let f(x) = ((x+2)/(x^3 −4x +3)) 1) calculate f^((n)) (x) 2) developp f at integr serie.


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Question Number 35988 by abdo mathsup 649 cc last updated on 26/May/18

$l e t f (x) = \frac{x + 2}{x^{3} - 4 x + 3}$ $1) c a l c u l a t e f^{(n)} (x)$ $2) d e v e l o p p f a t i n t e g r s e r i e .$
Commented by abdo mathsup 649 cc last updated on 27/May/18
$1) we have x^3 −4x +3 =x^3 −x −3x +3 =x(x^2 −1) −3(x−1) = (x−1)( x^2 +x −3) ⇒ f(x) = ((x+2)/((x−1)(x^2 +x−3))) = (a/(x−1)) +((bx +c)/(x^2 +x−3)) a=lim_(x→1) (x−1)f(x) = (3/(−1)) =−3 lim_(x→+∞) x f(x)=0 = a +b ⇒b=−a = 3 ⇒ f(x) = ((−3)/(x−1)) + ((3x+c)/(x^(2 ) +x −3)) f(0) = (2/3) = 3 −(c/3) ⇒2= 9 −c ⇒ c =7 ⇒ f(x) = ((−3)/(x−1)) +((3x+7)/(x^2 +x−3)) let find the roots of x^2 +x−3 , Δ =1−4(−3) =13 x_(1 ) = ((−1+(√(13)))/2) , x_2 = ((−1−(√(13)))/2) so ((3x+7)/(x^2 +x−3)) = ((3x+7)/((x−x_1 )(x−x_2 ))) =((3(x−x_1 ) +3x_1 +7)/((x−x_1 )(x−x_2 ))) = (3/(x−x_2 )) + ((3x_(1 ) +7)/((x−x_1 )(x−x_2 ))) (3/(x−x_2 )) + ((3x_1 +7)/(x_1 −x_2 )) { (1/(x−x_1 )) −(1/(x−x_2 ))} ⇒ f(x) = −(3/(x−1)) +(3/(x−x_2 )) + (λ/(x−x_1 )) −(λ/(x−x_2 )) with λ = ((3x_(1 ) +7)/(x_1 −x_2 )) ⇒ f^((n)) (x) = −3 (((−1)^n n!)/((x−1)^(n+1) )) + (3−λ)(((−1)^n n!)/((x−x_2 )^(n+1) )) +λ (((−1)^n n!)/((x−x_1 )^(n+1) ))$
$1) w e h a v e x^{3} - 4 x + 3 = x^{3} - x - 3 x + 3$ $= x (x^{2} - 1) - 3 (x - 1) = (x - 1) (x^{2} + x - 3) \Rightarrow$ $f (x) = \frac{x + 2}{(x - 1) (x^{2} + x - 3)} = \frac{a}{x - 1} + \frac{b x + c}{x^{2} + x - 3}$ $a = {l i m}_{x \to 1} (x - 1) f (x) = \frac{3}{- 1} = - 3$ ${l i m}_{x \to + \infty} x f (x) = 0 = a + b \Rightarrow b = - a = 3 \Rightarrow$ $f (x) = \frac{- 3}{x - 1} + \frac{3 x + c}{x^{2} + x - 3}$ $f (0) = \frac{2}{3} = 3 - \frac{c}{3} \Rightarrow 2 = 9 - c \Rightarrow c = 7 \Rightarrow$ $f (x) = \frac{- 3}{x - 1} + \frac{3 x + 7}{x^{2} + x - 3} l e t f i n d t h e r o o t s o f$ $x^{2} + x - 3, Δ = 1 - 4 (- 3) = 13$ $x_{1} = \frac{- 1 + \sqrt{13}}{2}, x_{2} = \frac{- 1 - \sqrt{13}}{2} s o$ $\frac{3 x + 7}{x^{2} + x - 3} = \frac{3 x + 7}{(x - x_{1}) (x - x_{2})}$ $= \frac{3 (x - x_{1}) + 3 x_{1} + 7}{(x - x_{1}) (x - x_{2})} = \frac{3}{x - x_{2}} + \frac{3 x_{1} + 7}{(x - x_{1}) (x - x_{2})}$ $\frac{3}{x - x_{2}} + \frac{3 x_{1} + 7}{x_{1} - x_{2}} {\frac{1}{x - x_{1}} - \frac{1}{x - x_{2}}} \Rightarrow$ $f (x) = - \frac{3}{x - 1} + \frac{3}{x - x_{2}} + \frac{λ}{x - x_{1}} - \frac{λ}{x - x_{2}} w i t h$ $λ = \frac{3 x_{1} + 7}{x_{1} - x_{2}} \Rightarrow$ $f^{(n)} (x) = - 3 \frac{{(- 1)}^{n} n!}{{(x - 1)}^{n + 1}} + (3 - λ) \frac{{(- 1)}^{n} n!}{{(x - x_{2})}^{n + 1}}$ $+ λ \frac{{(- 1)}^{n} n!}{{(x - x_{1})}^{n + 1}}$
Commented by abdo mathsup 649 cc last updated on 27/May/18

