Bounds for $\log(1-x)$
I would like to show the following
$$-x-x^2 \le \log(1-x) \le -x, \quad x \in [0,1/2].$$
I know that for $|x|<1$, we have $\log(1-x)=-\left(x+\frac{x^2}{2}+\cdots\right)$. The inequality on the right follows because the difference is $\frac{x^2}{2}+ \frac{x^3}{3} + \cdots \ge 0$.
For the inequality on the left, the difference is $\frac{x^2}{2}-\left(\frac{x^3}{3}+\frac{x^4}{4} + \cdots\right)$. How do I show this is nonnegative?
$\endgroup$ 22 Answers
$\begingroup$$$\frac{d}{dx}\left(\log(1-x)+x+x^2\right) = 1+2x-\frac{1}{1-x}=\frac{x(1-2x)}{1-x} $$ is a non-negative function on $\left[0,\frac{1}{2}\right]$, hence the LHS-inequality follows.
$\endgroup$ 1 $\begingroup$$\begin{array}\\ -\ln(1-x) &=\sum_{k=1}^{\infty} \frac{x^k}{k}\\ &=x+\sum_{k=2}^{\infty} \frac{x^k}{k}\\ &=x+x^2\sum_{k=2}^{\infty} \frac{x^{k-2}}{k}\\ &=x+x^2\sum_{k=0}^{\infty} \frac{x^{k}}{k+2}\\ &\le x+x^2\sum_{k=0}^{\infty} \frac{x^{k}}{2}\\ &\le x+\frac{x^2}{2}\sum_{k=0}^{\infty} x^{k}\\ &= x+\frac{x^2}{2}\frac1{1-x}\\ &= x+\frac{x^2}{2(1-x)}\\ &\le x+x^2 \quad \text{if $0 \le x \le \frac12$}\\ \end{array} $
$\endgroup$ 2
