Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 29(b)/Solution 1

Example of Newton's method.

In part (a) we computed $x_{2}=x_{0}=0$ and $x_{1}=x_{3}=-1$. These are labelled on the $x$-axis of the adjacent graph (the points $(x_{n},f(x_{n}))$ and the linear approximations are indicated by the red lines). Since Newton's method for a particular function only uses the number $x_{n}$ to produce $x_{n+1}$, it follows from part (a) that if $x_{n}=0$, then $x_{n+1}=-1$ and, if $x_{n}=-1$, then $x_{n+1}=0$. Therefore, the sequence of estimates is given by

$x_{n}={\begin{cases}0&{\text{ if }}n{\text{ is even}}\\-1&{\text{ if }}n{\text{ is odd}}\end{cases}}$

and the sequence will not converge to a root.