this question branches off another question that can be seen here
Now we begin be taking a look at the following expressions: $$ \sum_{k=1}^{n-l} \sum_{j-0}^m \frac{\ln(g)^{m-j}}{g^k} \frac{d^j}{df^j}[(f)_k] X = \sum_{p=1}^n \sum_{q=1}^{p} \sum_{k=q}^{n-l} (-1)^{k-q}\frac{q!}{(p-m)!}{m \choose p-q}{k \brack q} f^{p-m} \frac{\ln(g)^{m-j}}{g^k} X $$ Note that ${k \brack q}$ is stirling's number and $(f)_k$ is the falling factorial The identity is valid for $n \geq 2$, $1 \leq l \leq n-1$ and $1 \leq m \leq n$.
Now i will present my work i have done so far: $$ (f)_n = \sum_{k=0}^n (-1)^{n-k}{n\brack k} f^k $$
Therefore
$$ \frac{d^j}{df^j}[(f)_k] = \sum_{v=j}^k (-1)^{k-v}{k \brack v} \frac{v!}{(v-j)!} f^{v-j} $$
Therefore by placing this in the formula presented, (Using $x_{n-l,k}$ for $B_{n-l,k}^g$)
$$ \sum_{k=1}^{n-l} \sum_{j=0}^m {m \choose j} \frac{d^j}{df^j}[(f)_k] \frac{\ln(g)^{m-j}}{g^k} X_{n-l,k} $$
Then turns into
$$ \sum_{k=1}^{n-l} \sum_{j=0}^m \sum_{v=j}^k {m \choose j} (-1)^{k-v}{k \brack v} \frac{v!}{(v-j)!} f^{v-j} \frac{\ln(g)^{m-j}}{g^k} X_{n-l,k} $$
But here i become wary as i do not know what to do about the variables, the exression has been shown to work with small number of $n$'s but i don't know what kind of transformation is needed for this to work. Any help would be greatly appreciated!
