Lower bound of log(1-x)

Question

I've been thinking about this problem for several weeks, and I can't seem to come to an answer. I've tried using Taylor's Series with Taylor's Remainder Theorem, but I've hit a dead end in that direction. I've been told the lower bound is $$-x-\frac{x^2}{2}-...-\frac{x^k}{k}-\frac{x^k}{k} \leq log(1-x) \leq -x-\frac{x^2}{2}-...-\frac{x^k}{k}$$ where the bounds of x are $0 \leq x \leq \frac{1}{2}$. Does anyone have a clue how to prove this?

Answer 1

Consider $$\frac{d}{dx}\left(\log(1-x)+x+x^2\right) = 1+2x-\frac{1}{1-x}=\frac{x(1-2x)}{1-x} $$ now the above is a non-negative function on $\left[0,\frac{1}{2}\right]$, hence $$-x-x^2 \le \log(1-x) \le -x, \quad x \in [0,1/2].$$

Answer 2

Let $f(x)=\log(1-x)$ for $x \in I:=[0, \frac{1}{2}]$. For $x \in I$ we have

$f'(x)=\frac{1}{x-1}<0$, hence $f$ is decreasing on $I$. Therefore

$- \log 2=f(1/2) \le f(x) \le f(0)=0$ for all $x \in I$.