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Question Number 40105 by maxmathsup by imad last updated on 15/Jul/18

let f(x) = x^n e^(−2nx) with n integr natural calculate f^((n)) (0).

letf(x)=xne−2nxwithnintegrnaturalcalculatef(n)(0).

Commented by maxmathsup by imad last updated on 17/Jul/18

leibniz formula give f^((p)) (x)=Σ_(k=0) ^p C_p ^k (x^n )^((k)) (e^(−2nx) )^((p−k)) but (x^n )^((k)) = n(n−1)....(n−k+1) x^(n−k) if k≤n and (x^n )^((k)) =0 if k>n ⇒ (x^n )^((k)) =((n!)/((n−k)!)) x^(n−k) also we have {e^(−2n) }^((p−k)) =(−2n)^(p−k) e^(−2nx) ⇒ f^((p)) (x)=Σ_(k=0) ^p C_p ^k ((n!)/((n−k)!)) x^(n−k) (−2n)^(p−k) e^(−2nx) let suppose n≤p f^((p)) (x)= Σ_(k=0) ^(n−1) C_p ^k ((n!)/((n−k)!)) x^(n−k) (−2n)^(p−k) e^(−2nx ) + C_p ^n ((n!)/(0!))(−2n)^(p−n) e^(−2nx) +Σ_(k=n+1) ^p C_p ^k ((n!)/((n−k)!)) x^(n−k) (−2n)^(p−k) e^(−2nx) ⇒ f^((p)) (0) = C_p ^n n! (−2n)^(p−n) ⇒ f^((n)) (0) =n! C_n ^n =n!

leibnizformulagivef(p)(x)=∑k=0pCpk(xn)(k)(e−2nx)(p−k)but(xn)(k)=n(n−1)....(n−k+1)xn−kifk⩽nand(xn)(k)=0ifk>n⇒(xn)(k)=n!(n−k)!xn−kalsowehave{e−2n}(p−k)=(−2n)p−ke−2nx⇒f(p)(x)=∑k=0pCpkn!(n−k)!xn−k(−2n)p−ke−2nxletsupposen⩽pf(p)(x)=∑k=0n−1Cpkn!(n−k)!xn−k(−2n)p−ke−2nx+Cpnn!0!(−2n)p−ne−2nx+∑k=n+1pCpkn!(n−k)!xn−k(−2n)p−ke−2nx⇒f(p)(0)=Cpnn!(−2n)p−n⇒f(n)(0)=n!Cnn=n!

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