Trigonometric inequality

Question

For odd and coprime positive integers $p$ and $q$, the following inequality holds: $$\sum_{m=1}^{p} \sum_{n=1}^{q} \frac{2}{\cos(\frac{2m\pi}{p})+\cos(\frac{2n\pi}{q})} \le pq(|p-q|+1)$$

Unfortunately, the problem has no solution at MSE.

Answer 1

I will continue from the point where Christophe Leuridan's answer stopped. Denote our sum $S$, then $$ S=2p\sum_{n=1}^q \frac{\sin(2\pi np/q)}{(1+\cos(2πnp/q))\sin(2\pi n/q)}. $$ For $n=q$ this is undefined, but the limit of our summand is $p/2$. Hence $S=p^2+2pT$ with $$ T=\sum_{n=1}^{q-1} \frac{\sin(2\pi np/q)}{(1+\cos(2\pi np/q))\sin(2\pi n/q)}. $$ Next, assume that $p<q$ and $q-p=2h$. We want to prove $S\leq pq(2h+1)$. This is the same as $p^2+2pT\leq pq(2h+1)$, i.e. $T\leq h(q+1)$. Since $\sin(2u)=2\sin(u)\cos(u)$, $1+\cos(2u)=2\cos^2(u)$, we see that $$ T=\sum_{n=1}^{q-1} \frac{\tan(\pi np/q)}{\sin(2\pi n/q)}. $$ Next, $\tan(\pi np/q)=-\tan(2\pi hn/q)$. Let $\zeta_q=\exp(2\pi i/q)$. Then we get $$ \frac{\tan(\pi np/q)}{\sin(2\pi n/q)}=-\frac{\zeta_q^{nh}-\zeta_q^{-nh}}{\zeta_q^n-\zeta_q^{-n}}\cdot\frac{2}{\zeta_q^{nh}+\zeta_q^{-nh}}. $$ Expand the two fractions on the right-hand side into finite geometric series: \begin{align*} \frac{\zeta_q^{nh}-\zeta_q^{-nh}}{\zeta_q^n-\zeta_q^{-n}} &=\zeta_q^{-n(h-1)}\sum_{a=0}^{h-1}\zeta_q^{2na},\\ \frac{2}{\zeta_q^{nh}+\zeta_q^{-nh}} &=\zeta_q^{nh}\sum_{b=0}^{q-1}(-1)^b\zeta_q^{2bnh}. \end{align*} It follows that $$ T=-\sum_{n=1}^{q-1}\zeta_q^{n}\sum_{\substack{0\leq a\leq h-1\\ 0\leq b\leq q-1}}(-1)^b\zeta_q^{2na+2bnh}. $$ Switching the order of summation and using the fact that sum of $\zeta_q^{nk}$ over $1\leq n\leq q-1$ is $-1$ for $q\nmid k$ and $q-1$ for $q\mid k$, we see that $$ T=-q\sum_{\substack{0\leq a\leq h-1\\0\leq b\leq q-1\\2a+2bh\equiv -1 \pmod q}}(-1)^b+h\sum_{0\leq b\leq q-1}(-1)^b. $$ The second sum is equal to $1$ and the first one is at least $-h$, since $h$ and $q$ are coprime (hence for fixed $a$ there is a unique $b$). Therefore, $T\leq (-q)(-h)+h=q(h+1)$, as needed.

Added by GH from MO. The result of the above calculation can be expressed in the more symmetric form $$\frac{1}{pq}\sum_{m=1}^{p} \sum_{n=1}^{q} \frac{2}{\cos(\frac{2m\pi}{p})+\cos(\frac{2n\pi}{q})}=1-2\sum_{\substack{0\leq a\leq h-1\\0\leq b\leq q-1\\pb\equiv 2a+1 \pmod q}}(-1)^b.$$

