Does there exist a formula to expand

\begin{align} \left[h(p)\frac{d^m}{d p^m}\right]^n\Big(f(p)g(p)\Big)? \end{align}

I know that the Leibniz identity is

\begin{align} \frac{d^n}{d p^n}fg=\sum_{k=0}^n{n\choose k}\frac{d^k f}{dp^k}\frac{d^{n-k} g}{dp^{n-k}}\,, \end{align}

but I am not how to extend it. I would expect that I would have the same sort of form, but instead of ${n \choose k}$, I would have a function of $p$, $a_k(p)$, such that

\begin{align} \left[h(p)\frac{d^m}{d p^m}\right]^n\Big(f(p)g(p)\Big)=\sum_{k=0}^na_k(p)\frac{d^k f}{dp^k}\frac{d^{n-k} g}{dp^{n-k}}\,. \end{align}

Is there a way to evaluate what $a_k(p)$ would be?

  • $\begingroup$ What happened to $m$ ? Did you decide that $m=1$ ? $\endgroup$
    – WW1
    Mar 19, 2017 at 18:44
  • $\begingroup$ I was not sure what to do with $m$. I was thinking that it might be absorbed into $a_k(p)$ somehow. Maybe, instead of $a_k(p)$, it should be $a_k(p)\frac{d^{km}}{d^{km}}$? If that is the case, then $\sum_{n=0}^{\infty}a_k(p)\frac{d^{km}}{d^{km}}\frac{d^kf}{dp^k}\frac{d^{n-k}g}{dp^{n-k}}$ could then be evaluated with the Leibniz identity. $\endgroup$
    – user85503
    Mar 19, 2017 at 18:56


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