Prove $\sum^{n+1}_{j=1}\left|\cos\left(j\cdot x\right)\right|\geqslant \frac n4$

Question

How do you prove that $\displaystyle\sum^{n+1}_{j=1}\left|\cos\left(j\cdot x\right)\right|\geqslant \dfrac{n}{4}$, where $x\in\mathbb{R}$?

I tried mathematical induction, but it doesn't work. I also have some ideas on complex number solution, such as setting $z=\cos\left(x\right)+i\sin\left(x\right)$ and working on $Re\left(z\right)$, but neither does that work.

Can $x$ be any number? It does look like induction may be helpful but I haven't tried it myself. — SBK, Oct 18 '22 at 14:14
This may help $\sum^n_{j=1}\cos jx =-\tfrac12 +\frac12\frac{\sin(2n+1)x/2}{\sin x/2}$ — Mittens, Oct 18 '22 at 14:31
@Bruno your counterexample fails because for example we can already have $\sum_{j=1}^{n+1} |\cos(jx)| \geq \frac{n+1}{4}$; from the inequality $A(n) \ge n/4$ and $A(n)+a_{n+1} \ge (n+1)/4$ we cannot deduce anything really about $a_{n+1}$ since $A(n)$ can be much bigger than $n/4$ — Conrad, Oct 18 '22 at 15:03
For the one who upvoted, I just found a silly blunder in my answer so I deleted it. — Math The Novice, Oct 18 '22 at 16:02
Just for closure: Oliver Diaz's approach would sadly not work since you basically show the reverse, as in, you show that for $x \not\in \pi\mathbb{Z}$, $(\sum^n_{j=1}\cos(jx))_n$ is bounded, which is true but doesn't make progress for our result. — Bruno B, Oct 18 '22 at 16:03
@MathTheNovice I think it was only a $2$ being too much in the second sum, otherwise it was fine? — Bruno B, Oct 18 '22 at 16:04
@BrunoB $\sum^{n+1}{j=1}\vert \cos(jx)\vert + \frac{1}{2} \sum^{n+1}{j=1} \vert \cos(2jx)\vert$ may not $\geq 3M_n/2$ since their minimums are at different points — Math The Novice, Oct 18 '22 at 16:07
I have posted a question about a conjecture which (if true) would lead to a simple solution to this problem: https://math.stackexchange.com/q/4556366/42969 — Martin R, Oct 19 '22 at 07:55

score 15 · Accepted Answer · edited Oct 20 '22 at 13:36

Note that $|\cos x|+|\cos 2x| \geqslant \dfrac{\sqrt{2}}{2} $ with equality at $x=\pi/4$ (splitting into $0 \leqslant x \leqslant \pi/4, \pi/4 \le x \le \pi/2$ and expliciting one gets a quadratic in $\cos x$ etc) so we can assume $n \ge 3$ since $\sqrt 2/2 \ge n/4$ for $n=1,2$

Now $$\sum^{n+1}_{j=1}|\cos(jx)| \ge \sum^{n+1}_{j=1}\cos^2(jx)=\frac{n+1}{2}+\frac{1}{2}\sum^{n+1}_{j=1}\cos(2jx)$$

Now $\sum^{n+1}_{j=1}\cos(2jx)=(\frac{\sin ((2n+3)x)}{\sin x}-1)/2$ and we know that $\sin x \ge \frac{2x}{\pi}$ so $$|(\frac{\sin ((2n+3)x)}{\sin x}-1)/2| \le \frac{1}{2}(\frac{\pi}{2x}+1)$$

Hence if $\frac{\pi}{2x}+1 \le n+2$ or $x \ge \frac{\pi}{2n+2}$ we have $$\sum^{n+1}_{j=1}|\cos(jx)| \ge \frac{n+1}{2}-\frac{1}{2}|(\frac{\sin ((2n+3)x)}{\sin x}-1)/2| \ge \frac{n+1}{2}-\frac{n+2}{4}=\frac{n}{4}$$

