Bisection method

Exercises · Derivatives · with worked solutions · PDF

Exercise 1

Using the bisection method, find the solutions of the equation

\[ x^3+2x-1= 0 \]

correct to two decimal places.

(First of all, one must determine how many solutions there are.)

Solution

Written in the form

\[ x^3=1-2x, \]

the equation suggests first comparing the graphs of the functions \(y=x^3\) and \(y=1-2x\), in order to determine beforehand the number of solutions of the equation. Comparing the two graphs, we deduce that there exists a unique solution \(\alpha \in (0, 1)\), i.e., a unique zero of the function \(f(x)=x^3+2x-1\) in that interval. More formally:

\[ \exists ! \; \alpha \in (0,1): \; f(\alpha)=0 \]

We therefore proceed by applying the bisection method to the interval \([a,b]=[0,1]\), with \(f(0)=-1<0\) and \(f(1)=2>0\).

Iteration 1 \((n=0)\)

\[\begin{align*} [a_0,b_0]=[0,1] \\ c_0=\frac{a_0+b_0}{2}=\frac{1}{2} \end{align*}\]

with \(f(c_0)=\frac{1}{8}>0\). Then, in the next iteration, \(b_1=c_0\).

Solution

Iteration 2 \((n=1)\)

\[ [a_1,b_1]=\left[0,\frac{1}{2}\right], \; f(0)<0, \; f\left(\frac{1}{2}\right)>0 \]

\[ c_1=\frac{a_1+b_1}{2}=\frac{1}{4} \]

with \(f(c_1)<0\). Then, in the next iteration, \(a_2=c_1\).

Iteration 3 \((n=2)\)

\[ [a_2,b_2]=\left[\frac{1}{4},\frac{1}{2}\right], \; f\left(\frac{1}{4}\right)<0, \; f\left(\frac{1}{2}\right)>0 \]

\[ c_2=\frac{a_2+b_2}{2}=\frac{3}{8} \]

with \(f(c_2)<0\). Then, in the next iteration, \(a_3=c_2\). To reach the desired level of approximation, we will have to wait until iteration 9 \((n=8)\), where we will obtain

\[ a_8=\frac{29}{64}\simeq0,\mathbf{45}3 \hspace{1cm} b_8=\frac{117}{256}\simeq 0,\mathbf{45}7 \]

We can finally state that \(\alpha \in \left(\frac{29}{64},\frac{117}{256}\right)\) correct to two decimal places.

Exercise 2

Using the bisection method, find the solutions of the equation

\[ x+\log x = 0 \]

correct to two decimal places.

(First of all, one must determine how many solutions there are.)

Solution

Written in the form

\[ \log x=-x, \]

the equation suggests first comparing the graphs of the functions \(y=\log x\) and \(y=-x\), in order to determine beforehand the number of solutions of the equation. Comparing the two graphs, we deduce that there exists a unique solution \(\alpha \in (0, 1)\), i.e., a unique zero of the function \(f(x)=x+\log x\) in that interval. More formally:

\[ \exists ! \; \alpha \in (0,1): \; f(\alpha)=0 \]

We therefore proceed by applying the bisection method to the interval \([a,b]=[0+\varepsilon,1], \; \varepsilon>0\), with \(f(1)=1>0\) and \(f<0\) in a right neighborhood of the point \(x_0=0\) of arbitrarily small radius \(\varepsilon\) (note that \(y=\log x\) is not defined at \(x=0\)).

Iteration 1 \((n=0)\)

\[\begin{align*} [a_0,b_0]=[0+\varepsilon,1] \\ c_0=\frac{a_0+b_0}{2}=\frac{1+\varepsilon}{2}\simeq \frac{1}{2} \end{align*}\]

with \(f(c_0)=\frac{1}{2}-\log 2<0\). Then, in the next iteration, \(a_1=c_0\).

Iteration 2 \((n=1)\)

\[ [a_1,b_1]=\left[\frac{1}{2},1\right], \; f\left(\frac{1}{2}\right)<0, \; f(1)>0 \]

\[ c_1=\frac{a_1+b_1}{2}=\frac{3}{4} \]

with \(f(c_1)>0\). Then, in the next iteration, \(b_2=c_1\).

Iteration 3 \((n=2)\)

\[ [a_2,b_2]=\left[\frac{1}{2},\frac{3}{4}\right], \; f\left(\frac{1}{2}\right)<0, \; f\left(\frac{3}{4}\right)>0 \]

\[ c_2=\frac{a_2+b_2}{2}=\frac{5}{8} \]

with \(f(c_2)>0\). Then, in the next iteration, \(b_3=c_2\).

Solution

Iteration 4 \((n=3)\)

\[ [a_3,b_3]=\left[\frac{1}{2},\frac{5}{8}\right], \; f\left(\frac{1}{2}\right)<0, \; f\left(\frac{5}{8}\right)>0 \]

\[ c_3=\frac{a_3+b_3}{2}=\frac{9}{16} \]

with \(f(c_3)<0\). Then, in the next iteration, \(a_4=c_3\). Observe that

\[ a_3=0,5, \; b_3=0,625 \]

and we are therefore still far from having two correct decimal places. To reach the desired level of approximation, we will have to wait until iteration 10 \((n=9)\), where we will obtain

\[ a_9=\frac{145}{256}\simeq0,\mathbf{56}6 \hspace{1cm} b_9=\frac{291}{512}\simeq 0,\mathbf{56}8 \]

We can finally state that \(\alpha \in \left(\frac{145}{256},\frac{291}{512}\right)\) correct to two decimal places.

Exercise 3

Using the bisection method, find the solutions of the equation

\[ 2\sin x= x \]

correct to one decimal place.

(First of all, one must determine how many solutions there are.)

Exercise 4

Using the bisection method, find the solutions of the equation

\[ x^2-2-\log x=0 \]

correct to one decimal place.

(First of all, one must determine how many solutions there are.)