Past exam of the mathematical methods II course of the University of Oxford / 2014 / 2

The Taylor expansions at zero give $1-e^x=-x-x^2/2+O(x^3)$ and $1-\cos x=x^2/2+O(x^4)$. Squaring the first expression gives $x^2+O(x^3)$, so canceling the leading quadratic power yieldsBoth numerator and denominator vanish, but their leading nonzero coefficients have ratio two.

Expand before taking the fourth derivative:The apparent singularity at zero is removable. The convergent power series may be differentiated termwise, givingAt $x=0.1$, the successive contributions are $-2$, $0.000252$ and approximately $-3.405\times 10^{-9}$. ThusEven the first two nonzero terms give $-1.999748$; the next retained correction quantifies that estimate's accuracy.

Use $t=\tan u$ with the inverse tangent branch $-\pi /2<u<\pi /2$. Then $dt/(1+t^2)=du$, soFor $|t|<1$, the geometric series gives $(1+t^2{)}^{-1}=1-t^2+t^4-t^6+\cdots$. Integrating term by term on compact subintervals of $(-1,1)$ producesThe integral identity itself holds for every real $x$; the displayed local power-series argument uses $|x|<1$.