Fibonacci as iteration of fractional linear function

[JmsNxn]

Is there a way to perform, let's say, "a crescent iteration" on the fibonacci sequence, so that we somehow map it to the reals?

We can always multiply by a 1-periodic function \(\theta(z)\), would it be possible to make \(\theta(z)F(z)\) real valued?

I've always wondered that, but I could never think of a solution; now seems as good a time to ask as any.

I thought Gottfried has the answer to that?

 Oh yes, my mistake. I didn't even see Gottfried's post. I just saw your reply to me. I'm curious what \(\theta\) would look like though. I apologize again.

But there's still a bit of work involved, here. We have a real iterate of \(1/(1+x)\), but how exactly would we pull out the Fibonacci formula?

Would we have to solve:

\[
f_d(t) = \frac{F_{\text{Gottfried}}(t) - F_{\text{Gottfried}}(t-1)z}{F_{\text{Gottfried}}(t+1) - F_{\text{Gottfried}}(t)z}\\
\]

Or can you see a simpler way that I'm not seeing, where \(F_{\text{Gottfried}}(t) = \theta(t)\phi(t)\)?

[Gottfried]

Would we have to solve:

\[
f_d(t) = \frac{F_{\text{Gottfried}}(t) - F_{\text{Gottfried}}(t-1)z}{F_{\text{Gottfried}}(t+1) - F_{\text{Gottfried}}(t)z}\\
\]

Or can you see a simpler way that I'm not seeing, where \(F_{\text{Gottfried}}(t) = \theta(t)\phi(t)\)?

Hmm, first remark: the "dual-" function \( f_d(x) \) is identical to \( f(x)\) for real \(x\) and even iteration height - this is simply because it is the other functional root of \( f(f(x)) \) where the (Schroeder-) multiplier \( \Psi \) is positive instead negative as it is with \(f(x)\) .   Update perhaps it is just that for in \( f^{oh}(x) \) the real height is just replaced by imaginary height - with some scaling factor, didn't check this at the moment end update.

second remark: didn't think of more properties of \( f_d(x) \) because I thought, it is simply a cheap exemplar of a real-to-real function, where Henryk could apply his derivations and which is still a near relative to the rational \( 1/(1+x) \) - function. (The Schroeder-function, for instance is of course the same - except the sign change on the \( \Psi \))

Cannot say more at the moment, especially nothing about a \( \theta()\)-component, sorry...

[JmsNxn]

Would we have to solve:

\[
f_d(t) = \frac{F_{\text{Gottfried}}(t) - F_{\text{Gottfried}}(t-1)z}{F_{\text{Gottfried}}(t+1) - F_{\text{Gottfried}}(t)z}\\
\]

Or can you see a simpler way that I'm not seeing, where \(F_{\text{Gottfried}}(t) = \theta(t)\phi(t)\)?

Hmm, first remark: the "dual-" function \( f_d(x) \) is identical to \( f(x)\) for real \(x\) and even iteration height - this is simply because it is the other functional root of \( f(f(x)) \) where the (Schroeder-) multiplier \( \Psi \) is positive instead negative as it is with \(f(x)\) .  

second remark: didn't think of more properties of \( f_d(x) \) because I thought, it is simply a cheap exemplar of a real-to-real function, where Henryk could apply his derivations and which is still a near relative to the rational \( 1/(1+x) \) - function. (The Schroeder-function, for instance is of course the same - except the sign change on the \( \Psi \))

Cannot say more at the moment, especially nothing about a \( \theta()\)-component, sorry...

OHHHHH

Okay, so you've found the other functional square root of \(f(f(x))\) where \(f = \frac{1}{1+x}\). Yes, this probably wouldn't produce the Fibonacci sequence in the same natural way. \(f\) wouldn't even be a linear fractional transformation would it?

Yes, so these are two different questions. I think Daniel was much closer to what I was asking, but I'm still unsure of how he's pulling out a real valued Fibonacci iteration.

[Gottfried]

Okay, so you've found the other functional square root of \(f(f(x))\) where \(f = \frac{1}{1+x}\). Yes, this probably wouldn't produce the Fibonacci sequence in the same natural way. \(f\) wouldn't even be a linear fractional transformation would it?

