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

let f(x)=(e^(−2x) /(x+1)) 1) calculate f^((n)) (x) and f^((n)) (0) . 2) develop f at integr serie .

$${let}\:{f}\left({x}\right)=\frac{{e}^{−\mathrm{2}{x}} }{{x}+\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right)\:. \\ $$$$\left.\mathrm{2}\right)\:{develop}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$

Commented by maxmathsup by imad last updated on 27/Nov/18

1) leibniz formula give f^((n)) (x) =Σ_(k=0) ^n C_n ^k ((1/(x+1)))^((k)) (e^(−2x) )^((n−k)) ⇒ f^((n)) (x) =Σ_(k=0) ^n C_n ^k (((−1)^k k!)/((x+1)^(k+1) )) (−2)^(n−k) e^(−2x) and f^((n)) (0) =Σ_(k=0) ^n C_n ^k (−1)^k k! (−2)^(n−k) =Σ_(k=0) ^n ((n!)/(k!(n−k)!))(−2)^n k! =(−2)^n n!Σ_(k=0) ^n (1/((n−k)!)) 2) we have f(x)=Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n ⇒f(x)=Σ_(n=0) ^∞ (−2)^n {Σ_(k=0) ^n (1/((n−k)!))}x^n ⇒f(x)=Σ_(n=0) ^∞ (Σ_(k=0) ^n (((−2)^n )/((n−k)!)))x^n .

$$\left.\mathrm{1}\right)\:{leibniz}\:{formula}\:{give}\:{f}^{\left({n}\right)} \left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)^{\left({k}\right)} \left({e}^{−\mathrm{2}{x}} \right)^{\left({n}−{k}\right)} \:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\frac{\left(−\mathrm{1}\right)^{{k}} {k}!}{\left({x}+\mathrm{1}\right)^{{k}+\mathrm{1}} }\:\left(−\mathrm{2}\right)^{{n}−{k}} \:{e}^{−\mathrm{2}{x}} \:\:{and}\: \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\left(−\mathrm{1}\right)^{{k}} {k}!\:\left(−\mathrm{2}\right)^{{n}−{k}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{{n}!}{{k}!\left({n}−{k}\right)!}\left(−\mathrm{2}\right)^{{n}} \:{k}! \\ $$$$=\left(−\mathrm{2}\right)^{{n}} {n}!\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{\left({n}−{k}\right)!} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:\Rightarrow{f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{2}\right)^{{n}} \left\{\sum_{{k}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\left({n}−{k}\right)!}\right\}{x}^{{n}} \\ $$$$\Rightarrow{f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \left(\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\:\frac{\left(−\mathrm{2}\right)^{{n}} }{\left({n}−{k}\right)!}\right){x}^{{n}} \:\:\:. \\ $$

Commented by Abdulhafeez Abu qatada last updated on 28/Nov/18

very beautiful

$$ \\ $$$$ \\ $$$${very}\:{beautiful} \\ $$

Commented by maxmathsup by imad last updated on 28/Nov/18

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

Answered by Abdulhafeez Abu qatada last updated on 26/Nov/18

Let y = f(x) = (e^(−2x) /(x + 1)) y′ = ((−2(x+1)e^(−2x) − e^(−2x) )/((x + 1)^2 )) y′ = −2y − (y/((x+1))) y′(x + 1) + y = −2y(x + 1) y′(x + 1) + y = −2y(x + 1) y′(x + 1) + (1+2(x + 1))y = 0 y′(x + 1) + (2x + 3)y = 0 Applying Leibnitz theorem to obtain the nth derivative y^((n+1)) (x + 1) + ny^((n)) + y^((n)) (2x + 3) + 2ny^((n−1)) = 0 at x = 0 (y^((n+1)) )_0 + n(y^((n)) )_0 + 3(y^((n)) )_0 + 2n(y^((n−1)) )_0 = 0 (y^((n+1)) )_0 + 4(y^((n)) )_0 + 2n(y^((n−1)) )_0 = 0 This is a recurrence relation, so we apply the Leibnitz−maclaurin method to obtain a series (y)_0 = f(0) = (e^(−2(0)) /(0 + 1)) = 1 y′(x + 1) + (2x + 3)y = 0 (y′)_0 + 3(y)_0 = 0 (y′)_0 + 3(y)_0 = 0 (y′)_0 + 3 = 0 (y′)_0 = −3 (y′′)_0 + 4(−3) + 2 = 0 (y′′)_0 = 10 (y′′′)_0 + 4(y′′)_0 + 2(2)(y′)_0 = 0 (y′′′)_0 + 4(10) + 2(2)(−3) = 0 (y′′′)_0 = −28 (y^((iv)) )_0 = 52 (y^((v)) )_0 = 16 y = (y)_0 + x(y′)_0 + ((x^2 (y′′)_0 )/(2!)) + ((x^3 (y′′′)_0 )/(3!)) + ((x^4 (y^((iv)) )_0 )/(4!)) + ((x^5 (y^((v)) )_0 )/(5!)) +... y = 1 − 3x + 5x^2 − ((14x^3 )/3) + ((13x^4 )/6) + ((2x^5 )/(15)) +... y = f(x) = (e^(−2x) /(x + 1)) = 1 − 3x + 5x^2 − ((14x^3 )/3) + ((13x^4 )/6) + ((2x^5 )/(15)) +... Hafeez Ayinde(Abu qatada)

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

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