Answer to Question #109856 in Real Analysis for Pappu Kumar Gupta

Question

Use Lagrange’s mean value theorem to prove that
x-x^2/2<log(1+x)<x-x^2/2(1+x) if x>0

Answer 1

According to Lagrange’s Theorem, there exists at least one point ‘c’ on the open interval (a,b), such as

$f'(c)=\frac{f(b)-f(a)}{b-a}$  

Let's prove that $\log(x+1)>x-\frac{x^2}{2}$

Let's denote $f=\log(x+1)+\frac{x^2}{2}$

Using Lagrange’s Mean Value Theorem for function f, will lead us to

$\exist c\in (0,b)$ such as $f'(c)=\frac{f(b)-f(0)}{b-0}=\frac{\log(b+1)+\frac{b^2}{2}}{b}$

Therefore

$f'(c)=\frac{1}{c+1}+c=\frac{\log(b+1)+\frac{b^2}{2}}{b}$

$\frac{1}{c+1}+c=1+\frac{c^2}{1+c}>1$

Hence

$\frac{\log(b+1)+\frac{b^2}{2}}{b}>1$

$\log(b+1)+\frac{b^2}{2}>b$

So $\log(x+1)>x-\frac{x^2}{2}$

Let's prove that $\log(x+1)<x-\frac{x^2}{2(1+x)}$

Let's denote $f=\log(x+1)+\frac{x^2}{2(x+1)}$

Using Lagrange’s Mean Value Theorem for function f, will lead us to $\exist c\in (0,b)$ such as

$\frac{\log(b+1)+\frac{b^2}{2(b+1)}}{b}=f'(c)=\frac{1}{c+1}+\frac{2c2(c+1)-2c^2}{4(1+c)^2}=1-\frac{2c^2}{4(1+c)^2}<1$

Hence $\frac{\log(b+1)+\frac{b^2}{2(b+1)}}{b}<1$

$\log(b+1)+\frac{b^2}{2(b+1)}<b$

So $\log(x+1)<x-\frac{x^2}{2(1+x)}$