As many of you probably know, the cycloid is given by the parametric equation: \begin{equation} x= t-\sin t\tag{1} \label{eq:x} \end{equation} \begin{equation} y= 1- \cos t\tag{2}\label{eq:y}. \end{equation} I would like to have an equation for the cycloid that does not depend on the parameter $t$. If we could just solve \eqref{eq:x} for $t$ we could just plug that expression into \ref{eq:y} and I would be happy. But, as far as I know we can't solve \ref{eq:x} for $t$. However, it is easy enough to find that $t= x+ \sin t $. We know that $\sin t$ must be somewhere between $-1,1$, so for an approximation, we might decide to simply drop this term and substitute $t = x$ into equation \ref{eq:y} to get $1- \cos(x)$. Looking at a graph this looks like a reasonable but very rough approximation of the actual cycloid. I noticed that there is a simple way to improve this approximation.
Recall that $t= x+ \sin t $. What happens if we just plug in the expression $x+ \sin t$ for $t$ into the right hand side to get $t= x+ \sin( x+ \sin t)$. Again, since we are approximating, we can just let $\sin t = 0$. Substitute $t= x+ \sin x$ into \ref{eq:y} to obtain the much better approximation to the cycloid $$ y= 1-\cos( x+ \sin(x)).$$
This has lead me to conjecture the following: Let $f_1(x) = x $ and $f_n(x) = x + \sin \big( f_{n-1} (x) \big)$. Define $g_n(x) = 1- \cos f_n(x)$. Then, $g(x)= \lim_{n \to \infty} g_n(x)$ is the cycloid.
I do not have a proof of this and could not find this description of the cycloid anywhere (maybe because it is a pain to work with). So, I have two questions:
How can I prove my conjecture?
Is the recursive method to "solve" equation \ref{eq:x} for $x$ used elsewhere?
