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I'm looking for lower bounds on the function $x\mapsto \log(1-x)$ with $0\leq x<1$. Actually, I only need $x\in [0,c]$, $c<1$.

So far, I've tried using Taylor expansion with integral remainder but since all derivatives are negative, I can only get an upper-bound.

Teddy38
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anonymus
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  • There is no lower bound. log(1- x) goes to negative infinity as x goes to 1. For your other question, with c< 1 (which is not at all the same as your first question) the lower bound is log(1- c). – user247327 Nov 13 '17 at 18:37
  • @user247327: what do you have against functions with the same behaviour which are less than $\log(1-x)$ on $[0,1)$? – Jack D'Aurizio Nov 13 '17 at 18:39
  • $$\sqrt{1-x}-\frac{1}{\sqrt{1-x}}$$ is a chance, for instance. There are better ones, among rational or algebraic approximations. – Jack D'Aurizio Nov 13 '17 at 18:41
  • @JackD'Aurizio : do you think it's possible to get a lower bound with polynomial/Taylor-like form when $x$ is close to $0$ (or less than $\frac{1}{2}$ let's say) ? Some kind of dual bound of this one : $x-x^2/2 \leq \log(1+x)\leq x-x^2/2 + x^3/3$. – anonymus Nov 13 '17 at 18:45
  • @anonymus: if you fix $c\in(0,1)$, you may achieve something similar over $[0,c]$, with the involved coefficients depending on $c$. Just bound $\frac{1}{1+x}$ by convexity and apply termwise integration. What is the purpose of producing such inequality, if I may ask? – Jack D'Aurizio Nov 13 '17 at 18:48
  • @JackD'Aurizio : I'm studying the asymptotic behaviour of a big sum. Since it is a sum, I can't just substitute the terms by their asymptotic approximations. I need to (lower+upper) bound the terms of my sum "by hand". – anonymus Nov 13 '17 at 18:53
  • @anonymus: If I am right about the problem you are currently approaching, I guess you may find this answer inspiring: https://math.stackexchange.com/a/1409131/44121 – Jack D'Aurizio Nov 13 '17 at 18:55
  • @JackD'Aurizio : thank you for the suggestion. I encounter binomial coefficient (so naturally, factorial approximation). But so far, I can't really see how it is going to help me. Anyway, I'm pretty sure that my question has some answer. – anonymus Nov 13 '17 at 19:03
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    How good a bound do you need? For every $c \in (0,1)$ there is a $K_c$ with $-K_c x \leqslant \log (1-x) \leqslant -x$ on $[0,c]$. For many things, that's good enough. If you need something better, $$-\sum_{k = 1}^{m-1} \frac{x^k}{k} - \frac{x^m}{m(1-x)}$$ may do it for a suitable $m$. – Daniel Fischer Nov 13 '17 at 19:12
  • @anonymus Did the comments and/or the answer settle your question? – Clement C. Nov 16 '17 at 18:02

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From this very handy cheatsheet of inequalities: for $0\leq x<1$, $$\begin{align} \ln(1-x) \geq \frac{-x}{\sqrt{1-x}} \tag{$\dagger$} \end{align}$$

Proof. let $f$ be defined on $(-1,1)$ by $f(x) = \ln(1-x) + \frac{x}{\sqrt{1-x}}$. It is $\mathcal{C}^1$ on its domain, $f(0)=0$, and $$ f'(x) = \frac{2-x-2\sqrt{1-x}}{2(1-x)^{3/2}}\,. $$ It is easy to see that, for $-1< x<1$, $f'(x)\geq 0$ (it suffices to analyze the numerator, as the denominator is positive). This gives the inequality.

Clement C.
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