 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/10/solution

For this polynomial,

f^{'} (x) = 2 x

, so the specified Newton root-finding iteration simplifies to

x_{n + 1} = \frac{1}{2} (x_{n} + \frac{2}{x _{n}}) .

Starting at one gives

x_{1} = (1 + 2) /2 = 3/2

, followed by

x_{2} = (3/2 + 4/3) /2 = 17/12

. Consequently

x_{1} = 1.5, x_{2} \approx 1.42 to two decimal places .

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