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?
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1What is $k$? is the last term in the LHS a mistake? – Nathaniel Bubis Feb 23 '18 at 12:24
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The inequality you propose should be reversed. See this plot where I use the partial expension $-x - x^2/2$. – Zubzub Feb 23 '18 at 12:27
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@nbubis The last term in the LHS was not a mistake. Let me edit and show the RHS for clarification. – noah.c Feb 23 '18 at 12:34
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@Zubzub you are correct to say $-x- \frac{x^2}{2}$ should be reversed. However, on the RHS, it would be $-x- \frac{2x^2}{2}$, which would not be reversed. – noah.c Feb 23 '18 at 12:37
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^ I meant LHS. Oops. This is what we want. – noah.c Feb 23 '18 at 12:48
2 Answers
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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 $$-x-x^2 \le \log(1-x) \le -x, \quad x \in [0,1/2].$$
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This is good. Out of curiosity, is there a way to prove the lower bound without knowing $[0,\frac{1}{2}]$ are the bounds of $x$, and to prove the bounds along the way? – noah.c Feb 23 '18 at 13:00
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Never mind! I used similar logic as you and figured out my own question. Thanks. – noah.c Feb 23 '18 at 14:46
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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 $x \in I$.
Fred
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