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Question Number 98943 by mathmax by abdo last updated on 17/Jun/20

let f (x) = \frac{1}{{(1 + x^{2})}^{3}} developp f at integr serie

Answered by maths mind last updated on 17/Jun/20

f (x) = \frac{1}{(1 + x^{2})}

f^{'} (x) = \frac{- 2 x}{{(1 + x^{2})}^{2}}

f^{″} (x) = \frac{- 2}{{(1 + x^{2})}^{2}} + \frac{8 x^{2}}{{(1 + x^{2})}^{3}}

f^{″} (x) = \frac{f^{'} (x)}{x} + 8 x^{2} . \frac{1}{{(1 + x^{2})}^{3}}

f (x) = Σ {(- 1)}^{n} x^{2 n}

f^{'} (x) = \sum_{n ⩾ 1} 2 n {(- 1)}^{n} x^{2 n - 1}

f^{″} (x) = \sum_{n ⩾ 1} . 2 n (2 n - 1) {(- 1)}^{n} x^{2 n - 2}

\frac{1}{{(1 + x^{2})}^{3}} = \frac{1}{8 x^{2}} {\sum_{n ⩾ 0} 2 (n + 1) (2 n + 1) {(- 1)}^{n + 1} x^{2 n} - \sum_{n ⩾ 0} (2 n + 2) {(- 1)}^{n + 1} x^{2 n}}

= \frac{1}{2} . \sum_{n ⩾ 1} {(- 1)}^{n + 1} (n^{2} + n) x^{2 n - 2}

= \frac{1}{2} \sum_{n ⩾ 0} {(- 1)}^{n} (n^{2} + 3 n + 2) x^{2 n}

Commented by mathmax by abdo last updated on 17/Jun/20

thanks sir .

Answered by mathmax by abdo last updated on 17/Jun/20

we have \frac{1}{1 + x^{2}} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n} by derivation we get

- \frac{2 x}{{(1 + x^{2})}^{2}} = \sum_{n = 1}^{\infty} 2 n {(- 1)}^{n} x^{2 n - 1} \Rightarrow \frac{- 2}{{(1 + x^{2})}^{2}} = \sum_{n = 1}^{\infty} 2 n {(- 1)}^{n} x^{2 n - 2}

\Rightarrow 2 \times \frac{2 (2 x) (1 + x^{2})}{{(1 + x^{2})}^{4}} = \sum_{n = 2}^{\infty} 2 n (2 n - 2) {(- 1)}^{n} x^{2 n - 3} \Rightarrow

\frac{8 x}{{(1 + x^{2})}^{3}} = \sum_{n = 2}^{\infty} 2 n (2 n - 2) {(- 1)}^{n} x^{2 n - 3} \Rightarrow

\frac{4}{{(1 + x^{2})}^{3}} = \sum_{n = 2}^{\infty} n (2 n - 2) {(- 1)}^{n} x^{2 n - 4}

= \sum_{n = 0}^{\infty} (n + 2) (2 n + 2) {(- 1)}^{n + 2} x^{2 n} = 2 \sum_{n = 0}^{\infty} (n + 1) (n + 2) {(- 1)}^{n} x^{2 n} \Rightarrow

\frac{1}{{(1 + x^{2})}^{3}} = \frac{1}{2} \sum_{n = 0}^{\infty} (n^{2} + 3 n + 2) {(- 1)}^{n} x^{2 n} .