In Mathematics, the bisection method is a straightforward technique to find the numerical solutions to an equation in one unknown. Among all the numerical methods, the bisection method is the simplest one to solve the transcendental equation.
The bisection method is used to find the roots of a polynomial equation. It separates the interval and subdivides the interval in which the root of the equation lies. The principle behind this method is the intermediate theorem for continuous functions. It works by narrowing the gap between the positive and negative intervals until it closes in on the correct answer. This method narrows the gap by taking the average of the positive and negative intervals. It is a simple method, and it is relatively slow. The bisection method is also known as interval halving method, root-finding method, binary search method or dichotomy method.
Let, consider a continuous function “f” which is defined on the closed interval [a, b], is given with f(a) and f(b) of different signs. Then by intermediate theorem, there exists a point x belong to (a, b) for which f(x) =0.
Follow the below procedure to get the solution for the continuous function:
For any continuous function f(x),
The bisection method is an approximation method to find the roots of the given equation by repeatedly dividing the interval. This method will divide the interval until the resulting interval is found, which is extremely small.
Question: Determine the root of the given equation x^{2}-3 = 0 for x ∈ [1,2]
Solution:
Given: x^{2}-3 = 0
Let f(x) = x^{2}-3
Now, find the value of f(x) at a= 1 and b=2.
f(x=1) = 1^{2}-3 = 1 – 3 = -2 < 0
f(x=2) = 2^{2}-3 = 4 – 3 = 1 > 0
The given function is continuous and the root lies in the interval [1, 2].
Let “t” be the midpoint of the interval.
I.e., t = (1+2)/2
t =3 / 2
t = 1.5
Therefore, the value of the function at “t” is
f(t) =f(1.5) = (1.5)^{2}-3 = 2.25 – 3 = -0.75 < 0
f(t) is negative, so b is replaced with t= 1.5 for the next iterations.
The iterations for the given functions are:
Iterations |
a |
b |
t |
f(a) |
f(b) |
f(t) |
1 |
1 |
2 |
1.5 |
-2 |
1 |
-0.75 |
2 |
1.5 |
2 |
1.75 |
-0.75 |
1 |
0.062 |
3 |
1.5 |
1.75 |
1.625 |
-0.75 |
0.0625 |
-0.359 |
4 |
1.625 |
1.75 |
1.6875 |
-0.3594 |
0.0625 |
-0.1523 |
5 |
1.6875 |
1.75 |
1.7188 |
-01523 |
0.0625 |
-0.0457 |
6 |
1.7188 |
1.75 |
1.7344 |
-0.0457 |
0.0625 |
0.0081 |
7 |
1.7188 |
1.7344 |
1.7266 |
-0.0457 |
0.0081 |
-0.0189 |
So, at the seventh iteration, we get the final interval [1.7266, 1.7344]
Hence, 1.7344 is the approximated solution.