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B3001/UNIT 10/1
SOLVING NON LINEAR EQUATIONS
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1
Unit 10 SOLVING NON LINEAR EQUATION
General Objective : To solve non linear equations.
Specific Objective : Upon completion of this module, you will be
able to:
1. Solve non linear equationS using fixed point
iteration.
2. solve non linear equationS using Newton-Raphson
method
Objectives
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B3001/UNIT 10/2
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10.0 INTRODUCTION
In this module we consider one of the most basic problems of
numerical
approximation, the root finding problem, It involves finding the
root x of an equation of
the form f(x) = 0, for a given function f. We will solve non
linear equations using the
Fixed Point Iteration and Newton-Raphson method. Sometimes the
equations given to
us are in a form of polynomials.
10.1 FIXED POINT ITERATION
A fixed point for a given function g is the number p for which
g(p) = p. In thus
section we will consider the problem of finding solutions to
fixed-point problems and the
connection between these problems and the root-finding problems
we wish to solve.
Root finding problems and fixed-point problems are equivalent
classes. Given a
root finding problem f(p) = 0 we can define a function g with a
fixed point at p in a
number of ways. i.e.. g(x) = x - f(x) or as g(x) = x + 3
f(x).
Our first task is to become comfortable with this new type of
problem and to
decide when a function has fixed points and how the fixed points
can be determined.
INPUT
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Let have a polynomial f(x). We rewrite in the form of x = p(x)
to find
approximate iteration that focus to a root (solution)
Iteration steps:
1. Find an approximate root. Put this value to the right to find
new root,
x1 = p(xo).
2. This new approximation solution is substitute to the right to
find another
approximate root
x2 = p(x1).
3. This process is repeatedly as to find the most appropriate
root
Xn+1 = p(xn)..( 10.1 )
(10.1) is a fixed-point iteration. If xn+1 xn diminished when n
increases.
The root will converge to = p (),
Example 10.1:
Find the root for 5xex = 1 . Give the root to the nearest
0.0001, using fixed point
iteration method.
Solution:
Write the equation in the form of
)(5
xpe
xx
Let xo = 1, fixed point iteration will list the results as;
x1 = 0.07357
x2 = 0.18581
x3 = 0.16608
x4 = 0.16939
x5 = 0.16883
x6 = 0.16892
After the sixth iteration we found that .0001.01 nn xx
0.16892 0.16883
= 0.00009 < 0.0001
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B3001/UNIT 10/4
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The first three iteration shows that the points are
approximately nearing the real
root
Fig 10.1
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B3001/UNIT 10/5
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ACTIVITY 10.1
10.1 Use fixed point iteration, solve for the equation to the
nearest 10-3
.
a. x sin x = 0
b. x2 + 5x 3 = 0
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B3001/UNIT 10/6
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FEEDBACK 10.1
10.1 a. 0.0
b. 0.541
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B3001/UNIT 10/7
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10.2 NEWTON-RAPHSON METHOD
The Newton-Raphson or simply Newtons method is one of the most
powerful
and well known numerical methods for solving a root-finding
problem f(x) = 0. There are
many ways of introducing Newtons method, we are going to start
with a fixed point
nearest to the root, and we have to draw a tangent from that
fixed point to the function.
That tangent cuts the x-axis at a second fixed point called x.
We have to repeat the
process until it is nearest to the exact root. (Refer to fig
10.2),
Rajah 10.2
Let the first point (x0,y0) to the function y0 = f(x0). The
tangent equation will be
y - f(x0) = f(x0) (x x0 ).
If tangent cuts the x-axis at x1, then
0 - f(x0) = f(x0) (x1 x0 )
INPUT
Refer Fig. 10.2 :
When x = x1 ; y = 0, then
equations can be find as
0 - f(x0) = f(x0) (x x0 )
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B3001/UNIT 10/8
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010
0
)('
)(xx
xf
xf
Then
)('
)(
0
001
xf
xfxx .(10.2)
From fig.7.2, we find that is the real root, then x1 is between
and x0. In this
context x1 is approximately a better root than x0. This process
repeats itself when
x1 substitute x0 to give a root x2, this new root is between and
x1. These
processes will be repeated until too approximate roots having
the same degree of
wanted place value.
Example 10.2:
Solve the following equation
3x3 + 2x 4 = 0
Correct to the third decimal place..
Solution:
Rewrite the equation
y = 3x3 = 4 - 2x
Given the the functions of y = 3x3
and y = 4 2x (Fig. 9.3), we will have to draw
two graphs. The point of intersection.is the root
(solution).
