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

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15  
The number of the bored student (with a cos-button on his calculator) :-) –  Myself Jun 22 '11 at 17:07
    
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 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 at 13:34
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3 Answers 3

up vote 11 down vote accepted

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

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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
    
I have just seen here another discussion on what "closed form" means. –  corto Jun 22 '11 at 17:30
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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.

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Thanks for the interesting link! –  corto Jun 22 '11 at 18:18
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The solution for $cos x = x$ must exist between $-1$ and $1$

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

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$$f'(x) = -\sin x -1$$

Aand $f'(x)$ is negative for all $x$ in $[-1, 1]$

$ $

SO there can be at most one solution for the equation $$\cos x = x$$

$ $

As you know one solution, you can rest assured there is no other solution.

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2  
This does not answer the question. –  MJD Jul 2 at 1:50
    
by the way, what do u mean by "closed form solution" –  Holy cow Jul 2 at 3:16
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