Numerical Analysis (Solution of Non-Linear Equations)

Solution of
Non-Linear
Equations
Chapter 2

Introduction
Bisection Method
Regula-Falsi Method
Method of iteration
Newton - Raphson Method
Muller’s Method
Graeffe’s Root Squaring
Method

ExampleExample
SolveSolve xx33
– 9– 9xx + 1 = 0+ 1 = 0
for the root betweenfor the root between
xx = 2 and= 2 and xx = 4 by the= 4 by the
bisection method.bisection method.

Solution GivenSolution Given ff ((xx) =) = xx33
– 9– 9xx + 1.+ 1.
HereHere ff (2) = -9,(2) = -9, ff (4) = 29.(4) = 29.
Therefore,Therefore, ff (2)(2) ff (4) < 0 and hence(4) < 0 and hence
the root lies between 2 and 4.the root lies between 2 and 4.
LetLet xx00 = 2,= 2, xx11 = 4. Now, we define= 4. Now, we define
as a first approximation to a root ofas a first approximation to a root of
ff ((xx) = 0 and note that) = 0 and note that ff (3) = 1, so(3) = 1, so
thatthat ff (2)(2) ff (3) < 0.(3) < 0. Thus the root liesThus the root lies
between 2 and 3between 2 and 3
0 1
2
2 4
3
2 2
x x
x
+ +
= = =

We further define,We further define,
and note thatand note that ff ((xx33) =) = ff (2.5) < 0, so that(2.5) < 0, so that
ff (2.5)(2.5) ff (3) < 0. Therefore, we define the(3) < 0. Therefore, we define the
mid-point,mid-point,
Similarly,Similarly, xx55 = 2 . 875 and= 2 . 875 and xx66 = 2.9375= 2.9375
and the process can be continued untiland the process can be continued until
the root is obtained to the desiredthe root is obtained to the desired
accuracy.accuracy.
0 2
3
2 3
2.5
2 2
x x
x
+ +
= = =
3 2
4
2.5 3
2.75, etc.
2 2
x x
x
+ +
= = =

These results are presented in the table.These results are presented in the table.
nn xxnn f ( xf ( xnn ))
22 33 1.01.0
33 2.52.5 -5.875-5.875
44 2.752.75 -2.9531-2.9531
55 2.8752.875 -1.1113-1.1113
66 2.93752.9375 -0.0901-0.0901

Regula-Falsi
Method
Method of falseMethod of false
positionposition
Method of falseMethod of false
positionposition

Here, we choose two points xn
and xn -1 such that f (xn) and f
(xn-1) are of opposite signs.
Intermediate value property
suggests that the graph of
y = f (x) crosses the x-axis
between these two points and
therefore, a root say lies
between these two points.

Thus, to find a real root of
f (x) = 0 using Regula-Falsi
method, we replace the part of
the curve between the points
A[xn, f(xn)] and B[xn-1, f (xn-1)] by
a chord in that interval and we
take the point of intersection of
this chord with the x-axis as a
first approximation to the root.

Now, the equation of the chord
joining the points A and B is
(2.1)
Setting y = 0 in Eq. (2.1), we get
1 1
( )
( ) ( )
n n
n nn n
y f x x x
f x f x x x− −
− −
=
− −
1
1
( )
( ) ( )
n n
n n
n n
x x
x x f x
f x f x
−
−
−
= −
−

Hence, the first approximation to the root
of f (x) = 0 is given by
(2.2)
we observe that f (xn-1) and f (xn+1) are of
opposite sign. Thus, it is possible to
apply to above procedure, to determine
the line through B and A1 and so on.
Hence, the successive approximations to
the root of f (x) = 0 is given by Eq. (2.2).
1
1
1
( )
( ) ( )
n n
n n n
n n
x x
x x f x
f x f x
−
+
−
−
= −
−

Example
Use the Regula-Falsi method to
compute a real root of the equation
x3
– 9x + 1 = 0,
(i) if the root lies between 2 and 4
(ii) if the root lies between 2 and 3.
Comment on the results.

