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Question Number 94338 by mathmax by abdo last updated on 18/May/20

d e v e l o p p a t i n t e r g r s e r i e f (x) = \frac{1}{(x + 3) (x^{2} + 4)}

Answered by mathmax by abdo last updated on 18/May/20

fist let find f^{(n)} (0), f (x) = \frac{1}{(x + 3) (x - 2 i) (x + 2 i)} = \frac{a}{x + 3} + \frac{b}{x - 2 i} + \frac{c}{x + 2 i}

a = \frac{1}{13} b = \frac{1}{(2 i + 3) 4 i} and c = \frac{1}{- 4 i (- 2 i + 3)} \Rightarrow

f (x) = \frac{1}{13 (x + 3)} + \frac{1}{i (8 i + 12) (x - 2 i)} + \frac{1}{4 i (2 i - 3) (x + 2 i)} \Rightarrow

f^{(n)} (x) = \frac{1}{13} \times \frac{{(- 1)}^{n} n!}{{(x + 3)}^{n + 1}} - \frac{i}{8 i + 12} \times \frac{{(- 1)}^{n} n!}{{(x - 2 i)}^{n + 1}} - \frac{i {(- 1)}^{n} n!}{(8 i - 12) {(x + 2 i)}^{n + 1}}

\Rightarrow f^{(n)} (0) = \frac{1}{13} \times \frac{{(- 1)}^{n} n!}{3^{n + 1}} - \frac{i}{8 i + 12} \frac{{(- 1)}^{n} n!}{{(- 2 i)}^{n + 1}} - \frac{i {(- 1)}^{n} n!}{(8 i - 12) {(2 i)}^{n + 1}}

f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} \Rightarrow f (x) = \frac{1}{13} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{3^{n + 1}} x^{n}

- \frac{i}{8 i + 12} Σ \frac{{(- 1)}^{n}}{{(- 2 i)}^{n + 1}} x^{n} - \frac{i}{8 i - 12} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 i)}^{n + 1}} x^{n}