let f(x)= (e^(−3x) /(x^2 +4)) developp f at integr serie.


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Question Number 39214 by math khazana by abdo last updated on 03/Jul/18

$l e t f (x) = \frac{e^{- 3 x}}{x^{2} + 4}$ $d e v e l o p p f a t i n t e g r s e r i e .$
Commented by math khazana by abdo last updated on 04/Jul/18
$f(x)=Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n let find f^((n)) (0) f(x)=(e^(−3x) /((x−2i)(x+2i))) =(e^(−3x) /(4i)){ (1/(x−2i)) −(1/(x+2i))} =(1/(4i)){ (e^(−3x) /(x−2i)) −(e^(−3x) /(x+2i))} ⇒ f^((n)) (x)=(1/(4i)){ ( (e^(−3x) /(x−2i)))^((n)) −((e^(−3x) /(x+3i)))^((n)) } but leibniz formula give ((e^(−3x) /(x−2i)))^((n)) =Σ_(k=0) ^n C_n ^k ((1/(x−2i)))^((k)) (e^(−3x) )^(n−k)) =Σ_(k=0) ^n C_n ^k (((−1)^k k!)/((x−2i)^(k+1) )) (−3)^(n−k) e^(−3x ) also ( (e^(−3x) /(x+2i)))^((n)) = Σ_(k=0) ^n C_n ^k (((−1)^k k!)/((x+2i)^(k+1) )) (−3)^(n−k) e^(−3x) f^((n)) (x)=(1/(4i)) Σ_(k=0) ^n (−1)^k k!(−3)^(n−k) C_n ^k { (1/((x−2i)^(k+1) ))−(1/((x+2i)^(k+1) ))} f^((n)) (0)=(1/(4i)) Σ_(k=0) ^n (−1)^k k!(−3)^(n−k) C_n ^n { (1/((−2i)^(k+1) )) −(1/((2i)^(k+1) ))} but (1/((−2i)^(k+1) )) + (1/((2i)^(k+1) )) =(((2i)^(k+1) −(−2i)^(k+1) )/4^(k+1) ) =((2iIm( (2i)^(k+1) ))/4^(k+1) ) = ((2i 2^(k+1) e^((i(k+1)π)/2) )/(2^(k+1) 2^(k+1) )) =(i/2^k ) e^((i(k+1)π)/2) ⇒ f^((n)) (0) =(1/4) Σ_(k=0) ^n (−1)^k k!(−3)^(n−k) C_n ^k (1/2^k ) e^(i(((k+1)π)/2)) and f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n$
$f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} l e t f i n d f^{(n)} (0)$ $f (x) = \frac{e^{- 3 x}}{(x - 2 i) (x + 2 i)} = \frac{e^{- 3 x}}{4 i} {\frac{1}{x - 2 i} - \frac{1}{x + 2 i}}$ $= \frac{1}{4 i} {\frac{e^{- 3 x}}{x - 2 i} - \frac{e^{- 3 x}}{x + 2 i}} \Rightarrow$ $f^{(n)} (x) = \frac{1}{4 i} {{(\frac{e^{- 3 x}}{x - 2 i})}^{(n)} - {(\frac{e^{- 3 x}}{x + 3 i})}^{(n)}} b u t$ $l e i b n i z f o r m u l a g i v e$ ${(\frac{e^{- 3 x}}{x - 2 i})}^{(n)} = \sum_{k = 0}^{n} C_{n}^{k} {(\frac{1}{x - 2 i})}^{(k)} {(e^{- 3 x})}^{n - k)}$ $= \sum_{k = 0}^{n} C_{n}^{k} \frac{{(- 1)}^{k} k!}{{(x - 2 i)}^{k + 1}} {(- 3)}^{n - k} e^{- 3 x} a l s o$ ${(\frac{e^{- 3 x}}{x + 2 i})}^{(n)} = \sum_{k = 0}^{n} C_{n}^{k} \frac{{(- 1)}^{k} k!}{{(x + 2 i)}^{k + 1}} {(- 3)}^{n - k} e^{- 3 x}$ $f^{(n)} (x) = \frac{1}{4 i} \sum_{k = 0}^{n} {(- 1)}^{k} k! {(- 3)}^{n - k} C_{n}^{k} {\frac{1}{{(x - 2 i)}^{k + 1}} - \frac{1}{{(x + 2 i)}^{k + 1}}}$ $f^{(n)} (0) = \frac{1}{4 i} \sum_{k = 0}^{n} {(- 1)}^{k} k! {(- 3)}^{n - k} C_{n}^{n} {\frac{1}{{(- 2 i)}^{k + 1}} - \frac{1}{{(2 i)}^{k + 1}}}$ $b u t \frac{1}{{(- 2 i)}^{k + 1}} + \frac{1}{{(2 i)}^{k + 1}} = \frac{{(2 i)}^{k + 1} - {(- 2 i)}^{k + 1}}{4^{k + 1}}$ $= \frac{2 i I m ({(2 i)}^{k + 1})}{4^{k + 1}} = \frac{2 i 2^{k + 1} e^{\frac{i (k + 1) π}{2}}}{2^{k + 1} 2^{k + 1}}$ $= \frac{i}{2^{k}} e^{\frac{i (k + 1) π}{2}} \Rightarrow$ $f^{(n)} (0) = \frac{1}{4} \sum_{k = 0}^{n} {(- 1)}^{k} k! {(- 3)}^{n - k} C_{n}^{k} \frac{1}{2^{k}} e^{i \frac{(k + 1) π}{2}}$ $a n d f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}$
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