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


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Question Number 42689 by prof Abdo imad last updated on 31/Aug/18

$l e t f (x) = \frac{x}{x^{3} - 2 x + 1}$ $1) f i n d D_{f}$ $2) f i n d f^{(n)} (x) t h e n f^{(n)} (0)$ $3) d e v e l o p p f a t i n t e g r s e r i e .$
Commented by maxmathsup by imad last updated on 01/Sep/18
$1) x∈ D_f ⇔ x^3 −2x +1 ≠0 but x^3 −2x +1 =x^3 −1 −2x +2 =(x−1)(x^2 +x+1)−2(x−1) =(x−1)(x^2 +x+1 −2) =(x−1)(x^2 +x−1) roots of x^2 +x−1? Δ=1−4(−1)=5 ⇒α =((−1+(√5))/2) and β =((−1−(√5))/2) ⇒ D_f =R−{1,((−1+(√5))/2) ,((−1−(√5))/2)}$
$1) x \in D_{f} \Leftrightarrow x^{3} - 2 x + 1 \neq 0 b u t$ $x^{3} - 2 x + 1 = x^{3} - 1 - 2 x + 2 = (x - 1) (x^{2} + x + 1) - 2 (x - 1)$ $= (x - 1) (x^{2} + x + 1 - 2) = (x - 1) (x^{2} + x - 1) r o o t s o f x^{2} + x - 1 ?$ $Δ = 1 - 4 (- 1) = 5 \Rightarrow α = \frac{- 1 + \sqrt{5}}{2} a n d β = \frac{- 1 - \sqrt{5}}{2} \Rightarrow$ $D_{f} = R - {1, \frac{- 1 + \sqrt{5}}{2}, \frac{- 1 - \sqrt{5}}{2}}$
Commented by maxmathsup by imad last updated on 01/Sep/18
$2) we have f(x)= (x/((x−1)(x−α)(x−β))) =(a/(x−1)) +(b/(x−α)) +(c/(x−β)) a =lim_(x→1) (x−1)f(x) = (1/((1−α)(1−β))) b =lim_(x→α) (x−α)f(x) = (α/((α−1)(α−β))) c =lim_(x→β) (x−β)f(x) = (β/((β−1)(β−α))) but α =((−1+(√5))/2) and β=((−1−(√5))/2) ⇒ a = (1/((1−((−1+(√5))/2))(1−((−1−(√5))/2)))) = (4/((3−(√5))(3+(√5)))) =(4/4) =1 b = ((−1+(√5))/(2(1−((−1+(√5))/2))(−(√5)))) =((−1+(√5))/((√5)(3−(√5)))) =((−1+(√5))/(3(√5)−5)) . c = ((−1−(√5))/(2(1−((−1−(√5))/2))((√5)))) =((−1−(√5))/((√5)(3−(√5)))) =((−1−(√5))/(3(√5)−5)) ⇒ f(x)= (1/(x−1)) +(1/(3(√5)−5)){ ((−1+(√5))/(x−α)) −((1+(√5))/(x−β))} ⇒ f^((n)) (x) = (((−1)^n n!)/((x−1)^(n+1) )) + ((−1+(√5))/(3(√5)−5)) (((−1)^n n!)/((x−α)^(n+1) )) −((1+(√5))/(3(√5)−5)) (((−1)^n n!)/((x−β)^(n+1) )) ⇒ f^((n)) (0) =−n! +((−1+(√5))/(3(√5)−5)) (((−1)^n n!)/((−α)^(n+1) )) −((1+(√5))/(3(√5)−5)) (((−1)^n n!)/((−β)^(n+1) )) =−n! −((−1+(√5))/(3(√5)−5)) ((n!)/α^(n+1) ) +((1+(√5))/(3(√5)−5)) ((n!)/β^(n+1) ) ⇒ f^((n)) (0) =−n! +((n!)/(3(√5)−5)){ ((1+(√5))/β^(n+1) ) +((1−(√5))/α^(n+1) )}$
