let f(x)=x^n arctan(x^2 ) with n integr natural 1) calculate f^((n)) (x) and f^((n)) (0) 2)developp f at integr serie .


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Question Number 55270 by maxmathsup by imad last updated on 20/Feb/19

$l e t f (x) = x^{n} a r c t a n (x^{2}) w i t h n i n t e g r n a t u r a l$ $1) c a l c u l a t e f^{(n)} (x) a n d f^{(n)} (0)$ $2) d e v e l o p p f a t i n t e g r s e r i e .$
Commented by maxmathsup by imad last updated on 02/Mar/19
$1) leibniz formula give f^((n)) (x)=Σ_(k=0) ^n C_n ^k (x^n )^((k)) (arctan(x^2 ))^((n−k)) but (x^n )^((k)) =n(n−1)...(n−k+1)x^(n−k) =((n!)/((n−k)!)) x^(n−k) if k≤n ⇒ f^((n)) (x) =Σ_(k=0) ^n ((n!)/(k!(n−k)!)) ((n!)/((n−k)!)) x^(n−k) {arctan(x^2 )}^()n−k)) =Σ_(k=0) ^n (((n!)^2 )/(k!{(n−k)!}^2 )) { arctan(x^2 )}^((n−k)) let find (arctan(x^2 ))^((m)) we have {arctan(x^2 )}^((1)) =((2x)/(1+x^4 )) =((2x)/((x^2 −i)(x^2 +i))) =((2x)/((x−(√i))(x+(√i))(x−(√(−i)))(x+(√(−i))))) =((2x)/((x−e^((iπ)/4) )(x+e^((iπ)/4) )(x−e^(−((iπ)/4)) )(x+e^(−((iπ)/4)) ))) =F(x)=(a/(x−e^((iπ)/4) )) +(b/(x+e^((iπ)/4) )) +(c/(x−e^(−((iπ)/4)) )) +(d/(x+e^(−((iπ)/4)) )) a =((2 e^((iπ)/4) )/(2e^((iπ)/4) (2i sin((π/4)))(2cos((π/4))))) = (1/(4i(1/2))) =(1/(2i)) b =((−2 e^((iπ)/4) )/((−2e^((iπ)/4) )(−2cos((π/4)))(−2i sin((π/4))))) =(1/(2i)) c =((2e^(−((iπ)/4)) )/((−2isin((π/4)))(2cos((π/4)))(2e^(−((iπ)/4)) ))) =−(1/(2i)) d =((−2e^(−((iπ)/4)) )/((−2cos((π/4)))(2i sin((π/4)))(−2e^(−((iπ)/4)) ))) =−(1/(2i)) ⇒ (arctan(x^2 ))^((1)) =(1/(2i)){(1/(x−e^((iπ)/4) )) +(1/(x+e^((iπ)/4) )) −(1/(x−e^(−((iπ)/4)) )) −(1/(x+e^(−((iπ)/4)) ))} ⇒ (arctan(x^2 ))^((m)) =(1/(2i))(−1)^(m−1) (m−1)!{ (1/((x−e^((iπ)/4) )^m )) +(1/((x+e^((iπ)/4) )^m )) −(1/((x−e^(−((iπ)/4)) )^m ))−(1/((x+e^(−((iπ)/4)) )))} ⇒ f^((n)) (x) =(1/(2i))Σ_(k=0) ^n (((n!)^2 )/(k!{(n−k)!}^2 ))(−1)^(n−k−1) (n−k−1)!{(1/((x−e^((iπ)/4) )^(n−k) ))+(1/((x+e^((iπ)/4) )^(n−k) ))−(1/((x−e^(−((iπ)/4)) )^(n−k) )) −(1/((x+e^(−((iπ)/4)) )^(n−k) ))}$
