define an operator

Question

How is it possible to define the operator $(x+\frac{d}{dx})^n$ as a function of $n$? I use

 op[x_] = (x + D[#, x]) &;

with the action on, for example, $\cos(x)$

 op[x][Cos[x]]

for $n=1$. How is it possible to extend the definition for an arbitrary $n$?

Answer 1

I believe that you want to apply the following operator $n$ times:

 op[x_] = (x*# + D[#, x]) &

Otherwise you add the $x$ rather than apply it as an operator.

I do these in a recursive function way:

 Clear[op]
 op[x_, n_] := op[x, n] = (x*# + D[#, x]) &[op[x, n - 1]];
 op[x, 0] = Cos[x];

Answer 2

Well, apparently using the Nest command you can do it even more clearly as I have seen in "another post". Here is the more simple way:

  op[x_,n_,input_]:=Expand@Nest[(x*# + D[#, x]) &,input,n]

I have added the Expand to simplify the final expression. You can just replace the input with the desired function when calling the function or I believe with your syntax you can use:

  op[x,3,#]&[Cos[x]]

I have added this because with recursive functions you have to remember to clear the definition.

Answer 3

What about

op[f_, i_] := Nest[x # + D[#, x] &, f[x], i]

Answer 4

What about

op[x_, n_] = (x + D[#, {x, n}]) &;

??

For example,

op[x, 3][x^3]

(*  6 + x  *)

Have fun!