# Polynomial functional equation

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?

• I assume that the given equation must hold for all real numbers $x \neq -1$? Mar 26 at 18:51
• Yes of course. I should have specified that $x \neq -1$ and $f(x) \neq 1$ similarly. Mar 26 at 18:55
• Mostly asking because the question becomes somewhat different if the equation must hold for all complex values of $x$ since one can show that the only possible root of $f$ that is in the domain over which the equation has to hold is $0$, which if the equation holds for all complex numbers would imply that $f(x) = cx^n$ for some $c$ and $n$. But over the reals all we can say based on this is that $f$ is $cx^n$ times some product of quadratic factors with negative discriminant. Mar 26 at 18:58
• Of course other arguments would allow use to narrow down $f$ in the real case. I'm not claiming that irreducible quadratics are solutions; only that they have no real roots. I suspect that actually the only solution is $f(x) = -x$. Mar 26 at 18:58
• Another solution is the constant polynomial $f(x) = c$ for all $x$, where $c=1/(c-1)$. Mar 26 at 19:15

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 $$n and $$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 $$f(x)=-x$$.

• There must be an error here, as this misses @Goncalo's solution, $f(x)=c$ where $c^2-c-1=0$. I haven't spotted where yet! Mar 26 at 19:22
• Make $d=0$ in your equation $a^2 x^d - a = (x+1)^d$. Mar 26 at 19:37
• @Joshua Tilley I am confused about where you got the step of $f(x)=ax^d$ and $a^2x^d-a=(x+1)^d$. I feel like I am missing something obvious. Mar 30 at 20:59
• It was shown that $\deg((x+1)^df(\frac{1}{x+1}))=0$. This implies that $f(x)=ax^d$. The following equation is found by substituting $f(x)=ax^d$ back in. Mar 30 at 21:16

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.

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 $$p$$: $$\begin{equation} p \in \{-x, \frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2}\} \end{equation}$$

• Maybe a bit complicated for this type of problem, but I like the fact that you can use tools from analysis (limits) and the neat fact that any non-constant polynomial $p$ goes to $\pm\infty$ as $x \rightarrow \pm\infty$, so I will leave it up for any interested... Mar 27 at 20:09