Is there a closed form solution for $\frac {d^n}{dx^n} \frac {1}{1-ae^x}$?

Question

I am trying to find a closed form solution for $$\frac {d^n}{dx^n} \frac {1}{1-ae^x}$$ I have tried recognizing a pattern by computing specific cases but so far no luck.
I do know that the function is a composition of $x^{-1}$ and $1-ae^x$. The $k$th derivative of the latter is the same for all $k\ge 1$ and there is a closed form of the $k$th derivative of the former. Yet I am still not finding any answer.
However, when computing the derivatives manually I can see that in the numerator there is always a monic polynomial in $ae^x$ with constant term equal to one, all multiplied by $-ae^x$ and the denominator is always raised to the $n+1$ for the $n$th derivative.
I tried arranging the coefficients of the polynomial in the numerator in a triangle like that of Pascal's but I only recognized a pattern for the coefficients of the first and last two terms.
I tried Faa di Bruno's Formula but frankly it is over my head. Is there a clever solution or even a solution at all?

Answer 1

By explicitly computing the first few derivatives it looks like the $n$'th derivative of $\frac{1}{1-ae^x}$ is on the form

$$\color{red}{\frac{d^n}{dx^n}\left(\frac{1}{1-ae^x}\right) = \frac{\sum_{k=1}^n f(n,k) a^ke^{kx}}{(1-ae^x)^{n+1}}}$$

where $f(n,k)$ are constants. This functional form can formally be proven by induction. By taking the derivative of the expression above and equating it to the same formula with $n\to n+1$ we get $$\frac{d}{dx}\left(\frac{\sum_{k=1}^n f(n,k) a^ke^{kx}}{(1-ae^x)^{n+1}}\right) = \frac{\sum_{k=1}^{n+1} f(n+1,k) a^ke^{kx}}{(1-ae^x)^{n+2}} $$

which when expanded gives us a recursion formula for $f(n,k)$:

$$f(n+1,k) = k f(n,k) + (n+2-k) f(n,k-1)~~~\text{for}~~~k=2,3,\ldots,n$$

with $f(n+1,1) = f(n+1,n+1) = 1$. This is the known recursion formula for the (shifted) Eulerian numbers $f(n,k) = A(n,k-1) \equiv \left\langle {n \atop k-1} \right\rangle$.

For more information about the Eulerian numbers see e.g. MathWorld : Euler's number triangle, Wiki : Eulerian Numbers, OEIS : Eulerian Numbers. The first few values are (each row gives $f(n,k)$ for $k=1,2,\ldots,n$):

$$\matrix{1 & n=1\\1~~~1 & n=2\\1~~~4~~~1 & n=3\\1~~~11~~~11~~~1 & n=4\\1~~~26~~~66~~~26~~~1 & n=5\\1~~~57~~~302~~~302~~~57~~~1 & n=6}$$

Answer 2

One can start by applying the identity $\frac{1}{1-y}=1+y+y^2+\ldots$ setting $y=ae^x$. This gives* $$\frac{1}{1-ae^x}=1+ae^x+a^2e^{2x}+a^3e^{3x}+\ldots.$$ This makes differentiating really easy: $$\frac{d^n}{dx^n}\frac{1}{1-ae^x}=ae^x+2^na^2e^{2x}+3^na^3e^{3x}+\ldots.$$ Now, it's a reasonably standard fact about generating functions that, for any polynomial $P$ of degree $n$, there is another one $Q$ of degree no more than $n$ such that $$\frac{Q(y)}{(1-y)^{n+1}}=P(0)+P(1)y+P(2)y^2+P(3)y^3+\ldots$$ and we see that we have basically this situation here, where $P(m)=m^n$. How do we find $Q(y)$? Well, we just multiply both sides of the expression through by $(1-y)^{n+1}=\sum_{i=0}^{n+1}(-1)^i{n+1\choose i}y^i$. Then, we collect terms, and find that if $Q(y)=\sum_{i=0}^{n}a_iy^i$ we have $$a_i=\sum_{j=0}^i(-1)^j{n+1\choose j}P(i-j)=\sum_{j=0}^i(-1)^j{n+1\choose j}(i-j)^n.$$ As is noted in @Winther's answer, these coefficients are Eulerian numbers. That is, $a_i=A(n,i-1)$ in the terminology given on the linked page. Wikipedia notes under the identities section that they are the generating function for the sequence of $n^{th}$ powers, which is exactly why they show up here. If we take this polynomial $Q$, we get that $$\frac{d^n}{dx^n}\frac{1}{1-ae^{x}}=\frac{Q(ae^x)}{(1-ae^x)^{n+1}}$$

(*Warning: Everything works out okay in the general case, but the logic here only technically applies for $|ae^x|<1$. More or less, we can figure that, since the answer is at least correct on some open set and everything behaves well everywhere (e.g. is holomorphic), things will work out. Some more deep explanation may exist)