Evaluating the recurrence $f_k(x)= f'_{k-1}(x)+f_{k-1}(x) f_1(x)$ with $f_0(x)=1$, $f_1(x)=e^x$ Is it possible to recover $f_n(x)$ for any $n$ based off this question? The relation is:
$ \frac{d^n}{dx^n} f_1(x)$ from $\frac{d}{dx} f_{k-1}(x)=f_k(x)-f_{k-1}(x) f_1(x)$
$f_0(x)=1$ and I add $f_1(x)=e^x$
Using this you can find $f_n(x)$ for all $n$ when $n$ is a whole number.

*

*$f_0(x)=1,f_1(x)=e^x$

*$f_2(x)=e^x+e^{2x}$

*$f_3(x)=e^x+3e^{2x}+e^{3x}$

*$f_4(x)=e^x+7e^{2x}+6e^{3x}+e^{4x}$, etc.

My question is can you generalize $f_n(x)$ get $n$ to be any number. If so how would you solve $f_{1/2} (x)$ or $f_{i}(x)$
 A: It looks like
$$
f_n(x) = \sum_{k=0}^{n} \left\{{n \atop k}\right\} e^{kx},
$$where $\left\{{n\atop k}\right\}$ are the Stirling numbers of the second kind. You can show these satisfy the usual recurrence using basic differentiation rules.
As far as generalizing, i.e. determining $f_{1/2}(x)$, this can be done several ways, for instance in this paper. Hope that helps!
A: More explicitly,
$f_{k-1}'(x)
=f_k(x)-f_{k-1}(x) f_1(x)
=f_k(x)-f_{k-1}(x)e^x
$
so
$f_k(x)
=f_{k-1}'(x)+f_{k-1}(x)e^x
$
or,
shifting the index and
removing $(x)$ for conciseness,
$f_{k+1}
=f_{k}'+e^xf_{k}
$.
If
$f_k(x)
=\sum_{j=1}^k c_{k, j}e^{jx}
$,
with
$c(1, 1) = 1$,
then
$f_k'(x)
=\sum_{j=1}^k jc_{k, j}e^{jx}
$
so
$\begin{array}\\
f_{k+1}
&=f_{k}'+e^xf_{k}\\
&=\sum_{j=1}^k c_{k, j}(e^{jx})'+e^x\sum_{j=1}^k c_{k, j}e^{jx}\\
&=\sum_{j=1}^k jc_{k, j}e^{jx}+\sum_{j=1}^k c_{k, j}e^{(j+1)x}\\
&=\sum_{j=1}^k jc_{k, j}e^{jx}+\sum_{j=2}^{k+1} c_{k, j-1}e^{jx}\\
&=c(k, 1)e^x+\sum_{j=1}^k (jc_{k, j}+c(k, j-1))e^{jx}+c_{k, k}e^{(k+1)x}\\
&=\sum_{j=1}^{k+1} c_{k+1, j}e^{jx}\\
\end{array}
$
so
$c(k+1, 1) 
= c(k, 1)
= 1
$,
$c(k+1, k+1) 
= c(k, k)
=1 $,
and,
for $j = 2$ to $k$,
$c(k+1, j)
=(jc_{k, j}+c(k, j-1))
$
and these turn out to be
the Stirling numbers
of the second kind
as Integrand wrote.
