Method for the minimum of $x\cdot\sin(1/x)$? I have the function $f(x)=x\cdot \sin\Bigl(\dfrac{1}{x}\Bigr)$ $(\mathbb R^*\rightarrow \mathbb R)$.
Is there a method to find the global minimum of $f$?
Thanks
 A: $f$ is an even function, so it suffices to consider $x > 0$. We have
$$\lim_{x\to\infty} f(x) = 1, \text{ and } \lim_{x \searrow 0} f(x) = 0.$$
Also, $f$ is continuous and attains negative values, for example $f\left(\frac{2}{3\pi}\right) = -\frac{2}{3\pi} < 0$, hence it attains a global minimum (not a priori necessarily in a unique point).
For $x \geqslant \frac{1}{\pi}$, we have $f(x) \geqslant 0$, and for $0 < x < \frac{2}{3\pi}$, we have $$f(x) \geqslant - \left\lvert x\sin \frac1x\right\rvert \geqslant - \lvert x\rvert > - \frac{2}{3\pi},$$ so a point where a global minimum is attained must lie between $\frac{2}{3\pi}$ and $\frac{1}{\pi}$.
To find the global minimum, we can (numerically) solve $f'(x) = 0$, or
$$\sin \frac1x = \frac{1}{x}\cos \frac1x$$
in the interval $\left[\frac{2}{3\pi},\frac1\pi\right)$.
Since $f'\left(\frac{2}{3\pi}\right) = -1 < 0$, the minimum is actually attained in the interval $\left(\frac{2}{3\pi},\frac1\pi\right)$.
So let's use $y = \frac1x$ for ease of notation and solve
$$\sin y = y\cos y \iff y = \tan y$$
in the interval $(\pi, \frac{3\pi}{2})$. In that interval, we can also write it as $y = \arctan y + \pi$, where $\arctan$ denotes the principal branch. That gives a reasonably quickly converging approximation (of course, the Newton method converges faster, but the derivative of $\arctan$ is $\approx \frac{1}{20}$ near the fixed point, so the convergence is fast enough for double precision)
$$y \approx 4.493409457909064;\quad x \approx 0.22254815844566586; \quad f(x) \approx -0.21723362821122166,$$
and the global minimum is (except for the sign, since $f$ is even) indeed unique.
A: Let $F(x)=\frac{\sin x} x$. Consider $x>0$  is sufficient
the only number $x_0\in (\frac\pi 2,\frac{3\pi}2)$ such that $\tan x_0= x_0$
I think the  global minimum of $F(x)$ is $F(x_0)$
If $x\le\pi$, then $F(x)\ge0$;
If $x\ge2\pi$, then $F(x)\ge -\frac1x\ge-\frac1{2\pi}$.
so Consider $x\in (\pi,2\pi)$  is sufficient.
$$F^\prime (x)=\frac{x\cos x-\sin x}{x^2}=\frac{\cos x}{x^2}(x-\tan x)$$
if $x\in (\pi,x_0), x-\tan x>0, \cos x <0$, so $F^\prime (x)<0$;
if $x\in (x_0,\frac{3\pi}2), x-\tan x<0, \cos x <0$, so $F^\prime (x)>0$;
if $x\in (\frac{3\pi}2,2\pi), \sin x<0, \cos x >0$, so $F^\prime (x)>0$;
we get that the  global minimum of $F(x)$ is $F(x_0)$
