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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 $$ \left( (2 n - 1) \frac{\pi}{2},(2 n + 1) \frac{\pi}{2} \right), \quad n \in \mathbb{Z}. $$

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    $\begingroup$ 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) $\endgroup$ Jul 12, 2013 at 20:58
  • $\begingroup$ @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. $\endgroup$
    – user5402
    Jul 12, 2013 at 21:17
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    $\begingroup$ 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. $\endgroup$ Jul 12, 2013 at 22:59
  • $\begingroup$ 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. $\endgroup$ Jul 12, 2013 at 23:00
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    $\begingroup$ This question is related: Derivation of asymptotic solution of $\tan(x) = x$. $\endgroup$ Jul 13, 2013 at 0:37

4 Answers 4

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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).

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  • $\begingroup$ 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. $\endgroup$
    – user5402
    Jul 13, 2013 at 7:39
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One closed-form solution is $0$.

1)

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 algebraically 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:

$$it\left(e^t\right)^2+it-i\left(e^t\right)^2+i=0.$$

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.

2)

Maybe the equation is solvable by generalized Lambert W, see
[Mezö 2017] Mezö, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553
[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)
[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018

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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 $-x_k$.

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    $\begingroup$ The is the weirdest answer I've ever up-voted. $\qquad$ $\endgroup$ Dec 22, 2020 at 16:54
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    $\begingroup$ $$ \begin{align} \tan x = {} & \tan\left( \left( \tfrac\pi2 + k\pi \right) - \tfrac 1 {\pi/2 + k\pi} \right) \\ {} \\ = {} & \frac{\tan\left( \frac\pi2 + k\pi \right) - \tan \left( \frac 1 {\pi/2 + k\pi} \right)}{1 + \tan\left( \frac\pi2 + k\pi \right)\tan \left( \frac 1 {\pi/2 + k\pi} \right)} \\ {} \\ = {} & \frac{\infty - \frac 1 {\pi/2 + k\pi}}{1 + \infty\cdot\frac 1 {\pi/2 + k\pi}} \\{} \\ & \text{(since $\tan=\infty$ at certain points)} \\ {} \\ = {} & \lim_{t\,\to\,\infty} \frac{t-\frac1{\pi/2 + k\pi}}{1+t\cdot \frac 1 {\pi/2 + k\pi}} = \frac\pi2 + k\pi \approx x. \end{align} $$ $\endgroup$ Dec 22, 2020 at 16:54
  • $\begingroup$ ${} \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. $\qquad$ $\endgroup$ Dec 22, 2020 at 16:57
  • $\begingroup$ For $k=13$ I get $x\approx42.38792$ and $\tan x\approx42.40364. \qquad$ $\endgroup$ Dec 22, 2020 at 17:00
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    $\begingroup$ For clarification: The solution of Quanto isn't an exact solution, it's an approximate solution. $\endgroup$
    – IV_
    Mar 12, 2022 at 18:38
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The Closed Form:

I have derived a closed form, using Mathematica functions, for the inverse of the function, based on this series definition. First, let’s find the closed form for:

$$\tan(x)=x\iff \frac{\sin(x)}x=\cos(x)\iff\sqrt\frac{2}{\pi x}\left(\frac{\sin(x)}x-\cos(x)\right)=\text J_\frac32(x)=0$$

where appears the Bessel J function mentioned in OEIS A$115365$. Now use the Bessel J zero function $\text j_{n,x}$:

$$\text J_\frac32(x)=0 \implies x=\text j_{\frac32,n\in\Bbb N}\implies \tan(\text j_{\frac32,n} )= \text j_{\frac32,n }$$

Test the formula here with the $n$th real zero. See here for a sum of Bessel J zeros

Therefore:

$$\boxed{\tan(x)=x\implies x=\text j_{\frac32, n\in\Bbb N}\sim\pi n}:$$

enter image description here

The “closed form” of the inverse of $\tan(x)-x$:

Our goal is to use The series definition of Inverse Beta Regularized $\text I^{-1}_z(a,b)$ about $z=0$ to extend it’s domain and technically get a closed form. Here are some definitions including the Regularized Beta function $\text I_z(a,b)$ where it’s inverse calculates the quantile:

$$\text I_z(a,b)=\frac{\int_0^z t^{a-1} (1-t)^{b-1}dt}{\int_0^1 t^{a-1} (1-t)^{b-1}dt }=x \mathop\implies^{0\le x\le 1}_{a,b>0}z=\text I^{-1}_x(a,b)$$

The restrictions on $z$ make there be an inverse. Now observe that:

$$-\frac\pi2 \text I_{\sin^2(z)}\left(\frac32,-\frac12\right)=-\frac\pi2\int_0^{\sin^2(z)} \frac {\sqrt z}{2(1-z)^\frac32}dz=\tan(z)-z=x\implies z\mathop=^{0\le z\le \frac\pi2}\sin^{-1}\sqrt{\text I^{-1}_{-\frac{2x}\pi}\left(\frac32,-\frac12\right)}$$

but $b<0$, so Wolfram Alpha will not evaluate the function. However, there is a series expansion:

$$\text I^{-1}_z(a,b)\mathop=^{a>0}(az\text B(a,b))^\frac1a+\frac{b-1}{a+1} (az\text B(a,b))^\frac 2a+ \frac{(b-1)(a^2+3ab-a+5b-4)}{2(a+1)^2(a+2)}(az\text B(a,b))^\frac 3a +…$$

Here is a plot of the expansion up to $4$ terms with the principal root. The inverse is accurate enough for $|z|<\frac12$. Using the link’s further terms is hard in Wolfram Alpha since the coefficients are too long to type out:

$$\boxed{z=\tan(y)-y\implies y\mathop=^\text{series}_\text{extension}\sin^{-1}\sqrt{\text I^{-1}_{-\frac{2z}\pi}\left(\frac32,-\frac12\right)} = \sin^{-1}\sqrt{3^\frac23x^\frac23-\frac95\sqrt[3]3x^\frac43+\frac{432}{175}x^2-\frac{821}{875}3^\frac23 x^\frac83+…}}$$

enter image description here

Test the series here. Evaluation of more series terms are needed.

Actual Closed forms for the inverse:

See here for a 2 closed form answers to the inverse function of $x-\tan(x)$ where one is with a limit.

Please correct me and give me feedback!

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  • $\begingroup$ When you refer to the DLMF, you should refer to a formula or at least a subsection. "for a sum of Bessel J zeros" What sum are you talking about? I cannot see any sums of Bessel zeros there, but I can see asymptotic expansions for them. $\endgroup$
    – Gary
    Mar 14, 2022 at 1:08
  • $\begingroup$ None of those are closed forms in the commonly accepted meaning of that term. $\endgroup$ Apr 25, 2022 at 1:45
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    $\begingroup$ That depends. Some people consider Bessel functions acceptable in a closed form, some don't. $\endgroup$ Apr 25, 2022 at 7:58
  • $\begingroup$ Expression with a limit or series is not usually considered a "closed form". $\endgroup$
    – Anixx
    Jun 13, 2022 at 20:44
  • $\begingroup$ @Anixx Yes, but that result is relevant. Also, that formula is in this link. $\endgroup$ Jun 13, 2022 at 21:43

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