Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 19

For every integer $n$, $\cos ((2n-1)\pi )=-1$. The series is therefore the negative of the harmonic series:For a direct divergence argument, each block $2^{j-1}<n\le 2^j$ contributes at least $2^{j-1}/2^j=1/2$ to the positive sum. Its partial sums grow without bound, soDespite the odd multiples of $\pi$, the signs do not alternate.

For each fixed $n$, the inner geometric series sums to $2^{-n}/(1-1/2)=2^{1-n}$. Summing this second geometric series givesAll terms are nonnegative, so there is no conditional-rearrangement issue; the iterated sum is finite and the double series converges. Equivalently, collecting equal values of $m$ gives ${\sum }_{m=0}^{\infty }(m+1)2^{-m}=4$.

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.