Pari/GP "formulates" the function (reconstructed from the powerseries for f(x,h)_iteratable and f_d(x,h)_iteratable )

Code:

f_it(x,1)
%546 = ( 3.09016994375*x + 5.00000000000)/(3.09016994375*x^2 + 8.09016994375*x + 5.00000000000)

fd_it(x,1)
%548 = (2.76393202250*x^2 + 5.85410196625*x + 2.23606797750)/(1.38196601125*x^2 + 6.38196601125*x + 6.70820393250)

This can likely be simplified more, because Pari/GP could not safely cancel multiples of \( \phi \) (golden mean) due to internal digits-representation (no formal expression with \( \phi \) as an indeterminate so far, tomorrow I might have it done)  
At least the f_d(x) function seems to have quadratics in numerator and denominator, so I'm not sure whether it can be a fractional linear function in disguise...

Update Script of a q&d Pari/GP session:

Code:

[1,2]*[0,1;1,1]    \\ test Fibonacci-2x2-matrix in action [1,2] --> [2,3]
%560 = [2, 3]

[1,2]*[0,1;1,1]^2 \\ test Fibonacci-2x2-matrix in action [1,2] --> --> [3,5]
%562 = [3, 5]

[1,2]*[0,1;1,1]^3 \\ test Fibonacci-2x2-matrix in action [1,2] --> -->  --> [5,8]
%564 = [5, 8]

Fib2=[0,1;1,1]  \\ give it a name
%566 =
[0 1]
[1 1]

\\ diagonalize to get the 2x2-matrix for the dual
tmpM=mateigen(Fib2);
tmpW=tmpM^-1;
tmpD=diag(tmpW*Fib2*tmpM)
%568 = [-0.618033988750, 1.61803398875]~  \\ we see the negative eigenvalue

Fib2d=tmpM*matdiagonal(abs(tmpD))*tmpW  \\ create the fd_2-matrix using the positive values as eigenvalues
%570 =
[0.894427191000 0.447213595500]
[0.447213595500  1.34164078650]

[1,2]*Fib2d     \\ test Fib_d-2x2-matrix in action [1,2] --> [1.7888...,3.13049...]
%572 = [1.78885438200, 3.13049516850]

[1,2]*Fib2d^2  \\ test Fib_d-2x2-matrix in action [1,2] --> --> [3,5]
%574 = [3.00000000000, 5.00000000000]

Don't read the symbolic description of Fib2d from the numbers at the moment... with \( a = 0.447213595500.... = 1/\sqrt 5 \) (by W|A) it is then \( \begin{pmatrix} 2 & 1  \\ 1 & 3 \end{pmatrix} / \sqrt 5 \)

Using the iterpretation that \( r= a = 1/ \sqrt 5\) we have the 2x2-matrix iterations

Code:

Fib2d = [2,1;1,3]* 'r       \\ defining the matrix with symbolic entry for the imprecise real value 1/sqrt(5)
%600 =
[2*r   r]
[  r 3*r]

[1,2]*Fib2d
%602 = [4, 7] *r

[1,2]*Fib2d^2
%604 = [15, 25]*r^2   \\ here we can cancel by r^2=1/5: -->  -->  [3,5]

[1,2]*Fib2d^3
%606 = [55, 90]*r^3    \\ here we can cancel by r^2=1/5: --> --> --> [11,18]*r

[1,2]*Fib2d^4
%608 = [200, 325]*r^4   \\ here we can cancel by r^4=1/25: --> --> --> --> [8,13]

Hmmm, having a 2x2-transfermatrix Fib2d this should actually be a fractional linear transformation.

[JmsNxn]

This can likely be simplified more, because Pari/GP could not safely cancel multiples of \( \phi \) (golden mean) due to internal digits-representation (no formal expression with \( \phi \) as an indeterminate so far, tomorrow I might have it done)  
At least the f_d(x) function seems to have quadratics in numerator and denominator, so I'm not sure whether it can be a fractional linear function in disguise...

Hmmm, if the quadratic term were smaller I'd be inclined to entertain that it's not a quadratic, but as Sheldon hit me pretty hard with "follow the fucking numbers"--I'm inclined to disagree.

[Daniel]

Is there a way to perform, let's say, "a crescent iteration" on the fibonacci sequence, so that we somehow map it to the reals?

