I am trying to find the solution to $$\left(x\cdot\frac{d}{dx}\right)^n\cdot f(x)$$ I first assumed the solution to be of the form $$\left(x\cdot\frac{d}{dx}\right)^n\cdot f(x)=\sum\limits_{i=0}^\infty\sum\limits_{j=0}^\infty a_{n,i,j}\cdot x^i\cdot\frac{d^j}{dx^j}f(x)$$ Multiplying both sides with $x\cdot\frac{d}{dx}$ gives $$\left(x\cdot\frac{d}{dx}\right)^{n+1}\cdot f(x)=\sum\limits_{i=0}^\infty\sum\limits_{j=0}^\infty a_{n,i,j}\cdot\left(i\cdot x^{i-1}\cdot\frac{d^j}{dx^j}f(x)+x^i\cdot\frac{d^{j+1}}{dx^{j+1}}f(x)\right),$$ which gives a recurrence relation for $a_{n,i,j}$: $$a_{n+1,i,j}=a_{n,i+1,j}\cdot(i+1)+a_{n,i,j-1},$$ with initial condition $a_{0,i,j}=\begin{cases}1&i=0\wedge j=0\\0&\mbox{otherwise}\end{cases}$.
Does anybody know how to solve such a 3 dimensional recurrence relation? Any help would be greatly appreciated!
UPDATE: Substituting $a_{n,i,j}=i!\cdot b_{n,i,j}$, simplifies the recurrence relation to: $$b_{n+1,i,j}=b_{n,i+1,j}+b_{n,i,j-1},$$ with initial condition $b_{0,i,j}=\begin{cases}1&i=0\wedge j=0\\0&\mbox{otherwise}\end{cases}$.
Now I tried seperation of variables, substituting $b_{n,i,j}=N(n)\cdot I(i)\cdot J(j)$: $$N(n+1)\cdot I(i)\cdot J(j)=N(n)\cdot I(i+1)\cdot J(j)+N(n)\cdot I(i)\cdot J(j-1)$$ Upon division by $N(n)\cdot I(i)\cdot J(j)$ this gives the following ordinary differential equations: $$\lambda_n=\frac{N(n+1)}{N(n)}=\frac{I(i+1)}{I(i)}+\frac{J(j-1)}{J(j)}=\lambda_i+\lambda_j$$ with solutions $N(n)=\lambda_n^n$, $I(i)=\lambda_i^i$ and $J(j)=\lambda_j^{-j}=(\lambda_n-\lambda_i)^{-j}$, and total solution $$b_{n,i,j}=\lambda_n^n\cdot\lambda_i^i\cdot(\lambda_n-\lambda_i)^{-j}$$ I just don't know what to do with the boundary condition, does anybody know what to do with the boundary conditions?