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I'm reading an article which states the following:

For summands such that $x < \epsilon$, we have the inequality $\ln(1-x) > -\frac{x}{1-\epsilon}.$

Why is this true?

I could rewrite the following (note $x$ and $\epsilon$ are both positive integers and $x < 1$.) $$\begin{align*} x&<\epsilon\\ 1-x&>1-\epsilon\\ \ln(1-x)&>\ln(1-\epsilon) \end{align*}$$ Now if, $f(\epsilon) = \ln(1-\epsilon)$, it looks like I need some term $xf'(\epsilon)$ on the right to get the statement from the article. Like some sort of Taylor approximation? Why is the statement true?

Michael Rozenberg
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Di-lemma
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2 Answers2

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For $0<x<\varepsilon$ you have: $$\ln (1-x) = -\int_0^x\frac{1}{1-t}dt > -\int_0^x\frac{1}{1-\varepsilon}dt = -\frac{x}{1-\varepsilon}$$

Bill O'Haran
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trancelocation
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Assume $x>0$, The taylor expansion of $\ln(1-x)$ is (taylor series of ln(1+x)?) $$ \ln(1-x) = (-x) - (- x)^2/2 + (-x)^3/n+\cdots $$ because $x$ is positive this is less than equal of $$ -(x + x^2/2+x^3/3+ \cdots \geq -x(1+x+x^2+\cdots)=-\frac{x}{1-x}. $$ Making the denominator as small as possible by setting $x=\epsilon$, we make the fraction as larger and the minus means that we make the whole expression smaller we find that $$ -\frac{x}{1-x} \geq -\frac{x}{1-\epsilon} $$ and we proven $$ \ln(1-x) \geq -\frac{x}{1-\epsilon} $$

Stefan
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