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Iteration exercises: f(x)=x^2 - 0.5 ; Fixpoint-irritation...
bo198214 Wrote:Interesting phenomenon, I guess it has to do with that the function is not strictly increasing in the vicinity of the second fixed point. If you develop the regular half iterate at the left fixed point it gives a non-real function.

right. but is that the only neccessary and sufficient reason ?

bo198214 Wrote:If I remember correctly the matrix power approach even yields a non-real solution if applied at 0 between both fixed point *when the function is strictly increasing*.

in general yes.

bo198214 Wrote:For complex fixed points, its anyway (mostly) not the case that the regular iteration at one fixed point has the other fixed point as fixed point.

i think you can even say for any 3 real fixed points that are not attractive , it is not the case that regular iteration at one fixed point has the other 2 real fixed points as fixed points.

anyways , i think we would all benefit by being carefull with fixpoint ideas.

the following looks intresting to me :

if g(g(x)) = f(x)

f(x) has 2 fixpoints of which 1 in commen with g(x).

how many other fixpoints can g(x) have , and what determines the amount and position ?

i think the radius of g(x) and f(x) are important in all this.

also if the fixpoints are attractive or not seems important.

i think in case f(x) and g(x) are both entire , the number of their fixpoints are - in general - as good as unbounded.

it might be intresting to consider :

fixpoint of f( inverse f(x) + 1 ) = fixpoint of g(x) ?

( i think its clear why ? )

furthermore analytic fixpoint shift might be another " tetration uniqueness condition " ?!?

just some ideas ...



Messages In This Thread
RE: Iteration exercises: f(x)=x^2 - 0.5 ; Fixpoint-irritation... - by tommy1729 - 03/02/2009, 08:48 PM

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