Find the Maximum Trigonometric polynomial coefficient $A_{k}$

Solution 1:

Since $\cos{kx} =\frac { e^{ikx} + e^{-ikx}}{2}$ we can restate the condition in this form let $$P(x) = 1 - \sum_{k=-n}^{k=n} \frac{A_{k}}{2}*e^{ikx}$$ where $A_{-k} = A_{k}$. If $P(x) >= 0$ find the upper bound on $A_{k}$. Now, there is a lemma due to Riesz which says that a real positive polynomial is a square. In words, that there is a trigonometric polynomial $Q(x) = \sum_{k=-n}^{k=n}\frac{1}{2\pi}a_{k}e^{ikx}$ such that $P(x) = |Q(x)|^{2} = Q(x)\overline Q(x)$ Multiplying out and comparing coefficients, one gets $$\sum |a_{k}|^{2} = (2\pi)^{2}$$ and $$\frac{1}{2}A_{k} = (\frac{1}{2\pi})^{2}\sum a_{l}*\overline a_{l + k}$$ Then, if one applies Cauchy-Schwartz inequality to the last expression and uses the previous estimate on sum of squares of $a_{k}$'s one gets: $$A_{k} <= 2$$ Not the optimal estimation (as the case n=1 shows) but a uniform one with respect to $n$.

Solution 2:

NTstrucker@AoPS's answer:

It is a known result: $$\max A_k = 2 \cos \frac{\pi}{\lfloor \frac{n}{k} \rfloor + 2}, \ 1\le k \le n.$$ See e.g. Theorem 16.2.4 in [1].

So, $\max A_1 = 2\cos \frac{\pi}{n + 2}$,
$\max A_m = 1$ for $\lfloor \frac{n}{2}\rfloor + 1 \le m \le n$,
$\max A_{\lfloor \frac{n}{2}\rfloor} = \sqrt{2}$.

[1] Qazi Ibadur Rahman and Gerhard Schmeisser, “Analytic Theory of Polynomials”, 2002.