local inverse of polynomial Is there a possibility to invert a polynomial locally? I've got the following problem, concerning control theory: Imagine a ideal amplifier with a feedback loop:

Let firstly A be not dependent on frequency. If A is linear we can write the following: $v_o = (v_1-v_2) \cdot A$ where $v_2=v_o \cdot F$
hence $v_o= (v_1-F \cdot v_o)A$
since A is linear we obtain:
$v_o=v_1 \cdot A- v_o \cdot A \cdot F$
hence
$v_o-AF v_o=v_1 \cdot A \rightarrow v_o=v_1\frac{ A}{1+A \cdot F}$
But now let A a polynomial, say of degree 3 and no offset: 
$A(x)= a_1 x+a_2 x^2+ a_3 x^3$. Then the equation above does not hold anymore. 
But we can assume we can invert the polynomial locally around zero, then the equation becomes (after some karate):
\begin{equation}
v_o= A(v_1-F \cdot v_o)\\
A^{-1}(v_o)= v_1 -F \cdot v_o \\ 
\text{assuming A is invertible locally} \\
F \cdot v_o + A^{-1}(v_o) = v_1\\
(F \cdot I + A^{-1})(v_o) = v_1 \\ \text{(is that correct?)}\\
v_o = (F \cdot I + A^{-1})^{-1}(v_1)\\
\end{equation}
Lastly we assume that the input signal $v_i$ is of the form $sin(\omega x)$ and we want to know how much the output sine is distorted (means the percentage of higher harmonics like $sin(2 \omega x)$ etc.) 
If we had found the operator $B = (F \cdot I + A^{-1})^{-1}$ we could run the input signal through it and do an FFT to find out about the distortion. 
We assume the coefficients $a_1 ... a_3 $to be reasonably arbitrary (so I can play around with them if possible), as is F. 
How do I find the operator $B$?
Is this somehow solvable by functional-analysis (since it is a non linear operator we cannot use the Neumann-series, or can we?)
 A: You don't need linearity to write down Neumann series. Neumann series is just a fancy name for the geometric series: as long as the common ratio is less than 1 (in some norm), then it does converge (in that same norm).
Namely, if $\ ||AF||_*<1$ (for some subadditive norm $\|\cdot\|_*$), then
$$
\left\|(I-AF)^{-1} - \sum_{k=0}^n (AF)^k\right\|_* \to 0 \mbox{ as } n\to\infty.
$$
This is because $\|(AF)^k\|\leq\|AF\|^k\to 0$ as $k\to \infty$, and
$$
\sum_{k=0}^n (AF)^k = (I - AF)^{-1} (I - (AF)^{n+1}).
$$
A: I found the answer to my specific problem: the keywords are Taylor series and Implicit Function Theorem. Imagine the x the input, y the output, f a non-linear function:
\begin{equation}
y= f(x-b \cdot y)\\
\end{equation}
(this is as above, just renamed). Then we can do the following: \begin{equation}
0= f(x-b \cdot y)-y\\
F(x,y(x))=0
\end{equation}
Means we have an implicit function. The implicit function theorem allows us under some assumptions that we can make to find the derivative of $y$ in a given point, we choose (x,y)=(0,0). So we have:
\begin{equation}
y^\prime = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}
\end{equation} 
As $f(x)$ we have $a_1x+a_2x^2+a_3x^3$ so substituting $x$ by $x-by$ and differentiating we obtain (after setting x to zero in the counter and y to zero in denominator!):
\begin{equation}
 -\frac{a_1-2a_2by+3b^2y^2a_3}{-a_1b-2a_2bx-3x^2b-1}
\end{equation} 
Furthermore we set the remaining x's and y's to zero (which we could have done already) and obtain:
\begin{equation}
 \frac{a_1}{a_1b+1}
\end{equation} 
for $y'$ in the point (0,0). Further differentiating gives higher derivatives of y and we can use them as coefficients of a Taylor polynomial. This is somehow tiresome but will lead to exactly what I wanted. Thanks anyway.
