Prove that for $0

Question

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.

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

score 5 · Accepted Answer · answered Jan 14 '13 at 15:57

5

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

answered Jan 14 '13 at 15:57

Zarrax

44,950

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

Did · Answer 2 · 2013-01-14T16:03:45.337

4

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 $u(x)-u(0)=u'(c)x\gt2x$.

edited Jan 14 '13 at 16:03

answered Jan 14 '13 at 15:43

Did

279,727

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
Ahhh... right. $ $ – Did Jan 14 '13 at 16:01
1

Why should it be $1\over{1+c}$? – Harold Jan 14 '13 at 16:11
@Harold Sorry, ignore my previous comment. – David Mitra Jan 14 '13 at 16:24
@DavidMitra it's ok. Is there then any chance my original direction for an answer is correct or I should use the other solutions suggested here? – Harold Jan 14 '13 at 16:27
IMHO, it all depends on the function one applies mean value theorem (MVT) to: MVT on $u$ is my answer. MVT on $x\mapsto\log(1+x)$ alone does not work (as you showed). – Did Jan 14 '13 at 17:34

Arin Chaudhuri · Answer 3 · 2013-01-14T18:18:27.330

1

[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. $

edited Jan 14 '13 at 18:18

answered Jan 14 '13 at 15:59

Arin Chaudhuri

6,349

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

score 0 · Answer 4 · answered Jan 14 '13 at 20:52

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, $$ \log(1+x)-\log(1-x)>2x $$

Prove that for $0

4 Answers4