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Need an algorithm to calculate a function

  • Thread starter Mentz114
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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:

Answers and Replies

arildno
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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:
Mentz114
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Thanks, arildno. Food for thought.
 

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