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let-f-x-1-1-x-2-1-calculate-f-n-x-2-find-f-n-0-3-developp-f-x-at-integr-serie-

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Question Number 33256 by prof Abdo imad last updated on 14/Apr/18
let f(x) = (1/(1+x^2 )) 1) calculate f^((n)) (x) 2) find f^((n)) (0) 3) developp f(x) at integr serie.
letf(x)=11+x21)calculatef(n)(x)2)findf(n)(0)3)developpf(x)atintegrserie.
Commented by prof Abdo imad last updated on 15/Apr/18
we have f(x)= (1/((x−i)(x+i))) = (1/(2i))((1/(x−i)) −(1/(x+i))) ⇒ f^((n)) (x)= (1/(2i))( (((−1)^n n!)/((x−i)^(n+1) )) −(((−1)^n n!)/((x+i)^(n+1) ))) = (((−1)^n n!)/(2i)) ( (((x+i)^(n+1) −(x−i)^(n+1) )/((x^2 +1)^(n+1) )) ) ⇒ f^((n)) (0) = (((−1)^n n!)/(2i)) ( i^(n+1) −(−i)^(n+1) ) ⇒ f^((2n)) (0)= (((2n)!)/(2i)) ( i^(2n+1) −(−i)^(2n+1) ) = (((−1)^n (2n)!)/(2i)) (2i) = (−1)^n (2n)! f^((2n+1)) (0) =0 3) f(x)=Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n = Σ_(n=0) ^∞ (−1)^n (((2n)!)/((2n)!)) x^(2n) ⇒f(x) = Σ_(n=0) ^∞ (−1)^n x^(2n) and radius of convergenve is R =1 .
wehavef(x)=1(x−i)(x+i)=12i(1x−i−1x+i)⇒f(n)(x)=12i((−1)nn!(x−i)n+1−(−1)nn!(x+i)n+1)=(−1)nn!2i((x+i)n+1−(x−i)n+1(x2+1)n+1)⇒f(n)(0)=(−1)nn!2i(in+1−(−i)n+1)⇒f(2n)(0)=(2n)!2i(i2n+1−(−i)2n+1)=(−1)n(2n)!2i(2i)=(−1)n(2n)!f(2n+1)(0)=03)f(x)=∑n=0∞f(n)(0)n!xn=∑n=0∞(−1)n(2n)!(2n)!x2n⇒f(x)=∑n=0∞(−1)nx2nandradiusofconvergenveisR=1.

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