let f(x)= (x/(x^2 +x+1)) 1) calculate f^((n)) (0) 2) developp f at integr serie .


Question and Answers Forum

All Questions Topic List
Relation and Functions Questions
Previous in All Question Next in All Question
Previous in Relation and Functions Next in Relation and Functions
Question Number 36010 by abdo mathsup 649 cc last updated on 27/May/18

$l e t f (x) = \frac{x}{x^{2} + x + 1}$ $1) c a l c u l a t e f^{(n)} (0)$ $2) d e v e l o p p f a t i n t e g r s e r i e .$
Commented by prof Abdo imad last updated on 28/May/18
$let decompose f(x)inside C(x) the roots of p(x)=x^2 +x+1 are are j =e^(i((2π)/3)) and j^− =e^(−i((2π)/3)) so f(x)= (x/((x−j)(x−j^− ))) = (a/(x−j)) +(b/(x−j^− )) a=lim_(x→j) (x−j)f(x) = (j/(j−j^− )) = (j/(2 ((√3)/2))) = (j/(√3)) b =lim_(x→j^− ) (x−j^− )f(x)= (j^− /(j^− −j)) =−(j^− /(√3)) ⇒ f(x) = (j/((√3)(x−j))) −(j^− /((√3)(x−j^− ))) ⇒ f^((n)) (x) = (j/(√3)){ (((−1)^n n!)/((x−j)^(n+1) ))} −(j^− /(√3)) { (((−1)^n n!)/((x−j^− )^(n+1) ))} f^((n)) (x) = (((−1)^n n!)/(√3)){ (j/((x−j)^(n+1) )) −(j^− /((x−j^− )^(n+1) ))} =(((−1)^n n!)/(√3)){ ((j(x−j^− )^(n+1) −j^− (x−j)^(n+1) )/((x^2 +x+1)^(n+1) ))}$
$l e t d e c o m p o s e f (x) i n s i d e C (x) t h e r o o t s o f$ $p (x) = x^{2} + x + 1 a r e a r e j = e^{i \frac{2 π}{3}} a n d \bar{j} = e^{- i \frac{2 π}{3}} s o$ $f (x) = \frac{x}{(x - j) (x - \bar{j})} = \frac{a}{x - j} + \frac{b}{x - \bar{j}}$ $a = {l i m}_{x \to j} (x - j) f (x) = \frac{j}{j - \bar{j}} = \frac{j}{2 \frac{\sqrt{3}}{2}} = \frac{j}{\sqrt{3}}$ $b = {l i m}_{x \to \bar{j}} (x - \bar{j}) f (x) = \frac{\bar{j}}{\bar{j} - j} = - \frac{\bar{j}}{\sqrt{3}} \Rightarrow$ $f (x) = \frac{j}{\sqrt{3} (x - j)} - \frac{\bar{j}}{\sqrt{3} (x - \bar{j})} \Rightarrow$ $f^{(n)} (x) = \frac{j}{\sqrt{3}} {\frac{{(- 1)}^{n} n!}{{(x - j)}^{n + 1}}} - \frac{\bar{j}}{\sqrt{3}} {\frac{{(- 1)}^{n} n!}{{(x - \bar{j})}^{n + 1}}}$ $f^{(n)} (x) = \frac{{(- 1)}^{n} n!}{\sqrt{3}} {\frac{j}{{(x - j)}^{n + 1}} - \frac{\bar{j}}{{(x - \bar{j})}^{n + 1}}}$ $= \frac{{(- 1)}^{n} n!}{\sqrt{3}} {\frac{j {(x - \bar{j})}^{n + 1} - \bar{j} {(x - j)}^{n + 1}}{{(x^{2} + x + 1)}^{n + 1}}}$
Commented by prof Abdo imad last updated on 28/May/18
$so f^((n)) (0) = (((−1)^n n!)/(√3)){ ((j(−j^− )^(n+1) −j^− (−j)^(n+1) )/1)} =(((−1)^n n!)/(√3)){ (−1)^(n+1) j(j^− )^(n+1) −(−1)^(n+1) j^− j^(n+1) } =(((−1)^n n!)/(√3)) (−1)^(n+1) { j^−^n −j^n } = (1/(√3)){ 2Im(j^n )) = (2/(√3)) sin(((2nπ)/3)) f^((n)) (0) = (2/(√3))sin(((2nπ)/3)). 2) we have f(x) = Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n ⇒ f(x) = Σ_(n=0) ^∞ (2/(√3)) ((sin(((2nπ)/3)))/(n!)) x^n . f(x) = (2/(√3)) Σ_(n=0) ^∞ ((sin(((2nπ)/3)))/(n!)) x^n .$
$s o f^{(n)} (0) = \frac{{(- 1)}^{n} n!}{\sqrt{3}} {\frac{j {(- \bar{j})}^{n + 1} - \bar{j} {(- j)}^{n + 1}}{1}}$ $= \frac{{(- 1)}^{n} n!}{\sqrt{3}} {{(- 1)}^{n + 1} j {(\bar{j})}^{n + 1} - {(- 1)}^{n + 1} \bar{j} j^{n + 1}}$ $= \frac{{(- 1)}^{n} n!}{\sqrt{3}} {(- 1)}^{n + 1} {\overset{-^{n}}{j} - j^{n}}$ $= \frac{1}{\sqrt{3}} {2 I m (j^{n})) = \frac{2}{\sqrt{3}} s i n (\frac{2 n π}{3})$ $f^{(n)} (0) = \frac{2}{\sqrt{3}} s i n (\frac{2 n π}{3}) .$ $2) w e h a v e f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} \Rightarrow$ $f (x) = \sum_{n = 0}^{\infty} \frac{2}{\sqrt{3}} \frac{s i n (\frac{2 n π}{3})}{n!} x^{n} .$ $f (x) = \frac{2}{\sqrt{3}} \sum_{n = 0}^{\infty} \frac{s i n (\frac{2 n π}{3})}{n!} x^{n} .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum