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I noticed something odd while messing around on my calculator.

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

With $$\cos^n(c) =\underbrace{\cos \circ\cos \circ\cos \circ \cdots \circ \cos \circ \cos}_{n \text{ times}}(c)$$

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?

Guy Fsone
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Zach Sugano
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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 $\cos^x(c)$. –  Nov 02 '12 at 04:50
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    See Exercise 5.22 in Principles of Mathematical Analysis for a more general result. – wj32 Nov 02 '12 at 04:52
  • Sorry, what would be the proper format for that? – Zach Sugano Nov 02 '12 at 04:53
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    @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 $$\underbrace{\cos(\cos(\cdots \cos(}{x \text{ times}}c)))$$ –  Nov 02 '12 at 05:46
  • @user17762 I'd argue that the proper notation for exponents is $(\cos \theta)^n$, since $f^n$ does denote the $n$-fold composition of $f$ in many contexts. To disambiguate, one can also use $f^{\circ n}$. – user76284 Feb 10 '20 at 19:16

6 Answers6

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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).

Ben Grossmann
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Benjamin Dickman
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4

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.

Community
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giorgiomugnaini
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  • @Anixx : your elegant integral seems a particular case of (15) in

    http://www.researchgate.net/publication/257293973_On_an_integral_representation_of_a_class_of_Kapteyn_%28FourierBessel%29_series_Keplers_equation_radiation_problems_and_Meissels_expansion

     – giorgiomugnaini Mar 04 '15 at 20:12
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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.

Robert Israel
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  • Why does that inequality imply that $r$ is a global attractor? – wj32 Nov 02 '12 at 05:03
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    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 02 '12 at 05:33
  • @copper.hat: Thanks. – wj32 Nov 02 '12 at 05:50
  • 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 02 '12 at 07:11
  • 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 02 '12 at 07:16
  • @RobertIsrael: So is continuity the key requirement there? Why couldn't I set $r=0$ and choose some $f$ that makes $x_n$ a positive, decreasing sequence with a positive $\inf$? – wj32 Nov 03 '12 at 01:19
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its the solution to cos(x)=x, also sometimes known as the dottie number

boby
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0

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

Narasimham
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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.

Trismegistos
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