Commuting functions

Question

I encountered the following exercice:

Suppose $f$ is a quadratic function of the form $f(x)=ax^2+b$ with $a\neq 0$. Show that if $g(x)=c x^2+d$ with $c\neq 0$ is another function commuting with $f$ then $f(x)=g(x)$ for all $x$.

Starting with the definition of commuting function

$$(f\circ g)(x)=(g\circ f)(x),\quad \textrm{for all } x$$

and using the fact that two polynomials are equal as functions if all corresponding coefficients are equal I obtain the relation:

$$ad+b=cb+d$$

But I cannot figure out how to proceed. Thanks for any ideas.

Answer 1

Note that$$f\bigl(g(x)\bigr)-g\bigl(f(x)\bigr)=-a^2 c x^4-2 a b c x^2+a c^2 x^4+2 a c d x^2+a d^2-b^2 c+b-d.$$So, since $f$ and $g$ commute, you have$$\left\{\begin{array}{l}a c^2-a^2 c=0\\2 a c d-2 a b c=0\\a d^2-cb^2+b-d=0.\end{array}\right.$$The first equation is equivalent at $ac(a-c)=0$; since $a,c\neq0$, it follows that $a=c$. On the other hand, the second equation is $2ac(d-b)=0$. Again, since $a,c\neq0$, this implies that $b=d$.

Answer 2

$f(g(x)) = a(cx^2+d)^2 + b = ac^2x^4 + 2acdx^2 + ad^2 + b\\ g(f(x)) = c(ax^2+b)^2 + d = a^2cx^4 + 2abcx^2 + b^2c + d$

If $f(g(x))=g(f(x)) \space \forall x$ then the co-efficients of $x^4$ and $x^2$ and the constant term must be the same in both expressions, so

$ac^2 = a^2c \\ acd = abc \\ ad^2+b = b^2c+d$

I'll let you take it from there.