8
$\begingroup$

EDIT:

My question was poorly worded.

I wasn't asking about showing $\ln(1+x) > \frac{2x}{2+x}$ for $x>0$.

What I wanted to know is why the lower bound provided by $ \frac{2x}{2+x}$ was so tight for small positive values of $x$. This is addressed in robjohn's answer.

$\endgroup$
8
  • 2
    $\begingroup$ Have you tried putting both on the same side and computing the derivative? $\endgroup$ – xavierm02 Jul 20 '13 at 16:07
  • 2
    $\begingroup$ Additinonaly, we know that $\ln(x+1)>x-x^2/2$. $\endgroup$ – Mikasa Jul 20 '13 at 16:35
  • 1
    $\begingroup$ @Laura: No. :) It is in $x>0$ indeed. $\endgroup$ – Mikasa Jul 20 '13 at 16:45
  • 1
    $\begingroup$ @Laura: $\log(1+x)=x-x^2/2+x^3/3-\dots$ and $\frac{2x}{2+x}=x-x^2/2+x^3/4-\dots$. Thus, $\log(1+x)-\frac{2x}{2+x}=x^3/12+\dots$. Thus, the inequality breaks down for $x\le0$. $\endgroup$ – robjohn Jul 20 '13 at 23:23
  • 1
    $\begingroup$ @Laura: what I am saying is that the inequality $\log(1+x)\gt\frac{2x}{2+x}$ cannot hold in any (open) neighborhood of $0$. $\endgroup$ – robjohn Jul 21 '13 at 7:29
3
$\begingroup$

As I mentioned in a comment, the power series near $x=0$ are $$ \begin{align} \log(1+x)&=x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4)\\ \frac{2x}{2+x}&=x-\frac{x^2}{2}+\frac{x^3}{4}+O(x^4)\\ \log(1+x)-\frac{2x}{2+x}&=\frac{x^3}{12}+O(x^4) \end{align} $$ So near $x=0$, the value, and the first and second derivatives match. That means the functions $\log(1+x)$ and $\frac{2x}{2+x}$ match to second order. Not quite as simple, $\frac{x(6+x)}{6+4x}$ matches $\log(1+x)$ to third order. Rational approximations to functions are called Padé Approximations.

Furthermore, $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\left(\log(1+x)-\frac{2x}{2+x}\right) &=\frac{\mathrm{d}}{\mathrm{d}x}\left(\log(1+x)+\frac{4}{2+x}-2\right)\\ &=\frac1{1+x}-\frac4{(2+x)^2}\\ &=\frac{x^2}{(1+x)(2+x)^2} \end{align} $$ So, for $x\gt-1$, $\log(1+x)-\frac{2x}{2+x}$ is an increasing function. At $x=0$, $\log(1+x)-\frac{2x}{2+x}=0$.

Therefore, for $x\gt0$, $$ \log(1+x)\gt\frac{2x}{2+x} $$ and for $-1\lt x\lt0$, $$ \log(1+x)\lt\frac{2x}{2+x} $$

$\endgroup$
13
$\begingroup$

Hint: Let $f(x)=\ln(1+x)-\frac{2x}{2+x}$, and show that $f(0)=0$, and that $f'(x)>0$ for all $x>0$.

$\endgroup$
1
  • $\begingroup$ So you are assuming that $x\gt0$. $\endgroup$ – robjohn Jul 21 '13 at 7:48
3
$\begingroup$

Proof by contradiction: $$\ln(1+x) \le \frac{2x}{2+x}$$ $$\ln(1+x) \le 2 - \frac{4}{2+x}$$ $$\ln(1+x) - 2 < -\frac{4}{2+\ln(1+x)}$$ $$\ln^2(1+x) < 0$$

$\endgroup$
2
  • $\begingroup$ Line 3 does not follow from line 2. Line 4 does not follow from line 3. $\endgroup$ – robjohn Jul 21 '13 at 7:47
  • $\begingroup$ Yep, forgot the minus in the line 3 $\endgroup$ – gukoff Jul 21 '13 at 8:21
1
$\begingroup$

To show that the inequality holds for $x \geq 0$ note that

$$ \log(1+x) = \int_1^{1+x} \frac{1}{t}dt. $$

Approximate the function $t \mapsto \frac{1}{t}$ by its tangent at $t = 1 + \frac{x}{2}$ to get

$$ \frac{1}{t} \geq -\frac{1}{(1+\frac{x}{2})^2}(t - 1 - \frac{x}{2}) + \frac{1}{1 + \frac{x}{2}} = -\frac{4 t}{(x + 2)^2} + \frac{4}{x+2}$$

for all $t \geq 1$. In particular

$$ \log(1+x) \geq \int_1^{1+x} \left(-\frac{4 t}{(x + 2)^2} + \frac{4}{x+2}\right) dt = \frac{2x}{x+2}.$$

This lower bound is optimal among all those that can be obtained by using such a tangent approximation. In general it depends what you mean by "optimal". For example

$$ \log(1+x) - \frac{2x}{a x + 2} = \frac{a-1}{2} x^2 + O(x^3) $$

which implies that $a \geq 1$ to get a possible lower bound. But for $a > 1$ this lower bound is worse than for $a=1$. On the other hand

$$ \log(1+x) \geq \frac{8x}{3(x+4)} $$

for all $x\geq 0$ and this is a better lower bound for $x \geq 4$.

$\endgroup$
0
$\begingroup$

At zero both are zero. Their derivatives are $$\frac 1 {1+x},\qquad \frac 4 {(2+x)^2}=\frac 1 {(1+x/2)^2}$$

$\endgroup$
0
$\begingroup$

Also for sufficiently large values of $x$ it is quite obvious since \begin{equation} \frac{2x}{2+x} = 2 - \frac{4}{2 + x} \end{equation} and therefore $\frac{2x}{2+x}$ has upper boundary while $\ln(x+1)$ is unbounded.

$\endgroup$
0
$\begingroup$

We change the question to find an upperbound:$$\ln(1+x) > \frac{nx}{n+x}$$ Let's do the following substitution: $\left( 1+x \right) ={ e }^{ h }.$ The question becomes, proving $$h>\frac { n\left( { e }^{ h }-1 \right) }{ n+{ e }^{ h }-1 } ,$$ where $h>0.$

We search for $n$: $$n<\frac { h{ e }^{ h }-h }{ { e }^{ h }-h-1 } $$. For larger values of $h$ we know that the equation will satisfy therefore we have to prove for smaller values of $h$. $$\lim _{ h\rightarrow o }{ \left( \frac { h{ e }^{ h }-h }{ { e }^{ h }-h-1 } \right) } =2.$$ Therefore the upper bound of $n$ is $2$.

$\endgroup$
1
  • $\begingroup$ The inequality doesn't hold: $2-\frac{4}{e^h+1}>2-\frac{4}{2+h} \ \forall \ h$ $\endgroup$ – Alex Jul 20 '13 at 17:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.