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?

Solution 1:

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 $f(x)=f(\cos x)\ \forall x \in \Bbb{R}$.

Solution 2:

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.

Solution 3:

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 $0 < x < \pi/2$

Solution 4:

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!

Solution 5:

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 $\bar x-\cos \bar x=0$.