let f(x)=arctan(1+x^2 ) 1) calculate f^((n)) (x) and f^((n)) (0) 2) developpf 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 81720 by mathmax by abdo last updated on 14/Feb/20

$l e t f (x) = a r c t a n (1 + x^{2})$ $1) c a l c u l a t e f^{(n)} (x) a n d f^{(n)} (0)$ $2) d e v e l o p p f a t i n t e g r s e r i e$
Commented by mathmax by abdo last updated on 15/Feb/20
$1) we have f^′ (x)=((2x)/(1+(1+x^2 )^2 )) =((2x)/(1+x^4 +2x^2 +1)) =((2x)/(x^4 +2x^2 +2)) x^4 +2x^2 +2=0 →t^2 +2t +2=0 (t=x^2 )→Δ^′ =1−2=−1 ⇒ x_1 =−1+i =(√2)e^(i((3π)/4)) and x_2 =−1−i =(√2)e^(−((i3π)/4)) ⇒ f^′ (x)=((2x)/((x^2 −(√2)e^((i3π)/4) )(x^2 −(√2)e^(−((i3π)/4)) ))) =((2x)/((x−αe^((i3π)/8) )(x+α e^((i3π)/8) )(x−α e^(−((i3π)/8)) )(x+α e^(−((i3π)/8)) 3)) =(a/(x−α e^((i3π)/8) )) +(b/(x+α e^((i3π)/8) )) +(c/(x−α e^(−i((3π)/8)) )) +(d/((x +e^(−((i3π)/8)) ))) (α=(√(√2))) a =((2α e^((i3π)/8) )/(2α e^((i3π)/8) (√2) (2i)×((√2)/2))) =(1/(2i)) b=((−2α e^((i3π)/8) )/(−2α e^((i3π)/8) (√2)(2i)×((√2)/2))) =(1/(2i)) c =((2α e^(−((i3π)/8)) )/(2α e^(−((i3π)/8)) (√2)(−2i)×((√2)/2))) =−(1/(2i)) d =((−2α e^(−((i3π)/8)) )/(−2α e^(−((i3π)/8)) (√2)(−2i)×((√2)/2))) =−(1/(2i)) ⇒ f^′ (x) =(1/(2i)){ (1/(x−α e^((i3π)/8) )) +(1/(x+α e^((i3π)/8) )) −(1/(x−α e^(−((i3π)/8)) ))−(1/(x+α e^(−((i3π)/8)) ))} ⇒ f^((n)) (x) =(1/(2i)){ (((−1)^(n−1) (n−1)!)/((x−α e^((i3π)/8) )^n )) +(((−1)^(n−1) (n−1)!)/((x+α e^((i3π)/8) )^n )) −(((−1)^(n−1) (n−1)!)/((x−αe^(−((i3π)/8)) )^n ))−(((−1)^(n−1) (n−1)!)/((x+αe^(−((i3π)/8)) )^n ))} =(((−1)^(n−1) (n−1)!)/(2i)){(1/((x−α e^((i3π)/8) )^n ))−(1/((x−α e^(−((i3π)/8)) )^n )) +(1/((x+α e^(−((i3π)/8)) )^n )) −(1/((x +α e^(−((i3π)/8)) )^n ))} =(((−1)^(n−1) (n−1)!)/(2i)){((−2i Im(x−α e^(−((i3π)/8)) )^n )/((x^2 −2α cos(((3π)/8))x +α^2 )^n )) +((−2i Im(x+α e^(−((i3π)/8)) )^n )/((x^2 +2α cos(((3π)/8))x +α^2 )^n ))} f^((n)) (x)=(−1)^n (n−1)!{((Im(x−α e^(−((i3π)/8)) )^n )/((x^2 −2α cos(((3π)/8))x +α^2 )^n )) +((Im(x+αe^(−((i3π)/8)) )^n )/((x^2 +2α cos(((3π)/8))x+α^2 )^n ))}$
$1) w e h a v e f^{^{'}} (x) = \frac{2 x}{1 + {(1 + x^{2})}^{2}} = \frac{2 x}{1 + x^{4} + 2 x^{2} + 1} = \frac{2 x}{x^{4} + 2 x^{2} + 2}$ $x^{4} + 2 x^{2} + 2 = 0 \to t^{2} + 2 t + 2 = 0 (t = x^{2}) \to Δ^{^{'}} = 1 - 2 = - 1 \Rightarrow$ $x_{1} = - 1 + i = \sqrt{2} e^{i \frac{3 π}{4}} a n d x_{2} = - 1 - i = \sqrt{2} e^{- \frac{i 3 π}{4}} \Rightarrow$ $f^{^{'}} (x) = \frac{2 x}{(x^{2} - \sqrt{2} e^{\frac{i 3 π}{4}}) (x^{2} - \sqrt{2} e^{- \frac{i 3 π}{4}})} = \frac{2 x}{(x - α e^{\frac{i 3 π}{8}}) (x + α e^{\frac{i 3 π}{8}}) (x - α e^{- \frac{i 3 π}{8}}) (x + α e^{- \frac{i 3 π}{8}} 3}$ $= \frac{a}{x - α e^{\frac{i 3 π}{8}}} + \frac{b}{x + α e^{\frac{i 3 π}{8}}} + \frac{c}{x - α e^{- i \frac{3 π}{8}}} + \frac{d}{(x + e^{- \frac{i 3 