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


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Question Number 37838 by math khazana by abdo last updated on 18/Jun/18

$${let}\:{f}\left({x}\right)=\:{e}^{−\mathrm{2}{x}} \:{arctan}\left({x}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{1}\right){calculate}\:{f}^{\left({n}\right)} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie} \\ $$
Commented by math khazana by abdo last updated on 20/Jun/18
$Leiniz formula give f^((n)) (x)= Σ_(k=0) ^n C_n ^k (arctan(x^2 ))^((k)) (e^(−2x) )^((n−k)) we have (e^(−2x) )^((1)) =−2 e^(−2x) (e^(−2x) )^((2)) =(−2)^2 e^(−2x) ⇒ (e^(−2x) )^((p)) =(−2)^p e^(−2x) also we have (arctan(x^2 ))^((1)) = ((2x)/(1+x^4 )) ⇒ (arctan(x^2 ))^((n)) =(((2x)/(1+x^4 )))^((n−1)) let w(x) = (x/(1+x^4 )) w(x)= (x/((x^2 −i)(x^2 +i))) =(x/((x−(√i))(x+(√i))(x−(√(−i)))(x+(√(−i))))) = (x/((x −e^((iπ)/4) )(x+ e^((iπ)/4) )( x−e^(−((iπ)/4)) )(x+e^(−((iπ)/4)) ))) =(a/(x−e^((iπ)/4) )) +(b/(x +e^((iπ)/4) )) +(c/(x−e^(−((iπ)/4)) )) +(d/(x + e^(−((iπ)/4)) )) λ_i = (z_i /(4z_i ^3 )) = (1/(4z_i ^2 )) ⇒a= (1/(4(e^((iπ)/4) )^2 )) = (1/(4i)) =((−i)/4) b =(1/(4i))=((−i)/4) , c=−(1/(4i)) =(i/4) , d=−(1/(4i)) =(i/4) ⇒ w^((n−1)) (x) =((−i)/4) (((−1)^(n−1) (n−1)!)/((x−e^((iπ)/4) )^n )) −(i/4) (((−1)^(n−1) (n−1)!)/((x +e^((iπ)/4) )^n )) +(i/4) (((−1)^(n−1) (n−1)!)/((x−e^(−((iπ)/4)) )^n )) +(i/4) (((−1)^(n−1) (n−1)!)/((x +e^(−((iπ)/4)) )^n )) ⇒ (arctan(x^2 ))^((n)) =(i/2)(−1)^(n−1) (n−1)!{ ((−1)/((x −e^((iπ)/4) )^n )) +((−1)/((x +e^((iπ)/4) )^n )) + (1/((x −e^(−((iπ)/4)) )^n )) + (1/((x +e^(−((iπ)/4)) )^n )) } ⇒ f^((n)) (x) = (i/2)(−1)^(n−1) (n−1)!Σ_(k=0) ^n (−2)^(n−k) C_n ^k {((−1)/((x−e^((iπ)/4) )^k )) +((−1)/((x +e^((iπ)/4) )^k )) + (1/((x−e^(−((iπ)/4)) )^k )) +(1/((x +e^(−((iπ)/4)) )^k ))} e^(−2x) .$
$${Leiniz}\:{formula}\:{give} \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\left({arctan}\left({x}^{\mathrm{2}} \right)\right)^{\left({k}\right)} \:\left({e}^{−\mathrm{2}{x}} \right)^{\left({n}−{k}\right)} \\ $$$${we}\:{have}\:\left({e}^{−\mathrm{2}{x}} \right)^{\left(\mathrm{1}\right)} =−\mathrm{2}\:{e}^{−\mathrm{2}{x}} \\ $$$$\left({e}^{−\mathrm{2}{x}} \right)^{\left(\mathrm{2}\right)} =\left(−\mathrm{2}\right)^{\mathrm{2}} \:{e}^{−\mathrm{2}{x}} \:\Rightarrow\:\left({e}^{−\mathrm{2}{x}} \right)^{\left({p}\right)} =\left(−\mathrm{2}\right)^{{p}} \:{e}^{−\mathrm{2}{x}} \\ $$$${also}\:{we}\:{have}\: \\ $$$$\left({arctan}\left({x}^{\mathrm{2}} \right)\right)^{\left(\mathrm{1}\right)} =\:\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{4}} }\:\Rightarrow \\ $$$$\left({arctan}\left({x}^{\mathrm{2}} \right)\right)^{\left({n}\right)} =\left(\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{4}} }\right)^{\left({n}−\mathrm{1}\right)} \\ $$$${let}\:{w}\left({x}\right)\:=\:\frac{{x}}{\mathrm{1}+{x}^{\mathrm{4}} } \\ $$$${w}\left({x}\right)=\:\frac{{x}}{\left({x}^{\mathrm{2}} \:−{i}\right)\left({x}^{\mathrm{2}} +{i}\right)}\:=\frac{{x}}{\left({x}−\sqrt{{i}}\right)\left({x}+\sqrt{{i}}\right)\left({x}−\sqrt{−{i}}\right)\left({x}+\sqrt{−{i}}\right)} \\ $$$$=\:\frac{{x}}{\left({x}\:−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({x}+\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left(\:{x}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({x}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$$$=\frac{{a}}{{x}−{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:\:+\frac{{b}}{{x}\:+{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:\:+\frac{{c}}{{x}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\:+\frac{{d}}{{x}\:+\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} } \\ $$$$\lambda_{{i}} =\:\frac{{z}_{{i}} }{\mathrm{4}{z}_{{i}} ^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{4}{z}_{{i}} ^{\mathrm{2}} }\:\Rightarrow{a}=\:\frac{\mathrm{1}}{\mathrm{4}\left({e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{4}{i}}\:=\frac{−{i}}{\mathrm{4}} \\ $$$${b}\:=\frac{\mathrm{1}}{\mathrm{4}{i}}=\frac{−{i}}{\mathrm{4}}\:\:\:,\:\:\:{c}=−\frac{\mathrm{1}}{\mathrm{4}{i}}\:=\frac{{i}}{\mathrm{4}}\:\:\:\:,\:\:\:{d}=−\frac{\mathrm{1}}{\mathrm{4}{i}}\:=\frac{{i}}{\mathrm{4}}\:\Rightarrow \\ $$$${w}^{\left({n}−\mathrm{1}\right)} \left({x}\right)\:=\frac{−{i}}{\mathrm{4}}\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} } \\ $$$$−\frac{{i}}{\mathrm{4}}\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}\:+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }\:\:+\frac{{i}}{\mathrm{4}}\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} } \\ $$$$+\frac{{i}}{\mathrm{4}}\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}\:+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }\:\Rightarrow \\ $$$$\left({arctan}\left({x}^{\mathrm{2}} \right)\right)^{\left({n}\right)} \\ $$$$=\frac{{i}}{\mathrm{2}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\left\{\:\:\frac{−\mathrm{1}}{\left({x}\:−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }\:+\frac{−\mathrm{1}}{\left({x}\:+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }\right. \\ $$$$\left.+\:\frac{\mathrm{1}}{\left({x}\:−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }\:\:+\:\frac{\mathrm{1}}{\left({x}\:+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }\:\right\}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right) \\ $$$$=\:\frac{{i}}{\mathrm{2}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{2}\right)^{{n}−{k}} \:{C}_{{n}} ^{{k}} \left\{\frac{−\mathrm{1}}{\left({x}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{k}} }\right. \\ $$$$\left.\:+\frac{−\mathrm{1}}{\left({x}\:+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{k}} }\:\:+\:\frac{\mathrm{1}}{\left({x}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{k}} }\:+\frac{\mathrm{1}}{\left({x}\:+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{k}} }\right\}\:{e}^{−\mathrm{2}{x}} \:. \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 20/Jun/18
$error of calculus at the final line f^((n)) (x) =(i/2) Σ_(k=0) ^n (−2)^(n−k) (−1)^(k−1) (k−1)! C_n ^k { ((−1)/((x−e^((iπ)/4) )^k )) +((−1)/((x +e^((iπ)/4) )^k )) + (1/((x−e^(−((iπ)/4)) )^k )) +(1/((x+e^(−((iπ)/4)) )^k ))}e^(−2x)$
$${error}\:{of}\:{calculus}\:{at}\:{the}\:{final}\:{line} \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\frac{{i}}{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{2}\right)^{{n}−{k}} \:\:\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!\:{C}_{{n}} ^{{k}} \left\{\right. \\ $$$$\left.\frac{−\mathrm{1}}{\left({x}−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{k}} }\:+\frac{−\mathrm{1}}{\left({x}\:+{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{k}} }\:\:+\:\frac{\mathrm{1}}{\left({x}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{k}} }\:+\frac{\mathrm{1}}{\left({x}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{k}} }\right\}{e}^{−\mathrm{2}{x}} \\ $$
