Find all pairs of polynomials $f(x),g(x)$ that satisfy $f(xy)=f(x)+g(x)f(y)$ for all $x,y$.
Problem
Source: Ukraine 1998 Grade 10 P8
Tags: Polynomials, fe, algebra, functional equation, Functional Equations
Math4Life2020
06.06.2021 03:10
We can start by thinking about the degrees of $f$ and $g$. Case 1. $\deg(f)=0$ Then, the degree of $g$ is also $0$. Writing $f(x)=a, g(x)=b$, we obtain the following solutions: $f(x)=0, g(x)=b, b\in \mathbb{R}$; $g(x)=-1, f(x)=a, a \in \mathbb{R}$. Case 2. $\deg(f) \geq 1$ Then, $\deg(f)=\deg(g)$. I do not know how to proceed, can someone build on my ideas?
jasperE3
06.06.2021 03:20
Seting $y=0$, we get that $f(x)=a-ag(x)$, where $a=f(0)$. If $f(0)=0$, then we get $f\equiv0$. If $f(0)\ne0$, then the FE becomes $g(xy)=g(x)g(y)$, and the only multiplicative polynomial is $x^n$.
rama1728
06.08.2021 11:37
jasperE3 wrote: Find all pairs of polynomials $f(x),g(x)$ that satisfy $f(xy)=f(x)+g(x)f(y)$ for all $x,y$.
Let \(P(x,y)\) be the assertion that \(f(xy)=f(x)+g(x)f(y)\). Then, \(P(x,0)\) gives \[f(x)=c-cg(x)\]where \(c=f(0)\). If \(c=0\), then \(f(x)=0\) for all \(x\), and \(g\) can be any polynomial. If \(f\) is constant, then \(f(x)=a\) and \(g(x)=0\) works, no matter what \(a\) is. Finally, if \(f\) is non-constant and \(c\neq 0\), then substituting \(f(x)=a-ag(x)\) in the original functional equation gives \(g(xy)=g(x)g(y)\), implying that \(g(x)=x^n\) as that is the only multiplicative polynomial. Thus, we get another solution \(f(x)=c\left(1-x^n\right)\).