Question

Suppose you wish to find a root of $f\left(x\right)=x^3+x^2+x+1$in the interval [−3, 2]. Compare the accuracy of Newton’s method as applied in Example 8 with the accuracy of the Bisection Method used in Exercise 80 of Section 1.4 for the same function. Which method gets closest to the root in three iterations?

Short answer

Newton's method gets closest to the root in 3 iterations.

Step-by-step solution

Step 1. Given information.

We have been given that the root of the function $f\left(x\right)=x^3+x^2+x+1$is to be found in the interval [−3, 2].

We have to compare the accuracy of Newton’s method as applied in Example 8 with the accuracy of the Bisection Method used in Exercise 80 of Section 1.4 for the same function. Which method gets closest to the root in three iterations?

Step 2. Newton's method to approximate the roots

Initial guess is $\left(x_0,f\left(x_0\right)\right)=(0,1)$

Find the derivative

$f^{\mathrm{\prime }}\left(x\right)=3x^2+2x+1$

at $x_0=0$

$\begin{array}{r}{f}^{\mathrm{\prime }}(-3)=3(0{)}^2+2(0)+1\\ =3\left(0\right)+2\left(0\right)+1\\ =1\end{array}$

Find the tangent line to the curve,

$\begin{array}{r}y-1=1(x-0)\\ y-1=x\\ y=x+1\end{array}$

Thus, the root of the tangent line is :

$x=-1$

Step 3. Find the derivative for the new guess.

New guess is :

$\left(x_1,f\left(x_1\right)\right)=(-1,0)$

Find the derivative

$f^{\mathrm{\prime }}\left(x\right)=3x^2+2x+1$

at $x_0=0$

$\begin{array}{r}{f}^{\mathrm{\prime }}(-1)=3(-1{)}^2+2(-1)+1\\ =3(-1)+2(-1)+1\\ =-4\end{array}$

Find the tangent line to the curve,

$\begin{array}{r}y-0=-4(x+1)\\ y=-4x-4\end{array}$

The root of the tangent line is :

$x=-1$

Step 4. repeating the process

New guess is :

$\left(x_2,f\left(x_2\right)\right)=(-1,0)$

Find the derivative :

$f^{\mathrm{\prime }}\left(x\right)=3x^2+2x+1$

at $x_0=0$

$\begin{array}{r}{f}^{\mathrm{\prime }}(-1)=3(-1{)}^2+2(-1)+1\\ =3(-1)+2(-1)+1\\ =-4\end{array}$

Find the tangent line to the curve,

$\begin{array}{r}y-0=-4(x+1)\\ y=-4x-4\end{array}$

The root of the tangent line is :

$x=-1$

Thus, the second and third approximations are equal.

Step 5. Using bisection method

$\begin{array}{l}f\left(x\right)=x^3+x^2+x+1\\ a=-3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}f(-3)=-20<0\\ b=2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}f\left(2\right)=15>0\end{array}$

iteration 1,

$\begin{array}{r}m=\frac{a+b}{2}\\ =\frac{-3+2}{2}\\ =-\frac{1}{2}=-0.5\\ f\left(m\right)={\left(-\frac{1}{2}\right)}^3+{\left(-\frac{1}{2}\right)}^2+\left(-\frac{1}{2}\right)+1\\ =\frac{5}{8}>0\end{array}$

because f(m)>0, replace b with m,

$\begin{array}{l}a=-3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}f(-3)=-20<0\\ b=\frac{-1}{2}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}f\left(\frac{-1}{2}\right)=\frac{5}{8}>0\end{array}$

Step 6. Using the bisection method (iteration 2).

Iteration 2,

$\begin{array}{r}m=\frac{a+b}{2}\\ =\frac{-3-\frac{1}{2}}{2}\\ =-\frac{7}{4}=-1.75\\ f\left(m\right)={\left(-\frac{7}{4}\right)}^3+{\left(-\frac{7}{4}\right)}^2+\left(-\frac{7}{4}\right)+1\\ =\frac{5}{8}>0\end{array}$

Thus, the newton's method get closest to the root in 3 iterations.