Question Number 48668 by maxmathsup by imad last updated on 26/Nov/18

$${let}\:{f}\left({x}\right)=\frac{{e}^{−\mathrm{2}{x}} }{{x}+\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right)\:. \\ $$$$\left.\mathrm{2}\right)\:{develop}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$
Commented by maxmathsup by imad last updated on 27/Nov/18

$$\left.\mathrm{1}\right)\:{leibniz}\:{formula}\:{give}\:{f}^{\left({n}\right)} \left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)^{\left({k}\right)} \left({e}^{−\mathrm{2}{x}} \right)^{\left({n}−{k}\right)} \:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\frac{\left(−\mathrm{1}\right)^{{k}} {k}!}{\left({x}+\mathrm{1}\right)^{{k}+\mathrm{1}} }\:\left(−\mathrm{2}\right)^{{n}−{k}} \:{e}^{−\mathrm{2}{x}} \:\:{and}\: \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\left(−\mathrm{1}\right)^{{k}} {k}!\:\left(−\mathrm{2}\right)^{{n}−{k}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{{n}!}{{k}!\left({n}−{k}\right)!}\left(−\mathrm{2}\right)^{{n}} \:{k}! \\ $$$$=\left(−\mathrm{2}\right)^{{n}} {n}!\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{\left({n}−{k}\right)!} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:\Rightarrow{f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{2}\right)^{{n}} \left\{\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\left({n}−{k}\right)!}\right\}{x}^{{n}} \\ $$$$\Rightarrow{f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \left(\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\:\frac{\left(−\mathrm{2}\right)^{{n}} }{\left({n}−{k}\right)!}\right){x}^{{n}} \:\:\:. \\ $$
Commented by Abdulhafeez Abu qatada last updated on 28/Nov/18

$$ \\ $$$$ \\ $$$${very}\:{beautiful} \\ $$
Commented by maxmathsup by imad last updated on 28/Nov/18

$${thank}\:{you}\:{sir}. \\ $$
Answered by Abdulhafeez Abu qatada last updated on 26/Nov/18

