BiSection Method In Numerical Techniques

What is Bisection Method?

In mathematics, the bisection method in numerical techniques is used to find the actual root of a function. That applies to any function for finding the root.

To find the root we have to know about two values and their sign which is (positive)+ or -(negative).

The method repeatedly bisects the interval defined by these values and then selects the point at which the function changes the sign. It is a very simple and easy method but relatively slow. Because of this, it is used to obtain an approximation to a solution.

The bisection method is also called bracketing because it brackets the root within the interval.

The Method Of Bisection Rules

• The Bisection method is used 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].
• At each step, the method divides the interval point into two parts by computing the midpoint c = (a+b)/2 of the interval.

• After that, we put the c value at the function f(x).
• if the f(c) value is a positive value then the (b) value is replaced by the (c) value and if the f(c) value is negative then the (a) value is replaced by (c).

• if the f(c) value is a positive value then the f(b) value is replaced by the f(c) value and if the f(c) value is negative then the f(a) value is replaced by f(c).

Bisection Method Algorithm

• There are steps to find the root of a continuous function.
• For any continuous function f(x),
• Find two points a and b as an approximation root by applying values to the function.
• Find the midpoint c=a+b/2. Midpoint says c.
• C is the root of the given function if f(c) = 0;
•  else follow the next step.
• Divide the interval [a, b] – If f(c)*f(a) <0,
• there exist a root between c and a– else if f(c) *f (b) < 0, there exist a root between c and b
• Repeat above three steps until f(c) = 0.

Example:

Finding the root of a polynomial, the approximate root of f(x) = x³ + 5x + 1=0 , correct upto 3 decimal place.

Let the function is continous

F(x) = x³ + 5x + 1

Let put x=0 in f(x)

Then f(x)=1

Let put x=1 in f(x)

Then f(x)=7

Let put x=-1 in f(x)

Then f(x)=-5

Then us asign a value as (-1) and b value as (o)

It is the basic root

Mind the mid point c= a+b/c

                            C = -0.5

The find F(c)= x³ + 5x + 1

              F(c) = -1.62

The f(c) value is negative so we replace the a value with c.

The steps are repeated up to correct 3 decimal points.

n	a	b	C	F(a)	F(b)	F(c)
0	-1	0	-0.5	-3	1	-1.625
1	-0.5	0	-0.25	-1.625	1	-0.265
2	-0.25	0	-0.125	-0.265	1	0.373
3	-0.25	-0.125	-0.1875	-0.265	0.373	0.0559
4	-0.25	-0.1875	-0.21875	-0.265	0.0559	-0.1039
5	-0.2187	-0.1875	-0.2031	-0.1039	0.0559	-0.0238
6	-0.2031	-0.1875	-0.1953	-0.0238	0.0559	0.0160
7	-0.2031	-0.1953	-0.1992	-0.0238	0.0160	-3.9043
8	-0.1992	-0.1953	-0.1972	-3.9043	0.0160	6.0754
9	-0.1992	-0.1972	-0.1982	-3.9043	6.0754	1.2140
10	-0.1992	-0.1982	-0.1987	-3.9043	1.2140	0.0459
11	-0.1992	-0.1987	-0.1989	-3.9043	0.0459	-2.6246

From the above calculation, we come to a conclusion that the reading c9 c10 c11 agree with one another for 3 decimal places is -0.198

  Hence the approximation root of the function f(x)=x³ + 5x + 1  is -0.198 correct upto 3 decimal place by using bisection method.