Past exam of the mathematics course of the University of Cambridge / 2020 / ia / Paper 1 / 9D (Analysis I)

Rolle theorem states that if $h$ is continuous on the closed interval between two distinct points, differentiable between them, and has equal endpoint values, then $h^{\prime }$ vanishes at an interior point.

Let $P_N(t)={\sum }_{j=0}^N{f}^{(j)}(0)t^j/j!$ and, for $x\ne 0$, putThen $h(x)=0$ and $h^{(j)}(0)=0$ for $0\le j\le N$. Applying Rolle's theorem successively $N+1$ times gives a point $\xi =\theta x$, $0<\theta <1$, where $h^{(N+1)}(\xi )=0$. Hencewhich is the stated Taylor theorem with Lagrange remainder.

If $f^{\prime }=0$, the $N=0$ case gives $f(x)=f(0)+x{f}^{\prime }(\theta x)=f(0)$. Thus $f$ is constant; equivalently this is the immediate corollary of the mean value theorem.

DifferentiateUsing $s^{\prime }=c$ and $c^{\prime }=-s$ givesBy the zero derivative on a connected open set, $F$ is independent of $x$.

The constant from part (i) is found by setting $x=0$:Taking $a=x+y$ therefore gives the sine addition formula

Differentiation givesThe initial conditions set the constant to one, sothe Pythagorean trigonometric identity.

Since $|c|\le 1$, Taylor theorem with Lagrange remainder givesfor some $0<\xi <1$, whilefor some $0<\eta <2$. The intermediate value theorem supplies $k\in (1,2)$ with $c(k)=0$. The sine addition formula gives $s(2k)=2s(k)c(k)=0$. The analogous cosine addition formula gives $c(2k)=c(k{)}^2-s(k{)}^2=-1$, soThus $s$ is a periodic function.