Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 15

The finite form of Taylor's theorem with the first $n$ terms iswhere $\xi$ lies strictly between $a$ and $x$ for $x\ne a$. A sufficient hypothesis is that $f$ has continuous derivatives through order $n-1$ on the interval joining the endpoints and an $n$th derivative in its interior. The displayed polynomial has degree at most $n-1$; its Taylor remainder uses the $n$th derivative, not the $(n+1)$st. At $x=a$ the remainder is zero. This is a finite Taylor polynomial identity and does not require an infinite Taylor series to converge.

For $f(x)=x^{1/5}$, the first two derivatives are $f^{\prime }(x)=x^{-4/5}/5$ and $f^{\prime \prime }(x)=-4x^{-9/5}/25$. Since $32=2^5$,The first three terms of the Taylor expansion, with $x-32=1$, therefore giveFor a check on accuracy, $f^{(3)}(x)=36x^{-14/5}/125$ is positive and decreasing on $[32,33]$. The Taylor remainder is consequently positive and less than $f^{(3)}(32)/6=3/1024000<3\times 10^{-6}$. The approximation is a slight underestimate.

Divide numerator and denominator by $x$ before taking their Maclaurin series:For the reciprocal denominator, the geometric series gives $1-x^2/6+7x^4/360+O(x^6)$. Multiplying gives the Taylor expansionStrictly, the original quotient has a removable singularity at zero; this is the Taylor series of its continuous, analytic extension with value one there. The absence of odd powers also follows because the ratio of these two odd functions is even.

The definition of hyperbolic sine simplifies the composition exactly for $x>0$:Set $h=x-1$. The geometric series $1/(1+h)=1-h+h^2-h^3+h^4+O(h^5)$, valid for $|h|<1$, then gives