Commented by math khazana by abdo last updated on 20/Jun/18
$f^((n)) (0) =(i/2)Σ_(k=1) ^n (−2)^(n−k) (−1)^(k−1) (k−1)!{ −(−e^((iπ)/4) )^(−k) − ( e^((iπ)/4) )^(−k) +(−e^(−((iπ)/4)) )^(−k) +(e^(−((iπ)/4)) )^(−k) } =(i/2)Σ_(k=1) ^n (−2)^(n−k) (−1)^(k−1) (k−1)!{−(−1)^k e^(−((ikπ)/4)) −e^(−((ikπ)/4)) +(−1)^k e^((ikπ)/4) + e^((ikπ)/4) } =(i/2) Σ_(k=1) ^n (−2)^(n−k) (−1)^(k−1)) (k−1)!{((−1)^k +1)(e^((ikπ)/4) −e^(−((ikπ)/4)) ) =(i/2)Σ_(k=1) ^n (−1)^(k−1) (k−1)!(−2)^(n−k) {1+(−1)^k }2isin(((kπ)/4)) =Σ_(k=1) ^n (−1)^k (k−1)!(−2)^(n−k) (1+(−1)^k )sin(((kπ)/4))$
$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{{i}}{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{2}\right)^{{n}−{k}} \left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \:\left({k}−\mathrm{1}\right)!\left\{\right. \\ $$$$\left.−\left(−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{−{k}} \:−\:\left(\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{−{k}} \:\:+\left(−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{−{k}} \:+\left({e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{−{k}} \right\} \\ $$$$=\frac{{i}}{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\left(−\mathrm{2}\right)^{{n}−{k}} \left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!\left\{−\left(−\mathrm{1}\right)^{{k}} \:{e}^{−\frac{{ik}\pi}{\mathrm{4}}} \right. \\ $$$$\left.−{e}^{−\frac{{ik}\pi}{\mathrm{4}}} \:\:+\left(−\mathrm{1}\right)^{{k}} \:{e}^{\frac{{ik}\pi}{\mathrm{4}}} \:\:+\:{e}^{\frac{{ik}\pi}{\mathrm{4}}} \right\} \\ $$$$=\frac{{i}}{\mathrm{2}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{2}\right)^{{n}−{k}} \left(−\mathrm{1}\right)^{\left.{k}−\mathrm{1}\right)} \left({k}−\mathrm{1}\right)!\left\{\left(\left(−\mathrm{1}\right)^{{k}} +\mathrm{1}\right)\left({e}^{\frac{{ik}\pi}{\mathrm{4}}} −{e}^{−\frac{{ik}\pi}{\mathrm{4}}} \right)\right. \\ $$$$=\frac{{i}}{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!\left(−\mathrm{2}\right)^{{n}−{k}} \left\{\mathrm{1}+\left(−\mathrm{1}\right)^{{k}} \right\}\mathrm{2}{isin}\left(\frac{{k}\pi}{\mathrm{4}}\right) \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \left({k}−\mathrm{1}\right)!\left(−\mathrm{2}\right)^{{n}−{k}} \left(\mathrm{1}+\left(−\mathrm{1}\right)^{{k}} \right){sin}\left(\frac{{k}\pi}{\mathrm{4}}\right) \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 20/Jun/18
$3) f(x)=Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n = 2 Σ_(n=1) ^∞ (1/(n!)){Σ_(p=1) ^([(n/2)]) (2p−1)!(−2)^(n−2p) sin(((pπ)/2))}x^n$
$$\left.\mathrm{3}\right)\:{f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \\ $$$$=\:\mathrm{2}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}!}\left\{\sum_{{p}=\mathrm{1}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \left(\mathrm{2}{p}−\mathrm{1}\right)!\left(−\mathrm{2}\right)^{{n}−\mathrm{2}{p}} {sin}\left(\frac{{p}\pi}{\mathrm{2}}\right)\right\}{x}^{{n}} \\ $$
Commented by math khazana by abdo last updated on 20/Jun/18

$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\sum_{{p}=\mathrm{1}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\left(\mathrm{2}{p}−\mathrm{1}\right)!\left(−\mathrm{2}\right)^{{n}−\mathrm{2}{p}} \:{sin}\left(\frac{{p}\pi}{\mathrm{2}}\right). \\ $$
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