The inequality $$3 + 4 \cos \theta + \cos 2 \theta \geq 0$$ plays a key role in the proof of the classical zero-free region of the Riemann zeta function. Are there other inequalities of the form $$\sum_{i=0}^k a_i \cos b_i \theta \geq 0,\;\;\;\;\;a_\geq 0$$ such that $a_{i_0} = \sum_{i\ne i_0} a_i$ for some $0\leq i_0\leq k$ and $a_{0} < \frac{3}{4} a_{i_0}$, $b_0=0$?
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asked Feb 5, 2020 at 17:27
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3$\begingroup$ $1+\cos \theta \ge 0$; $99+100 \cos \theta +\cos 2\theta \ge 0$; $98+100\cos \theta +\cos (2\theta) + \cos (4\theta) \ge 0$ (didn't check the last one, but surely it's correct). Depends on what you're after? For zero free regions this is not the relevant criterion (one coefficient being sum of the others) -- presumably you've already looked at Kadiri, Stechkin ... $\endgroup$– LuciaCommented Feb 5, 2020 at 18:58
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$\begingroup$ I should have added that $a_0<a_{i_0}$. What would be useful would be $a_0< 3 a_{i_0}/4$, actually. $\endgroup$– H A HelfgottCommented Feb 5, 2020 at 19:12
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$\begingroup$ apologies, but I do not understand what difference the condition you added makes; in the example I gave in the answer box I have $k=i_0=3$, $a_0,a_1,a_2=1,2,3$, $b_0,b_1,b_2=1,2,3$, and $a_{3}=6$, $b_3=0$, so $a_0=1<3a_{i_0}/4=9/2$. $\endgroup$– Carlo BeenakkerCommented Feb 5, 2020 at 19:29
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$\begingroup$ I should have said $b_0=0$. $\endgroup$– H A HelfgottCommented Feb 5, 2020 at 19:29
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$\begingroup$ @Lucia: what I have in mind is actually just to improve explicit bounds on $1/\zeta(\sigma+it)$ for $\sigma>1$ (so as to improve explicit bounds on $1/\zeta(1+it)$). Or would a better zero-free region necessarily follow from an inequality such as the one I request? $\endgroup$– H A HelfgottCommented Feb 5, 2020 at 19:46
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2 Answers
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Assuming the $b_i$ are all distinct (or at least non-zero for $i \neq 0$), this is not possible. (Otherwise there are trivial examples, e.g. $1 + 2 \cos(0 \theta)+ \cos(0 \theta) \geq 0$ or $1 + 4 \cos \theta + \cos(2\theta) + 2 \cos(0 \theta) \geq 0$.)
Suppose that $\sum_{i=0}^k a_i \cos b_i \theta \geq 0$. Since $a_{i_0} = \sum_{i \neq i_0} a_i$, this implies that whenever $\cos b_{i_0} \theta = -1$, one must have $\cos b_i \theta = +1$ for all other $i$. In particular, the other $b_i$ must be integer multiples of $2b_{i_0}$. We now have
$$ a_0 + a_{i_0} \cos b_{i_0} \theta + \sum_{i \neq 0, i_0} a_i \cos b_i\theta \geq 0$$
with the $b_i$ in the sum nonzero integer multiples of $2b_{i_0}$. Performing a Taylor expansion around $\theta = \pi / b_{i_0}$ to second order, we conclude that
$$ - a_{i_0} \frac{b_{i_0}^2}{2} + \sum_{i \neq 0, i_0} a_i \frac{b_i^2}{2} \geq 0$$
and hence (since $b_i^2 \geq 4 b_{i_0}^2$ and $b_{i_0} \neq 0$)
$$ \sum_{i \neq 0, i_0} a_i \leq \frac{1}{4} a_{i_0}$$
or equivalently
$$ a_0 \geq \frac{3}{4} a_{i_0}.$$
Thus one cannot have $a_0 < \frac{3}{4} a_{i_0}$. This argument also shows that up to rescaling and other trivial rearrangements, Mertens' inequality $3 + 4 \cos(\theta)+\cos(2\theta) \geq 0$ is the unique inequality that attains $a_0 = \frac{3}{4} a_{i_0}$.
At a more metamathematical level, if there were a variant of Mertens' trigonometric inequality that gave superior numerical results towards the classical zero free region, I would imagine that this would already have been noticed by now. :-)
answered Feb 5, 2020 at 20:43
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1$\begingroup$ Nice. I had a different and much more modest possible application in mind, but, as I made my question more precise, I started seeing that it could give a numerically better zero-free region (and hence the answer was almost certainly "no", for meta-mathematical reasons, as you put it). $\endgroup$ Commented Feb 6, 2020 at 8:12
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this is an answer to the question as originally posed, without the additional conditions on $a_0$ and $b_0$
for example, $$6+\cos \theta+2 \cos 2 \theta+3 \cos 3 \theta\geq 0,$$ or more generally $$\tfrac{1}{2}k(k+1)+\sum_{n=1}^k n \cos n\theta\geq 0.$$
answered Feb 5, 2020 at 19:00