# Math Help - Real Analysis

1. ## Real Analysis

If x>0, show that 1 + x + x^2/2 < e^x < 1 + x + x^2/2 e^x

2. Originally Posted by dhhnerd
If x>0, show that 1 + x + x^2/2 < e^x
I do the easier inequality, hopefully, the second one is similar.

First we will prove that $1+x < e^x$ for $x>0$.
Define $f(x) = e^x - x - 1$.
Then $f(0) = 0$ but $f'(x) = e^x - 1 > 0$ for $x>0$.
Therefore, $f$ is increasing for $x>0$.
But since we have $f(0)=0$ it must mean that $f(x) > 0$ for $x>0$.
Therefore, $e^x - x - 1 > 0 \implies e^x > 1+x$.

Second define $g(x) = e^x - \tfrac{1}{2}x^2 - x - 1$.
Then $g(0)=0$ but $g'(x) = e^x - x - 1>0$ for $x>0$ by above.
Therefore, $g$ is increasing for $x>0$.
But since we have $g(0)=0$ it must mean that $g(x)>0$ for $x>0$.
Therefore, $e^x > \tfrac{1}{2}x^2 + x + 1$

3. Originally Posted by dhhnerd
If x>0, show that 1 + x + x^2/2 < e^x < 1 + x + x^2/2 e^x
What is the second inequality? $e^x<1+x+\frac{x^2}{2}+e^x$?

Or $e^x<1+x+\frac{x^2}{2}\cdot{e^x}$

Both are very simple