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Question Number 67187 by mathmax by abdo last updated on 23/Aug/19
let f(x) =arctan(x^3 ) 1)calculate f^((n)) (x)and f^((n)) (0) 2) developp f at integr serie 3) calculate ∫_0 ^1 arctan(x^3 )dx
letf(x)=arctan(x3)1)calculatef(n)(x)andf(n)(0)2)developpfatintegrserie3)calculate∫01arctan(x3)dx
Commented by mathmax by abdo last updated on 28/Aug/19
1) f(x) =arctan(x^3 ) ⇒f^′ (x) =((3x^2 )/(1+x^6 )) ⇒f^((n)) (x)=(((3x^2 )/(1+x^6 )))^((n−1)) let decompose inside C(x) F(x) =((3x^2 )/(x^6 +1)) z^6 +1 =0 ⇒z^6 =−1 =e^(i(2k+1)π) so if z=r e^(iθ) we get r=1 and 6θ =e^(i(2k+1)π) ⇒ θ =e^((i(2k+1)π)/6) so the poles of F are Z_k =e^((i(2k+1)π)/6) k∈[[0,5]] Z_0 =e^((iπ)/6) , Z_1 =e^(i(π/2)) =i Z_2 =e^((i5π)/6) , Z_3 = e^((i7π)/6) , Z_4 = e^((i9π)/6) , Z_5 = e^((i11π)/6) f^((n)) (x) =Σ_(k=0) ^(n−1) C_(n−1) ^k (3x^2 )^((k)) ((1/(x^6 +1)))^((n−1−k)) (leibniz) =3x^2 ×{(1/(x^6 +1))}^((n−1)) +(n−1)(6x){(1/(x^6 +1))}^((n−2)) +6 C_(n−1) ^2 {(1/(x^6 +1))}^((n−3)) let find {(1/(x^6 +1))}^((k)) with k from N wr have G(x) =(1/(x^6 +1)) =Σ_(i=0) ^5 (λ_i /(x−z_i )) λ_i =(1/(6z_i ^5 )) =−(1/6)z_i ⇒G(x) =−(1/6)Σ_(i=0) ^5 (z_i /(x−z_i )) ⇒ G^((k)) (x) =−(1/6)Σ_(i=0) ^5 z_i ×(((−1)^k k!)/((x−z_i )^(k+1) )) ⇒ G^((n−1)) (x) =−(1/6)Σ_(i=0) ^5 z_i ×(((−1)^(n−1) (n−1)!)/((x−z_i )^n )) G^((n−2)) (x) =−(1/6)Σ_(i=0) ^5 z_i × (((−1)^(n−2) (n−2)!)/((x−z_i )^(n−1) )) G^((n−3)) (x) =−(1/6)Σ_(i=0) ^5 z_i ×(((−1)^(n−3) (n−3)!)/((x−z_i )^(n−2) )) so the value of f^((n)) (x) is determined.
1)f(x)=arctan(x3)⇒f′(x)=3x21+x6⇒f(n)(x)=(3x21+x6)(n−1)letdecomposeinsideC(x)F(x)=3x2x6+1z6+1=0⇒z6=−1=ei(2k+1)πsoifz=reiθwegetr=1and6θ=ei(2k+1)π⇒θ=ei(2k+1)π6sothepolesofFareZk=ei(2k+1)π6k∈[[0,5]]Z0=eiπ6,Z1=eiπ2=iZ2=ei5π6,Z3=ei7π6,Z4=ei9π6,Z5=ei11π6f(n)(x)=∑k=0n−1Cn−1k(3x2)(k)(1x6+1)(n−1−k)(leibniz)=3x2×{1x6+1}(n−1)+(n−1)(6x){1x6+1}(n−2)+6Cn−12{1x6+1}(n−3)letfind{1x6+1}(k)withkfromNwrhaveG(x)=1x6+1=∑i=05λix−ziλi=16zi5=−16zi⇒G(x)=−16∑i=05zix−zi⇒G(k)(x)=−16∑i=05zi×(−1)kk!(x−zi)k+1⇒G(n−1)(x)=−16∑i=05zi×(−1)n−1(n−1)!(x−zi)nG(n−2)(x)=−16∑i=05zi×(−1)n−2(n−2)!(x−zi)n−1G(n−3)(x)=−16∑i=05zi×(−1)n−3(n−3)!(x−zi)n−2sothevalueoff(n)(x)isdetermined.

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