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.

nabCF(a)F(b)F(c)
0-10-0.5-31-1.625
1-0.50-0.25-1.6251-0.265
2-0.250-0.125-0.26510.373
3-0.25-0.125-0.1875-0.2650.3730.0559
4-0.25-0.1875-0.21875-0.2650.0559-0.1039
5-0.2187-0.1875-0.2031-0.10390.0559-0.0238
6-0.2031-0.1875-0.1953-0.02380.05590.0160
7-0.2031-0.1953-0.1992-0.02380.0160-3.9043
8-0.1992-0.1953-0.1972-3.90430.01606.0754
9-0.1992-0.1972-0.1982-3.90436.07541.2140
10-0.1992-0.1982-0.1987-3.90431.21400.0459
11-0.1992-0.1987-0.1989-3.90430.0459-2.6246

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.

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