Let $M>0$ be an integer, $c\in(0,\frac{1}{2})$ a real number, $$ a_{m,n}:=\frac{2n}{\pi}\sin\frac{m\pi}{2n}, $$ $$ A_n:=\left\{a_{m,n}:~m=1,\ldots,n-1\right\}, \text{ and} $$ $$ d_n:=\operatorname{dist}\left(M-\frac{1}{2},A_n\right):=\min_{m=1,..,n-1}\left|M-\frac{1}{2}-a_{m,n}\right|. $$ Find as small as possible an integer $N=N(M,c)$ such that for all $n\ge N$ it holds that $d_n\ge c$.

Note that the distance $d_n=|M-\frac{1}{2}-a_{p,n}|$ for some $p\in\{1,\ldots,n-1\}$ satisfying $$ M-1\le a_{p,n}\le M. $$ Note also that if $\frac{m}{n}$ is sufficiently small then $a_{m,n}\approx m$. So if $\frac{M}{N}$ is sufficiently small then $a_{M,n}\approx M$ and since $a_{M,n}\le M$ we can make $0\le M-a_{M,n}\le \frac{1}{2}-c$ and $d_n=|M-\frac{1}{2}-a_{M,n}|=\frac{1}{2}-(M-a_{M,n})\ge c$.

Guided by the above asymptotic analysis, we can find the following $N$. Note that $\sin x> x - \frac{x^3}{3!}$ for $x\in(0,\frac{\pi}{2})$. So $$ M- a_{M,n}\le M-\frac{2n}{\pi}\left(\frac{M\pi}{2n} - \frac{M^3\pi^3}{48n^3}\right). $$ So to make $M-a_{M,n}\le \frac{1}{2}-c$, it suffices to make $$ M-\frac{2n}{\pi}\left(\frac{M\pi}{2n} - \frac{M^3\pi^3}{48n^3}\right)\le \frac{1}{2}-c, $$ that is, $$ \frac{M^3\pi^2}{24n^2}\le \frac{1}{2}-c. $$ Solving for $n$ gives $$ n\ge \sqrt{\frac{M^3\pi^2}{12-24c}} $$ which can be taken as a feasible $N$. But my key question is to find as small as possible $N$. For example, can we show that the smallest $N$ is of the order $$ M^{\frac{3}{2}}(1-2c)^{-\frac{1}{2}}, $$ considering $M$ is large and $1-2c>0$ is small?

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    $\begingroup$ This is a nice (and well presented) problem. (+1) $\endgroup$ Commented Jul 10 at 11:00

1 Answer 1


If you want to explore tighter bounds, you could try $$\sin(x) \geq \pi ^{-\frac{\pi ^2}{3}} x \left(\pi ^2-x^2\right)^{\frac{\pi ^2}{6}} \quad \quad\text{for} \quad x\in (0,\pi)$$ the maximum difference being $0.13$ around $x=2.71$.

Another one is $$\sin(x) \geq \frac{x \left(60-7 x^2\right)}{3\left(x^2+20\right)}$$


From this paper, $$\sin(x) > x\, \left(\frac{2}{\pi }\right)^{\frac{4x^2}{\pi ^2}}\quad\quad\text{for} \quad x\in \left(0,\frac \pi 2\right)$$ the maximum difference being $0.01$ around $x=1.19$.

This could lead to interesting equations.

  • $\begingroup$ Your first inequality is marvelous, may I have a name or source? $\endgroup$ Commented Jul 10 at 13:25
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    $\begingroup$ @Nuke_Gunray. If my memory is good (who knows ?), I think that I made it more than fifty years ago for a problem in physics. In fact, I think that is could still be improved. I shall try and, if anything comes out, I shall let you know. Cheers :-) $\endgroup$ Commented Jul 10 at 13:38
  • $\begingroup$ Thank you very much for your answer. Do you still remember how you found this specifc choice of exponents? They are beautiful. $\endgroup$ Commented Jul 10 at 13:57
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    $\begingroup$ @Nuke_Gunray. I started from Wallis product; this is sure. Now, the remaining is still fuzzy. If you look at my old posts, you will notice that I have be doing many things with the sine function. $\endgroup$ Commented Jul 10 at 14:09
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    $\begingroup$ @Nuke_Gunray Have a look at arxiv.org/pdf/2011.04430 I did not know it. Very interesting bounds. $\endgroup$ Commented Jul 11 at 5:22

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