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How can I calculate $\displaystyle\sup_{x\in [-1,1]}\{ |\sin(nx)- \sin(mx)| : m,n\in \mathbb{N}\}$ ?

This is what i have tried

$\sin(nx)- \sin(mx)= 2\cos (\frac{nx+mx}{2})\sin(\frac{nx-mx}{2})$

$\Rightarrow$

$|\sin(nx)- \sin(mx)|= 2|\cos (\frac{nx+mx}{2})||\sin(\frac{nx-mx}{2})|$

But I do not know what else to do...

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    What exactly is your supremum referring to? Is it the supremum over $x$ for fixed $m.n$ or is it with respect to $x,m,n$? – 2017-01-15
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    Supreme over x and m, n fixed – 2017-01-15
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    I think the interval $[-1,1]$ is very ugly and maybe the problem has an error... would make more sense to calculate the supreme in $[-\pi,\pi]$? – 2017-01-15

1 Answers 1

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An idea: First, you can suppose that there is(are) solution(s) for $\sin{(nx)} = \pm1$ and $\sin{(mx)} = \pm1$. For find the sup, you need x such that $\sin{nx} = 1$(or -1, whatever) and $\sin{mx}= -1$(or 1, if first is 1). So, we have

$$nx = \frac{\pi}{2} + 2 k \pi \;\;\&\,\,mx = \frac{3\pi}{2} + 2 \tilde{k} \pi$$

With $k,\tilde{k}\in \mathbb{Z}$.

add to it, that $x\in [-1,1]$, And you will find a explicit formula for sup.

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    _If_ those equations have a solution between $-1$ and $1$, then sure. But what if they don't? Say, $n=1$, $m=2$? – 2018-09-14
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    Henning, you are correct. Is a simple idea for how to work. – 2018-09-14