Question Number 170198 by Mastermind last updated on 18/May/22
$${Find}\:{the}\:{first}\:{three}\:{term}\:{of}\:{the}\:{series} \\ $$$${for}\:{e}^{{x}} {ln}\left(\mathrm{1}+{x}\right) \\ $$$$ \\ $$$${Mastermind} \\ $$
Answered by Mathspace last updated on 18/May/22
$${e}^{{x}} =\sum_{{n}=\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{{n}!}=\mathrm{1}+{x}\:+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+{o}\left({x}^{\mathrm{4}} \right) \\ $$$${ln}^{\left(\mathrm{1}\right)} \left(\mathrm{1}+{x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}}=\mathrm{1}−{x}+{x}^{\mathrm{2}} −{x}^{\mathrm{3}} +{o}\left({x}^{\mathrm{4}} \right)\:\Rightarrow \\ $$$${e}^{{x}} {ln}\left(\mathrm{1}+{x}\right)=\left(\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+{o}\left({x}^{\mathrm{4}} \right)\right)\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} −{x}^{\mathrm{3}} +{o}\left({x}^{\mathrm{4}} \right)\right) \\ $$$$=\mathrm{1}−{x}+{x}^{\mathrm{2}} −{x}^{\mathrm{3}} +{x}−{x}^{\mathrm{2}} +{x}^{\mathrm{3}} −{x}^{\mathrm{4}} \\ $$$$+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{3}} }{\mathrm{2}}+\frac{{x}^{\mathrm{4}} }{\mathrm{2}}−\frac{{x}^{\mathrm{5}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{6}}−\frac{{x}^{\mathrm{4}} }{\mathrm{6}}+\frac{{x}^{\mathrm{5}} }{\mathrm{6}}−\frac{{x}^{\mathrm{6}} }{\mathrm{6}}+{o}\left({x}^{\mathrm{6}} \right) \\ $$$$=.... \\ $$
Commented by Mastermind last updated on 20/May/22
$${That}'{s}\:{last}\:{answer}? \\ $$
Commented by Mastermind last updated on 19/May/22
$${thanks}\:{sir} \\ $$