L.D. College of Engineering

Prepared By: Divya Bhatia

Topics to be Covered• Introduction of Bisection method

• Graphical Representation Of Bisection Method

• Finding Roots of Equations

• Classification of Equations

• Algorithm

• Flowchart

• C program

• Examples

• Introduction of Regula Falsi method

• Finding roots

• False Position formulae

• Algorithm

• C program

• Examples

• Merits Demerits

• Conclusion

Finding Roots of Equations

• In this topic we are examining equations with one independent variable.

• These equations may be linear or non-linear• Non-linear equations may be polynomials or

generally non-linear equations• A root of the equation is simply a value of the

independent variable that satisfies the equation

Classification of Equations

• Linear: independent variable appears to the first power only, either alone or multiplied by a constant

• Nonlinear:– Polynomial: independent variable appears

raised to powers of positive integers only– General non-linear: all other equations

BISECTION METHOD

The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. It is a very simple and robust method, but it is also relatively slow. Because of this, it is often used to obtain a rough approximation to a solution which is then used as a starting point for more rapidly converging methods. The method is also called the interval halving method.

Graphical Representation Of Bisection Method

.

Continue...

The method is applicable for numerically solving the equation f(x) = 0 for the real variable x, where f is a continuous function defined on an interval [a, b] and where f(a) and f(b) have opposite signs. In this case a and b are said to bracket a root since, by the intermediate value theorem, the continuous function f must have at least one root in the interval (a, b).

Algorithm • 'Bisection Method

• 'Start loop

• Do While (abs(right - left) !=0.000)

• 'Calculate midpoint of domain

• midpoint = (right + left) / 2

• 'Find f(midpoint)

• If ((f(left) * f(midpoint)) < 0) Then

• 'Throw away right half

• right = midpoint

• Else If ((f(right) * f(midpoint)) < 0)

• 'Throw away left half

• left = midpoint

• Else

• 'Our midpoint is exactly on the root

• Exit Do

• End If

• Loop

• Return midpoint

Input lower and upper limits low and high

Define tolerance tol

while high-low > 2*tol

mid = (high+low)/2Evaluate function at

lower limit and midpoint:

fl = f(low), fm = f(mid)

fl*fm > 0?

Keep upper half of range:

low = mid

Keep lower half of range:

high = mid

Display root (mid)

YESNO

C Program For Bisection Method

#include<stdio.h>

#include<conio.h>

# define eqn x*x-5

float f(float x)

{

float ans;

ans=eqn;

return(ans);}

Continue…

void main()

{

float a,b,x=0,x1=0;

int i=0;

printf("Enter the a and b\n");

scanf("%f%f",&a,&b);

if(f(a)*f(b)>=0)

{

printf("The interval is wrong\n");

getch();

return ;}

Continue... do

{

x=(a+b)/2;

printf("\ta=%f \tb=%f \tX%d=%f \tf(x%d)=%f\n",a,b,i,x,i,f(x));

if(x==x1)

{

printf("\n\nThe root is %f\n",x);

break;

}

x1=x;

Continue... i++;

if(f(a)*f(x)<0)

b=x;

else

a=x;

}while(f(a)-f(b)!=0.000);

printf("\nThe final ans is %f",x);

getch();

}

Enter the a and b22.5 a=2.000000 b=2.500000 X0=2.250000 f(x0)=0.062500 a=2.000000 b=2.250000 X1=2.125000 f(x1)=-0.484375 a=2.125000 b=2.250000 X2=2.187500 f(x2)=-0.214844 a=2.187500 b=2.250000 X3=2.218750 f(x3)=-0.077148 a=2.218750 b=2.250000 X4=2.234375 f(x4)=-0.007568 a=2.234375 b=2.250000 X5=2.242188 f(x5)=0.027405 a=2.234375 b=2.242188 X6=2.238281 f(x6)=0.009903 a=2.234375 b=2.238281 X7=2.236328 f(x7)=0.001163

Output of the Bisection Method

a=2.235962 b=2.236084 X12=2.236023 f(x8)=-0.000201 a=2.236023 b=2.236084 X13=2.236053 f(x9)=-0.000065 a=2.236053 b=2.236084 X14=2.236069 f(x10)=0.000003 a=2.236053 b=2.236069 X15=2.236061 f(x11)=-0.000031 a=2.236061 b=2.236069 X16=2.236065 f(x12)=-0.000014 a=2.236065 b=2.236069 X17=2.236067 f(x13)=-0.000005 a=2.236067 b=2.236069 X18=2.236068 f(x14)=-0.000001 a=2.236068 b=2.236069 X19=2.236068 f(x15)=0.000001 a=2.236068 b=2.236068 X20=2.236068 f(x16)=0.000000

The final ans is 2.236068

Continue…

Bisection Method Example - Polynomial

• Now consider this example:

• Use the bisection method, with allowed error of 0.0001

Bisection Method Example - Polynomial

• If limits of -10 to 0 are selected, the solution converges to x = -2

Bisection Method Example - Polynomial

• If limits of 0 to 10 are selected, the solution converges to x = 4

Bisection Method Example - Polynomial

• If limits of -10 to 10 are selected, which root is found?

