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let-f-x-x-2n-e-3x-find-f-n-o-and-calculate-f-2021-0-

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Question Number 140637 by Mathspace last updated on 10/May/21
let f(x)=x^(2n) e^(−3x) find f^((n)) (o) and calculate f^((2021)) (0)
letf(x)=x2ne−3xfindf(n)(o)andcalculatef(2021)(0)
Answered by mathmax by abdo last updated on 12/May/21
by leibniz f^((p)) (x)=Σ_(k=0) ^p C_p ^k (x^(2n) )^((k)) (e^(−3x) )^((p−k)) (x^(2n) )^((1)) =2n x^(2n−1) ,(x^(2n) )^((2)) =2n(2n−1)x^(2n−2) .... (x^(2n) )^((k)) =2n(2n−1)....(2n−k+1)x^(2n−k) ⇒ f^((p)) (x)=Σ_(k=0) ^p C_p ^k 2n(2n−1)2n−2)....(2n−k+1)x^(2n−k) (−3)^(p−k) e^(−3x) ⇒f^((n)) (x)=Σ_(k=0) ^n C_n ^k 2n(2n−1)....(2n−k+1)x^(2n−k) (−3)^(n−k) e^(−3x) ⇒ f^((n)) (0)=0 ⇒f^((2021)) (0)=0
byleibnizf(p)(x)=∑k=0pCpk(x2n)(k)(e−3x)(p−k)(x2n)(1)=2nx2n−1,(x2n)(2)=2n(2n−1)x2n−2….(x2n)(k)=2n(2n−1)….(2n−k+1)x2n−k⇒f(p)(x)=∑k=0pCpk2n(2n−1)2n−2)….(2n−k+1)x2n−k(−3)p−ke−3x⇒f(n)(x)=∑k=0nCnk2n(2n−1)….(2n−k+1)x2n−k(−3)n−ke−3x⇒f(n)(0)=0⇒f(2021)(0)=0

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