I was in the middle of an exercise, and I now have to simplify/determine a general formula for the following derivative:
\begin{equation*}
\frac{d^{n-1}}{dz^{n-1}}\left[(z+i)^{-n}\right]
\end{equation*}
What I have done: After giving some values to $n$, I noticed the following pattern:
\begin{equation*}
n =2: -2(z+i)^{-3} \\
n=3: 12(z+i)^{-5} \\
n =4: -120(z+i)^{-7}
\end{equation*}
Which made me write the following general formula:
\begin{equation*}
\frac{d^{n-1}}{dz^{n-1}}\left[(z+i)^{-n}\right] = (-1)^{n+1}(z+i)^{-(2n-1)}\prod_{k=0}^{n-2}(n+k)
\end{equation*}
And this is where my problems begin. I am having trouble proving this formula through mathematical induction and I don't even know if the productory part is correct at all (it seems correct for the cases I stated above).
Best Answer
For fixed integer $k$ is $$ \frac{d^{n-1}}{dz^{n-1}}\left[(z+i)^{-k}\right] = (-k)(-k-1)\cdots(-k-(n-2))(z+i)^{-k-(n-1)} \\ = (-1)^{n-1}k(k+1)\cdots(k+n-2)(z+i)^{-k-n+1} $$ which can be proven with induction over $n$. Setting $k=n$ gives $$ \frac{d^{n-1}}{dz^{n-1}}\left[(z+i)^{-n}\right] = (-1)^{n-1}n(n+1)\cdots(n+(n-2))(z+i)^{-2n+1} \, , $$ confirming your result.