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Question Number 90960 by mathmax by abdo last updated on 27/Apr/20

let f(x) =x^3 cos(2x)  calculate f^((n)) (x) and f^((n)) (0)

letf(x)=x3cos(2x)calculatef(n)(x)andf(n)(0)

Commented by mathmax by abdo last updated on 27/Apr/20

leibniz give f^((n)) (x) =Σ_(k=0) ^n C_n ^k  (x^3 )^((k)) ( cos(2x))^((n−k))   first let find  (cos(2x))^((p))   (cos(2x))^((1))  =−2sin(2x) =2 cos(2x+(π/2))  (cos(2x))^((2))  =4cos(2x+2(π/2)) let suppose  (cos(2x))^((p))  =2^p  cos(2x+((pπ)/2)) ⇒(cos(2x))^((p+1)) =−2^(p+1)  sin(2x+((pπ)/2))  =2^(p+1)  cos(2x+(((p+1)π)/2)) ⇒  f^((n)) (x) =C_n ^0 x^3  2^n  cos(2x+((nπ)/2))+C_n ^1 3x^2  2^(n−1)  cos(2x+(((n−1)π)/2))  +C_n ^2  (6x) 2^(n−2)  cos(2x+(((n−2)π)/2)) +C_n ^3 6 2^(n−3)  cos(2x+(((n−3)π)/2))  f^((n)) (x)=2^n  x^3  cos(2x+((nπ)/2))+3n2^(n−1) x^2  cos(2x+(((n−1)π)/2))  +3n(n−1) 2^(n−2)  x cos(2x+(((n−2)π)/2))+n(n−1)(n−2)2^(n−3)  cos(2x+(((n−3)π)/2))

leibnizgivef(n)(x)=k=0nCnk(x3)(k)(cos(2x))(nk)firstletfind(cos(2x))(p)(cos(2x))(1)=2sin(2x)=2cos(2x+π2)(cos(2x))(2)=4cos(2x+2π2)letsuppose(cos(2x))(p)=2pcos(2x+pπ2)(cos(2x))(p+1)=2p+1sin(2x+pπ2)=2p+1cos(2x+(p+1)π2)f(n)(x)=Cn0x32ncos(2x+nπ2)+Cn13x22n1cos(2x+(n1)π2)+Cn2(6x)2n2cos(2x+(n2)π2)+Cn362n3cos(2x+(n3)π2)f(n)(x)=2nx3cos(2x+nπ2)+3n2n1x2cos(2x+(n1)π2)+3n(n1)2n2xcos(2x+(n2)π2)+n(n1)(n2)2n3cos(2x+(n3)π2)

Commented by mathmax by abdo last updated on 27/Apr/20

f^((n)) (0) =n(n−1)(n−2)2^(n−3)  cos((((n−3)π)/2))

f(n)(0)=n(n1)(n2)2n3cos((n3)π2)

Answered by MWSuSon last updated on 27/Apr/20

f^((n)) (x)=[2^n cos (2x+((nπ)/2))x^3 +n2^(n−1) cos (2x+(((n−1)π)/2))3x^2   +((n(n−1)2^(n−2) cos (2x+(((n−2)π)/2))6x)/2)+((n(n−1)(n−2)2^(n−3) cos (2x+(((n−3)π)/2))6)/6).  f^((n)) (0)=[n(n−1)(n−2)2^(n−3) cos ((((n−3)π)/2))]

f(n)(x)=[2ncos(2x+nπ2)x3+n2n1cos(2x+(n1)π2)3x2+n(n1)2n2cos(2x+(n2)π2)6x2+n(n1)(n2)2n3cos(2x+(n3)π2)66.f(n)(0)=[n(n1)(n2)2n3cos((n3)π2)]

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