# Distance of a real number to a discrete set of scaled sine values

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?

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

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

Edit

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.

• Your first inequality is marvelous, may I have a name or source? Commented Jul 10 at 13:25
• @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 :-) Commented Jul 10 at 13:38
• Thank you very much for your answer. Do you still remember how you found this specifc choice of exponents? They are beautiful. Commented Jul 10 at 13:57
• @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. Commented Jul 10 at 14:09
• @Nuke_Gunray Have a look at arxiv.org/pdf/2011.04430 I did not know it. Very interesting bounds. Commented Jul 11 at 5:22