# What is the solution of $\cos(x)=x$?

There is an unique solution with $$x$$ being approximately $$0.739085$$. But is there also a closed-form solution?

• I wonder if it can be expressed in terms of the Lambert $W$-function, which some people would like to add to the list of closed forms. Commented Jun 23, 2011 at 1:43
• The solution is unique since $y=x$ intersects $y= \cos x$ only one time.
– user57052
Commented Jan 12, 2013 at 18:33
• I believe there is no known way to write this solution in terms of the Lambert W function. Commented Jul 2, 2014 at 2:45
• Where there is a will there is usually a way.
– Jay
Commented Nov 29, 2018 at 18:13
• It has a unique real solution. Does it have a unique complex solution? Commented Feb 23, 2022 at 16:34

The equation in question is a transcendental equation. Apart of guessing, numerical or analytical methods, there is no way of solving the equation without using another transcendental function, and therefore argue in circles.

In this case, denote $$g(x)=\cos x -x$$, see that its derivative is negative with countable many zeros, and therefore $$g$$ is strictly decreasing, yielding that there is at most one solution to $$g(x)=0$$. Since $$g(0)g(\pi/2)<0$$ there is such a solution. Arbitrary precise approximations can be found using Newton, bisection, or false position method.

As user Myself commented, it is a challenge (not so hard) to prove that the sequence $$x_{n+1}=\cos x_n, x_0 \in \Bbb{R}$$ converges to the unique solution to $$\cos x=x$$.

Another related problem which I encountered last week when trying to help one of my friends for an exam is to find all continuous functions $$f : \Bbb{R} \to \Bbb{R}$$ with the property that $$f(x)=f(\cos x)\ \forall x \in \Bbb{R}$$.

• But, in my understanding, being transcendental does not imply that the equation cannot have a closed form solution? Commented Jun 22, 2011 at 17:19
• What do you mean with a closed form?
– SBF
Commented Jun 22, 2011 at 17:20
• Some equations like $\sin x=x$ have solutions you can "put your hand on", in this case $x=0$. But this is just because we can guess them. Commented Jun 22, 2011 at 17:25
• Something like "An equation is said to be a closed-form solution if it solves a given problem in terms of functions and mathematical operations from a given generally accepted set." mathworld.wolfram.com Commented Jun 22, 2011 at 17:26
• If you search "transcendental equations" on the Internet, I guess things will become more clear (look at the Wikipedia link). The idea is that in a transcendental equation you cannot derive a formula for the solution $x$ using only basic arithmetic operations. You may express $x$ using arithmetic operations involving series, Lambert's function, or Bessel functions, but these functions are by definition solutions to some equations we cannot solve explicitly. Commented Mar 4, 2015 at 10:40

Mathworld calls this the Dottie Number. The page makes no mention of existence/non-existence of "closed" form and I would guess it is still open.

Remembering the Kepler equation and its solution, the Dottie number can be analytically written as:

$$D = 2\sum_{n=0}^\infty \left( \frac{J_{4n+1}(4n+1)}{4n+1} - \frac{J_{4n+3}(4n+3)}{4n+3}\right)$$

where $J_{n}$ are the Bessel functions. Such series is convergent and can be evaluated numerically.

A proof and numerical evaluations are provided in :

Solving $2x - \sin 2x = \pi/2$ for $0 < x < \pi/2$

Here is a closed form I just found. Using Kepler’s Equation, the median of a beta distribution, and Inverse Beta Regularized $$\text I^{-1}_z(a,b)$$. Here is the closed form:

$$\boxed{\text{Dottie Number}=\text D=\sin^{-1}\left(1-2\text I^{-1}_\frac12\left(\frac 12,\frac 32\right)\right)}$$

$$\text B_{\sin^2(z)}\left(\frac 12,\frac32\right)=z+\cos(z)\sin(z)$$

taking the inverse, and a bit of algebra to get the form above. Also using the Half Covered Sine:

$$\text I^{-1}_\frac12\left(\frac 12,\frac 32\right)=\text{hacoversin}(\text {D)}=\frac12(1-\sin(\text D))\implies \text D= {\text{hacoversin}}^{-1} \text I^{-1}_\frac12\left(\frac 12,\frac 32\right)$$

With an error of $$10^{-179}$$ in this numerical evaluation. This is a side post since there already is an accepted answer. Please correct me and give me feedback!

