Find Taylor polynomial at $x=0$ which approximates $e^x$ with accuracy at least $1$ for each $x \in [-2,2]$.
I dont undestand these questions that involve the $n^{th}$ remainder. I know I need to find $R_n(x)$ for $e^x$ where it is less than $1$. Once I know what $n$ gives a remainder less than one I then find the $n^{th}$ degree Taylor polynomials.
Can someone help me understand this concept and the process of solving these kind of questions.
If this is a fairly standard calculus course, you are working with the Lagrange form of the remainder. Notations tend to differ, but if $R_m(x)$ is the remainder when we truncate the Maclaurin series just after the $x^m$ term, then $$|R_m(x)|=\frac{|f^{(m+1)}(\xi)|}{(m+1)!} |x|^{m+1},$$ where $\xi$ is a number between $0$ and $x$.
In our case, all derivatives are $e^x$. If our $x$ is in the interval $[-2,2]$, then $|f^{(m+1)}(\xi)|\lt e^2$. And since $|x|\le 2$, we have $|x|^{m+1}\le 2^{m+1}$.
So in our situation, we have $$|R_m(x)|\lt \frac{e^2 2^{m+1}}{(m+1)!}.$$ Now you need to hunt around for an $m$ such that the above quantity is less than $1$. The expression is not a very nice one, so calculator experimentation is probably the best approach.
Remark: The problem is rather artificial, since in real work one would use the Maclaurin series for $x$ reasonably close to $0$, certainly not as far out as $x=2$.