We can always multiply by a 1-periodic function \(\theta(z)\), would it be possible to make \(\theta(z)F(z)\) real valued?

I've always wondered that, but I could never think of a solution; now seems as good a time to ask as any.

See Fibonacci almost to the bottom of the page for a real iteration of the Fibonacci series.

Could you elaborate, Daniel? Sorry, I'm not too sure what's going on here.

I understand you are writing:

\[
f(z) = \sum_{n=1}^\infty f_n \frac{z^n}{n!}\\
\]

Where now we are taking a parabolic iteration:

\[
f^{\circ t}(z)\\
\]

about \(0\), but how does this produce a fractional fibonacci that is real valued?

Not doubting you, just curious.

Sorry, I was on a combinatorial kick at the time. Due to the close connection between integer sequences and generating functions I displayed the iterated Fibonacci series in terms of an integer sequence. Let \(f(x)\) be the generating function for the Fibonacci series. The series associated with \(f^n(x)\) is then
\(\{0,1,n,n+n^2, -\frac{n}{2}+\frac{5n^2}{2}+n^3, -\frac{n}{3}+\frac{13n^3}{3}+n^4,\ldots\}\)

Fibonacci series
Let \(n=1\:\textrm{then}\:\{0,1,1,2,3,5,8 \ldots\}\)

OEIS A007440  Reversion of g.f. for Fibonacci numbers
Let \(n=-1\:\textrm{then}\:\{0,1,-1,0,2,-3,-2 \ldots\}\)

[JmsNxn]

....

Sorry, I was on a combinatorial kick at the time. Due to the close connection between integer sequences and generating functions I displayed the iterated Fibonacci series in terms of an integer sequence. Let \(f(x)\) be the generating function for the Fibonacci series. The series associated with \(f^n(x)\) is then
\(\{0,1,n,n+n^2, -\frac{n}{2}+\frac{5n^2}{2}+n^3, -\frac{n}{3}+\frac{13n^3}{3}+n^4,\ldots\}\)

Fibonacci series
Let \(n=1\:\textrm{then}\:\{0,1,1,2,3,5,8 \ldots\}\)

OEIS A007440  Reversion of g.f. for Fibonacci numbers
Let \(n=-1\:\textrm{then}\:\{0,1,-1,0,2,-3,-2 \ldots\}\)

Not to sound wrong, how do we convert this into \(F(t)\) such that:

\[
F(t+2) = F(t+1) + F(t)\\
\]

Such that \(F(t)\) is real analytic (analytic for \(t\in \mathbb{R}\) and maps to \(\mathbb{R}\)). As much as I see where you're going, I'm confused about how you map it to a fibonacci solution. It seems close to what you are saying though.

[JmsNxn]

Okay, so you've found the other functional square root of \(f(f(x))\) where \(f = \frac{1}{1+x}\). Yes, this probably wouldn't produce the Fibonacci sequence in the same natural way. \(f\) wouldn't even be a linear fractional transformation would it?

Pari/GP "formulates" the function (reconstructed from the powerseries for f(x,h)_iteratable and f_d(x,h)_iteratable )

Code:

f_it(x,1)
%546 = ( 3.09016994375*x + 5.00000000000)/(3.09016994375*x^2 + 8.09016994375*x + 5.00000000000)

fd_it(x,1)
%548 = (2.76393202250*x^2 + 5.85410196625*x + 2.23606797750)/(1.38196601125*x^2 + 6.38196601125*x + 6.70820393250)

This can likely be simplified more, because Pari/GP could not safely cancel multiples of \( \phi \) (golden mean) due to internal digits-representation (no formal expression with \( \phi \) as an indeterminate so far, tomorrow I might have it done)  
At least the f_d(x) function seems to have quadratics in numerator and denominator, so I'm not sure whether it can be a fractional linear function in disguise...

