 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/19/c/solution

The supplied bracket gives

0 < e_{0} = 1.5 - 2 < 0.1

. Part (b) ensures

x_{n} > 2 > 1.4

, so each error satisfies

0 < e_{n + 1} < \frac{e _{n}^{2}}{2.8} .

d_{n} = e_{n} /2.8

gives

d_{n + 1} < d_{n}^{2}

and

d_{0} < 0.1/2.8 = 1/28

. After three steps the exponent has doubled three times:

0 < e_{3} < \frac{2.8}{2 8 ^{8}} \approx 7.4113 \times 1 0^{- 12} < 1 0^{- 11} .

This rigorous error bound uses only the given bracket, not advance knowledge of the digits of

2

The actual rational iterates are

x_{1} = 17/12

x_{2} = 577/408

, and

x_{3} = 665857/470832 \approx 1.414213562374690

. They illustrate the bound, but the proof comes from the exact error identity and the positive lower bound on the denominators.

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