1 + frac{(log_e n)^2}{2!} + frac{(log_e n)^4} | vedclass

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Question

(C) We know the expansion of the hyperbolic cosine function: $\cosh(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots = \frac{e^x + e^{-x}}{2}$.Substit

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We know the expansion of the hyperbolic cosine function: $\cosh(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dots = \frac{e^x + e^{-x}}{2}$.Substituting $x = \log_e n$,we get:$1 + \frac{(\log_e n)^2}{2!} + \frac{(\log_e n)^4}{4!} + \dots = \frac{e^{\log_e n} + e^{-\log_e n}}{2}$.Since $e^{\log_e n} = n$ and $e^{-\log_e n} = e^{\log_e(n^{-1})} = n^{-1}$,the expression becomes:$\frac{n + n^{-1}}{2}$.

$1 + \frac{(\log_e n)^2}{2!} + \frac{(\log_e n)^4}{4!} + \dots = $

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