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Question Number 55270 by maxmathsup by imad last updated on 20/Feb/19
let f(x)=x^n arctan(x^2 )  with n integr natural  1) calculate f^((n)) (x) and f^((n)) (0)  2)developp f at integr serie .
letf(x)=xnarctan(x2)withnintegrnatural1)calculatef(n)(x)andf(n)(0)2)developpfatintegrserie.
Commented by maxmathsup by imad last updated on 02/Mar/19
1) leibniz formula give  f^((n)) (x)=Σ_(k=0) ^n  C_n ^k   (x^n )^((k))   (arctan(x^2 ))^((n−k))   but (x^n )^((k)) =n(n−1)...(n−k+1)x^(n−k)  =((n!)/((n−k)!)) x^(n−k)    if k≤n ⇒  f^((n)) (x) =Σ_(k=0) ^n  ((n!)/(k!(n−k)!)) ((n!)/((n−k)!)) x^(n−k)  {arctan(x^2 )}^()n−k))   =Σ_(k=0) ^n    (((n!)^2 )/(k!{(n−k)!}^2 )) { arctan(x^2 )}^((n−k))    let find (arctan(x^2 ))^((m))   we have  {arctan(x^2 )}^((1)) =((2x)/(1+x^4 )) =((2x)/((x^2 −i)(x^2  +i))) =((2x)/((x−(√i))(x+(√i))(x−(√(−i)))(x+(√(−i)))))  =((2x)/((x−e^((iπ)/4) )(x+e^((iπ)/4) )(x−e^(−((iπ)/4)) )(x+e^(−((iπ)/4)) ))) =F(x)=(a/(x−e^((iπ)/4) )) +(b/(x+e^((iπ)/4) )) +(c/(x−e^(−((iπ)/4)) )) +(d/(x+e^(−((iπ)/4)) ))  a =((2 e^((iπ)/4) )/(2e^((iπ)/4) (2i sin((π/4)))(2cos((π/4))))) = (1/(4i(1/2))) =(1/(2i))  b =((−2 e^((iπ)/4) )/((−2e^((iπ)/4) )(−2cos((π/4)))(−2i sin((π/4))))) =(1/(2i))  c =((2e^(−((iπ)/4)) )/((−2isin((π/4)))(2cos((π/4)))(2e^(−((iπ)/4)) ))) =−(1/(2i))  d =((−2e^(−((iπ)/4)) )/((−2cos((π/4)))(2i sin((π/4)))(−2e^(−((iπ)/4)) ))) =−(1/(2i)) ⇒  (arctan(x^2 ))^((1))  =(1/(2i)){(1/(x−e^((iπ)/4) )) +(1/(x+e^((iπ)/4) )) −(1/(x−e^(−((iπ)/4)) )) −(1/(x+e^(−((iπ)/4)) ))} ⇒  (arctan(x^2 ))^((m)) =(1/(2i))(−1)^(m−1) (m−1)!{ (1/((x−e^((iπ)/4) )^m )) +(1/((x+e^((iπ)/4) )^m )) −(1/((x−e^(−((iπ)/4)) )^m ))−(1/((x+e^(−((iπ)/4)) )))} ⇒  f^((n)) (x) =(1/(2i))Σ_(k=0) ^n   (((n!)^2 )/(k!{(n−k)!}^2 ))(−1)^(n−k−1) (n−k−1)!{(1/((x−e^((iπ)/4) )^(n−k) ))+(1/((x+e^((iπ)/4) )^(n−k) ))−(1/((x−e^(−((iπ)/4)) )^(n−k) )) −(1/((x+e^(−((iπ)/4)) )^(n−k) ))}
1)leibnizformulagivef(n)(x)=k=0nCnk(xn)(k)(arctan(x2))(nk)but(xn)(k)=n(n1)(nk+1)xnk=n!(nk)!xnkifknf(n)(x)=k=0nn!k!(nk)!n!(nk)!xnk{arctan(x2)})nk)=k=0n(n!)2k!{(nk)!}2{arctan(x2)}(nk)letfind(arctan(x2))(m)wehave{arctan(x2)}(1)=2x1+x4=2x(x2i)(x2+i)=2x(xi)(x+i)(xi)(x+i)=2x(xeiπ4)(x+eiπ4)(xeiπ4)(x+eiπ4)=F(x)=axeiπ4+bx+eiπ4+cxeiπ4+dx+eiπ4a=2eiπ42eiπ4(2isin(π4))(2cos(π4))=14i12=12ib=2eiπ4(2eiπ4)(2cos(π4))(2isin(π4))=12ic=2eiπ4(2isin(π4))(2cos(π4))(2eiπ4)=12id=2eiπ4(2cos(π4))(2isin(π4))(2eiπ4)=12i(arctan(x2))(1)=12i{1xeiπ4+1x+eiπ41xeiπ41x+eiπ4}(arctan(x2))(m)=12i(1)m1(m1)!{1(xeiπ4)m+1(x+eiπ4)m1(xeiπ4)m1(x+eiπ4)}f(n)(x)=12ik=0n(n!)2k!{(nk)!}2(1)nk1(nk1)!{1(xeiπ4)nk+1(x+eiπ4)nk1(xeiπ4)nk1(x+eiπ4)nk}

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