Let $n,k$ be given positive integers and $n\ge k$. Let $A_i, i=1, 2, \cdots, n$ be given real numbers. If for all real numbers $x$ we have $$A_{1}\cos{x}+A_{2}\cos{(2x)}+\cdots+A_{n}\cos{(nx)}\le 1$$ Find the maximum value of $A_{k}$.
I don't know if this question has been studied
If $n=2$ it is easy to solve it.
On
I'll just do a subcase of the $n=3$ case.
$A_{k}=1$ is the maximum for k=2, 3.
First notice that letting $A_{k}=1$ and taking the other values of $A_{j}=0$ we have
$$cos(kx)\leq 1$$
which holds for all x. So each $A_{k}$ is at least 1. Now we will show that each $A_{k}$ is at most 1 for $k=2,3$.
Plugging in $x=0$ gives us $A_{1}+A_{2}+A_{3} \leq 1$.
If $A_{3}>1$ then this forces $A_{1}+A_{2} \leq 1-A_{3} < 0$. However, plugging in $x=\frac{2\pi}{3}$ we have $A_{3} + \frac{-A_{1}-A_{2}}{2} \leq 1$
So $A_{3}\leq 1+\frac{A_{1}+A_{2}}{2} < 1$
Therefore $A_{3}\leq 1$ thus $A_{3}=1$.
If $A_{2}>1$ then $A_{1}+A_{3} \leq 1-A_{2} < 0$.
Plugging in $x=\pi$ we have $A_{2} + -A_{1}-A_{3} \leq 1$.
So $A_{2}\leq 1+ A_{1} + A_{3} < 1$.
Therefore $A_{2}\leq 1$ thus $A_{2}=1$.
On
Update (2021/02/05)
We need to solve the optimization problem: \begin{align} &\max\quad A_m\\ &\mathrm{s.t.}\quad \sum_{k=1}^n A_k\cos (kx) \le 1, \ \forall x\in [0, 2\pi], \\ &\qquad\quad A_k \in \mathbb{R}, \forall k. \end{align} It can be viewed as a "Linear Programming" with uncountable many linear constraints. However, as in the special cases ($n=2, 3, 4, 5, 6$ etc. see below), perhaps we can choose a few linear constraints such that $\max A_m$ is the same. In other words, the following optimization problem has the same maximum as the previous one. \begin{align} &\max\quad A_m\\ &\mathrm{s.t.}\quad \sum_{k=1}^n A_k\cos (kx_j) \le 1, \ j = 1, 2, \cdots, M, \\ &\qquad\quad A_k \in \mathbb{R}, \forall k. \end{align} where $x_j$'s are chosen in advance (see the special cases below).
How to choose a set of $x_1, x_2, \cdots, x_M$?
Conjecture 1: $\max A_m = 1$ for $m = \lfloor \frac{n}{2}\rfloor + 1, \cdots, n$.
(Edit (2021/02/08): I have proved Conjecture 1 in my another answer.)
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Some special cases $n=3, 4$
Remark: I will add more cases if possible.
Case $n=3$
Letting $x = 0$ and $x = \frac{2\pi}{5}$ respectively, we have \begin{align} A_1 + A_2 + A_3 &\le 1, \tag{1}\\ A_1\frac{\sqrt{5}-1}{4} - (A_2 + A_3)\frac{\sqrt{5}+1}{4} &\le 1. \tag{2} \end{align} $(1)\times \frac{\sqrt{5} + 1}{4} + (2)$ gives $A_1 \le \frac{\sqrt{5}+1}{2}$.
Also, when $A_1 = \frac{\sqrt{5}+1}{2}, A_2 = - \frac{2}{\sqrt{5}}, A_3 = \frac{1}{2} - \frac{\sqrt{5}}{10}$, we have \begin{align} &A_1\cos x + A_2\cos 2x + A_3 \cos 3x - 1\\ =\ & \frac{5 - \sqrt{5}}{40}(1-\cos x)(\sqrt{5} - 1 - 4\cos x)^2 \le 0. \end{align} The desired result follows.
