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 - ...#