let f(x)=(e^(−x) /(x^2 +4)) 1) calculate f^((n)) (x) and f^((n)) (0) 2) developp f at integr serie .


Question and Answers Forum

All Questions Topic List
Relation and Functions Questions
Previous in All Question Next in All Question
Previous in Relation and Functions Next in Relation and Functions
Question Number 46424 by maxmathsup by imad last updated on 25/Oct/18

$l e t f (x) = \frac{e^{- x}}{x^{2} + 4}$ $1) c a l c u l a t e f^{(n)} (x) a n d f^{(n)} (0)$ $2) d e v e l o p p f a t i n t e g r s e r i e .$
Commented by maxmathsup by imad last updated on 27/Oct/18
$1) we have f(x)=(e^(−x) /((x−2i)(x+2i))) =(1/(4i))e^(−x) { (1/(x−2i)) −(1/(x+2i))} =(1/(4i)){(e^(−x) /(x−2i)) −(e^(−x) /(x+2i))} ⇒f^((n)) (x)=(1/(4i)){ ((e^(−x) /(x−2i)))^((n)) −((e^(−x) /(x+2i)))^((n)) } leibniz formulae give ((e^(−x) /(x−2i)))^((n)) =Σ_(k=0) ^n C_n ^k ((1/(x−2i)))^((k)) (e^(−x) )^((n−k)) =Σ_(k=0) ^n C_n ^k (((−1)^k k!)/((x−2i)^(k+1) )) (−1)^(n−k) e^(−x ) also we have ((e^(−x) /(x+2i)))^((n)) =Σ_(k=0) ^n C_n ^k (((−1)^k k!)/((x+2i)^(k+1) )) (−1)^(n−k) e^(−x) ⇒ f^((n)) (x)=(1/(4i)) Σ_(k=0) ^n (−1)^n C_n ^k k!{(1/((x−2i)^(k+1) )) −(1/((x+2i)^(k+1) ))}e^(−x) =(1/(4i)) Σ_(k=0) ^n (−1)^n ((n!)/((n−k)!)) (((x+2i)^(k+1) −(x−2i)^(k+1) )/((x^2 +4)^(k+1) )) e^(−x)$
$1) w e h a v e f (x) = \frac{e^{- x}}{(x - 2 i) (x + 2 i)} = \frac{1}{4 i} e^{- x} {\frac{1}{x - 2 i} - \frac{1}{x + 2 i}}$ $= \frac{1}{4 i} {\frac{e^{- x}}{x - 2 i} - \frac{e^{- x}}{x + 2 i}} \Rightarrow f^{(n)} (x) = \frac{1}{4 i} {{(\frac{e^{- x}}{x - 2 i})}^{(n)} - {(\frac{e^{- x}}{x + 2 i})}^{(n)}}$ $l e i b n i z f o r m u l a e g i v e$ ${(\frac{e^{- x}}{x - 2 i})}^{(n)} = \sum_{k = 0}^{n} C_{n}^{k} {(\frac{1}{x - 2 i})}^{(k)} {(e^{- x})}^{(n - k)}$ $= \sum_{k = 0}^{n} C_{n}^{k} \frac{{(- 1)}^{k} k!}{{(x - 2 i)}^{k + 1}} {(- 1)}^{n - k} e^{- x} a l s o w e h a v e$ ${(\frac{e^{- x}}{x + 2 i})}^{(n)} = \sum_{k = 0}^{n} C_{n}^{k} \frac{{(- 1)}^{k} k!}{{(x + 2 i)}^{k + 1}} {(- 1)}^{n - k} e^{- x} \Rightarrow$ $f^{(n)} (x) = \frac{1}{4 i} \sum_{k = 0}^{n} {(- 1)}^{n} C_{n}^{k} k! {\frac{1}{{(x - 2 i)}^{k + 1}} - \frac{1}{{(x + 2 i)}^{k + 1}}} e^{- x}$ $= \frac{1}{4 i} \sum_{k = 0}^{n} {(- 1)}^{n} \frac{n!}{(n - k)!} \frac{{(x + 2 i)}^{k + 1} - {(x - 2 i)}^{k + 1}}{{(x^{2} + 4)}^{k + 1}} e^{- x}$
Commented by maxmathsup by imad last updated on 27/Oct/18
$x=0 ⇒f^((n)) (0) =(1/(4i)) Σ_(k=0) ^n (((−1)^n n!)/((n−k)!)) (((2i)^(k+1) −(−2i)^(k+1) )/4^(k+1) ) =(1/(4i)) Σ_(k=0) ^n (((−1)^n n!)/((n−k)!)) ((2i Im{(2i)^(k+1) })/4^(k+1) ) =(1/2) Σ_(k=0) ^n (((−1)^n n!)/((n−k)!)) ((2^(k+1) sin((((k+1)π)/2)))/2^(2k+2) ) = Σ_(k=0) ^n (((−1)^n n!)/((n−k)!)) ((sin((((k+1)π)/2)))/2^(k+2) )$
$x = 0 \Rightarrow f^{(n)} (0) = \frac{1}{4 i} \sum_{k = 0}^{n} \frac{{(- 1)}^{n} n!}{(n - k)!} \frac{{(2 i)}^{k + 1} - {(- 2 i)}^{k + 1}}{4^{k + 1}}$ $= \frac{1}{4 i} \sum_{k = 0}^{n} \frac{{(- 1)}^{n} n!}{(n - k)!} \frac{2 i I m {{(2 i)}^{k + 1}}}{4^{k + 1}} = \frac{1}{2} \sum_{k = 0}^{n} \frac{{(- 1)}^{n} n!}{(n - k)!} \frac{2^{k + 1} s i n (\frac{(k + 1) π}{2})}{2^{2 k + 2}}$ $= \sum_{k = 0}^{n} \frac{{(- 1)}^{n} n!}{(n - k)!} \frac{s i n (\frac{(k + 1) π}{2})}{2^{k + 2}}$
Commented by maxmathsup by imad last updated on 27/Oct/18

$2) w e h a v e f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} \Rightarrow$ $f (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} (\sum_{k = 0}^{n} \frac{1}{(n - k)!} \frac{s i n (\frac{(k + 1) π}{2})}{2^{k + 2}}) x^{n} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum