Proving the Inequality of e^x Using Taylor's Theorem

[AaronEx]

Homework Statement

Show that if [itex]0 \le x \le a[/itex], and [itex]n[/itex] is a natural number, then [tex]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)!}[/tex]

Homework Equations

I used Taylor's theorem to prove [itex]e^x[/itex] is equal to the LHS of the inequality plus an error term [itex]\frac{e^cx^{n+1}}{(n+1)!}[/itex], where the series is of order [itex]n[/itex] centered at [itex]x=0[/itex].

The Attempt at a Solution

With the expansion of [itex]e^x[/itex] 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 [itex]e^x[/itex] as the expansion again, but this time stating that the expansion is on the interval [itex][0,a][/itex], and so I argued that the RHS was larger than [itex]e^x[/itex] since [itex]e^c \le e^a[/itex]. Is this true as well? Have I left anything relevant out?

 

[mfb]

Looks fine.