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Values of x1 for which an iterative formula converges or diverges?

  1. Aug 28, 2014 #1
    I was wondering if there is a way of finding out the values of x1 for which an iterative formula converges.
    Specifically, I am looking at the formula xn+1=0.25(xn3+1) and I am thinking about it in the context of finding roots of a function, although it would be great if there was a general way of finding whether a sequence converges or diverges.

    I haven't had any ideas yet, so thank you in advance for any help!! :)
  2. jcsd
  3. Aug 28, 2014 #2
    I found the answer: it converges if the absolute value of the derivative of the function f(x) where your formula is xn+1=f(xn) must be less 1. I get that, but now II have another question...

    For the function x3-4x+1, if you want to find the roots using this iterative process you can use xn+1=0.25(xn3+1) or xn+1=cube root(4xn-1). I know that there are 3 roots. I have only considered the two in the interval 0-1 and 1-2, but it seems like for each of these roots only one of the formulae converges, the other diverges. Is it the case that there will always be one fomrula which converges, or are there some roots that can never be found by this process? Can this be shown mathematically?

    Thank you :)
  4. Aug 28, 2014 #3

    Ray Vickson

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    You will gain a lot of insight into your problem if you look at it using a "cobweb diagram" methodology. See, eg., http://en.wikipedia.org/wiki/Cobweb_plot or
    https://www.math.ubc.ca/~andrewr/620341/pdfs/ga_sum.pdf . This last one has examples similar to yours.
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