let-g-x-1-1-x-4-1-find-g-n-x-2-calculate-g-n-0-3-if-g-x-u-n-x-n-find-the-sequence-u-n-

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

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}}) \dots

b e c o n t i n u e d \dots