Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 15

For $n\ge 3$, Taylor's theorem after $n$ terms, counting the constant term, isUnder the usual hypothesis that $f$ is $n-1$ times continuously differentiable on the interval joining $a$ and $x$, and $n$ times differentiable in its interior, the Taylor theorem with Lagrange remainder givesThis is a finite expansion with a controlled remainder; it does not assume an infinite Taylor series converges. For $x=a$ the remainder is zero.

Let $f(x)=\cosh \sqrt{x}$ near $x=2$. The chain rule givesSubstitute $x=2$ into the first three terms of the Taylor expansion:The quadratic coefficient is $f^{\prime \prime }(2)/2$, not $f^{\prime \prime }(2)$. The apparent square root creates no local difficulty at the positive expansion point.

Use the Maclaurin series $\sin x=x-x^3/6+O(x^5)$ and $(1+x{)}^{-2}=1-2x+3x^2+O(x^3)$. Multiplying through the cubic order givesThese are the first three nonzero terms; the constant term vanishes.

Both numerator and denominator vanish quadratically. Factor out $x^2$ before dividing the power series:The reciprocal of the denominator is $1+x^2/2-x^4/12+O(x^6)$. HenceThe expansion describes the continuous extension at the removable singularity $x=0$, where the original quotient is undefined but its limit is one. Odd powers vanish because the quotient is even.