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

We know the global maxima 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

• Why are you interested in this? And what makes you think you can expect a closed form? Jan 13 at 18:18
• @RajaKrishnappa This site has expansions for the multivalued inverse of $\frac{\tan(x)}x$ as your equation is $\frac{\tan(nx)}{nx}=\frac{\tan(x)}x$ or convert to a sine equation like here. However, these are series solutions. Are you looking for an approximation or an exact form solution? Jan 13 at 19:53
• @K.defaoite, a) I would like to know the points where the intensity of a narrow slit diffraction grating is maximum/minimum and b) It seems like there is a pattern (see plot in desmos) and the function looks simple Jan 16 at 3:03
• @ТymaGaidash, I am looking for exact form solution because the function seems so simple yet complex. I am certain I am missing something and wanted some insights from the experienced. If it can be proved that there is no close form solution, an approximation is the next best thing and is welcomed :) Jan 16 at 3:05
• crossposted at mathoverflow.net/q/462270/11260 Jan 16 at 9:58

$$\tan(nx)=n\tan(x)\iff n\cos(nx)\sin(x)-\sin(nx)\cos(x)$$

Now apply product to sum formulas and double both sides:

$$n\cos(nx)\sin(x)-\sin(nx)\cos(x)=0\iff(1+n)\sin((1-n)x)-(1-n)\sin((n+1)x)=0$$

We have a series solution to:

Is there any way to solve for $$k$$, given $$\beta \sin (k-k N)-\sin (k N+k)=0$$?

for the real roots. Here, $$\beta=\frac{1+n}{1-n}$$, so:

$$\bbox[2px,border:2.5px solid #0D98CA]{\tan(nx)=n\tan(x)\implies x_j=\frac{\pi j}{n+1}+\frac1{2i(n+1)}\sum_{k=1}^\infty\sum_{m=0}^k\frac{(-1)^m}{(k-m)!m!}\left(m+\frac{k-2m}{n+1}-1\right)^{(n-1)}\left(\frac{1+n}{1-n}\right)^k\exp\left(\frac{2\pi i j(m(n-1)+k)}{n+1}\right)}$$

shown here:

using

ReplaceAll[{j->J,n->N}][\[Pi] j/(n+1)+1/(2 I (n+1))Sum[Sum[(-1)^m ((1+n)/(1-n))^k/(m! (k-m)!) Exp[(2 \[Pi] I j ((n-1) m+k))/(n+1)] FactorialPower[m+(k-2 m)/(n+1)-1,k-1],{m,0,k}],{k,1,a}]]

for given $$a,J,N$$. Some cases have closed elementary forms like $$n=0,\dots 7$$ and for $$n=8$$, a quintic, has single series or special function solutions. However, for harder cases, the boxed result is a solution. Either sum has a closed form with the Fox H or Fox Wright function too.