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- by Nirajan Sah 4 months ago

**Bisection Method**

Theorem :

An equation f (x) = 0 , where f (x) is a real continuous function, has at least one root between x_{/} and x_{u} if **f(x _{l}) .f(x_{u}) < 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(x_{l})f(x_{u}) > 0, there may or may not be any root between x_{l} and x_{u}.

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

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, x_{l} and x_{u}, one can find the midpoint, x_{m} between x_{l} and x_{u}

. This gives us two new intervals (x_{l},x_{m}), and (x_{m} ,x_{u}).

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( x_{l}).f(x_{m} ) and if f(x_{l}).f(x_{m} ) < 0 then the new bracket is between x_{l} and x_{m} , otherwise, it is between x_{m} and x_{u}.

5.So, you can see that you are literally halving the interval. As one repeats this process, the width of the interval [x_{l},x_{u}] 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.

Nirajan Sah
answered
about 1 month
ago

The function involved is f(x) = x^{2}-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