For $n\geq1$, choose the integer $m$ with
$$
3^m\leq n\lt3^{m+1}
$$
and define the partial sum
$$
p_n(x)=\sum_{k=0}^ma_kT_{3^k}(x).
$$
It belongs to $\mathcal P_n$, and the <triangle inequality> gives
$$
\|f_0-p_n\|_\infty\leq\sum_{k=m+1}^\infty a_k.
$$
At the $3^{m+1}+1$ points
$$
x_j=\cos\frac{j\pi}{3^{m+1}},
\qquad j=0,\ldots,3^{m+1},
$$
every tail term has the same alternating sign because
$$
T_{3^k}(x_j)
=\cos\left(j\pi3^{k-m-1}\right)=(-1)^j,
\qquad k\geq m+1.
$$
Thus
$$
f_0(x_j)-p_n(x_j)=(-1)^j
\sum_{k=m+1}^\infty a_k.
$$
There are at least $n+2$ such points because $n\lt3^{m+1}$. The <Chebyshev alternation theorem> proves
$$
\boxed{p_n=\sum_{k=0}^ma_kT_{3^k}},
\qquad
\boxed{E_n(f_0)=\sum_{k=m+1}^\infty a_k},
\qquad 3^m\leq n\lt3^{m+1}.
$$
For $n=0$, the partial sum is empty and the same argument at $x=\pm1$ gives $p_0=0$ and $E_0(f_0)=\sum_{k=0}^\infty a_k$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 318 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-318/2/d/solution