula for the derivative of a
function, calculate numerically the derivatives of the following functions at
the specified points. Make a table for each case showing the variation of the
exact derivative, the numerical derivative, and the absolute true error with
the step size h. Let h vary between 1.0 and 1.E-20 decreasing each time by
a factor of 10 times.
(a.) (x) = Cos(-10x2) at x = 0
(b.) (x) = e-ln(1/x) at x = 1.0
(c.) (x) = x/(5+3x-5) at x = 10
Note: (x0)  [(x0 +h) -(x0)]/h.
36
Case Study (Cont’d)
Let us first determine analytical derivatives:
(a)  '(x) = 20xSin(-10x);  '(0) = 0.
(b)  '(x) = [e-ln(1/x)]/x = 1 ;  '(1) = 1.0
(c)  '(x) = (5 + 18x-5)/(5 + 3x-5)2 ;  '(10) = 0.2000048
Numerical results for parts (a.), (b.), and (c.) are tabulated in Tables C1.7.1a-c, and shown in
Figures C1.7.1a-c. For case (b.), note that (x) = eln(x) = x. Hence, df/dx = 1.0. We obtain
the exact derivative with any value of h that is not very small so that the round off error
becomes significant. When h becomes very small, we are adding a small number h to a
large number that cannot be handled by the computer. For example, for small h:
[(1 + h) - (1)]/h  [(1) - (1)]/h = 0.0
(wrong !)
37
Case Study (Cont’d)
Table: Variation of the True Error with Step Size for Calculating the Derivative of (x) = Cos(-10x2)
Exact Derivative = 0.000000E+00
Numerical Derivative at x = 0.000000E+00
h
df/dx
True Error
1.000000E+00
-1.839072E+00
1.839072E+00
1.000000E-01
-4.995835E-02
4.995835E-02
1.000000E-02
-4.999999E-05
4.999999E-05
9.999999E-04
-4.999999E-08
4.999999E-08
9.999999E-05
-5.000016E-11
5.000016E-11
9.999999E-06
-4.878910E-14
4.878910E-14
9.999999E-07
0.000000E+00
0.000000E+00
9.999999E-08
0.000000E+00
0.000000E+00
9.999999E-09
0.000000E+00
0.000000E+00
9.999999E-10
0.000000E+00
0.000000E+00
9.999999E-11
0.000000E+00
0.000000E+00
9.999999E-12
9.999999E-13
9.999999E-14
9.999999E-15
9.999999E-16
9.999999E-17
9.999999E-18
9.999999E-19
1.000000E-19
1.000000E-20
0.000000E+00
0.000000E+00
0.000000E+00
0.000000E+00
0.000000E+00
0.000000E+00
0.000000E+00
0.000000E+00
0.000000E+00
0.000000E+00
0.000000E+00
0.000000E+00
0.000000E+00
0.000000E+00
0.000000E+00
0.000000E+00
0.000000E+00
0.000000E+00
0.000000E+00
0.000000E+00
38
Case Study (Cont’d)
Figure C1.7.1a Step size h vs. true error
2
for (x) = Cos(-x ).
Figure C1.7.1c Step size h vs. true error for
-5
(x) = x/(5+3x ).
39
•
End of Chapter 1
40
References
•
•
•
•
•
Celik, Ismail, B., “Introductory Numerical Methods for Engineering Applications”,
Ararat Books & Publishing, LCC., Morgantown, 2001
Fausett, Laurene, V. “Numerical Methods, Algorithms and Applications”, Prentice Hall,
2003 by Pearson Education, Inc., Upper Saddle River, NJ 07458
Rao, Singiresu, S., “Applied Numerical Methods for Engineers and Scientists, 2002
Prentice Hall, Upper Saddle River, NJ 07458
Mathews, John, H.; Fink, Kurtis, D., “Numerical Methods Using MATLAB” Fourth
Edition, 2004 Prentice Hall, Upper Saddle River, NJ 07458
Varol, A., “Sayisal Analiz (Numerical Analysis), in Turkish, Course notes, Firat
University, 2001
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