Assume that f is a function with $|f^{(n)}(x)| \le 11,$ for all n and all real x. Let $T_n(x)$ denote the Taylor polynomial of degree n for f(x) about the point $x=0$.
What is the least integer n for which you can be sure that $T_n(2)$ approximates f(2) accurately to 3 decimal places? i.e., so that the absolute error $\displaystyle \left|R_n\left(2\right)\right|$ is less than $\displaystyle \frac{5}{10^{4}} = 0.0005$ ? Here $\displaystyle R_n\left(2\right)$ denotes the Lagrange form of the Remainder, evaluated at $\displaystyle 2$.
So I'm guessing that for whatever derivative $n$, my function will always be less than 11.
So the remainder:
$$ \frac{f^{n+1} (c) \cdot x^{n+1}}{(n+1)!} < 0.0005 $$ should be less than 3 decimal places
->
$$ \frac{11 \cdot 2^{n+1}}{(n+1)!} < \frac{5}{10^4} $$
->
$$ \frac{(n+1)!}{2^{n+1}} > 22000$$
but how do I solve for $n$ manually?
By trial and error n = 11.