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Question Number 38521 by math khazana by abdo last updated on 26/Jun/18

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

1) c a l c u l a t e f^{(n)} (x)

2) f i n d 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 .

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Commented by abdo mathsup 649 cc last updated on 28/Jun/18

1) l e t d r c o m p o s e f i n s i d e C

f (x) = \frac{a}{x - 2} + \frac{b}{x - j} + \frac{c}{x - \bar{j}} w i t h j = e^{\frac{i 2 π}{3}}

f (x) = \frac{2 x + 1}{(x - 2) (x - j) (x - \bar{j})}

a = {l i m}_{x \to 2} (x - 2) f (x) = \frac{5}{7}

b = {l i m}_{x \to j} (x - j) f (x) = \frac{2 j + 1}{(j - 2) 2 i \frac{\sqrt{3}}{2}} = \frac{2 j + 1}{i \sqrt{3} (j - 2)}

c = {l i m}_{x \to \bar{j}} (x - \bar{j}) f (x) = \frac{2 \bar{j} + 1}{(\bar{j} - 2) (- 2 i \frac{\sqrt{3}}{2})}

= - \frac{2 \bar{j} + 1}{i \sqrt{3} (\bar{j} - 2)} \Rightarrow f (x) = \frac{5}{7 (x - 2)} + \frac{2 j + 1}{i \sqrt{3} (j - 2) (x - j)}

- \frac{2 \bar{j} + 1}{i \sqrt{3} (\bar{j} - 2) (x - \bar{j})} \Rightarrow

f^{(n)} (x) = \frac{5}{7} \frac{{(- 1)}^{n} n!}{{(x - 2)}^{n + 1}} + \frac{2 j + 1}{i \sqrt{3} (j - 2)} \frac{{(- 1)}^{n} n!}{{(x - j)}^{n + 1}}

- \frac{2 \bar{j} + 1}{i \sqrt{3} (\bar{j} - 2)} \frac{{(- 1)}^{n} n!}{{(x - \bar{j})}^{n + 1}} b u t

2) f^{(n)} (0) = \frac{5 {(- 1)}^{n} n!}{7 {(- 2)}^{n + 1}} + \frac{2 j + 1}{i \sqrt{3} (j - 2)} \frac{{(- 1)}^{n} n!}{{(- j)}^{n + 1}}

- \frac{2 \bar{j} + 1}{i \sqrt{3} (\bar{j} - 2)} \frac{{(- 1)}^{n} n!}{{(- \bar{j})}^{n + 1}}

= - \frac{5}{7} \frac{n!}{2^{n + 1}} - \frac{2 j + 1}{i \sqrt{3} (j - 2)} \frac{n!}{j^{n + 1}} + \frac{2 \bar{j} + 1}{i \sqrt{3} (\bar{j} - 2)} \frac{n!}{{(\bar{j})}^{n + 1}}

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

= \sum_{n = 0}^{\infty} {- \frac{5}{7 . 2^{n + 1}} - \frac{2 j + 1}{i \sqrt{3} (j - 2) j^{n + 1}} + \frac{2 \bar{j} + 1}{i \sqrt{3} (\bar{j} - 2) {(\bar{j})}^{n + 1}}} x^{n}