A function in terms of another function I have two functions that are polynomials. For example:
$$F=x^2+2x+1  \hspace{5mm} \text{ and } \hspace{5mm} G=2x^2-x+2$$
I need to write one of these two functions in terms of the other one. For the example above the answer would be:
$G$ as a function of $F$ is: $G=2F-5F^{1/2}+5$
As you know, depending on the complexity of $F$ and $G$, doing such a manipulation can be very difficult. Therefore, I am wondering if there is a known mathematical way to do this algebraic manipulation. Please note that the general format of $F$ and $G$ is always as below:
$$F=\sum(ai × xi^i)|i=0,1,2,…,N$$
$$G=\sum(bi × xi^i)|i=0,1,2,…,N$$ 
Thank you!
 A: I think this is not always possible. For your example, note that:
$$H(x):=2F(x)-5\sqrt{F(x)}+5=2x^2+4x+2-5\sqrt{(x+1)^2}+5=2x^2+4x-5|x+1|+7$$
is different form $G(x)$, since, for $x=-2$ we have:
$$H(-2)=2\neq12=G(-2)$$
In general, $\sqrt{F(x)}$ is not differentiable at every zero of $f$, so it cannot be a polynomial function, since all polynomial function are infinitely differenitable. So, we cannot include such expressions in our final formula.
Now, let $F(x),G(x)$ be two polynomials of degrees $n,m$ respectively. Then, form the Eucledian division algorithm we have that there exists two, unique, polynomials $p(x),q(x)$ such that:
$$G(x)=p(x)F(x)+q(x),\ \deg(q(x))<\deg(F(x))\text{ or }q(x)=0.$$
If $n\neq m$, pparently, since $\deg(q(x))<\deg(F(x))$ we cannot write $q(x)$ in terms of $F(x)$ so our problem does not have a solution in this case. If $n=m$ (which is the case you describe), we have that $p(x)=c$, for some $c\in\mathbb{R}$, so:
$$G(x)=cF(x)+q(x),\ \deg(q(x))<\deg(F(x))\text{ or }q(x)=0.$$
For $n=1$, apparently the problem has a solution, since $q(x)=r\in\mathbb{R}$, so:
$$G(x)=cF(x)+r.$$
For $n=2$, we have that $\deg(q(x))<2$ so, in the case when $\deg(q(x))=1$ we cannot solve the problem since a linear polynomial cannot be written in terms of a higher degree polynomial without involving non-polynomial functions. For instance, take $F,G$ as given in your example:
$$G(x)=2x^2-x+2=2F(x)-5x.$$
It is evident now that one can write $G$ in terms of $F$ if and only if $q(x)=0$.

Attention!
On all the above it is assumed that the only functions allowed are polynomials and that we want two express $G$ in terms of $F$ using only scalars, summation and multiplication. If one lets, for instance, function synthesis you have several ways to write it. The first one is to consider the two-variable polynomial:
$$T_q(x,y)=cy+q(x)$$
(using the previous notation). Now it is evident that:
$$G(x)=T_q(x,F(x)).$$
Another way is - using your example - to note that:
$$G(x)=2F\left(x-\frac{1}{4}\right)+\frac{7}{16}.$$
The latter has also a nice geometric meaning - for $n=2$: every parabola - that is secribed by a polynomial - can be transformed to any other parabola through a vertical translation $\left(\frac{7}{16}\right)$, a horizontal trnaslation $\left(\frac{1}{4}\right)$ and a scale transformation $(2)$.
Let us now try to generalise this result to every polynomial. So, we would like to prove that, if $F,G$ are two polynomials of the same degree, then there exist $a,b,c\in\mathbb{R}$ such that:
$$G(x)=aF(x-b)+c.$$
At first, it is evident that, if $a=\frac{b_n}{a_n}$, where $a_n,b_n$ are the coefficients of the highest powered term of $F,G$ respectively and $n$ is their degree. But, for our frustration, we cannot go further than this, since, if, for instance $G(x)=x^3+x+1$ and $F(x)=x^3$, we may choos $a=1$ but, for every choice of $b,c$, such that $c-b^3=1$, we have:
$$F(x-b)+c=(x-b)^3+c=x^3-3bx^3+3b^2x-b^3+c=x^3-3bx^2+3b^2x+1.$$
Since we demand that $F(x-b)+c=G(x)$, ont the one hand we should have $b=\pm\frac{1}{\sqrt{3}}$ but, on the other hand, there should be no quadratic term, so we should have $b=0$, which is a contradiction.
Intuitively, this was also expected since $yx^3$ has no extrema, but, on the other hand, there do exist third degree curves that have extrema and translations and scale transformations do not affect the existence and number of extrema. The same applies to higher degree curves that have diffent numbers of extrema, not "strictly" dependent on their degree, as well (think of $y=x^4$ and $y=x^4-x+1$).
Also, another way to prove that, even if such numbers existed, we could not always calculate them, at least for $n>4$, is that, if that was true, then, for every polynomial $P$ with degree $n\geq5$ there would exist $a,b,c\in\mathbb{R}$, such that:
$$P(x)=a(x-b)^n+c$$
so, the polynomial equation
$$P(x)=0$$
would be equivalent to the folowing:
$$a(x-b)^n+c=0\Leftrightarrow(x-b)^n=-\frac{c}{a}$$
which we can easily determine if it has (any) solutions and, moreover, this solution is algebraic, which contradicts the Abel-Ruffini Theorem if we suupose that we can derive the coefficients $a,b,c$ from $P$.
