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Fix integers $m,n\geq0$.

Do we have the inequality $\displaystyle\sum_{a=0}^m\sum_{b=0}^n\cos(abx)>0$ for all $x\in\mathbb{R}$?

We can also write this function as \begin{align*} \sum_{a=0}^m\sum_{b=0}^n\cos(abx)&=m+n+1+\sum_{a=1}^m\sum_{b=1}^n\cos(abx)\\ &=m+n+1+\sum_{a=1}^m\frac{1}{2}\left(\frac{\sin((n+1/2)ax)}{\sin(ax/2)}-1\right)\\ &=\frac{m}{2}+n+1+\frac{1}{2}\sum_{a=1}^mD_n(ax), \end{align*} where $$D_n(x)=\frac{\sin((n+1/2)x)}{\sin(x/2)}$$ is the Dirichlet kernel (up to a factor of $2\pi$, depending on your convention).

Using this formula, it is easy to check the conjecture for small values of $m$ and $n$ (desmos link).

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  • $\begingroup$ Did you try to obtain a formula for the sum of Dirichlet kernels using $\sin t = \Re e^{it}$ and the formula for geometric sums? $\endgroup$
    – Gary
    Oct 18, 2021 at 3:53
  • $\begingroup$ I don't think that there is a closed form formula for the sum of Dirichlet kernels. The denominators make things tricky. You could phrase this whole question in terms of the real part of $\sum\sum e^{abx}$, which makes it easier to see why the denominators prevent you from applying the geometric series formula twice. $\endgroup$ Oct 18, 2021 at 4:02
  • $\begingroup$ Yes, sorry I missed the denominator. $\endgroup$
    – Gary
    Oct 18, 2021 at 4:16
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    $\begingroup$ Any information about this question (e.g., source)? $\endgroup$
    – River Li
    Oct 18, 2021 at 6:16
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    $\begingroup$ @RiverLi It arose in the context of this mathoverflow question: mathoverflow.net/questions/405593 $\endgroup$ Oct 18, 2021 at 16:41

1 Answer 1

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No! Consider the case $m = n$ and $x = \frac{8π}{4n + 3}$. We have

\begin{split} \sum_{a=0}^n \sum_{b=0}^n \cos(abx) &= \frac{3n}{2} + 1 + \frac12 \sum_{a=1}^n \frac{\sin\left(\left(n + \frac12\right)ax\right)}{\sin\left(\frac12ax\right)} \\ &= \frac{3n}{2} + 1 + \frac12 \sum_{a=1}^n \frac{\sin\left(2πa - \frac{2πa}{4n + 3}\right)}{\sin\left(\frac{4πa}{4n + 3}\right)} \\ &= \frac{3n}{2} + 1 + \frac12 \sum_{a=1}^n \frac{\sin\left(-\frac{2πa}{4n + 3}\right)}{\sin\left(\frac{4πa}{4n + 3}\right)} \\ &= \frac{3n}{2} + 1 - \frac14 \sum_{a=1}^n \sec\left(\frac{2πa}{4n + 3}\right) \\ &< \frac{3n}{2} + 1 - \frac14 \int_0^n \sec\left(\frac{2πa}{4n + 3}\right)\,da \\ &= \frac{3n}{2} + 1 + \frac{4n + 3}{8π} \ln \tan \frac{3π}{4(4n + 3)} \\ &\sim \frac{3n}{2} + 1 + \frac{4n + 3}{8π} \ln \frac{3π}{4(4n + 3)} \\ &→ -∞ \quad \text{as $n → ∞$}. \end{split}

We can confirm this by plotting the exact sum at $m = n$ and $x = \frac{8π}{4n + 3}$ (blue); it first becomes negative at $n = 3286$. By optimizing $x$ near $\frac{8π}{4n + 3}$ to minimize the sum (orange), we find a slightly earlier negative value at $m = n = 3161$ and $x = 0.001987239$.

plot

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  • $\begingroup$ excellent counterexample! $\endgroup$
    – River Li
    Oct 24, 2021 at 1:03
  • $\begingroup$ Incredible! I had no idea! $\endgroup$ Oct 24, 2021 at 1:12
  • $\begingroup$ From this approach, we can derive an infinite family of counterexamples when $m=n$. In particular, any function $f(n)$ such that $1/f(n)\in(n+1/2,n+1)$ for all positive $n$ yields a counterexample when $x=2\pi f(n)$. $\endgroup$
    – TheSimpliFire
    Oct 24, 2021 at 11:19
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    $\begingroup$ Wow. ${}{}{}{}{}$ $\endgroup$ Oct 24, 2021 at 12:31
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    $\begingroup$ @TheSimpliFire Not quite. $x = \frac{4π}{2n + 1}$ and $x = \frac{2π}{n + 1}$ do not lead to counterexamples, so neither does any function sufficiently close to them. You need a stronger condition like $\frac{2π}{x} ∈ (n + \frac12 + ε, n + 1 - ε)$ for some constant $ε > 0$. (But the math is uniquely simple for $x = \frac{8π}{4n + 3}$, which allows the double-angle cancellation on line 4.) $\endgroup$ Oct 24, 2021 at 18:56

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