let f(x)= (x^2 /(1+x^4 )) 1) calculate f^((n)) (x) 2) find f^((n)) (0) 3) developp f at integr serie 4) calculate ∫


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Question Number 37820 by prof Abdo imad last updated on 17/Jun/18

$l e t f (x) = \frac{x^{2}}{1 + x^{4}}$ $1) c a l c u l a t e f^{(n)} (x)$ $2) f i n d f^{(n)} (0)$ $3) d e v e l o p p f a t i n t e g r s e r i e$ $4) c a l c u l a t e \int_{0}^{1} f (x) d x .$
Commented by prof Abdo imad last updated on 18/Jun/18
$2) f^((n)) (0)=(((−1)^n n!)/4) e^(−((iπ)/4)) ((−1)^(n+1) −1) e^(−i((n+1)/4)π) +(((−1)^n n!)/4) e^((iπ)/4) ((−1)^(n+1) −1) e^((i(n+1)π)/4) =(((−1)^n n!)/4)((−1)^(n+1) −1){ e^(−(((n+2))/4)π) +e^((i(n+2)π)/4) } =(((−1)^n n!)/2){(−1)^(n+1) −1)cos((((n+2)π)/4)) ⇒f^((2n)) (0)= (((2n)!)/2)(−2)cos (((2n+2)π)/4) =−(2n)! cos( ((nπ)/2) +(π/2))=(2n)! sin(((nπ)/2)) and f^((2n+1)) (0)=0 3) we have f(x)=Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n =Σ_(n=0) ^∞ ((f^((2n)) (0))/((2n)!)) x^(2n) =Σ_(n=0) ^∞ sin(((nπ)/2)) x^(2n) .but sin(((nπ)/2))=0 ifn=2p and sin(((nπ)/2))=sin((((2p+1)π)/2))=sin(pπ+(π/2)) =(−1)^(p ) if n=2p+1 ⇒ f(x)= Σ_(n=0) ^∞ (−1)^n x^(2(2n+1)) =Σ_(n=0) ^∞ (−1)^n x^(4n+2) .the radius of convergence is R=1 .$
$2) f^{(n)} (0) = \frac{{(- 1)}^{n} n!}{4} e^{- \frac{i π}{4}} ({(- 1)}^{n + 1} - 1) e^{- i \frac{n + 1}{4} π}$ $+ \frac{{(- 1)}^{n} n!}{4} e^{\frac{i π}{4}} ({(- 1)}^{n + 1} - 1) e^{\frac{i (n + 1) π}{4}}$ $= \frac{{(- 1)}^{n} n!}{4} ({(- 1)}^{n + 1} - 1) {e^{- \frac{(n + 2)}{4} π} + e^{\frac{i (n + 2) π}{4}}}$ $= \frac{{(- 1)}^{n} n!}{2} {{(- 1)}^{n + 1} - 1) c o s (\frac{(n + 2) π}{4})$ $\Rightarrow f^{(2 n)} (0) = \frac{(2 n)!}{2} (- 2) c o s \frac{(2 n + 2) π}{4}$ $= - (2 n)! c o s (\frac{n π}{2} + \frac{π}{2}) = (2 n)! s i n (\frac{n π}{2})$ $a n d f^{(2 n + 1)} (0) = 0$ $3) w e h a v e f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n}$ $= \sum_{n = 0}^{\infty} \frac{f^{(2 n)} (0)}{(2 n)!} x^{2 n}$ $= \sum_{n = 0}^{\infty} s i n (\frac{n π}{2}) x^{2 n} . b u t s i n (\frac{n π}{2}) = 0 i f n = 2 p$ $a n d s i n (\frac{n π}{2}) = s i n (\frac{(2 p + 1) π}{2}) = s i n (p π + \frac{π}{2})$ $= {(- 1)}^{p} i f n = 2 p + 1 \Rightarrow$ $f (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 (2 n + 1)}$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{4 n + 2} . t h e r a d i u s o f c o n v e r g e n c e$ $i s R = 1 .$
