The derivative of a recurrence relation of functions

Question

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

Answer 1

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 $f_n(x)$.

Answer 2

$$\log f_n (x) = \log a_n + 2 \log f_{n-1} (x) $$ Let us denote $\log f_n(x)$ by $g_n(x)$, and also $\log a_n$ by $b_n$, for brevity. We have $$ g_n(x)=b_n + 2 g_{n-1}(x) $$ This is readily solved: $$ g_n(x)= 2^n g_0(x) + \sum_{k=1}^{n } 2^{n-k} b_k $$ OR, equivalently, $$ g_n(x)= 2^n \log x + \sum_{k=1}^{n } 2^{n-k} b_k $$

This can be utilized to calculate the derivative $$ \frac{\partial g_n(x)}{\partial b_m}= 2^{n-m} \\ \Longrightarrow \frac{1}{f_n(x)}\frac{\partial f_n(x)}{\partial b_m}= 2^{n-m} \\ \Longrightarrow \frac{a_m}{f_n(x)}\frac{\partial f_n(x)}{\partial a_m}= 2^{n-m} \\ \Longrightarrow \frac{\partial f_n(x)}{\partial a_m}=f_n(x) \frac{2^{n-m}}{a_m} $$ For $n=L$, we arrive at the following closed-form solution $$ f_L(x)= x^{2^L} \prod_{k=1}^L a_k^{2^{L-k}} \\ \Longrightarrow \frac{\partial f_L(x)}{\partial a_m}= \left(x^{2^L} \prod_{k=1}^L a_k^{2^{L-k}} \right)\times \frac{2^{L-m}}{a_m} $$