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:
$n>{\log }_2\frac{b-a}{ϵ}={\log }_2\frac{2-1}{10^{-4}}={\log }_2\frac{1}{10^{-4}}={\log }_{2}10^4=\frac{\log 10^4}{\log 2}=\frac{4}{.3010}=13.29$.
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.