Download Numerical Method

LECTURE 2 OF 4
7.0 NUMERICAL METHODS
7.2
Solutions of Non-Linear
Equations
LEARNING OUTCOMES:
At the end of the lesson, students should be
able to:
a) To discuss the iteration method by
writing f(x)=0 in the form of x=g(x). The
iteration scheme is n=1,2,3….
xn1  g( xn )
b) Identify that the iteration method fails
'
g
when ( x1 )  1 in the neighborhood of
the roots of the equation
Example:
(1) Solve x2 – 4x – 5 = 0
=> ( x + 1 )( x – 5) = 0
x = -1 or x = 5
The roots of
the equation
(2) Solve x3 – ex = 0
We cannot find the exact root of the
above equation
Numerical Method ( ITERATION METHOD)
can be used to find approximate roots /
solutions.
7.2 (1) : ITERATION METHOD
The steps are :
1. Find an initial approximate value x1 .
2. Rewrite the equation in the form : x = g(x),
g(x) is called the iteration function
3. Find g’(x)
4. This method fails if
So we are looking for
g ( x1 )  1
'
g ( x1 )  1
'
How to find an initial approximate value ?
2 Methods
Graphical
Algebraic
Graphical Method
Sketch the graph y = f(x) .The real root
the point where the graph intercept at x-axis
OR
Rewrite f(x) = 0 to a new form : F(x) = G(x)
Sketch the graph y = F(x) and y = G(x). The
real root
the point of intersection
between the two graphs.
Algebraic Method
•Find two values a and b such that f(a) and
f(b) have different signs
•At least one root must lie between a and b if
f(x) is continuous.
Example
Locate the approximate value of the equation
ln x  x  4  0 by using the graphical
method.
Solution
y
The intersection is
approximately at x = 2.9
3 4
y = ln x
x
y=-x+4
Example 1
Show that the equation x3 – 6x + 7 = 0 has a
root between -3 and -2. Find the root of the
equation correct to 3 d.p. using the Iteration
Method.
Solution
Let f(x) = x3 – 6x + 7
f(-3) = (-3)3 -6(-3) + 7 = -2 < 0
f(-2) = (-2)3 – 6(-2) + 7 = 11 > 0



different
sign
Since f(-3) < 0 and f(-2) > 0 , thus one of
the the roots is between -3 and -2
Let x1 = -2.5
the initial approximate value
Determine g(x) by writing f(x) = 0 to a
new form : x = g(x)
x  6x  7  0
3
6x  x  7
3
x 7
x
6
3
x 7
g( x ) 
6
3
x  6x  7
3
x  6x  7
3
g( x ) 
3
6x  7
g( x ) 
x1  2.5
CHECK
x 7
g( x ) 
6
2
1
x
'
2
g ( x)  3 x 
6
2
x1  2.5
3


( 2.5)
g ( x) 
2
2
'
 3.125  1
x2 
3
3
6x  7
6( 2.5)  7
Therefore, g(x) is
2.80204
an
iteration
x function
 3 6( 2.80204)  7
3
 2.87696
 an
2.89495
g(x)xis4 not
iteration
x5function
 2.89923
x6  2.90025
x7  2.90049
 The required
solution is x= - 2.900
USING THE CALCULATOR
To find the approximate root for x3 – 6x + 7 = 0
x1 = -2.5 ,
iteration function : g ( x ) 
3
6x  7
Key in
MODE 1 (COMP)
ALPHA Y ALPHA = 3 (6 ALPHA X – 7)
CALC (X?)
-2.5 = (-2.80204 ..)
CALC Ans = ( -2.87696…)
CALC Ans = ( -2.89495…)
CONTINUE THE PROCESS UNTILL YOU GET
THE SAME ANSWER FOR AT LEAST TWICE
Example 2
1
Show that the equation x   4  0
x
Has a root between 0.2 and 0.3. Taking 0.2
2
as the first approximation, find the root of
the equation correct to 3d.p .
Solution
At least one root
1
must lie between
f ( x)  x   4
0.2 and 0.3
x
1
2
f (0.2)  (0.2) 
 4  0.96   
0.2
1
2
f (0.3)  (0.3) 
 4  0.756   
0.3
2
1
1
2
x  40  x 4
x
xTherefore, g(x) is
1
an iteration
x
function
2
x 4
1
1
2
g( x )  2
 x 4
x 4
2 x
'
g ( x) 
2
2
( x  4)
2


2(0.2)
 0.0245  1
g (0.2) 
2
2
((0.2)  4)
'

x  g( x )  x  4
2

1
x1  0.2
1
x2 
 0.24752
2
(0.2)  4
1
x3 
 0.24623
2
(0.24752)  4
1
x4 
 0.24627
2
(0.24623)  4
 The required
solution is 0.246
Example 3
Using the iteration method, find the
x
solution of f ( x )  x  e near x1  0.5
correct to 3 d.p.
Solution:
x
x  e
x=g(x)
g( x )   e x
g ( x )  e
'
x
Given x1 = -0.5,
0.5
g ( 0.5)  e
 0.6065  1
'
x  g( x )   e
0.56844
x1  0.5
x9   e
 0.56641
0.56641
x2   e 0.5  0.60653
x


e


0.56756
10
x3   e 0.60653  0.54524
0.56756
x


e
 0.56691
0.54524
11
x4   e
 0.57970
0.56691
x12   e
 0.56728
0.57970
x5   e
 0.56006
0.56006
x6   e
 0.57117
x7   e 0.57117  0.56486
0.56486
x8   e
 0.56844
x

The required solution is - 0.567