= Solution
Since $|T_j(x)|\leq1$ on $[-1,1]$ and the <coefficients> are summable, the <Weierstrass M-test> gives <uniform convergence> of this <positive lacunary Chebyshev series> and makes its sum continuous.
Let $r$ be the largest integer with $3^r\leq n$, and set $r=-1$ when $n=0$. The candidate and its tail size are
$$
\boxed{p_n(x)=\sum_{k=0}^ra_kT_{3^k}(x),\qquad E_n(f)=\sum_{k=r+1}^\infty a_k.}
$$
The empty sum for $n=0$ is zero. The candidate has degree at most $n$. The <triangle inequality> bounds its error by the proposed tail sum, and equality holds at $x=1$, since every <Chebyshev polynomial> has $T_j(1)=1$ and all tail <coefficients> are positive. It remains to prove that another <polynomial> cannot improve this error.
Put $L=3^{r+1}$, the first omitted degree, so $L>n$. At the points $y_j=\cos(j\pi/L)$ for $j=0,\ldots,L$, every omitted term has
$$
T_{3^k}(y_j)=\cos\left(3^{k-r-1}j\pi\right)=(-1)^j,\qquad k\geq r+1,
$$
because $3^{k-r-1}$ is odd. Uniform convergence therefore gives
$$
f(y_j)-p_n(y_j)=(-1)^j\sum_{k=r+1}^\infty a_k.
$$
These $L+1$ distinct points run in decreasing order; reversing their order still gives alternating signs. Since $L+1\geq n+2$, select $n+2$ consecutive ones and apply the <Chebyshev alternation theorem>. It proves both optimality and uniqueness of the displayed candidate. Thus the best <polynomial> and the error stay unchanged between successive powers of three; this exact truncation property depends on the synchronized tail signs, not on a general rule that truncating a Chebyshev expansion is optimal.
Back to article page