Commented by prof Abdo imad last updated on 18/Jun/18
$1) we have f(x)= (x^2 /((x^2 )^2 −i^2 )) =(x^2 /((x^2 −i)(x^2 +i))) = (x^2 /((x−e^(i(π/4)) )(x+e^((iπ)/4) )(x−e^(−((iπ)/4)) )(x+e^(−((iπ)/4)) ))) = (a/(x−e^((iπ)/4) )) +(b/(x +e^(i(π/4)) )) +(c/(x−e^(−((iπ)/4)) )) +(d/(x(1/)+e^(−((iπ)/4)) )) let p(x)=1+x^4 we have λ_i = (z_i ^2 /(4z_i ^3 )) = (1/(4 z_i )) ⇒ a= (1/4) e^(−((iπ)/4)) b =−(1/4)e^(−((iπ)/4)) c = (1/4) e^((iπ)/4) d=−(1/4)e^((iπ)/4) ⇒ f(x)=(1/4)e^(−((iπ)/4)) (1/(x−e^((iπ)/4) )) −(1/4)e^(−((iπ)/4)) (1/(x +e^((iπ)/4) )) +(1/4)e^((iπ)/4) (1/(x−e^(−((iπ)/4)) )) −(1/4) e^((iπ)/4) (1/(x+e^(−((iπ)/4)) )) ⇒ f^((n)) (x)=(1/4)e^(−((iπ)/4)) (−1)^n n!{ (1/((x−e^((iπ)/4) )^(n+1) ))−(1/((x+e^(i(π/4)) )^(n+1) ))} +(1/4)e^((iπ)/4) (−1)^n n!{ (1/((x−e^(−((iπ)/4)) )^(n+1) )) −(1/((x+e^(−((iπ)/4)) )^(n+1) ))}$
$1) w e h a v e f (x) = \frac{x^{2}}{{(x^{2})}^{2} - i^{2}}$ $= \frac{x^{2}}{(x^{2} - i) (x^{2} + i)} = \frac{x^{2}}{(x - e^{i \frac{π}{4}}) (x + e^{\frac{i π}{4}}) (x - e^{- \frac{i π}{4}}) (x + e^{- \frac{i π}{4}})}$ $= \frac{a}{x - e^{\frac{i π}{4}}} + \frac{b}{x + e^{i \frac{π}{4}}} + \frac{c}{x - e^{- \frac{i π}{4}}} + \frac{d}{x \frac{1}{} + e^{- \frac{i π}{4}}}$ $l e t p (x) = 1 + x^{4} w e h a v e$ $λ_{i} = \frac{z_{i}^{2}}{4 z_{i}^{3}} = \frac{1}{4 z_{i}} \Rightarrow$ $a = \frac{1}{4} e^{- \frac{i π}{4}}$ $b = - \frac{1}{4} e^{- \frac{i π}{4}}$ $c = \frac{1}{4} e^{\frac{i π}{4}}$ $d = - \frac{1}{4} e^{\frac{i π}{4}} \Rightarrow$ $f (x) = \frac{1}{4} e^{- \frac{i π}{4}} \frac{1}{x - e^{\frac{i π}{4}}} - \frac{1}{4} e^{- \frac{i π}{4}} \frac{1}{x + e^{\frac{i π}{4}}}$ $+ \frac{1}{4} e^{\frac{i π}{4}} \frac{1}{x - e^{- \frac{i π}{4}}} - \frac{1}{4} e^{\frac{i π}{4}} \frac{1}{x + e^{- \frac{i π}{4}}} \Rightarrow$ $f^{(n)} (x) = \frac{1}{4} e^{- \frac{i π}{4}} {(- 1)}^{n} n! {\frac{1}{{(x - e^{\frac{i π}{4}})}^{n + 1}} - \frac{1}{{(x + e^{i \frac{π}{4}})}^{n + 1}}}$ $+ \frac{1}{4} e^{\frac{i π}{4}} {(- 1)}^{n} n! {\frac{1}{{(x - e^{- \frac{i π}{4}})}^{n + 1}} - \frac{1}{{(x + e^{- \frac{i π}{4}})}^{n + 1}}}$
	Answered by behi83417@gmail.com last updated on 18/Jun/18