Answer 2

There is a short way to take out trigonometry and remain with the sum of $\pm 1$, similar (but not immediately equivalent, see later) to these appearing in the Christophe Leuridan and Alexander Kalmynin's solution: \begin{align*}\sum_{m=1}^{p} \sum_{n=1}^{q} \frac{2}{\cos(\frac{2m\pi}{p})+\cos(\frac{2n\pi}{q})} &=\sum_{x^q=y^p=1}\frac4{x+1/x+y+1/y}\\[4pt] &=\sum_{x^q=y^p=1}\frac{4x}{(1+x/y)(1+xy)}\\[4pt] &=\sum_{x^q=y^p=1}\frac{x(1+(x/y)^{pq})(1+(xy)^{pq})}{(1+x/y)(1+xy)}\\[4pt] &=\sum_{x^q=y^p=1}x\sum_{a=0}^{pq-1}(-1)^a(x/y)^a\sum_{b=0}^{pq-1}(-1)^b(xy)^b\\[4pt] &=pq\sum_{\substack{0\leqslant a,b\leqslant pq-1\\q|1+a+b,\ p|b-a}}(-1)^{a+b}, \end{align*} since the summation of a single monomial $x^A y^B$ over $\{x^q=y^p=1\}$ gives $pq$ when $q|A$ and $p|B$ and 0 otherwise.

Thus, an equivalent reformulation of the inequality is $$ M:=\sum_{\substack{0\leqslant a,b\leqslant pq-1\\q|1+a+b,\ p|b-a}}(-1)^{a+b}\leqslant |p-q|+1.\tag{$\heartsuit$} $$ I include the proof of $(\heartsuit)$, it is short and elementary but, as for me, not trivial. Without Sasha's answer I would think longer about this. We may assume (from the very beginning) that $q>p$. Call a pair $(a,b)\in \{0,1,\ldots,pq-1\}^2$ appropriate, if $q|1+a+b$ and $p|b-a$, and denote the set of appropriate pairs by $\Theta$.

We partition the sum over appropriate pairs in $(\heartsuit)$ onto three parts, over the sets $\Omega_1,\Omega_2,\Omega_3$, where \begin{align*} \Omega_1&:=\{(a,b)\in \Theta:(q-p)/2\leqslant a\leqslant pq-1,(q+p)/2\leqslant b\leqslant pq-1\}\\ \Omega_2&:=\{(a,b)\in \Theta:0\leqslant a<(q-p)/2\}\\ \Omega_3&:=\{(a,b)\in \Theta:0\leqslant b<(q+p)/2\}. \end{align*} Note that the sets $\Omega_1$, $\Omega_2$ and $\Omega_3$ are disjoint (for a pair of integers $(a,b)$ enjoying both inequalities $0\leqslant a<(q-p)/2$ and $0\leqslant b<(q+p)/2$ we have $0<a+b+1<q$, thus it is not appropriate).

The main idea is that on $\Omega_1$ there exists a sign-reversing involution $$(a,b)\mapsto (pq-1+(q-p)/2-a,pq-1+(q+p)/2-b),$$ which preserves being an appropriate pair but changes the parity of $a+b$.

On $\Omega_2$, we must have $b=q-1-a+m\cdot q$ for $m\in \{0,1,\ldots,p-1\}$ (since $b\equiv q-1-a\pmod q$) and $m=m(a)$ is chosen so that $q-1-2a+m(a)q$ is divisible by $p$, there exists unique such $m(a)$, and $(-1)^{a+b}=(-1)^{m(a)}$.

Analogously, on $\Omega_3$, we must have $a=q-1-b+m(b)\cdot q$, and $(-1)^{a+b}=(-1)^{m(b)}$. Thus, the sum over $\Omega_2$ and $\Omega_3$ is nothing but $$\sum_{a=0}^{(q-p)/2-1}(-1)^{m(a)}+\sum_{b=0}^{(q+p)/2-1}(-1)^{m(b)}.$$ Now note that $m((q-1)/2)=0$ and for $0\leqslant a<q$, $a\ne (q-1)/2$, we have $m(a)+m(q-1-a)=p$. Thus, the sum of $(-1)^{m(a)}$ for $a\in [(q-p)/2,(q+p)/2-1]$ (this set is symmetric with respect to $(q-1)/2$, so we may partition it onto $\{(q-1)/2\}$ and pairs with sum $q-1$) equals $1$. Therefore, $$ M=1+2\sum_{a=0}^{(q-p)/2-1} (-1)^{m(a)}\leqslant 1+(q-p), $$ with equality of and only if $m(a)$ is even for all $a=0,1,\ldots,(q-p)/2-1$.

Answer 3

I give a way that may work for odd numbers. This is too long for a comment.

