 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/19/b/iii/solution

Use the Exact Newton error identity for a quadratic to expose the cancellation without a long quotient expansion. Put

e_{k} = x_{k} - x_{*}

. Since

f (x_{*}) = 0

, direct expansion gives

f (x_{k}) = 2 (x_{*} - ε) e_{k} + e_{k}^{2}, e_{k + 1} = e_{k} - \frac{2 ( x _{*} - ε ) e _{k} + e _{k}^{2}}{2 ( x _{*} - ε + e _{k} )} = \frac{e _{k}^{2}}{2 ( x _{k} - ε )} .

From part (ii),

e_{1} = \frac{1}{2} ε^{2} + O (ε^{3})

and

x_{1} - ε \to 1

e_{2} = \frac{\frac{1}{4} ε ^{4} + O ( ε ^{5} )}{2 ( 1 + O ( ε ^{2} ))} = \frac{1}{8} ε^{4} + O (ε^{5}) .

Thus the stronger result is

∣ x_{2} - x_{*} ∣ \sim \frac{1}{8} ∣ ε ∣^{4}, n_{2} = 4 > 3.

The vanishing lower-order terms are a concrete instance of quadratic convergence: to leading order, this step squares the previous error.

 New to topics? Read the docs here!