# Better trigonometrical inequalities for $\zeta(s)$?

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 $$a_{0} < \frac{3}{4} a_{i_0}$$, $$b_0=0$$?

• $1+\cos \theta \ge 0$; $99+100 \cos \theta +\cos 2\theta \ge 0$; $98+100\cos \theta +\cos (2\theta) + \cos (4\theta) \ge 0$ (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 ... – Lucia Feb 5 '20 at 18:58
• I should have added that $a_0<a_{i_0}$. What would be useful would be $a_0< 3 a_{i_0}/4$, actually. – H A Helfgott Feb 5 '20 at 19:12
• 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 $a_0=1<3a_{i_0}/4=9/2$. – Carlo Beenakker Feb 5 '20 at 19:29
• I should have said $b_0=0$. – H A Helfgott Feb 5 '20 at 19:29
• @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? – H A Helfgott Feb 5 '20 at 19:46

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. :-)

• 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). – H A Helfgott Feb 6 '20 at 8:12

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 $$\tfrac{1}{2}k(k+1)+\sum_{n=1}^k n \cos n\theta\geq 0.$$