Question

Given a function `h` defined by `h(x) = (x-4)/(x+4)`, how do you find a function `f` such that `h = f@f` ?

Answer 1

`f(x)=((1 + 2 sqrt[2]) x-4)/(x+2 sqrt[2] + 4)`

Explanation:

Proposing `f(x)` such that

`f(x)=(a x + b)/(c x + d)` and after doing

`f(f(x))=(b (a + d) + (a^2 + b c) x)/(b c + d^2 + c (a + d) x)`

and then identifying

`{(b (a + d) = -4), ((a^2 + b c) = 1), (b c + d^2 = 4), (c (a + d) = 1):}`

and solving for `a,b,c,d` obtaining

#((a = sqrt[1/7 (16 sqrt[2]-13)]),( b = -4 sqrt[ 1/7 (4 sqrt[2]-5)]), (c = sqrt[1/7 (4 sqrt[2]-5)]), (d = sqrt[ 8/7 + (16 sqrt[2])/7]))#

There are four solutions, with two real and two complex. We decided to show only the most simple.

With this solution we have

`f(x)=((1 + 2 sqrt[2]) x-4)/((x+2 sqrt[2] + 4)`

Note. The relations associated to the coefficients for `f, h` can be obtained by doing

`((a, b),(c, d))^2=((a^2 + b c, a b + b d),(a c + c d, b c + d^2))`

For `f@f@f` the relations are given by

`((a, b),(c, d))^3`

so generally

`f@f @ cdots @ f` for a `n`-composition we have

`((a, b),(c, d))^n`

so this bilinear transformation follows the composition rules of linear operators.