Is there a finite expression for $k$-th derivative of \begin{align} f(x)={x^n}{e^{ - x}} \end{align}
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1$\begingroup$ Have you tried differentiating a number of times, then guessing the pattern: e.g., $f^{(n)} = P_n (x)e^{-x}$, and working out the recursive condition on the polynomials $P_n$ (which depend on $N$ as well)? $\endgroup$– David HandelmanCommented Aug 9, 2020 at 14:54
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2$\begingroup$ This is not research level, so it doesn't belong at MO. It could probably do well at MSE, although (a) I'd be surprised if it's not there already, and (b) as @DavidHandelman says, it's something for which you can at least gather experimental data (and show that you have done so before asking others to do it). $\endgroup$– LSpiceCommented Aug 9, 2020 at 15:15
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$\begingroup$ Have u tried to use Leibniz's formula? am affraid that is not a question for MO website ,MO website is not for standard questions but for question in high level of research $\endgroup$– zeraoulia rafikCommented Aug 9, 2020 at 17:26
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1 Answer
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The command of Maple
diff(x^n*exp(-x), x $ k);
produces $$\sum _{k_1=0}^{k}{k\choose {k_1}}{\it pochhammer} \left( n -{k_1}+1,{k_1} \right) {x}^{n-{k_1}}{{\rm e}^{-x}} \left( -1 \right) ^{k-{k_1}} $$ and the code of Mathematica
D[x^n*Exp[-x], {x, k}]
performs $$e^{-x} k! \binom{n}{k} x^{n-k} \, _1F_1(-k;-k+n+1;x) .$$
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3$\begingroup$ Just use Leibniz's formula $(fg)^{(k)}= \sum_{i=0}^k \binom{k}{i} f^{(i)} g^{(k-i)}$. $\endgroup$ Commented Aug 9, 2020 at 15:01
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$\begingroup$ @Ira Gessel: Disagreed since CASes simplify it, e.g. for $k=5$ we have $${{\rm e}^{-x}} \left( \left( {n}^{5}-10\,{n}^{4}+35\,{n}^{3}-50\,{n}^ {2}+24\,n \right) {x}^{n-5}+ \left( -5\,{n}^{4}+30\,{n}^{3}-55\,{n}^{2 }+30\,n \right) {x}^{n-4}+ \left( 10\,{n}^{3}-30\,{n}^{2}+20\,n \right) {x}^{n-3}+ \left( -10\,{n}^{2}+10\,n \right) {x}^{n-2}+5\,{x} ^{n-1}n-{x}^{n} \right) .$$ $\endgroup$ Commented Aug 9, 2020 at 15:08
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1$\begingroup$ With what are you disagreeing? Your first formula, from Maple, is literally just @IraGessel's suggestion. $\endgroup$– LSpiceCommented Aug 9, 2020 at 15:14
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$\begingroup$ @LSpice, I repeat that both CASes may find simplified expressions for concrete values of $k$. Those expressions are too long to do it by hand. $\endgroup$ Commented Aug 9, 2020 at 15:18