If $x \le \frac{\pi}{2n+3}$ then $(2n+3)x \le \pi$ so $\frac{\sin ((2n+3)x)}{\sin x} \ge 0$ hence $$\sum^{n+1}_{j=1}|\cos(jx)| \ge \frac{n+1}{2}-\frac{1}{4}\geqslant \frac{n}{4}$$

hence assume $\frac{\pi}{2n+3} \le x \le \frac{\pi}{2(n+1)}$ then all $\cos jx \geqslant 0, j=1,..n+1$ since $(n+1)x \leqslant \pi/2$ hence:

$$\sum^{n+1}_{j=1}|\cos(jx)|=\sum^{n+1}_{j=1}\cos(jx)=(\frac{\sin ((n+3/2)x)}{\sin x/2}-1)/2$$

$\frac{\pi}{2}\le (n+3/2)x \le \frac{\pi}{2}+\frac{\pi}{4n+4}$ so $\sin ((n+3/2)x) \ge \cos \frac{\pi}{4n+4} \ge 1- (\frac{\pi}{4n+4})^2/2 $ while $\sin x/2 \le x/2 \le \frac{\pi}{4n+4}$ so $\frac{1}{ \sin x/2} \ge \frac{4n+4}{\pi}$, hence

$$2\sum^{n+1}_{j=1}|\cos(jx)| \ge \frac{4n+4}{\pi}(1- (\frac{\pi}{4n+4})^2/2 ) -1 \ge \frac{4n}{\pi}+4/\pi - \pi/32-1\ge \frac{n}{2}$$ since $n \ge 3$, so we are done!

Now $\int_0^{\pi/2}|\cos jx|dx=1, j \ge 1$ so $\int_0^{\pi/2}\sum^{n+1}_{j=1}|\cos(jx)|dx=n+1$ which suggests that the average sum is about $\frac{2n+2}{\pi}$ hence the lower bound $n/4$ most likely can be improved

It may be completely unrelated, but the problem (and your final remark) reminds me about https://math.stackexchange.com/q/2476812/42969, where an averaging technique is used to prove an inequality about complex numbers (which also involves the factor $1/\pi$). — Martin R, Oct 18 '22 at 17:39
@Martin I agree with that and maybe something along those lines can be obtained; if one puts $z_k=e^{ikx}$ one gets that a subsum of $|\sum \cos jx + i \sum \sin jx|$ for some indices is at least $\frac{n+1}{\pi}$ so if we can arrange that the $\cos$ part is large, it would work — Conrad, Oct 18 '22 at 17:43
For $n>3$, it appears $\sum_{j=1}^{2n}|\cos jx|\ge n$ where equality is achieved at odd multiples of $\pi/2$. — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Oct 18 '22 at 17:44
@TheSimpliFire wouldn't surprise me in the least as the sum is definitely bigger than $n/4$; maybe a careful analysis of the Chebyshev polynomials works to give the better bound (since the average is large $2(n+1)/\pi$ which is more than $5n/8$ and the Chebyshev polynomials are fairly flat) — Conrad, Oct 18 '22 at 17:45
@Conrad I've edited my comment to reflect the observation that the $-1/2$ term doesn't seem necessary for $n>3$. It seems Chebyshev is the way to go (maybe we could also bound using Weierstrass approximation and compare Fourier coefficients). — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Oct 18 '22 at 17:56
For $x=\frac{c \pi}{2n+3}$ with $c>1$ and close to $1$ fixed clearly gives $\frac{\sin ((2n+3)x)}{\sin x}$ on the order of $-\frac{2n}{c\pi}$ so the method above cannot do better for large $n$ than $n(1/2-1/(2\pi))$ if we let $c \to 1$ — Conrad, Oct 18 '22 at 18:06

Martin R · Answer 2 · 2022-10-22T15:03:38.107

It has been shown in

A lower bound on the count of numbers in an arithmetic progression which are close to an integer

that for fixed $x \in \Bbb R$, at least $\lfloor (n+1)/2 \rfloor$ indices $j \in \{ 1, 2, \ldots, n+1\}$ have the property that $$ |j \frac x\pi - k| \le \frac 13 \text{ for some integer $k$}. $$ For those indices $j$ is $|\cos(j x) | \ge \frac 12$. It follows that $$ \sum_{j=1}^{n+1} |\cos(j x) | \ge \left\lfloor \frac{n+1}2 \right\rfloor \frac 1 2 \ge \frac n 4 \, . $$

Prove $\sum^{n+1}_{j=1}\left|\cos\left(j\cdot x\right)\right|\geqslant \frac n4$

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