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

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.