Calculer les de^� rive^� es n-ie^� mes en 0 de la fonction de^� finie par f(x)=(x^2 /(1+x^4 ))


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Question Number 129815 by stelor last updated on 19/Jan/21

$Calculer les d \overset{´}{e r i v} \overset{´}{e e s} n - i \overset{`}{e m e s} en 0 de$ $la fonction d \overset{´}{e f i n i e} par f (x) = \frac{x^{2}}{1 + x^{4}}$
	Answered by mathmax by abdo last updated on 19/Jan/21
	$first let decompose f(x)=(x^2 /(x^4 +1)) z^4 +1=0 ⇒z^4 =−1=e^(i(2k+1)π) ⇒z_k =e^((i(2k+1)π)/4) and k∈[[0,3]] f(x)=(x^2 /(Π_(k=0) ^3 (x−z_k ))) =Σ_(k=0) ^3 (a_k /(x−z_k )) a_k =(z_k ^2 /(4z_k ^3 )) =(1/(4z_k )) ⇒f(x)=(1/4)Σ_(k=0) ^3 (1/(z_k (x−z_k )))=(1/4)Σ_(k=0) ^3 (e^(−i(((2k+1)π)/4)) /(x−z_k )) ⇒ f^((n)) (x)=(1/4)Σ_(k=0) ^3 e^(−((k(2k+1)π)/4)) ×(((−1)^n n!)/((x−z_k )^(n+1) )) ⇒ f^((n)) (0) =(1/4)Σ_(k=0) ^3 e^(−((i(2k+1)π)/4)) (((−1)^n n!)/((−1)^(n+1) z_k ^(n+1) )) =−(1/4)Σ_(k=0) ^3 ((n!)/z_k ^(n+2) ) =−((n!)/4)Σ_(k=0) ^3 (e^(−((i(2k+1)π)/4)) )^(n+2) =−((n!)/4)Σ_(k=0) ^3 e^(−((i(2k+1)(n+2)π)/4)) =−((n!)/4){e^(−i(n+2)(π/4)) +e^(−((3i(n+2)π)/4)) +e^(−((5i(n+2)π)/4)) + e^(−((7i(n+2)π)/4)) }$
	$first let decompose f (x) = \frac{x^{2}}{x^{4} + 1}$ $z^{4} + 1 = 0 \Rightarrow z^{4} = - 1 = e^{i (2 k + 1) π} \Rightarrow z_{k} = e^{\frac{i (2 k + 1) π}{4}} and k \in [[0, 3]]$ $f (x) = \frac{x^{2}}{\prod_{k = 0}^{3} (x - z_{k})} = \sum_{k = 0}^{3} \frac{a_{k}}{x - z_{k}}$ $a_{k} = \frac{z_{k}^{2}}{{4 z}_{k}^{3}} = \frac{1}{{4 z}_{k}} \Rightarrow f (x) = \frac{1}{4} \sum_{k = 0}^{3} \frac{1}{z_{k} (x - z_{k})} = \frac{1}{4} \sum_{k = 0}^{3} \frac{e^{- i \frac{(2 k + 1) π}{4}}}{x - z_{k}} \Rightarrow$ $f^{(n)} (x) = \frac{1}{4} \sum_{k = 0}^{3} e^{- \frac{k (2 k + 1) π}{4}} \times \frac{{(- 1)}^{n} n!}{{(x - z_{k})}^{n + 1}} \Rightarrow$ $f^{(n)} (0) = \frac{1}{4} \sum_{k = 0}^{3} e^{- \frac{i (2 k + 1) π}{4}} \frac{{(- 1)}^{n} n!}{{(- 1)}^{n + 1} z_{k}^{n + 1}}$ $= - \frac{1}{4} \sum_{k = 0}^{3} \frac{n!}{z_{k}^{n + 2}} = - \frac{n!}{4} \sum_{k = 0}^{3} {(e^{- \frac{i (2 k + 1) π}{4}})}^{n + 2}$ $= - \frac{n!}{4} \sum_{k = 0}^{3} e^{- \frac{i (2 k + 1) (n + 2) π}{4}}$ $= - \frac{n!}{4} {e^{- i (n + 2) \frac{π}{4}} + e^{- \frac{3 i (n + 2) π}{4}} + e^{- \frac{5 i (n + 2) π}{4}} + e^{- \frac{7 i (n + 2) π}{4}}}$
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