π}{8}})} (α = \sqrt{\sqrt{2}})$ $a = \frac{2 α e^{\frac{i 3 π}{8}}}{2 α e^{\frac{i 3 π}{8}} \sqrt{2} (2 i) \times \frac{\sqrt{2}}{2}} = \frac{1}{2 i}$ $b = \frac{- 2 α e^{\frac{i 3 π}{8}}}{- 2 α e^{\frac{i 3 π}{8}} \sqrt{2} (2 i) \times \frac{\sqrt{2}}{2}} = \frac{1}{2 i}$ $c = \frac{2 α e^{- \frac{i 3 π}{8}}}{2 α e^{- \frac{i 3 π}{8}} \sqrt{2} (- 2 i) \times \frac{\sqrt{2}}{2}} = - \frac{1}{2 i}$ $d = \frac{- 2 α e^{- \frac{i 3 π}{8}}}{- 2 α e^{- \frac{i 3 π}{8}} \sqrt{2} (- 2 i) \times \frac{\sqrt{2}}{2}} = - \frac{1}{2 i} \Rightarrow$ $f^{^{'}} (x) = \frac{1}{2 i} {\frac{1}{x - α e^{\frac{i 3 π}{8}}} + \frac{1}{x + α e^{\frac{i 3 π}{8}}} - \frac{1}{x - α e^{- \frac{i 3 π}{8}}} - \frac{1}{x + α e^{- \frac{i 3 π}{8}}}} \Rightarrow$ $f^{(n)} (x) = \frac{1}{2 i} {\frac{{(- 1)}^{n - 1} (n - 1)!}{{(x - α e^{\frac{i 3 π}{8}})}^{n}} + \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + α e^{\frac{i 3 π}{8}})}^{n}}$ $- \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x - α e^{- \frac{i 3 π}{8}})}^{n}} - \frac{{(- 1)}^{n - 1} (n - 1)!}{{(x + α e^{- \frac{i 3 π}{8}})}^{n}}}$ $= \frac{{(- 1)}^{n - 1} (n - 1)!}{2 i} {\frac{1}{{(x - α e^{\frac{i 3 π}{8}})}^{n}} - \frac{1}{{(x - α e^{- \frac{i 3 π}{8}})}^{n}} + \frac{1}{{(x + α e^{- \frac{i 3 π}{8}})}^{n}}$ $- \frac{1}{{(x + α e^{- \frac{i 3 π}{8}})}^{n}}}$ $= \frac{{(- 1)}^{n - 1} (n - 1)!}{2 i} {\frac{- 2 i I m {(x - α e^{- \frac{i 3 π}{8}})}^{n}}{{(x^{2} - 2 α c o s (\frac{3 π}{8}) x + α^{2})}^{n}} + \frac{- 2 i I m {(x + α e^{- \frac{i 3 π}{8}})}^{n}}{{(x^{2} + 2 α c o s (\frac{3 π}{8}) x + α^{2})}^{n}}}$ $f^{(n)} (x) = {(- 1)}^{n} (n - 1)! {\frac{I m {(x - α e^{- \frac{i 3 π}{8}})}^{n}}{{(x^{2} - 2 α c o s (\frac{3 π}{8}) x + α^{2})}^{n}} + \frac{I m {(x + α e^{- \frac{i 3 π}{8}})}^{n}}{{(x^{2} + 2 α c o s (\frac{3 π}{8}) x + α^{2})}^{n}}}$
Commented by mathmax by abdo last updated on 15/Feb/20
$f^((n)) (0) =(−1)^n (n−1)!{((Im(−α)^n e^(−((i3nπ)/8)) )/α^(2n) ) +((Im(α^n e^(−((i3nπ)/8)) ))/α^(2n) ) =(((−1)^n (n−1)!)/2^(n/2) ){−(−α)^n sin(((3nπ)/8))−α^n sin(((3nπ)/8))} =(−1)^(n+1) (n−1)!(({2^(n/4) +(−1)^n 2^(n/4) }sin(((3nπ)/8)))/2^(n/2) ) =(((−1)^(n+1) (n−1)!)/2^(n/4) )(1+(−1)^n )sin(((3nπ)/8))$
$f^{(n)} (0) = {(- 1)}^{n} (n - 1)! {\frac{I m {(- α)}^{n} e^{- \frac{i 3 n π}{8}}}{α^{2 n}} + \frac{I m (α^{n} e^{- \frac{i 3 n π}{8}})}{α^{2 n}}$ $= \frac{{(- 1)}^{n} (n - 1)!}{2^{\frac{n}{2}}} {- {(- α)}^{n} s i n (\frac{3 n π}{8}) - α^{n} s i n (\frac{3 n π}{8})}$ $= {(- 1)}^{n + 1} (n - 1)! \frac{{2^{\frac{n}{4}} + {(- 1)}^{n} 2^{\frac{n}{4}}} s i n (\frac{3 n π}{8})}{2^{\frac{n}{2}}}$ $= \frac{{(- 1)}^{n + 1} (n - 1)!}{2^{\frac{n}{4}}} (1 + {(- 1)}^{n}) s i n (\frac{3 n π}{8})$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum