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 $f_{i}(x)$
Best Answer
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.