Solution
Let f (x) = x3
– 9x + 1.
f (2) = – 9 and f (4) = 29.
Since f (2) and f (4) are of
opposite signs, the root of
f (x) = 0 lies between 2 and 4.

Taking x1 = 2, x2 = 4 and
using Regula-Falsi method,
the first approximation is
given by
2 1
3 2 2
2 1
( )
( ) ( )
2 29
4 2.47368
38
x x
x x f x
f x f x
−
= −
−
×
= − =

Now f (x3) = –6.12644.
Since f (x2) and f (x3) are of
opposite signs, the root lies
between x2 and x3.
The second approximation to the
root is given as
3 2
4 3 3
3 2
( )
( ) ( )
2.73989
x x
x x f x
f x f x
−
= −
−
=

Therefore f (x4) = – 3. 090707.
Now, since f (x2) and f (x4) are
of opposite signs, the third
approximation is obtained
from
and f (x5) = – 1.32686.
4 2
5 4 4
4 2
( ) 2.86125
( ) ( )
x x
x x f x
f x f x
−
= − =
−

This procedure can be continued till
we get the desired result. The first
three iterations are shown as in the
table.
n xn+1 f (xn+1)
2 2.47368 -6.12644
3 2.73989 -3.090707
4 2.86125 -1.32686

(ii) f (2) = – 9 and f (3) = 1. Since
f (2) and f (3) are of opposite signs,
the root of f (x) = 0 lies between 2
and 3. Taking x1 = 2, x2 = 3 and
using Regula-Falsi method, the
first approximation is given by
3
2 1
3 2 2
2 1
( )
( ) ( )
1
3 2.9
10
( ) 0.711
x x
x x f x
f x f x
f x
−
= −
−
= − =
=−

Since f (x2) and f (x3) are of
between x2 and x3.
The second approximation
to the root is given as
3 2
4 3 3
3 2
4
( ) 2.94156
( ) ( )
( ) 0.0207
x x
x x f x
f x f x
f x
−
= − =
−
= −

Now, we observe that f (x2)
and f (x4) are of opposite
signs, the third approximation
is obtained from
4 2
5 4 4
4 2
( )
( ) ( )
2.94275
( ) 0.0011896
x x
x x f x
f x f x
f x
−
= −
−
=
=−

This procedure can be
continued till we get the desired
result. The first three iterations
are shown as in the table.
n xn+1 f (xn+1)
2 2.9 -0.711
3 2.94156 -0.0207
4 2.94275 -0.0011896

We observe that the value of the
root as a third approximation is
evidently different in both the
cases, while the value of x5, when
the interval considered is (2,
3), is closer to the root.
Important observation: The initial
interval (x1, x2) in which the root of
the equation lies should be
sufficiently small.

Example
Use Regula-Falsi method to
find a real root of the equation
log x – cos x = 0
accurate to four decimal places
after three successive
approximations.

Solution
Given f (x) = log x – cos x.
We observe that
f (1) = 0-0.5403,and
f (2)=0.69315+0.41615=1.1093
Since f (1) and f (2) are of
between x1 = 1, x2 = 2.

The first approximation is obtained
from
2 1
3 2 2
2 1
3
( )
( ) ( )
1.1093
2 1.3275
1.6496
( ) 02833 0.2409 0.0424
x x
x x f x
f x f x
f x
−
= −
−
= − =
= − =

Now, since f (x1) and f (x3) are
of opposite signs, the second
approximation is obtained as
4
3
4
(.3275)(.0424)
1.3275
0.0424 0.5403
1.3037
( ) 1.24816 10
x
f x −
= −
+
=
= ×

Similarly, we observe that f (x1)
and f (x4) are of opposite signs,
so, the third approximation is
given by
The required real root is 1.3030.
5
4
5
(0.3037)(0.001248)
1.3037
0.001248 0.5403
1.3030
( ) 0.6245 10
x
f x −
= −
+
=
= ×

Numerical Analysis (Solution of Non-Linear Equations)

More Related Content

What's hot

Similar to Numerical Analysis (Solution of Non-Linear Equations)

More from Asad Ali

Recently uploaded

Numerical Analysis (Solution of Non-Linear Equations)