• In this case f(-10) and f(10) are both positive, and f(0) is negative

Bisection Method Example - Polynomial

• Which half of the interval is kept?• Depends on the algorithm used – in our example,

if the function values for the lower limit and midpoint are of opposite signs, we keep the lower half of the interval

Bisection Method Example - Polynomial

• Therefore, we converge to the negative root

Example 2

•  Find the root of cos(x) - x * exp(x) = 0

The graph of this equation is given in the figure. 

Let a = 0 and b = 1   

.Iteration 

No.a b c f(a) * f(c)

1 0 1 0.5 0.053 (+ve)

2 0.5 1 0.75 -0.046 (-ve)

3 0.5 0.75 0.625 -0.357 (-ve)

4 0.5 0.625 0.562 -7.52 * 10-3(-ve)

5 0.5 0.562 0.531 -2.168 * 10-3 (-ve)

6 0.5 0.531 0.516 3.648 * 10-4 (+ve)

7 0.516 0.531 0.524 -9.371 * 10-5(-ve)

8 0.516 0.524 0.520 -3.649 * 10-5(-ve)

9 0.516 0.520 0.518 -3.941 * 10-6(-ve)

10 0.516 0.518 0.517 1.229 * 10-5(+ve)

.

•So one of the roots of cos(x) - x * exp(x) = 0 is approximately 0.517

Example 3

•Find the root of    x - sin(x) - (1/2)= 0

•  The graph of this equation is given in the figure. 

•    Let a = 1 and b = 2 

.Iteration No. a b c f(a) * f(c)

1 1 2 1.5 -8.554*10-4 (-ve)

2 1 1.5 1.25 0.068 (+ve)

3 1.25 1.5 1.375 0.021 (+ve)

4 1.375 1.5 1.437 5.679*10-3 (+ve)

5 1.437 1.5 1.469 1.42*10-3 (+ve)

6 1.469 1.5 1.485 3.042*10-4 (+ve)

7 1.485 1.5 1.493 5.023*10-5 (+ve)

8 1.493 1.5 1.497 2.947*10-6 (+ve)

.

•So one of the roots of x-sin(x)-(1/2) = 0 is approximately 1.497. 

FALSE POSITION METHOD

Finding roots / solving equations● The given quadratic formula provides a quick answer to all

quadratic equations:● Easy

But, not easy

● No exact general solution (formula) exists for equations with exponents greater than 4.

Finding roots…● For this reason, we have to find out the root to

solve the equation.

● However we can say how accurate our solution is as compared to the “exact” solution.

● One of the method is FALSE POSITION.

To refine the bisection method, we can choose a ‘false-position’ instead of the midpoint.The false-position is defined as the x position where a line connecting the two boundary points crosses the axis.

The False-Position Method (Regula-Falsi)

Regula Falsi• For example, if f(xlow) is much closer to zero than

f(xup), it is likely that the root is closer to xlow than to xup.

• False position method is an alternative approach where f(xlow) and f(xup) are joined by a straight line; the intersection of which with the x-axis represents and improved estimate of the root.

• The intersection of this line with the x axis represents an improved estimate of the root.

False Position formulae● Using similar triangles, the intersection of the straight line

with the x axis can be estimated as

● This is the False Position formulae. The value of x then replaces whichever of the two initial guesses, low x or up x , yields a function value with the same sign as f (x) .

