calculateA-n-1-2i-0-1-1-ix-n-1-ix-n-dx-

Question Number 49661 by maxmathsup by imad last updated on 08/Dec/18

{c a l c u l a t e A}_{n} = \frac{1}{2 i} \int_{0}^{1} {{(1 + i x)}^{n} - {(1 - i x)}^{n}} d x

Commented by maxmathsup by imad last updated on 09/Dec/18

w e h a v e f i r s t {(1 + i x)}^{n} - {(1 - i x)}^{n} = \sum_{k = 0}^{n} C_{n}^{k} {(i x)}^{k} - \sum_{k = 0}^{n} C_{n}^{k} {(- i x)}^{k}

= \sum_{k = 0}^{n} C_{n}^{k} (i^{k} - {(- i)}^{k}) x^{k} = \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} (i^{2 p + 1} - {(- i)}^{2 p + 1}) x^{2 p + 1}

= 2 i \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} {(- 1)}^{p} x^{2 p + 1} \Rightarrow

A_{n} = \sum_{p = 0}^{[\frac{n - 1}{2}]} {(- 1)}^{p} C_{n}^{2 p + 1} \int_{0}^{1} x^{2 p + 1} d x = \sum_{p = 0}^{[\frac{n - 1}{2}]} {(- 1)}^{p} \frac{C_{n}^{2 p + 1}}{2 p + 2} .

Commented by maxmathsup by imad last updated on 09/Dec/18

a n o t h e r m e t h o d

w e h a v e A_{n} = \frac{1}{2 i} {[\frac{1}{i (n + 1)} {(1 + i x)}^{n + 1} + \frac{1}{i (n + 1)} {(1 - i x)}^{n + 1}]}_{0}^{1}

= \frac{- 1}{2 (n + 1)} {{(1 + i)}^{n + 1} + {(1 - i)}^{n + 1} - 2}

= \frac{- 1}{2 (n + 1)} {2 R e ({(1 + i)}^{n + 1}) - 2} b u t 1 + i = \sqrt{2} e^{i \frac{π}{4}} \Rightarrow {(1 + i)}^{n + 1} = 2^{\frac{n + 1}{2}} e^{i (n + 1) \frac{π}{4}} \Rightarrow

A_{n} = - \frac{1}{n + 1} {2^{\frac{n + 1}{2}} c o s (n + 1) \frac{π}{4} - 1}

= \frac{1}{n + 1} {1 - 2^{\frac{n + 1}{2}} c o s ((n + 1) \frac{π}{4})} .

Answered by Smail last updated on 09/Dec/18

A_{n} = \frac{1}{2 i} {[\frac{1}{i (n + 1)} {(1 + i x)}^{n + 1} - \frac{1}{- i (n + 1)} {(1 - i x)}^{n + 1}]}_{0}^{1}

= - \frac{1}{2} (\frac{{(1 + i)}^{n + 1}}{n + 1} + \frac{{(1 - i)}^{n + 1}}{n + 1} - \frac{1}{n + 1} - \frac{1}{n + 1})

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

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

= - \frac{1}{(n + 1)} ({(\sqrt{2})}^{n + 1} c o s (\frac{π (n + 1)}{4}) - 2)

Commented by maxmathsup by imad last updated on 09/Dec/18

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y u p

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