let g(x)= (1/(1+x^4 )) 1) find g^((n)) (x) 2) calculate g^((n)) (0) 3) if g(x)=Σ u_n x^n find the sequence u


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 33257 by prof Abdo imad last updated on 14/Apr/18

$l e t g (x) = \frac{1}{1 + x^{4}}$ $1) f i n d g^{(n)} (x)$ $2) c a l c u l a t e g^{(n)} (0)$ $3) i f g (x) = Σ u_{n} x^{n} f i n d t h e s e q u e n c e u_{n}$
Commented by prof Abdo imad last updated on 15/Apr/18
$let decompose g(x) inside C(x) g(x)= (1/((x^2 )^2 −i^2 )) = (1/((x^2 −i)(x^2 +i))) = (1/((x−(√i))(x+(√i))(x −(√(−i)))(x+(√(−i))))) = (1/((x −e^(i(π/4)) )(x +e^(i(π/4)) )(x −e^(−i(π/4)) )( x +e^(−i(π/4)) ))) = (a/(x −e^(i(π/4)) )) +(b/(x +e^(i(π/4)) )) + (c/(x −e^(−i(π/4)) )) + (d/(x + e^(−i(π/4)) )) a = (1/(p^′ ( e^(i(π/4)) ))) with p(x)= 1+x^4 ⇒p^′ (x)=4x^3 if z_k pole p^′ (z_k ) = 4z_k ^3 =((−4)/z_k ) ⇒a =−(e^(i(π/4)) /4) p^′ (−e^(i(π/4)) ) =((−4)/(−e^(i(π/4)) )) = (4/e^(i(π/4)) ) ⇒ b = (1/4) e^(i(π/4)) p^′ ( e^(−i(π/4)) ) =((−4)/e^(−i(π/4)) ) ⇒ c =−(1/4) e^(−i(π/4)) p^′ ( −e^(−i(π/4)) ) = ((−4)/(−e^(−i(π/4)) )) ⇒ d= (1/4) e^(−i(π/4)) g(x) =−(e^(i(π/4)) /(4(x −e^(i(π/4)) ))) +(e^(i(π/4)) /(4( x +e^(i(π/4)) ))) −(e^(−i(π/4)) /(4(x −_ e^(−i(π/4)) ))) + (e^(−i(π/4)) /(4( x +e^(−i(π/4)) ))) ⇒ g^((n)) (x) = (e^(i(π/4)) /4) ( (((−1)^n n!)/((x +e^(i(π/(4{))) )^(n+1) )) −(((−1)^n n!)/((x −e^(i(π/4)) )^(n+1) ))) + (e^(−i(π/4)) /4) ( (((−1)^n n!)/((x + e^(−i(π/4)) )^(n+1) )) − (((−1)^n n!)/((x −e^(−i(π/4)) )^(n+1) ))) g^((n)) (0) =(((−1)^n n! e^(i(π/4)) )/4)( e^(−i(((n+1)π)/4)) −(−1)^(n+1) e^(−i(((n+1)π)/4)) ) + (((−1)^n n! e^(−i(π/4)) )/4)( e^(i(((n+1)π)/4)) −(−1)^(n+1) e^(i(((n+1)π)/4)) )... be continued...$
$l e t d e c o m p o s e g (x) i n s i d e C (x)$ $g (x) = \frac{1}{{(x^{2})}^{2} - i^{2}} = \frac{1}{(x^{2} - i) (x^{2} + i)}$ $= \frac{1}{(x - \sqrt{i}) (x + \sqrt{i}) (x - \sqrt{- i}) (x + \sqrt{- i})}$ $= \frac{1}{(x - e^{i \frac{π}{4}}) (x + e^{i \frac{π}{4}}) (x - e^{- i \frac{π}{4}}) (x + e^{- i \frac{π}{4}})}$ $= \frac{a}{x - e^{i \frac{π}{4}}} + \frac{b}{x + e^{i \frac{π}{4}}} + \frac{c}{x - e^{- i \frac{π}{4}}} + \frac{d}{x + e^{- i \frac{π}{4}}}$ $a = \frac{1}{p^{^{'}} (e^{i \frac{π}{4}})} w i t h p (x) = 1 + x^{4} \Rightarrow p^{^{'}} (x) = 4 x^{3}$ $i f z_{k} p o l e p^{^{'}} (z_{k}) = 4 z_{k}^{3} = \frac{- 4}{z_{k}} \Rightarrow a = - \frac{e^{i \frac{π}{4}}}{4}$ $p^{^{'}} (- e^{i \frac{π}{4}}) = \frac{- 4}{- e^{i \frac{π}{4}}} = \frac{4}{e^{i \frac{π}{4}}} \Rightarrow b = \frac{1}{4} e^{i \frac{π}{4}}$ $p^{^{'}} (e^{- i \frac{π}{4}}) = \frac{- 4}{e^{- i \frac{π}{4}}} \Rightarrow c = - \frac{1}{4} e^{- i \frac{π}{4}}$ $p^{^{'}} (- e^{- i \frac{π}{4}}) = \frac{- 4}{- e^{- i \frac{π}{4}}} \Rightarrow d = \frac{1}{4} e^{- i \frac{π}{4}}$ $g (x) = - \frac{e^{i \frac{π}{4}}}{4 (x - e^{i \frac{π}{4}})} + \frac{e^{i \frac{π}{4}}}{4 (x + e^{i \frac{π}{4}})} - \frac{e^{- i \frac{π}{4}}}{4 (x -_{} e^{- i \frac{π}{4}})}$ $+ \frac{e^{- i \frac{π}{4}}}{4 (x + e^{- i \frac{π}{4}})} \Rightarrow$ $g^{(n)} (x) = \frac{e^{i \frac{π}{4}}}{4} (\frac{{(- 1)}^{n} n!}{{(x + e^{i \frac{π}{4 {}})}^{n + 1}} - \frac{{(- 1)}^{n} n!}{{(x - e^{i \frac{π}{4}})}^{n + 1}})$ $+ \frac{e^{- i \frac{π}{4}}}{4} (\frac{{(- 1)}^{n} n!}{{(x + e^{- i \frac{π}{4}})}^{n + 1}} - \frac{{(- 1)}^{n} n!}{{(x - e^{- i \frac{π}{4}})}^{n + 1}})$ $g^{(n)} (0) = \frac{{(- 1)}^{n} n! e^{i \frac{π}{4}}}{4} (e^{- i \frac{(n + 1) π}{4}} - {(- 1)}^{n + 1} e^{- i \frac{(n + 1) π}{4}})$ $+ \frac{{(- 1)}^{n} n! e^{- i \frac{π}{4}}}{4} (e^{i \frac{(n + 1) π}{4}} - {(- 1)}^{n + 1} e^{i \frac{(n + 1) π}{4}}) . . .$ $b e c o n t i n u e d . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum