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Question Number 33257 by prof Abdo imad last updated on 14/Apr/18

let g(x)= (1/(1+x^4 )) 1) find g^((n)) (x) 2) calculate g^((n)) (0) 3) if g(x)=Σ u_n x^n find the sequence u_n

$${let}\:{g}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{4}} } \\ $$$$\left.\mathrm{1}\right)\:{find}\:{g}^{\left({n}\right)} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{g}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right)\:{if}\:{g}\left({x}\right)=\Sigma\:{u}_{{n}} \:{x}^{{n}} \:\:\:{find}\:{the}\:{sequence}\:{u}_{{n}} \\ $$

Commented by prof Abdo imad last updated on 15/Apr/18

let decompose g(x) inside C(x) g(x)= (1/((x^2 )^2 −i^2 )) = (1/((x^2 −i)(x^2 +i))) = (1/((x−(√i))(x+(√i))(x −(√(−i)))(x+(√(−i))))) = (1/((x −e^(i(π/4)) )(x +e^(i(π/4)) )(x −e^(−i(π/4)) )( x +e^(−i(π/4)) ))) = (a/(x −e^(i(π/4)) )) +(b/(x +e^(i(π/4)) )) + (c/(x −e^(−i(π/4)) )) + (d/(x + e^(−i(π/4)) )) a = (1/(p^′ ( e^(i(π/4)) ))) with p(x)= 1+x^4 ⇒p^′ (x)=4x^3 if z_k pole p^′ (z_k ) = 4z_k ^3 =((−4)/z_k ) ⇒a =−(e^(i(π/4)) /4) p^′ (−e^(i(π/4)) ) =((−4)/(−e^(i(π/4)) )) = (4/e^(i(π/4)) ) ⇒ b = (1/4) e^(i(π/4)) p^′ ( e^(−i(π/4)) ) =((−4)/e^(−i(π/4)) ) ⇒ c =−(1/4) e^(−i(π/4)) p^′ ( −e^(−i(π/4)) ) = ((−4)/(−e^(−i(π/4)) )) ⇒ d= (1/4) e^(−i(π/4)) g(x) =−(e^(i(π/4)) /(4(x −e^(i(π/4)) ))) +(e^(i(π/4)) /(4( x +e^(i(π/4)) ))) −(e^(−i(π/4)) /(4(x −_ e^(−i(π/4)) ))) + (e^(−i(π/4)) /(4( x +e^(−i(π/4)) ))) ⇒ g^((n)) (x) = (e^(i(π/4)) /4) ( (((−1)^n n!)/((x +e^(i(π/(4{))) )^(n+1) )) −(((−1)^n n!)/((x −e^(i(π/4)) )^(n+1) ))) + (e^(−i(π/4)) /4) ( (((−1)^n n!)/((x + e^(−i(π/4)) )^(n+1) )) − (((−1)^n n!)/((x −e^(−i(π/4)) )^(n+1) ))) g^((n)) (0) =(((−1)^n n! e^(i(π/4)) )/4)( e^(−i(((n+1)π)/4)) −(−1)^(n+1) e^(−i(((n+1)π)/4)) ) + (((−1)^n n! e^(−i(π/4)) )/4)( e^(i(((n+1)π)/4)) −(−1)^(n+1) e^(i(((n+1)π)/4)) )... be continued...

$${let}\:{decompose}\:{g}\left({x}\right)\:{inside}\:{C}\left({x}\right) \\ $$$${g}\left({x}\right)=\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} −{i}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} −{i}\right)\left({x}^{\mathrm{2}} +{i}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\left({x}−\sqrt{{i}}\right)\left({x}+\sqrt{{i}}\right)\left({x}\:−\sqrt{−{i}}\right)\left({x}+\sqrt{−{i}}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\left({x}\:−{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\left({x}\:+{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\left({x}\:\:−{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)\left(\:{x}\:+{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)} \\ $$$$=\:\frac{{a}}{{x}\:−{e}^{{i}\frac{\pi}{\mathrm{4}}} }\:\:+\frac{{b}}{{x}\:+{e}^{{i}\frac{\pi}{\mathrm{4}}} }\:+\:\frac{{c}}{{x}\:−{e}^{−{i}\frac{\pi}{\mathrm{4}}} }\:+\:\frac{{d}}{{x}\:+\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} } \\ $$$${a}\:=\:\frac{\mathrm{1}}{{p}^{'} \left(\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)}\:\:{with}\:{p}\left({x}\right)=\:\mathrm{1}+{x}^{\mathrm{4}} \:\Rightarrow{p}^{'} \left({x}\right)=\mathrm{4}{x}^{\mathrm{3}} \\ $$$${if}\:{z}_{{k}} \:{pole}\:{p}^{'} \left({z}_{{k}} \right)\:=\:\mathrm{4}{z}_{{k}} ^{\mathrm{3}} \:\:=\frac{−\mathrm{4}}{{z}_{{k}} }\:\Rightarrow{a}\:=−\frac{{e}^{{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}} \\ $$$${p}^{'} \left(−{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)\:=\frac{−\mathrm{4}}{−{e}^{{i}\frac{\pi}{\mathrm{4}}} }\:=\:\frac{\mathrm{4}}{{e}^{{i}\frac{\pi}{\mathrm{4}}} }\:\Rightarrow\:{b}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \\ $$$${p}^{'} \left(\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)\:=\frac{−\mathrm{4}}{{e}^{−{i}\frac{\pi}{\mathrm{4}}} }\:\Rightarrow\:{c}\:=−\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \\ $$$${p}^{'} \:\left(\:−{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)\:\:=\:\:\frac{−\mathrm{4}}{−{e}^{−{i}\frac{\pi}{\mathrm{4}}} }\:\:\Rightarrow\:{d}=\:\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \\ $$$${g}\left({x}\right)\:=−\frac{{e}^{{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}\left({x}\:−{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)}\:\:+\frac{{e}^{{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}\left(\:{x}\:+{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)}\:−\frac{{e}^{−{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}\left({x}\:−_{} {e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)} \\ $$$$+\:\frac{{e}^{−{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}\left(\:\:{x}\:+{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)}\:\Rightarrow \\ $$$${g}^{\left({n}\right)} \left({x}\right)\:=\:\frac{{e}^{{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}}\:\left(\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}\:+{e}^{{i}\frac{\pi}{\mathrm{4}\left\{\right.}} \right)^{{n}+\mathrm{1}} }\:−\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}\:−{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)^{{n}+\mathrm{1}} }\right) \\ $$$$+\:\frac{{e}^{−{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}}\:\left(\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}\:+\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)^{{n}+\mathrm{1}} }\:−\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} \boldsymbol{{n}}!}{\left({x}\:−{e}^{−{i}\frac{\pi}{\mathrm{4}}} \right)^{{n}+\mathrm{1}} }\right) \\ $$$${g}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!\:{e}^{{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}}\left(\:\:{e}^{−{i}\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{4}}} \:−\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{e}^{−{i}\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{4}}} \right) \\ $$$$+\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{4}}\left(\:\:{e}^{{i}\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{4}}} \:−\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \:{e}^{{i}\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{4}}} \right)... \\ $$$${be}\:{continued}... \\ $$

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