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Bisection method

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Bisection Method 

Theorem :

An equation f (x) = 0 , where f (x) is a real continuous function, has at least one root between x/ and xu if f(xl) .f(xu) < 0.

What is the bisection method and what is it based on? 

One of the first numerical methods developed to find the root of a nonlinear equation f (x) = 0 was the bisection method (also called binary-search method). 

The method is based on the following theorem:

Note that :if f(xl)f(xu) > 0, there may or may not be any root between xl and xu.

[ fig (2) and (3)] suggests If f(xl)f(xu) < 0 then there may be more than one root between xl and xu, [ fig (1) and (4)]. so theorem guarantees one root between xl and xu 

Fig1: at least one root between two points if the function is real, continuous and changes sign.


Fig2 : If the function doesn't change sign between two points, the root of the function f(x)=0 may still exist between two points.


Fig4 : If the function f(x) changes sign between the two points, more than one root fo the equation f(x)=0 may exist between the two points.




 Understanding to Bisection method :

1. Since the method is based on finding the root between two points, the method falls under the category of bracketing methods.

2.Since the root is bracketed between two points, xl and xu, one can find the midpoint, xm between xl and xu

. This gives us two new intervals (xl,xm), and (xm ,xu).

 3.Now, our problem is in which interval root of the equation f(x)= 0 lies?

4.Well, one can find the sign of f( xl).f(xm ) and if f(xl).f(xm ) < 0 then the new bracket is between xl and xm , otherwise, it is between xm and xu.

5.So, you can see that you are literally halving the interval. As one repeats this process, the width of the interval [xl,xu] becomes smaller and smaller, and you can zero in to the root of the equation f(x)=0.


Important while solving problem

(1) An algebraic equation f(x) = 0 can have at most as many as positive roots as the 
number of changes of sign of f(x).
(2) An algebraic equation f(x) = 0 can have at most as many as negative roots as the 
number of changes of sign of f(-x).


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Nirajan Sah asked about 1 month ago

Find the square root of 2 with an error less than 10-4 by bisection method.

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Nirajan Sah answered about 1 month ago

The function involved is f(x) = x2-2. The following table steps through the iteration until the size of the interval, given in the last column is less than 0.01.

Hence, the answer is (1.41406+1.42187)/2= 1.417