Letting $x=0$, $x = \frac{2}{3}\pi$ and $x = \pi$ respectively, we have \begin{align} A_1 + A_2 + A_3 &\le 1, \tag{3}\\ -\frac{1}{2}A_1 - \frac{1}{2}A_2 + A_3 &\le 1, \tag{4}\\ -A_1 + A_2 - A_3 &\le 1. \tag{5} \end{align} $(3) + (5)$ gives $A_2 \le 1$, and $(3) + 2\times (4)$ gives $A_3 \le 1$.
The remaining is easy.
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Case $n=4$
Letting $x = \frac{\pi}{6}$ and $x = \frac{\pi}{2}$ respectively, we have \begin{align} \frac{\sqrt{3}}{2} A_1 + \frac{1}{2}A_2 - \frac{1}{2}A_4 &\le 1, \tag{6}\\ -A_2 + A_4 &\le 1.\tag{7} \end{align} $(6)\times 2 + (7)$ gives $A_1 \le \sqrt{3}$.
Also, when $A_1 = \sqrt{3}, A_2 = - \frac{7}{6}, A_3 = \frac{1}{\sqrt{3}}, A_4 = - \frac{1}{6}$, we have \begin{align} &A_1\cos x + A_2\cos 2x + A_3 \cos 3x + A_4\cos 4x - 1\\ =\ & -\frac{\cos^2 x}{3} (\sqrt{3} - 2\cos x)^2\\ \le\ & 0. \end{align} The desired result follows.
Letting $x = \frac{\pi}{8}$ and $x = \frac{7\pi}{8}$ respectively, we have \begin{align} A_1 \cos \frac{\pi}{8} + \frac{A_2}{\sqrt{2}} + A_3\cos \frac{3\pi}{8} &\le 1, \tag{8} \\ -A_1 \cos \frac{\pi}{8} + \frac{A_2}{\sqrt{2}} - A_3\cos \frac{3\pi}{8} &\le 1. \tag{9} \end{align} $(8) + (9)$ gives $A_2\le \sqrt{2}$.
Also, when $A_1 = 0, A_2 = \sqrt{2}, A_3 = 0, A_4 = -\frac{1}{2}$, we have \begin{align} &A_1\cos x + A_2\cos 2x + A_3 \cos 3x + A_4\cos 4x - 1\\ =\ & -\frac{1}{4} (2 + \sqrt{2} - 4\cos^2 x)^2\\ \le\ & 0. \end{align} The desired result follows.
Letting $x = 0$, $x = \frac{\pi}{2}$, $x = \frac{2\pi}{3}$ and $x=\pi$ respectively, we have \begin{align} A_1 + A_2 + A_3 + A_4 &\le 1, \tag{10}\\ -A_2 + A_4 &\le 1, \tag{11}\\ -\frac{1}{2}A_1 - \frac{1}{2}A_2 + A_3 - \frac{1}{2}A_4 &\le 1, \tag{12}\\ -A_1 + A_2 - A_3 + A_4 &\le 1. \tag{13} \end{align} $(10) + (12)\times 2$ gives $A_3 \le 1$, and $(10) + (11)\times 2 + (13)$ gives $A_4 \le 1$. The remaining is easy.
On
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 $n$.
On
My second answer:
Edit 2021/03/11
According to NTstrucker@AoPS's result (https://artofproblemsolving.com/community/c6h2477208), I give the following conjecture.
Conjecture 2: $\max A_1 = 2\cos \frac{\pi}{n+2}$.
When $n = 2$, $\max A_1 = \sqrt{2} = 2 \cos \frac{\pi}{4}$.
When $n = 3$, $\max A_1 = \frac{\sqrt{5} + 1}{2} = 2\cos \frac{\pi}{5}$.
When $n = 4$, $\max A_1 = \sqrt{3} = 2\cos \frac{\pi}{6}$.
When $n = 5$, $\max A_1 = 2\cos \frac{\pi}{7}$.
When $n = 6$, $\max A_1 \approx 1.8478$ (I only get numerical result) should be $2\cos \frac{\pi}{8} \approx 1.847759065$.
