Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 70 / 3

Use real-valued continuous functions and real trigonometric polynomials. The trigonometric Chebyshev alternation theorem says that $p\in {\mathcal{T}}_n$ is the unique best uniform approximation to $f$ if and only if there are distinct points, ordered within one period,Thus the required number of alternating extrema is $2n+2$. When the error is zero, the assertion is interpreted trivially and $p=f$ is unique. The count reflects $\dim {\mathcal{T}}_n=2n+1$ and the fact that a nonzero degree-$n$ trigonometric polynomial has at most $2n$ zeros in one period. In particular a strictly better candidate would force its difference from $p$ to have alternating signs at these points and hence at least $2n+2$ zeros, counting the closing interval around the circle. The trigonometric Chebyshev alternation theorem also gives necessity and uniqueness.

The Weierstrass M-test gives uniform convergence and continuity of the series. For the proposed trigonometric polynomial, set $q=5^{m+1}$ and $A={\sum }_{k=m+1}^{\infty }{a}_k$. The tail satisfiesAt $x_j=j\pi /q$, for $j=0,\dots ,2q-1$, the integer $5^k/q$ is odd for every omitted frequency. Thus $\cos (5^k{x}_j)=(-1{)}^j$ and $r(x_j)=(-1{)}^{j}A$. In particular $\Vert r{\Vert }_{\infty }=A$.

Since $n<q$, these $2q$ points include at least $2n+2$ consecutive alternating extrema in a single period. Also $5^m\le n$, so $t_n\in {\mathcal{T}}_n$. The trigonometric Chebyshev alternation theorem proves that the proposed partial sum is the unique best approximant, withThis is an instance of positive lacunary trigonometric series: all omitted odd frequency ratios align their extrema.

The geometric series formula and part (b) givePut $\alpha =\log 3/\log 5$. Then $5^{\alpha }=3$, so $n^{\alpha }<3^{m+1}$ andThis is the largest possible decay exponent: at $n=5^m$, an estimate with exponent $\beta >\alpha$ would require $\frac{1}{2}\le c(3/5^{\beta }{)}^m$, whose right side tends to zero. Any smaller exponent also gives a valid weaker estimate. In fact $\frac{1}{2}{n}^{-\alpha }\le E_n(f)<\frac{3}{2}{n}^{-\alpha }$ for every $n\ge 1$.