Finding the degree so that a Taylor polynomial yields an approximation within a specified accuracy. So, for the function $f(x)=e^x$ centered around $x_0=0$, I need to find a Taylor polynomial of degree $n$ such that it approximates $f(x)$ within $10^{-6}$ on the interval $[0,0.5]$.
I took this to mean that the error (|$R_n(x)$|) has to be less than or equal to $10^{-6}$.
The formula for the error would be $|R_n(x)|=|\frac{1}{(n+1)!}f^{(n+1)}(z)(x-x_0)^{n+1}$|.
In trying to find the degree of the Taylor polynomial, we are essentially trying to solve for $n$ with the upper-bound being $10^{-6}$.
This means $|R_n(x)|=|\frac{1}{(n+1)!}f^{(n+1)}(z)(x-x_0)^{n+1}|=|\frac{1}{(n+1)!}e^zx^{n+1}| \leq 10^{-6}$. Using a CAS, I brute-forced a numerical solution and got that the first positive solution is approximately $n=6.34354$. 
Since the Taylor polynomial must be of integer degree, I rounded up and concluded that a seventh-degree polynomial would be enough to give an estimate with the requested accuracy. Is there any other way of doing this though? I feel like this is a very ad hoc solution that wouldn't be appropriate in all cases.
 A: As Ian commented, if you look at this question of mine, there is  analytical solutions to the problem of solving for $n$ the equation
$$\frac{e^x\,x^{n+1}}{(n+1)!} = 10^{-k}$$ Ignoring the $\sqrt p$ in Stirling approximation of $p!$ (as I did as a shortcut), this would lead to the overestimate
$$n=-1+\frac{\log (a)}{W\left(\frac{\log (a)}{e x}\right)}\qquad \text{where}\qquad a=\frac{e^x\, 10^k}{\sqrt{2 \pi }}$$ where appears Lambert function.
Applied to your case $(k=6,x=\frac12)$, this will give $n=6.72$ which is not far from the exact solution you found.
Computing for a few values of $k$, we should get
$$\left(
\begin{array}{ccc}
 k & \text{approximation} & \text{exact} \\
 1 & 1.718 & 1.426 \\
 2 & 2.936 & 2.608 \\
 3 & 3.990 & 3.643 \\
 4 & 4.952 & 4.594 \\
 5 & 5.856 & 5.490 \\
 6 & 6.715 & 6.344 \\
 7 & 7.541 & 7.165 \\
 8 & 8.340 & 7.959 \\
 9 & 9.116 & 8.732 \\
 10 & 9.872 & 9.486 \\
 11 & 10.61 & 10.22 \\
 12 & 11.34 & 10.95
\end{array}
\right)$$
A: Since $e^x$ and $x^{n+1}$ are increasing on $[0,0.5]$, you have 
$$|R_n(x)|\leq \underbrace{\frac{1}{(n+1)!}\sqrt{e}\frac{1}{2^{n+1}}}_{a_n:=} \stackrel{!}{\leq }10^{-6}$$ 
Furthermore, you may estimate $\sqrt{e} < 1.65$.
As $a_n$ is strictly decreasing you just find the first $n \in \mathbb{N}$ for which the inequality is satisfied. There is no solving for $n$ using $\Gamma$ necessary.
