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let-f-x-e-3x-x-2-4-developp-f-at-integr-serie-




Question Number 39214 by math khazana by abdo last updated on 03/Jul/18
let f(x)= (e^(−3x) /(x^2  +4))  developp f  at integr serie.
letf(x)=e3xx2+4developpfatintegrserie.
Commented by math khazana by abdo last updated on 04/Jul/18
f(x)=Σ_(n=0) ^∞  ((f^((n)) (0))/(n!)) x^n    let find f^((n)) (0)  f(x)=(e^(−3x) /((x−2i)(x+2i))) =(e^(−3x) /(4i)){  (1/(x−2i)) −(1/(x+2i))}  =(1/(4i)){  (e^(−3x) /(x−2i)) −(e^(−3x) /(x+2i))} ⇒  f^((n)) (x)=(1/(4i)){ ( (e^(−3x) /(x−2i)))^((n))  −((e^(−3x) /(x+3i)))^((n)) } but  leibniz formula give  ((e^(−3x) /(x−2i)))^((n)) =Σ_(k=0) ^n   C_n ^k   ((1/(x−2i)))^((k))  (e^(−3x) )^(n−k))   =Σ_(k=0) ^n   C_n ^k   (((−1)^k  k!)/((x−2i)^(k+1) )) (−3)^(n−k)  e^(−3x )   also  ( (e^(−3x) /(x+2i)))^((n)) = Σ_(k=0) ^n   C_n ^k     (((−1)^k  k!)/((x+2i)^(k+1) )) (−3)^(n−k)  e^(−3x)   f^((n)) (x)=(1/(4i)) Σ_(k=0) ^n   (−1)^k k!(−3)^(n−k)  C_n ^k { (1/((x−2i)^(k+1) ))−(1/((x+2i)^(k+1) ))}  f^((n)) (0)=(1/(4i)) Σ_(k=0) ^n (−1)^k  k!(−3)^(n−k)  C_n ^n  {  (1/((−2i)^(k+1) )) −(1/((2i)^(k+1) ))}  but  (1/((−2i)^(k+1) )) + (1/((2i)^(k+1) )) =(((2i)^(k+1)  −(−2i)^(k+1) )/4^(k+1) )  =((2iIm( (2i)^(k+1) ))/4^(k+1) ) = ((2i 2^(k+1)  e^((i(k+1)π)/2) )/(2^(k+1)   2^(k+1) ))  =(i/2^k ) e^((i(k+1)π)/2)  ⇒  f^((n)) (0) =(1/4) Σ_(k=0) ^n   (−1)^k  k!(−3)^(n−k)  C_n ^k   (1/2^k ) e^(i(((k+1)π)/2))   and f(x) =Σ_(n=0) ^∞   ((f^((n)) (0))/(n!)) x^n
f(x)=n=0f(n)(0)n!xnletfindf(n)(0)f(x)=e3x(x2i)(x+2i)=e3x4i{1x2i1x+2i}=14i{e3xx2ie3xx+2i}f(n)(x)=14i{(e3xx2i)(n)(e3xx+3i)(n)}butleibnizformulagive(e3xx2i)(n)=k=0nCnk(1x2i)(k)(e3x)nk)=k=0nCnk(1)kk!(x2i)k+1(3)nke3xalso(e3xx+2i)(n)=k=0nCnk(1)kk!(x+2i)k+1(3)nke3xf(n)(x)=14ik=0n(1)kk!(3)nkCnk{1(x2i)k+11(x+2i)k+1}f(n)(0)=14ik=0n(1)kk!(3)nkCnn{1(2i)k+11(2i)k+1}but1(2i)k+1+1(2i)k+1=(2i)k+1(2i)k+14k+1=2iIm((2i)k+1)4k+1=2i2k+1ei(k+1)π22k+12k+1=i2kei(k+1)π2f(n)(0)=14k=0n(1)kk!(3)nkCnk12kei(k+1)π2andf(x)=n=0f(n)(0)n!xn

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