Update Script of a q&d Pari/GP session:

Code:

[1,2]*[0,1;1,1]    \\ test Fibonacci-2x2-matrix in action [1,2] --> [2,3]
%560 = [2, 3]

[1,2]*[0,1;1,1]^2 \\ test Fibonacci-2x2-matrix in action [1,2] --> --> [3,5]
%562 = [3, 5]

[1,2]*[0,1;1,1]^3 \\ test Fibonacci-2x2-matrix in action [1,2] --> -->  --> [5,8]
%564 = [5, 8]

Fib2=[0,1;1,1]  \\ give it a name
%566 =
[0 1]
[1 1]

\\ diagonalize to get the 2x2-matrix for the dual
tmpM=mateigen(Fib2);
tmpW=tmpM^-1;
tmpD=diag(tmpW*Fib2*tmpM)
%568 = [-0.618033988750, 1.61803398875]~  \\ we see the negative eigenvalue

Fib2d=tmpM*matdiagonal(abs(tmpD))*tmpW  \\ create the fd_2-matrix using the positive values as eigenvalues
%570 =
[0.894427191000 0.447213595500]
[0.447213595500  1.34164078650]

[1,2]*Fib2d     \\ test Fib_d-2x2-matrix in action [1,2] --> [1.7888...,3.13049...]
%572 = [1.78885438200, 3.13049516850]

[1,2]*Fib2d^2  \\ test Fib_d-2x2-matrix in action [1,2] --> --> [3,5]
%574 = [3.00000000000, 5.00000000000]

Don't read the symbolic description of Fib2d from the numbers at the moment... with \( a = 0.447213595500.... = 1/\sqrt 5 \) (by W|A)  it is then \( \begin{pmatrix} 2  & 1  \\ 1 & 3 \end{pmatrix} / \sqrt 5 \) 

Using the iterpretation that  \( r= a = 1/ \sqrt 5\) we have the 2x2-matrix iterations

Code:

Fib2d = [2,1;1,3]* 'r       \\ defining the matrix with symbolic entry for the imprecise real value 1/sqrt(5)
%600 =
[2*r   r]
[  r 3*r]

[1,2]*Fib2d
%602 = [4, 7] *r

[1,2]*Fib2d^2
%604 = [15, 25]*r^2   \\ here we can cancel by r^2=1/5: -->  -->  [3,5]

[1,2]*Fib2d^3
%606 = [55, 90]*r^3    \\ here we can cancel by r^2=1/5: --> --> --> [11,18]*r

[1,2]*Fib2d^4
%608 = [200, 325]*r^4   \\ here we can cancel by r^4=1/25: --> --> --> --> [8,13]

Hmmm, having a 2x2-transfermatrix Fib2d this should actually be a fractional linear transformation.

OKAY! Absolutely beautiful! Yes this should be a LFT! I think this implicitly solves \(\theta(t)\), I'm just not sure how yet. Definitely agree this will get it though!

I think this is very important, because it shows how the iteration of an LFT is Fibonacci, and shows how a restriction to reals of an LFT creates a real Fibonacci.

[bo198214]

OKAY! Absolutely beautiful! Yes this should be a LFT! I think this implicitly solves \(\theta(t)\), I'm just not sure how yet. Definitely agree this will get it though!

I think this is very important, because it shows how the iteration of an LFT is Fibonacci, and shows how a restriction to reals of an LFT creates a real Fibonacci.

It is not a Fibonacci! As Gottfried pointed out already it only agrees on the even Fibonacci indexes.
Hence it is not a theta transforming the non-realness (of the original Fibonacci interpolation) into a realness - which honestly I can not imagine that it exists, because of the pole in the middle between the fixed points - but one never knows. That's why Gottfried was excluding these poles in the middle (which only occurs for uneven iterates) and banning it to the left side of the left fixed point (where it occurs for even iterates, or to the right side of the right fixed point when we consider negative even iterates) by demanding that it only agrees with Fibonacci on even iterates.
Even you pointed out to me already once that the iterative root of the iterative square is probably not the original function anymore.
But the trick with changing the sign of the eigenvalue is quite naughty, Gottfried!
But yes, it boils down to taking the regular iteration of \(f^{\circ 2}\) and just taking 1/2 of it \((f^{\circ 2})^{\mathfrak{R} t/2}\).

[tommy1729]

see also my comment here :

https://math.eretrandre.org/tetrationfor...=fibonacci

and 

https://math.eretrandre.org/tetrationfor...=fibonacci

So basically an old idea you are reviving.


I could say more but im in a hurry.

regards

tommy1729