[bleagos]

Hey All,

I am looking for examples of iterative functions that have the property for all possible real initial values we have



The only ones I could think of were:




Can anybody think of any more?

Thanks

[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.

[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.

[bleagos]

(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

[bleagos]

(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

[bleagos]

(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:



Then 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 which gives

[DFranklin]

It's actually a bit easier if you know t_0 > 1, I think.

[DFranklin]

As a "something to think about" function:

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