Show that $e^x=1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}+R_{n+1}$

Question

Show that
$\qquad$ $\qquad$ $e^x=1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}+R_{n+1}$
with
$\qquad \qquad$ $0 \lt R_{n+1} \lt e^x \frac{x^{n+1}}{(n+1)!}$ if $0 \lt x$
and
$\qquad \qquad$ $|R_{n+1}| \lt \frac{|x|^{n+1}}{(n+1)!}$ if $x \lt 0$

I have no idea how to do this or where to start?
I know that is the taylor expansion of $e^x$ and I know that $R_{n+1}=(-1)^n\frac{x^{n+1}(1-\theta)^n}{(1+\theta x)^{n+1}}$

Answer 1

You can use the Lagrange form of the remainder. It can be shown that if $f:\mathbb{R} \to \mathbb{R}$ is $C^{n+1}$, then

$$f(x)=\sum_{k=0}^{n}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k+\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{(n+1)}$$

for $\xi \in (x,x_0)$.

$f(x)=e^x$ is a $C^{\infty}$ function; let $x_0=0.$ For every $n\in \mathbb{N}$ we have

$$e^x=\sum_{k=0}^{n}\frac{e^{x_0}}{k!}(x-x_0)^k+\frac{e^{~\xi}}{(n+1)!}(x-x_0)^{(n+1)}=\sum_{k=0}^{n}\frac{x^k}{k!}+\frac{e^{~\xi}}{(n+1)!}(x)^{(n+1)}$$

for $\xi \in (0,x)$. Since $e^{~\xi}$ is a monotonically increasing function the remainder can easily be estimated observing that:

• If $x>0$, $\sup \{e^{~\xi}|\xi\in (0,x)\}=e^x$
• If $x\leq0$, $\sup \{e^{~\xi}|\xi\in (-|x|,0)\}=e^0=1$

This leads to the conclusion: in fact

$$|R_{n+1}|=\left |\frac{e^{~\xi}}{(n+1)!}(x)^{(n+1)}\right |< \cases{ \frac{e^{x}}{(n+1)!}(x)^{(n+1)} & \text{if } x\ge 0\cr \frac{1}{(n+1)!}|x|^{(n+1)} & \text{if } x\lt 0}$$

You can find more information on the Lagrange remainder here.