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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?

  • 3
    @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)))$2012-11-02

7 Answers 7

14

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).

4

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.

  • 0
    @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 $\inf$?2012-11-03
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.

  • 0
    @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_expansion2015-03-04
3

The number is

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

  • 0
    @Blue: Misunderstanding. When I wrote "discontinuity exactly in $ t-\sin(t)=\pi/2$", I wanted to mean "discontinuity exactly WHEN $t-\sin(t)=\pi/2$".2015-03-05
1

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

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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.

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