Consider the operator given by,

$$ P = x \frac{d}{dx}$$


$$ P^2 = x \frac{d}{dx} ( x\frac{d}{dx}) = x \frac{d}{dx} + x^2 \frac{d^2}{dx^2}$$


$$ P^2 = x \frac{d}{dx} ( x\frac{d}{dx}) = x \frac{d}{dx} + x^2 \frac{d^2}{dx^2}$$

and on another application of the operator,

$$ P^3 = [x \frac{d}{dx}] P^2 = x\frac{d}{dx}(x \frac{d}{dx}) + x \frac{d}{dx}( x^2 \frac{d^2}{dx^2}) = x \frac{d}{dx} + 3x^2 \frac{d^2}{dx^2} + x^3 \frac{d^3}{dx^3} $$

I tried writing more iterations but I can't find / a general form of what $P^k$ should be. So my question is if you are given an operator as the one shown, is there any standard procedure to find the a formula for the kth iteration of the operator?

The actual reason why I want to know about this

Any help would be appreciated :D

  • $\begingroup$ What exactly do you mean by $x\frac{d}{dx}$? What is $\frac{d}{dx}$ the derivative of? Do you mean sometthing like $x\frac{dy}{dx}$? Also, I'm not quite sure what you mean by $P^2$ etc. Do you mean differentiating $P$ with respect to $x$ and you're simply writing it as $P^2$? $\endgroup$ Commented Aug 25, 2020 at 12:00
  • 1
    $\begingroup$ It is inspired by heaviside's operator method of solving differential equations math24.net/differential-operators/…. $\endgroup$ Commented Aug 25, 2020 at 12:05
  • 1
    $\begingroup$ @A-levelStudent It seems clear to me that $P=xD$ is the variable $x$ right-multiplied by the associated differential operator. This is a well-defined differential operator. $\endgroup$ Commented Aug 25, 2020 at 12:06
  • $\begingroup$ @PeterForeman ok, I obviously haven't encountered it yet, apologies for my ignorance :) $\endgroup$ Commented Aug 25, 2020 at 12:07
  • 1
    $\begingroup$ Related: Exponential of a function times derivative. Especially: $$ \left[1 + x\frac{d}{dx} + \frac{1}{2!}\left(x\frac{d}{dx}\right)^2 + \frac{1}{3!}\left(x\frac{d}{dx}\right)^3 + \cdots\right] f(x) = e^{x\frac{d}{dx}} f(x) = f(e\,x) $$ $\endgroup$ Commented Sep 27, 2021 at 11:08

1 Answer 1


The coefficients are Stirling numbers of the second kind.

We can write $$P^n=\sum_{k=1}^n a_{n,k}x^k D^k$$ where $D=\frac d{dx}$. Then $$P^{n+1}=\sum_{k=1}^n a_{n,k}(xD)(x^k D^k) =\sum_{k=1}^n a_{n,k}(kx^k D^k+ x^{k+1} D^{k+1})$$ so that $$a_{n+1,1}=a_{n,1},$$ $$a_{n+1,n+1}=a_{n,n}$$ and $$a_{n+1,k}=a_{n,k-1}+ka_{n,k}$$ These recurrences are the same as for the Stirling numbers, so $$a_{n,k}=S(n,k).$$

  • 1
    $\begingroup$ I think you mean $x^k$ $\endgroup$ Commented Aug 25, 2020 at 12:46
  • 1
    $\begingroup$ (+1) Very interesting that these numbers appear here! $\endgroup$ Commented Aug 25, 2020 at 12:46
  • $\begingroup$ That is very interesting indeed, but what does it mean? Like, what does it mean that the co-efficents are stirling numbers, is there some deeper meaning or is this an abstract result $\endgroup$ Commented Aug 26, 2020 at 8:20
  • $\begingroup$ How did you know that such a series existed for the jth composition of that operator $\endgroup$ Commented Nov 23, 2020 at 16:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .