ds.dynamical systems - Uniqueness in Composition of Polynomials

Your special case is right. More generally:

Let $fleft(xright)=x+b$ with $bneq 0$.
Let $gleft(xright)=cx^2+dx+e$ with $c>0$, $dinmathbb R$ and $einmathbb R$.

In fact, it is clear that every composition of $f$'s and $g$'s is a polynomial of positive degree and with positive leading coefficient (since $c>0$).

If we have a polynomial $Pinmathbb Rleft[Xright]$ which is a composition of $f$'s and $g$'s, we can always reconstruct the last step of the composition. Namely, we search for a nonnegative real $u$ such that $P-u=cQ^2+dQ+e$ for some polynomial $Qin mathbb Rleft[Xright]$ of positive degree and with positive leading coefficient. If the last step has been a $g$, then $u=0$ must work; if the last step was an $f$, but some $g$ occured in the composition, then we must have a solution with $uneq 0$ (in fact, if the last steps were $g$, $f$, $f$, ..., $f$ in this order, with $f$ occuring $k$ times, then $u$ must be $kbneq 0$); if the composition consists of $f$'s only, then there is no solution (because $P$ must have degree $1$). The important thing is that the $u$, if it exists, is unique. In fact, if there would be two different $u$'s, then the two corresponding $Q$'s - let's call them $Q_1$ and $Q_2$ - would satisfy $left(cQ_1^2+dQ_1+eright)-left(cQ_2^2+dQ_2+eright)=w$ for some nonzero real $w$ (here, $w$ is the difference of the two $u$'s). This equation rewrites as $cleft(Q_1-Q_2right)left(Q_1+Q_2+1right)=left(c-dright)Q_1-left(c-dright)Q_2+w$. Thus, (remembering that $c>0$) we conclude that

$degleft(Q_1-Q_2right)+degleft(Q_1+Q_2+1right)=degleft(cleft(Q_1-Q_2right)left(Q_1+Q_2+1right)right)$
$=degleft(left(c-dright)Q_1-left(c-dright)Q_2+wright)leqmaxleftlbrace deg Q_1,deg Q_2rightrbrace$.

But at least one of the two degrees $degleft(Q_1-Q_2right)$ and $degleft(Q_1+Q_2+1right)$ must actually be equal to $maxleftlbrace deg Q_1,deg Q_2rightrbrace$ (because $Q_1$ and $Q_2$ are linear combinations of $Q_1-Q_2$ and $Q_1+Q_2$), and thus the other one must be zero or $-infty$ (the degree of the zero polynomial). In other words, one of the polynomials $Q_1-Q_2$ and $Q_1+Q_2+1$ is constant. But the polynomial $Q_1+Q_2+1$ cannot be constant (because $Q_1$ and $Q_2$ have positive degree and positive leading coefficients). Hence, the polynomial $Q_1-Q_2$ is constant.

So let $Q_1-Q_2=k$ for $kinmathbb R$. Then, $left(cQ_1^2+dQ_1+eright)-left(cQ_2^2+dQ_2+eright)=w$ rewrites as $ckleft(Q_1+Q_2right)+dk=0$ (since $Q_1-Q_2=k$). Hence, the polynomial $ckleft(Q_1+Q_2right)$ also must be constant, so that $ck=0$ (since the polynomial $Q_1+Q_2$ is not constant, because $Q_1$ and $Q_2$ are two polynomials with positive degree and positive leading terms). Since $c>0$, this yields $k=0$, and thus $Q_1-Q_2=k=0$, so that $Q_1=Q_2$, and therefore $0=left(cQ_1^2+dQ_1+eright)-left(cQ_2^2+dQ_2+eright)=w$, contradicting $wneq 0$.