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Question Number 37282 by abdo.msup.com last updated on 11/Jun/18

l e t f (x) = \frac{x}{1 + x^{2} + x^{4}}

1) f i n d f^{(n)} (x)

2) c a l c u l a t e f^{(n)} (0)

3) d e v e l o p p f a t i n t e g r s e r i e .

Commented by abdo.msup.com last updated on 17/Jun/18

f^{(n)} (0) = \frac{{(- 1)}^{n} n!}{\sqrt{3}} ({(- 1)}^{n} - 1)) s i n ((n + 1) \frac{π}{3}

Commented by abdo.msup.com last updated on 17/Jun/18

\Rightarrow f^{(2 n)} (0) = 0 a n d

f^{(2 n + 1)} (0) = \frac{2 (2 n + 1)!}{\sqrt{3}} s i n {(2 n + 2) \frac{π}{3}}

3) f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}

= \sum_{n = 0}^{\infty} \frac{f^{(2 n + 1)} (0)}{(2 n + 1)!} x^{2 n + 1}

= \sum_{n = 0}^{\infty} \frac{2}{\sqrt{3}} s i n {(2 n + 2) \frac{π}{3}} x^{2 n + 1} .

Commented by prof Abdo imad last updated on 17/Jun/18

1) p o l e s o f f ?

1 + x^{2} + x^{4} = 0 \Rightarrow z^{2} + z + 1 = 0 (z = x^{2}) \Rightarrow

z = j o r z = \bar{j}

x^{2} = e^{i \frac{2 π}{3}} \Rightarrow x = \bar{+} e^{\frac{i π}{3}}

x^{2} = e^{- \frac{i 2 π}{3}} \Rightarrow x = \bar{+} e^{- \frac{i π}{3}} s o

f (x) = \frac{x}{(x - e^{\frac{i π}{3}}) (x + e^{\frac{i π}{3}}) (x - e^{- \frac{i π}{3}}) (x + e^{- \frac{i π}{3}})}

= \frac{a}{x - e^{\frac{i π}{3}}} + \frac{b}{x + e^{\frac{i π}{3}}} + \frac{c}{x - e^{- \frac{i π}{3}}} + \frac{d}{x + e^{- \frac{i π}{3}}}

b u t f (z_{k}) = \frac{z_{k}}{4 z_{k}^{3} + 2 z_{k}} = \frac{1}{4 z_{k}^{2} + 2}

a = \frac{1}{4 e^{\frac{2 i π}{3}} + 2} = \frac{1}{4 (- \frac{1}{2} + i \frac{\sqrt{3}}{2}) + 2} = \frac{1}{2 i \sqrt{3}}

b = \frac{1}{4 e^{\frac{2 i π}{3}} + 2} = \frac{1}{2 i \sqrt{3}}

c = \frac{1}{4 e^{- \frac{2 i π}{3}} + 2} = \frac{1}{4 (- \frac{1}{2} - i \frac{\sqrt{3}}{2}) + 2} = - \frac{1}{2 i \sqrt{3}}

d = - \frac{1}{2 i \sqrt{3}} \Rightarrow

f (x) = \frac{1}{2 i \sqrt{3}} {\frac{1}{x - e^{\frac{i π}{3}}} + \frac{1}{x + e^{\frac{i π}{3}}} - \frac{1}{x - e^{- \frac{i π}{3}}} - \frac{1}{x + e^{- \frac{i π}{3}}}}

f^{(n)} (x) = \frac{{(- 1)}^{n} n!}{2 i \sqrt{3}} {\frac{1}{{(x - e^{\frac{i π}{3}})}^{n + 1}} + \frac{1}{{(x + e^{\frac{i π}{3}})}^{n + 1}}

- \frac{1}{{(x - e^{- \frac{i π}{3}})}^{n + 1}} - \frac{1}{{(x + e^{- \frac{i π}{3}})}^{n + 1}}}

Commented by prof Abdo imad last updated on 17/Jun/18

2) f^{(n)} (0) = \frac{{(- 1)}^{n} n!}{2 i \sqrt{3}} {\frac{{(- 1)}^{n + 1}}{e^{\frac{i (n + 1) π}{3}}} + \frac{1}{e^{\frac{i (n + 1) π}{3}}}

- \frac{{(- 1)}^{n + 1}}{e^{- \frac{i (n + 1) π}{3}}} - \frac{1}{e^{- \frac{i (n + 1) π}{3}}}}

= \frac{{(- 1)}^{n} n!}{2 i \sqrt{3}} {{(- 1)}^{n + 1} e^{- \frac{i (n + 1) π}{3}} + e^{- \frac{i (n + 1) π}{3}}

- {(- 1)}^{n + 1} e^{\frac{i (n + 1) π}{3}} - e^{\frac{i (n + 1) π}{3}}}

= \frac{{(- 1)}^{n} n!}{2 i \sqrt{3}} {{(- 1)}^{n} 2 i s i n ((n + 1) \frac{π}{3}) - 2 i s i n (n + 1) \frac{π}{3}}

Commented by prof Abdo imad last updated on 17/Jun/18

λ_{k} = \frac{z_{k}}{p^{^{'}} (z_{k})} w i t h p (x) = 1 + x^{2} + x^{4} \Rightarrow

λ_{k} = \frac{z_{k}}{4 z_{k}^{3} + 2 z_{k}} s o c h a n g e f (z_{k}) b y λ_{k}