# $\frac{1}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{1-e^{-\frac{ik\pi}{n+1}}}=1$?

I'm working on an assignment where part of it is showing that $S_k=0$ for even $k$ and $S_k=1$ for odd $k$, where

$$S_k:=\sum_{j=0}^{n}\cos(k\pi x_j)= \frac{1}{2}\sum_{j=0}^{n}(e^{ik\pi x_{j}}+e^{-ik\pi x_{j}})$$

Here $x_j=j/(n+1)$.

So, working through the algebra:

$$\frac{1}{2}\sum_{j=0}^{n}(e^{ik\pi x_{j}} +e^{-ik\pi x_{j}}) =\dots =\frac{1}{2}\cdot\frac{1-e^{ik\pi}}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{2}\cdot\frac{1-e^{-ik\pi}}{1-e^{-\frac{ik\pi}{n+1}}}$$

Obviously $S_k=0$ for even $k$'s, since $e^{i\pi\cdot\text{even integer}}=1$. But when $k$ is odd we get $$\frac{1}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{1-e^{-\frac{ik\pi}{n+1}}}$$ which isn't obviously one to me, at least. Wolfram alpha confirms it equals 1.

My question: How does one see that it equals 1?

$$\frac{1}{1-e^{-a}} = \frac{1}{1-e^{-a}}\cdot\frac{e^a}{e^a} = \frac{e^a}{e^a-1} = \frac{-e^a}{1-e^a}.$$

Hence $$\frac{1}{1-e^a} + \frac{1}{1-e^{-a}} = 1.$$

• +1. You know what, before reading your answer, I was thinking of posting this computation, calling $z$ your $\mathrm e^a$ and $1/z$ your $\mathrm e^{-a}$.
– Did
Feb 22, 2014 at 17:40
• @Did : That works too. Feb 22, 2014 at 17:42

Using trigonometric identities

\begin{align} \cos (k \pi x_j) &= \frac{\sin (k \pi x_{j+1}) - \sin (k \pi x_{j-1})}{2 \sin (\frac{k \pi}{n+1})} \\ \sum_{j=0}^{n} \cos (k \pi x_j) &= 1+ \frac{1}{2 \sin (\frac{k \pi}{n+1})} \sum_{j=1}^{n} {\sin (k \pi x_{j+1}) - \sin (k \pi x_{j-1})} \\ \end{align} In the summation on the RHS, the terms that remain are

$-\sin (\frac{k \pi}{n+1}) + \sin (\frac{k \pi n}{n+1}) + \sin (k \pi)$ which clearly equals $0$ when $k$ is even, and $-2 \sin(\frac{k \pi}{n+1})$ when $k$ is odd.

To see that the given expression is equal to $1$, set

$\omega = e^{\frac{ik\pi}{n + 1}}, \tag{1}$

then the expression becomes

$\dfrac{1}{1 - \omega} + \dfrac{1}{1 - \bar{\omega}} = \dfrac{1 - \bar{\omega} + 1 - \omega}{(1 - \omega)(1 - \omega)} = \dfrac{2 - (\omega + \bar{\omega})}{(1 - \omega)(1 - \bar{\omega})}$ $= \dfrac{2 - (\omega + \bar{\omega})}{(1 + \omega \bar{\omega}) - (\omega + \bar{\omega})} = \dfrac{2 - (\omega + \bar{\omega})}{2 - (\omega + \bar{\omega})} = 1, \tag{2}$

since $\omega \bar{\omega} = 1$.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

• Can be made much simpler using $1/\omega$ instead of $\bar\omega$.
– Did
Feb 22, 2014 at 17:41

Reminders first:

$$\forall w,z \in \mathbb C, \overline{\left(\frac{w}{z}\right)}=\frac{\overline w}{\overline z}$$ $$\forall w,z \in \mathbb C, \overline{w+z}=\overline w +\overline z$$ $$\forall a\in \mathbb R, \overline{e^{ia}}=e^{-ia},$$ $$\forall z \in \mathbb C, z+ \overline z=2 \Re( z)$$ Apply these rules to $$\frac{1}{1-e^{\frac{ik\pi}{n+1}}}$$

$$\overline {\left(\frac{1}{1-e^{\frac{ik\pi}{n+1}}}\right)}=\frac{\overline1}{\overline1- \overline {e^{\frac{ik\pi}{n+1}}}}=\frac{1}{1-e^{-\frac{ik\pi}{n+1}}}$$

Then let $\alpha:=\Re\left(\frac{1}{1-e^{\frac{ik\pi}{n+1}}} \right)$ The following identity holds: $$\frac{1}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{1-e^{-\frac{ik\pi}{n+1}}}=2 \alpha$$

Now if we let $a:=e^{\frac{k\pi}{n+1}}$,

$$\alpha=\Re(\frac{1}{1-\cos(a)-i\sin(a)})= \frac{1-\cos(a)}{(1-\cos(a))^2+\sin^2(a)} = 1/2$$

Hence $$\frac{1}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{1-e^{-\frac{ik\pi}{n+1}}}=1$$

• I'm having a little trouble understanding your formula, how did you derive it? Are you taking $Re$ of one of the fractions?... Feb 22, 2014 at 18:32
• Indeed. These fractions are conjugates so they have the same real part. I took the real part of, say the first one. To do so I rewrote the first fraction as $$\frac{1-cos(a)+isin(a)}{(1-cos(a))^2+sin^2(a)}$$. The computation of the real part of this is straightforward. Feb 22, 2014 at 18:48
• How do you see that the fractions are conjugates, even though they're not on the form $a+bi,a-bi$? How is it that the expression in your answer evaluates to $1/2$? Won't the $\frac{1}{n+1}$ potentially mess up any nice integer output? Feb 23, 2014 at 8:15
• I added all the details I could possibly add. Feb 23, 2014 at 11:06
• Thanks for taking the time! Feb 23, 2014 at 11:22