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

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