Find $f(x)$ such that $f(x)+f\left(\frac{1}{x}\right)=f(x)\cdot f\left(\frac{1}{x}\right)$

Question

Given a polynomial $f(x)$ of n degree such that

$$f(x)+f\left(\frac{1}{x}\right)=f(x)\cdot f\left(\frac{1}{x}\right)$$

Find the polynomial

I've tried considering $f(x)=\sum \limits_{i=1}^{n} a_{i}x^{i}$ and after a number of steps I finally arrived at the answer.However I am not convinced.Is there a better way to solve this functional equation.

Thanks in advanced!

Answer 1

Hint: $$f(x) = 1 + \frac{1}{f(\frac{1}{x})-1}$$

So $f(x) - 1 = \frac{x^n}{P(x)}$, where P(x) is a polynomial of degree $\le n$. The left-hand side is a polynomial, so $P(x) = a x^k$ for some $k$. Now $a$ and $k$ can be found from the functional equation itself. But, as Barry Cipra notes in comments, since the left-hand side of the equality that defines $P$ has degree $n$ the only possible value for $k$ is $0$.

Answer 2

$$f(y) = 1+y$$

$$f(x)+f\left(\frac{1}{x}\right) = 1+x+1+\frac{1}{x} = 2 + x + \frac{1}{x}$$

$$f(x) f\left(\frac{1}{x}\right) = (1+x) \left ( 1+\frac{1}{x}\right) = 1 + x + \frac{1}{x} + x \frac{1}{x} = 2 + x + \frac{1}{x}$$

In fact, $f(y) = 1+y^k$ for $k \in \mathbb{N}$ seems to work for the same reason.

Answer 3

Suppose that the degree of $f$ is greater or equal to $1$. If $g(x)=f(x)-1$, then $g$ is a polynomial, and $$g(x)g\left(\frac{1}{x}\right)=(f(x)-1)\left(f\left(\frac{1}{x}\right)-1\right)=f(x)f\left(\frac{1}{x}\right)-f(x)-f\left(\frac{1}{x}\right)+1=1.$$ Then $g(0)=0$: this can be seen from the equation above after taking the limit as $x\to\infty$. So, $\frac{g(x)}{x}$ is a polynomial that satisfies the same equation, with degree one less than the degree of $g$. So, if $g$ is of degree $n$, by induction you have that $\frac{g(x)}{x^n}$ should be constant, and equal to $\pm1$. So, $f(x)=1\pm x^n$.

Answer 4

Assume $f(x)=c$ is constant. Then we arrive at $c+c=c\cdot c$, i.e. $c=0$ or $c=2$.

Otherwise write $f(x)=c+x^kg(x)$ where $k\ge1$ and $g(0)\ne0$. Then for $x\ne0$ $$2c+x^kg(x)+x^{-k}g(x^{-1})=f(x)+f(\frac1x)=f(x)f(\frac1x)=cx^kg(x)+c^2+cx^{-k}g(x^{-1})+g(x)g(\frac1x)$$ From looking at the highest power occuring in $g$ we see $c=1$, hence this simplifies to $$ g(x)g(\frac1x)=1$$ so $g$ is a monomial, hence constant. Thus $f(x)=1+ax^k$ and we further check that $a=\pm1$. Thus the complete list of solutions is $$ f(x)=0,\qquad f(x)=2,\qquad f(x)=1\pm x^k\text{ with }k\ge1$$

Answer 5

A slight variation of @detnvpp using the fundamental theorem of algebra. Set $g(x)=f(x)-1$ then $$g(x)g\left(\frac{1}{x}\right)=(f(x)-1)\left(f\left(\frac{1}{x}\right)-1\right)=f(x)f\left(\frac{1}{x}\right)-f(x)-f\left(\frac{1}{x}\right)+1=1, \forall x\ne 0 $$

if $\alpha \ne 0$ is a zero of the polynomial $g(x)$ then $g(\alpha)g(\frac{1}{\alpha})=0$. That contradicts $g(x)g\left(\frac{1}{x}\right)=1$. Therefore $g(x)$ has no zeros and is constant or $g(x)$ has exactly one zero $\alpha=0$ for. In the first case $g(x)=\pm 1$ and therefore $f(x)=0$ or $f(x)=2$. In the second case $g(x)=x^n$ because of $\left(g(1)\right)^2=1$ and therefore $f(x)=1 \pm x^n$