The sum to infinity of the series 1 + frac{x^ | vedclass

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(B) We know the expansion of $e^x$ and $e^{-x}$ as follows:$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$$e^{-x} = 1 - x + \

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$\frac{e^x + e^{-x}}{2}$

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(B) We know the expansion of $e^x$ and $e^{-x}$ as follows:$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$$e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \dots$Adding these two series:$e^x + e^{-x} = (1 + 1) + (x - x) + \left(\frac{x^2}{2!} + \frac{x^2}{2!}\right) + \left(\frac{x^3}{3!} - \frac{x^3}{3!}\right) + \left(\frac{x^4}{4!} + \frac{x^4}{4!}\right) + \dots$$e^x + e^{-x} = 2 + 2\frac{x^2}{2!} + 2\frac{x^4}{4!} + \dots$$e^x + e^{-x} = 2 \left(1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots\right)$Therefore,$1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots = \frac{e^x + e^{-x}}{2}$.

The sum to infinity of the series $1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots$ is

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$\frac{\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \infty}{1 + \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + \dots \infty} = $

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