Re: " Wanted solutions to f ( (3x- 5) / (x +1) ) = 1/ f (x) ; x , f real "

"G. A. Edgar" <edgar@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message news:<d46bft$kk$1@xxxxxxxxxxxxxxxxxxxxxxxxx>...
> In article <d45fbu$2km$1@xxxxxxxxxxxxxxxxxxxxxxxxx>, Alain Verghote
> <alainverghote@xxxxxxxx> wrote:
>
> > f(x) is not a constant function .
> >
> > Once more a functional equation ?
> > YES,but this one does request I believe other
> > methods than causal process, Abel counting functions
> > phi(x)on R such as phi((3x- 5)/(x +1)) = phi(x) + 1,
> > or continuously iterated fonctions.
> >
> > I would like to know your ideas on this matter.
> >
> > Friendly,Alain.
>
> Consider the function S(x) = (3*x-5)/(x+1).
> Note that SSSS(x)=x for all x. Also SS(x)=x has no real solutions.
> So the (projective) real line is partitioned into cycles of size 4
> by this map. Our funcition f alternates between a value and its
> reciprocal as we go around one of these cycles. To make things more
> explicit: S maps interval [-infinity,-1) onto [3,infinity),
> maps [3,infinity) onto [1,3), maps [1,3) onto [-1,1) and maps
> [-1,1) onto [-infinity,-1). So define f arbitrarily (nonzero)
> on [-infinity,-1), then use S to get the values of f on the rest of
> the real line:
> for x in [-1,1), define f(x) = 1/f(S(x));
> for x in [1,3), define f(x) = 1/f(S(x)) = f(SS(x));
> for x in [3,infinity), define f(x) = 1/f(S(x)) = 1/f(SSS(x)).

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f((3*x-5)/(x+1))= 1/f(x) .

Dear Friends,

I am mainly interested in solving methods and steps to walk upward.
I propose the following schedule:

1°)ANALYZING THE CAUSE:
=> homographic function ,two kinds:one or two fixed points,
here h(x)=(3x-5)/(x+1)=x gives x1=1+2I ,x2=1-2I
then there exists a simple iterated relation:
(h^[r](x)-x1)/(h^[r](x)-x2)= a^r*(x-x1)/(x-x2),r integer.
for r=1 a=-I imaginary , a^4 = 1 ,giving the smallest period
of h(x) 4 .

2°)DESCRIBING THE PROCESS:
=> when x-> (3x-5)/(x+1) , f -> 1/f .
if needed,

3°)STUDYING A SIMPLER NABE CASE:
=> case x -> h(x) , f -> f or f is a 'h' invariant function,
using period 4 ,h^[4](x)= x we may have:
f(x) = g(x + h(x)+h^2(x)+h^3(x)) 1°
m(x + h^2(x)*m(h(x)+h^3(x)) 2°
p(x)*p(h(x))*p(h^2(x))*p(h^3(x)) 3°

4°)GIVING SOLUTIONS:
we must take into account the inversion of f(x) =>1/f(x) ,
'splitting' 1,2,3° gives direct solutions:
f(x) ={(x + h(x)+h^2(x)}/{h(x)+h^2(x)+h^3(x)} 1°
m(x + h^2(x))/ m(h(x)+h^3(x)) 2°
p(x)*p(h^2(x))/{p(h(x))*p(h^3(x))} 3°

5°)AN ILLUSTRATING EXAMPLE:
numerous solutions may be built ...
f(x)=(x + h^2(x))/ {(h(x)+h^3(x)}=
{x^2-5)(x+1)(x-3)}/{2(x^2-10x+5)(x-1)} ,

Sincerely,Alain.

.