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

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Answer 1 · 2016-01-29T13:04:52

$ln(1-x) = - x - x^2/2 - x^3/3 - x^4/4 - ...$

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

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

$frac{-1}{1-x} = -(1-x)^{-1}$

$= -( 1 + x + x^2 + x^3 + ... )$

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

$ln(1-x) = - x - x^2/2 - x^3/3 - x^4/4 - ...$

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