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In case you're wondering how to get differentials to act like operators in Mathematica, I stumbled across a package Carl Woll made to solve this issue in this question. There's a a more recent version of the package in his github than is linked in his comment. Here's how to write it using the package:

$\left(f\left(x\right)\frac{\partial}{\partial x}\right)^nf\left(x\right)$=Simplify[((Subscript[operator[DifferentialOperator[]], x] f[x])^n)[f[x]]]

What I want is a summation which can give the same result. I tried telling Mathematica to assume n was a positive integer but it didn't simplify any more. Can anyone help me?

Example for n=3:

$f(x)(f'(x)^3+4f(x)f'(x)f''(x)+f(x)^2f'''(x))$

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    $\begingroup$ Based on the comments below Bob's answer, I believe you are asking a Math question rather than a Mathematica question. The answer is likely related to the Faa Di Bruno formula. $\endgroup$
    – QuantumDot
    Commented Sep 17, 2019 at 23:07
  • $\begingroup$ Oh... when I was looking up how to use differentials as operators in Mathematica I was looking in this stackexchange. I guess I absent mindedly decided to use this stackexchange when I asked this question... My bad. Also, thanks for the help, I'll look into it. $\endgroup$
    – Laff70
    Commented Sep 18, 2019 at 4:43

1 Answer 1

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You can use a recursive definition

Clear["Global`*"]

d[0] = f[x];

d[n_Integer?Positive] := d[n] =
  f[x]*D[d[n - 1], x] // Simplify

Column[d /@ Range[0, 4], Dividers -> All]

enter image description here

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  • $\begingroup$ While that is a simpler new way to do it which only requires vanilla Mathematica, it isn't quite what I'm looking for unfortunately. I appreciate it though. EDIT: I should clarify a bit, I'm going to be using this to find the value of these odd derivatives of f[x] at a specific value. So each $f^n(x)$ should simplify down to a single value rather than a full fledged function. The final value would then be inserted into a Taylor series expansion. Having to have my computer find the derivative at each step in the calculation might eat up too much time. I think a sum would work better. $\endgroup$
    – Laff70
    Commented Sep 17, 2019 at 0:18
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    $\begingroup$ @Laff70 Can you give an example what you are expecting. From what I can tell, 4th line of the table reproduces your formula. $\endgroup$
    – yarchik
    Commented Sep 17, 2019 at 0:36
  • $\begingroup$ I want something along the lines of:$$\sum_{a=0}^?c_a\prod_{b=0}^? f^b\left(x_0\right)$$I hope that's helpful enough. $\endgroup$
    – Laff70
    Commented Sep 17, 2019 at 1:11

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