$f(x)=e^x$ be approximated by Taylor's polynomial of degree $n$ at the point ... I am stuck on the following problem: 

Let $f(x)=e^x$ be approximated by Taylor's polynomial of degree $n$ at the point $x=\frac12$ and on the entire interval $[0,1]$. If the absolute error in this approximation does not exceed $10^{-2},$ then the value of $n$ should be taken as 
  
  
*
  
*$0$  
  
*$1$  
  
*$2$  
  
*$3$ 

I am not sure how to tackle it. Can someone explain? Thanks in advance for your time. 
What I know is given below: 
 
Here I have to determine the value of $n$ so that $R_n(x)<10^{-2}$.That is where I am stuck and could not do my calculations right.
Given that $f(x) =e^x$. Here, we obtain:
\begin{align*}
f(x) &\approx p_3(x) \\
&= f(0) + f'(0)(x - 0) + \frac{f''(0)}{2!}(x-0)^2 + \frac{f'''(0)}{3!}(x-0)^3 \\
\end{align*}
Thus, substituting $x = 0.5$ and taking $n=3$ yields:
\begin{align*}
 f(0.5) 
&\approx p_3(0.5) \\
&= 1 + 0.5+\frac{1}{2}(0.5)^2 + \frac{1}{6}(0.5)^3  \\
&\approx  1+0.5+0.125+0.0208 \\
&=1.6458 
\end{align*}
 A: Observe, you can add zero and use the known MacClaurin series for the exponential:
$$e^x =  e^{x-1/2+1/2} =e^{x-1/2}e^{1/2}= \sqrt{e}(e^{x-1/2})$$
hence
$$ e^x = \sqrt{e}\left(1+(x-1/2)+\frac{1}{2}(x-1/2)^2+ \cdots +\frac{1}{k!}(x-1/2)^k+\cdots \right)$$
Now, you just need to determine which remainder theorem you wish to apply given the interval $[0,1]$ is the desired interval of application. The coefficient of the $k$-th term in the Taylor expansion is seen to be $\sqrt{e}/k!$
A Theorem (from Ash's Real Analysis) which I'll use:
If $f$ has $k$ derivatives on a closed interval $I$ with $\partial I = \{a,b \}$ then 
$$ f(b) = T_k(b) +R_{k}(b) = \sum_{j=0}^k \frac{f^{(j)}(a)}{j!}(b-a)^j  +R_{k}(b) $$
where $R_{k}(b) = f(b)-T_k(b)$ is the $k$-th remainder. Moreover, there exists $c \in int(I)$ such that
$$ R_k(b) = \frac{f^{(k+1)}(c)}{(k+1)!}(b-a)^{k+1}. $$
Ok, so think of $a=1/2$ and $b=x \in [0,1]$ you see that $(b-a) \leq 1/2$. Moreover, we also observe 
$$\frac{f^{(k+1)}(c)}{(k+1)!} = \frac{\sqrt{e}}{(k+1)!}$$ 
Therefore,
$$ |R_k(x)| \leq \frac{\sqrt{e}}{(k+1)!}\left( \frac{1}{2} \right)^{k+1} $$
You wish to find $k$ for which $|R_k(x)| < 0.01$. I found with wolfram alpha that $k=3$ gives $|R_k(x)| < 0.0043$.
