How does multiplying $\sin(x)$ by $x$ shift the extrema points of the $\sin(x)$ function from periodic intervals of $\frac{n\pi}{2}$, where $n \in 2\mathbb{Z}+1$ (odd integers) to the offsets of
$x = 2.0287578,\ $
$x = 4.9131804,\ $
$x = 7.9786657,\ $
$x = 11.085538,\ $
$x = 14.207436,…\ $
and how does it change the regular sine functions $-1,1$ amplitude to the new amplitudes of $x\sin(x)$
Is it possible to create a formula to hits all these $x$ values (don't care how complex it may be, so long as it takes the integer n incrementally and returns one of those x values above), something that contains $\frac{n\pi}{2}$ and doesn't involve numerical approximations or using it's derivative at 0?
EDIT: In terms of numerical approximations, I would be happy with the limit approach such as using the example of $\tan(e^{x})$, where using the function $\ln(\arctan(999999999)\cdot n)$ as $\lim_{x \rightarrow \infty }$ or in this case 999999999 gets to a close approximation like using a sequence formula of $\ln(\frac{n\pi}{2})$.
note all you have to do it increment the value $n$, leaving $99999999$ alone
And if it's not possible, why so? What are the properties of the function $x\sin(x)$ that doesn't allow this to be so?
Best Answer
The extrema occur at the roots of
$$\sin(x)+x\cos(x)=0,$$ or $$\tan x=-x.$$
As the tangent has vertical asymptotes at $x=\left(n+\frac12\right)\pi$, the solutions are close to these values. More precisely, $x\approx\left(2n+\frac12\right)\pi$ in the positives.
Now using Taylor,
$$\tan x=\frac1{\tan\left(\left(2n+\frac12\right)\pi-x\right)}\approx\frac1{\left(2n+\frac12\right)\pi-x}$$
giving the quadratic equation
$$x^2-\left(2n+\frac12\right)\pi x-1=0$$ that you solve for $x$.
$$x=\frac{\sqrt{n^2+n+\frac{17}4}+2n+1}4$$
You can improve on this solution by means of one Newton's iteration,
$$x\leftarrow x-\frac{\tan x+x}{\tan^2x+2}.$$
The analytical expression in terms of $n$ is heavy:
$$\frac{\sqrt{n^2+n+\frac{17}4}+2n+1}4-\frac{\tan \dfrac{\sqrt{n^2+n+\frac{17}4}+2n+1}4+\dfrac{\sqrt{n^2+n+\frac{17}4}+2n+1}4}{\tan^2\dfrac{\sqrt{n^2+n+\frac{17}4}+2n+1}4+2}.$$
The first root, corresponding to $n=0$ is $2.056952$, and the improved value $2.027446$. Then $\tan 2.027446=-2.0354865$.
The larger $n$, the better the approximations.
Update:
The last formulas are wrong as the factor $\pi$ was dropped. But the numerical values are correct.