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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 $(*)$?

J. Chapman
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    Not sure what you are asking. Newton's method, starting from the known root to the nearby equation, should work here. – lulu Oct 18 '22 at 13:53
  • I was ultimately hoping to find a closed form approximation for the solution of the second polynomial – J. Chapman Oct 18 '22 at 14:05
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    I expect that numerical schemes are the best you'll get. Happily, those tend to be very efficient. – lulu Oct 18 '22 at 14:11
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    You say "the solution" where I'm guessing you mean either a linear factor of the polynomial in $x$ or else the set of roots in $x$. One difficulty is that a product of roots having more than one or two members typically has no simple mechanisms for absorbing a constant offset. – abiessu Oct 18 '22 at 14:11
  • @lulu numerical schemes are closed form approximations. They are sometimes exact in the limit, but simply truncating them at an arbitrary depth is closed-form and also an approximation. NN2's answer illustrates this. Also "two iterations of Newton's method" is a closed form approximation. – Him Oct 18 '22 at 22:07

2 Answers2

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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 $(\bar{x}+\epsilon)$.

NN2
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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 $$\Delta x=-\frac {\Delta c_0}{\sum_ {k=1}^n k\,c_k\, x^{k-1} }$$

Claude Leibovici
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