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let-f-x-artan-x-1-1-2x-developp-f-at-integr-serie-

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Question Number 34863 by a.i msup by abdo last updated on 12/May/18
let f(x)= ((artan(x+1))/(1+2x)) developp f at integr serie .
letf(x)=artan(x+1)1+2xdeveloppfatintegrserie.
Commented by math khazana by abdo last updated on 13/May/18
for ∣x∣<(1/2) we have (1/(1+2x)) = Σ_(n=0) ^∞ (−2x)^n =Σ_(n=0) ^∞ (−2)^n x^n let put w(x)=arctan(x+1) w^′ (x) = (1/(1+(x+1)^2 )) = (1/((x+1)^2 −i^2 )) = (1/((x+1−i)(x+1+i))) =(1/(2i)) ( (1/(x+1−i)) −(1/(x+1+i))) ⇒ ⇒w^((n+1)) (x) =(1/(2i)){ ((1/(x+1−i)))^((n)) −((1/(x+1+i)))^((n)) } =(1/(2i)) (((−1)^n n!)/((x+1−i)^(n+1) )) −(1/(2i)) (((−1)^n n!)/((x+1+i)^(n+1) )) ⇒ w^((n+1)) (0) = (((−1)^n n!)/(2i(1−i)^(n+1) )) −(((−1)^n n!)/(2i(1+i)^(n+1) )) ⇒w^((n)) (0)= (((−1)^(n−1) (n−1)!)/(2i)){ (((1+i)^n −(1−i)^n )/2^n )} = (((−1)^(n−1) (n−1)!)/(2i .2^n )) 2iIm(1+i)^n =(((−1)^(n−1) (n−1)!)/2^n ) ((√2))^n sin(n(π/4)) we have w(x)= Σ_(n=0) ^∞ ((w^((n)) (0))/(n!)) x^n = 1+Σ_(n=1) ^∞ (((−1)^(n−1) (n−1)! ((√2))^n )/(2^n n!)) sin(n(π/4))x^n f(x)=( Σ_(n=0) ^∞ a_n x^n ) (Σ_(n=0) ^∞ b_n x^n ) =Σ c_n x^n with c_n = Σ_(i+j=n) a_i bj so tbe developpement of f(x) is known .
for∣x∣<12wehave11+2x=∑n=0∞(−2x)n=∑n=0∞(−2)nxnletputw(x)=arctan(x+1)w′(x)=11+(x+1)2=1(x+1)2−i2=1(x+1−i)(x+1+i)=12i(1x+1−i−1x+1+i)⇒⇒w(n+1)(x)=12i{(1x+1−i)(n)−(1x+1+i)(n)}=12i(−1)nn!(x+1−i)n+1−12i(−1)nn!(x+1+i)n+1⇒w(n+1)(0)=(−1)nn!2i(1−i)n+1−(−1)nn!2i(1+i)n+1⇒w(n)(0)=(−1)n−1(n−1)!2i{(1+i)n−(1−i)n2n}=(−1)n−1(n−1)!2i.2n2iIm(1+i)n=(−1)n−1(n−1)!2n(2)nsin(nπ4)wehavew(x)=∑n=0∞w(n)(0)n!xn=1+∑n=1∞(−1)n−1(n−1)!(2)n2nn!sin(nπ4)xnf(x)=(∑n=0∞anxn)(∑n=0∞bnxn)=Σcnxnwithcn=∑i+j=naibjsotbedeveloppementoff(x)isknown.

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