Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com