Question Number 166814 by mathocean1 last updated on 28/Feb/22
$${f}\left({x}\right)={x}^{{n}} {e}^{−{nx}} \\ $$$${Determinate}\:{f}^{\left({n}\right)} \left({x}\right). \\ $$
Answered by TheSupreme last updated on 28/Feb/22
$${f}\left({x}\right)=\left({xe}^{−{x}} \right)^{{n}} = \\ $$$${f}^{{n}} \left({x}\right)={D}^{{n}} \left({t}^{{n}} \right)\mid_{{t}={xe}^{−{x}} } {D}^{{n}} \left({xe}^{−{x}} \right) \\ $$$${D}^{\mathrm{1}} =\left(\mathrm{1}−{x}\right){e}^{−{x}} \\ $$$${D}^{\mathrm{2}} =\left(−\mathrm{1}−\mathrm{1}+{x}\right){e}^{−{x}} =\left(−\mathrm{2}+{x}\right){e}^{−{x}} \\ $$$${D}^{\mathrm{3}} =\left(\mathrm{1}+\mathrm{2}−{x}\right){e}^{−{x}} =\left(\mathrm{3}−{x}\right){e}^{−{x}} \\ $$$${D}^{{n}} =\left(−\mathrm{1}\right)^{{n}} \left({x}−{n}\right){e}^{−{x}} \\ $$$${f}^{{k}} \left({x}\right)=\left(−\mathrm{1}\right)^{{n}} {x}^{{n}−{k}} {e}^{−\left({n}−{k}\right){x}} \frac{{n}!}{\left({n}−{k}\right)!}\left({x}−{k}\right){e}^{−{x}} \\ $$
Answered by Mathspace last updated on 28/Feb/22
$${f}^{\left({p}\right)} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{p}} {C}_{{p}} ^{{k}} \left({x}^{{n}} \right)^{\left({k}\right)} \left({e}^{−{nx}} \right)^{\left({p}−{k}\right)} \\ $$$$\left.{but}\:\left({x}^{{n}} \right)^{\left({k}\right)} ={n}\left({n}−\mathrm{1}\right)…\right)\left(\right. \\ $$$$\left.{n}−{k}+\mathrm{1}\right){x}^{{n}−{k}} =\frac{{n}!}{\left({n}−{k}\right)!}{x}^{{n}−{k}} \\ $$$$\left({e}^{−{nx}} \right)^{\left({p}−{k}\right)} =\left(−{n}\right)^{{p}−{k}} \:{e}^{−{nx}} \:\Rightarrow \\ $$$${f}^{\left({p}\right)} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{p}} \:{C}_{{p}} ^{{k}} \:\frac{{n}!}{\left({n}−{k}\right)!}\left(−{n}\right)^{{p}−{k}} {e}^{−{nx}} \\ $$$$\Rightarrow{f}^{\left({n}\right)} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\frac{{n}!}{\left({n}−{k}\right)!}\left(−{n}\right)^{{n}−{k}} \:{e}^{−{nx}} \\ $$