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I was considering the Operator $$ x\,\frac{\rm d}{{\rm d}x} $$ and applying it $n$ times to an arbitrary function $f(x)$. Is there a general formula for it?

I started with the first few \begin{align} \left(x\,\frac{\rm d}{{\rm d}x}\right)^{1} f(x) &= x f' \\ \left(x\,\frac{\rm d}{{\rm d}x}\right)^{2} f(x) &= x f' + x^2 f'' \\ \left(x\,\frac{\rm d}{{\rm d}x}\right)^{3} f(x) &= x f' + 3x^2 f'' + x^3 f''' \\ \left(x\,\frac{\rm d}{{\rm d}x}\right)^{4} f(x) &= x f' + 7 x^2 f'' + 6x^3 f''' + x^4 f'''' \\ \vdots \end{align} but I don't see any pattern yet. Obviously the first and last coefficients are always $1$.

I figured if I start with any $n$ of the form $$ \left(x\,\frac{\rm d}{{\rm d}x}\right)^{n} f(x) = \sum_{k=1}^n a_k \, x^k f^{(k)}(x) \tag{1} $$ where $a_1=a_n=1$, then recursively \begin{align} \left(x\,\frac{\rm d}{{\rm d}x}\right)^{n+1} f(x) &= xf^{(1)}(x) + \sum_{k=2}^n \left(k \, a_k + a_{k-1}\right) x^k f^{(k)}(x) + x^{n+1} f^{(n+1)}(x) \\ &= \sum_{k=1}^{n+1} \left(k \, a_k + a_{k-1}\right) x^k f^{(k)}(x) \tag{2} \end{align} where $a_0=a_{n+1}=0$.

From (1) and (2) we get for fixed $k$ the coupled recurrence $$ a_k(n+1) = k\, a_k(n) + a_{k-1}(n) $$ which can be put in matrix form \begin{align} \begin{pmatrix} a_1(n+1) \\ a_2(n+1) \\ a_3(n+1) \\ \vdots \\ a_{n-1}(n+1) \\ a_n(n+1) \end{pmatrix} &= \begin{pmatrix} 1 & 0 & 0 & \dots & 0 & 0 \\ 1 & 2 & 0 & \dots & 0 & 0 \\ 0 & 1 & 3 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \dots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & n-1 & 0 \\ 0 & 0 & 0 & \dots & 1 & n \end{pmatrix} \begin{pmatrix} a_1(n) \\ a_2(n) \\ a_3(n) \\ \vdots \\ a_{n-1}(n) \\ a_n(n) \end{pmatrix} \\ &=\begin{pmatrix} 1 & 0 & 0 & \dots & 0 & 0 \\ 1 & 2 & 0 & \dots & 0 & 0 \\ 0 & 1 & 3 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \dots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & n-1 & 0 \\ 0 & 0 & 0 & \dots & 1 & n \end{pmatrix}^n \begin{pmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \\ 0 \end{pmatrix} \, . \end{align} So either I'm too stupid or there is no obvious one.

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  • $\begingroup$ This might be relevant to your interests: Weyl Algebra. I once had to do something like this, and I was just as confused as you are. However, in my problem I only needed to know information about the first and last terms. You can make inductive arguments about those, I think. $\endgroup$ – Prototank Sep 4 '18 at 12:57
  • $\begingroup$ Well the first and last terms are always one, so there is not much to do or? $\endgroup$ – Diger Sep 4 '18 at 13:01
  • $\begingroup$ I guess my question is: what do you need the general form for? It is fine if you are just curious, I was assuming you were trying to show something else and that this is a piece of the puzzle. $\endgroup$ – Prototank Sep 4 '18 at 13:01
  • $\begingroup$ @Diger, I don't think you're stupid. However you transformed your problem into a boundary problem, if you can solve that you can find a closed form solution. $\endgroup$ – user8469759 Sep 4 '18 at 13:02
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    $\begingroup$ I want to know the growth of the coefficients in order to estimate a coefficient of a series. I need radius of convergence. $\endgroup$ – Diger Sep 4 '18 at 13:03
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The coefficients are the Stirling numbers of the second kind.

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  • $\begingroup$ Must have been solved before ;) Thanks. How did you know? $\endgroup$ – Diger Sep 4 '18 at 13:15
  • $\begingroup$ I recognized 1, 7, 6, 1. :-) $\endgroup$ – Hans Lundmark Sep 4 '18 at 13:16
  • $\begingroup$ You could also compute a bit further and enter the coefficients of a later row into the OEIS. $\endgroup$ – Hans Lundmark Sep 4 '18 at 13:19
  • $\begingroup$ Because you look at these numbers everyday? I would not even remember one of them... $\endgroup$ – Diger Sep 4 '18 at 13:19
  • $\begingroup$ Thanks. Didn't know of this side. Very helpful. $\endgroup$ – Diger Sep 4 '18 at 13:21

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