Lecture 4

Numerical Analysis

Solution of Non-Linear Equations

Chapter 2

IntroductionBisection MethodRegula-Falsi MethodMethod of iterationNewton - Raphson MethodMuller’s MethodGraeffe’s Root Squaring Method

Bisection Method

(Bolzano)

Example Example

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

Solution Given Solution Given f f ((xx) = ) = xx33 – 9 – 9xx + 1. + 1. Here Here 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. Let Let xx00 = 2, = 2, xx11 = 4. Now, we define = 4. Now, we define

as a first approximation to a root of as a first approximation to a root of ff ( (xx) = 0 and note that ) = 0 and note that ff (3) = 1, so (3) = 1, so that that ff (2) (2) ff (3) < 0. (3) < 0. Thus the root lies Thus the root lies between 2 and 3between 2 and 3

0 12

2 43

2 2

x xx

+ += = =

We further define,We further define,

and note that and 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 until and the process can be continued until the root is obtained to the desired the root is obtained to the desired accuracy. accuracy.

0 23

2 32.5

2 2

x xx

+ += = =

3 24

2.5 32.75, etc.

2 2

x xx

+ += = =

These results are presented in the table.These results are presented in the table.

nn xxnn f ( xf ( xn n ))

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 false Method of false positionposition

Method of false Method 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 xf x f x x x− −

− −=− −

1

1

( )( ) ( )

n nn n

n n

x xx 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).

11

1

( )( ) ( )

n nn n n

n n

x xx 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 13 2 2

2 1

( )( ) ( )

2 294 2.47368

38

x xx 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 24 3 3

3 2

( )( ) ( )

2.73989

x xx 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 25 4 4

4 2

( ) 2.86125( ) ( )

x xx 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 13 2 2

2 1

( )( ) ( )

13 2.9

10( ) 0.711

x xx x f x

f x f x

f x

−= −−

= − =

=−

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 24 3 3

3 2

4

( ) 2.94156( ) ( )

( ) 0.0207

x xx 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 25 4 4

4 2

( )( ) ( )

2.94275

( ) 0.0011896

x xx 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.

ExampleUse 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 opposite signs, the root lies between x1 = 1, x2 = 2.

The first approximation is obtained from

2 13 2 2

2 1

3

( )( ) ( )

1.10932 1.3275

1.6496( ) 02833 0.2409 0.0424

x xx 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

34

(.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

45

(0.3037)(0.001248)1.3037

0.001248 0.54031.3030

( ) 0.6245 10

x

f x −

= −+

== ×

Lecture 4

Numerical Analysis