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

  1. Jul 30, 2016 #1
    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. jcsd
  3. Jul 30, 2016 #2

    fresh_42

    Staff: Mentor

    There is no closed form you are looking for. What is it all about and where are ##x## and ##t## from?
     
  4. Jul 30, 2016 #3
    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
     
  5. Jul 30, 2016 #4

    fresh_42

    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?
     
  6. Jul 30, 2016 #5
    Yea you got it. I was able to write in terms of x too but not y which is what I need to graph it properly without parametric which is sort of the reason. The real reason is because I wanted to learn something new but it seems like my suspicion is being confirmed that it can't really be done unless there is some crazy math out there we don't know about to reduce or simplify...
     
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