Smoothness of the radius of convergence

Question

Let $(x\mapsto a_n(x))_n$ be a sequence of smooth functions defined on some fixed interval $I$. Consider the power series $\sum_{n\geq 0}a_n(x)t^n$ and denote by $R(x)$ its radius of convergence. Does there exist references in litterature dealing with continuity and smoothness of $x\mapsto R(x)$ ? Examples of questions I want to ask are as follows :

• Assuming $0<R(x)<\infty$ for every $x\in I$, under what conditions is $x\mapsto R(x)$ continuous ? smooth ? analytic ?
• Is there a formula for the derivative $\frac{d}{dx}R(x)$ ?
• If not, would there be a simple way to prove that this derivative is non-zero ?

EDIT taking into account comments and answers below.

Of course I'm aware of Cauchy-Hadamard formula $1/R(x)=\limsup a_n(x)^{1/n}$. This was actually one of the starting points of my question, so I think I should clarify a bit.

In very specific cases, these kind of limits can be smooth functions. For instance, consider a positive matrix $A(x)$ and let $a_n(x)$ be its norm. Then, the limit $1/R(x)=\lim a_n(x)^{1/n}$ exists and is the spectral radius of $A$. Since $A$ is positive, the spectral radius also is the Perron-Frobenius dominant eigenvalue of $A$, so it is an analytic function of $A$. Assuming that $x\mapsto A(x)$ is analytic, you get that $R(x)$ is an anaylitic function of $x$. Also, even if you don't get formulae for the derivatives of $R$, in more specific situations, for example when this Perron-Frobenius eigenvalue is a strictly convex function (which is actually studied in litterature), then you know exactly when the derivative of $R(x)$ is zero.

This is why I added the tag reference request. I agree with the comment of Anthony Quas and I'm not expecting some very general result. I'm asking if these questions are dealt with in some particular situations in litterature.

Answer 1

You have the explicit Hadamard formula $$ \frac{1}{R(x)}=\limsup_n\vert a_n(x)\vert^{1/n}=\inf_n\bigl(\sup_{k\ge n}\vert a_k(x)\vert^{1/k}\bigr), $$ triggering semi-continuity properties for $1/R$: setting $b_n(x)=\sup_{k\ge n}\vert a_k(x)\vert^{1/k}$ and assuming that $a_k$ is continuous of $x$, you get that $b_n$ is lower semi-continuous. Then what you could say is that $1/R$ is the infimum of a sequence of lower semi-continuous functions.