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!

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    $\begingroup$ 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. $\endgroup$ – bonsoon Feb 8 at 0:36
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    $\begingroup$ @bonsoon That looks like an answer. Please make it into one? $\endgroup$ – 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}

  • $\begingroup$ So, in other words, $C_n(k)={n\brace k}$? $\endgroup$ – clathratus Feb 8 at 1:13
  • $\begingroup$ @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 ${n \brace k}$. $\endgroup$ – 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

$$ x^n = \sum_{k=0}^{n} \left\{ {n \atop k} \right\} (x)_k, $$


  • $\left\{{n \atop k}\right\}$ 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.

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    $\begingroup$ This is really, really cool. Thank you $\endgroup$ – clathratus Feb 8 at 0:55
  • $\begingroup$ @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 :) $\endgroup$ – Sangchul Lee Feb 8 at 0:58
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    $\begingroup$ Thanks for filling the details :) $\endgroup$ – bonsoon Feb 8 at 0:59
  • $\begingroup$ 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. $\endgroup$ – clathratus Feb 8 at 1:10
  • $\begingroup$ @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. $\endgroup$ – Sangchul Lee Feb 8 at 1:17

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