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

The positive polynomial root is $x_{\ast }=\epsilon +\sqrt{1+{\epsilon }^2}$. Its Taylor expansion near zero isThe starting error is therefore $x_0-x_{\ast }=-\epsilon -\frac{1}{2}{\epsilon }^2+O({\epsilon }^4)$. ConsequentlyFor signed $\epsilon$, the magnitude scaling should be written with $|\epsilon |$; this is the precise interpretation of the proportionality in the question.

The Newton root-finding iteration for this quadratic polynomial isAt $x_0=1$, it gives $x_1=1/(1-\epsilon )$. Its geometric series expansion is $1+\epsilon +{\epsilon }^2+{\epsilon }^3+O({\epsilon }^4)$. Subtracting the expansion of the polynomial root yields

Use the Exact Newton error identity for a quadratic to expose the cancellation without a long quotient expansion. Put $e_k=x_k-x_{\ast }$. Since $f(x_{\ast })=0$, direct expansion givesFrom part (ii), $e_1=\frac{1}{2}{\epsilon }^2+O({\epsilon }^3)$ and $x_1-\epsilon \to 1$. It follows thatThus the stronger result isThe vanishing lower-order terms are a concrete instance of quadratic convergence: to leading order, this step squares the previous error.