See explanation here

• I check numerically is true with error of $10^{-100000}$. Commented Feb 23, 2022 at 16:54
• The factor 2 before the root is wrong (not needed) in your last comment. It should be $\sqrt{1-\left(2\text I^{-1}_\frac12\left(\frac 12,\frac 32\right)-1\right)^2}$ Commented May 20, 2022 at 10:42
• @Anixx Correct. I did not want to retype the mathjax, but you already did. Thanks. Commented May 20, 2022 at 20:47
• Try to plot Log[InverseBetaRegularized[1/2, x, 3/2]] and -1/x^0.889, the graphics nearly coincide. Commented Oct 24, 2022 at 7:13

As I know, there is no exact way to get the solution for $$\cos(x)=x$$. But, you can use Newton's Method to get an approximate answer:

Consider the function $$f(x)=\cos x−x$$

This gives us that $$f'(x)=-\sin x-1$$

Newton's Method states that $$x_{\text{n+1}} = x_{\text{n}} - \dfrac{f(x_{\text{n}})}{f'(x_{\text{n}})}$$

Then, just start at $$x_{\text{0}}=1$$ and repeat this method all over again, until you are satisfied. Don't forget that rounding numbers might lead to wrong answers!

• Best answer as in easiest Commented Apr 17, 2020 at 8:22
• So this is noncomputable, then. Commented Mar 2 at 6:54

It is very easy to show that the equation $\cos x = x$ has a unique solution. For example take $f(x) = x - \cos x$ and notice that $f'(x) = 1+\sin x \ge 0$ (equality holding in isolated points) so $f(x)$ is strictly increasing and hence the equation can have at most one solution. Since $f(x)>0$ for $x\ge 1$ and $f(x)<0$ for $x\le 0$, and the function is continuous, by the intermediate values theorem there exists one and only one solution $\bar x \in [0,1]$.

For this particular equation there is also a very nice numeric approximation. In fact $\bar x = \lim x_n$ where $x_{n+1} = \cos (x_n)$ is any iteration of the function $\cos x$. You can easily find the numeric value for $\bar x$ simply putting any number in your pocket calculator and pressing repeatedly the $\cos$ button. In fact $\bar x$ is the fixed point of the $\cos$ function and, (at least in $[0,1]$) the $\cos$ function is a contraction hence every iterated sequence converges to the unique fixed point.

I can also convince you that $\bar x$ is an exact solution to the equation $\cos x = x$. I think that you agree that $\sqrt[3]2$ is an exact solution of the equation $x^3=2$, don't you? Now notice what's going on here... one notices that the function $x^3$ is strictly increasing hence invertible. You give a name to the inverse function and call it: cubic root. Then you find an algorithm to compute the cubic root on your calculator. Isn't this the same thing we did with the function $f(x) = x-\cos x$?

By definition the number $q = \sqrt[3]2$ is the only real number such that $q^3=2$. Analogously, the number $\bar x$ is the only number such that $\bar x-\cos \bar x=0$.

Your equation cannot be solved in terms of elementary functions, elementary functions and Lambert W or elementary functions and Generalized Lambert W of Mezö et al.. It can be solved in terms of "Leal-functions" and possibly by Generalized Lambert W of [Castle 2018].

1.) Elementary functions, elementary numbers

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

The function on the left-hand side of the latter equation is an algebraic function in dependence of both $$t$$ and $$e^t$$. Liouville proved that such kind of functions (over a complex domain without isolated points) don't have (partial) inverses that are elementary functions.

The equation is also an algebraic equation in dependence of both $$t$$ 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.

How can we show that $A(z,e^z)$ and $A(\ln (z),z)$ have no elementary inverse?

2.) Lambert W, Generalized Lambert W

The latter equation also shows that the equation cannot be solved in terms of elementary functions and Lambert W or Generalized Lambert W of Mezö et. al. either. But possibly is it solvable in terms of Generalized Lambert W of [Castle 2018].

[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

3.) "Leal-functions"

$$\cos(x)=x$$ $$\cos(x)-x=0$$ $$x\to-t$$: $$t+\cos(t)=0$$ $$t=\text{Lcos}_2(0)$$ $$x=-\text{Lcos}_2(0)$$

[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

• @TymaGaidash For mathematics, it's enough to describe a new function. So we can work with the cited representations now. I wrote what I found so far. My aim was to make these inverses better known. That could help to find someone who implements it. At the moment, everyone has to make an own implementation for themselves.
– IV_
Commented Dec 25, 2022 at 16:49
• @TymaGaidash You could help to try to find the relations between "Leal-functions" and InverseBetaRegularized function.
– IV_
Commented Dec 25, 2022 at 16:59

I'm truth, there the above answers are correct, stating that the number currently has no closed form solution.

An old approximation that I came up with to $\cos$ can offer a decent approximation to the Dottie number however.

$$\sqrt[3.911]{\frac{1-x^2}{1+x^2}} \approx \cos{x}$$

Setting the above approximation $=x$ and solving can give a closed form solution that lies within 4 decimal points of the solution.

Not a closed form but a way to generate very good approximations of the Dottie number.

