What are the local maxima and minima of $\frac{\sin(nx)}{\sin(x)}$

Question

FYI: I asked this question here couple of days ago but got no answer yet.

$n$ is an integer

We know the global maximum of the function $\sin(nx)/\sin(x)$ is $n$ (thanks to this question), but what are the local maxima and minima points of the function in $(-\pi,\pi)$? (it represents the intensity of a narrow slit diffraction grating)

I tried equating the derivative to $0$ and the equation to be solved is $n\tan(x) = \tan(nx)$. I am unable to progress from here. I also tried expanding $\sin(nx)$ using multiple-angle formula but made no progress.

For convenience: 1st derivative $$ = \frac{n\cos(nx)\sin(x) - \sin(nx)\cos(x)}{\sin(x)^2}$$and 2nd derivative = $$\frac{2\sin(nx)\cos(x)^2 + \sin(nx)\sin(x)^2 - n^2\sin(nx)\sin(x)^2 - 2n\cos(nx)\cos(x)\sin(x)}{\sin(x)^3}$$

Here is a plot of the function in Desmos: https://www.desmos.com/calculator/kt0hntsbcb

Answer 1

Have a look at my answer to this question.

For a given $n$, excluding the two trivial solutions $x=0$ and $x=\pi$, using $$x= \frac {\pi\, y}{n} \qquad\text{for}\qquad y \in (0,n)$$

a rather good first estimate of the $k^{\text{th}}$ root is given by $$y_k^{(0)} = k+\frac 12\qquad \implies \qquad \large\color{blue}{x_k^{(0)}=\frac {(2k+1)\, \pi}{2n}}$$ Trying for $n=10$ as in your plot

$$\left( \begin{array}{ccc} k & y_k^{(0)} & \text{solution} \\ 1 & 1.5 & 1.43516 \\ 2 & 2.5 & 2.46763 \\ 3 & 3.5 & 3.48359 \\ 4 & 4.5 & 4.49491 \\ 5 & 5.5 & 5.50509 \\ 6 & 6.5 & 6.51641 \\ 7 & 7.5 & 7.53237 \\ 8 & 8.5 & 8.56484 \\ \end{array} \right)$$

The first iterate of Newton method already improves

$$y_k^{(1)} =y_k^{(0)}-\frac{n}{\pi }\,\frac{\cot \left(\frac{\pi }{n}y_k^{(0)}\right)}{n^2-2 \csc ^2\left(\frac{\pi }{n}y_k^{(0)}\right)+1}$$ which would imply $$\large\color{blue}{x_k^{(1)} =x_k^{(0)}-\frac{\cot \bigg(x_k^{(0)}\bigg)}{n^2-\csc ^2\bigg(x_k^{(0)}\bigg)}}$$

Repeating the calculations for $n=10$, the results $$\left( \begin{array}{ccc} k & y_k^{(1)} & \text{solution} \\ 1 & 1.43157 & 1.43516 \\ 2 & 2.46718 & 2.46763 \\ 3 & 3.48353 & 3.48359 \\ 4 & 4.49490 & 4.49491 \\ 5 & 5.50510 & 5.50509 \\ 6 & 6.51647 & 6.51641 \\ 7 & 7.53282 & 7.53237 \\ 8 & 8.56843 & 8.56484 \\ \end{array} \right)$$

Repeating the calculations in terms of $x$

$$\left( \begin{array}{cccc} k & x_k^{(0)} & x_k^{(1)} & \text{solution} \\ 1 & 0.47124 & 0.45061 & 0.45087 \\ 2 & 0.78540 & 0.77519 & 0.77523 \\ 3 & 1.09956 & 1.09440 & 1.09440 \\ 4 & 1.41372 & 1.41212 & 1.41212 \\ 5 & 1.72788 & 1.72948 & 1.72948 \\ 6 & 2.04204 & 2.04720 & 2.04719 \\ 7 & 2.35619 & 2.36640 & 2.36636 \\ 8 & 2.67035 & 2.69098 & 2.69072 \\ \end{array} \right)$$

This means that using the $\color{red}{\text{first}}$ iterate of an higher order method (Halley, Householder or higher), we can have better and better explicit approximations.

This is illustrated below $$\left( \begin{array}{ccccc} k & \text{Newton}& \text{Halley}& \text{Householder}& \text{solution} \\ 1 & 0.4506120 & 0.4510008 & 0.4508669 & 0.4508697 \\ 2 & 0.7751941 & 0.7752448 & 0.7752278 & 0.7752279 \\ 3 & 1.0943972 & 1.0944039 & 1.0944016 & 1.0944016 \\ 4 & 1.4121164 & 1.4121166 & 1.4121166 & 1.4121164 \\ 5 & 1.7294762 & 1.7294760 & 1.7294761 & 1.7294762 \\ 6 & 2.0471955 & 2.0471888 & 2.0471910 & 2.0471910 \\ 7 & 2.3663986 & 2.3663478 & 2.3663649 & 2.3663648 \\ 8 & 2.6909806 & 2.6905919 & 2.6907258 & 2.6907230 \\ \end{array} \right)$$

Answer 2

You need to solve the following equation for $z=e^{ix}$ $$\left(z^2+1\right) \left(z^{2 n}-1\right)-n \left(z^2-1\right) \left(z^{2 n}+1\right)=0.$$ Wolfram Alpha can do that for you. The answer is in radicals for $n\leq 7$,for example $$n=7\Rightarrow z=\sqrt{\frac{1}{6} \left(\sqrt{7}+i \sqrt{2 \left(\sqrt{7}+14\right)}-1\right)}.$$ Note that $|z|=1$, so $x$ is real. This is one of the solutions, for the others you vary the sign of the square roots (and there are also roots at $\pm 1$ and $\pm i$, see figure).

