# Freaky Polynomial: $P_n(x)=\left(x\frac{d}{dx}\right)^n f(x)$

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!

• Look up Stirling numbers of the second kind: oeis.org/A008277 This $C_n(k)$ counts the number of ways to partition a set of $n$ distinct things into $k$ nonempty piles. – bonsoon Feb 8 at 0:36
• @bonsoon That looks like an answer. Please make it into one? – Pedro Tamaroff Feb 8 at 0:44

For integers $$n,k$$ with $$n\ge 0$$, let $${n\brace k}$$ denote the Stirling numbers of the second kind, which count the number of partitions of a set of size $$n$$ into $$k$$ nonempty, non-distinct parts. These satisfy the following recurrence, which can be taken as their definition: $${n\brace k}={n-1 \brace k-1}+k{n-1\brace k},\\ {0\brace 0}=1,{0\brace k}=0\text{ for }k\neq 0$$

Now, let $$D$$ be the differential operator, and let $$X$$ be the operator which takes in a function $$f$$ and returns $$xf$$. As a special case of the product rule, we have the operator identity $$DX=XD+1$$ where $$1$$ is the identity, $$1f=f$$. Indeed, applying both sides to some function $$f$$, you get $$D(xf)=x(Df)+f$$. More generally, $$DX^k=X^kD+kX^{k-1}$$ You can now prove by induction that $$(XD)^n=\sum_k {n \brace k}X^kD^k$$ where the sum ranges over all integral $$k$$ (but is effectively finite since $${n \brace k}$$ is zero for $$k$$ outside $$[0,n]$$) as follows:

\begin{align} (XD)^n &=(XD)(XD)^{n-1} \\&=(XD)\sum_k {n-1 \brace k}X^k D^k \\&=\sum_k {n-1\brace k}XDX^kD^k \\&=\sum_k {n-1\brace k}X(X^kD+kX^{k-1})D^k \\&=\sum_k {n-1 \brace k}(X^{k+1}D^{k+1}+kX^kD^k) \\&=\sum_k \Big(k{n-1\brace k}+{n-1\brace k-1}\Big)X^kD^k \\&=\sum_k {n\brace k}X^kD^k \end{align}

• So, in other words, $C_n(k)={n\brace k}$? – clathratus Feb 8 at 1:13
• @clathratus That's right. To relate this back to your question, take the operators on both sides of $(XD)^n=\sum_{k=0}^n {n \brace k}X^kD^k$, and apply them to $f$. Since $D^kf=f_k$, you get that $P_n(x)=\sum_{k=1}^{n}{n \brace k}x^kf_k(x)$, so that the $C_n(k)$ you want are precisely ${n \brace k}$. – Mike Earnest Feb 8 at 1:17

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.

• This is really, really cool. Thank you – clathratus Feb 8 at 0:55
• @clathratus, Glad it helped. This answer still leaves the question as to why the identity at the beginning holds, but the proof should be available from any reasonably well-versed textbook in combinatorics or from googling :) – Sangchul Lee Feb 8 at 0:58
• Thanks for filling the details :) – bonsoon Feb 8 at 0:59
• I'm sorry to do this, but I am un-accepting your answer for the time being, until someone shows me a proof for my identity. – clathratus Feb 8 at 1:10
• @clathratus, User Mike Earnest gave a nice, self-contained answer for the identity $$\left(x\frac{d}{dx}\right)^n = \sum_{k=0}^{n} \left\{ {n \atop k} \right\} x^k \left(\frac{d}{dx}\right)^k$$ which also appears at the end of my answer. Now comparing what we get from this with your equality, it is evident that $C_n(k) = \left\{{n \atop k}\right\}$, so $C_n(k)$ is exactly the Stirling numbers of the second kind. – Sangchul Lee Feb 8 at 1:17