If S = sumlimits_{n = 0}^infty frac{(log x)^{ | vedclass

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(C) We know that the Taylor series expansion for $\cosh(y)$ is given by $\sum\limits_{n=0}^\infty \frac{y^{2n}}{(2n)!} = \frac{e^y + e^{-y}}{2}$.Given

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$\frac{1}{2}(x + x^{-1})$

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(C) We know that the Taylor series expansion for $\cosh(y)$ is given by $\sum\limits_{n=0}^\infty \frac{y^{2n}}{(2n)!} = \frac{e^y + e^{-y}}{2}$.Given $S = \sum\limits_{n=0}^\infty \frac{(\log x)^{2n}}{(2n)!}$,we substitute $y = \log x$.Thus,$S = \frac{e^{\log x} + e^{-\log x}}{2}$.Since $e^{\log x} = x$ and $e^{-\log x} = e^{\log(x^{-1})} = x^{-1}$,we get $S = \frac{x + x^{-1}}{2}$.

If $S = \sum\limits_{n = 0}^\infty \frac{(\log x)^{2n}}{(2n)!}$,then $S$ =

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