	$f (x) = \frac{1}{2} . \frac{2 x^{2}}{(x^{2} + i) (x^{2} - i)} = \frac{1}{2} . \frac{(x^{2} + i) + (x^{2} - i)}{(x^{2} + i) (x^{2} - i)} =$ $= \frac{1}{2} [\frac{1}{x^{2} + i} + \frac{1}{x^{2} - i}] = \frac{1}{2} [\frac{1}{(x + i \sqrt{i}) (x - i \sqrt{i})} +$ $+ \frac{1}{(x - \sqrt{i}) (x + \sqrt{i})}] = \frac{1}{2} [\frac{1}{2 i \sqrt{i}} [\frac{(x + i \sqrt{i}) - (x - i \sqrt{i})}{(x + i \sqrt{i}) (x - i \sqrt{i})}] +$ $+ [\frac{1}{2 \sqrt{i}} [\frac{(x + \sqrt{i}) - (x - \sqrt{i})}{(x + \sqrt{i}) (x - \sqrt{i})}]] \overset{\sqrt{i} = a}{=}$ $= - \frac{a}{4} [\frac{1}{x - a^{3}} - \frac{1}{x + a^{3}} + \frac{i}{x - a} - \frac{i}{x + a}]$ $f^{1} (x) = - \frac{a}{4} [\frac{- 1}{{(x - a^{3})}^{2}} + \frac{1}{{(x + a^{3})}^{2}} - \frac{i}{{(x + a)}^{2}} + \frac{i}{{(x - a)}^{2}}]$ $f^{n} (x) = \frac{n! a {(- 1)}^{n}}{4} [\frac{1}{{(x - a^{3})}^{n + 1}} - \frac{1}{{(x - a^{3})}^{n + 1}} +$ $+ \frac{i}{{(x - a)}^{n + 1}} - \frac{i}{{(x - a)}^{n + 1}}] .$ $2) f^{n} (0) = 0$ $4) \int \frac{x^{2} d x}{1 + x^{4}} = - \frac{a}{4} [l n \frac{x + a^{3}}{x - a^{3}} + i . l n \frac{x + a}{x - a}]$ $F (1) = - \frac{a}{4} [l n \frac{1 + a^{3}}{1 - a^{3}} + i . l n \frac{1 + a}{1 - a}] =$ $- \frac{\sqrt{i}}{4} [i \sqrt{i} + \sqrt{i} - i + i (i \sqrt{i} + \sqrt{i} + i)] =$ $= - \frac{\sqrt{i}}{4} [i \sqrt{i} + \sqrt{i} - i - \sqrt{i} + i \sqrt{i} - 1] =$ $= - \frac{\sqrt{i}}{4} [2 i \sqrt{i} - i - 1] = - \frac{1}{4} (- 2 - i \sqrt{i} - \sqrt{i})$ $F (0) = - \frac{a}{4} [l n (- 1) + i . l n (- 1)] =$ $= - \frac{a}{4} (\frac{i π}{2} + i . \frac{i π}{2}) = - \frac{π \sqrt{i}}{8} (i - 1) = \frac{π \sqrt{i}}{8} (1 - i) .$ $I = F (1) - F (0) = \frac{1}{8} [4 + (i + 1) \sqrt{i} + π \sqrt{i} (i - 1)] .$ $i = c o s \frac{π}{2} + i s i n \frac{π}{2} = e^{\frac{i π}{2}} \Rightarrow \sqrt{i} = e^{\frac{i π}{4}}$ $\frac{1 + a^{3}}{1 - a^{3}} = \frac{{(1 + a^{3})}^{2}}{1 - a^{6}} = \frac{(1 + 2 i \sqrt{i} - i)}{1 + i} . \frac{1 - i}{1 - i} =$ $= \frac{1}{2} (1 - i + 2 i \sqrt{i} + 2 \sqrt{i} - i - 1) = i \sqrt{i} + \sqrt{i} - i$ $\frac{1 + a}{1 - a} = \frac{{(1 + a)}^{2}}{1 - a^{2}} = \frac{1 + 2 \sqrt{i} + i}{1 - i} . \frac{1 + i}{1 + i} =$ $= \frac{1}{2} (1 + i + 2 \sqrt{i} + 2 i \sqrt{i} + i - 1) = i \sqrt{i} + \sqrt{i} + i$
	Commented by math khazana by abdo last updated on 18/Jun/18

	$s i r B e h i i f f (x) = \frac{1}{x + a} a f r o m C w e h a v e$ $f^{(n)} (x) = \frac{{(- 1)}^{n} n!}{{(x + a)}^{n + 1}}!$
	Commented by behi83417@gmail.com last updated on 18/Jun/18

	$c o r r e c t e d! t h a n k s .$
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