Geometric constant

The geometric constant is a special number which may be important in the development of a theory of space-time in pilitron physics.

The constant is approximately:


 * $$\lambda \approx 0.739085133215$$

It is defined such that:


 * $$\lambda = \cos \lambda$$

Derivation
Here are some ways to find the value of the constant.

Infinite-order cosine
If we start with some real number $$n$$, take its cosine, then take the cosine of the answer, and keep going, you'll be approaching the geometric constant:


 * $$\lambda = \lim_{x \rightarrow \infty} \cos^{x}(n)$$

Trigonometry
On a right-angle triangle, with some angle $$\theta$$, opposite to a side of length $$O$$, adjacent to a side of length $$A$$, and with a hypotenuse of length $$H$$, the cosine is given by:


 * $$\cos \theta = \frac{A}{H}$$

If the triangle is put inside of a circle, such that the hypotenuse of the trangle is the radius of the circle, and we have an arc length between one vertex of the triangle to the other, $$\alpha$$, we have:


 * $$\cos \frac{\alpha}{H} = \frac{A}{H}$$

If the angle $$\theta$$ is in fact the geometric constant, then $$\cos \theta = \theta$$ which means that:


 * $$\lambda = \frac{\alpha}{H} = \frac{A}{H}$$

Which also proves that $$\alpha = A$$. Additionally, when $$H=1$$, we get:


 * $$\lambda = \alpha = A$$

Which means that:


 * $$\lambda^2 + A^2 = 1$$

If we subtract $$A^2$$ from both sides, we get:


 * $$\lambda^2 = 1 - A^2$$

Now if we square root both sides, we get:


 * $$\lambda = \sqrt{1 - A^2}$$

Or, equivalently:


 * $$\lambda = \sqrt{1 - \alpha^2}$$