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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.

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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 $x \to 0$.

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  • $\begingroup$ 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? $\endgroup$ – Lepnak Apr 19 at 2:04
  • $\begingroup$ 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. $\endgroup$ – Minus One-Twelfth Apr 19 at 2:28
  • $\begingroup$ Hmm I think I see what I did wrong. Thanks for all your answers. $\endgroup$ – Lepnak Apr 19 at 2:37
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$$y=\ln(1-x)$$ $$y'=-\frac{1}{1-x}=-\sum_{n=0}^{\infty}x^n$$ so $$\ln(1-x)=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=-\sum_{n=1}^{\infty}\frac{x^{n}}{n}$$

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