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Iteration Method(theory):

Let f(x)=0 -------------(1)

Be the given equation can be written as

x= (x) …………..(2) ,where |'(x) |<

consider the first approximation root is 𝑥

1

=a

and the second approximation root 𝑥

2

putting x= 𝑥

1

in right hand side of

equation (2),we get

2

1

Similarly, 𝑥

3

2

2

𝑛− 1

Iteratiom method(problem):

Find the root of the equation 𝑥

3

• 𝑥 − 1 = 0 by using the iteration method.

Solution:

Let f(x)= 𝑥

3

Here, f(0)= - 1<

f(1)= 1>

so the root lies between 0 and 1

then the first approximation

0

0 + 1

2

Now from equation (1)

Iteration method(program):

clc

all clear

g=@(x) x.^3+x-1;

f=@(x) 1/(x.^2+1);

df=@(x) - 2*x/(x.^2+1)^2;

x=input('Enter the initial value: ');

t=input('Tolerence:');

n=input('Enter the iteration number:');

while(df(x)>1)

fprintf('There is no root');

end

for i=1:n

x1=f(x);

fprintf('\n x%d= %.4f\n',i,x1);

if abs(x1-x)<t

break;

end

x=x1;

end

fprintf('\n\n The root is %f',x1);

x=-20:0.01:20;

plot(x,g(x),'b','linewidth',2)

hold on

plot(x1,f(x1),'or')

hold on

plot([-5,5],[0,0])

hold on

plot([0,0],[-5,5])

legend('f(x)=x.^3+x-1','root');

xlabel('x-axis','fontsize',14);

ylabel('y-axis','fontsize',14);

grid on

hold off

output:

ans =

logical

The curve AB cut the x-axis at Q point.The approximation root equation (1) is

OQ.

Let A[a,f(a)] B[b,f(b)] be the extremities of the chord AB.Now the equation of

the chord AB.

y-f(a)=

𝑓(𝑏)−𝑓(𝑎)

( 𝑏−𝑎

)

To find OQ put y=

0 - f(a)=

𝑓(𝑏)−𝑓(𝑎)

( 𝑏−𝑎

)

(x-a)= -

𝑓(𝑎)(𝑏−𝑎)

𝑓(𝑏)−𝑓(𝑎)

(x-a)=

𝑓(𝑎)(𝑎−𝑏)

𝑓(𝑏)−𝑓(𝑎)

x=a+

𝑓(𝑎)(𝑎−𝑏)

𝑓(𝑏)−𝑓(𝑎)

x=

𝑎𝑓(𝑏)−𝑏𝑓(𝑎)

𝑓

( 𝑏

) −𝑓(𝑎)

This is the required equation.

False position method(math):

Find a root of the equation 𝑥

3

− 3 𝑥 − 5 = 0 by the method of false position.

Solution:

Let f(x)=𝑥

3

Since f(2)= - 3<

f(3)= 13>

The root lies between 2 and 3.

0

2 × 13 − 3 × (− 3 )

Now f (𝑥

0

) = f (2.18756) = - 1.095<

False position method(program):

clc

clear all

f=@(x)x.^3-3*x-5;

x1=input('Enter the upper interval a=');

x2=input('Enter the lower interval b=');

t=input('Tollerence T=');

fx1=f(x1);

fx2=f(x2);

if (fx1*fx2>0)

disp('There are no roots');

elseif(fx1*fx2<0)

disp('There is a root');

for i=1:

m=((x1fx2)-(x2fx1))/(fx2-fx1);

fm=f(m);

fprintf('\n %f %f %f \n',i,m,fm);

if abs(fm)<t

break

end

if(fx1*fm<0)

x2=m;

else

x1=m;

end

end

end

fprintf('\n\n The root is %f',x1)

hold on

grid on

plot(x,f(x),'m','linewidth',5)

plot(m,f(m),'o')

plot([-5,5],[0,0])

plot([0,0],[-5,5])

legend('f(x)=x.^3-3*x-5;');

xlabel('x-axis','fontsize',14);

ylabel('y-axis','fontsize',14);

hold off

output:

• Enter the initial value:
• Tolerence:0.
• Enter the iteration number:
  • x1= 1.
  • x2= 0.
  • x3= 0.
  • x4= 0.
  • x5= 0.
  • x6= 0.
  • x7= 0.
  • x8= 0.
  • x9= 0.
  • x10= 0.
  • x11= 0.
  • x12= 0.
  • x13= 0.
  • x14= 0.
  • x15= 0.
  • x16= 0.
  • x17= 0.
• Enter the upper interval a=
• Enter the lower interval b=
• Tollerence T=0.
  • 1.000000 2.187500 - 1. There is a root
  • 2.000000 2.339844 0.
  • 3.000000 2.216064 - 0.
  • 4.000000 2.239273 - 0.
  • 5.000000 2.258130 - 0.
  • 6.000000 2.273451 - 0.
  • 7.000000 2.285900 0.
  • 8.000000 2.275785 - 0.
  • 9.000000 2.277682 - 0.
  • 10.000000 2.279223 0.
  • 11.000000 2.277971 - 0.
  • 12.000000 2.278 206 - 0.
  • 13.000000 2.278396 - 0.
  • 1 4.000000 2.278551 - 0.
  • 15.000000 2.278677 - 0.
  • 16.000000 2.278780 - 0.
  • 17.000000 2.278863 - 0.
  • 18.000000 2.278930 - 0.
  • 19.000000 2.278985 - 0.

