Fibonacci as iteration of fractional linear function

[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.