Absolute value of an exponential sum

Question

Consider the following sequence $$ x_n=\Bigl|\sum_{t=1}^n\exp\Bigl\{2\pi i\Bigl[\frac1d-\frac {\lfloor n/d\rfloor}n\Bigr]t\Bigr\}\Bigr| $$ with some fixed positive integer $d$ such that $d<n$. I am trying to establish a lower bound for $x_n$ for all large values of $n$.

Is it true that $x_n\ge cn$ with some absolute positive constant $c$ for large values of $n$?

If $n$ is an integer multiple of $d$, then $x_n=n$ since $1/d-\lfloor n/d\rfloor/n=0$. In the general case, we have that $$ \frac1d-\frac{\lfloor n/d\rfloor}n =\frac{n-d\lfloor n/d\rfloor}d\cdot\frac1n $$ and $$ 0\le\frac{n-d\lfloor n/d\rfloor}{d}<1 $$ so it seems that $x_n$ should be close to $n$ for large values of $n$, but I have no idea if it is possible to show this rigorously.

Any help is much appreciated!

Answer 1

(I found my error and corrected it. I also made the proof simpler.)

$x_n=\Bigl|\sum_{t=1}^n\exp\Bigl\{2\pi i\Bigl[\frac1d-\frac {\lfloor n/d\rfloor}n\Bigr]t\Bigr\}\Bigr| $.

I will show that

$\begin{array}\\ x_n &=\dfrac{|\sin( \pi\left\{\frac{n}{d}\right\})|}{|\sin( \frac{\pi}{n}\left\{\frac{n}{d}\right\})|}\\ &\ge n\frac{|\sin( \pi\left\{\frac{n}{d}\right\})|}{|(\pi\left\{\frac{n}{d}\right\})|}\\ &\ge n(1-\frac16(\pi\left\{\frac{n}{d}\right\})^2)\\ \end{array} $

---

Let $a =2\pi(\dfrac1d-\frac {\lfloor \frac{n}{d}\rfloor}{n}) =\dfrac{2\pi}{n}(\dfrac{n}{d}-\lfloor \dfrac{n}{d}\rfloor) =\dfrac{2\pi}{n}\left\{\dfrac{n}{d}\right\} $.

Then

$\begin{array}\\ x_n &=\Bigl|\sum_{t=1}^ne^{ i at}\Bigr|\\ &=\Bigl|e^{ i a}\sum_{t=0}^{n-1}e^{ i at}\Bigr|\\ &=\Bigl|e^{ i a}\dfrac{1-e^{ i a n}}{1-e^{ i a}}\Bigr|\\ &=|e^{ i a}|\Bigl|\dfrac{1-e^{ i a n}}{1-e^{ i a}}\Bigr|\\ &=\Bigl|\dfrac{1-e^{ i a n}}{1-e^{ i a}}\Bigr| \qquad\text{since } |ab| = |a|\,|b|\\ &=\dfrac{|2\sin( n a/2)|}{2|\sin( a/2)|} \qquad\text{since } |1-e^{ix}| = 2\sin(x/2) \text{ (see below)}\\ &=\dfrac{|\sin( n a/2)|}{|\sin( a/2)|}\\ &=\dfrac{|\sin( n \frac{2\pi}{n}\left\{\frac{n}{d}\right\}/2)|}{|\sin(\frac{2\pi}{n}\left\{\frac{n}{d}\right\}/2)|}\\ &=\dfrac{|\sin( \pi\left\{\frac{n}{d}\right\})|}{|\sin( \frac{\pi}{n}\left\{\frac{n}{d}\right\})|}\\ &=\dfrac{|\sin( \pi\left\{\frac{n}{d}\right\})( \frac{\pi}{n}\left\{\frac{n}{d}\right\})|}{|(\frac{\pi}{n}\left\{\frac{n}{d}\right\})\sin( \frac{\pi}{n}\left\{\frac{n}{d}\right\})|}\\ &=n\dfrac{|\sin( \pi\left\{\frac{n}{d}\right\})( \frac{\pi}{n}\left\{\frac{n}{d}\right\})|}{|(\pi\left\{\frac{n}{d}\right\})\sin(\frac{\pi}{n}\left\{\frac{n}{d}\right\})|}\\ &\ge n\dfrac{|\sin( \pi\left\{\frac{n}{d}\right\})|}{|(\pi\left\{\frac{n}{d}\right\})|} \qquad \text{since } \sin(x) \le x\\ &\ge n(1-\frac16(\pi\left\{\frac{n}{d}\right\})^2) \qquad \text{since } \sin(x) \ge x-x^3/6\\ \end{array} $

Auxiliary results.

$\begin{array}\\ |1-e^{ix}| &=|1-(\cos(x)+i\sin(x))|\\ &=|1-\cos(x)-i\sin(x)|\\ &=\sqrt{(1-\cos(x))^2+\sin^2(x)}\\ &=\sqrt{1-2\cos(x)+\cos(x)^2+\sin^2(x)}\\ &=\sqrt{2-2\cos(x)}\\ &=\sqrt{2}\sqrt{1-\cos(x)}\\ &=2\sqrt{(1-\cos(x))/2}\\ &=2|\sin(x/2)|\\ |1+e^{ix}| &=|1+(\cos(x)+i\sin(x))|\\ &=|1-\cos(x)-i\sin(x)|\\ &=\sqrt{(1+\cos(x))^2+\sin^2(x)}\\ &=\sqrt{1+2\cos(x)+\cos(x)^2+\sin^2(x)}\\ &=\sqrt{2+2\cos(x)}\\ &=\sqrt{2}\sqrt{1+\cos(x)}\\ &=2\sqrt{(1+\cos(x))/2}\\ &=2|\cos(x/2)|\\ (a+ib)(c+id) &=(ac-bd)+i(ad+bc)\\ |(a+ib)(c+id)| &=\sqrt{(ac-bd)^2+(ad+bc)^2}\\ &=\sqrt{a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2}\\ &=\sqrt{a^2c^2+b^2d^2+a^2d^2+b^2c^2}\\ |(a+ib)||(c+id)| &=\sqrt{a^2+b^2}\sqrt{c^2+d^2}\\ &=\sqrt{a^2c^2+a^2d^2+b^2c^2+b^2d^2}\\ &=|(a+ib)(c+id)|\\ \end{array} $

Answer 2

Let $\{y\}=y-\lfloor y\rfloor$ denote the fractional part of $y$. Inside the modulus signs we have the geometric series $$ \sum_{t=1}^n \exp\bigg( \frac{2\pi i}n \Big\{ \frac nd \Big\} t \bigg) = \sum_{t=1}^n \exp(\alpha t) = \exp(\alpha) \frac{\exp(n\alpha)-1}{\exp(\alpha)-1}, $$ where $\alpha = \frac{2\pi i}n \big\{ \frac nd \big\}$ (and we are assuming that $\alpha\ne0$); therefore $$ x_n = \frac{|\exp(n\alpha)-1|}{|\exp(\alpha)-1|}. $$ Since $\alpha$ is very small, the denominator is $\sim|\alpha|$ by the mean value theorem, and so $$ x_n \sim |\alpha|^{-1} |\exp(n\alpha)-1|. $$ The first term does have size close to a constant times $n$. The second term, however, can be quite small if $\{\frac nd\}$ is close to $0$ or $1$, that is, if $n$ is just more than or just less than a multiple of $d$. In particular, if $d$ divides $n\pm1$, then the order of magnitude of $x_n$ will be roughly $\frac nd$ rather than $n$.