Simplify recurrence $\frac{d}{dx} f_{n-1}(x)= f_n(x)- f_{n-1}(x) f_1(x)$

Question

Suppose we have a sequence of infinitely differentiable functions $ \{ f_k(x) \}$. Now suppose that these functions satisfy the following recursion: \begin{align} \frac{d}{dx} f_{n-1}(x)= f_n(x)- f_{n-1}(x) f_1(x) \end{align}

where $f_0(x)=1$ for all $x$. Can we re-write $f_n(x)$ only in terms of $f_1(x)$ and derivatives of $f_1(x)$?

It is not difficult to see that this is possible. The difficulty is to create the exact formula.

Answer 1

I don't know if this is what you want. It is easy to see $$f_n(x) = f_{n-1}^\prime(x) + f_{n-1}(x)f_1(x). \tag{1}$$ Multiplying both sides of (1) by $e^{\int f_1(x)dx}$ gives $$f_n(x)e^{\int f_1(x)dx} = \bigg[f_{n-1}(x) e^{\int f_1(x)dx}\bigg]'. \tag{2}$$ Let $$ g_n(x)=f_n(x)e^{\int f_1(x)dx}. $$ Then (2) gives $$ g_n(x)=g_{n-1}'(x) $$ Using this recursion $(n-1)$-times, one has $$ g_n(x)=g_1^{(n-1)}(x)=\frac{d^{n-1}}{dx^{n-1}}\bigg[f_1(x)e^{\int f_1(x)dx}\bigg] $$ and hence $$ f_n(x)=e^{-\int f_1(x)dx}\frac{d^{n-1}}{dx^{n-1}}\bigg[f_1(x)e^{\int f_1(x)dx}\bigg] $$

Answer 2

$$\frac{d}{dx}f_{n-1}(x)=f_n(x)-f_{n-1}(x)f_1(x),\ \ \ \ f_0(x)=1$$

$$f_n(x)=\frac{d}{dx}f_{n-1}(x)+f_{n-1}(x)f_1(x)$$ $\ $

$$f_1(x)=\frac{d}{dx}f_0(x)+f_0(x)f_1(x)=f_1(x)$$

$$f_2(x)=\frac{d}{dx}f_1(x)+f_1(x)f_1(x)$$

$$f_3(x)=\frac{d}{dx}f_2(x)+f_2(x)f_1(x)$$

$$...$$ $\ $

$f_n(x)$ is the following special complete exponential Bell polynomial:

$$f_n(x)=B_n(f_1^{(0)}(x),f_1^{(1)}(x),...,f_1^{(n-1)}(x))=\sum_{\sum_{t=1}^{n}tk_t=n}\frac{n!}{\prod_{i=1}^{n}i!^{k_{i}}k_{i}!}\prod_{i=1}^{n}{f_1^{(i-1)}(x)}^{k_{i}}$$

The vectors of running indices $(k_1,...,k_n)$ in the sum expression represent the integer partitions of $n$.

Applying Faà di Bruno's Formula (Higher chain rule), we get:

$$f_n(x)=e^{-\int f_1(x)dx}\frac{d^n}{dx^n}e^{\int f_1(x)dx}=e^{-\int f_1(x)dx}\frac{d^{n-1}}{dx^{n-1}}\left(f_1(x)e^{\int f_1(x)dx}\right)$$