Perturbing pole of Laurent polynomial/series in a single summand

Question

I am working with the ring of Laurent polynomials $\mathbb{F}[X,X^{-1}]$ over $\mathbb{F}$ for some algebraically closed field $\mathbb{F}$ of any characteristic. I encountered a problem emerging from extended generating functions (giving rise to Laurent series) in the context of interacting particle systems, where the poles are not constant. See here for my question on the name of such objects: https://math.stackexchange.com/questions/4847680/name-of-laurent-series-with-countably-many-poles

Now my question on the structure for a simplified case. Given a Lauren polynomial $L\in\mathbb{F}[X,X^{-1}]$ with $L=\sum_{k\in\mathbb{Z}}a_k X^k$, only finitely many $a_k$ non-zero, and some $b\in \mathbb{F}\setminus\{0\}$ as well as $l\in\mathbb{Z}$ with $l<0$, such that $a_l\neq 0$ is there a way to compare $L$ and the new polynomial $$L_b'=\sum_{k\in\mathbb{Z}\setminus\{l\}}a_k X^k + a_k(X+b)^l?$$

Even in the case of $\mathbb{F}=\mathbb{C}$ I am struggeling to say anything about the impact of this perturbation of one summand in terms of norms or scalar products, exploiting the vector space structure of $\mathbb{C}$.

In the general ring case, I thought about comparing the ideal $\langle (X+b)^l \rangle$ to "something else" in $\mathbb{F}[X,X^{-1}]$ interpreting the expression $ (X+b)^l $ as the multiplicative inverse of $(X+b)^{(|l|)}$ which gives by the binomial theorem $$(X+b)^{(|l|)}=\sum_{k=0}^{|l|}\binom{|l|}{k}b^k X^{|l|-k}.$$

Consequently, I concluded that $$\langle (X+b)^l \rangle=\left\langle \left(\sum_{k=0}^{|l|}\binom{|l|}{k}b^k X^{|l|-k}\right)^{-1}\right\rangle$$ but here I get stuck trying to move forward.

---

more explanations included here

Intuitively, I would like to have some notion of "small perturbation, small impact" for which I need also a topology and continuous maps on $\mathbb{F}[X,X^{-1}]$. In fact, for a given topology on $\mathbb{F}$ and one on $\mathbb{F}[X,X^{-1}]$ I consider for $L\in\mathbb{F}[X,X^{-1}]$ the maps $$(L-L_{\cdot}')(X):\mathbb{F}\to \mathbb{F}[X,X^{-1}].$$

Now a comparison of $L$ and $L_b'$ as a function of $b$ means to me:
Is $(L-L_{\cdot}')(X)$ continuous (this depends on the chosen topologies and should reflect in the case $\mathbb{F}=\mathbb{C}$ the notion of complex functions being close)? Without continuity the idea of small perturbations does not seem to be sensible at all.
And on the other hand, is there a way to extend the ring $\mathbb{F}[X,X^{-1}]$ in such a way to a ring $\mathcal{R}$ such that I can say something about $L$ and $L'_b$ being coprime in $\mathcal{R}$ or the equivalence class $[L_b']\in\mathcal{R}/\langle L \rangle$?

All ressources concerning this type of comparison, which I have found, assume that all summands with negative exponent have the same pole (as for example discussed here https://math.stackexchange.com/questions/1116393/prove-that-space-of-laurent-series-is-not-hilbert).

In the end, $b$ will be replaced by a continuous function and in the spirit of the first link, all summands should have "their" designated poles.

Could someone help me in finding or showing, even for $\mathbb{F}=\mathbb{C}$, such a notion of similarity between Laurent Polynomials?

Answer 1

This is too long for a comment, as I'm still not completely sure of the shape of the answer you're looking for. Please correct me if I'm wrong, but from your verbal description the correct formula should be $$ L_b'= a_l(X+b)^l + \sum_{k\in\mathbb{Z}\setminus\{l\}}a_k X^k . $$ Then the difference you are interested in is just $$\begin{aligned} L - L_b' &= a_k \left(X^l - (X+b)^l\right) = a_k X^l \left(1 - (1+b X^{-1})^l\right) \\ &= -a_k \sum_{n=1}^\infty \binom{l}{n} b^n X^{-n} , \end{aligned}$$ using the binomial series, which applies also to negative exponents.

As you can see, the coefficients of the difference are continuous in $b$, non-vanishing only for $n>0$, but the coefficients grow geometrically when $|b|>1$. If you just take the product topology $\mathbb{F}[X,X^{-1}] \cong \mathbb{F}^\mathbb{Z}$, then that's sufficient for the dependence on $b$ to be continuous (no matter how fast the coefficients grow as $n\to \infty$). If you want to require some summability (say with $\mathbb{F} = \mathbb{C}$) of your Laurent series, then you have to be careful with the radius of convergence. If you require summability in a punctured disk around $X=\infty$ of radius $|X| = B > 0$. Then the dependence on $b$ will still be continuous until it hits $|b| = B$.