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The Problem

Let $r,q,m$ be positive integers such that $4 \leq r$ and $1<m,q\leq r/2$. Is it the case that

$$\left | \sum_{k=0}^{q-1} \zeta^{km}\right | < \left |\sum_{k=0}^{q-1} \zeta^{k}\right |,$$

where $\zeta$ is a primitive $r^{th}$ root of unity.

Continuous Variant

Through use of a non-standard trigonometric identity which may be found here we can change the previous inequality to

$$|\sin(mq\pi/r)\sin(\pi/r)| < |\sin(m\pi/r)\sin(q\pi/r)|.$$

Then we may substitute real variables. Let $\theta \in (0,\pi/4]$ and $a,b \in \mathbb R$ so that $0< a\theta, b \theta \leq \pi$. The we have the inequality

$$| \sin(ab \theta) \sin(\theta) |< | \sin(a \theta)\sin(b \theta) |. $$

Fix $\theta$ and consider the function

$$f(x,y)=\frac{\sin(xy\theta)\sin(\theta)}{\sin(x\theta)\sin(y\theta)}$$

where $(x,y) \in [1,\pi/(2\theta)]^2=X$. So if we can show that $f$ has a unique maximum on $X$ at $(1,1)$ we would also have our result. One can show with a somewhat painful computation that the critical points of $f$ are on the line $x=y$, so we can consider instead the function

$$g(x)=\frac{\sin(x^2\theta)\sin(\theta)}{\sin^2(x\theta)}$$

and the corresponding inequality

$$|\sin(x^2\theta)\sin(\theta)|<|\sin^2(x\theta)|.$$

I have some partial results in this case where essentially I can control the inequality depending on where $x^2\theta$ is modulo $2\pi$. Which comes down to everything except when the residue of $x^2\theta$ modulo $2\pi$ is in the interval $(2x\theta-\theta,2x\theta)$.

Motivation

Let the generalized binomial coefficient $C_q(n,k)$ be the coefficient of $x^k$ in the polynomial $(1+x+\cdots+x^{q-1})^k$, sometimes this appears in the literature as $\binom{n}{k}_q$. Define

$$PC_q(n,r,k)=\sum_{j \in \mathbb Z} C_q(n,k+rj)$$

then the sums in the statement of the problem arise naturally as the coefficients of a discrete Fourier expansion of $PC_q$. In particular proving the desired inequality shows that $PC_q$ behaves as a sine function as $n \rightarrow \infty$. There's some more background on $C_q$ and $PC_q$ is this answer.

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1 Answer

up vote 3 down vote accepted
+75

Actually, I believe something a bit stronger is true: For any $S \subseteq \{1,\dots,r\}$ with $|S|=q$, we have $$ (*) |\sum_{j \in S} \zeta^j| \leq |\sum_{k=0}^{q-1} \zeta^k|.$$

Consider an arbitrary $S$, and let $v$ be the unit vector in the direction of $\sum_{j \in S} \zeta^j$ (here and later on we're thinking of the complex numbers as a $2$-dimensional real vectors). Let $S'$ be the set consisting of the $q$ roots of unity having largest (positive) projection onto $v$. Then we have \begin{eqnarray*} |\sum_{j \in S} \zeta^j| &=& proj_{v}\left(\sum_{j \in S} \zeta^j\right) \\ &=& \sum_{j \in S} proj_v \left(\zeta^j\right) \\ &\leq& \sum_{j \in S'} proj_v \left(\zeta^j\right) \\ &=& proj_v \left(\sum_{j \in S'} \zeta^j \right) \\ &\leq& | \sum_{j \in S'} \zeta^j | \end{eqnarray*} But for any $v$ the set $S'$ consists of $q$ consecutive roots of unity, so the last expression is the right hand side of (*).


If equality holds, then $S$ must consist of $q$ consecutive residues modulo $r$. In other words, the set $\{1, m, 2m, \dots, (q-1)m\}$ is some permutation of $\{a,a+1,\dots,a+q-1\}$ (modulo $r$) for some $a$ and $q$. Let us fix some $m,a,q$ for which this occurs.

For $1 \leq j \leq q$, let $f(j)$ be the integer between $0$ and $q-1$ such that $mj$ is congruent to $a+f(j)$ modulo $r$. One of two things must happen:

-If $f(j)=j$ for all $j$, then $a=m=1$, not allowed.

-Otherwise, there is some $j$ for which $f(j)>f(j+1)$. But this would imply $m(j+1)-mj \geq a+r-(a+q-1)$, or, rearranging, $m+q \geq r+1$, which is also not allowed.

Essentially what the second case corresponds to is that your condition on $m$ and $q$ versus $r$ makes it impossible to "wrap around" from one end of the interval to the other between two consecutive multiples of $m$.

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Could you prove your last statement for the benefit of slow readers like me? (that $S^prime$ consists of $q$ consecutive roots) –  Alexander Gruber Feb 15 '13 at 0:20
    
Since all roots of unity have the same magnitude, the roots with largest projection onto $v$ are those whose direction lies closest to $v$. So if you take an arc centered at $v$ and growing in size, then $S'$ will consist of all the roots of unity in that arc when it first contains $q$ points (possibly leaving a single endpoint out if it jumps from $q-1$ to $q+1$). –  Kevin Costello Feb 15 '13 at 1:32
    
One other note: I've been assuming that actually $\zeta=\exp\left(\frac{2 \pi i}{r}\right)$ (since it seems like if you allow $\zeta$ to be an arbitrary primitive root of unity than, for example the inequality with $\zeta=\exp\left(2 \pi i \frac{3}{11}\right)$ and $m=4$ is exactly the reverse of the one with $\zeta=\exp\left(2 \pi i \frac{1}{11}\right)$ and $m=3$). If I'm misunderstanding something, please let me know. –  Kevin Costello Feb 15 '13 at 1:38
    
How do we know that the first $q$ roots of unity are closest to the direction of $v$? $v$ should be able to be in just about any direction right? –  Alexander Gruber Feb 15 '13 at 7:12
    
Right, but the absolute value of the sum any $q$ consecutive roots of unity is the same as the sum of the first $q$ (since they differ only by a rotation). –  Kevin Costello Feb 15 '13 at 7:55
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