let f(x)= e^(−2x) arctanx 1) calculate f^((n)) (x) 2) find f^((n)) (0) 3) developp f at integr serie


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Question Number 35235 by abdo.msup.com last updated on 17/May/18

$l e t f (x) = e^{- 2 x} a r c t a n x$ $1) c a l c u l a t e f^{(n)} (x)$ $2) f i n d f^{(n)} (0)$ $3) d e v e l o p p f a t i n t e g r s e r i e$
Commented by abdo mathsup 649 cc last updated on 18/May/18
$leibniz formula give f^()n)) (x)= Σ_(k=0) ^n C_n ^k (arctanx)^((k)) (e^(−2x) )^((n−k)) we have (arctanx)^((1)) = (1/(1+x^2 )) ⇒ (arctanx)^((k)) = ((1/(1+x^2 )))^((k−1)) =((1/((x−i)(x+i))))^((k−1)) =(1/(2i)){ (1/(x−i)) −(1/(x+i))}^((k−1)) = (1/(2i)){ (((−1)^(k−1) (k−1)!)/((x−i)^k )) −(((−1)^(k−1) (k−1)!)/((x+i)^k ))} =(((−1)^(k−1) (k−1)!)/(2i)) {(((x+i)^k −(x−i)^k )/((x^2 +1)^k ))} let find (e^(−2x) )^((p)) we have (e^(−2x) )^((1)) =−2 e^(−2x) , (e^(−2x) )^((2)) =(−2)^2 e^(−2x) ....(e^(−2x) )^((p)) = (−2)^p e^(−2p) ⇒ f^((n)) (x) =arctanx (−2)^n e^(−2x) +Σ_(k=1) ^n C_n ^k (((−1)^(k−1) (k−1)!)/(2i)){ (((x+i)^k −(x−i)^k )/((x^2 +1)^k ))}(−2)^(n−k) e^(−2x) 2) f^((n)) (0)= Σ_(k=1) ^n C_n ^k (((−1)^(k−1) (k−1)!)/(2i)){((i^k −(−i)^k )/1)}(−2)^(n−k) but i^k −(−i)^k =e^(i((kπ)/2)) −e^(−i((kπ)/2)) =2i sin(((kπ)/2)) ⇒ f^((n)) (0)=Σ_(k=1) ^n C_n ^k (k−1)!(−1)^(k−1) (−2)^(n−k) sin(((kπ)/2))$
$l e i b n i z f o r m u l a g i v e f^{) n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(a r c t a n x)}^{(k)} {(e^{- 2 x})}^{(n - k)}$ $w e h a v e {(a r c t a n x)}^{(1)} = \frac{1}{1 + x^{2}} \Rightarrow$ ${(a r c t a n x)}^{(k)} = {(\frac{1}{1 + x^{2}})}^{(k - 1)} = {(\frac{1}{(x - i) (x + i)})}^{(k - 1)}$ $= \frac{1}{2 i} {\frac{1}{x - i} - \frac{1}{x + i}}^{(k - 1)}$ $= \frac{1}{2 i} {\frac{{(- 1)}^{k - 1} (k - 1)!}{{(x - i)}^{k}} - \frac{{(- 1)}^{k - 1} (k - 1)!}{{(x + i)}^{k}}}$ $= \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} {\frac{{(x + i)}^{k} - {(x - i)}^{k}}{{(x^{2} + 1)}^{k}}}$ $l e t f i n d {(e^{- 2 x})}^{(p)} w e h a v e$ ${(e^{- 2 x})}^{(1)} = - 2 e^{- 2 x}, {(e^{- 2 x})}^{(2)} = {(- 2)}^{2} e^{- 2 x}$ $. . . . {(e^{- 2 x})}^{(p)} = {(- 2)}^{p} e^{- 2 p} \Rightarrow$ $f^{(n)} (x) = a r c t a n x {(- 2)}^{n} e^{- 2 x}$ $+ \sum_{k = 1}^{n} C_{n}^{k} \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} {\frac{{(x + i)}^{k} - {(x - i)}^{k}}{{(x^{2} + 1)}^{k}}} {(- 2)}^{n - k} e^{- 2 x}$ $2) f^{(n)} (0) =$ $\sum_{k = 1}^{n} C_{n}^{k} \frac{{(- 1)}^{k - 1} (k - 1)!}{2 i} {\frac{i^{k} - {(- i)}^{k}}{1}} {(- 2)}^{n - k}$ $b u t i^{k} - {(- i)}^{k} = e^{i \frac{k π}{2}} - e^{- i \frac{k π}{2}} = 2 i s i n (\frac{k π}{2}) \Rightarrow$ $f^{(n)} (0) = \sum_{k = 1}^{n} C_{n}^{k} (k - 1)! {(- 1)}^{k - 1} {(- 2)}^{n - k} s i n (\frac{k π}{2})$
Commented by abdo mathsup 649 cc last updated on 18/May/18
$3) we have C_n ^k (k−1)! =((n!)/(k!(n−k)!))(k−1)! = ((n!)/(k(n−k)!)) the developpent of f is f(x) = Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n =Σ_(n=0) ^∞ { Σ_(k=1) ^n (1/(k(n−k)!))(−1)^(k−1) (−2)^(n−k) sin(((kπ)/2))}x^n$
$3) w e h a v e C_{n}^{k} (k - 1)! = \frac{n!}{k! (n - k)!} (k - 1)!$ $= \frac{n!}{k (n - k)!} t h e d e v e l o p p e n t o f f i s$ $f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}$ $= \sum_{n = 0}^{\infty} {\sum_{k = 1}^{n} \frac{1}{k (n - k)!} {(- 1)}^{k - 1} {(- 2)}^{n - k} s i n (\frac{k π}{2})} x^{n}$
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