Construct a table, for a few values of f(x).
x 0 1
f(x) -4 1
The solution or the root is between x = 0 and x = 1.
Newton-Raphson
Method
For x = 1,
f(x) = 3x3 + 2x 4
f(1) = 3(1) + 2(1) 4 = 1
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B3001/UNIT 10/9
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Rajah 10.3
We first choose initial approximations x0 through method of
False Position as below:
11
40
41
10
x
= 0.8
This value xo is substituted using Newton-Raphson method
(10.2)
f(x0) = f(0.8)
= 3(0.8)3 + 2(0.8) 4
= -0.864
f(x) = 9x2 + 2
f(x0) = f(0.8)
= 9(0.8)2 + 2
= 7.76
Using false position
formulae :
22
11
12
0
1
yx
yx
yyx
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B3001/UNIT 10/10
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Using (10.1) , we will find
76.7
864.08.01
x
= 0.91134
Successive of approximations are
4748.9
09339.091134.02 x
= 0.90148
and
3140.9
000767.090148.03 x
= 0.90139
When x2 and x3 are rounded to the third decimal place both x2
and x3 is the same
number. When this is true thats mean the root of the equation is
x = 0.901
(correct to third decimal place). In other words, the iteration
stops when
.0001.023 xx
Example 10.3:
Calculate estimation of a positive root correct to the third
decimal place for the
following equation.
Kos x x2 = 0
Solution :
f(x) = kos x x2
x 0.0 0.5 1.0
f(x) 1.0 0.6275 -0.4597
3(0.91134)3 + 2(0.91134) 4
= 0.09339
9(0.91134)2 + 2 = 9.4748
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B3001/UNIT 10/11
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Using False Position method,.
6275.00.1
4597.05.0
4597.0627.0
10
x
0872.1
7735.0
= 0.7114
the root is in between 0.5 dan 1
Using Newton-Raphson iteration,
f(0.7114) = kos (0.7114) 0.71142
= 0.2513
f(x) = -sin x 2x
f(0.7114) = -sin (0.7114) 2(0.7114)
= -2.0757
The next approximation is
0757.2
2513.07114.01
x
= 0.8325
f(0.5903) = -0.0200
f(0.5903) = -2.4046
4046.2
0200.08325.02 x
= 0.8242
f(0.8242) = -0.00016
f(0.8242) = -2.3824
3824.2
00016.08242.03 x
= 0.8241
Therefore, the real root correct to three decimal place is
0.824.
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B3001/UNIT 10/12
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ACTIVITY 10.2
10.1 Use Newton-Raphson method to find solutions accurate to
within 10-3
for
the following problems.
a. 3 kos x 5x + 4 = 0
b. x2 2x 3 + ex = 0
c. sin x x2 + 3 = 0
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B3001/UNIT 10/13
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FEEDBACK 10.2
1 a. 1.0818
b. 1.3544
c. 1.4183
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B3001/UNIT 10/14
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SELF ASSESSMENT
10.1 The equation x3 + 2x
2 5x 1 = 0 has a approximation solution x = 1.4. Use
Newton-Raphson method to find solution accurate to three decimal
place.
10.2 The equation 2x3 7x2 x + 12 = 0 has a approximation
solution x = 1.5. Use
Newton-Raphson method to find solution accurate to three decimal
place
10.3 Use Newton-Raphsons method to approximate, within 10-4, the
value of x that
produces the point on the graph of ex + x 2 = 0
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B3001/UNIT 10/15
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FEEDBACK
10.1
x xn f(xn) f(xn) )('
)(1
n
nn
xf
xfx
0 1.4 -1.336 6.48 1.606
1 1.61 0.3075 9.2163 1.5766
2 1.577 0.01075 8.7688 1.5758
3 1.5758 0.00023 8.7526 1.57577
The solution correct to 10-3
is x = 1.576.
10.2
x xn f(xn) f(xn) )('
)(1
n
nn
xf
xfx
0 1.5 1.5 -8.5 1.676
1 1.676 0.0769 -7.610 1.6861
2 1.6861 0.000307 -7.5478 1.68614
The solution correct to 10-4
is x = 1.686.
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B3001/UNIT 10/16
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10.3
x xn f(xn) f(xn) )('
)(1
n
nn
xf
xfx
0 0.4 -0.108 2.49 0.4434
1 0.443 0.000372 2.557 0.44286
2 0.4429 0.000117 2.5572 0.44285
The solution correct to 10-4
is x = 0.4429.