The value of 1 - log 2 + frac{(log 2)^2}{2!} | vedclass

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(B) The given series is $1 - \log 2 + \frac{(\log 2)^2}{2!} - \frac{(\log 2)^3}{3!} + \dots$ This is the expansion of the exponential series $e^x = 1

Vedclass · Accepted Answer

$0.5$

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(B) The given series is $1 - \log 2 + \frac{(\log 2)^2}{2!} - \frac{(\log 2)^3}{3!} + \dots$ This is the expansion of the exponential series $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$ where $x = -\log 2$. Substituting $x = -\log 2$ into the series,we get $e^{-\log 2}$. Using the property of logarithms,$e^{-\log 2} = e^{\log(2^{-1})} = e^{\log(1/2)}$. Since $e^{\log_e y} = y$,we have $e^{\log(1/2)} = \frac{1}{2} = 0.5$.

The value of $1 - \log 2 + \frac{(\log 2)^2}{2!} - \frac{(\log 2)^3}{3!} + \dots$ is

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