Menu Close

let-give-f-x-x-n-1-e-x-with-n-from-N-find-f-n-0-

Tinku Tara 0 Comments
Pin




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

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *