Roots of $x \cos{x}-1$

Question

Edit:I'm not interested in solving it numerically

I'm trying to find an equation of roots of

$x \cos{x}-1$

I've looked at this post, but doesn't answer what I'm looking for

To be clear, as an example, the roots of $\cos{x}-.5$ are:

$x = (6 π n + π)/3, n \in \mathbb{Z}$

Is there no defined way to do the same thing with $x\cos{x}-1$? If so, is there a proof this can't be done? Or is there something else I should know about this problem?

Answer 1

If you look for the zeros of function $$f(x)=x\cos(x)-1$$ they will be closer and closer to $(2n+1)\frac \pi2$.

Expanded as an infinite series around $\color{red}{x_0=(2n+1)\frac \pi2}$ $$f(x)=-1+\sum_{k=0}^\infty (-1)^n \,\frac{ k \cos \left(\frac{\pi k}{2}\right)-x_0 \sin \left(\frac{\pi k}{2}\right)}{k!}\,(x-x_0)^k$$ Truncate to some order and using power series reversion $$\color{blue}{x_{(n)}=x_0-\left(\frac{1}{x_0^3}+\frac{2}{3 x_0^5}+\frac{83}{15 x_0^7}+O\left(\frac{1}{x_0^9}\right)\right)+}$$ $$\color{blue}{(-1)^{n+1}\left(\frac{1}{x_0}+\frac{1}{6 x_0^3}+\frac{83}{40 x_0^5}+\frac{5}{2 x_0^7}+O\left(\frac{1}{x_0^9}\right)\right)}$$

which is quite accurate even for very small values of $n$ $$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution}\\ -4 & -10.90373354 & -10.90373353 \\ -3 & -7.979630987 & -7.979631071 \\ -2 & -4.487698238 & -4.487669604 \\ -1 & -2.011120445 & -2.073932809 \\ +1 & +4.917179487 & +4.917185925 \\ +2 & +7.724153466 & +7.724153192 \\ +3 & +11.08590172 & +11.08590173 \\ +4 & +14.06601357 & +14.06601357 \\ \end{array} \right)$$

Answer 2

The keyword you need is: closed form.

$$x\cos(x)-1=0$$

The left-hand side of your equation is the function term of an elementary function.

$$\cos(x)=\frac{1}{2}e^{ix}+\frac{1}{2}e^{-ix}$$ (Wikipedia: Trigonometric functions)

The following solution theory is, unfortunately, not yet generally known.

$$x\left(\frac{1}{2}e^{ix}+\frac{1}{2}e^{-ix}\right)-1=0$$

This equation defines an algebraic relation between more than one algebraically independent monomials ($x$, $e^{ix}$). The left-hand side of this equation is therefore not in a form to read its inverse as an elementary function which we can read from the equation. That means we cannot rearrange the equation for $x$ by applying only finite numbers of elementary functions/operations from the left to the left-hand side and to the right-hand side which we can read from the equation.

If elementary partial inverses exist, is a different mathematical problem.

$$\frac{1}{2}x(e^{ix})^2-e^{ix}+\frac{1}{2}x=0$$ $x\to\frac{t}{i}$: $$-\frac{1}{2}it(e^t)^2-e^t-\frac{1}{2}it=0$$

The left-hand side of this equation is an algebraic function of both $t$ and $e^t$. Liouville proved that such kind of functions (over a complex domain without isolated points) doesn't have elementary partial inverses: How can we show that $A(z,e^z)$ and $A(\ln (z),z)$ have no elementary inverse?

If the equation has solutions that are elementary numbers, is a different mathematical problem.

The equation is also an irreducible algebraic equation of $t$ and $e^t$. Lin proved, assuming Schanuel's conjecture is true, that such kind of equations don't have solutions except $0$ that are elementary numbers. You can read this also from Chow.

Closed-form inverses can give hints for properties and calculation of the inverses. The inverse relation, Lcos, of the functions with the function term $x\cos(x)$ is mentioned in [Vazquez-Leal et al. 2020] table 11 and 13 and can be represented in terms of the inverse relation Lcosh as presented there.

We can take these inverse relations as closed forms because some of their algebraic properties and their applicability for some other kinds of equations are presented in the cited article.

[Vazquez-Leal et al. 2020] Vazquez-Leal, H.; Sandoval-Hernandez, M. A.; Filobello-Ninoa, U.: The novel family of transcendental Leal-functions with applications to science and engineering. Heliyon 6 (2020) (11) e05418
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see for example i.a.:

What does closed form solution usually mean?

How to know if I can't solve an equation with "standard" methods?

Why some inverse functions do not have a closed form

Answer 3

After transforming $x\cos(x)$ into $\left(\frac\pi2\text{sgn}(x)+x\right)\sin|x|$ , there is an explicit solution using Fourier sine series of an inverse function. Note the scaling was adjusted for a more accurate graph:

setting $x=-1$ gives:

$$\boxed{\sec(a)=a\implies a=-\frac\pi2-\frac1\pi\sum_{n=1}^\infty \sin(n)\int_0^\frac\pi2 t\sin\left(n\left(t+\frac\pi2\right)\sin(t)\right)(2t\cos(t)+\pi\cos(t)+2\sin(t))dt=-2.073932809\dots}$$

Which converges slowly here. A basic attempt at the coefficients expands $\sin(y)$ as a series and the inner $\sin(t)$ with exponential functions. Then we used the binomial theorem:

$$\int_0^\frac\pi2 t\sin\left(n\left(t+\frac\pi2\right)\sin(t)\right)(2t\cos(t)+\pi\cos(t)+2\sin(t))dt=\sum_{k=0}^\infty\sum_{m=0}^{2k+1\text{ or }\infty}\frac{(-1)^k}{m!(2k-m+1)!}\left(\frac{in}2\right)^{2k+1} \int_0^\frac\pi2 t((2t+\pi)\cos(t)+2\sin(t))\left(t+\frac\pi2\right)^{2k+1}e^{i((2(m-k)-1)t-m\pi} dt $$

The indefinite integral is a sum of four incomplete gamma functions which are too long to write out as a final answer. Hopefully, there is a simpler expansion. Similar methods will probably find more fixed points.

Answer 4

COMMENT.- Each time that $f(x)=x\cos(x)$ is equal to $1$ you have a root of $g(x)=x\cos(x)-1$ so you do have the equation $$\cos(x)=\frac1x$$ In other words you have the solutions are the intersections of the hiperbola $y=\dfrac1x$ and the curve cosinus $y=\cos(x)$. You don't have periodicity of these zeros and for each root you must to solve a trascendental equation to which apply numerical methods in order to get an approximated solution.

There are no closed form for all the roots.