0

(𝑎+𝑏)

2

In this case two possibilities will b

(1).f (𝑥

0

(2). f (𝑥

0

Now if (1) occure then 𝑥

0

is the root of the equation f(x)=0 and the

process will be terminated.

If (2) occure the root either lies between the 𝑥

0

and a or 𝑥

0

and b

Accordingly f(𝑥

0

) .f(a)<0 or 𝑓(𝑥

0

). 𝑓(b)<0 .So we have the second

approxition ,

1

(𝑥

0

+𝑎)

2

or 𝑥

1

( 𝑥

0

+𝑏

)

2

similar process of this method

Bisection method(math):

Find the root of the equation 𝑥

3

− 𝑥 − 1 = 0 by using bisection method.

Solution:

Let f(x)= 𝑥

3

f(1)=1- 1 - 1=-1<

and

f(2)=8- 2 - 1=5>

f(𝑥

6

)= f(1. 3203 ) =

3

The root lies between 1.3203 and 1. 3281

7

f(𝑥

7

)= f(1. 3242 ) = ( 1. 3242 )

3

The root lies between 1.3242 and 1. 3281

8

f(𝑥

8

)= f(1. 3261 ) = ( 1. 3261 )

3

The required root lies between 1.3242 and 1.3261.

Bisection method(program):

clc

clear all

f=@(x)x.^3-x-1;

x1=input('Enter the upper interval a=');

x2=input('Enter the lower interval b=');

t=input('Tollerence T=');

fx1=f(x1);

fx2=f(x2);

if (fx1*fx2>0)

disp('There are no roots');

else

disp('There is a root');

for i=1:

m=(x1+x2)/2;

fm=f(m);

fprintf('\n %f %f %f\n',i,m,fm);

if abs(fm)<t

break

end

if(fx1*fm<0)

x2=m;

else

x1=m;

end

end

end

fprintf('\n\n The root is %f',x1)

x1=-20:0.01:20;

hold on

grid on

plot(x1,f(x1),'m','linewidth',5)

plot(m,f(m),'o')

plot([-5,5],[0,0])

plot([0,0],[-5,5])

legend('f(x)=3x.^3+8x.^2+8*x+5;');

xlabel('x-axis','fontsize',14);

ylabel('y-axis','fontsize',14);

hold off

output:

Enter the upper interval a=

Enter the lower interval b=

Tollerence T=0.

There is a root

f(a) + hf'(a)=

hf'(a) = - f(a)

h= -

𝑓(𝑎)

𝑓′(𝑎)

So, the first approximation

1

𝑓(𝑎)

𝑓

′

( 𝑎

)

2

1

1

𝑓(𝑎

1

)

𝑓′(𝑎

1

)

𝑛+ 1

𝑛

𝑛

𝑓(𝑎

𝑛

)

𝑓′(𝑎

𝑛

)

By repeating this method we get the close approximation deseired root.

Newton-Raphson method(problem):

Solve for a real root of the eqution 𝑥

4

2

− 80 = 0 by using Newton -

Rophson method.

Solution:

Given equation

4

2

Let f(x)= 𝑥

4

2

f'(x)= 4𝑥

3

Here f(2) = - 60 <

f(3) = 10 >

The root lies between 2 and 3

The initial approximation value of the root ,

0

2 + 3

2

Now ,we get from Newton-Rophson method

1

0

𝑓(𝑥

0

)

𝑓′(𝑥

0

)

2

1

𝑓(𝑥

1

)

𝑓′(𝑥

1

)

3

2

𝑓(𝑥

2

)

𝑓′(𝑥

2

)

4

3

𝑓(𝑥

3

)

𝑓′(𝑥

3

)

5

4

𝑓(𝑥

4

)

𝑓′(𝑥

4

)

The required root is 2.

Newton-Rophson method(program)

clc

clear all

f=@(x)x.^4+x.^2-80;

df=@(x)4x.^3-2x;

x=input('Intial value=');

t=input('tollerence=');

n=input('Number of iteration=');

if df(x)==

fprintf('there is no root');

else

fprintf('there is root');

for i=1:n

x1=x-f(x)/df(x);

fprintf('\n x%d = %.4f\n',i,x1);

if (df(x1)==0)

fprintf('there is no root');

end

if(abs(x-x1)<t)