When $n = 7$, $\max A_1 \approx 1.8794$ (I only get numerical result) should be $2\cos \frac{\pi}{9} \approx 1.879385242$.
When $n = 8$, $\max A_1 \approx 1.9021$ (I only get numerical result) should be $2\cos \frac{\pi}{10} \approx 1.902113033$.
When $n = 9$, $\max A_1 \approx 1.9190$ (I only get numerical result) should be $2\cos \frac{\pi}{11} \approx 1.918985947$.
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Let us prove that $\max A_m = 1$ if $\lfloor \frac{n}{2}\rfloor + 1 \le m \le n$.
We need Theorem 1. The proof is given at the end.
Theorem 1: Let $n\ge 2$ be a given integer. Let $A_i, i=1, 2, \cdots, n$ be given real numbers such that $A_1\cos x + A_2 \cos 2x + \cdots + A_n \cos n x \le 1, \ \forall x\in \mathbb{R}$. Then $A_m \le 1$ for $m = \lfloor \frac{n}{2}\rfloor + 1, \cdots, n$.
By Theorem 1, we have immediately $\max A_m = 1$ if $\lfloor \frac{n}{2}\rfloor + 1 \le m \le n$ (simply let $A_m = 1$ and $A_k = 0$ for all $k\ne m$ and the condition $A_m \cos m x \le 1, \forall x\in \mathbb{R}$ is clearly satisfied.).
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Proof of Theorem 1: First, we have Fact 1. The proof of Fact 1 is easy and thus omitted.
Fact 1: Let $M\ge 2$ be an integer. Let $K$ be a positive integer. Then $$\frac{1}{M} + \frac{\cos K\pi}{M}\cdot \frac{1 + (-1)^M}{2} + \frac{2}{M}\sum_{j=1}^{\lfloor \frac{M+1}{2}\rfloor - 1} \cos \frac{2jK\pi}{M} = \left\{\begin{array}{cc} 0 & M \nmid K \\[5pt] 1 & M \mid K. \end{array} \right.$$
Second, we have Fact 2.
Fact 2: Let $n\ge 2$ be a given integer. Let $m$ be an integer with $\lfloor \frac{n}{2}\rfloor + 1 \le m \le n$. Let $A_i, i= 1, 2, \cdots, n$ be given real numbers. Let $f(x) = \sum_{k=1}^n A_k\cos (kx)$. Then $$\frac{f(0)}{m} + \frac{f(\pi)}{m}\cdot \frac{1 + (-1)^m}{2} + \frac{2}{m}\sum_{j=1}^{\lfloor \frac{m+1}{2}\rfloor - 1} f\left(\frac{2j\pi}{m}\right) = A_m. $$ (Proof of Fact 2: It suffices to prove that $$\frac{1}{m} + \frac{\cos k\pi}{m}\cdot \frac{1 + (-1)^m}{2} + \frac{2}{m}\sum_{j=1}^{\lfloor \frac{m+1}{2}\rfloor - 1} \cos \frac{2j k \pi}{m} = \left\{\begin{array}{cc} 0 & k \ne m \\[5pt] 1 & k = m. \end{array} \right. $$ This is true by Fact 1.)
By Fact 2, Theorem 1 is proved. We are done.
On
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.
COMMENT
May be this idea can help:
We use following identity:
$$\sin^2(x)+\sin^2(2x)+\sin^2 (3x)+ . . .+\sin^2(nx)=\frac{n\sin(x)-\sin(nx)\cos(n+1)x}{2\sin x}$$
Taking derivative of both sides we get:
$$2\sin(x)\cos(x)+4\sin(2x)\cos(2x)+6\sin(3x)\cos(3x)+ . . . 2n\sin(nx)\cos(nx)=\big(\frac{n\sin(x)-\sin(nx)\cos(n+1)x}{2\sin x}\big)'\leq 1$$
We solve derivative $\leq 1$, we get a relation between n and x which may be used to find maximum value of $A_k$. In fact the proble can be reduced to: If:
$\big(\frac{n\sin(x)-\sin(nx)\cos(n+1)x}{2\sin x}\big)'\leq 1$
Find maximum of $A_n=f(x)= 2n\sin(nx)$