Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 1 / 11E (Analysis)

The intermediate value theorem states that if $f:[\alpha ,\beta ]\to \mathbb{R}$ is continuous and $y$ lies between $f(\alpha )$ and $f(\beta )$, then some $c\in [\alpha ,\beta ]$ satisfies $f(c)=y$.

It is enough to prove the caseLetChoose $x_n\in E$ with $x_n\to c$. Continuity gives $f(c)\le y$. If $f(c)<y$, continuity would make $f(x)<y$ for some $x>c$, contradicting that $c$ is an upper bound of $E$. Thus $f(c)=y$. The case with reversed endpoint inequalities follows by replacing $f$ with $-f$.

Choose $z_{+},z_{-}\in \mathbb{C}$ withRestrict $f$ to the line segment between them:This is continuous, with $h(0)<0<h(1)$. The intermediate value theorem gives a $t_0$ with $h(t_0)=0$. Thereforeis the required zero on a line segment.

Define the continuous functionAt the mesh points its values telescope:If one term is zero, the result follows immediately. Otherwise the terms cannot all have the same sign, so there are mesh points at which $h$ is positive and negative. The intermediate value theorem between those points gives an $x\in [0,(n-1)a]$ with $h(x)=0$. Thus the telescoping increment lemma yields