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Question Number 46424 by maxmathsup by imad last updated on 25/Oct/18
let f(x)=(e^(−x) /(x^2 +4)) 1) calculate f^((n)) (x) and f^((n)) (0) 2) developp f at integr serie .
$${let}\:{f}\left({x}\right)=\frac{{e}^{−{x}} }{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$
Commented by maxmathsup by imad last updated on 27/Oct/18
1) we have f(x)=(e^(−x) /((x−2i)(x+2i))) =(1/(4i))e^(−x) { (1/(x−2i)) −(1/(x+2i))} =(1/(4i)){(e^(−x) /(x−2i)) −(e^(−x) /(x+2i))} ⇒f^((n)) (x)=(1/(4i)){ ((e^(−x) /(x−2i)))^((n)) −((e^(−x) /(x+2i)))^((n)) } leibniz formulae give ((e^(−x) /(x−2i)))^((n)) =Σ_(k=0) ^n C_n ^k ((1/(x−2i)))^((k)) (e^(−x) )^((n−k)) =Σ_(k=0) ^n C_n ^k (((−1)^k k!)/((x−2i)^(k+1) )) (−1)^(n−k) e^(−x ) also we have ((e^(−x) /(x+2i)))^((n)) =Σ_(k=0) ^n C_n ^k (((−1)^k k!)/((x+2i)^(k+1) )) (−1)^(n−k) e^(−x) ⇒ f^((n)) (x)=(1/(4i)) Σ_(k=0) ^n (−1)^n C_n ^k k!{(1/((x−2i)^(k+1) )) −(1/((x+2i)^(k+1) ))}e^(−x) =(1/(4i)) Σ_(k=0) ^n (−1)^n ((n!)/((n−k)!)) (((x+2i)^(k+1) −(x−2i)^(k+1) )/((x^2 +4)^(k+1) )) e^(−x)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)=\frac{{e}^{−{x}} }{\left({x}−\mathrm{2}{i}\right)\left({x}+\mathrm{2}{i}\right)}\:=\frac{\mathrm{1}}{\mathrm{4}{i}}{e}^{−{x}} \left\{\:\frac{\mathrm{1}}{{x}−\mathrm{2}{i}}\:−\frac{\mathrm{1}}{{x}+\mathrm{2}{i}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{i}}\left\{\frac{{e}^{−{x}} }{{x}−\mathrm{2}{i}}\:−\frac{{e}^{−{x}} }{{x}+\mathrm{2}{i}}\right\}\:\Rightarrow{f}^{\left({n}\right)} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}{i}}\left\{\:\left(\frac{{e}^{−{x}} }{{x}−\mathrm{2}{i}}\right)^{\left({n}\right)} \:−\left(\frac{{e}^{−{x}} }{{x}+\mathrm{2}{i}}\right)^{\left({n}\right)} \right\} \\ $$$${leibniz}\:{formulae}\:{give}\: \\ $$$$\left(\frac{{e}^{−{x}} }{{x}−\mathrm{2}{i}}\right)^{\left({n}\right)} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(\frac{\mathrm{1}}{{x}−\mathrm{2}{i}}\right)^{\left({k}\right)} \:\left({e}^{−{x}} \right)^{\left({n}−{k}\right)} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\frac{\left(−\mathrm{1}\right)^{{k}} {k}!}{\left({x}−\mathrm{2}{i}\right)^{{k}+\mathrm{1}} }\:\left(−\mathrm{1}\right)^{{n}−{k}} \:{e}^{−{x}\:} \:\:{also}\:{we}\:{have}\: \\ $$$$\left(\frac{{e}^{−{x}} }{{x}+\mathrm{2}{i}}\right)^{\left({n}\right)} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} \:{k}!}{\left({x}+\mathrm{2}{i}\right)^{{k}+\mathrm{1}} }\:\left(−\mathrm{1}\right)^{{n}−{k}} \:{e}^{−{x}} \:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}{i}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{n}} \:{C}_{{n}} ^{{k}} \:{k}!\left\{\frac{\mathrm{1}}{\left({x}−\mathrm{2}{i}\right)^{{k}+\mathrm{1}} }\:−\frac{\mathrm{1}}{\left({x}+\mathrm{2}{i}\right)^{{k}+\mathrm{1}} }\right\}{e}^{−{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{i}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{n}} \:\:\frac{{n}!}{\left({n}−{k}\right)!}\:\frac{\left({x}+\mathrm{2}{i}\right)^{{k}+\mathrm{1}} −\left({x}−\mathrm{2}{i}\right)^{{k}+\mathrm{1}} }{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)^{{k}+\mathrm{1}} }\:{e}^{−{x}} \\ $$
Commented by maxmathsup by imad last updated on 27/Oct/18
x=0 ⇒f^((n)) (0) =(1/(4i)) Σ_(k=0) ^n (((−1)^n n!)/((n−k)!)) (((2i)^(k+1) −(−2i)^(k+1) )/4^(k+1) ) =(1/(4i)) Σ_(k=0) ^n (((−1)^n n!)/((n−k)!)) ((2i Im{(2i)^(k+1) })/4^(k+1) ) =(1/2) Σ_(k=0) ^n (((−1)^n n!)/((n−k)!)) ((2^(k+1) sin((((k+1)π)/2)))/2^(2k+2) ) = Σ_(k=0) ^n (((−1)^n n!)/((n−k)!)) ((sin((((k+1)π)/2)))/2^(k+2) )
$${x}=\mathrm{0}\:\Rightarrow{f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\mathrm{1}}{\mathrm{4}{i}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({n}−{k}\right)!}\:\frac{\left(\mathrm{2}{i}\right)^{{k}+\mathrm{1}} \:−\left(−\mathrm{2}{i}\right)^{{k}+\mathrm{1}} }{\mathrm{4}^{{k}+\mathrm{1}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{i}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({n}−{k}\right)!}\:\frac{\mathrm{2}{i}\:{Im}\left\{\left(\mathrm{2}{i}\right)^{{k}+\mathrm{1}} \right\}}{\mathrm{4}^{{k}+\mathrm{1}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({n}−{k}\right)!}\:\:\frac{\mathrm{2}^{{k}+\mathrm{1}} \:{sin}\left(\frac{\left({k}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)}{\mathrm{2}^{\mathrm{2}{k}+\mathrm{2}} } \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({n}−{k}\right)!}\:\frac{{sin}\left(\frac{\left({k}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)}{\mathrm{2}^{{k}+\mathrm{2}} } \\ $$
Commented by maxmathsup by imad last updated on 27/Oct/18
2) we have f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n ⇒ f(x) =Σ_(n=0) ^∞ (−1)^n ( Σ_(k=0) ^n (1/((n−k)!)) ((sin((((k+1)π)/2)))/2^(k+2) ))x^n .
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:\Rightarrow \\ $$$${f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \left(\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\left({n}−{k}\right)!}\:\frac{{sin}\left(\frac{\left({k}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)}{\mathrm{2}^{{k}+\mathrm{2}} }\right){x}^{{n}} \:. \\ $$

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