# $e^x$ Inequality

1. Dec 7, 2013

### AaronEx

1. The problem statement, all variables and given/known data
Show that if $0 \le x \le a$, and $n$ is a natural number, then $$1+\frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^n}{n!} \le e^x \le 1+\frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^n}{n!}+\frac{e^ax^{n+1}}{(n+1)!}$$
2. Relevant equations
I used Taylor's theorem to prove $e^x$ is equal to the LHS of the inequality plus an error term $\frac{e^cx^{n+1}}{(n+1)!}$, where the series is of order $n$ centered at $x=0$.

3. The attempt at a solution
With the expansion of $e^x$ in the relevant equations section, I proved that, since the error term is greater than or equal to zero, the inequality holds true. Is this correct? As for the RHS inequality, I represented $e^x$ as the expansion again, but this time stating that the expansion is on the interval $[0,a]$, and so I argued that the RHS was larger than $e^x$ since $e^c \le e^a$. Is this true as well? Have I left anything relevant out?

2. Dec 7, 2013

Looks fine.