Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 177957 by infinityaction last updated on 11/Oct/22

$$\mathrm{let}\mathrm{a}>0\mathrm{find}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{infinite}$$$$\mathrm{series}$$$$1+\frac{{\left(\mathrm{loga}\right)}^2}{2!}+\frac{{\left(\mathrm{loga}\right)}^4}{4!}+\frac{{\left(\mathrm{loga}\right)}^6}{6!}...$$

Answered by mr W last updated on 11/Oct/22

$$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$$$e^{-x}=1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+...$$$$e^x+e^{-x}=2(1+\frac{x^2}{2!}+\frac{x^4}{4!}+...)$$$$\frac{e^x+e^{-x}}{2}=1+\frac{x^2}{2!}+\frac{x^4}{4!}+...$$$$withx=\log a$$$$\frac{a+\frac{1}{a}}{2}=1+\frac{{\left(\log a\right)}^2}{2!}+\frac{{\left(\log a\right)}^4}{4!}+...$$$$\Rightarrow 1+\frac{{\left(\log a\right)}^2}{2!}+\frac{{\left(\log a\right)}^4}{4!}+...=\frac{a^2+1}{2a}$$

Commented by infinityaction last updated on 11/Oct/22

$$thankyousir$$

Commented by Tawa11 last updated on 11/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by Ar Brandon last updated on 11/Oct/22

$$1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+\cdot \cdot \cdot =\cosh \left(x\right)$$$$\Rightarrow 1+\frac{{\left(\log a\right)}^2}{2!}+\frac{{\left(\log a\right)}^4}{4!}+\frac{{\left(\log a\right)}^6}{6!}+\cdot \cdot \cdot =\cosh \left(\log a\right)$$$$=\frac{1}{2}(e^{\log a}+e^{-\log a})=\frac{1}{2}(a+\frac{1}{a})=\frac{a^2+1}{2a}$$

Commented by Tawa11 last updated on 11/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com