Convergence of Fourier Series — Question 3

Question 3 – Solution

Strategy. Positive coefficients make the usual triangle bound attain its value at a common phase.

Step 1: Turn the tail estimate into an equality. The Weierstrass test applies because $\sum n^{-2}<\infty$. Put $T_N=\sum_{n>N}n^{-2}$. For every $x$, $|F(x)-P_N(x)|\leq T_N$, while at $x=0$ every cosine equals one. Hence $$\boxed{\|F-P_N\|_\infty=T_N.}$$ This is an exact error, not merely a sufficient upper bound.

Step 2: Identify every equality point. For the tail to equal $T_N$, every positive summand must have $\cos nx=1$. Using two consecutive integers $n>N$ forces $x=0$ modulo $2\pi$. For the tail to equal $-T_N$, every $\cos nx$ would have to be $-1$. But then $\cos(2nx)=1$ for a frequency also in the tail, a contradiction. Thus the maximum absolute error occurs exactly at $x=0$ modulo $2\pi$.

Step 3: Certify the minimal truncation. Strict monotonicity of $t^{-2}$ gives $$\boxed{\frac 1{N+1}<T_N<\frac 1N.}$$ At $N=100$, $T_N<1/100$. At $N=99$, $T_N>1/100$; monotonicity excludes every smaller $N$. Therefore the exact smallest choice is $\boxed{N=100}$. The two integral bounds straddle the threshold tightly enough to decide it.

Step 4: Compare the two norms. Orthogonality gives $$\boxed{\|F-P_N\|_2^2=\pi\sum_{n>N}\frac 1{n^4},\qquad \frac{\pi}{3(N+1)^3}<\|F-P_N\|_2^2<\frac{\pi}{3N^3}.}$$ The uniform error is of order $N^{-1}$, whereas the $L^2$ error is of order $N^{-3/2}$ (its square is of order $N^{-3}$). Distinguish the norm from its square.