Regula Falsi Method Algorithm:

1.Start2.Read values of x0, x1 and *Here x0 and x1 are the two initial guesses3.Computer function values f(x0) and f(x1)4.Check whether the product of f(x0) and f(x1) is negative or not. If it is positive take another initial guesses. If it is negative then goto step 5.5.Determine: x = [x0*f(x1) – x1*f(x0)] / (f(x1) – f(x0))6.Check whether the product of f(x1) and f(x) is negative or not. If it is negative, then assign x0 = x; If it is positive, assign x1 = x;7.Check whether the value of f(x) is greater than 0.00001 or not. If yes, goto step 5. If no, goto step 8.8.Display the root as x.9.Stop

#include<stdio.h>

#include<conio.h>

# define eqn x*x-5

float f(float x)

{

float ans;

ans=eqn;

return(ans);

}

void main()

{

float a,b,x=0,x1=0;

int i=0;

printf("Enter the a and b\n");

scanf("%f%f",&a,&b);

C Program For Regula Falsi

if(f(a)*f(b)>=0)

{

printf("The interval is wrong\n");

getch();

return ;}

do

{

x=a-((b-a)/(f(b)-f(a)))*f(a);

printf("\ta=%f\tb=%f\tX%d=%f\tf(x%d)=%f\n",a,b,i,x,i,f(x));

if(x==x1)

{

printf("\n\nThe root is %f\n",x);

break;

}

Continue…

x1=x;

i++;

if(f(a)*f(x)<0)

b=x;

else

a=x;

}while(f(a)-f(b)!=0.000);

printf("\nThe final ans is %f",x);

getch();

}

Continue…

Enter the a and b22.5 a=2.000000 b=2.500000 X0=2.222222 f(x0)=-0.061728 a=2.222222 b=2.500000 X1=2.235294 f(x1)=-0.003460 a=2.235294 b=2.500000 X2=2.236025 f(x2)=-0.000193 a=2.236025 b=2.500000 X3=2.236066 f(x3)=-0.000010 a=2.236066 b=2.500000 X4=2.236068 f(x4)=-0.000001 a=2.236068 b=2.500000 X5=2.236068 f(x5)=0.000000

The final ans is 2.236068

Output of Regula Falsi Program

EXAMPLE

places. decimal threecorrect to 0.9 and 0.8between lies which02)(equation ofroot real theFind• xxexf

Solution

851.0

)2196.0(2136.0)2196.0(9.02136.08.0

)()()()(

2136.0)9.0()( 2196.0)8.0()(

0.9 and 0.8between liesroot that theknow We

2

01

01102

1

0

x

xfxfxfxxfxx

fxffxf

Solution

900.0 851.0 take weNow

900.0 xand 0.851 xbetween less 0f(x) ofroot

0 )00697.0(2136.0)f(x)f(x

00697.0)851.0()(

10

12

21

2

xandx

The

Sincefxf

Solution

• Therefore, the root is 0.8526,corret to three decimal places.

85260 )00697.0(2136.0

)00697.0(900.02136.0851.0

)()()()(

3

01

01103

.x

xfxfxfxxfxx

EXAMPLE

places. decimalfour correct to 02.1log ofroot real theCompute• 10 xx

Solution

.23136.0)f(x ,59794.0)0(,3 xand 2 xTaking

3. and 2between liesroot A 23136.02.143196.12.13log3)3(

59794.02.160206.02.12log2)2(2.1)1(

2.1log)(

1

10

10

10

10

f

vefvef

vefxxxf

Solution

3. and 2.72102between liesroot The 01709.0)72102.2()f(x

72102.2 )59794.0(23136.0

)59794.0(323136.02

).......(.................... )()(

)()(

2

2

01

01102

vefNow

x

ixfxfxfxxfxx• .

Hence the root is 2.70465 correct to four decimal places.

Solution

etc. 2.74065 x, 2.74064 x

areion approximat successive theprocess, this72102.2

)59794.0(23136.0)59794.0(323136.02 x

get, We

.(i),.......... 0.23136)f(x and 01709.0)f(x ,3 ,72102.2 xTaking

5

4

3

3

10

10

repeatingx

x

ExampleLets look for a solution to the equation x3-2x-3=0. We consider the function f(x)=x3-2x-3On the interval [0,2] the function is negative at 0 and positive at 2. This means that a=0 and b=2 (i.e. f(0)f(2)=(-3)(1)=-3<0, this means we can apply the algorithm).

This is negative and we will make the a =3/2 and b is the same and apply the same thing to the interval [3/2,2].

This is negative and we will make the a =54/29 and b is the same and apply the same thing to the interval [54/29,2].

Merits & Demerits● Merits As the interval becomes small, the

interior point generally becomes much closer to root.

Faster convergence than bisection. Often superior to bisection.

Demerits● Demerits It can’t predict number of

iterations to reach a give precision. It can be less precise than

bisection – no strict precision guarantee.

● Though the difference between Bisection and False Position Method is little but for some cases False Position Method is useful and for some problems Bisection method is effective….

● In fact they both are necessary to solve any equation by ‘Bracketing method’.