Zeta iterations

[Balarka Sen]

Consider the (real) attractive fixed point of zeta z = -0.29590500557... In A069857, the author claims that this is also the limit of zeta^[m](z) for complex values of z as m tends to be large (or perhaps infinity, not sure if it exists). Can we show that the limit exists and -0.29590500557 IS the limit?

Balarka
.

[Gottfried]

Consider the (real) attractive fixed point of zeta z = -0.29590500557... In A069857, the author claims that this is also the limit of zeta^[m](z) for complex values of z as m tends to be large (or perhaps infinity, not sure if it exists). Can we show that the limit exists and -0.29590500557 IS the limit?

Balarka
.

I don't have a final answer, but the following observation are perhaps helpful to find such an answer.
A fixpoint is attracting, if the absolute value of its derivative is smaller than 1. So to see, whether z is attracting also for the complex plane its absolute value must be smaller in any direction of h in the derivative-formula (zeta(z+h/2)-zeta(z-h/2))/h .
To do this in Pari/GP h can be defined as h=exp(I*phi)*1e-20 for varying phi. For all phi tested the absolute of the derivative at z was smaller than 1. However - a heuristic is no proof...

Gottfried

[Balarka Sen]

to see, whether z is attracting also for the complex plane its absolute value must be smaller in any direction of h in the derivative-formula (zeta(z+h/2)-zeta(z-h/2))/h .

Yes, it's obviously a question whether it's actually a fixed point in the whole complex plane but that wasn't the thing I've asked, actually. I want to know why zeta^[m](x) for complex x's tends to that unique fixed point for large m's? Is it just a mathematical coincidence or has a reason behind it?

Thank you for your time,

Balarka
.

[Gottfried]

want to know why zeta^[m](x) for complex x's tends to that unique fixed point for large m's? Is it just a mathematical coincidence or has a reason behind it?

Thank you for your time,

Balarka
.

Hi Belarka - possibly I just misunderstand your question. But if not: just because the absolute value of the derivative at z is smaller than 1 is that reason - at least as far as I understand that thing (I may be wrong, though).

Gottfried

[Balarka Sen]

just because the absolute value of the derivative at z is smaller than 1 is that reason

I see, but how did you derived that? Is it very straightforward from the definition of attractive fixed point but I cannot see it? I am having a feeling that I possibly asked a pretty stupid question

[Gottfried]

I see, but how did you derived that? Is it very straightforward from the definition of attractive fixed point but I cannot see it?

Hmm, perhaps you'll find better/more complete/more helpful explanations when googling for "basins of attraction" or "petals of attraction". Did you try our hyperoperation-wiki already?
Perhaps it is also helpful to remember, that the function zeta1(s) = zeta(s)+1/(1-s) is entire (and has a power series which is easily derivable with Pari/GP). Then also the derivative must be entire; and also the derivative of the remaing part zeta2(s) = -1/(1-s) is zeta2(s)'=1/(1-s)^2 which is finite only except at s=1.

But well, I think I'd stop here and confirm my much freehanded assumtions myself now...

Gottfried

[tommy1729]

just because the absolute value of the derivative at z is smaller than 1 is that reason

I see, but how did you derived that? Is it very straightforward from the definition of attractive fixed point but I cannot see it? I am having a feeling that I possibly asked a pretty stupid question

I wanted to comment on that to give some kind of intuitive explaination.

For simplicity recenter the fixpoint at 0.

Now if f(z) is analytic and f(0)=0 and there is no fixpoint near 0 then consider writing f(z) = 0 + a z + b z^2 + c z^3 + ...

Now from calculus you know that if z gets very small ( where very depends on the size of the taylor coefficients and the remainder ) then f(z) gets well approximated by f(z) = a z + O(z^2)

Now if a is 1 then f(z) approximates id(z) and hence the fixpoint 0 is " neutral " also called " invariant " or " parabolic ".

Now clearly if a = i ( imaginary unit) then by f(z) = a z + O(z^2) we get a simple rotation but not a strong attraction or repellation.

This already hints that looking at the complex number a in polar coordinates makes more intuitive sense.

To visualize the 'small' z take a disk around z=0 with a small radius.

Now because f(z) approximates a z , AND the theta of a determines how much we rotate , the absolute value must determine if it is attracting or repelling.

By attracting we mean that any point in the small disk must be mapped closer to the fixpoint , and by repelling we mean that any point must be mapped further away from the fixpoint.

For instance if f(z) is of the form 0 + 0.5 z + 0.0001 z^2 + ... then it is clear that f(z) in the small radius ( = z small => higher powers of z have lesser influence ) behaves like f(z)^[t] = 0.5^t z which is clearly attracting.

By analogue if a = 2 it is clearly repelling.

if a = 2 i , then that is equivalent to repelling + rotating.

This more or less concludes my personal nonformal intuitive explaination ( for complex values ) and I hope it is clear and helpfull to you.

Btw , Note that I did not discuss what happens when a = 0. I only gave a sketch of how and why when the absolute value of a satisfies 0 < abs(a) < oo.

When a=0 many formula's and properties fail anyway and for the case a=0 you need to understand more of the Fatou-flower.

I could say alot more but I do not want to confuse you with harder stuff and you can read up on the subject too. Hence I did not want to post more than that what occurs sponteanously to me.

Together with some basic complex analysis and basic real calculus
I think this will get you much further in the subjects and discussions about tetration and their cruxial details.


regards

tommy1729

[bo198214]

Perhaps formal derivation:
Attractive fixpoint means
.
By continuity of the left side in there is a disk with center such that for all . ()

So for any we have
, particularly is again in .

and so on
.

The right sight converges to 0 for , hence by definition of convergence: for .

[tommy1729]

How many fixpoints does zeta have actually ?

If a fractal converges to a single fixpoint then that is the only fixpoint.

So I guess the question is how many fixpoints does zeta have ?

afterall if x is distinct from y and zeta(x)=x and zeta(y)=y then taking zeta zeta zeta ... zeta(y) = y ( and not x ! ).

attractive or not is another matter.

Notice that zeta(x+yi) for x >> y gives 1 and zeta(...zeta(1)...) = x

I also note that zeta(z) = 1 + 1/2^z + ... so for Re(z) > 1 and abs(z) = oo we get the same limit as for zeta (...zeta (1)...).

regards

tommy1729

[Gottfried]

How many fixpoints does zeta have actually ?

Because the bernoulli-numbers grow faster than their index, they have alternating signs and finally, because the bernoulli-numbers are interpolated by the zeta at negative arguments we have, that the negative part of the x=y-diagonal is crossed infinitely many times. So we have (at least countably) infinitely many real fixpoints.


Gottfried