$2) w e h a v e f^{(n)} (0) = 3 n! + (3 - λ) \frac{n!}{- x_{2}^{n + 1}}$ $+ λ \frac{n!}{- x_{1}^{n + 1}}$ $f^{(n)} (0) = 3 (n!) + (λ - 3) \frac{n!}{x_{2}^{n + 1}} - λ \frac{n!}{x_{1}^{n + 1}}$ $f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}$ $= \sum_{n = 0}^{\infty} (3 + \frac{λ - 3}{x_{2}^{n + 1}} - \frac{λ}{x_{1}^{n + 1}}) x^{n} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 27/May/18

	$D r = x^{3} - 4 x + 4 - 1$ $= x^{3} - 1 - 4 (x - 1)$ $= (x - 1) (x^{2} + x + 1) - 4 (x - 1)$ $(x - 1) (x^{2} + x + 1 - 4)$ $= (x - 1) (x^{2} + x - 3)$ $\frac{x + 2}{(x - 1) (x^{2} + x - 3)} = \frac{A}{x - 1} + \frac{B x + C}{(x^{2} + x - 3)}$ $x + 2 = A (x^{2} + x - 3) + (x - 1) (B x + C)$ $p u t x = 1$ $3 = - A s o A = - 3$ $x + 2 = - 3 x^{2} - 3 x + 9 + {B x}^{2} + C x - B x - C$ $x + 2 = x^{2} (- 3 + B) + x (- 3 - B + C) + 9 - C$ $- 3 + B = 0 B = 3$ $- 3 - B + C = 1$ $- 3 - 3 + C = 1 C = 7$ $f (x) = \frac{- 3}{x - 1} + \frac{3 x + 7}{x^{2} + x - 3}$ $c o n t f$ $f (x) = u + v$ $\frac{u}{- 3} = {(x - 1)}^{- 1}$ $\frac{u_{1}}{- 3} = (- 1) {(x - 1)}^{- 2}$ $\frac{u_{2}}{- 3} = (- 1) (- 2) {(x - 1)}^{- 3}$ $\frac{u_{3}}{- 3} = (- 1) (- 2) (- 3) {(x - 1)}^{- 4}$ $. . . . .$ $. . . . .$ $\frac{u_{n}}{- 3} = (- 1) (- 2) (- 3) . . . (- n) {(x - 1)}^{- n - 1}$ $\frac{u_{n}}{- 3} = \frac{{(-)}^{n} \times n!}{{(x - 1)}^{n + 1}}$ $u_{n} = (- 3) \times \frac{{(- 1)}^{n} \times n!}{{(x - 1)}^{n + 1}}$ $p l s c h e c k c o n t d$
	Commented by behi83417@gmail.com last updated on 26/May/18

	$d e a r m r t a n m a y a n d m r R a s h e e d!$ $a c o m m e n t h a v e p o s t e d t o Q . 35844$ $p l e a s e c h e c k i t . t h a n k s .$
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