Soft question: I was curious as to how one could measure the degree with which a polynomial is perturbed. More formally, let $P(x) \in \mathbb{C}$ be a polynomial and $\epsilon$ be a very small number, how could one effectively measure the 'difference' between $P(x+\epsilon)$ and $P(x)$? I guess I'm looking for some definitions of such 'difference', and for that to be 'theoretically fruitful' (meaning something that can be manipulated easily to yield expressions and different forms). What first came to mind was using the Taylor series of $P(x+\epsilon)$ to get the following$$P(x+\epsilon) = P(x)+\epsilon P'(x)+\frac{\epsilon^2}{2}P''(x)+...$$Let $\delta(P)$ denote the degree with which $P(x)$ perturbs. A suitable start would be perhaps the simplest one$$P(x+\epsilon) - P(x) \approx \epsilon P'(x)$$however, this is unsatisfactory as it is an approximation and doesn't exactly answer the question as it is a polynomial, so therefore something like$$\delta(P) = \int_a^b\mid P(x+\epsilon) - P(x) \mid dx$$is better. To evaluate this we would need to split the integral up into two parts: one that considers the negative values, and the other the positive. I will show my working for the positive values just to show what I have so far. For $x \geq 0$,$$\delta(P) = \int_a^b P(x+\epsilon) - P(x) dx$$letting $P(x) = c_0+c_1x+...+c_nx^n$ we get$$ = \int_a^b \left(\sum_{k=0}^n c_k (x+\epsilon)^k - c_k x^k\right) dx$$arriving subsequently at the final$$\delta(P) = \sum_{k=0}^n c_k \left(\frac{(b+\epsilon)^{k+1}-(a+\epsilon)^{k+1}+a^{k+1}-b^{k+1}}{k+1}\right)$$whilst this measures some notion of the degree with which a polynomial differs due to perturbation, it doesn't seem very theoretically fruitful (as described above as being something easy to manipulate to yield expressions, and different forms), and therefore I was wondering if anyone can demonstrate something fitting the requirements. It can be a known metric, or even something you've invented yourself. Note: it is intended as a soft question and will be tagged as one. Thank you
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