Demonstration

**dhiab** · June 26th 2009, 12:05 AM

Prove :

**Twig** · June 26th 2009, 12:31 AM

Hi

There might be a better way to do this, but I´ll give one solution.
Note: $x-\frac{x^{2}}{2}$ is the first two terms of the maclaurin series of $ln(1+x)$ .

Anyway, let $f(x)=x-\frac{x^{2}}{2}-ln(1+x)$ , then we want to show that $f(x)<0$ for all $x>0$ .

Solution: $f$ is continous and differentiable for all $x>-1$ , with $f'(x)=\frac{-x^{2}}{1+x}$ . (I have already simplified the derivative here).

Note tha we are only interested in $x>0$ , so we don´t have to worry about the denominator of the derivative becoming zero.

Looking at $f'(x)$ we see that the only stationary point is $x=0$ . And for all $x>0$ we have that $f'(x)<0$ .

And because $f(0)=0$ , and the function is strictly decreasing for all $x>0$ , it immediately follows that $f(x)<0$ .

Now for the right inequality:

You could do the same thing for $g(x)=x-ln(1+x)$ , and show that $g(x)>0$ .

**red_dog** · June 26th 2009, 12:32 AM

Let $f(x)=\ln(1+x)-x, \ f<img src=$ 0,\infty)\to\mathbf{R}" alt="f(x)=\ln(1+x)-x, \ f

0,\infty)\to\mathbf{R}" />

$f'(x)=-\frac{x}{1+x}<0, \ \forall x>0$ , so f is strictly decreasing.

Then, $x>0\Rightarrow f(x)<f(0)=0\Rightarrow\ln(1+x)<x$

Try to prove the other inequality in a similar way by using the function

$g(x)=x-\frac{x^2}{2}-\ln(1+x), \ x>0$

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