Q:

Taylor expansion of $\ln(1-x)$

I was just wondering where the minus sign in the first term of the Taylor expansion of $ \ln(1-x) $ comes from? In wikipedia page and everywhere else $\ln(1-x)$ is given by 
$$
\ln(1-x) = -x-\dots
$$
But assuming $x$ is small and expand around $1$, I got
$$
\ln(1-x) \approx \ln(1) + \frac{d(\ln(1-x))}{dx}\bigg\vert_{x=0}[(1-x)-1] \approx 0 + \frac{1}{1-x}\bigg\vert_{x=0}(-1)(-x) = x.
$$
Using the definition of Taylor expansion $f(z) \approx f(a) + \frac{df(z)}{dz}\bigg\vert_{z=a}(z-a) $, where here $z=1-x$, $f(z) = \ln(1-z)$ and $a=1$.
I know you can get $\ln(1-x) \approx -x$ by e.g. substitute $x\rightarrow -x$ into the expansion of $\ln(1+x)$ and through other methods etc. But I still don't quite get how you can get the minus sign from Taylor expansion alone. Thanks. 

A:

$$y=\ln(1-x)$$
$$y'=-\frac{1}{1-x}=-\sum_{n=0}^{\infty}x^n$$
so
$$\ln(1-x)=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=-\sum_{n=1}^{\infty}\frac{x^{n}}{n}$$

A:

If one considers
$$
f(x)=\ln (1-x),\qquad |x|<1,
$$one has
$$
f(0)=0,\quad f'(x)=-\frac{1}{1-x},\quad f'(0)=-1,\quad f''(x)=-\frac{1}{(1-x)^2},\quad f''(0)=-1,
$$ giving, by the Taylor expansion,
$$
f(x)=0-x-\frac{x^2}2+O(x^3)
$$as $x \to 0$.

