Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 15

The finite form of Taylor's theorem iswhere $\xi$ lies between $a$ and $x$ when $x\ne a$. Sufficient hypotheses are continuous derivatives up to order $n-1$ on the interval joining the endpoints and an $n$th derivative in its interior. The first $n$ terms end at degree $n-1$; the Taylor remainder uses the $n$th derivative. At $x=a$ the remainder is zero. This finite Taylor polynomial identity does not require convergence of an infinite Taylor series.

Write the sine argument as $\pi /2+\delta$, with $\delta =\frac{\pi }{2}(e^x-1)=\frac{\pi }{2}x+\frac{\pi }{4}{x}^2+O(x^3)$. The sine addition formula gives $\sin (\pi /2+\delta )=\cos \delta$. Its Maclaurin series is $1-{\delta }^2/2+O({\delta }^4)$, soThere is no linear term because the outer sine has zero derivative at $\pi /2$.

Set $h=x+1$, the displacement from the expansion point. The numerator has Taylor series $\sinh h=h+h^3/6+O(h^5)$, while the reciprocal denominator has geometric series $(1+h{)}^{-1}=1-h+h^2-h^3+h^4+O(h^5)$. Multiplication givesThe local power series converges for $|x+1|<1$, the distance to the denominator's zero at $x=-2$.

Use the Maclaurin series of the logarithm and hyperbolic cosine:To retain terms through degree six, only the constant and quadratic terms of the reciprocal factor can multiply the leading cubic term. ThusThe numerator in the original PDF is $\ln (1+x^3)$; the cubic power is missing from the local TeX transcription.