How can you find the taylor expansion of ln(1-x) about x=0?

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Jan 29, 2016

$\ln \left(1 - x\right) = - x - {x}^{2} / 2 - {x}^{3} / 3 - {x}^{4} / 4 - \ldots$

Explanation:

Note that $\frac{d}{\mathrm{dx}} \left(\ln \left(1 - x\right)\right) = \frac{- 1}{1 - x}$, $x < 1$.

You can express $\frac{- 1}{1 - x}$ as a power series using binomial expansion (for $x$ in the neighborhood of zero).

$\frac{- 1}{1 - x} = - {\left(1 - x\right)}^{- 1}$

$= - \left(1 + x + {x}^{2} + {x}^{3} + \ldots\right)$

To get the Maclaurin Series of $\ln \left(1 - x\right)$, integrate the above "polynomial". You will get

$\ln \left(1 - x\right) = - x - {x}^{2} / 2 - {x}^{3} / 3 - {x}^{4} / 4 - \ldots$

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