A Nested Functional Equation

Algebra Level 5

A strictly monotonic function $f(x)$ defined at $(0,+∞)$ satisfies the equation: $f(x)f \left(f(x)+\dfrac{1}{x}\right)=1$ .

What is the value of $f(1)$ ?

0

\dfrac{1+\sqrt{5}}{2}

1

\dfrac{1-\sqrt{5}}{2}

\dfrac{1+\sqrt{5}}{2}

\dfrac{1-\sqrt{5}}{2}

1 solution

Alex Burgess
Jul 8, 2019

Let $f(x) = Ax^n, g(f(x)) = f(x) f(f(x)+\frac{1}{x}) = Ax^n A^{n+1}x^{n^2} + O(x^{n^2-n-1}) = A^{n+2}x^{n^2+n} + O(x^{n^2-n-1})$ .

In order to equal $1$ for all $x$ , we either need a $x^{-(n+1)}$ term or $n = 0, -1$ .

Ignoring options with a $x^{-(n+1)}$ term for now (too much algebra, and then need to see if it's monotonic):

Let $f(x) = \frac{A}{x} + B$ , this is monotonic, and strictly monotonic if $A \neq 0$ .

$g(f(x)) = (\frac{A}{x} + B)f(\frac{A+1}{x} + B) = (\frac{A}{x} + B)(\frac{Ax}{Bx + (A+1)} + B) = \frac{A^2+ABx}{Bx + (A+1)} + \frac{AB}{x} + B^2$ .

$g(f(x)) = 1 \implies 1 - B^2 = \frac{A^2+ABx}{Bx + (A+1)} + \frac{AB}{x}$ ,

so $(1-B^2)(Bx^2 + (A+1)x) = A^2x+ABx^2 + AB(Bx + (A+1))$ .

$B(A + B^2 - 1)x^2 + (A^2 + AB^2 + (B^2-1)(A+1))x + AB(A+1) = 0$ .

Evaluating terms of orders $O(x^2)$ and $O(1)$ , we get:

$B(A + B^2 - 1) = 0$ and $AB(A+1) = 0$ .

These have solutions:

1 The term of order $O(x)$ becomes:

$A^2 - A - 1 - 0$ , so $A = \frac{1 \pm \sqrt5}{2}$ .

2 The term of order $O(x)$ becomes:

$B^2 - 1 = 0$ , so both are solutions.

3 The term of order $O(x)$ becomes:

$1 - B^2 = 0$ , which is false.

Hence the solutions are:

$f(x) = \frac{1 \pm \sqrt5}{2x}$ OR $\pm 1$ .

$f(1) = \frac{1 \pm \sqrt5}{2}$ OR $\pm 1$

If the question asked for strictly monotonic functions, then the answer would be $f(1) = \frac{1 \pm \sqrt5}{2}$ .

Let $f(x)$ be a polynomial from $x^n$ to $x^{-m}$ .

Define the polynomial of non-negative exponents, of order $N = n+m$ , $F = (f + \frac{1}{x})x^m$ $(x^{m-1}$ may be negative).

$g(f(x)) = \frac{F-x^{m-1}}{x^m} (A_N \frac{F}{x^m}^n + ... + A_0 \frac{F}{x^m}^{-m}) = 1$ .

CASE 1: $n = 0: O(\frac{F}{x^m}) = O(1)$ :

$\frac{F-x^{m-1}}{x^m} = O(1)$ , so $m = 0, 1$ ,

$m = 0$ leads to 2 solutions above.

$m = -1$ leads to $n = -1$ which is a contradiction in this section.

CASE 2: $n>0: A_N \frac{F}{x^m}^n$ provides the highest order in the brackets:

Hence: $A_N(A_N x^N)^{n+1} = x^{m(n+1)}$ , so $n + m = N = m$ . Contradiction.

CASE 3: $n<0: A_0 \frac{F}{x^m}^{-m}$ provides the highest order in the brackets:

Hence: $O(x^{m^2+m-1}) = O( (x^{N+1})^m )$

$m^2+m-1 = m^2 + mn + m$ , so $mn = -1$ . So $n = -1, m = 1$ , relates to 2 solutions above.

Alex Burgess - 1 year, 11 months ago

Edit: $f(x)$ isn't valid as $f(x)$ is only defined on the positives.

Alex Burgess - 1 year, 11 months ago