Is there a way to use the solution of a polynomial to approximate the solution to a similar polynomial? Suppose I know the solution to the polynomial
$$
c_0 + c_1 x + c_2 x^2 + ... + c_n x^n = 0 \hspace{15mm} (*)
$$
If I wanted to approximate the solution to the polynomial
$$
(c_0 - \delta) + c_1 x + c_2 x^2 + ... + c_n x^n = 0
$$
for some value of delta that was small relative to $c_0$, is there a way to find the approximate solution by using the known solution to polynomial $(*)$?
 A: Consider the implicit function $$f(x)=c_0 +\sum_ {k=1}^n c_k\,  x^k  = 0$$
$$\frac{\partial f(x)}{\partial c_0}=1$$
$$\frac{\partial f(x)}{\partial x}=\sum_ {k=1}^n k\,c_k\, x^{k-1}$$
$$\frac {dx}{dc_0}=-\frac{\frac{\partial f(x)}{\partial c_0} } {\frac{\partial f(x)}{\partial x} }=-\frac 1{\sum_ {k=1}^n k\,c_k\, x^{k-1} }$$ Going to $\Delta$'s, for any root
$$\Delta x=-\frac {\Delta c_0}{\sum_ {k=1}^n k\,c_k\, x^{k-1} }$$
A: Let's define the function
$$f(x) = c_1 x + c_2 x^2 + ... + c_n x^n$$
And $\bar{x}$ is a solution of the equation
$$-c_0 = f(x) \tag{1}$$
With $\delta \ll c_0$, we expect that the solution of
$$-c_0 - \delta= f(x)\tag{2}$$
is very closed to $\bar{x}$. Then, we can suppose that the solution of $(2)$ is $(\bar{x}+\epsilon)$ with $\epsilon$ is small ( $\epsilon \sim o(1)$ ).
From $(1),(2)$, we have
$$-\delta= f(\bar{x}+\epsilon)-f(\bar{x}) \tag{3}$$
With the first order Taylor approximation, we have
$$\begin{align}
-\delta &= f(\bar{x}+\epsilon)-f(\bar{x}) \\
 &\approx (\bar{x}+\epsilon) -\bar{x}) f'(\bar{x}) )+ o(\epsilon^2) \\
 &\approx \epsilon f'(\bar{x}) ) \\
\end{align}
$$
So,
$$\epsilon  \approx \frac{-\delta}{f'(\bar{x})} = \frac{-\delta}{\sum_{i=1}^n ic_i\bar{x}^{(i-1)}}$$
and the closed-form solution of $(2)$ is $(\bar{x}+\epsilon)$.
