nnth derivative of e1/xe^{1/x}

I am trying to find the n‘th derivative of f(x)=e1/x. When looking at the first few derivatives I noticed a pattern and eventually found the following formula

\frac{\mathrm d^n}{\mathrm dx^n}f(x)=(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 n+k}

I tested it for the first 20 derivatives and it got them all. Mathematica says that it is some hypergeometric distribution but I don’t want to use that. Now I am trying to verify it by induction but my algebra is not good enough to do the induction step.

Here is what I tried for the induction (incomplete, maybe incorrect)

\begin{align*}
\frac{\mathrm d^{n+1}}{\mathrm dx^{n+1}}f(x)&=\frac{\mathrm d}{\mathrm dx}(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 n+k}\\
&=(-1)^n e^{1/x} \cdot \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} (-2n+k) x^{-2 n+k-1}\right)-e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 (n+1)+k}\\
&=(-1)^n e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k}((-2n+k) x^{-2 n+k-1}-x^{-2 (n+1)+k)})\\
&=(-1)^{n+1} e^{1/x} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k}(2n x-k x+1) x^{-2 (n+1)+k}
\end{align*}

I don’t know how to get on from here.

Answer

How’s this?

\left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} \left(-2n+k\right) x^{-2 n+k-1}\right) – \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 \left(n+1\right)+k}\right) =

= \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} \left(-2n+k\right) x^{-2\left(n+1\right)+k+1}\right) – \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 \left(n+1\right)+k}\right) =

= \left(\sum _{k’=1}^{n} \left(k’-1\right)! \binom{n}{k’-1} \binom{n-1}{k’-1} \left(-2n+k’-1\right) x^{-2\left(n+1\right)+k’}\right) – \left(\sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{-2 \left(n+1\right)+k}\right) =

= \left(\sum _{k’=0}^{n} \left(k’-1\right)! \binom{n}{k’-1} \binom{n-1}{k’-1} \left(-2n+k’-1\right) x^{-2\left(n+1\right)+k’}\right) – \left(\sum _{k=0}^{n} k! \binom{n}{k} \binom{n-1}{k} x^{-2 \left(n+1\right)+k}\right) =

= \sum _{k=0}^{n} \left(\left(k-1\right)! \binom{n}{k-1} \binom{n-1}{k-1} \left(-2n+k-1\right) – k! \binom{n}{k} \binom{n-1}{k}\right) x^{-2 \left(n+1\right)+k}

Then
\left(k-1\right)! \binom{n}{k-1} \binom{n-1}{k-1} \left(-2n+k-1\right) – k! \binom{n}{k} \binom{n-1}{k} =

= \frac{\left(k-1\right)!n!\left(n-1\right)!\left(-2n+k-1\right)}{\left(n-k+1\right)!\left(k-1\right)!\left(n-k\right)!\left(k-1\right)!} – \frac{k!n!\left(n-1\right)!}{\left(n-k\right)!k!k!\left(n-k-1\right)!} =

= \frac{n!\left(n-1\right)!\left(-2n+k-1\right)k}{\left(n-k+1\right)!\left(n-k\right)!k!} – \frac{n!\left(n-1\right)!\left(n-k\right)\left(n-k+1\right)}{\left(n-k\right)!k!\left(n-k+1\right)!} =

= \frac{n!\left(n-1\right)!}{\left(n-k+1\right)!\left(n-k\right)!k!} \left(\left(-2n+k-1\right)k – \left(n-k\right)\left(n-k+1\right)\right) =

= \frac{-n\left(n+1\right)n!\left(n-1\right)!}{\left(n-k+1\right)!\left(n-k\right)!k!} =

= -\frac{\left(n+1\right)!}{\left(n-k+1\right)!} \binom{n}{k} =

= -k! \binom{n+1}{k} \binom{n}{k}

Attribution
Source : Link , Question Author : Listing , Answer Author : J. M. ain’t a mathematician

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