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Prove the inequality: $[x-(x^2)/2 < \ln(1+x) < x]$ , $x>0$


The right side is easy, I used taylor expansion to show that $e^x > 1+x$ since $e^x = 1 + x + x^2/2 + x^3/3! +\cdots $

The left side I expanded $\ln(1+x)$ to $x-x^2/2 + x^3/3 - x^4/4 +\cdots$ moved $(x-x^2/2)$ to the left and got this:

$$0 < x^3/3 - x^4/4 \cdots $$

The only thing I can think of here is that since the first term is positive and this series goes on forever the first term will always stay positive since the other terms cancel out... Not sure this is mathematically coherent or even correct.

thanks.

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I've simplified it to this: $ x(1/4 + x^2/6 + ...) < (1/3 + x^2/5 + ...)$ Still doesn't make sense since the left side is being multiplied by x so why whould it be smaller? –  nofe Feb 3 '12 at 11:13
    
You forgot the exponents on the taylor expansion of $e^x$. –  Alex Becker Feb 3 '12 at 11:30
    
you're right but only when copying it onto here...the left side of the equation is what's giving me trouble not the right. –  nofe Feb 3 '12 at 11:34
    
Another way without using Taylor expansion. Consider the function $f(x)= \ln (1+x) -x +x^{2} /2 ,\quad x\geq 0$. It is $f'(x)=\frac{x^2}{1+x} \geq 0$ with equality iff $x=0$.So $f$ is stricly increasing and for $x>0$ it is $f(x)>f(0)=0$, hence the conclusion –  passenger Feb 3 '12 at 11:44
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3 Answers

up vote 2 down vote accepted

For the left inequality, since you know that $x-\frac{x^2}{2}=\ln(1+x)$ when $x=0$ you need only check the derivatives: $\frac{d}{dx}(x-\frac{x^2}{2})=1-\frac{x}{2}$ while $\frac{d}{dx}(\ln(1+x))=\frac{1}{1+x}$, and so $$\frac{d}{dx}(\ln(1+x))-\frac{d}{dx}(x-\frac{x^2}{2})=\frac{1}{1+x}-(1-\frac{x}{2})=\frac{1-(1+x)(1-\frac{x}{2})}{1+x}=\frac{-x^2}{2(1+x)}$$ which is negative for positive $x$, thus for $x>0$, $\ln(1+x)$ is always growing faster than $x-\frac{x^2}{2}$ so it must be greater than $x-\frac{x^2}{2}$.

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Maybe we can let $f(x)=(x-x^2/2)-\ln(1+x)$,and prove the maximum of $f(x)$ is $0$ by using the derivative of $f(x)$.

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Define function $f(x) $ as :

$f(x)=\frac{x^2}{2}+\ln(1+x)-x$ . Note that : $f(0)=0$

Let us find first derivative of function $f(x)$ :

$f'(x)=\frac{x^2}{x+1}$ , Note that $f'(x) > 0$ for all $x>0$

Since $f(0)=0$ and $f'(x) > 0$ for all $x>0$ it follows that :

$\frac{x^2}{2}+\ln(1+x)-x >0 \Rightarrow \ln(1+x) >x- \frac{x^2}{2}$ , for all $x>0$

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