Approximate $e^x$ with bounded-size sliding window over its Taylor series? Suppose I want to approximate $e^x$ for very large $x$. Suppose my tolerance were constant, for example, within a factor of $e$ (between $e^{x-1}$ and $e^{x+1}$). Are theres constants $k$ and $n$ such that for all $x > n$, I could approximate $e^x$ by just evaluating $k$ of its Taylor series terms? I don't mean starting from the beginning. I can evaluate any $k$ consecutive terms. How would I find this out?
What if, instead of a constant error tolerance, I can tolerate $O(\log(x))$ error? What is the error size like if I use $\log(x)$ terms?
To further expand, given any function $f(x)$ and its Taylor series, how many terms of its Taylor series, $T(x)$, would I need to approximate $f(x)$ with an error $E(x)$?

I do not know if this will answer the question.
Consider first the problem of
$$e^{-x}=\sum_{n=0}^\infty (-1)^n \frac {x^n}{n!}=\sum_{n=0}^p (-1)^n \frac {x^n}{n!}+\sum_{n=p+1}^\infty (-1)^n \frac {x^n}{n!}$$ and you want to know $p$ such that
$$\sum_{n=p+1}^\infty (-1)^n \frac {x^n}{n!} \leq 10^{-k}$$ Since this is an alternating series it suffices to find $p$ such that
$$\frac {x^{p+1}}{(p+1)!}\leq 10^{-k}\quad \implies \quad (p+1)! \geq x^{p+1} \,10^k$$
If you look at this question, you will find that
$$p \sim x \, e^{1+W(t)}-\frac 32 \quad \text{where} \qquad t=\frac{1}{2 e x}\log \left(\frac{10^{2 k}}{2 \pi  x}\right)$$ $W(t)$ being Lambert function.
For example, using $x=5$ and $k=20$, this would give, as a real $p=39.0514$ (the exact solution being $p=39.0519$); so, using the ceiling function $p=40$.
Now, compute
$$\sum_{n=0}^{40} \frac {5^n}{n!}=\color{red}{148.41315910257660342111}40$$ while
$$e^5=\color{red}{148.4131591025766034211156}$$
