I was shock puzzled when Leo.M mentioned the possibility that a superfunction might not result

from or not have an iteration group.

This question arose from actual real valued superfunctions of \(b^z\) with b around eta minor.

There the fixed point derivative is negative and so would imaginably not allow for e.g. a half iterate.

However the given superfunctions are valid superfunctions S, satisfying

$$ S(x+1) = f(S(x)) $$

however they show oscillatory behaviour are hence not injective as we are used to from positive fixed point multipliers/derivatives.

In those cases you can just define

$$ f^{\circ t}(x) = S(t+S^{-1}(x)) $$

and you have an iteration (semi) group particularly satisfying

$$ f^{\circ s+t }(x) = f^{\circ s}(f^{\circ t}(x)),\quad f^{\circ 1} (x)=f(x) $$

So how can this be done in case S in not invertible?! Using Multivalued functions!

As a research object I started with the edge case \(f(x)=-x\) (multiplier -1).

This case is so edge that it often is not even called parabolic fixed point, because \(f(f(x))=x\).

However it serves our research purpose.

One can easily give the regular iteration group:

$$ f^{\mathfrak{R} t}(z) = (-1)^t z = e^{\pi i t}z $$

which is complex valued, the superfunction would be \(S^{\mathfrak{R}}(z)=e^{\pi i z}\).

Now we know from the real valued Fibonacci extension that this strange "realifyer" is used as a real replacement for \((-c)^t\)

$$ \rho(t) = \cos(\pi t) c^t $$

It satisfies \(\rho(t+1)=(-c)\rho(t)\) and \(\rho(1)=(-c)^1\) but is completely real valued.

And typically it can be used in conjuction with the inverse Schröder function, or the regular Superfunction to get a real valued solution, when we replace \((-c)^t \) with it:

So in our case we can define a superfunction for \(f(x)=-x\) by:

$$S(x)=\cos(\pi x) $$

I started with this edge case c=1 because I can easily invert this function, while it much more difficult when \(c\neq 1\).

So our iteration semigroup can be defined as

$$f^{\circ t}(x)= \cos\left(\pi \left(t\pm\frac{\arccos(x)}{\pi}\right)\right) = \cos(\pi t \pm \arccos(x)) $$

we need to choose the right branch of \(\arccos\).

Visually this gives an Aha! We have a circle as half iterate and and an ellipse as 1/3 iterate.

But we can check (pink line) that it indeed are proper iterates, as long as we choose the right branch,

though it does not even have the fixed point a 0 - very edge case.

And these are ellipses in the geometrical sense, the real thing - not such something that looks like it.

You can convince yourself when expanding \( \cos(\pi t \pm \arccos(x))\).

So this gives already a rich result for such a simple function, but it becomes even better for \(c\neq 1\).

Here instead of cumbersomely trying to compute inverse we just use a parametrized version,

which is quite simple to obtain an iteration group from a given Superfunction S:

$$ (x,y) = (S(s),S(s+t)) $$

that's it already.

This time the iterates have the same fixed point, as the original function, but - as they are not regular at the fixed point - they behave wildly around the fixed point (not too wild though).

So, conclusion:

Oscillating superfunctions are still associated with continuous iteration (semi)groups, only that they are multivalued now and one has to be careful with choosing the right branch (which is that way with most complex functions anyways).

Later I want to transfer that to our Fibonacci LFT - so stay tuned

from or not have an iteration group.

This question arose from actual real valued superfunctions of \(b^z\) with b around eta minor.

There the fixed point derivative is negative and so would imaginably not allow for e.g. a half iterate.

However the given superfunctions are valid superfunctions S, satisfying

$$ S(x+1) = f(S(x)) $$

however they show oscillatory behaviour are hence not injective as we are used to from positive fixed point multipliers/derivatives.

In those cases you can just define

$$ f^{\circ t}(x) = S(t+S^{-1}(x)) $$

and you have an iteration (semi) group particularly satisfying

$$ f^{\circ s+t }(x) = f^{\circ s}(f^{\circ t}(x)),\quad f^{\circ 1} (x)=f(x) $$

So how can this be done in case S in not invertible?! Using Multivalued functions!

As a research object I started with the edge case \(f(x)=-x\) (multiplier -1).

This case is so edge that it often is not even called parabolic fixed point, because \(f(f(x))=x\).

However it serves our research purpose.

One can easily give the regular iteration group:

$$ f^{\mathfrak{R} t}(z) = (-1)^t z = e^{\pi i t}z $$

which is complex valued, the superfunction would be \(S^{\mathfrak{R}}(z)=e^{\pi i z}\).

Now we know from the real valued Fibonacci extension that this strange "realifyer" is used as a real replacement for \((-c)^t\)

$$ \rho(t) = \cos(\pi t) c^t $$

It satisfies \(\rho(t+1)=(-c)\rho(t)\) and \(\rho(1)=(-c)^1\) but is completely real valued.

And typically it can be used in conjuction with the inverse Schröder function, or the regular Superfunction to get a real valued solution, when we replace \((-c)^t \) with it:

So in our case we can define a superfunction for \(f(x)=-x\) by:

$$S(x)=\cos(\pi x) $$

I started with this edge case c=1 because I can easily invert this function, while it much more difficult when \(c\neq 1\).

So our iteration semigroup can be defined as

$$f^{\circ t}(x)= \cos\left(\pi \left(t\pm\frac{\arccos(x)}{\pi}\right)\right) = \cos(\pi t \pm \arccos(x)) $$

we need to choose the right branch of \(\arccos\).

Visually this gives an Aha! We have a circle as half iterate and and an ellipse as 1/3 iterate.

But we can check (pink line) that it indeed are proper iterates, as long as we choose the right branch,

though it does not even have the fixed point a 0 - very edge case.

And these are ellipses in the geometrical sense, the real thing - not such something that looks like it.

You can convince yourself when expanding \( \cos(\pi t \pm \arccos(x))\).

So this gives already a rich result for such a simple function, but it becomes even better for \(c\neq 1\).

Here instead of cumbersomely trying to compute inverse we just use a parametrized version,

which is quite simple to obtain an iteration group from a given Superfunction S:

$$ (x,y) = (S(s),S(s+t)) $$

that's it already.

This time the iterates have the same fixed point, as the original function, but - as they are not regular at the fixed point - they behave wildly around the fixed point (not too wild though).

So, conclusion:

Oscillating superfunctions are still associated with continuous iteration (semi)groups, only that they are multivalued now and one has to be careful with choosing the right branch (which is that way with most complex functions anyways).

Later I want to transfer that to our Fibonacci LFT - so stay tuned