I need to find the smallest value of n such that the remainder estimate $|R_n| \le \frac{M}{(n+1)!} (x-a)^{n+1}$, where M is the maximum value of $|f^{(n+1)}(z)|$ on the interval between a and the indicated point, yields $|R_n| \le \frac{1}{1000}$ on the interval $f(x) = e^{-2x}, [-1,1]$. $a$ is expanded around $0$.
The whole concept of Taylor series is completely new to me. I don't like posting questions without some type of attempt to answer, but I have no idea where to start.
Too advanced but written for your curiosity.
The series expansion of $e^{-2x}$ around $x=0$ is given by $$e^{-2x}=\sum_{n=0}^\infty (-1)^n\frac {2^n}{n!}x^n$$ which is an alternating series. So, making the problem more general,you want to know the smallest value of $n$ such that $$\frac {2^{n+1}}{(n+1)!} \leq 10^{-k}$$ For the time being, let $m=n+1$ and the problem is the same as finding the solution of $$m!=2^m \,10^k$$ If you look at this question of mine, you could see a magnificent approximation provided by @robjohn $$m!=a^m 10^k\implies m\sim ea\exp\left(\operatorname{W}\left(\frac k{ea}\log(10)-\frac1{2ea}\log(2\pi a)\right)\right)-\frac12$$ where appears Lambert function.
Applied to your case $(a=2)$ and taking into account that $t=W(t)\,e^{W(t)}$, this would simplify to $$n\sim\frac{k \log (100)-\log (4 \pi )}{2 W\left(\frac{k \log (100)-\log (4 \pi )}{4 e}\right)}-\frac32$$
You can easily compute $W(t)$ using the series expansions given in the linked Wikipedia page.
For $k=3$ , this would give in the real domain $n=8.22$ then, for your question $n=9$.
Let us check $$\frac{2^9}{9!}=\frac{4}{2835}\approx 0.00141 > 0.001\qquad \frac{2^{10}}{10!}=\frac{4}{14175}\approx 0.00028 < 0.001$$
I put below a table of the solutions for different values of $k$ (in blue the exact solution in the real domain) $$\left( \begin{array}{cccc} k & n \simeq & \lceil n \simeq \rceil & \color{blue}{n} \\ 1 & 4.89390 & 5 & \color{blue}{ 4.89949 } \\ 2 & 6.67841 & 7 & \color{blue}{ 6.68202 } \\ 3 & 8.21655 & 9 & \color{blue}{ 8.21926 } \\ 4 & 9.61264 & 10 & \color{blue}{ 9.61483 } \\ 5 & 10.9126 & 11 & \color{blue}{ 10.9144 } \\ 6 & 12.1416 & 13 & \color{blue}{ 12.1432 } \\ 7 & 13.3153 & 14 & \color{blue}{ 13.3167 } \\ 8 & 14.4442 & 15 & \color{blue}{ 14.4455 } \\ 9 & 15.5358 & 16 & \color{blue}{ 15.5369 } \\ 10 & 16.5956 & 17 & \color{blue}{ 16.5966 } \\ 11 & 17.6279 & 18 & \color{blue}{ 17.6289 } \\ 12 & 18.6361 & 19 & \color{blue}{ 18.6370 } \\ 13 & 19.6228 & 20 & \color{blue}{ 19.6237 } \\ 14 & 20.5904 & 21 & \color{blue}{ 20.5912 } \\ 15 & 21.5406 & 22 & \color{blue}{ 21.5414 } \\ 16 & 22.4751 & 23 & \color{blue}{ 22.4758 } \\ 17 & 23.3951 & 24 & \color{blue}{ 23.3957 } \\ 18 & 24.3018 & 25 & \color{blue}{ 24.3024 } \\ 19 & 25.1962 & 26 & \color{blue}{ 25.1968 } \\ 20 & 26.0792 & 27 & \color{blue}{ 26.0797 } \end{array} \right)$$
$M$ is the maximum value of $|f^{(n+1)}(z)|$ on the interval between $a$ and the indicated point ($x$)
That is where you start. Can you find a formula for $f^{(n)}(z)$? Can you use that formula to find an upper bound for $|f^{(n+1)}(z)|$?
