Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 75 / 4

For a function on an interval, the modulus of continuity is $\omega (f,t)={\sup }_{|x-y|\le t}|f(x)-f(y)|$, with the pair restricted to that interval. For a periodic function the equivalent definition is ${\sup }_{\theta ,|h|\le t}|g(\theta +h)-g(\theta )|$. The mean value theorem and $|\sin \theta |\le 1$ giveBoth cosine values lie in $[-1,1]$. Hence for $|h|\le t$,Taking the supremum provesIf the periodic modulus of continuity is defined using circular distance instead, choose lifts at that shortest distance and apply the same argument. The cosine substitution for polynomial approximation therefore does not enlarge this modulus.

The first Jackson theorem for periodic approximation states that an absolute constant $C_J$ exists such that, for every continuous $2\pi$-periodic function $g$ and integer $n\ge 1$, there is a trigonometric polynomial $q_n$ of degree at most $n$ satisfyingApply it to the continuous periodic function $g(\theta )=f(\cos \theta )$, which is even. Replace the approximant by its even partSince $g(-\theta )=g(\theta )$, the triangle inequality gives $\Vert g-q_n^{\mathrm{ev}}{\Vert }_{\infty }\le \Vert g-q_n{\Vert }_{\infty }$. This symmetrization cancels all sine terms, so $q_n^{\mathrm{ev}}={\sum }_{j=0}^n{\alpha }_j\cos (j\theta )$. The Chebyshev polynomials satisfy $T_j(\cos \theta )=\cos (j\theta )$ and have degree $j$; the latter follows inductively from $T_0=1$, $T_1=x$ and $T_{j+1}=2x{T}_j-T_{j-1}$. Thereforeis an algebraic polynomial of degree at most $n$, and $p_n(\cos \theta )=q_n^{\mathrm{ev}}(\theta )$. Because cosine maps onto $[-1,1]$, the supremum norms agree exactly:Finally, the definition of the best error, periodic Jackson approximation, and part (a) give the successive boundsThusNeither the constant nor the choice of approximation space depends on $f$. The cosine substitution for polynomial approximation and even symmetrization are what preserve the required degree bound.