Suppose the following recursive equation holds:
\begin{align}
\frac{d}{dx} f_{k-1}(x)=f_k(x)-f_{k-1}(x) f_1(x)
\end{align}
where $f_0=1$.
Question: Can we use this recursion to find
\begin{align}
f^{(n)}_1= \frac{d^n}{dx^n} f_1(x)=\,?
\end{align}
I have used it to find the first four derivatives (hopefully correct), but I couldn't find the pattern.
\begin{align}
f^{(1)}_1&=f_2-f_1^2\\
f^{(2)}_1&=f_3-3f_2f_1+2f_1\\
f^{(3)}_1&=f_4-4f_3f_1+15f_1^2f_2-9f_1^4\\
f^{(4)}_1&=f_5-5f_4f_1+23f_1^2f_3-f_3f_2-81f_1^3f_2+30f_1f_2^2+36f_1^5
\end{align}
This question is related to something that I asked here.
Best Answer
I get:
$$f_1^{(1)}(x)=f_2(x)-f_1^2(x)$$
$$f_1^{(2)}(x)=2f_1^3(x)-3\,f_1(x)f_2(x)+f_3(x)$$
$$f_1^{(3)}(x)=-6f_1^4(x)+12f_1^2(x)f_2(x)-4f_1(x)f_3(x)-3f_2^2(x)+f_4(x)$$
$$f_1^{(4)}(x)=24f_1^5(x)-60f_1^3(x)f_2(x)+20f_1^2(x)f_3(x)+30f_1(x)f_2^2(x)-5f_1(x)f_4(x)-10f_2(x)f_3(x)+f_5(x)$$
$$f_1^{(5)}(x)=-120f_1^6(x)+360f_1^4(x)f_2(x)-120f_1^3(x)f_3(x)-270f_1^2(x)f_2^2(x)+30f_1^2(x)f_4(x)+120f_1(x)f_2(x)f_3(x)+30f_2^3 (x)-6f_1(x)f_5(x)-15f_2(x)f_4(x)-10f_3^2(x)+f_6(x)$$
$$...$$
Could you please correct and edit this in your question?
Clearly we could laborious try to derive a pattern first. But let's look first at The Online Encyclopedia of Integer Sequences (OEIS).
We get the integer sequence OEIS: A263634:
$$f_1^{(n)}(x)=\sum_{k=1}^{n+1}(-1)^{k-1}(k-1)!B_{n+1,k}(f_1(x),...,f_{n-k+2}(x)$$
$\ $
Now you can start proving this result.
To prove the result, you could use e.g. the answers at Simplify recurrence $\frac{d}{dx} f_{n-1}(x)= f_n(x)- f_{n-1}(x) f_1(x)$ :
Prove that the result are the logarithmic polynomials above. See Comtet, L.: Advanced Combinatorics. Reidel, 1974. p. 140: 3.5. Logarithmic and Potential Polynomials