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Question Number 49804 by maxmathsup by imad last updated on 10/Dec/18

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

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

Commented by Abdo msup. last updated on 11/Dec/18

1) letfirst find  f^((n)) (x)  f^((n)) (x) =((1/(x+1)) e^(−x) )^((n))  leibniz formula give  f^((n)) (x) = Σ_(k=0) ^n  C_n ^k  ((1/(x+1)))^((k{) (e^(−x) )^((n−k))   =Σ_(k=0) ^n  C_n ^k   (((−1)^k k!)/((x+1)^(k+1) )) (−1)^(n−k)  e^(−x)   =(−1)^n  Σ_(k=0) ^n  ((k! C_n ^k )/((x+1)^(k+1) )) e^(−x)   =(−1)^n  Σ_(k=0) ^n     ((n!)/((n−k)!))  (e^(−x) /((x+1)^(k+1) )) ⇒  f^((n)) (0) =n!(−1)^n  Σ_(k=0) ^n  (1/((n−k)!))  and  f^((n)) (1) =n!(−1)^n  Σ_(k=0) ^n    (1/((n−k)!)) (e^(−1) /2^(k+1) ) .  2) f(x) =Σ_(n=0) ^∞    ((f^((n)) (0))/(n!)) x^n   =Σ_(n=0) ^∞    (−1)^n (Σ_(k=0) ^n  (1/((n−k)!:)))x^n   but   Σ_(k=0) ^n    (1/((n−k)!)) =_(n−k=p)    Σ_(p=0) ^n  (1/(p!)) ⇒  f(x) =Σ_(n=0) ^∞  (−1)^n (Σ_(p=0) ^n  (1/(p!))) x^(n )   .

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

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