Numerical Analysis/Bisection Method Worked Example

X^3=20 ,f(X),   X4=4, X1=1 take four iterations.

Example
Find an approximation of $$\sqrt {3}$$ correct to within 10-4 by using the bisection method on $$ f(x) = x^2 - 3 \ $$. starting on [1, 2].

Analysis of the Problem
The number of iterations we will use, n, must satisfy the following formula: $$ n > \log_2\frac{b-a}{\epsilon} = \log_2\frac{2-1}{10^{-4}} = \log_2\frac{1}{10^{-4}} = \log_2{10^4} = \frac{\log{10^4}}{\log{2}} = \frac{4}{.3010} = 13.29$$. Thus, we will use 14 iterations of the bisection method.

Iteration #1
First, we find the midpoint of the interval [1, 2]: $$ c_1 = \frac{1 + 2}{2} = 1.5 \ $$  Then we check if $$ f(c_1) \ $$ is positive or negative:  $$ f(c_1) = 1.5^2 - 3 = -0.75 \ $$  The value of $$ f(c_1) \ $$ is negative, which means that $$ c_1 \ $$ is less than $$ \sqrt{3} $$. We therefore use $$ c_1 \ $$ as the left endpoint of our new interval and keep 2 as the right endpoint. yes

Iteration #2
Repeat the process from Iteration #1 to do Iteration #2: Solution:

We continue by finding the midpoint of our new interval, [1.5, 2]: $$ c_2 = \frac{1.5 + 2}{2} = 1.75 \ $$  Then we check if $$ f(c_2) \ $$ is positive or negative:  $$ f(c_2) = 1.75^2 - 3 = 0.0625 \ $$  The value of $$ f(c_2) \ $$ is positive, which means that $$ c_2 \ $$ is greater than $$ \sqrt{3} $$. We therefore keep 1.5 as the left endpoint of our new interval and use $$ c_2 \ $$ as the right endpoint.

Iteration #3 - Iteration #14
Complete Iteration #3 - Iteration #14: Solution: the final answer is follow or describe below

Conclusion
Thus, we have found that Solution: 1.73201  is an approximation of $$\sqrt {3}$$ correct to within 10-4.