Higher order derivatives of the binomial factor

Question

Let $p$,$l$ be positive integers and $\theta$ be a parameter. The question is to compute the following quantity: \begin{equation} \kappa^{(p)}_l := \left. \frac{\partial^p}{\partial \theta^p} \binom{\theta}{l} \right|_{\theta=0} \end{equation} With the help of Mathematica we found the result for consecutive values of p. We have: \begin{equation} \kappa^{(p)}_l = \left\{ \begin{array}{rr} \delta_{l,0} & \quad \mbox{if $p=0$} \\ \frac{(-1)^{l+1}}{l} & \quad \mbox{if $p=1$} \\ 2 (-1)^l \cdot \frac{H_{l-1}}{l} & \quad \mbox{if $p=2$} \\ 3 (-1)^l \cdot \frac{H^{(2)}_{l-1} - H_{l-1}^2}{l} & \quad \mbox{if $p=3$} \\ 4 (-1)^l \cdot \frac{2 H^{(3)}_{l-1} - 3 H^{(2)}_{l-1} H_{l-1}+H_{l-1}^3}{l} & \quad \mbox{if $p=4$} \\ 5 (-1)^l \cdot \frac{6 H^{(4)}_{l-1} - 8 H^{(3)}_{l-1} H_{l-1}-3 H^{(2)}_{l-1} H^{(2)}_{l-1}+6 H^{(2)}_{l-1} H_{l-1}^2 - H_{l-1}^4}{l} & \quad \mbox{if $p=5$} \\ \end{array} \right. \end{equation} if $l\ge 1$ and $\kappa^{(p)}_0=\delta_{p,0}$. Here $H^{(p)}_l$ is the harmonic number of order $p$. Now, the obvious question would be to find the result for generic values of $p$.

Answer 1

To find the requested derivatives I would follow a different scheme, starting by rewriting the binomial in the version that uses the falling factorial $$ \left( \begin{gathered} \theta \\ l \\ \end{gathered} \right) = \frac{{\theta ^{\,\underline {\,l\,} } }} {{l!}} $$ Now $\theta ^{\,\underline {\,l\,} } $ is a polynomial in $\theta$ of degree $l$, which can be written in terms of powers of $\theta$ using the (unsigned) Stirling Numbers of 1st kind as $$ \theta ^{\,\underline {\,l\,} } = \theta \left( {\theta - 1} \right)\; \cdots \;\left( {\theta - l + 1} \right) = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,l} \right)} {\left( { - 1} \right)^{\,l - k} \left[ \begin{gathered} l \\ k \\ \end{gathered} \right]\theta ^{\,k} } $$ Since the polynomial coincides with its MacLaurin series, it comes that: $$ \left. {\frac{{\partial ^{\,p} }} {{\partial \theta ^{\,p} }}\left( \begin{gathered} \theta \\ l \\ \end{gathered} \right)} \right|_{\,\theta = 0} = \left. {\frac{{\partial ^{\,p} }} {{\partial \theta ^{\,p} }}\frac{{\theta ^{\,\underline {\,l\,} } }} {{l!}}} \right|_{\,\theta = 0} = \left( { - 1} \right)^{\,l - p} \frac{{p!}} {{l!}}\left[ \begin{gathered} l \\ p \\ \end{gathered} \right] $$