# Thread: Polynomial f(x) such that f(x)f(1/x)=f(x)+f(1/x)

1. ## Polynomial f(x) such that f(x)f(1/x)=f(x)+f(1/x)

My teacher said that the polynomial function such that $f(x)f\left(\frac{1}{x}\right)=f\left(x \right )+f\left(\frac{1}{x} \right )$ exists and that must be of the form $f(x)=1 \pm x^n$. But he didnt give any proof. Can anyone help me with a proof that the function satisfying the above condition can only be $f(x)=1 \pm x^n$

2. ## Re: Polynomial f(x) such that f(x)f(1/x)=f(x)+f(1/x)

Originally Posted by NameIsHidden
My teacher said that the polynomial function such that $f(x)f\left(\frac{1}{x}\right)=f\left(x \right )+f\left(\frac{1}{x} \right )$ exists and that must be of the form $f(x)=1 \pm x^n$. But he didnt give any proof. Can anyone help me with a proof that the function satisfying the above condition can only be $f(x)=1 \pm x^n$
$f(x)=\frac{f(\frac{1}{x})}{f(\frac{1}{x})-1}$

$f(\tfrac{1}{x})=\frac{f(x)}{f(x)-1}$

\begin{align*}f(x)\cdot f(\tfrac{1}{x}) &= \frac{f(\frac{1}{x})}{f(\frac{1}{x})-1} \cdot \frac{f(x)}{f(x)-1} \\ &= \frac{1}{f(\tfrac{1}{x})-1} \cdot \frac{1}{f(x)-1}\end{align*}

Note that
$f(\tfrac{1}{x})-1 = \frac{f(x)-[f(x)-1]}{f(x)-1} = \frac{1}{f(x)-1}$
so that
$\frac{1}{f(\tfrac{1}{x})-1} = f(x)-1$.

So,
$f(x)\cdot f(\tfrac{1}{x}) = [f(x)-1] \cdot \frac{1}{f(x)-1} = 1$.

Now let
$g(x)=f(x)-1$
and thus
$g(\tfrac{1}{x}) = f(\tfrac{1}{x}) -1$.

Finally,
$g(x)\cdot g(\tfrac{1}{x}) = [f(x)-1] \cdot [f(\tfrac{1}{x})-1] = \frac{f(x)-1}{f(x)-1} = 1$.

(Of course $g(x)\cdot g(x^{-1}) = 1$!)

If $g(x)$ is polynomial, then $g(x) = \pm x^n \implies f(x)=1\pm x^n$.