I am investigating the polynomial $$P_n(x)=\left(x\frac{d}{dx}\right)^n f(x)=xP_{n-1}'(x)$$
for some known function $f$. I defined $f_n(x)=\frac{d}{dx}f_{n-1}(x)$ with $P_0=f_0=f$. And I also defined
$$P_n(x)=\sum_{k=1}^{n}C_n(k)x^kf_k(x)$$
And I am interested in finding an explicit form, or at least a recurrence relation for $C_n(k)$. With manual calculation, I was able to find up through $n=6$, but I failed to recognize any pattern, so I thought I'd ask for help. For those interested, a 'table' of values:
$n=1$:
$$C_1(1)=1$$
$n=2$:
$$C_2(1)=1,\quad C_2(2)=1$$
$n=3$:
$$C_3(1)=1,\quad C_3(2)=3,\quad C_3(3)=1$$
$n=4$:
$$C_4(1)=1,\quad C_4(2)=7,\quad C_4(3)=6,\quad C_4(4)=1$$
$n=5$:
$$C_5(1)=1,\quad C_5(2)=15,\quad C_5(3)=25,\quad C_5(4)=10,\quad C_5(5)=1$$
$n=6$:
$$C_6(1)=1,\quad C_6(2)=31,\quad C_6(3)=90,\quad C_6(4)=65,\quad C_6(5)=15,\quad C_6(6)=1$$
The only pattern I can see is $C_n(1)=C_n(n)=1$. Also, it is easily shown that, since $P_n=xP_{n-1}'$,
$$\sum_{k=1}^{n}C_n(k)x^kf_k(x)=\sum_{k=1}^{n-1}C_{n-1}(k)x^k\left[f_k(x)+xf_{k+1}(x)\right]$$
Although I'm not sure that helps. I am very lost, please help. Thanks!
Best Answer
Here is a supplementary to user @bonsoon's comment as to why the Stirling numbers of the second kind pop up. This begins with the identity
$$ x^n = \sum_{k=0}^{n} \left\{ {n \atop k} \right\} (x)_k, $$
where
$\left\{{n \atop k}\right\}$ is the Stirling number of the second kind, which counts the number of ways of partitioning the set $\{1,\cdots,n\}$ into $k$ parts, and
$(x)_k = x(x-1)\cdots(x-k+1)$ is the falling factorial.
(See the Wikipedia article, for instance.) Now if we introduce two operators $D = \frac{d}{dx}$ and $L = x\frac{d}{dx}$, then they satisfy $ L^n(x^a) = a^n x^a $ and $ D^n(x^a) = (a)_n x^{a-n} $, and so,
$$ L^n(x^a) = a^n x^a = \sum_{k=0}^{n} \left\{ {n \atop k} \right\} (a)_k x^a = \sum_{k=0}^{n} \left\{ {n \atop k} \right\} x^k D^k (x^a). $$
Since both $L$ and $D$ are linear, the same identity holds for any polynomial $f(x)$ in place of $x^a$, yielding
$$ L^n f = \sum_{k=0}^{n} \left\{ {n \atop k} \right\} x^k D^k f. $$
Of course, this extends to any $C^n$-function $f$ by polynomial approximation.