# Find the sum of the maximum of the trigonometric polynomial coefficient

somedays ago,I have studying and discussing this Maximum Trigonometric polynomial coefficient problem

Now I want consider a simple problem:

(The sum of the maximum of the trigonometric polynomial coefficient:) Let $$n$$ be give postive integers If for all real numbers $$x$$ we have $$f(x)=A_{1}\cos{x}+A_{2}\cos{(2x)}+\cdots+A_{n}\cos{(nx)}\ge -1$$ Find the maximum value of $$A_{1}+A_{2}+\cdots+A_{n}$$.

Here is what I tried:use this well kown : $$\sum_{j=0}^{n-1}\xi^k_{j}=\begin{cases} n&n|k\\ 0&otherwise \end{cases}$$ where $$\xi_{i}$$be the $$n$$th roots of unity let $$x_{k}=\dfrac{2k\pi}{n+1},k=0,1,\cdots,n$$,so we have $$\sum_{k=0}^{n}\cos{(mx_{k})}=0,m=1,2,\cdots,n$$

so $$\sum_{k=0}^{n}f(x_{k})=\sum_{m=1}^{n}\sum_{k=0}^{n}A_{m}\cos{(mx_{k})}=0$$ then we have $$A_{1}+A_{2}+\cdots+A_{n}=f(0)=-\sum_{k=1}^{n}f(x_{k})\le n$$

I Conjecture the maximum is $$n$$.But I can't find example such reach this maximum

I have find example for $$n=2$$,then let $$A_{1}=\dfrac{4}{3},A_{2}=\dfrac{2}{3},A_{1}+A_{2}=2$$,then $$f(x)=\dfrac{1}{3}(2\cos{x}+1)^2-1\ge -1$$ for all postive $$n$$,I can't find it.Thanks

• For $1\le k\le n$, if we define $$A_k=\frac{2(n+1-k)}{n+1}$$ then we have $A_1+\cdots +A_n=n$, and the constraint $$\sum_{k=1}^n A_k \cos(kx) \ge -1\;\text{for all}\;x\in\mathbb{R}$$ appears to hold (verified for $1\le n\le 50$). Jan 24, 2021 at 15:05
• How to find this $A_{k}$?and How to prove this $\sum_{k=1}^{n}A_{k}\cos{(kx)}\ge -1$ Jan 24, 2021 at 23:52

Solution due to @quasi:

Let $$A_k = 2 - \frac{2k}{n+1}$$. We have $$A_1 + A_2 + \cdots + A_n = n$$.

We have \begin{align} &\sum_{k=1}^n A_k \cos(kx) + 1\\ =\ & 2\sum_{k=1}^n \cos (kx) - \frac{2}{n+1}\sum_{k=1}^n k \cos (kx) + 1\\ =\ & 2\sum_{k=1}^n \cos (kx) - \frac{2}{n+1}\left(\sum_{k=1}^n \sin (kx)\right)' + 1\\ =\ & 2\left(\frac{-\sin \frac{x}{2} + \sin\frac{(2n+1)x}{2}}{2\sin \frac{x}{2}}\right) - \frac{2}{n+1}\left(\frac{\cos \frac{x}{2} - \cos \frac{(2n+1)x}{2}}{2\sin \frac{x}{2}}\right)' + 1\\ =\ & \frac{\sin x \cos(nx) + \cos x \sin (nx) - \sin x + \sin(nx)}{\sin x}\\ &\quad - \frac{2}{n+1}\left(\frac{1 + \cos x - \cos(nx) - \cos x \cos (nx) + \sin x \sin (nx)}{2\sin x}\right)' + 1\\ =\ & \frac{1 - \cos((n+1)x)}{(n+1)(1-\cos x)}\\ \ge\ & 0. \end{align}

Update: We can find $$A_k = 2 - \frac{2k}{n+1}$$ through the following steps.

Note that when $$A_k$$ is a polynomial in $$k$$, $$\sum_{k=1}^n A_k \cos(kx)$$ has a closed form.

We first try $$A_k = a_n + b_n k$$.

From $$\sum_{k=1}^n A_k = a_n n + \frac{b_n n (n+1)}{2} = n$$, we have $$a_n + \frac{b_n (n+1)}{2} = 1$$. Thus, $$b_n = \frac{2(1-a_n)}{n+1}$$.

We need \begin{align} \sum_{k=1}^n \left(a_n + \frac{2(1-a_n)}{n+1}k\right) \cos(kx) + 1\ge 0, \ \forall x \in (0, 2\pi). \tag{1} \end{align}

When $$n=2$$, (1) becomes $$\frac{1 - \cos x}{3}(1 + 2\cos x)\Big(a_2 + \frac{4\cos x - 1}{1 - \cos x}\Big) \ge 0, \ \forall x \in (0, 2\pi). \tag{2}$$ It is easy to prove that (2) holds if and only if $$a_2 = 2$$.

When $$n=3$$, (1) becomes $$2\cos x (1 - \cos^2 x)\Big(a_3 + \frac{3\cos x - 2}{1 - \cos x}\Big) \ge 0, \ \forall x \in (0, 2\pi). \tag{3}$$ It is easy to prove that (3) holds if and only if $$a_3 = 2$$.

We guess that when $$a_n = 2$$, (1) holds. Fortunately, it works.

• How to find this $A_{k}=?$ Feb 2, 2021 at 13:18
• @inequality I Guess: Note that when $A_k$ is a polynomial in $k$, the sum has a closed form, since it is well-known that $\sum_{k=1}^n \cos (kx)$ and $\sum_{k=1}^n \sin (kx)$ both have closed form, so do $\sum_{k=1}^n k^m\cos (kx)$ and $\sum_{k=1}^n k^m\sin (kx)$. We first try $A_k = a + bk$. From $\sum_{k=1}^n A_k = an + \frac{b(n+1)n}{2} = n$, we have $a + \frac{b(n+1)}{2} = 1$. $\cdots$ Feb 2, 2021 at 14:39