Polynomial functional equation

Question

Find all polynomials $f : R \rightarrow R$ such that $f\left(\frac{1}{x+1}\right)=\frac{1}{f(x)-1}$.

Since the functions must be polynomials, I tried using an argument by degrees, but this did not lead me anywhere. Can someone help with some ideas?

Answer 1

Suppose that the degree of $f$ is $d$ then

$$(x+1)^df\left(\frac{1}{x+1}\right)(f(x)-1)=(x+1)^d\tag 1$$

The point is that $(x+1)^df\left(\frac{1}{x+1}\right)$ is a polynomial in $x$. This is then an identity of polynomials.

$f(x)$ is not identically equal to $1$, and so $\deg(f(x)-1)=\deg(f(x))=d$. It then follows that $\deg((x+1)^df\left(\frac{1}{x+1}\right))=0$. Writing $f(x)=\sum_{n=0}^d a_nx^n$ we have $(x+1)^df\left(\frac{1}{x+1}\right)=\sum_{n=0}^da_n(x+1)^{d-n}$ so that we have $a_n=0$ for $n<d$ and $f(x)=ax^d$. Substituting $f(x)=ax^d$ into $(1)$,

$$a^2x^d-a=(x+1)^da(x+1)^{-d}(ax^d-1)=(x+1)^d$$

If $d=0$, then we get $f(x)=a$ for $a$ any solution to $a^2-a-1=0$. Otherwise, set $x=-1$ to get $a(\pm a-1)=0$, so $a\in \{0,1,-1\}$.

$a=0$ doesn't work.

$a=1$ gives $x^d-1=(x+1)^d$, which doesn't work.

$a=-1$ gives $x^d+1=(x+1)^d$ so that $d=1$.

Thus, the only other solution is $f(x)=-x$.

Answer 2

Here is a slightly different solution with some input from a comment by @dxiv.

As in the other answer, we note that if $d$ is the degree of $f$ then $$ p(x) = {(x + 1)}^d f\left(\frac{1}{x + 1}\right) (f(x) - 1) $$ is a polynomial. We have that $p(x) = {(x + 1)}^d$ for all real $x \neq 1$, and since $p$ and ${(x + 1)}^d$ agree at infinitely many points, we have that $$ {(x + 1)}^d f\left(\frac{1}{x + 1}\right) (f(x) - 1) = p(x) = {(x + 1)}^d $$ is an identity of polynomials that holds even over the complex numbers.

Let $r$ be any complex root of $f$. Suppose that $r \neq 0$. Then let $x = \frac{1}{r} - 1$. We get that $$ \frac{1}{r^d} f(r) \left(f\left(\frac{1}{r} - 1\right) - 1\right) = \frac{1}{r^d} $$ which is a contradiction since $f(r) = 0$ but $\frac{1}{r^d} \neq 0$.

It follows that $f(x) = cx^d$ for some constant $c$ and natural number $d$, where $d$ could be $0$ if $f$ is constant.

We thus have that $$ c^2 x^d - c = {(x + 1)}^d \frac{c}{{(x + 1)}^d} (cx^d - 1) = {(x + 1)}^d. $$

By expanding the right hand side and equating coefficients, we know that either $d = 0$ and $c$ satisfies $c^2 - c - 1 = 0$ (i.e. $c$ is the golden ratio or its conjugate) or we have that $d = 1$ and $c = -1$. In all other cases we get a $x^{d - 1}$ term on the right hand side but not on the left.

Answer 3

First, assume that $p$ is a non-constant real valued polynomial: $p(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_0, n \ge 1, a_n \ne 0 \implies lim_{x \to \pm \infty} \ p(x) = \pm\infty$. We also have $lim_{x \to -1} \ \frac{1}{x+1} = \infty \iff lim_{x \to -1} \ p(\frac{1}{x+1}) = \pm\infty$, which due to the given equality gives $\lim_{x \to -1} \ \frac{1}{p(x) - 1} = \pm \infty \iff \lim_{x \to -1} \ p(x) = 1$, and since $p$ is a polynomial, that is: $p \in C^\infty(D_p \equiv \mathbf{R})$, we must have

\begin{equation} p(-1) = 1 \end{equation}

Furthermore, we have $lim_{x \to \pm\infty} \frac{1}{1 + x} = 0$ and $lim_{x \to \pm\infty} \frac{1}{p(x) - 1} = 0$, and using the same argument of $p$ belonging to the class of smooth functions $C^\infty(\mathbf{R}$), we get

\begin{equation} p(0) = 0 \end{equation}

By the argument above, we see that this is the only root of $p$. Thus, we can write

\begin{equation} p(x) = a(x-0)^n = ax^n, a \in \mathbf{R} /0, n \in \mathbf{Z^+}. \end{equation}

Using the fact that $p(-1) = 1$, we can hence write $a(-1)^n = 1 \iff a = \pm 1$. Case $1$, $a = -1$:

\begin{equation} -\frac{1}{(x+1)^n} = \frac{1}{-x^n - 1} \iff (x+1)^n = x^n + 1 \end{equation}

Which holds $\forall x \in \mathbf{R}$ if and only if $n = 1$. Hence, $p(x) = -x$ is a solution. Case $2$, $a = 1$:

\begin{equation} \frac{1}{(x+1)^n} = \frac{1}{x^n - 1} \iff (x+1)^n = x^n - 1 \end{equation}

Which does not hold $\forall x \in \mathbf{R}$ regardless of the value of $n$. Hence, if $p$ is non-constant, we have $p(x) = -x$ as our only solution.

Now, consider the case of $p$ being a constant valued polynomial: $p \equiv c, c\in \mathbf{R}$. Using the given equality, we get:

\begin{equation} c = \frac{1}{c-1} \iff c^2 - c - 1 = 0 \iff c = \frac{1 \pm \sqrt{5}}{2} \end{equation}.

This finally gives all possible solutions for $p$: \begin{equation} p \in \{-x, \frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2}\} \end{equation}