Explaining $\cos^\infty$

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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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).

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.

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.

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

The number is

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

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.

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