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:

$i\$	$[a_{i},b_{i}]\$	$c_{i}\$	$f(c_{i})\$	$c_{i}=a_{i+1}\$ or $b_{i+1}\$?
3	[1.5, 1.75]	1.625	-0.35937	$c_{3}=a_{4}\$
4	[1.625, 1.75]	1.6875	-0.15235	$c_{4}=a_{5}\$
5	[1.6875, 1.75]	1.74875	-0.0459	$c_{5}=a_{6}\$
6	[1.71875, 1.75]	1.73438	0.00807	$c_{6}=b_{7}\$
7	[1.71875, 1.73438]	1.72657	-0.01896	$c_{7}=a_{8}\$
8	[1.72657, 1.73438]	1.73048	-0.0544	$c_{8}=a_{9}\$
9	[1.73048, 1.73438]	1.73243	.00131	$c_{9}=b_{10}\$
10	[1.73048, 1.73243]	1.73146	-0.00205	$c_{10}=a_{11}\$
11	[1.73146, 1.73243]	1.73195	-0.00035	$c_{11}=a_{12}\$
12	[1.73195, 1.73243]	1.73219	0.00048	$c_{12}=b_{13}\$
13	[1.73195, 1.73219]	1.73207	0.00007	$c_{13}=b_{14}\$
14	[1.73195, 1.73207]	1.73201	-0.00014	N/A

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.