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I noticed something odd while messing around on my calculator.

$$\lim_{x\to \infty} \cos^x(c)=0.7390851332$$ Where $c$ is a real constant.

My calculator is in radians and I got this number by simply taking the cosine of many numbers over and over again. No matter what number I use I always end up with that number. Why does this happen and where does this number come from?

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I suppose you mean $$\underbrace{\cos \circ\cos \circ\cos \circ \cdots \circ \cos \circ \cos}_{x \text{ times}}(c)$$ and not $\cos^x(c)$. – user17762 Nov 2 '12 at 4:50
See Exercise 5.22 in Principles of Mathematical Analysis for a more general result. – wj32 Nov 2 '12 at 4:52
Sorry, what would be the proper format for that? – Zach Sugano Nov 2 '12 at 4:53
@ZachSugano What you have written $$\cos^x(c) = \underbrace{\cos(c) \cdot \cos(c) \cdots \cos(c)}_{x \text{ times}}$$ whereas what you want is $$\underbrace{\cos(\cos(\cdots \cos(}_{x \text{ times}}c)))$$ – user17762 Nov 2 '12 at 5:46

7 Answers 7

up vote 12 down vote accepted

What you have found is the unique, attractive fixed point of $\cos(x)$.

For more on this point and these terms, see this (MathWorld) and this (Wikipedia).

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This is the unique real solution $r$ of $\cos(x) = x$.
For any $x \ne r$ we have $|\cos(x) - r| = \left|\int_{r}^x \sin(t)\ dt\right| < |x - r|$. This implies that $r$ is a global attractor for this iteration.

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Why does that inequality imply that $r$ is a global attractor? – wj32 Nov 2 '12 at 5:03
More is true, after the first iteration, all points lie in $[-1,1]$, hence $|\cos x - \cos y| = |\int_x^y \sin t \, dt| \leq \sin 1 |x-y|$. Hence it is a strict contraction (since $\sin 1 < 1$). – copper.hat Nov 2 '12 at 5:33
@copper.hat: Thanks. – wj32 Nov 2 '12 at 5:50
For the iteration $x_{n+1} = f(x_n)$, if $f$ is continuous and $|f(x) - r| < |x-r|$ for all $x \ne r$, then $r$ is a global attractor. Suppose $r$ is not the limit of the sequence $x_n$. Then this sequence has a limit point $s \ne r$. But $|f(s) - r| < |s - r|$. Let $\epsilon = |s - r| - |f(s) - r|$. By continuity of $|f(x) - r| - |x-r|$ there is $\delta > 0$ such that $|x - s| < \delta$ implies $|f(x) - r| < |x - r| - 2 \epsilon/3$. We can take $\delta < \epsilon/3$, and then this implies $|f(x) - r| < |s - r| - \epsilon/3$. (continued) – Robert Israel Nov 2 '12 at 7:11
But that says there is no more than one $x_n$ with $|x_n - s| < \delta$, as for all $k > n$ we have $|x_k - r| \le |f(x_n) - r| < |s-r| - \epsilon/3$ so that $|x_k - s| \ge |s - r| - |x_k - r| > \epsilon/3 > \delta$. And this contradicts the statement that $s$ is a limit point of $x_n$. – Robert Israel Nov 2 '12 at 7:16

As already discussed in other threads:

What is the solution of cos(x)=x?

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

fhe fixed point of $\cos(x)$ (i.e. the Dottie number) can be written as a particular solution of Kepler equation, therefore it can be also expressed as:

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

where $J_n(x)$ are Bessel functions.

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@Anixx : your elegant integral seems a particular case of (15) in… – giorgiomugnaini Mar 4 at 20:12

its the solution to cos(x)=x, also sometimes known as the dottie number

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The number is

$$\alpha=\frac1\pi \int_0^{\pi } \arctan\left(\tan \left(\frac{t-\sin t+\frac{\pi }{2}}2\right)\right) \, dt+\frac{1}{\pi }$$

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An expression holding $\arctan(\tan(\ldots))$? That does not make an elaborate impression. – GDumphart Mar 4 at 12:26
@GDumphart you can check it. If you eliminate arctan and tan, the result changes, becomes wrong. It you know other ways to simplify it, you are welcome. – Anixx Mar 4 at 12:30
@GDumphart : the $arctan(tan(...))$ in the integral has a discontinuity exactly in $t-\sin(t)=\pi/2$. Trivial simplifications can not be performed. It is interesting that $arctan(tan(x))$ is the sawtooth wave. See :… "On an integral representation of a class of Kapteyn (Fourier–Bessel) series: Kepler’s equation, radiation problems and Meissel’s expansion", A. Eisinberg,G. Fedele,A. Ferrise,D. Frascino – giorgiomugnaini Mar 4 at 20:25
@giorgiomugnaini: I was about to make the same observation about the discontinuity, except I was going to write the equation as $t-\pi/2 = \cos( t - \pi/2)$; that is, the jump happens when $t = \alpha + \pi/2$. – Blue Mar 4 at 20:36
@Blue: Misunderstanding. When I wrote "discontinuity exactly in $ t-\sin(t)=\pi/2$", I wanted to mean "discontinuity exactly WHEN $t-\sin(t)=\pi/2$". – giorgiomugnaini Mar 5 at 12:31

You may treat is as dynamical system with state transition function $f(x) = cos(x)$. After first two iterations $f^{n > 2}(x)$ will lay in interval $I = [cos(1), 1]$. Line $g(x) = x$ will intercept $cos(x)$ in interval $I$ exactly once so $cos(x)$ has unique fixed point in $I$. Because of unique fixed point and because $|f'(x)| < 1$ sequence $f^n(x)$ will converge to this fixed point.

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You found the real solution of $ \cos x = x $ through a fixed point converging process.

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