Prove $|\log(1-x)+x|\leqslant cx^2$

Question

Problem: Prove $|\log(1-x)+x|\leqslant cx^2$(c is a constant) holds for $|x|<\frac{1}{2}$.

My proof: $\log(x-1)=-x-O(x^2)$ then $\log(x-1)+x=-x-O(x^2)+x\implies \log(x-1)+x=O(x^2)\implies|\log(x-1)+x|=O(x^2)\implies |\log(x-1)+x|\leqslant cx^2$

However the author points out that the latter identity only holds for $|x|⩽\frac{1}{2}$. This later point did not show in my derivation.

Question:

What am I doing wrong?

Thanks in advance!

Answer 1

Kavi Rama Murthy has explained why the OP's work is invalid. I am presenting an explicit way to extract such a constant $c$.

By Taylor's Theorem, we have $$\ln(1-x)=-x-\frac{1}{2}\,x^2-\frac{1}{3\,\big(1-\xi(x)\big)^3}\,x^3\text{ for all }x<1\,,$$ where $\xi(x)$ is a number between $0$ and $x$. That is, $$-\ln(1-x)-x\leq \frac{1}{2}\,x^2+\frac{1}{3\,\left(1-\frac{1}{2}\right)^3}\,\left(x^2\cdot\frac{1}{2}\right)=\frac{11}{6}\,x^2\text{ if }0\leq x<\frac{1}{2}$$ and $$-\ln(1-x)-x\leq \frac{1}{2}x^2\text{ if }x<0\,.$$ That is, $$\big|\ln(1-x)+x\big|\leq \frac{11}{6}\,x^2\text{ for all }x\in\left[-\frac12,+\frac12\right]\,.$$

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In fact, let $f:\mathbb{R}_{<1}\to\mathbb{R}$ be the function defined by $$f(x):=\left\{\begin{array}{ll} \frac{-x-\ln(1-x)}{x^2}\,,&\text{if }x\in(-\infty,0)\cup(0,1)\,,\\ \frac{1}{2}\,,&\text{if }x=0\,. \end{array}\right.$$ Then, $f$ is an increasing positive function. Therefore, the best constant $c$ in this problem is $$f\left(\dfrac12\right)=4\,\ln(2)-2\approx 0.77259\,.$$

Answer 2

$\log (1-x)$ is defined only for $x<1$. As $x$ increases to $1$, $\log (1-x)+x \to =-\infty$ so the inequality cannot be true for $x$ near $1$. The question you have to think about is: what exactly is the meaning of $O(x^{2})$ here.

Answer 3

Rewrite your proof in this way: $$\log(1-x)+x=-x+O(x^2)+x=O(x^2)$$ so there is a constant $c>0$ such that $$|\log(1-x)+x|\leqslant cx^2$$

Answer 4

Let $f(x)=\frac{\log(1-x)+x}{x^2},x\in[-1/2,0)\cup(0,1/2]$ and $f(0)=\lim_{x\to 0}\frac{\log(1-x)+x}{x^2}$. So $f(x)$ is a continuous function defined on the compact set [-1/2,1/2]. Therefore, the function $f(x)$ has Maximum and minimum denoted by $M$ and $m$ respectively. So $|f(x)|\leq C$, where $C=max\{|M|,|m|\}$ for $x\in[-1/2,1/2]$. And then $$\left|\log(1-x)+x\right|\leq Cx^2.$$

Answer 5

For $|x|<1,$ $\ln (1-x) =-(x+x^2/2+x^3/3 + \cdots)$ Thus

$$\ln (1-x) + x = -(x^2/2+x^3/3 + \cdots)$$

This gives

$$|\ln (1-x) + x| \le |x^2/2+x^3/3 + \cdots)| \le x^2(1/2 + |x|/3 + |x|^2/4 +\cdots)$$ $$ \le x^2(1+|x|+|x|^2+ \cdots) = x^2/(1-|x|).$$

If we now use $|x|\le 1/2,$ we see the last expression is $\le x^2(1-1/2) = 2x^2.$ So we may take $c=2.$