The Problem:
If we consider approximating $e^x$ on $[-1,1]$, the Taylor Theorem for $x_0 =0$ says we can represent $e^x$ using a polynomial with a (known) remainder:
$e^x =\{1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \dots +\frac{1}{n!}x^n \} + \frac{1}{(n+1)!}x^{n+1}e^{C_x}$
Where $C_x$ is an unknown between $x $ and $0$. For simplicity, let us denote the polynomial (in curly backets) by $p_n(x)$ and the remainer (outside the curly brackets) by $R_n(x)$, such that
$e^x= p_n(x) + R_n(x)$
Suppose we want this approximation to be accurate to within $10^{-6}$ in absolute error, i.e. we want
$|e^x – p_n(x)| \leq10^{-6}$
For all $x $ on the interval $[-1,1]$.
Question:
For what value of $n$ will our accuracy condition be satisfied? How can this be done using the remainder of the approximation? What does this remainder represent?
Taylor's Thoerem:
Let $f(x)$ have $n+1$ continuous derivatives on $[a, b]$ for some $n \geq 0$, and let $x, x_0 \in [a, b]$. Then,
$f (x) = p_n(x) + R_n(x)$
for
$p_n(x) = \sum_{k=0}^n \frac{(x -x_0)^k}{k!} f^{(k)}(x_0)$
There exists a point $\zeta _x $ between $x$ and $x_0$ such that
$R_n(x) = \frac{(x -x_0)^{n+1}}{(n+1)!} f^{(n+1)}(\zeta_x)$
Best Answer
You have written everything down except for the last step: With $x=1$, we have $C_x\le1$. The most pessimistic case, the largest possible error, would be found at $C_x=1$ (more realistically, $C_x$ is located around the midpoint $\frac12$). With that you have to solve $$ \frac{e}{(n+1)!}<10^{-6} $$ for $n$. Either by computing a table or by using the Sterling formula for approximations.