# I X = t - sin(t) in terms of t

1. Jul 30, 2016

Sorry title should say x=t-sin(t) in terms of t...Is it possible? I get close.. I do t = x + sin(t) and use the relation again to get an infinite regression

T = x + sin(x +sin(x + sin(x + sin(x... ))))
I'm not sure if it has a real limit or simplification ... anyone know?

2. Jul 30, 2016

### Staff: Mentor

There is no closed form you are looking for. What is it all about and where are $x$ and $t$ from?

3. Jul 30, 2016

I also have y=1-cos(t).. it's a position vector for a Cal III course practice. I'm able to graph in parametric but rectangular has me stumped since I can only put it in terms of x and not y. I have the problem done and everything I'm just curious about this particular aspect

4. Jul 30, 2016

### Staff: Mentor

So, if I got you right (and made no mistake), you have a parametric curve $t \mapsto \begin{pmatrix} t - \sin t \\ 1 -\cos t \end{pmatrix}$, i.e. a map $I \rightarrow \mathbb{K}^2$ where $I$ is a real interval and $\mathbb{K}$ either the real or complex numbers. Why do you want to solve this for $t$?
Of course you could always write $t = \arccos(1-y) \; , \; x= \arccos(1-y) - \sqrt{2y - y^2}$ to get rid of the parameter $t$ and deal with the domains, where this is true. But why?

5. Jul 30, 2016