Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 12F / d / Solution

As printed, this part contains an indexing error. The displayed product ends at $m-1$, but the hypotheses contain $m+1$ alternating intervals indexed by $0,\dots ,m$. On the first $m$ intervals the printed polynomial has sign $(-1{)}^i$, while on the last interval it has sign $(-1{)}^{m-1}$ rather than the required $(-1{)}^m$. For example, when $m=1$ the printed product is the constant $Q=1$, which cannot reduce a negative extremum of $g$ on the final interval. The intended polynomial isWe prove the stated conclusion with this correction.

Each root $w_j$ lies in the gap between the $(j-1)$st and $j$th active intervals. Therefore $Q$ has sign $(-1{)}^i$ throughout $[u_i,v_i]$. In particular, wherever $|g|=M$, the numbers $Q$ and $g$ have the same sign, and $Q$ is nonzero.

LetThese are compact sets. On their union, the continuous function $|Q|$ has a positive minimum. Hence, throughout some open neighbourhood $U$ of $K_{+}\cup K_{-}$, subtracting a sufficiently small positive multiple $\eta Q$ moves $g$ strictly towards zero and gives $|g-\eta Q|<M$. On the compact complement $[0,1]\setminus U$, continuity gives a uniform margin $|g|\le M-ϵ$ for some $ϵ>0$. Taking alsoshows that $|g-etaQ|<M$ there. Thusfor every sufficiently small $\eta >0$.

It remains to prove necessity in the Chebyshev alternation theorem. Let $P$ be a best approximation of degree at most $n$, put $g=f-P$, and let $M=\Vert g{\Vert }_{\infty }$. If $g$ has no sequence of $n+2$ extrema with alternating signs, its sets of positive and negative extrema can be collected, from left to right, into $m+1$ alternating compact groups with $m\le n$. Choose disjoint intervals $[u_i,v_i]$ containing those groups and separated by regions on which $|g|<M$. After replacing $g$ by $-g$ if necessary, they satisfy the displayed hypotheses. The corrected construction produces a polynomial $Q$ of degree $m\le n$ and a small $\eta >0$ such thatThis contradicts the assumed optimality of $P$. Therefore every best polynomial has the required $n+2$ alternating extrema.

Solved by gpt-5.6-sol high.