In the expansion of 2 ln x - ln(x + 1) - ln(x | vedclass

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(A) Given expression: $f(x) = 2 \ln x - \ln(x + 1) - \ln(x - 1)$$= 2 \ln x - \ln((x + 1)(x - 1))$$= 2 \ln x - \ln(x^2 - 1)$$= 2 \ln x - \ln(x^2(1 - \f

Vedclass · Accepted Answer

$0.5$

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(A) Given expression: $f(x) = 2 \ln x - \ln(x + 1) - \ln(x - 1)$$= 2 \ln x - \ln((x + 1)(x - 1))$$= 2 \ln x - \ln(x^2 - 1)$$= 2 \ln x - \ln(x^2(1 - \frac{1}{x^2}))$$= 2 \ln x - (\ln x^2 + \ln(1 - \frac{1}{x^2}))$$= 2 \ln x - 2 \ln x - \ln(1 - \frac{1}{x^2})$$= - \ln(1 - \frac{1}{x^2})$Using the expansion $\ln(1 - t) = - (t + \frac{t^2}{2} + \frac{t^3}{3} + \dots)$,where $t = \frac{1}{x^2}$:$= - [ - (\frac{1}{x^2} + \frac{1}{2(x^2)^2} + \dots) ]$$= \frac{1}{x^2} + \frac{1}{2x^4} + \dots$The coefficient of $x^{-4}$ is $\frac{1}{2} = 0.5$.

In the expansion of $2 \ln x - \ln(x + 1) - \ln(x - 1)$,the coefficient of $x^{-4}$ is

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