# Finding result of composing operations many times

Consider the operator given by,

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

with,

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

or,

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

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?

Any help would be appreciated :D

• 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 $P^2$? Aug 25 '20 at 12:00
• It is inspired by heaviside's operator method of solving differential equations math24.net/differential-operators/…. Aug 25 '20 at 12:05
• @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. Aug 25 '20 at 12:06
• @PeterForeman ok, I obviously haven't encountered it yet, apologies for my ignorance :) Aug 25 '20 at 12:07
• Related: Exponential of a function times derivative. Especially: $$\left[1 + x\frac{d}{dx} + \frac{1}{2!}\left(x\frac{d}{dx}\right)^2 + \frac{1}{3!}\left(x\frac{d}{dx}\right)^3 + \cdots\right] f(x) = e^{x\frac{d}{dx}} f(x) = f(e\,x)$$ Sep 27 at 11:08

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 $$a_{n,k}=S(n,k).$$
• I think you mean $x^k$ Aug 25 '20 at 12:46