2 real fixpoints ... 3 iterations ?

[tommy1729]

ok , first i dont know if this has been asked or answered before , so forgive my ignorance if such is the case.

consider an analytic function f(z) , strictly rising on the reals with 2 real fixpoints.

now we can use regular iteration for each fixpoint and thus get two ' zones ' that are ' correct ' half-iterates. ( assuming radius = oo too )

sketch :

[ - oo .. zone 1 .. fixpoint 1 .. zone 2 .. fixpoint 2 .. zone 3 .. + oo ]

when expanded at fixpoint 1 and missing the fixpoint 2 ( and radius oo ) i consider that regular half-iterate as a correct solution to 'zone 1'

similarly :

when expanded at fixpoint 2 and missing the fixpoint 1 ( and radius oo ) i consider that regular half-iterate as a correct solution to 'zone 3'


So the question becomes , what gives zone 2 ?


Regards

Tommy1729

[bo198214]

now we can use regular iteration for each fixpoint and thus get two ' zones ' that are ' correct ' half-iterates. ( assuming radius = oo too )

sketch :

[ - oo .. zone 1 .. fixpoint 1 .. zone 2 .. fixpoint 2 .. zone 3 .. + oo ]

regular iteration at fp 1 is valid at zone 1&2 and regular iteration at fp 2 is valid at zone 2&3.

[tommy1729]

now we can use regular iteration for each fixpoint and thus get two ' zones ' that are ' correct ' half-iterates. ( assuming radius = oo too )

sketch :

[ - oo .. zone 1 .. fixpoint 1 .. zone 2 .. fixpoint 2 .. zone 3 .. + oo ]

regular iteration at fp 1 is valid at zone 1&2 and regular iteration at fp 2 is valid at zone 2&3.

no , you assume regular iteration at fp1 touches fp 2 , what if it doesnt ?

[jaydfox]

regular iteration at fp 1 is valid at zone 1&2 and regular iteration at fp 2 is valid at zone 2&3.

no , you assume regular iteration at fp1 touches fp 2 , what if it doesnt ?

But wouldn't that necessitate another fixed point between fp1 and fp2? Why else would regular iteration from fp1 not reach fp2? By your original stipulation, the function is strictly increasing (except, of course, at the fixed points). If it does not get from fp1 to fp2, there must be some point (which we can find by taking the limit of infinite iterations) which is a fixed point.

[bo198214]

no , you assume regular iteration at fp1 touches fp 2 , what if it doesnt ?

But wouldn't that necessitate another fixed point between fp1 and fp2? Why else would regular iteration from fp1 not reach fp2? By your original stipulation, the function is strictly increasing (except, of course, at the fixed points). If it does not get from fp1 to fp2, there must be some point (which we can find by taking the limit of infinite iterations) which is a fixed point.

Exactly.