# 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. – Gerry Myerson Jun 23 '11 at 1:43
• The solution is unique since $y=x$ intersects $y= \cos x$ only one time. – user57052 Jan 12 '13 at 18:33
• @JoelReyesNoche: The other is a duplicate of this one. Check the dates. – user88595 Jun 27 '14 at 8:33
• @user88595, sorry about that. I thought the proper procedure was to flag both; I should have just flagged the newer one. – Joel Reyes Noche Jun 27 '14 at 13:34
• I believe there is no known way to write this solution in terms of the Lambert W function. – GEdgar Jul 2 '14 at 2:45

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? – corto Jun 22 '11 at 17:19
• What do you mean with a closed form? – Ilya Jun 22 '11 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. – Beni Bogosel Jun 22 '11 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 – corto Jun 22 '11 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. – Beni Bogosel Mar 4 '15 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.

• Thanks for the interesting link! – corto Jun 22 '11 at 18:18

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 $0 < x < \pi/2$

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 – Squareoot Apr 17 '20 at 8:22

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 $\sqrt2$ 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 $q = \sqrt2$ is the only real number such that $q^3=2$. Analogously, the number $\bar x$ is the only number such that $\bar x-\cos \bar x=0$.

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 $$\left( \begin{array}{cc} n & x_{(n)} \\ 1 & \color{red}{0.739}536133515238 \\ 2 & \color{red}{0.739}100520482138 \\ 3 & \color{red}{0.739085}585917040 \\ 4 & \color{red}{0.7390851}49503943 \\ 5 & \color{red}{0.739085133}811963 \\ 6 & \color{red}{0.7390851332}38222 \\ 7 & \color{red}{0.73908513321}6073 \\ 8 & \color{red}{0.7390851332151}98 \\ 9 & \color{red}{0.73908513321516}2 \\ 10 & \color{red}{0.739085133215161} \end{array} \right)$$

• 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. – PM 2Ring Oct 28 '20 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 Jan 24 '20 at 15:34

The shortest correct answer to the question "What is the solution of cos(x)=x?" is

$$\huge {ա}$$

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.