$$ \\ $$$$ \\ $$$${Let}\:{y}\:=\:{f}\left({x}\right)\:=\:\frac{{e}^{−\mathrm{2}{x}} }{{x}\:+\:\mathrm{1}} \\ $$$${y}'\:=\:\frac{−\mathrm{2}\left({x}+\mathrm{1}\right){e}^{−\mathrm{2}{x}} \:−\:{e}^{−\mathrm{2}{x}} }{\left({x}\:+\:\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${y}'\:=\:−\mathrm{2}{y}\:−\:\frac{{y}}{\left({x}+\mathrm{1}\right)} \\ $$$${y}'\left({x}\:+\:\mathrm{1}\right)\:+\:{y}\:=\:−\mathrm{2}{y}\left({x}\:+\:\mathrm{1}\right) \\ $$$$ \\ $$$$\: \\ $$$${y}'\left({x}\:+\:\mathrm{1}\right)\:+\:{y}\:=\:−\mathrm{2}{y}\left({x}\:+\:\mathrm{1}\right) \\ $$$${y}'\left({x}\:+\:\mathrm{1}\right)\:+\:\left(\mathrm{1}+\mathrm{2}\left({x}\:+\:\mathrm{1}\right)\right){y}\:=\:\mathrm{0} \\ $$$${y}'\left({x}\:+\:\mathrm{1}\right)\:+\:\left(\mathrm{2}{x}\:+\:\mathrm{3}\right){y}\:=\:\mathrm{0} \\ $$$$ \\ $$$${Applying}\:{Leibnitz}\:{theorem}\:{to}\:{obtain}\:{the}\:{nth}\:{derivative} \\ $$$${y}^{\left({n}+\mathrm{1}\right)} \left({x}\:+\:\mathrm{1}\right)\:+\:{ny}^{\left({n}\right)} \:+\:{y}^{\left({n}\right)} \left(\mathrm{2}{x}\:+\:\mathrm{3}\right)\:+\:\mathrm{2}{ny}^{\left({n}−\mathrm{1}\right)} \:=\:\mathrm{0} \\ $$$${at}\:{x}\:=\:\mathrm{0} \\ $$$$\left({y}^{\left({n}+\mathrm{1}\right)} \right)_{\mathrm{0}} \:+\:{n}\left({y}^{\left({n}\right)} \right)_{\mathrm{0}} \:+\:\mathrm{3}\left({y}^{\left({n}\right)} \right)_{\mathrm{0}} \:+\:\mathrm{2}{n}\left({y}^{\left({n}−\mathrm{1}\right)} \right)_{\mathrm{0}} \:=\:\mathrm{0} \\ $$$$\left({y}^{\left({n}+\mathrm{1}\right)} \right)_{\mathrm{0}} \:+\:\mathrm{4}\left({y}^{\left({n}\right)} \right)_{\mathrm{0}} \:+\:\mathrm{2}{n}\left({y}^{\left({n}−\mathrm{1}\right)} \right)_{\mathrm{0}} \:=\:\mathrm{0} \\ $$$$ \\ $$$${This}\:{is}\:{a}\:{recurrence}\:{relation},\:{so}\:{we}\:{apply}\:{the}\:{Leibnitz}−{maclaurin}\:{method}\:{to}\:{obtain}\:{a}\:{series} \\ $$$$\left({y}\right)_{\mathrm{0}} \:=\:{f}\left(\mathrm{0}\right)\:=\:\frac{{e}^{−\mathrm{2}\left(\mathrm{0}\right)} }{\mathrm{0}\:+\:\mathrm{1}}\:=\:\mathrm{1} \\ $$$${y}'\left({x}\:+\:\mathrm{1}\right)\:+\:\left(\mathrm{2}{x}\:+\:\mathrm{3}\right){y}\:=\:\mathrm{0} \\ $$$$\left({y}'\right)_{\mathrm{0}} \:+\:\mathrm{3}\left({y}\right)_{\mathrm{0}} \:=\:\mathrm{0} \\ $$$$\left({y}'\right)_{\mathrm{0}} \:+\:\mathrm{3}\left({y}\right)_{\mathrm{0}} \:=\:\mathrm{0} \\ $$$$\left({y}'\right)_{\mathrm{0}} \:+\:\mathrm{3}\:=\:\mathrm{0} \\ $$$$\left({y}'\right)_{\mathrm{0}} \:=\:−\mathrm{3} \\ $$$$\left({y}''\right)_{\mathrm{0}} \:+\:\mathrm{4}\left(−\mathrm{3}\right)\:+\:\mathrm{2}\:=\:\mathrm{0}\: \\ $$$$\left({y}''\right)_{\mathrm{0}} \:\:=\:\mathrm{10} \\ $$$$\left({y}'''\right)_{\mathrm{0}} \:+\:\mathrm{4}\left({y}''\right)_{\mathrm{0}} \:+\:\mathrm{2}\left(\mathrm{2}\right)\left({y}'\right)_{\mathrm{0}} \:=\:\mathrm{0} \\ $$$$\left({y}'''\right)_{\mathrm{0}} \:+\:\mathrm{4}\left(\mathrm{10}\right)\:+\:\mathrm{2}\left(\mathrm{2}\right)\left(−\mathrm{3}\right)\:=\:\mathrm{0} \\ $$$$\left({y}'''\right)_{\mathrm{0}} \:=\:−\mathrm{28} \\ $$$$\left({y}^{\left({iv}\right)} \right)_{\mathrm{0}} \:=\:\mathrm{52} \\ $$$$\left({y}^{\left({v}\right)} \right)_{\mathrm{0}} \:=\:\mathrm{16} \\ $$$${y}\:=\:\left({y}\right)_{\mathrm{0}} \:+\:{x}\left({y}'\right)_{\mathrm{0}} \:+\:\frac{{x}^{\mathrm{2}} \left({y}''\right)_{\mathrm{0}} }{\mathrm{2}!}\:+\:\frac{{x}^{\mathrm{3}} \left({y}'''\right)_{\mathrm{0}} }{\mathrm{3}!}\:+\:\frac{{x}^{\mathrm{4}} \left({y}^{\left({iv}\right)} \right)_{\mathrm{0}} }{\mathrm{4}!}\:+\:\frac{{x}^{\mathrm{5}} \left({y}^{\left({v}\right)} \right)_{\mathrm{0}} }{\mathrm{5}!}\:+… \\ $$$${y}\:=\:\mathrm{1}\:−\:\mathrm{3}{x}\:+\:\mathrm{5}{x}^{\mathrm{2}} \:−\:\frac{\mathrm{14}{x}^{\mathrm{3}} }{\mathrm{3}}\:+\:\frac{\mathrm{13}{x}^{\mathrm{4}} }{\mathrm{6}}\:+\:\frac{\mathrm{2}{x}^{\mathrm{5}} }{\mathrm{15}}\:+… \\ $$$${y}\:=\:{f}\left({x}\right)\:=\:\frac{{e}^{−\mathrm{2}{x}} }{{x}\:+\:\mathrm{1}}\:=\:\mathrm{1}\:−\:\mathrm{3}{x}\:+\:\mathrm{5}{x}^{\mathrm{2}} \:−\:\frac{\mathrm{14}{x}^{\mathrm{3}} }{\mathrm{3}}\:+\:\frac{\mathrm{13}{x}^{\mathrm{4}} }{\mathrm{6}}\:+\:\frac{\mathrm{2}{x}^{\mathrm{5}} }{\mathrm{15}}\:+… \\ $$$$ \\ $$$${Hafeez}\:{Ayinde}\left({Abu}\:{qatada}\right) \\ $$