$2) w e h a v e f (x) = \frac{x}{(x - 1) (x - α) (x - β)} = \frac{a}{x - 1} + \frac{b}{x - α} + \frac{c}{x - β}$ $a = {l i m}_{x \to 1} (x - 1) f (x) = \frac{1}{(1 - α) (1 - β)}$ $b = {l i m}_{x \to α} (x - α) f (x) = \frac{α}{(α - 1) (α - β)}$ $c = {l i m}_{x \to β} (x - β) f (x) = \frac{β}{(β - 1) (β - α)} b u t α = \frac{- 1 + \sqrt{5}}{2} a n d β = \frac{- 1 - \sqrt{5}}{2} \Rightarrow$ $a = \frac{1}{(1 - \frac{- 1 + \sqrt{5}}{2}) (1 - \frac{- 1 - \sqrt{5}}{2})} = \frac{4}{(3 - \sqrt{5}) (3 + \sqrt{5})} = \frac{4}{4} = 1$ $b = \frac{- 1 + \sqrt{5}}{2 (1 - \frac{- 1 + \sqrt{5}}{2}) (- \sqrt{5})} = \frac{- 1 + \sqrt{5}}{\sqrt{5} (3 - \sqrt{5})} = \frac{- 1 + \sqrt{5}}{3 \sqrt{5} - 5} .$ $c = \frac{- 1 - \sqrt{5}}{2 (1 - \frac{- 1 - \sqrt{5}}{2}) (\sqrt{5})} = \frac{- 1 - \sqrt{5}}{\sqrt{5} (3 - \sqrt{5})} = \frac{- 1 - \sqrt{5}}{3 \sqrt{5} - 5} \Rightarrow$ $f (x) = \frac{1}{x - 1} + \frac{1}{3 \sqrt{5} - 5} {\frac{- 1 + \sqrt{5}}{x - α} - \frac{1 + \sqrt{5}}{x - β}} \Rightarrow$ $f^{(n)} (x) = \frac{{(- 1)}^{n} n!}{{(x - 1)}^{n + 1}} + \frac{- 1 + \sqrt{5}}{3 \sqrt{5} - 5} \frac{{(- 1)}^{n} n!}{{(x - α)}^{n + 1}} - \frac{1 + \sqrt{5}}{3 \sqrt{5} - 5} \frac{{(- 1)}^{n} n!}{{(x - β)}^{n + 1}} \Rightarrow$ $f^{(n)} (0) = - n! + \frac{- 1 + \sqrt{5}}{3 \sqrt{5} - 5} \frac{{(- 1)}^{n} n!}{{(- α)}^{n + 1}} - \frac{1 + \sqrt{5}}{3 \sqrt{5} - 5} \frac{{(- 1)}^{n} n!}{{(- β)}^{n + 1}}$ $= - n! - \frac{- 1 + \sqrt{5}}{3 \sqrt{5} - 5} \frac{n!}{α^{n + 1}} + \frac{1 + \sqrt{5}}{3 \sqrt{5} - 5} \frac{n!}{β^{n + 1}} \Rightarrow$ $f^{(n)} (0) = - n! + \frac{n!}{3 \sqrt{5} - 5} {\frac{1 + \sqrt{5}}{β^{n + 1}} + \frac{1 - \sqrt{5}}{α^{n + 1}}}$
Commented by maxmathsup by imad last updated on 01/Sep/18
$3) we have f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n ⇒ f(x) = Σ_(n=0) ^∞ {−1 + (1/(3(√5)−5))( ((1+(√5))/β^(n+1) ) +((1−(√5))/α^(n+1) ))}x^n .$
$3) w e h a v e f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} \Rightarrow$ $f (x) = \sum_{n = 0}^{\infty} {- 1 + \frac{1}{3 \sqrt{5} - 5} (\frac{1 + \sqrt{5}}{β^{n + 1}} + \frac{1 - \sqrt{5}}{α^{n + 1}})} x^{n} .$
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