Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 15

Taylor's theorem gives a finite Taylor polynomial of degree $n$, consisting of the first $n+1$ terms, and a remainder:In the Taylor theorem with Lagrange remainder form, assuming the required derivatives exist on an interval containing $a$ and $x$,for some $\xi$ strictly between $a$ and $x$ when $x\ne a$; at $x=a$ the remainder is zero. In particular, if $|f^{(n+1)}|\le M$ on that segment, then $|R_{n+1}(x)|\le M|x-a{|}^{n+1}/(n+1)!$. This is a finite expansion; it does not assume equality with an infinite Taylor series.

For the natural logarithm, $f(1)=0$, $f^{\prime }(1)=1$, $f^{\prime \prime }(1)=-1$, and $f^{(3)}(x)=2/x^3$. The quadratic Taylor polynomial therefore givesThe Taylor remainder is $R_3=2(0.1{)}^3/(6{\xi }^3)$ with $1<\xi <1.1$. It is positive, so the approximation is an underestimate. Moreover,Thus $1/3000\approx 0.0003334$ is a valid upper bound on the absolute error.

The Maclaurin series of the hyperbolic cosine is $\cosh x=1+x^2/2+x^4/24+O(x^6)$. The binomial series with exponent $-1/2$ gives $(1+u{)}^{-1/2}=1-u/2+3u^2/8+O(u^3)$. Substituting $u=x^2/2+x^4/24+O(x^6)$ yieldsThese are the first three nonzero terms; the remainder is written in Big O notation.

Put $h=x-\pi /2$. The sine addition identity gives $\sin x=\cos h=1-h^2/2+h^4/24+O(h^6)$. Applying the exponential series givesThus the first three nonzero terms in the Taylor series about $\pi /2$ are