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

1 Answer
Apr 1, 2017

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.