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SSCE 2393 NUMERICAL METHODS

CHAPTER 1

NONLINEAR EQUATIONS

Farhana Johar, Department of Mathematical Sciences, Faculty of Science, UTM.

[email protected]

Chapter 1: Nonlinear Equations

1.1 Introduction

1.2 Intermediate Value Theorem

1.3 Bisection Method

1.4 Fixed-point Iteration Method (simple iterative method)

1.5 Newton-Raphson Method

1.6 Summary

Example 2:

Find the interval [a, b] where the roots for

2

y = x − x − 3 lie.

Step 1: Breaking the original equation into few equations.

Set y = 0.

2

2

2

1 2

Set and 3

x x

x x

y x y x

Step 2: Plot y 1

and y 2

Step 3: Identify the intersection points.

Step 4: Determine the intervals.

∴ x * ∈[2,3] and [-2,-1].

1.2 Intermediate Value Theorem (IVT)

• to verify the existence of the roots.

Theorem:

If

f ( x )∈ C [ a , b ]

and K is any number between

f ( a )

and

f ( b )where f ( a ) f ( b )<0, then there exist a number c in

[a,b] for which f ( c )= K.

Example 3:

Verify that

5 3 2

x − 2 x + 3 x − 1 = 0

has a solution in the interval [0,1].

Solution:

f ( a ) = f ( 0 ) = − 1

f ( b ) = f ( 1 ) = 1

f ( a ) f ( b ) = − 1 < 0

∴ x *exists in[ 0 , 1 ]

f (b)

f (a)

K

a c b

15 2 therefore

1 1 1

a b c

i

1 1

1

1

f a f c

f c f

f a f

Therefore: 1 75 and 15

2 1 2 1

b = c =. a = a =.

Notice that ( )> ε

1

f c ,

so repeattheabovestepfor i = 2 , 3 ,..untilsatisfyingtheconditions.

From the table above, the algorithms stop at i = 6 since | ( ) |

i

f c < ε.

Therefore,

6

x * = c =.

i

i

a

i

b

i

c

i

f a ( )

i

f c

i

f c

i

f a ( )

i

f c

1.4 FIXED-POINT ITERATION METHOD

Ø Formula: ( )

i i

x = g x

• 1

Ø Convergence:

i

g x

(or

| g ( ) | 1 x

Ø Initial guess:

0

x

which is close to the root x*.

Ø Do the iteration up to i - th step until − < ε

i i − 1

x x for a given

value of ε and take

i

x * ≈ x

Example 5

Find a root for

3 2

f ( x )= x − x −

in the interval [1, 2], and take

Solution

Step 1: find g(x)

Set f(x) = 0 to get x = g(x)

3

2

2

3

1

2

3

1

2

3 2

3 2

2 and

−

x

x

g x x g x

x x

x x

x x

3

1

2

1

i + i

x x

Step 2: Iteration

Example 2:

Show that

2

x

x − x + e − =

can be manipulated to form

2

x

x e

x

=. Then, find a root, x * using fixed-point iteration

method. Use

0

x = − 1

Solution:

2

2

2

(shown).

x

x

x

x x e

x x e

x e

x

2

x

x g x

x e

1

2

i i

x

i i

x g x

x e

i

i

x

1

i i

x x

−

∴

x * ≈ x

4

1.5 NEWTON-RAPHSON METHOD

Formula :

1

i

i

i i

f x

f x

x x

Initial guess:

or sothat 0

0 0 0 0

x = a x = b f ( a ) f ( b )<

Do the iteration up to i - th step until

− 1

( ) 0 , ( ) or

i i i i

f x f x x x for a given value of ε

and take

i

x * ≈ x.

Example 1

Find a root for

3 2

f ( x )= x − x −

in the interval [1, 2] and take

. Stop when

i i 1

x x ε

−

Solution

***** you can choose any number between 1 and 2 to start,

x

0

3 2

f ( x )= x − x −

f x 3 x 2 x

2

1 so

0

0

0 1 0

f x

f x

x x x

i

Continue for next i = 1

1.6 Summary of Chapter 1 – Nonlinear Equations

Method Formula Algorithm Stopping Criteria

Bisection

i i

i

a b

c

i i i i i i

f a f c < → a = a b = c

• 1 + 1

If ( ) ( ) 0 ,

If f ( a

i

) f ( c

i

) > 0 → a

i + 1

= c

i

, b

i + 1

= b

i

f ( C

i

) = 0 , f ( C

i

or b

i

− a

i

Fixed point

iteration

i i

x = g x

• 1

− < ε

i i − 1

x x

Newton

Raphson

1

i

i

i i

f x

f x

x x

f ( x

i

) = 0 , f ( x

i

or x

i

− x

i − 1