# Express solutions of equation $\tan x= x$ in closed form

I know that the equation $$\tan(x) = x$$ can be solved using numerical methods, but I’m looking for a closed form of the solutions. In my opinion, having only numerical solutions means that we don’t know the problem, and sooner or later, we’ll be able to find a closed-form solution or at least a power-series solution.

I’m looking for an explicit form of a sequence $$(x_{n})_{n \in \mathbb{Z}}$$ in $$\mathbb{R}$$ such that

• $$0 < x_{0} < \dfrac{\pi}{2}$$,
• $$\tan(x_{n}) = x_{n}$$ for all $$n \in \mathbb{Z}$$, and
• $$(2 n - 1) \dfrac{\pi}{2} < x_{n} < (2 n + 1) \dfrac{\pi}{2}$$ for all $$n \in \mathbb{Z} \setminus \{ 0 \}$$.

The existence of such a sequence follows from the continuity of $$x \mapsto \tan(x) - x$$ over its domain. Its uniqueness follows from the monotonicity of $$\tan(x) - x$$ over the intervals $$\left( (2 n - 1) \frac{\pi}{2},(2 n + 1) \frac{\pi}{2} \right), \quad n \in \mathbb{Z}.$$

• Not every problem has a closeed form solution and in some cases it is even provable that no closed form exists (though that depends on what you'd be willing to accept as closed form) Jul 12 '13 at 20:58
• @HagenvonEitzen I think you have the quintic equation in mind right? by closed form I mean any expression of the form $x_n=f(n)$ whether $f(n)$ is an elementary function or a series; the important thing is that I don't want algorithms. Jul 12 '13 at 21:17
• I'd be surprised if this equation has a closed-form solution. It's obvious that $0$ is a solution, and that there is one positive solution slightly less than each of the points where the tangent function has a horizontal asymptote, except $\pm\pi/2$. The sum of the squares of the reciprocals of the solutions does have a simple closed form: It is $1/10$. People have posted proofs of that on m.s.e. if I'm not mistaken. Jul 12 '13 at 22:59
• Instead of looking at the graph of $y=\tan x-x$, just look at the graph of $y=\tan x$ and the graph of $y=x$, superimposed on each other. The $x$-coordinates of the places where they intersect are the solutions. Jul 12 '13 at 23:00
• This question is related: Derivation of asymptotic solution of $\tan(x) = x$. Jul 13 '13 at 0:37

As Hagen has succinctly mentioned in his comment above, whether an equation has a closed-form solution or not depends on the functions that you wish to admit as ‘elementary’. Questions about the existence of closed-form solutions are typically answered using differential Galois theory.

I thus cannot answer your question, but I can at least quote closed-form formulas for some infinite sums whose terms are fixed negative-integer powers of the positive real solutions of $\tan(x) = x$.

Theorem: If $(\lambda_{n})_{n \in \mathbb{N}}$ denotes the sequence of positive real solutions of $\tan(x) = x$ in increasing order, then \begin{align} \sum_{n=1}^{\infty} \frac{1}{\lambda_{n}} &= \infty, \\ \sum_{n=1}^{\infty} \frac{1}{\lambda_{n}^{2}} &= \frac{1}{10}, \\ \sum_{n=1}^{\infty} \frac{1}{\lambda_{n}^{4}} &= \frac{1}{350}. \end{align}

Reference

L. Hermia & N. Saito. On Rayleigh-Type Formulas for a Non-local Boundary Value Problem Associated with an Integral Operator Commuting with the Laplacian, preprint submitted to Journal of Mathematical Analysis and Applications (2010).

• That's helpful. It's good to see other people interested in such problems. I think this is a weakness in mathematics (the closed form solutions are rare whether for transcendental equations, differential equations and even worse for integro-differential equations) and some day (with an appropriate theory), we'll get over it. Jul 13 '13 at 7:39

One closed-form solution is $$0$$.

I can give an answer for the elementary functions.

