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I was wondering if there is an analytical solution to find the maxima of the following function in terms of $x$:

$$ f(x) = \frac{\sin(nx)}{\sin(x)} $$

where $n$ is any positive integer greater than $0$

I tried equating the partial derivative of $f(x)$ with respect to $x$ to $0$, but I can't seem to find any other solution than $x = \pi k$ to the resulting trigonometric equation. I was also not able to find a solution in existing posts.

Chappers
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Joshua Schroijen
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  • Are you looking for global maxima or all local maxima? – Daniel Fischer Apr 21 '17 at 20:25
  • Your function is discontinuous for all values of $x=πk$. If you decide to make it into a continuous function by adding the countable infinity of points needed to do so, then you will have a function with an infinity of maxima. – MasB Apr 21 '17 at 20:32
  • You should also try to plot this interesting function. – Jean Marie Apr 21 '17 at 20:39
  • I should have specified that I'm searching for local maximums, but global maximums are interesting too. ( I'm just intrigued by the function). – Joshua Schroijen Apr 21 '17 at 20:52
  • The plot depends greatly on the value of n, by the way. For some values, it reminds me of overlapping and added sinc functions .. – Joshua Schroijen Apr 21 '17 at 20:53

2 Answers2

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Hint $$ \left|1+e^{i2x}+e^{i4x}+\ldots + e^{i2(n-1)x}\right|= \left|\frac{1-e^{i2nx}}{1-e^{2ix}}\right| = \left|\frac{e^{inx}}{e^{ix}}\frac{\sin (nx)}{\sin(x)}\right|= \left|\frac{\sin (nx)}{\sin(x)}\right| $$

Also $$\left|1+e^{i2x}+e^{i4x}+\ldots + e^{i2(n-1)x}\right| \leq \left|1\right|+\left|e^{i2x}\right|+\left|e^{i4x}\right|+\ldots + \left|e^{i2(n-1)x}\right|=n$$ with equality when all terms are real and non-negative.

N. S.
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Write $f_n(x) = \dfrac{\sin nx}{\sin x}$. It is familiar from calculus that $\lim_{x \to 0} f_n(x) = n$. If we can prove $|f_n(x)| \le n$ for all $x$ that will imply that $f_n$ has maximum $n$.

Note that if $\sin x = 0$ then $\sin(nx) = 0$ too, so by L'Hospital's rule we find $f_n(x) = \pm n$ (at least in the limit). So assume $\sin x \not= 0$.

In case $n = 2$ we have $f_2(x) = 2 \cos x$ so that $|f_2(x)| \le 2$. Inductively if we assume $|f_k(x)| \le k$ then $$|f_{k+1}(x)| = \left| \frac{\sin(k+1)x}{\sin x} \right| = \left| \frac{\sin kx \cos x + \cos kx \sin x}{\sin x} \right|\le |f_k(x)||\cos x| + |\cos kx| \le k+1.$$

Thus the global maximum of $f_n$ is $n$.

Umberto P.
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