I was asked to use Taylor's Theorem to confirm that the Maclaurin polynomial $P_7(x)$ for $e^x$ is sufficient to give $e$ correct to $3$ decimal places. My calculus-book states:
"To get an approximation for $e=e^1$ correct to 3 decimal places, we need to have $E_n(1)<0.0005$"
QUESTION: I don't see why an error of $0.0005$ is sufficiently small, since $e+0.0005=2.7187...$ which rounds to $2.719$ which is not correct to $3$ decimal places. Any thoughts?
You could use the Lagrange formula for the remainder in the Taylor Polynomial development of order $\;n\;$ around $\;x_0=0\;$ to evaluate the error :
$$\text{For}\;c\in (0,x)\;:\;\;\left|\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\right|=\frac{e^c}{(n+1)!}|x|^{n+1}\le$$
$$\le\frac{e^c}{(n+1)!}\cdot\le\frac9{(n+1)!}$$
since we can choose $\;x\le1\;$ and estimating crudely $\;e^3\implies e^2<9\;$, and thus we want:
$$\frac9{(n+1)!}\cdot<10^{-3}\implies (n+1)!>9,000\implies n+1=8\implies n=7$$
and indeed:
$$e=1+1+\frac12+\frac16+\frac1{24}+\frac1{120}+\frac1{720}+\frac1{5,040}\approx2.71825$$
and then
$$2-2.71825=0.00002... <10^{-3}$$
It turns out that also $\;n=6\;$ would work, but we usually aren't that interested in the best or smallest possible $\;n\;$, but only in some value that will work and will be more or less close to the best one.