Answer 3

Following on Carlo's answer:

One has to solve $\Re(((a+ib)^2+1)((a+ib)^{2n}+1)+n((a+ib)^2-1)((a+ib)^{2n}+1)) = 0$

under the requirement that $a+ib$ is a point on the unit circle; i.e. $a^2+b^2=1$.

At this point you have to explicitly write the real part of the RHS of the equation at the top. This gives you an expression of the form $f(a^2,b) + a g(a^2,b) = 0$. Substitute $a^2=1-b^2$, and use the resultant to get rid of the singleton $a$. (I'm not doing this myself since it doesn't seem particularly fun.)

Answer 4

Going in a slightly different direction, let me give an approximate expression for the critical points.

From Carlo Beenakker's answer and my comment there, we know that the function only has extrema, and it has precisely one of those in every interval $[\frac\pi nk,\frac\pi n(k+1)]$. They are in the interior of the interval, except for $k$ in $n\mathbb Z$ or $-1+n\mathbb Z$, in which case they account for the global extrema at $\pi\mathbb Z$.

Let $x=x_{k,n}$ be the only extremum of your function in the interval above. From your expression for the derivative, we get $$ \tan(nx)=n\tan(x). $$ Set $u=nx-\pi k$; by definition $u\in(0,\pi)$. Expanding the above expression, we get $$ \tan(u)=n\tan\left(\frac{\pi k+u}n\right). $$

• If $k$ is bounded positive, then letting $f$ be the inverse of $v\mapsto\tan(v)-v=\pi k$ over $(0,\pi/2)$, we get $$ u = f\left(n\tan\left(\frac{\pi k+u}n\right)-u\right) = f(\pi k)+O(1/n^2). $$ With more effort, one gets an asymptotic expansion in $1/n$. A similar expression holds for $k$ negative by noting that in this case, $\tan(u)<0$ so $u=\pi+\arctan(\tan(u))$.
• If $k/n\in(\varepsilon,1/2]+\mathbb Z$, then $$ u = \frac\pi2-\arctan\left(\frac1n\cot\left(\frac{\pi k+u}n\right)\right) = \frac\pi2-\frac1n\cot\left(\frac{\pi k}n\right)+O(1/n^2) $$ as $n$ goes to infinity, where the constant in the $O$ term depends only on $\varepsilon$. As above, a similar expression holds for $k/n\in[-1/2,-\varepsilon)$.
• If $k/n\in[0,1/2-\varepsilon)+\mathbb Z$, then $$ u = \frac\pi2-\arctan\left(\frac1n\cot\left(\frac{\pi k+u}n\right)\right) = \frac\pi2-\arctan\left(\frac1{\pi k+u}\right)+O(1/n), $$ which gives $$ u=\frac\pi2-\frac1{\pi k}+O(1/k^2)+O(1/n) $$ if $k\to\infty$. The same thing goes for $k/n\in(-1/2+\varepsilon,0]+\mathbb Z$. This estimate is probably most useful when $k/n\to0$, in which case we can replace $O(1/n)$ by $O(k/n^2)$.

Answer 5

Symmetry can be used to halve the computational work.

For example, first display a graph of

$S(x) = \frac{\sin(N \pi x)}{\sin(\pi x)}$

by cutting and pasting (w/o the quotes) into the online graphing calculator Desmos

"S\left(x\right)\ =\frac{\sin\left(N\ \pi\ x\right)}{\sin\left(\pi\ x\right)}".

(Edit 1/24/24: I had to open the MO edit window to copy the later expressions in quotes to paste into Desmos for some strange reason, with Chrome.)

Note the period is $2$ for $N =10$ or any even $N > 1$. (The period is $1$ for odd $N$.)

Now look at the odd shifted version

$O(x) = S(x-.5) = -O(-x)$

over the interval $-.5 \leq x \leq .5$ with C&P of

"\left{-.5<x<.5:\frac{\sin\left(N\ \pi\ \left(x-.5\right)\right)}{\sin\left(\ \pi\ \left(x-.5\right)\right)}\right}".

Clearly we need only find four extrema (the global max at $x=.5$ is $O(.5) = S(0) = 10$) as depicted by

"\left{0<x<.5:\frac{\sin\left(N\ \pi\ \left(x-.5\right)\right)}{\sin\left(\ \pi\ \left(x-.5\right)\right)}\right}".

Superposing $y = \sin(N \pi x)$, we easily see that the ordinates of the four extrema are

near $x_0 = (0+1/2)/10 = .05$ with $O(.05) \approx 1.01247$ [graph $(.051,1.013)$];

near $x_1 = (1+1/2)/10 = .15$ with $O(.15) \approx -1.12233$ [graph $(.152,-1.124)$];

near $x_2 = (2+1/2)/10 = .25$ with $O(.25) \approx 1.41421$ [graph $(.253,1.421)$];

near $x_3 = (3+1/2)/10 = .35$ with $O(.35) \approx -2.20269$ [graph $(.356,-2.247)$].

Comparing with the extrema using the ordinates found by Claude Leibovici for $f(x) = \frac{\sin(N x)}{\sin(x)}$:

$f(0.4508697) \approx -2.24746$

$f(0.7752279) \approx 1.42144$

$f(1.0944016) \approx -1.1238$

$f(1.4121164) \approx 1.01259$

$f(1.7294762) \approx -1.01259$

$f(2.0471910) \approx 1.1238$

$f(2.3663648) \approx -1.42144$

$f(2.6907230) \approx 2.24746$,

which reflect the symmetries.

(For $N$ odd, $h(x) = S(x-.5) = S(-x-.5) = h(-x)$ is an even function and the symmetries can again be used to reduce the work.)