$1) l e i b n i z f o r m u l a g i v e f^{(n)} (x) = \sum_{k = 0}^{n} C_{n}^{k} {(x^{n})}^{(k)} {(a r c t a n (x^{2}))}^{(n - k)}$ $b u t {(x^{n})}^{(k)} = n (n - 1) . . . (n - k + 1) x^{n - k} = \frac{n!}{(n - k)!} x^{n - k} i f k ⩽ n \Rightarrow$ $f^{(n)} (x) = \sum_{k = 0}^{n} \frac{n!}{k! (n - k)!} \frac{n!}{(n - k)!} x^{n - k} {a r c t a n (x^{2})}^{) n - k)}$ $= \sum_{k = 0}^{n} \frac{{(n!)}^{2}}{k! {(n - k)!}^{2}} {a r c t a n (x^{2})}^{(n - k)} l e t f i n d {(a r c t a n (x^{2}))}^{(m)}$ $w e h a v e {a r c t a n (x^{2})}^{(1)} = \frac{2 x}{1 + x^{4}} = \frac{2 x}{(x^{2} - i) (x^{2} + i)} = \frac{2 x}{(x - \sqrt{i}) (x + \sqrt{i}) (x - \sqrt{- i}) (x + \sqrt{- i})}$ $= \frac{2 x}{(x - e^{\frac{i π}{4}}) (x + e^{\frac{i π}{4}}) (x - e^{- \frac{i π}{4}}) (x + e^{- \frac{i π}{4}})} = F (x) = \frac{a}{x - e^{\frac{i π}{4}}} + \frac{b}{x + e^{\frac{i π}{4}}} + \frac{c}{x - e^{- \frac{i π}{4}}} + \frac{d}{x + e^{- \frac{i π}{4}}}$ $a = \frac{2 e^{\frac{i π}{4}}}{2 e^{\frac{i π}{4}} (2 i s i n (\frac{π}{4})) (2 c o s (\frac{π}{4}))} = \frac{1}{4 i \frac{1}{2}} = \frac{1}{2 i}$ $b = \frac{- 2 e^{\frac{i π}{4}}}{(- 2 e^{\frac{i π}{4}}) (- 2 c o s (\frac{π}{4})) (- 2 i s i n (\frac{π}{4}))} = \frac{1}{2 i}$ $c = \frac{2 e^{- \frac{i π}{4}}}{(- 2 i s i n (\frac{π}{4})) (2 c o s (\frac{π}{4})) (2 e^{- \frac{i π}{4}})} = - \frac{1}{2 i}$ $d = \frac{- 2 e^{- \frac{i π}{4}}}{(- 2 c o s (\frac{π}{4})) (2 i s i n (\frac{π}{4})) (- 2 e^{- \frac{i π}{4}})} = - \frac{1}{2 i} \Rightarrow$ ${(a r c t a n (x^{2}))}^{(1)} = \frac{1}{2 i} {\frac{1}{x - e^{\frac{i π}{4}}} + \frac{1}{x + e^{\frac{i π}{4}}} - \frac{1}{x - e^{- \frac{i π}{4}}} - \frac{1}{x + e^{- \frac{i π}{4}}}} \Rightarrow$ ${(a r c t a n (x^{2}))}^{(m)} = \frac{1}{2 i} {(- 1)}^{m - 1} (m - 1)! {\frac{1}{{(x - e^{\frac{i π}{4}})}^{m}} + \frac{1}{{(x + e^{\frac{i π}{4}})}^{m}} - \frac{1}{{(x - e^{- \frac{i π}{4}})}^{m}} - \frac{1}{(x + e^{- \frac{i π}{4}})}} \Rightarrow$ $f^{(n)} (x) = \frac{1}{2 i} \sum_{k = 0}^{n} \frac{{(n!)}^{2}}{k! {(n - k)!}^{2}} {(- 1)}^{n - k - 1} (n - k - 1)! {\frac{1}{{(x - e^{\frac{i π}{4}})}^{n - k}} + \frac{1}{{(x + e^{\frac{i π}{4}})}^{n - k}} - \frac{1}{{(x - e^{- \frac{i π}{4}})}^{n - k}} - \frac{1}{{(x + e^{- \frac{i π}{4}})}^{n - k}}}$
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