Find periodic sequence formula for extrema of $x\sin(x)$, if possible

Question

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?

Answer 1

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.

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Update:

The last formulas are wrong as the factor $\pi$ was dropped. But the numerical values are correct.

Answer 2

In the same spirit as in @Yves Daoust's answer, instead of Taylor series, we could build Padé approximants of $\sin(x)+x\cos(x)$ around $x=\left(n+\frac12\right)\pi$.

In order to keep it simple (that is to say one linear equation), let us use for example the $[1,2]$ Padé approximant. I skip the intermediate calculations but we end with the simple $$x=t+\frac{6 t^2+9}{t(6 t^2+17)} \qquad \text{where} \qquad t=\left(n+\frac12\right)\pi$$

Below are given the results for the first solutions $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 0 & 2.047282372 & 2.028757838 \\ 1 & 4.913295940 & 4.913180442 \\ 2 & 7.978674318 & 7.978665712 \\ 3 & 11.08553997 & 11.08553841 \\ 4 & 14.20743717 & 14.20743673 \\ 5 & 17.33637808 & 17.33637792 \\ 6 & 20.46916747 & 20.46916740 \\ 7 & 23.60428481 & 23.60428477 \\ 8 & 26.74091603 & 26.74091601 \\ 9 & 29.87858652 & 29.87858651 \end{array} \right)$$

Using the $[1,3]$ Padé approximant, we should get $$x=t+\frac{4 t \left(6 t^2+17\right)}{24 t^4+100 t^2+49} \qquad \text{where} \qquad t=\left(n+\frac12\right)\pi$$ For the first root, this would give $2.02306$ (corresponding to a relative error of $0.28$%).