Since the solution is close to $$\frac \pi 4$$, we have for $$y=x-\cos(x)$$ $$y=\frac{\pi -2 \sqrt{2}}{4} +\left(1+\frac{1}{\sqrt{2}}\right) \left(x-\frac{\pi }{4}\right)+\frac{1}{\sqrt{2}}\sum_{n=2}^\infty\frac{\sin \left(\frac{\pi n}{2}\right)-\cos \left(\frac{\pi n}{2}\right)}{n!}\left(x-\frac{\pi }{4}\right)^n$$ Truncating to some order $$O\left(\left(x-\frac{\pi }{4}\right)^n\right)$$ and using series reversion, we should get things like $$x=\frac \pi 4+\frac{32 \left(11482+8119 \sqrt{2}\right) t}{\left(2+\sqrt{2}\right)^{11}}-\frac{16 \left(4756+3363 \sqrt{2}\right) t^2}{\left(2+\sqrt{2}\right)^{11}}+\frac{32 \left(5333+3771 \sqrt{2}\right) t^3}{3 \left(2+\sqrt{2}\right)^{11}}+O\left(t^4\right)$$ where $$t=\frac{1}{4} \left(\sqrt{2}-2\right) \left(4 y+\pi -2 \sqrt{2}\right)$$. Making $$y=0$$ and using this very truncated series would give $$x \sim 0.739085133238$$ to be compared to the exact $$0.739085133215$$

Playing with the $$n$$ of $$O\left(\left(x-\frac{\pi }{4}\right)^n\right)$$, we could get the following results $$\left( \begin{array}{cc} n & x_{(n)} \\ 1 & \color{red}{0.739}536133515238 \\ 2 & \color{red}{0.739}100520482138 \\ 3 & \color{red}{0.739085}585917040 \\ 4 & \color{red}{0.7390851}49503943 \\ 5 & \color{red}{0.739085133}811963 \\ 6 & \color{red}{0.7390851332}38222 \\ 7 & \color{red}{0.73908513321}6073 \\ 8 & \color{red}{0.7390851332151}98 \\ 9 & \color{red}{0.73908513321516}2 \\ 10 & \color{red}{0.739085133215161} \end{array} \right)$$

Edit

For the fun of it, using the $$\large 1,400$$ years old approximation $$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad \text{for} \qquad -\frac \pi 2 \leq x\leq\frac \pi 2$$ solving the cubic equation $$x^3+4 x^2+\pi ^2 x-\pi ^2=0$$ gives as an approximation $$x\sim -\frac{2}{3} \left(2+\sqrt{3 \pi ^2-16} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{128-63 \pi ^2}{2 \left(3 \pi ^2-16\right)^{3/2}}\right)\right)\right)=0.738305$$ that is to say a relative error of $$0.1$$%.

• That's pretty impressive for a polynomial approximation! True, Newton's method $x'=(\cos x+x\sin x)/(1+\sin x)$ converges much faster, but your solution avoids the use of trig function calls. Commented Oct 28, 2020 at 14:43

You could use cosine Taylor expansion and solve the polynomial $\sum_{i=0}^{n}(-1)^i \frac{x^{2i}}{(2i)!}-x=0$ For example; for n=1,2 you get the approximations $x=1,x=\sqrt{3}-1$.

• This method is impractical, as polynomial roots do not all have a closed-form solution. Therefore, other numerical approaches would be needed (e.g. Newton's method), which can equally well be done with the original formulation $\cos x-x$. So there is no benefit.
– Jam
Commented Jan 24, 2020 at 15:34

I would say that it IS already in closed form... if we follow the definition given here :https://en.m.wikipedia.org/wiki/Closed-form_expression

since trigonometric functions are considered "well-known"... don't you agree?

• what do you mean? Commented May 16, 2022 at 0:43
• You are right, $\cos(x)=x$ is an equation in closed form, it is an elementary equation. But the questioner asks for a closed-form solution (a solution in closed form) of that equation.
– IV_
Commented Aug 5, 2022 at 16:10

My AskConstants "constant recognition" program at http://AskConstants.org proposed the explicit exact closed form

    RealInverseSphericalBesselY [0, -1, 1],


where $$0$$ is the order and $$1$$ is the branch number.

This was subsequently proved in my article at https://arxiv.org/abs/2207.00707

• Sorry, you have not proved anything. Rayleigh's formulae (see Wikipedia page on Bessel functions) explicitly give $j_n(x)$ and $y_n(x)$ as rational functions of $\sin x,\cos x,x$, including $y_0(x)=-\frac{\cos x}x$. Commented Aug 5, 2022 at 5:13
• Please disclose your affiliation with AskConstants.org when referring to it, otherwise the post may be considered spam. See What makes something spam and when should I flag it?
– robjohn
Commented Aug 10, 2022 at 22:29
• What is this function? It is definitely not a function in Mathematica. Commented Aug 12, 2022 at 0:43

this can be really easy with the help of graphs. Draw the graphs of y= cosx and x=y. Intersection points on the graph will show the solutions.