let g(x) =((x−1)/(x^2 +x +1)) 1) find g^((n)) (x) 2)calculate g^((n)) (0) 3) developp g at integr serie.


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Question Number 42688 by prof Abdo imad last updated on 31/Aug/18

$l e t g (x) = \frac{x - 1}{x^{2} + x + 1}$ $1) f i n d g^{(n)} (x)$ $2) c a l c u l a t e g^{(n)} (0)$ $3) d e v e l o p p g a t i n t e g r s e r i e .$
Commented by maxmathsup by imad last updated on 02/Sep/18
$1) poles of g? roots of x^2 +x+1 are j=e^(i((2π)/3)) and j^− =e^(−((i2π)/3)) g(x)=((x−1)/((x−j)(x−j^− ))) =(a/(x−j)) +(b/(x−j^− )) a =lim_(x→j) (x−j)g(x) =((j−1)/(j−j^− )) =((j−1)/(2 i((√3)/2))) =((j−1)/(i(√3))) b =lim_(x→j^− ) (x−j^− )g(x) =((j^− −1)/((j^− −j))) =((j^− −1)/(−i(√3))) =((1−j^− )/(i(√3))) ⇒ g(x) = ((j−1)/(i(√3)(x−j))) −((j^− −1)/(i(√3)(x−j^− ))) ⇒ g^((n)) (x)=((j−1)/(i(√3))) (((−1)^n n!)/((x−j)^(n+1) )) −((j^− −1)/(i(√3))) (((−1)^n n!)/((x−j^− )^(n+1) )) =(((−1)^n n!)/(i(√3))){ ((j−1)/((x−j)^(n+1) )) −((j^− −1)/((x−j^− )^(n+1) ))} =(((−1)^n n!)/(i(√3))){ (((j−1)(x−j^− )^(n+1) −(j^− −1)(x−j)^(n+1) )/((x^2 +x+1)^(n+1) ))}$
$1) p o l e s o f g ? r o o t s o f x^{2} + x + 1 a r e j = e^{i \frac{2 π}{3}} a n d \bar{j} = e^{- \frac{i 2 π}{3}}$ $g (x) = \frac{x - 1}{(x - j) (x - \bar{j})} = \frac{a}{x - j} + \frac{b}{x - \bar{j}}$ $a = {l i m}_{x \to j} (x - j) g (x) = \frac{j - 1}{j - \bar{j}} = \frac{j - 1}{2 i \frac{\sqrt{3}}{2}} = \frac{j - 1}{i \sqrt{3}}$ $b = {l i m}_{x \to \bar{j}} (x - \bar{j}) g (x) = \frac{\bar{j} - 1}{(\bar{j} - j)} = \frac{\bar{j} - 1}{- i \sqrt{3}} = \frac{1 - \bar{j}}{i \sqrt{3}} \Rightarrow$ $g (x) = \frac{j - 1}{i \sqrt{3} (x - j)} - \frac{\bar{j} - 1}{i \sqrt{3} (x - \bar{j})} \Rightarrow$ $g^{(n)} (x) = \frac{j - 1}{i \sqrt{3}} \frac{{(- 1)}^{n} n!}{{(x - j)}^{n + 1}} - \frac{\bar{j} - 1}{i \sqrt{3}} \frac{{(- 1)}^{n} n!}{{(x - \bar{j})}^{n + 1}}$ $= \frac{{(- 1)}^{n} n!}{i \sqrt{3}} {\frac{j - 1}{{(x - j)}^{n + 1}} - \frac{\bar{j} - 1}{{(x - \bar{j})}^{n + 1}}}$ $= \frac{{(- 1)}^{n} n!}{i \sqrt{3}} {\frac{(j - 1) {(x - \bar{j})}^{n + 1} - (\bar{j} - 1) {(x - j)}^{n + 1}}{{(x^{2} + x + 1)}^{n + 1}}}$
