Upper bound on higher order derivatives of $\frac{1}{v(t)}$

Question

Suppose that $ v(t) >l>0$ and $$ \vert v^{(k)}(t) \vert \leq c \frac{k!}{r^k}. $$ Can we give an upper bound for $$ (\frac{1}{v(t)})^{(k)} $$ ?

Attempt: We first compute the first fourth order derivatives : $$ (v^{-1})'= (-1)v^{-2} v' \quad \quad \quad (v^{-1})''= (-1)(-2)v^{-3} (v')^2 + (-1)v^{-2} v''\\ (v^{-1})''' = (-1)(-2)(-3) v^{-4} (v')^3 + (-1)(-2)v^{-3} 3v' v'' + (-1)v^{-2} v''' \\ (v^{-1})^{(4)} = (-1)(-2)(-3)(-4) v^{-5} (v')^4 + (-1)(-2)(-3)v^{-4} 6 (v')^2 v'' \\ + (-1)(-2) v^{-3} ( 3 (v'')^2 + 4 v' v''') + (-1)v^{-2} v^{(4)} $$ With observation, we find that $$ (v^{-1})^{(n)} = \sum^n_{k=1} (-1)^k k! v^{-k-1} (a_{\mathbf{n}_{k,1}} v^{(\mathbf{n}_{k,1})} + a_{\mathbf{n}_{k,2}} v^{(\mathbf{n}_{k,2})} + \cdots + a_{\mathbf{n}_{k,A(n,k)}} v^{(\mathbf{n}_{k,A(n,k)})} ) $$ where $\mathbf{n}_{k,i}$,$i = 1, 2 \cdots, A(n,k)$ denotes the $k$ component vector $$ (n^{k,i}_1, n^{k,i}_2, \cdots, n^{k,i}_k), $$ which satisfies $ n^{k,i}_1 + n^{k,i}_2 + \cdots + n^{k,i}_k = n $. $$ v^{(\mathbf{n}_{k,i})} = v^{(n^{k,i}_1)} v^{(n^{k,i}_2)}\cdots v^{(n^{k,i}_k)} $$ $A(n,k)$ denotes the number of positive integer solutions of equation $$ x_1 + x_2 + \cdots + x_k = n $$ here notice that, for two solutions $(x_1,x_2,\cdots,x_n)$ $(y_1,y_2,\cdots, y_n)$ only with difference in order, we see them as the same solution and count only once, for example, $(1,1,2)$, $(1,2,1)$.

Now how can we represent the coefficients $$ a_{\mathbf{n}_{k,i}}, \quad k=1,\cdots,n, \quad i = 1, 2,\cdots, A(n,k) $$ and number $A(n,k)$?

(This is a transfer of the question https://math.stackexchange.com/questions/4998405/upper-bound-on-higher-order-derivatives-of-frac1vt)

Answer 1

To simplify writing and without loss of generality, consider the derivatives of $1/v$ at $t=0$. Your condition on the derivatives of $v$ implies that $v$ can be analytically extended to the open disk of radius $r$ centered at $0$. Moreover, $$|v(z)-v(0)|\le\sum_{k=1}^\infty\frac{|v^{(k)}(0)|}{k!}\,|z|^k \le c\sum_{k=1}^\infty\Big(\frac{|z|}r\Big)^k=c\frac{|z|}{r-|z|}\le l/2$$ and hence $$|v(z)|\ge l/2$$ for complex $z$ such that
$$|z|\le s:=\frac l{2c+l}\,r. \tag{1}\label{1}$$

So, for $u:=1/v$ we have $|u(z)|\le 2/l$ given \eqref{1}. So, by the Cauchy integral formula, $$|u^{(k)}(0)|\le \frac2l\frac{k!}{s^k} $$ for $k=0,1,\dots$.