Question

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

Answer 1

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

Explanation:

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 - ...`