 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/b/solution

Write

r = a > 0

and subtract it from the iteration:

e_{n + 1} = \frac{x _{n}^{2} + a - 2 r x _{n}}{2 x _{n}} = \frac{( x _{n} - r ) ^{2}}{2 x _{n}} = \frac{e _{n}^{2}}{2 x _{n}} .

x_{n} > r

, then

x_{n} > 0

and

e_{n} > 0

, so

e_{n + 1} > 0

. Induction from

x_{0} > r

proves $x_{n} > r$ for every $n$ . Also

x_{n + 1} - x_{n} = (a - x_{n}^{2}) / (2 x_{n}) < 0

, so the iterates decrease while remaining above the root. Their limit must satisfy

L = (L + a / L) /2

, hence

L = r

. The squared error numerator explains the ensuing quadratic convergence.

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