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Question Number 163467 by Zaynal last updated on 07/Jan/22

Answered by mr W last updated on 07/Jan/22

$$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}...=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{2n}}{\left(2n\right)!}+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{2n+1}}{(2n+1)!}...\left(i\right)$$$$e^{-x}=1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+...=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{2n}}{\left(2n\right)!}-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{2n+1}}{(2n+1)!}...\left(ii\right)$$$$\left(i\right)+\left(ii\right):$$$$e^x+e^{-x}=2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{2n}}{\left(2n\right)!}$$$$\Rightarrow \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{2n}}{\left(2n\right)!}=\frac{e^x+e^{-x}}{2}=\cosh x$$

Commented by Tawa11 last updated on 07/Jan/22

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

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