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Question Number 37269 by abdo.msup.com last updated on 11/Jun/18
let f(x)= e^(−2x) ln(1+x) developp f at integr serie .
$${let}\:{f}\left({x}\right)=\:{e}^{−\mathrm{2}{x}} {ln}\left(\mathrm{1}+{x}\right) \\ $$$${developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$
Commented by prof Abdo imad last updated on 24/Jun/18
f^((n)) (x)=Σ_(k=0) ^n (ln(1+x))^((k)) (e^(−2x) )^((n−k)) but (e^(−2x) )^((p)) =(−2)^p e^(−2x) ln(1+x)^((1)) = (1/(1+x)) ⇒(ln(1+x))^((p)) = (((−1)^(p−1) (p−1)!)/((1+x)^p )) f^((n)) (x)=(−2)^n e^(−2x) ln(1+x) +Σ_(k=1) ^n (((−1)^(k−1) (k−1)!)/((1+x)^k )) (−2)^(n−k) e^(−2x) ⇒ f^((n)) (0) =Σ_(k=1) ^n (−1)^(k−1) (k−1)!(−2)^(n−k) =(−2)^n Σ_(k=1) ^n (−1)^(k−1) (k−1)!(−2)^(−k) . f(x)=Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n =Σ_(n=0) ^∞ (((−2)^n )/(n!)){Σ_(k=1) ^n (−1)^(k−1) (k−1)!(−2)^(−k) }x^n
$${f}^{\left({n}\right)} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \left({ln}\left(\mathrm{1}+{x}\right)\right)^{\left({k}\right)} \left({e}^{−\mathrm{2}{x}} \right)^{\left({n}−{k}\right)} \:{but} \\ $$$$\left({e}^{−\mathrm{2}{x}} \right)^{\left({p}\right)} =\left(−\mathrm{2}\right)^{{p}} \:{e}^{−\mathrm{2}{x}} \\ $$$${ln}\left(\mathrm{1}+{x}\right)^{\left(\mathrm{1}\right)} =\:\frac{\mathrm{1}}{\mathrm{1}+{x}}\:\Rightarrow\left({ln}\left(\mathrm{1}+{x}\right)\right)^{\left({p}\right)} =\:\frac{\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \left({p}−\mathrm{1}\right)!}{\left(\mathrm{1}+{x}\right)^{{p}} } \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\left(−\mathrm{2}\right)^{{n}} \:{e}^{−\mathrm{2}{x}} {ln}\left(\mathrm{1}+{x}\right) \\ $$$$+\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left(\mathrm{1}+{x}\right)^{{k}} }\:\left(−\mathrm{2}\right)^{{n}−{k}} \:{e}^{−\mathrm{2}{x}} \:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!\left(−\mathrm{2}\right)^{{n}−{k}} \\ $$$$=\left(−\mathrm{2}\right)^{{n}} \:\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!\left(−\mathrm{2}\right)^{−{k}} \:. \\ $$$${f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{2}\right)^{{n}} }{{n}!}\left\{\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!\left(−\mathrm{2}\right)^{−{k}} \right\}{x}^{{n}} \\ $$
Commented by prof Abdo imad last updated on 24/Jun/18
another way but easy we have e^(−2x) =Σ_(n=0) ^∞ (((−2x)^n )/(n!)) and ln^′ (1+x)= (1/(1+x)) =Σ_(n=0) ^∞ (−1)^n x^n with ∣x∣<1⇒ ln(1+x)=Σ_(n=0) ^∞ (((−1)^n )/(n+1))x^(n+1) +c (c=0) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) x^n ⇒ f(x)=(Σ_(n=0) ^∞ (((−2)^n )/(n!))x^n ) (Σ_(n=1) ^∞ (((−1)^(n−1) )/n)x^n ) =(1+Σ_(n=1) ^∞ (((−2)^n )/(n!))x^n )(Σ_(n=1) ^∞ (((−1)^(n−1) )/n)x^n ) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n)x^n +Σ_(n=1) ^∞ c_n x^n with c_n = Σ_(i+j=n) a_i b_j =Σ_(i+j=n) (((−2)^i )/(i!)) (((−1)^(j−1) )/j) =Σ_(i=1) ^(n−1) a_i b_(n−i) =Σ_(i=1) ^(n−1) (((−2)^i )/(i!)) (((−1)^(n−i−1) )/(n−i))
$${another}\:{way}\:{but}\:{easy}\:\:{we}\:{have} \\ $$$${e}^{−\mathrm{2}{x}} =\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{2}{x}\right)^{{n}} }{{n}!}\:\:{and} \\ $$$${ln}^{'} \left(\mathrm{1}+{x}\right)=\:\frac{\mathrm{1}}{\mathrm{1}+{x}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \:\:{with}\:\mid{x}\mid<\mathrm{1}\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \:+{c}\:\left({c}=\mathrm{0}\right) \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{x}^{{n}} \:\Rightarrow \\ $$$${f}\left({x}\right)=\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{2}\right)^{{n}} }{{n}!}{x}^{{n}} \right)\:\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{x}^{{n}} \right) \\ $$$$=\left(\mathrm{1}+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{2}\right)^{{n}} }{{n}!}{x}^{{n}} \right)\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{x}^{{n}} \right) \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{x}^{{n}} \:\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:{c}_{{n}} {x}^{{n}} \:\:{with} \\ $$$${c}_{{n}} =\:\sum_{{i}+{j}={n}} {a}_{{i}} \:{b}_{{j}} \:\:=\sum_{{i}+{j}={n}} \:\frac{\left(−\mathrm{2}\right)^{{i}} }{{i}!}\:\:\frac{\left(−\mathrm{1}\right)^{{j}−\mathrm{1}} }{{j}} \\ $$$$=\sum_{{i}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:{a}_{{i}} {b}_{{n}−{i}} \:=\sum_{{i}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\:\frac{\left(−\mathrm{2}\right)^{{i}} }{{i}!}\:\frac{\left(−\mathrm{1}\right)^{{n}−{i}−\mathrm{1}} }{{n}−{i}} \\ $$

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