According to Liouville and Ritt, the elementary functions are those functions which are obtained in a finite number of steps by performing only algebraic operations and/or taking exponentials and/or logarithms (Wikipedia: Elementary function). Each elementary standard function, e.g. the trigonometric functions as in your equation, can be represented in this way.

Your equation $$\tan(x)-x=0$$ contains on its left-hand side an elementary function $$f$$ with $$f\colon x\mapsto\tan(x)-x$$. With $$\tan(x)=\frac{(e^{-ix}-e^{ix})i}{(e^{-ix}+e^{ix})}$$, your equation is $$\frac{(e^{-ix}-e^{ix})i}{(e^{-ix}+e^{ix})}-x=0$$. The elementary function on the left-hand side of the last equation is an algebraic function of the algebraic independent functions $$e^{ix}$$ and $$x$$. This kind of elementary functions cannot have an inverse that is an elementary function. This is a result of the theorem in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 that is proved also in Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math. 101 (1979) (4) 743-759. Therefore it is not possible to solve your equation by transforming it only by applying only elementary functions you can read from the equation.

The proof can be made also by another method, decribed in Khovanskii, A.: Topological Galois Theory. Solvability and Unsolvability of Equations in Finite Terms. Springer 2014 and articles of A. Khovanskii and Y. Burda. It is applied in the following article.
Belov-Kanel, A.; Malistov, A.; Zaytsev, R.: Solvability of equations in elementary functions. Journal of Knot Theory and Its Ramifications 29 (2020) (2) 204-205.
The article proves that $$\tan(x)-x$$ doesn't have an elementary inverse.

Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50 and Chow, T. Y.: What is a Closed-Form Number? Amer. Math. Monthly 106 (1999) (5) 440-448 prove that irreducible polynomial equations in dependence of both $$x$$ and $$e^x$$ with algebraic coefficients cannot have solutions $$x\neq 0$$ that are elementary numbers or explicit elementary numbers respectively.
Your equation can be transformed to that form.

Another method for solving certain ordinary equations in a differential field (like the elementary functions and the Liouvillian functions) is described in Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22.

Here is a simple, yet fairly accurate close-form solution to the equation $$\tan x=x$$

$$x_k = \frac{(1+2k)\pi}2 - \frac2{(1+2k)\pi}$$

with $$k=1,2,3\>...$$ for all the positive roots. Due to symmetry, the negative roots are $$-x_k$$.

• The is the weirdest answer I've ever up-voted. $\qquad$ Dec 22 '20 at 16:54
• \begin{align} \tan x = {} & \tan\left( \left( \tfrac\pi2 + k\pi \right) - \tfrac 1 {\pi/2 + k\pi} \right) \\ {} \\ = {} & \frac{\tan\left( \frac\pi2 + k\pi \right) - \tan \left( \frac 1 {\pi/2 + k\pi} \right)}{1 + \tan\left( \frac\pi2 + k\pi \right)\tan \left( \frac 1 {\pi/2 + k\pi} \right)} \\ {} \\ = {} & \frac{\infty - \frac 1 {\pi/2 + k\pi}}{1 + \infty\cdot\frac 1 {\pi/2 + k\pi}} \\{} \\ & \text{(since \tan=\infty at certain points)} \\ {} \\ = {} & \lim_{t\,\to\,\infty} \frac{t-\frac1{\pi/2 + k\pi}}{1+t\cdot \frac 1 {\pi/2 + k\pi}} = \frac\pi2 + k\pi \approx x. \end{align} Dec 22 '20 at 16:54
• ${} \qquad \uparrow \qquad {}$ The $\text{“}\infty\text{”}$ used above is not $+\infty$ or $-\infty,$ but rather it is the $\infty$ that is approached by going in either direction. $\qquad$ Dec 22 '20 at 16:57
• For $k=13$ I get $x\approx42.38792$ and $\tan x\approx42.40364. \qquad$ Dec 22 '20 at 17:00