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Iterative convergent functions

Hey All,

I am looking for examples of iterative functions tn+1=f(tn) t_{n+1} = f(t_n) that have the property for all possible real initial values t0>1 t_0 > 1 we have

tn→1,n→∞ t_n \rightarrow 1, n \rightarrow \infty

The only ones I could think of were:

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

Can anybody think of any more?

Thanks
Reply 1
Avatar for Bobifier
Bobifier
13
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.
(edited 12 years ago)
Reply 2
Avatar for DFranklin
DFranklin
18
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.
Reply 3
Avatar for bleagos
bleagos
OP
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 :smile:
Reply 4
Avatar for bleagos
bleagos
OP
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 :smile:
Reply 5
Avatar for bleagos
bleagos
OP
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)=t2 g(t) = \frac{t}{2}

Then tn+1=g(tn)+1 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 t3−t2+3t−2=0 t^3 - t^2 + 3t -2 = 0 which gives t≈1.35 t \approx 1.35
(edited 12 years ago)
Reply 6
Avatar for DFranklin
DFranklin
18
It's actually a bit easier if you know t_0 > 1, I think.
Reply 7
Avatar for DFranklin
DFranklin
18
As a "something to think about" function:

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

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