Show that $1 + \frac{x}{1!} + \dots + \frac{x^n}{n!} \leq e^x \leq 1 + \frac{x}{1!} + \dots + \frac{x^{n-1}}{(n-1)!} + \frac{e^{a}x^n}{n!}$

Question

Show that if $0 \leq x \leq a$ and $n \in \mathbb{N}$, then $$1 + \frac{x}{1!} + \dots + \frac{x^n}{n!} \leq e^x \leq 1 + \frac{x}{1!} + \dots + \frac{x^{n-1}}{(n-1)!} + \frac{e^{a}x^n}{n!}$$

Let $$e^x = 1 + x + \frac{x^2}{2!} + \dots + \frac{x^{n-1}}{(n-1)!} + \frac{x^n}{n!} + \frac{x^{n+1}}{(n+1)!}e^x$$ where $e^x$ is expanded by Taylor's theorem, but I am not sure if I am correctly applying Taylor's theorem because it isn't clear to me where the right side of the inequality comes from.

How does Taylor's theorem get used to complete the inequality? Is that even the right approach to this problem?

Answer 1

An idea: since$\;0\le x\le a\;$ , we have

$$e^x=\sum_{k=0}^\infty\frac{x^k}{k!}\ge \sum_{k=0}^n\frac{ x^k}{k!}$$

and you get the leftmost inequality. On the other hand $\;e^x\le e^a\;$ since the exponential function is monotonic ascending. Now take it from here, or using Taylor's Theorem that you just posted now:

$$e^x=\sum_{k=0}^{n-1}\frac{x^k}{k!}+\frac{e^c x^n}{n!}$$

But $\;e^x\le e^a\;$. End now the argument for the other inequality

Answer 2

By Taylor's theorem, $$ e^x = 1 + x + \frac{{x^2 }}{{2!}} + \cdots + \frac{{x^{n - 1} }}{{(n - 1)!}} + \frac{{x^n }}{{n!}}e^c $$ for all $x\geq 0$ with a suitable $0\leq c\leq x$. Now just note that if $0\leq x\leq a$ then $$ 1 \leq e^c \leq e^x \leq e^a . $$