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Proof of Equation (3.15)

\begin{equation*} \frac{{{\partial }^{\left( n \right)}}\left( h_{s}^{c} \right)}{\partial {{s}^{n}}}=\exp (-c)\sum\limits_{i=0}^{\infty }{\frac{{{\left( -1 \right)}^{n}}n!{{c}^{i}}}{i!{{\left( s+1+i \right)}^{n+1}}}} \end{equation*}
Proof:
\begin{align*} \frac{\partial \left( h_{s}^{c} \right)}{\partial s}=&{\exp(-c)}\sum\limits_{i=0}^{\infty }{\frac{\left( -1 \right){{c}^{i}}}{i!{{\left( s+1+i \right)}^{2}}}}, \\ \frac{{{\partial }^{2}}\left( h_{s}^{c} \right)}{\partial {{s}^{2}}}=&{\exp(-c)}\sum\limits_{i=0}^{\infty }{\frac{\left( -1 \right)\left( -2 \right){{c}^{i}}}{i!{{\left( s+1+i \right)}^{3}}}}, \\ & \vdots \\ \frac{{{\partial }^{n}}\left( h_{s}^{c} \right)}{\partial {{s}^{n}}}=&{\exp(-c)}\sum\limits_{i=0}^{\infty }{\frac{\left( -1 \right)\cdots \left( -n \right){{c}^{i}}}{i!{{\left( s+1+i \right)}^{n+1}}}}. \end{align*}

The above formulas conclude the first to nth partial derivative of h_{s}^{c} on s.
\square

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