 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability/5/b/i/solution

x_{k + 1} = x_{k} - \frac{x _{k}^{6} - 5 x _{k}^{4} + 6}{6 x _{k}^{5} - 20 x _{k}^{3}} .

It is defined when

x_{k} \in / {0, \pm 10/3}

. Finding an exact zero of a function is a stopping condition, rather than a reason to continue dividing.

 New to topics? Read the docs here!