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Question Number 52673 by maxmathsup by imad last updated on 11/Jan/19
let f(x)=(x^n −1)e^x determine f^((n)) (x) with n integr natural
$${let}\:{f}\left({x}\right)=\left({x}^{{n}} −\mathrm{1}\right){e}^{{x}} \:\:\:{determine}\:{f}^{\left({n}\right)} \left({x}\right)\:\:\:{with}\:{n}\:{integr}\:{natural} \\ $$
Commented by maxmathsup by imad last updated on 13/Jan/19
f^((n)) (x) =Σ_(k=0) ^n C_n ^k (x^n −1)^((k)) (e^x )^((n−k)) (leibniz formulae) =Σ_(k=0) ^n C_n ^k (x^n −1)^((k)) e^x =(x^n −1)e^x +e^x Σ_(k=1) ^n C_n ^k (x^n )^k but we have (x^n )^((1)) =nx^(n−1) ,(x^n )^((2)) =n(n−1)x^(n−2) ....(x^n )^((k)) =n(n−1)...(n−k+1)x^(n−k) =((n!)/((n−k)!)) x^(n−k) ⇒f^((n)) (x)=(x^n −1)e^x +e^x Σ_(k=1) ^n ((n!)/(k!(n−k)!)) ((n!)/((n−k)!)) x^(n−k) ⇒ f^((n)) (x)=(x^n −1)e^x +e^x Σ_(k=1) ^n (((n!)^2 )/(k!(n−k)!^2 )) x^(n−k) .
$${f}^{\left({n}\right)} \left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({x}^{{n}} −\mathrm{1}\right)^{\left({k}\right)} \:\left({e}^{{x}} \right)^{\left({n}−{k}\right)} \:\:\:\left({leibniz}\:{formulae}\right) \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({x}^{{n}} −\mathrm{1}\right)^{\left({k}\right)} {e}^{{x}} \:\:=\left({x}^{{n}} −\mathrm{1}\right){e}^{{x}} \:+{e}^{{x}} \sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({x}^{{n}} \right)^{{k}} \:\:\:\:{but}\:{we}\:{have} \\ $$$$\left({x}^{{n}} \right)^{\left(\mathrm{1}\right)} ={nx}^{{n}−\mathrm{1}} \:\:,\left({x}^{{n}} \right)^{\left(\mathrm{2}\right)} ={n}\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} ….\left({x}^{{n}} \right)^{\left({k}\right)} ={n}\left({n}−\mathrm{1}\right)…\left({n}−{k}+\mathrm{1}\right){x}^{{n}−{k}} \\ $$$$=\frac{{n}!}{\left({n}−{k}\right)!}\:{x}^{{n}−{k}} \:\Rightarrow{f}^{\left({n}\right)} \left({x}\right)=\left({x}^{{n}} −\mathrm{1}\right){e}^{{x}} \:+{e}^{{x}} \:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{{n}!}{{k}!\left({n}−{k}\right)!}\:\frac{{n}!}{\left({n}−{k}\right)!}\:{x}^{{n}−{k}} \:\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\left({x}^{{n}} −\mathrm{1}\right){e}^{{x}} \:+{e}^{{x}} \:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left({n}!\right)^{\mathrm{2}} }{{k}!\left({n}−{k}\right)!^{\mathrm{2}} }\:{x}^{{n}−{k}} \:\:. \\ $$
Answered by Smail last updated on 11/Jan/19
f′(x)=(nx^(n−1) +x^n −1)e^x f′′(x)=(n(n−1)x^(n−2) +2nx^(n−1) +x^n −1)e^x f^((3)) (x)=(n(n−1)(n−2)x^(n−3) +3n(n−1)x^(n−2) +3nx^(n−1) )e^x +f(x) f^((4)) (x)=(n→(n−3)x^(n−4) +4n→(n−2)x^(n−3) +6n(n−1)x^(n−2) +4nx^(n−1) )e^x +f(x) f^((n)) (x)=(P_(n−0) ^n C_0 ^n x^0 +P_(n−1) ^n C_1 ^n x^1 +P_(n−2) ^n C_2 ^n x^2 +...+P_0 ^n C_n ^n x^n −1)e^x f^((n)) (x)=e^x (Σ_(k=0) ^n P_(n−k) ^n C_k ^n x^k −1)
$${f}'\left({x}\right)=\left({nx}^{{n}−\mathrm{1}} +{x}^{{n}} −\mathrm{1}\right){e}^{{x}} \\ $$$${f}''\left({x}\right)=\left({n}\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} +\mathrm{2}{nx}^{{n}−\mathrm{1}} +{x}^{{n}} −\mathrm{1}\right){e}^{{x}} \\ $$$${f}^{\left(\mathrm{3}\right)} \left({x}\right)=\left({n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right){x}^{{n}−\mathrm{3}} +\mathrm{3}{n}\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} +\mathrm{3}{nx}^{{n}−\mathrm{1}} \right){e}^{{x}} +{f}\left({x}\right) \\ $$$${f}^{\left(\mathrm{4}\right)} \left({x}\right)=\left({n}\rightarrow\left({n}−\mathrm{3}\right){x}^{{n}−\mathrm{4}} +\mathrm{4}{n}\rightarrow\left({n}−\mathrm{2}\right){x}^{{n}−\mathrm{3}} +\mathrm{6}{n}\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{2}} +\mathrm{4}{nx}^{{n}−\mathrm{1}} \right){e}^{{x}} +{f}\left({x}\right) \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\left({P}_{{n}−\mathrm{0}} ^{{n}} {C}_{\mathrm{0}} ^{{n}} {x}^{\mathrm{0}} +{P}_{{n}−\mathrm{1}} ^{{n}} {C}_{\mathrm{1}} ^{{n}} {x}^{\mathrm{1}} +{P}_{{n}−\mathrm{2}} ^{{n}} {C}_{\mathrm{2}} ^{{n}} {x}^{\mathrm{2}} +…+{P}_{\mathrm{0}} ^{{n}} {C}_{{n}} ^{{n}} {x}^{{n}} −\mathrm{1}\right){e}^{{x}} \\ $$$${f}^{\left({n}\right)} \left({x}\right)={e}^{{x}} \left(\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{P}_{{n}−{k}} ^{{n}} {C}_{{k}} ^{{n}} {x}^{{k}} −\mathrm{1}\right) \\ $$

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