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Question Number 29349 by abdo imad last updated on 07/Feb/18
$${let}\:{give}\:{f}\left({x}\right)=\:\left({x}^{{n}} −\mathrm{1}\right)\:{e}^{−{x}} \:\:{with}\:{n}\:{from}\:{N}^{\bigstar} \: \\ $$$${find}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right)\:. \\ $$
Commented by abdo imad last updated on 09/Feb/18
$${the}\:{leibniz}\:{forula}\:{give}\:\:{f}^{\left({n}\right)} \left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \left({x}^{{n}} \:−\mathrm{1}\right)^{\left({k}\right)} \left({e}^{−{x}} \right)^{\left({n}−{k}\right)} \\ $$$$=\:\left({x}^{{n}} −\mathrm{1}\right)\left({e}^{−{x}} \right)^{\left({n}\right)} \:+\sum_{{k}=\mathrm{1}} ^{{n}} {C}_{{n}} ^{{k}} \left({x}^{{n}} −\mathrm{1}\right)^{\left({k}\right)} \:\left(\:{e}^{−{x}} \right)^{\left({n}−{k}\right)} \:\:{but} \\ $$$$\left({e}^{−{x}} \right)^{\left(\mathrm{1}\right)} =−\:{e}^{−{x}} \:,\:\:\left({e}^{−{x}} \right)^{\left(\mathrm{2}\right)} =\left(−\mathrm{1}\right)^{\mathrm{2}} \:{e}^{−{x}} \:...\left({e}^{−{x}} \right)^{\left({p}\right)} =\left(−\mathrm{1}\right)^{{p}} \:{e}^{−{x}} \:{so} \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\left(−\mathrm{1}\right)^{{n}} \left({x}^{{n}} −\mathrm{1}\right)\:{e}^{−{x}} \:+\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(−\mathrm{1}\right)^{{n}−{k}} \:{e}^{−{x}} \left({x}^{{n}} −\mathrm{1}\right)^{\left({k}\right)} \\ $$$$\left({x}^{{n}} −\mathrm{1}\right)^{\left(\mathrm{1}\right)} =\:{nx}^{{n}−\mathrm{1}} \:,\left({x}^{{n}} −\mathrm{1}\right)^{\left(\mathrm{2}\right)} ={n}\left({n}−\mathrm{1}\right)^{} {x}^{{n}−\mathrm{2}} .... \\ $$$$\left({x}^{{n}} −\mathrm{1}\right)^{\left({p}\right)} ={n}\left({n}−\mathrm{1}\right)...\left({n}−{p}+\mathrm{1}\right){x}^{{n}−{p}} \:=\frac{{n}!}{\left({n}−{p}\right)!}{x}^{{n}−{p}} \:{with}\:{p}\leqslant{n} \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\left(−\mathrm{1}\right)^{{n}} \left({x}^{{n}} −\mathrm{1}\right){e}^{−{x}} \:+\left(−\mathrm{1}\right)^{{n}} {e}^{−{x}} \:\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:{C}_{{n}} ^{{k}} \:\frac{{n}!}{\left({n}−{k}\right)!}{x}^{{n}−{k}} \\ $$$$=\left(−\mathrm{1}\right)^{{n}} \left({x}^{{n}} −\mathrm{1}\right){e}^{−{x}} \:+\left(−\mathrm{1}\right)^{{n}} {e}^{−{x}} \sum_{{k}=} ^{{n}−\mathrm{1}} \left(...\right){x}^{{n}−{p}} \\ $$$$+\left(−\mathrm{1}\right)^{{n}} \:{e}^{−{x}} \:\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} \:\boldsymbol{{C}}_{{n}} ^{{n}} \:{n}!\:\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)=\left(−\mathrm{1}\right)^{\boldsymbol{{n}}−\mathrm{1}} \:+\boldsymbol{{n}}! \\ $$