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Homework Help: Need an algorithm to calculate a function

  1. Jan 23, 2007 #1

    Mentz114

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    I'm trying to calculate a table of x vs t, with x as the dependent variable from
    this formula

    [tex]t + C = \frac{1}{2}(x\sqrt( 1 - x^2) + arcsin(x) ) [/tex]

    C=0.4783 is given when t=0 and x = 1/2

    I thought it would be simple but my code is giving nonsensical results, viz.
    a straight line ! I do get the correct answer when t =0 (1/2) and I can't find a fault in my code. My code also gives a straight line for

    t = arcsin(x) + 0.5 which obviously wrong, since x = sin( t - 1/2).

    Has anyone got a general algorithm for this ? I'm using simplex but I
    haven't tried Newton-Ralphson

    This is related to the 'Interesting Oscillator Potential' topic below.
     
    Last edited: Jan 23, 2007
  2. jcsd
  3. Jan 23, 2007 #2

    arildno

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    Dearly Missed

    You might use power series inversion:
    1. Introduce the "tiny" variable [itex]\epsilon=x-\frac{1}{2}[/itex]
    and find the first few Taylor series terms of the right-hand side about [itex]\epsilon=0[/itex]
    You have now effectively found t as a power series in [itex]\epsilon[/itex]

    2. Assume that this power series is invertible, i.e, it exists a function:
    [tex]\epsilon(t)=\sum_{i=0}^{\infty}a_{i}t^{i}[/tex]

    If you can find the coefficients [itex]a_{i}[/itex] you're finished! :smile:

    3.Insert this power series on the epsilon places, and collect terms of equal power in t. (For each finite power of t, only a finite number of terms need to be collected.*)

    4. Different powers of t are linearly independent functions, hence all coefficients of the powers must equal zero. This demand of zeroes furnishes you with the equations to determine the coefficients [itex]a_{i}[/itex]

    *EDIT:
    This requires that [itex]a_{0}=0[/itex].
    This, however, holds, since [itex]\epsilon=0[/itex] when t=0.
     
    Last edited: Jan 23, 2007
  4. Jan 23, 2007 #3

    Mentz114

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    Gold Member

    Thanks, arildno. Food for thought.
     
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