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?
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$\begingroup$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 $f(x)=f(\cos x)\ \forall x \in \Bbb{R}$.
$\endgroup$ 13 $\begingroup$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.
$\endgroup$ 1 $\begingroup$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 $0 < x < \pi/2$
$\endgroup$ 2 $\begingroup$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!
$\endgroup$ 1 $\begingroup$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:
$$\boxed{\text{Dottie Number}=\text D=\sin^{-1}\left(1-2\text I^{-1}_\frac12\left(\frac 12,\frac 32\right)\right)}$$
from The Incomplete Beta function $\text B_z(a,b)$:
$$\text B_{\sin^2(z)}\left(\frac 12,\frac32\right)=z+\cos(z)\sin(z)$$
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!
$\endgroup$ 4 $\begingroup$See explanation here
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 $\bar x-\cos \bar x=0$.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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 $0.1$%.
$\endgroup$ 1 $\begingroup$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 $x=1,x=\sqrt{3}-1$.
$\endgroup$ 1 $\begingroup$The shortest correct answer to the question "What is the solution of cos(x)=x?" is
$\huge {ա}$
(the Armenian letter that is used to denote the Dottie number.)
$\endgroup$ $\begingroup$Some simple rearrangements of the equation show that the equation is not in a form that is solvable in terms of Lambert W.
The main theorems in [Lin 1983] and [Chow 1999] imply, assuming Schanuel's conjecture is true, that your equation cannot have solutions that are elementary numbers or explicit elementary numbers respectively.
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[Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448
$\endgroup$ $\begingroup$I would say that it IS already in closed form... if we follow the definition given here :
since trigonometric functions are considered "well-known"... don't you agree?
$\endgroup$ 3 $\begingroup$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.
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