Isolate or approximate $x$ in $x \sin (\pi/x) \geq a$

Question

Is there a way to isolate or approximate $x$ in the following inequality?

$$x \sin (\pi/x) \geq a .$$

Answer 1

In the same spirit as @Jean Marie, let $x=\frac \pi y$ and consider the inequality as $$\frac{\sin(y)}y \geq \frac a \pi$$ Suppose that $\frac a \pi>0.128375$ which corresponds to the second maximun of $\text{sinc}(x)$, it is possible to find the range of $y$ where the inequality holds.

A very simple approximate solution uses the magnificent $$\sin(y) \simeq \frac{16 \,(\pi -y)\, y}{5 \pi ^2-4\, (\pi -y)\,y}$$ made, more than $1,400$ years ago, by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician.

This would give $$0 \leq y \leq \frac{ \left(2 \sqrt{-a^2+2 a+4}+a-4\right)}{2 a} \pi$$

Suppose $a=1$; the above formula will give as an upper bound $\left(\sqrt{5}-\frac{3}{2}\right) \pi \approx 2.31243$ while the "exact" solution, obtained using Newton method, would be $2.31373$.

What we could also do is to use the usual Taylor expansion of $\sin(y)$ and use series revsersion to get $$y=t+\frac{t^3}{40}+\frac{107 t^5}{67200}+\frac{3197 t^7}{24192000}+\frac{49513 t^9}{3973939200}+\frac{196309411 t^{11}}{154983628800000}+O\left(t^{13}\right)$$ where $t=\sqrt{6\left(1-\frac{a}{\pi }\right)}$. For the worked case, it would be worse (!!).

Answer 2

A first remark : if the inequality is verified for a certain $x$, it is true as well for $-x$, using parity Therefore, we can restrict our attention to positive values of $x$.

If the inequality is written under the form :

$$\frac{\sin(\tfrac{\pi}{x})}{\tfrac{\pi}{x}}\geq\tfrac{a}{\pi}$$

It can be transformed into :

$$\text{sinc}(\tfrac{1}{x})\geq b$$

where "sinc" is a well known function ("normalized" cardinal sine).

The solution $S$ is a union of intervals (possibly void if $b>1$ or $b<-0.2172...$ (first minimum of $\text{sinc}$ function).

Fig. 1 : The solution is featured on the $y$ axis as the union of intervals represented as thick red line segments.

Let us take an example (see figure), if $a=\pi$, i.e., $b=0.1$ (red line), we oberve that the inequality is verified iff :

$$\tfrac{1}{x}\in \ (0,0.91] \cup [2.25,2.68]$$

(approximate values), which itself can be converted into

$$0<\tfrac{1}{x}\leq 0.91 \ \ \ \ or \ \ \ \ 2.25 \leq \tfrac{1}{x} \leq 2.68$$

$$\iff \ \ x \geq 0.91 \ \ or \ \ \tfrac{1}{2.68}\leq x \leq \tfrac{1}{2.25}$$

and conclude $x \in [0.91, +\infty) \cup [0.37, 0.44].$

(don't forget to take the union with the opposite set).

Please note the presence of an unbounded interval.

The limits of these intervals can be refined at will (see in particular the approximation method given by @Claude Leibovici).

I have provided a graphical solution, using a kind of reflection on the hyperbola with equation $y=1/x$ : the solution is the union of intervals represented by thick red intervals.