Question Number 172455 by pablo1234523 last updated on 27/Jun/22
$$\mathrm{using}\:\mathrm{maclaurin}'\mathrm{s}\:\mathrm{series}\:\mathrm{exapand} \\ $$$$\frac{{x}}{\mathrm{2}}\left(\frac{{e}^{{x}} +\mathrm{1}}{{e}^{{x}} −\mathrm{1}}\right)\:\mathrm{upto}\:{x}^{\mathrm{4}} \:\mathrm{term} \\ $$
Answered by mr W last updated on 29/Jun/22
$$=\frac{{x}}{\mathrm{2}}\left(\frac{{e}^{\frac{{x}}{\mathrm{2}}} +{e}^{−\frac{{x}}{\mathrm{2}}} }{{e}^{\frac{{x}}{\mathrm{2}}} −{e}^{−\frac{{x}}{\mathrm{2}}} }\right) \\ $$$$=\frac{{x}}{\mathrm{2}}\mathrm{coth}\:\frac{{x}}{\mathrm{2}} \\ $$$$=\frac{{x}}{\mathrm{2}}\left(\frac{\mathrm{2}}{{x}}+\frac{{x}}{\mathrm{6}}−\frac{{x}^{\mathrm{3}} }{\mathrm{360}}+{O}\left({x}^{\mathrm{5}} \right)\right) \\ $$$$=\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{12}}−\frac{{x}^{\mathrm{4}} }{\mathrm{720}}+{O}\left({x}^{\mathrm{6}} \right) \\ $$
Commented by mr W last updated on 29/Jun/22