First, the quantity $$\cos\Big(\frac{2m\pi}{p}\Big)+\cos\Big(\frac{2n\pi}{q}\Big) = 2\cos\Big(\pi\Big(\frac{m}{p}+\frac{n}{q}\Big)\Big) \cos\Big(\pi\Big(\frac{m}{p}-\frac{n}{q}\Big)\Big)$$ cannot vanish since $p$ and $q$ are coprime. Hence the sum is well-defined.

Call $T_p$ the Chebyshev polynomial, defined by the equality $T_p(\cos(\theta)) = \cos(p\theta)$ for all real numbers $\theta$. Then for all integers $m$, $$T_p\Big(\cos\Big(\frac{2m\pi}{p}\Big)\Big) = T_p(\cos(2m\pi)) = T_p(\cos(0)) = 1.$$
If $p$ is odd, the real numbers $\cos(2m\pi/p)$ for $0 \le m \le (p-1)/2$ are distinct roots of $T_p$. The multiplicity is $2$ if $1 \le m \le (p-1)/2$, and $1$ if $m=0$. Therefore, we have all roots. Since $T_p$ has leading term $2^{p-1}X^p$, we derive $$T_p - 1 = 2^{p-1}(X-1)\prod_{m=1}^{(p-1)/2}\Big(X-\cos\Big(\frac{2m\pi}{p}\Big)\Big) = 2^{p-1} \prod_{m=1}^p\Big(X-\cos\Big(\frac{2m\pi}{p}\Big)\Big).$$ If $q$ is odd, a similar formula holds for $q$. Given an integer $n \in [1,q]$, one has $$T_p\Big(X-\cos\Big(\frac{2n\pi}{q}\Big)\Big) - 1 = 2^{p-1} \prod_{m=1}^p\Big(X-\cos\Big(\frac{2m\pi}{p}\Big)-\cos\Big(\frac{2n\pi}{q}\Big)\Big).$$ Thus $$\prod_{n=1}^q\Big( T_p \Big(X-\cos\Big(\frac{2n\pi}{q}\Big)\Big) - 1 \Big) = 2^{(p-1)q} \prod_{m=1}^p \prod_{n=1}^q \Big(X-\cos\Big(\frac{2m\pi}{p}\Big)-\cos\Big(\frac{2n\pi}{q}\Big)\Big).$$ Call $c_{pq}X^{pq} + \cdots + c_0$ this polynomial. The sum to be estimated is the sum of the reciprocal of its roots, which is $-c_1/c_0$. So we are lead to estimate the coefficients of this polynomial...

Another strategy is to write for each integer $n \in [1,q]$, $$\frac{T'_p\Big(X-\cos\Big(\frac{2n\pi}{q}\Big)\Big)}{T_p\Big(X-\cos\Big(\frac{2n\pi}{q}\Big)\Big) - 1} = \sum_{m=1}^p\frac{1}{X-\cos\Big(\frac{2m\pi}{p}\Big)-\cos\Big(\frac{2n\pi}{q}\Big)},$$ so $$\sum_{m=1}^p\frac{1}{\cos\Big(\frac{2m\pi}{p}\Big)+\cos\Big(\frac{2n\pi}{q}\Big)} = \frac{T'_p\Big(-\cos\Big(\frac{2n\pi}{q}\Big)\Big)}{1 - T_p\Big(-\cos\Big(\frac{2n\pi}{q}\Big)\Big)} = \frac{T'_p\Big(\cos\Big(\frac{2n\pi}{q}\Big)\Big)}{1 + T_p\Big(\cos\Big(\frac{2n\pi}{q}\Big)\Big)},$$ since $T_p$ is an odd function.

Using the equalities $T_p(\cos(\theta)) = \cos(p\theta)$ and $-T'_p(\cos(\theta))\sin\theta = - p \sin(p\theta)$, one derives $$\sum_{m=1}^p\frac{1}{\cos\Big(\frac{2m\pi}{p}\Big)+\cos\Big(\frac{2n\pi}{q}\Big)} = p \frac{\sin(2np\pi/q)}{\sin(2n\pi/q)} \times \frac{1}{1+\cos(2np\pi/q)}.$$ It remains to sum over all integers $n \in [1,q]$ and to bound above...