calculateA_n =(1/(2i)) ∫_0 ^1 {(1+ix)^n −(1−ix)^n }dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 49661 by maxmathsup by imad last updated on 08/Dec/18
$calculateA_n =(1/(2i)) ∫_0 ^1 {(1+ix)^n −(1−ix)^n }dx$
${c a l c u l a t e A}_{n} = \frac{1}{2 i} \int_{0}^{1} {{(1 + i x)}^{n} - {(1 - i x)}^{n}} d x$
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
$another method we have A_n =(1/(2i))[(1/(i(n+1)))(1+ix)^(n+1) +(1/(i(n+1)))(1−ix)^(n+1) ]_0 ^1 =((−1)/(2(n+1))){ (1+i)^(n+1) +(1−i)^(n+1) −2} =((−1)/(2(n+1))){ 2Re((1+i)^(n+1) )−2} but 1+i =(√2)e^(i(π/4)) ⇒(1+i)^(n+1) =2^((n+1)/2) e^(i(n+1)(π/4)) ⇒ A_n =−(1/(n+1)){ 2^((n+1)/2) cos(n+1)(π/4) −1} =(1/(n+1)){ 1−2^((n+1)/2) cos((n+1)(π/4))} .$
$a n o t h e r m e t h o d$ $w e h a v e A_{n} = \frac{1}{2 i} {[\frac{1}{i (n + 1)} {(1 + i x)}^{n + 1} + \frac{1}{i (n + 1)} {(1 - i x)}^{n + 1}]}_{0}^{1}$ $= \frac{- 1}{2 (n + 1)} {{(1 + i)}^{n + 1} + {(1 - i)}^{n + 1} - 2}$ $= \frac{- 1}{2 (n + 1)} {2 R e ({(1 + i)}^{n + 1}) - 2} b u t 1 + i = \sqrt{2} e^{i \frac{π}{4}} \Rightarrow {(1 + i)}^{n + 1} = 2^{\frac{n + 1}{2}} e^{i (n + 1) \frac{π}{4}} \Rightarrow$ $A_{n} = - \frac{1}{n + 1} {2^{\frac{n + 1}{2}} c o s (n + 1) \frac{π}{4} - 1}$ $= \frac{1}{n + 1} {1 - 2^{\frac{n + 1}{2}} c o s ((n + 1) \frac{π}{4})} .$
	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

	$w h e r e a r e y o u f r o m s i r s m a i l ?$
	Commented by Smail last updated on 09/Dec/18

	$I a m f r o m M o r o c c o, b u t I l i v e i n t h e U S A .$
	Commented by Smail last updated on 09/Dec/18

	$H o w a b o u t y o u ?$
	Commented by maxmathsup by imad last updated on 09/Dec/18

	$i a m a l s o m o r r o c a n t h e a c h e r f o r t h e h i g h a t l a s b u t i w o r k n o w i n c a s a b l a n c a .$
	Commented by Smail last updated on 09/Dec/18

	$W a w, w h a t a c o i n c i d e n c e .$
	Commented by Smail last updated on 09/Dec/18

	$W h i c h p a r t o f H i g h A t l a s a r e y o u f r o m ?$
	Commented by maxmathsup by imad last updated on 09/Dec/18

	$a n d w h a t d o y o u w o r k i n u s a s i r s m a i l ?$
	Commented by maxmathsup by imad last updated on 09/Dec/18

	$a a m f r o m A z i l a l c i t y .$
	Commented by Smail last updated on 09/Dec/18

	$A n a m e n D e m n a t e$
	Commented by maxmathsup by imad last updated on 09/Dec/18

	$r e a l l y a c o i n c i d e n c e w e a r e f r o m t h e s a m e r e g i o n .$
	Commented by Smail last updated on 09/Dec/18

	$y u p$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum