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

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Question

(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

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$0.5$

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

$1 + x \log_e a + \frac{x^2}{2!} (\log_e a)^2 + \frac{x^3}{3!} (\log_e a)^3 + \dots = $

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The coefficient of $x^n$ in $\frac{1-2x}{e^x}$ is:

$1 + \frac{{\log_e x}}{{1!}} + \frac{{(\log_e x)^2}}{{2!}} + \frac{{(\log_e x)^3}}{{3!}} + \dots \infty = $

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