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Question Number 40113 by maxmathsup by imad last updated on 15/Jul/18

let f(x)=ln(2+x) 1) give D_n (0) of f 2) developp f at integr serie

$${let}\:{f}\left({x}\right)={ln}\left(\mathrm{2}+{x}\right) \\ $$$$\left.\mathrm{1}\right)\:{give}\:{D}_{{n}} \left(\mathrm{0}\right)\:{of}\:{f} \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie} \\ $$

Commented by prof Abdo imad last updated on 17/Jul/18

1) we have f(x) =Σ_(k=0) ^n ((f^((k)) (0))/(k!)) x^k + (x^(n+1) /((n+1)!))θ(x) with lim_(x→0) θ(x)=0 but f^′ (x)= (1/(x+2)) ⇒f^((k)) (x)=((1/(x+2)))^((k−1)) =(((−1)^(k−1) (k−1)!)/((x+2)^k )) ⇒f^((k)) (0) = (((−1)^(k−1) (k−1)!)/(k!2^k )) for k≥1 ⇒ f(x) =ln(2) +Σ_(k=1) ^n (((−1)^(k−1) )/(k2^k )) x^k +(x^(n+1) /((n+1)!))θ(x) 2)f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n = f(0) +Σ_(n=1) ^∞ (1/(n!)){(((−1)^(n−1) (n−1)!)/2^n )}x^n =ln(2) + Σ_(n=1) ^∞ (((−1)^(n−1) )/(n 2^n )) x^n .

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \frac{{f}^{\left({k}\right)} \left(\mathrm{0}\right)}{{k}!}\:{x}^{{k}} \:\:+\:\frac{{x}^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)!}\theta\left({x}\right) \\ $$$${with}\:{lim}_{{x}\rightarrow\mathrm{0}} \theta\left({x}\right)=\mathrm{0}\:\:{but} \\ $$$${f}^{'} \left({x}\right)=\:\frac{\mathrm{1}}{{x}+\mathrm{2}}\:\Rightarrow{f}^{\left({k}\right)} \left({x}\right)=\left(\frac{\mathrm{1}}{{x}+\mathrm{2}}\right)^{\left({k}−\mathrm{1}\right)} =\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}+\mathrm{2}\right)^{{k}} } \\ $$$$\Rightarrow{f}^{\left({k}\right)} \left(\mathrm{0}\right)\:=\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{{k}!\mathrm{2}^{{k}} }\:{for}\:{k}\geqslant\mathrm{1}\:\Rightarrow \\ $$$${f}\left({x}\right)\:={ln}\left(\mathrm{2}\right)\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}\mathrm{2}^{{k}} }\:{x}^{{k}} \:\:+\frac{{x}^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)!}\theta\left({x}\right) \\ $$$$\left.\mathrm{2}\right){f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \\ $$$$=\:{f}\left(\mathrm{0}\right)\:+\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}!}\left\{\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}^{{n}} }\right\}{x}^{{n}} \\ $$$$={ln}\left(\mathrm{2}\right)\:+\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}\:\mathrm{2}^{{n}} }\:{x}^{{n}} \:. \\ $$$$ \\ $$

Commented by prof Abdo imad last updated on 17/Jul/18

f^((k)) (0)= (((−1)^(k−1) (k−1)!)/2^k ) for k≥1

$${f}^{\left({k}\right)} \left(\mathrm{0}\right)=\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\mathrm{2}^{{k}} }\:\:{for}\:{k}\geqslant\mathrm{1} \\ $$

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