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Iterative convergent functions

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    Hey All,

    I am looking for examples of iterative functions  t_{n+1} = f(t_n) that have the property for all possible real initial values  t_0 > 1 we have

     t_n \rightarrow 1, n \rightarrow \infty

    The only ones I could think of were:

     f(t) = t^x , 0<x<1
     f(t) = \frac{t+\epsilon}{\epsilon +1}, \epsilon > 0

    Can anybody think of any more?

    Thanks
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    Look up the Newton Raphson method, then use it on functions with a single root at 1. Most of the time it will converge to 1.

    Also consider how your values will approach 1 - for instance, construct a function such that the distance between the new value and 1 is half the distance between the old value and 1.

    Finally, consider any function you can think of that tends to a limit other than 0 and just multiply it by the reciprocal of the limit.
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    Your first suggestion doesn't work very well for t < 0.

    Note that your 2nd suggestion can be written as (t-1+1+e)/(e+1) = (t-1)/(e+1) + 1.

    That is, f(t) = g(t-1) + 1, where g(x) = x/(1+e). It's quite easy to see that the nth iterate f^n(t) is g^n(t-1) + 1.

    So you're basically looking for functions g(x) s.t. g^n(x) -> 0 as n->\infty

    I think this is a bit easier to deal with conceptually. E.g. g(x) = x /(2+x^2) should work.
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    (Original post by Bobifier)
    Look up the Newton Raphson method, then use it on functions with a single root at 1. Most of the time it will converge to 1.

    Also consider how your values will approach 1 - for instance, construct a function such that the distance between the new value and 1 is half the distance between the old value and 1.

    Finally, consider any function you can think of that tends to a limit other than 0 and just multiply it by the reciprocal of the limit.
    Thanks for the ideas, I never thought of the Newton Raphson method, that is really good!

    The second function I gave covers your second suggestion by letting epsilon = 1 and I like the third idea, it has given me something to think about.

    Thanks again
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    (Original post by DFranklin)
    Your first suggestion doesn't work very well for t < 0.

    Note that your 2nd suggestion can be written as (t-1+1+e)/(e+1) = (t-1)/(e+1) + 1.

    That is, f(t) = g(t-1) + 1, where g(x) = x/(1+e). It's quite easy to see that the nth iterate f^n(t) is g^n(t-1) + 1.

    So you're basically looking for functions g(x) s.t. g^n(x) -> 0 as n->\infty

    I think this is a bit easier to deal with conceptually. E.g. g(x) = x /(2+x^2) should work.
    I should have said this but I want t_n > 1 for all n so the first one would work.

    I never thought about it the way you did with g(x), I think that works well. Thank you
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    (Original post by DFranklin)
    Your first suggestion doesn't work very well for t < 0.

    Note that your 2nd suggestion can be written as (t-1+1+e)/(e+1) = (t-1)/(e+1) + 1.

    That is, f(t) = g(t-1) + 1, where g(x) = x/(1+e). It's quite easy to see that the nth iterate f^n(t) is g^n(t-1) + 1.

    So you're basically looking for functions g(x) s.t. g^n(x) -> 0 as n->\infty

    I think this is a bit easier to deal with conceptually. E.g. g(x) = x /(2+x^2) should work.
    Does the second idea neccessarily work? I am not sure for example:

     g(t) = \frac{t}{2}

    Then  t_{n+1} = g(t_n) + 1 has a fixed point at t = 2.

    Also for the example you gave we have a fixed point at the real solution of the cubic equatiobn  t^3 - t^2 + 3t -2 = 0 which gives  t \approx 1.35
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    It's actually a bit easier if you know t_0 > 1, I think.
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    As a "something to think about" function:

    f(t) = 1 (t < 2)
    f(t) = t-1 (t>=2).

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Updated: April 3, 2012
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