# 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.

## 2 Answers

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 $$x \to 0$$.

• Thanks for the answer but what about the $(z-a)$ part in the Taylor expansion $f(z) = f(a)+f^\prime(a)(z-a)$? Substitute $z=1-x$ and $a=1$ gives a $-x$ though? – Lepnak Apr 19 at 2:04
• The Taylor series centred at $0$ is $$f(x)=f(0)+f'(0)x +\cdots.$$ Use $f(0)$ and $f'(0)$ from Olivier Oloa's answer and you should get the right answer. In your OP, you are actually expanding $f(x)$ around $0$, not around $1$ (where $f(x)=\ln (1-x)$). So $a=0$. By the way, if you substitute $z=1-x$ where $f(z)=\ln (1-z)$, you would get $\ln(1-(1-x))=\ln x$, rather than $\ln(1-x)$ (which is what you want). So no need to do this substitution. – Minus One-Twelfth Apr 19 at 2:28
• Hmm I think I see what I did wrong. Thanks for all your answers. – Lepnak Apr 19 at 2:37

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