Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cycloid power series. problem from hell.

  1. Dec 9, 2007 #1
    :devil: Can you handle it?

    Find the first non-zero terms, the general term for the cycloid power series, and the interval of convergence for the cycloid power series.

  2. jcsd
  3. Dec 9, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    There seem to be a lot of "limits of death" and "problems from hell" today. I'm afraid you must show some work before anyone assists you on your homework questions. Unfortunately, daring the reader to answer for you doesn't work! Further, you should post in the homework forums next time.

    Welcome to PF!
  4. Dec 9, 2007 #3
    My bad. Well, I know that the basics for taylor polynomials. for a cycloid,
    I know that f(a)+ f'(a)(x-a)+ f''(a)(x-a)^2... will give me the right
    answer but I don't know how to do this in parametric mode or what
    really do other than what i've listed below. Thanks!


    so, dy/dx=-sin(theta)/1-cos(theta)

    I also know that d(dy/dx)/dx=d^2y/dx^2
  5. Jun 15, 2008 #4
    Problem from the sky

    Ignoring homotheties, let the following cycloid:
    x = t - sin t
    y = 1 - cos t,
    t in [0, 2pi].

    From this, we have y' = dy/dx = (dy/dt) / (dx/dt) = (sin t) / (1 - cos t) = ... = cot (t/2), and then y'' = d(y')/dx = ... = -(1/4) [ cosec (t/2) ]^4, ..., "AND SO ON" :rofl:

    Maybe it will be interesting calculate the series at t = pi. Doing that and putting y^(m) := (d^m)y /(dx)^m, it is easy to see that y^(2n+1)[pi] = 0, and y^(2n)[pi] < 0, n = 1, 2, 3, ... .

    The transcendental "cycloid function" y = cyc(x) is aproximately:
    cyc(x) =~
    -1/8 (x-pi)^2
    -1/384 (x-pi)^4
    -11/92160 (x-pi)^6
    -73/10321920 (x-pi)^8
    -887/1857945600 (x-pi)^10
    -136883/3923981107200 (x-pi)^12
    -7680089/2856658246041600 (x-pi)^14
    -26838347/124654178009088000 (x-pi)^16
    - ... "AND SO ON" :biggrin:

    This function satisfies the brachistochrone equation:

    y * (1 + y'^2) = 2 (*)​

    However not all curves satisfying this equation are cycloids. To see that, think about inserting a horizontal line segment between two arcs of the same cycloid. It will be a solution of the differential equation (*).

    Good luck.
  6. Jun 15, 2008 #5


    User Avatar
    Staff Emeritus
    Science Advisor

    The problem is that the phrase "cycloid power series" makes no sense to me! I know what the power series representation for a function is but what function are you talking about? Do you mean power series representations for x and y separately?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Cycloid power series. problem from hell.
  1. Optimization from hell (Replies: 1)

  2. Integral from Hell? (Replies: 3)

  3. Integrals from Hell (Replies: 4)

  4. Power Series (Replies: 4)

  5. Power Series Problem! (Replies: 1)