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I'm trying to solve this equation for the wavenumber $k$: $\sum_{n=1}^{N}e^{ikn}=\sqrt{N(k^2+a)}$ but I'm completely stuck. Any idea? (As you can imagine, I'm working with waves)

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  • $\begingroup$ I'm not even sure what you're supposed to solve for. $\endgroup$ – Fimpellizieri Aug 5 '16 at 2:23
  • $\begingroup$ Sorry! For k, that's my variable. I'm basically studying a wave and want to know for which wave number that equation holds. $\endgroup$ – Sebasen Aug 5 '16 at 2:25
  • $\begingroup$ Is anything about $a$ known? $\endgroup$ – Fimpellizieri Aug 5 '16 at 2:27
  • $\begingroup$ It's just a positive parameter independent of $N$ and $k$, nothing else $\endgroup$ – Sebasen Aug 5 '16 at 2:27
  • $\begingroup$ If it's easier to work with sines or cosines, that works for me too. Anything is useful haha $\endgroup$ – Sebasen Aug 5 '16 at 2:32
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HINT:

First, the summation can be evaluated in closed form as

$$\sum_{n=1}^N e^{ikn}=\frac{e^{ik(N+1)/2}\sin(kN/2)}{\sin(k/2)} \tag 1$$

Note that (i) the imaginary part of $(1)$ is $0$ and (ii) the real part is $\sqrt{N(k^2+a)}$.

Enforce (i) to find a relationship between $k$ and $N$. Then, using that relationship in (ii), eliminate $k$ in terms of $N$ obtain an transcendental family of equations for $N$.

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$\sum_{n=1}^{N}e^{ikn}=\sqrt{N(k^2+a)} $

Since

$\begin{array}\\ \sum_{n=1}^{N}e^{ikn} &=e^{ik}\sum_{n=0}^{N-1}e^{ikn}\\ &=e^{ik}\dfrac{e^{ikN}-1}{e^{ik}-1}\\ &=\dfrac{e^{ik(N+1)}-e^{ik}}{e^{ik}-1}\dfrac{e^{-ik}-1}{e^{-ik}-1}\\ &=\dfrac{(e^{ik(N+1)}-e^{ik})(e^{-ik}-1)}{2-(e^{ik}+e^{-ik})}\\ &=\dfrac{(e^{ikN}-1)(1-e^{ik})}{2-2\cos(k)}\\ \end{array} $

so you want $(e^{ikN}-1)(1-e^{ik}) =2(1-\cos(k))\sqrt{N(k^2+a)} $.

Since the left side is complex and the right side is real, I'm not sure how you want to go from here.

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Assume without loss of generality that $k \in [0,2\pi)$ (by the periodicity of the exponential).

Since the RHS of the equation is a positive real number and the LHS is a sum of the powers of $e^{ik}$, ie, a sum on $\mathbb{S}^1 \subset \mathbb{C}$, we need that sum to be symmetrical about the real line (so the imaginary parts cancel out). This happens only in the following cases (I think this is best seen geometrically):

  • $k=0$ or $k=\pi$ (and $N$ can be any positive integer)
  • $k=\frac{2\pi}{m}$ for some integer $m \geq 3$ and $N$ is a positive integer equivalent to either $m-1$ or $m$ modulo $m$.

If $k\neq 0$, it's easy to verify that these solutions cannot be positive. So the only possibility is $k=0$.

In this case, the LHS reduces to $N$, so another requirement for the solution is that $a=N$.

EDIT: I have a weird feeling so please someone correct me if I've written some absurdity. Is there a solution when $k$ is an irrational multiple of $2\pi$?

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Following up on Dr. MV's hint (it will absolutely work, and in fact the closed form for the sum will probably be easier to work with than what follows, but it may not give you as clear of an idea of what's happening here.)

First recall Euler's identity: $e^{ix}=\cos(x)+i\sin(x)$, thus the number $e^{ikn}$ (for $1 \leq n \leq N$) can be visualized as the point $(\sin(kn),\cos (kn))$

Consider for a moment the somewhat simpler case: if the left-hand side was written as $\sum_{n=1}^N e^{2\pi i n k}$, what does this sum look like? For now let's take $k=\frac{1}{8}$, then the first few points look like: $\{e^{i\pi /4}, e^{i\pi /2}, e^{ i3 \pi/ 4} \dots\}$. I recommend plotting these points, i.e $(\cos(\pi/4),\sin(\pi/4)) \dots $ in $\mathbb{R^2}$. What you will find is that when $k=\frac{1}{8}$, the points progress as follows, you begin at $(1,0)$, move counter-clockwise around the unit circle until you have traversed $1/8$ of its circumference and that's your first point $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$. Your next point is obtained by continuing this process (moving $1/8$ the circumference of the circle from this point to $(0,1)$, and so forth. So when will $\sum_{n=1}^N e^{2\pi i nk}$ be a real number? If and only if $kN=m$, some $m \in \mathbb{N}$ (why is this? If it's not clear, you haven't really drawn the picture!) In our toy example, if $N=8$, while your first point $\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$ certainly has an imaginary part, your 7th point will be $\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}$ and when you sum them they will cancel (seriously this is very clear to see by drawing it). Moreover it is very easy to compute in this case the contribution made to the right hand side from each pair of points!

The moral: the relation Dr. MV is talking about is that $\frac{1}{k}$ needs to divide $N$ (well, in my case, in your case you will have to play with $2\pi$ a little, but this is simply an exercise in patience and arithmetic and drawing), the number $m$ I was talking about is the number of times you will "traverse" the unit circle (in my example, by choosing $N=8$ you complete one full 'cycle'.) Note it suffices to solve the problem when $m=1$ (why?), so you may as well work with $N=\frac{1}{k}$.

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To the different contributors @Marty Cohen @Fimpellizieri @mb-

Following the remark of @Dr. MV , the LHS of the relationship is

$$F K_{N}(k)= e^{ik(N+1)/2} \frac{\sin(kN/2)}{\sin(k/2)}$$ where

$K_{N}(k):=\frac{\sin(kN/2)}{\sin(k/2)}$ is a "kernel", parent to Dirichlet kernel of order $N$ and $F$ is defined by $F:=e^{ik(N+1)/2}$.

Due to the fact that the LHS has to be real in order to match the RHS, $F$ should be equal to $1$ or $-1$, thus $k(N+1)/2$ should be of the form $M\pi$ for some integer $M$, i.e., one should have $$k=2M \pi/(N+1) \ \ \ (1).$$

Now, you are with equation $K_N(k)=\sqrt{N(k^2-1)} \ \ \ (2).$

Considering, for the moment, that parameter $k$ takes on real values, you can "graphically solve" (2) (note the quotes!) by having a look at the intersections of the curve of $y= K_N(k)$ (which looks a little like a Dirac comb) and the curve of $y=\sqrt{N(k^2+a)}$. Either two half components of a hyperbola if $a<0$ (colored in red) or a whole branch (colored in blue) if $a>0$. And then try to spot solutions that either fullfill condition (1)

In the figure below, the values of the parameters are $a=\pm12$ and $N=15$.

It remains to check that the abscissas $k$ of intersections points are integers or almost-integers.

enter image description here

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  • $\begingroup$ I have added a picture and corrected a small error. $\endgroup$ – Jean Marie Aug 5 '16 at 5:34
  • $\begingroup$ I have completely modified my answer in a "positive way" $\endgroup$ – Jean Marie Aug 8 '16 at 23:11

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