# Prove that for $0<x<1$, $\ln(1+x) - \ln(1-x)<2x$

Prove that for $0<x<1$, $\ln(1+x) - \ln(1-x)<2x$

My direction: $\ln(x)$ is differential and continuous at $(0,∞)$, and since $0<x<1$, clearly $[1-x,1+x]\subseteq(0,∞)$ and so $\ln(x)$ is continuous in $[1-x,1+x]$ and differential in $(1-x,1+x)$.

So, by Lagrange's theorem, there is a $c$ in $(1-x, 1+x)$, so that: $${{\ln(1+x)-\ln(1-x)}\over{1+x-1+x}}=\ln'(c)$$ $${{\ln(1+x)-\ln(1-x)}\over{2x}}=\ln'(c)$$ $$\ln(1+x)-\ln(1-x)=\ln'(c)2x$$

And here I got stuck... Would appreciate any help.

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What is $\ln(x-1)$ if $0<x<1$? It seems you mean $\ln(1+x)-\ln(1-x)$ and not $\ln(x+1)-\ln(x-1)$. – Hagen von Eitzen Jan 14 '13 at 15:45
Yes, thank you. – Harold Jan 14 '13 at 15:46
@Harold: Is the inequality $>$ or $<$? It should be $>$. – Mhenni Benghorbal Jan 14 '13 at 18:36

$\ln(1 + x) = x - x^2/2 + x^3/3 -...$, and $\ln(1 - x) = -x - x^2/2 - x^3/3 -...$, so you have $$\ln(1 + x) - \ln(1 - x) = 2x + {2x^3 \over 3} + {2x^5 \over 5} + ...$$ Looking at the first term you see that this is greater than $2x$.

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Nice idea. To make it work, one must show that the rest (everything in $+\cdot$) does not invert the comparison (and indeed this can be shown). – Did Jan 14 '13 at 17:31
It just follows from the fact that the coefficients are positive and $x > 0$. – Zarrax Jan 14 '13 at 17:49
I know. You might want to mention this in your answer. – Did Jan 14 '13 at 17:51

For every $0\lt x\lt1$, $\log(1+x)-\log(1-x)\gt2x$.

To show this, define $u(x)$ as the LHS and note that $u(0)=0$, $u'(0)=2$ and $u''(x)\gt0$ for every $x$ in $(0,1)$. Since the function $u$ is convex, the graph of $u$ lies above its tangent at $0$, which is the line $y=2x$. This is the result.

Edit: Alternatively, since $u''\gt0$, $u'\gt u'(0)$ on $(0,x)$ hence $u(x)-u(0)=u'(c)x\gt2x$.

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Thanks for this solution. Is there a way to solve it with LaGrange? – Harold Jan 14 '13 at 15:50
Lagrange multipliers? What for? There are no constraints here... – Did Jan 14 '13 at 15:56
No, the direction I tried in the original post. – Harold Jan 14 '13 at 15:57
By the way, when trying to prove such inequalities, a good idea is to try some values (this would have prevented you to propose the wrong inequality). – Did Jan 14 '13 at 15:57
Why should it be $1\over{1+c}$? – Harold Jan 14 '13 at 16:11

[Proof fixed after did's comment.]

Let $g(x) = \ln (1+ x) - \ln(1-x) - 2x.$

We have $g'(x) = \dfrac{1}{1+x} + \dfrac{1}{1-x} - 2$, hence $g{'}(0)= 0 = g(0)$, also, $g^{''}(x) = -\dfrac{1}{ ( 1 + x )^2 } + \dfrac{1}{ ( 1 - x )^2 } > 0,$ for any $x\in (0,1)$ as $\dfrac{1}{1-x} > 1$ while $0 < \dfrac{1}{1+x} < 1$.

From Taylor expansion, $$g(x) = g(0) + g^{'}(0)x + g''(\eta) x^2/2 = g''(\eta) x^2/2,$$

where $\eta \in (0,x).$

So $g(x) > 0$ which implies $\ln(1 + x) - \ln(1 -x) > 2x.$

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It seems that $g'(\eta)$ is not what you write. – Did Jan 14 '13 at 16:15

Note that for $x\in(0,1)$: $$\frac{\mathrm{d}}{\mathrm{d}x}\left(\vphantom{\frac{\mathrm{d}}{\mathrm{d}x}}\log(1+x)-\log(1-x)-2x\right)=\frac{2x^2}{1-x^2}\gt0$$ Therefore, the mean value theorem says that for some $\xi\in(0,x)$ $$\frac{(\log(1+x)-\log(1-x)-2x)-(\log(1+0)-\log(1-0)-2\cdot0)}{x-0}=\frac{2\xi^2}{1-\xi^2}\gt0$$ Therefore, $$\log(1+x)-\log(1-x)>2x$$

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