I am unsure of how to take the derivative of a recurrence relation of functions. For example consider the following recurrence relation:
\begin{equation}
\left\{ \begin{array}{cl} f_n(x) &= a_n\cdot f_{n-1}^2(x) \\ f_0(x) &= x
\end{array} \right.
\end{equation}
with $0 \leq n\leq L \in \mathbb{N}$ such that $F = f_L$. How do I notate, and thereby compute, the partial derivative of the relation with respect to some $a_m$; that is,
$$\frac{\partial F}{\partial a_m} =\;?$$
Does this result in a recursive definition of partial derivatives (i.e. $D_{a_m} f_n = c_n(x) D_{a_m}f_{n-1}$ where $D_x g$ is the partial derivative of $g$ with respect to $x$)? If so, can I solve this recurrence relation to a closed form of $m$ using the methods of homogeneous recurrence relations?
Lastly, can I construct a recurrence relation for $D_{a_m} F$ which acts in the reverse direction; that is, can I define a sequence such that $D_{a_m} F$ is defined by $D_{a_{m+1}}$, which is defined by $D_{a_{m+2}}, \dots,$ which is defined by $D_{a_{L}}$?
Thanks
Best Answer
Well, in case you have a general recurrence relation of the form $$f_n(x)=h_n(f_{n-1}(x))$$ you can write the partial derivatives as $$\partial_{a_m}f_n(x)=h_{n}'(f_{n-1}(x))\partial_{a_m}f_{n-1}(x)$$ I do not know if this can help but for sufficiently amicable form of $h_n(x)$ we might just be able to compute $h_n'(\cdot)$ to get a solvable recurrence relation for the partial derivatives of $f_n(x)$.