Dr.Shoeb_ME212_Lec-3_numerical methods_G

Introduction to
Numerical
Methods
Numerical Methods (ME 212)
Author: Dr. Shoeb Ahmed Syed
Papua New Guinea University of Technology
Department of Mechanical Engineering
1

How a Decimal Number is
Represented
2102
+ 5101
7100
+ 710−1
+ 610−2
= +
257.76
3

Base 2
23
22
0
21
+120
)
 (1 
+ 0 +1
 
=
(1011.0011)2
 + +1 2−4
)
2−1
2−2
2−3
(0 + +1
 10
=11.1875
4

Convert Base 10 Integer to
binary representation
Table 1 Converting a base-10 integer to binary representation.
Hence
(11)10 = (a3a2 a1a0 )2
= (1011)2
5
Quotient Remainder
11/2 5 1 = a0
5/2 2 1 = a1
2/2 1 0 = a2
1/2 0 1 = a3

Start
Input (N)10
Is Q = 0?
Yes
STOP
6
n = i
(N)10 = (an. . .a0)2
No
ai = R
i=i+1,N=Q
Divide N by 2 to get
quotient Q & remainder R
i = 0
Integer N to be
converted to binary
format

Fractional Decimal Number
to Binary
Table 2. Converting a base-10 fraction to binary representation.
decimal
Hence
(0.1875)10 = (a−1a−2a−3a−4 )2
= (0.0011)2
7
Number Number after
decimal
Number before
0.1875 2 0.375 0.375 0 = a−1
0.3752 0.75 0.75 0 = a−2
0.752 1.5 0.5 1 = a−3
0.52 1.0 0.0 1 = a−4

Start
Input (F)10
i = i −1, F = T
Is T =0?
Yes
STOP
8
n = i
(F)10 = (a-1. . .a-n)2
No
ai = R
Multiply F by 2 to get
number before decimal,
S and after decimal, T
i = −1
Fraction F to be
converted to binary
format

Decimal
(11.1875)
Number to Binary
( )
= ?.?
10 2
Since
(11)10
= (1011)2
and
(0.1875)10 =
we have
(11.1875)10
(0.0011)2
= (1011.0011)2
9

All Fractional Decimal Numbers
Cannot be Represented Exactly
Table 3. Converting a base-10 fraction to approximate binary representation.
before
(0.3)10  = (0.01001)2 =
(a−1a−2a−3a−4a−5 )2 0.28125
10
Number
Number
after
decimal
Number
Decimal
0.3 2 0.6 0.6 0 = a−1
0.62 1.2 0.2 1 = a−2
0.2 2 0.4 0.4 0 = a−3
0.4 2 0.8 0.8 0 = a−4
0.82 1.6 0.6 1 = a−5

Another Way to Look at
Conversion
(11.1875)10
Convert
(11)
to base 2
23
23
23
=
=
=
+ 3
10
21
21
+
+
+1
20
+
23
22
21
20
=1 + 0 +1 +1
=
(1011)2
11

(0.1875) 2−3
2−3
=
=
+ 0.0625
10
2−4
+
2−1
2−2
2−3
2−4
= 0 + 0 +1 +1
= (.0011)2
(11.1875)10 = (1011.0011)2
12

Floating Decimal Point : Scientific Form
2.5678×102
written as +
256.78 is
is written as +3.678×10−3
is written as − 2.5678×102
0.003678
− 256.78

Example
The form is
sign × mantissa ×10exponent
or
σ ×m×10e
Example: For
− 2.5678×102
σ = −1
m = 2.5678
e = 2

Floating Point Format for Binary
Numbers
2e
y =σ ×m×
(0 - ve)
σ = sign of number for + ve,1for
m = mantissa [(1)2
m < (10)2 ]
<
1 is not stored as it is always
e = integer exponent
given to be 1.

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) = (110110.11) = (1.1011011)× 25
10 2
≅ (1.1011)2 × (101)2
2
We have the representation as
mantissa exponent
Sign of the
number
Sign of the
exponent
0 0 1 0 1 1 1 0 1

Machine Epsilon
Defined as the measure of accuracy and found
by difference between 1 and the next number
that can be represented

Dr.Shoeb_ME212_Lec-3_numerical methods_G

IEEE 754 Standards for Single
Precision Representation

IEEE-754 Floating Point
Standard
Standardizes representation
floating point numbers on
different computers in single
double precision.
• of
and
• Standardizes representation
floating point operations on
different computers.
of

One Great Reference
What every computer scientist (and even if
you are not) should know about floating point
arithmetic!
http://www.validlab.com/goldberg/paper
.pdf

IEEE-754 Format Single
Precision
32 bits for single precision
Biased
Exponent (e’)
Sign
(s)
Mantissa (m)
1.m) ×
Value = (−1) ×( e'−127
s
2
2
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Exponent for 32 Bit IEEE-754
8 bits would represent
e′
0 ≤ ≤ 255
Bias is 127; so subtract
representation
127 from
−127 ≤ e ≤128

Exponent for Special Cases
e′
Actual range of
e′
1≤ ≤ 254
e′ and e′
= 0 = 255 are reserved for special numbers
e
Actual range of
−126 ≤ e ≤ 127

Special Exponents and Numbers
e′ = 0 all zeros
all ones
e′ = 255
s e′ m Represents
0 all zeros all zeros 0
1 all zeros all zeros -0
0 all ones all zeros ∞
1 all ones all zeros − ∞
0 or 1 all ones non-zero NaN

IEEE-754 Format
The largest number by magnitude
( ) ×
2
127 38
= 3.40×10
1.1........1 2
The smallest number by magnitude
( ) −126 −38
×2 = 2.18×10
1.00......0 2
Machine epsilon
2−23
= 1.19 ×10−7
ε =
mach

What
Some examples
seen
x2
is a Taylor series?
of T
aylor series which you must have
x4
x6
cos(x) =1− + − +
2!
x3
4!
x5
6!
x7
sin(x) = x − + − +
3!
x2
5!
x3
7!
x
e = 1+ x + + +
2! 3!

General Taylor Series
The general
f (x + h) =
form of the T
aylor series is given by
f (x)h +
f  (x) h2
+
f (x) h3
f (x)+ +
2! 3!
provided that all derivatives of f(x) are continuous
exist in the interval [x,x+h]
and
What does this mean in plain English?
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
function at
at that single point, and I can give you the value of the
any other point” (fine print excluded)

Example—Taylor Series
f (6) f (4)=125, f (4)= 74,
derivatives
Find the
f  (4)= 30,
value of
f (4)= 6
given that
and all other higher order
f (x)
of at are zero.
x = 4
Solution:
f (x + h)=
x =
h2
h3
f (x)+ f (x)h +
4
f  (x) f (x)
+ +
2! 3!
h = 6 − 4 = 2

1. http://numericalmethods.eng.usf.edu
2. http://numericalmethods.eng.usf.edu/topics/binary_repr
esentation.html
3.http://numericalmethods.eng.usf.edu/topics/floatingpoint_re
presentation.html
4. http://numericalmethods.eng.usf.edu/topics/taylor_seri
es.html
References

Dr.Shoeb_ME212_Lec-3_numerical methods_G

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