Commented by maxmathsup by imad last updated on 02/Sep/18
$2) g^((n)) (0) = (((−1)^n n!)/(i(√3))){ (j−1)(−j^− )^(n+1) −(j^− −1)(−j)^(n+1) } =−((n!)/(i(√3))){ (j−1)(j^− )^(n+1) −(j^− −1)j^(n+1) } =((n!)/(i(√3))) 2iIm((j^− −1)(j)^(n+1) ) but (j^− −1)j^(n+1) =( e^(−((i2π)/3)) −1)e^(i((2(n+1)π)/3)) =(−(1/2) −i((√3)/2))(cos(((2(n+1)π)/3))+i sin(((2(n+1)π)/3))) =−(1/2) cos(((2(n+1)π)/3))−(i/2) sin(((2(n+1)π)/3))−((i(√3))/2) cos(((2(n+1)π)/3)) +((√3)/2) cos(((2(n+1)π)/3))sin(((2(n+1)π)/3)) ⇒ g^((n)) (0) =((2n!)/(√3)) {−(1/2) sin(((2(n+1)π)/3))−((√3)/2) cos(((2(n+1)π)/3))} = −((n!)/(√3)){ sin(((2(n+1)π)/3)) +(√3)cos(((2(n+1)π)/3)) .$
$2) g^{(n)} (0) = \frac{{(- 1)}^{n} n!}{i \sqrt{3}} {(j - 1) {(- \bar{j})}^{n + 1} - (\bar{j} - 1) {(- j)}^{n + 1}}$ $= - \frac{n!}{i \sqrt{3}} {(j - 1) {(\bar{j})}^{n + 1} - (\bar{j} - 1) j^{n + 1}} = \frac{n!}{i \sqrt{3}} 2 i I m ((\bar{j} - 1) {(j)}^{n + 1}) b u t$ $(\bar{j} - 1) j^{n + 1} = (e^{- \frac{i 2 π}{3}} - 1) e^{i \frac{2 (n + 1) π}{3}}$ $= (- \frac{1}{2} - i \frac{\sqrt{3}}{2}) (c o s (\frac{2 (n + 1) π}{3}) + i s i n (\frac{2 (n + 1) π}{3}))$ $= - \frac{1}{2} c o s (\frac{2 (n + 1) π}{3}) - \frac{i}{2} s i n (\frac{2 (n + 1) π}{3}) - \frac{i \sqrt{3}}{2} c o s (\frac{2 (n + 1) π}{3})$ $+ \frac{\sqrt{3}}{2} c o s (\frac{2 (n + 1) π}{3}) s i n (\frac{2 (n + 1) π}{3}) \Rightarrow$ $g^{(n)} (0) = \frac{2 n!}{\sqrt{3}} {- \frac{1}{2} s i n (\frac{2 (n + 1) π}{3}) - \frac{\sqrt{3}}{2} c o s (\frac{2 (n + 1) π}{3})}$ $= - \frac{n!}{\sqrt{3}} {s i n (\frac{2 (n + 1) π}{3}) + \sqrt{3} c o s (\frac{2 (n + 1) π}{3}) .$
Commented by maxmathsup by imad last updated on 02/Sep/18
$3) we have g(x) =Σ_(n=0) ^∞ ((g^((n)) (0))/(n!)) x^n ⇒ g(x) =−(1/(√3)) Σ_(n=0) ^∞ { sin(((2(n+1)π)/3)) +(√3)cos(((2(n+1)π)/3))}x^n .$
$3) w e h a v e g (x) = \sum_{n = 0}^{\infty} \frac{g^{(n)} (0)}{n!} x^{n} \Rightarrow$ $g (x) = - \frac{1}{\sqrt{3}} \sum_{n = 0}^{\infty} {s i n (\frac{2 (n + 1) π}{3}) + \sqrt{3} c o s (\frac{2 (n + 1) π}{3})} x^{n} .$
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