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Can someone determine what this iteration works out to, where x'

  1. Apr 7, 2012 #1
    Can someone determine what this iteration works out to, where x' becomes x again each time, starting with x=1 and a and b are variables?

    x' = (1 + x) / (b (1 + x) + a)
     
  2. jcsd
  3. Apr 7, 2012 #2

    chiro

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    Re: iteration

    This looks like it is going to be some kind of continued fraction, but it might end up simplifying to be something less complicated. Are you aware of continued fractions?
     
  4. Apr 8, 2012 #3
    Re: iteration

    Yes, thanks, but it doesn't seem to work out that way.
     
  5. Apr 8, 2012 #4
    Re: iteration

    Here's something that might help. If I put in a=1 and b=1, it works out to x_infinity = (sqrt(5) - 1) / 2.
     
  6. Apr 8, 2012 #5
    Re: iteration

    Ahah. With b=1, the solution for x_infinity seems to be

    x_inf^2 + a x_inf - 1 = 0

    x_inf = a (sqrt(4 / a^2 + 1) - 1) / 2
     
  7. Apr 8, 2012 #6
    Re: iteration

    The solution for x_inf with a=1 is

    (x_inf^2 + x_inf) b - 1 = 0

    x_inf = [sqrt(b) sqrt(b + 4) - b] / (2 b)
     
  8. Apr 8, 2012 #7
    Re: iteration

    Okay, combining those two equations with (x_inf^2 + a x_inf) b - 1 = 0 still only gives the correct result for a=1 or b=1 only, so I figured there had to be something like (1 - a) (1 - b) in there somewhere to make the rest zero. The full solution for x_inf works out to be

    (x_inf^2 + a x_inf) b - 1 - (1 - a) (1 - b) x_inf = 0

    x_inf^2 b - (1 - a - b) x_inf - 1 = 0

    x_inf = [1 - a - b + sqrt((1 - a - b)^2 + 4 b)] / (2 b)

    Now I just need the solution for x_n with finite n.
     
  9. Apr 8, 2012 #8
    Re: iteration

    Um, yeah. It was brought to my attention elsewhere that where x_inf converges, we have x' = x at the limit, so we can just rearrange the original equation with

    x' = (1 + x) / (b (1 + x) + a)

    x (b (1 + x) + a) - (1 + x) = 0

    That works out just the same as the middle equation in the last post. Too bad I spent much of the day getting that far. :) Oh well. I might be able to use x_inf for my purposes, although finite x would still be very handy.
     
  10. Apr 8, 2012 #9

    HallsofIvy

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    Re: iteration

    That equation is just a quadratic:
    [tex]bx^2+ (a+ b- 1)x- 1= 0[/tex]
    and so, by the quadratic formula,
    [tex]x= \frac{1- a- b\pm\sqrt{(a+ b- 1)^2- 4b}}{2b}[/tex]
    If you can show that the sequence, starting with x= 1, is increasing, you know that that [itex]\pm[/itex] must be +.
     
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