Computing the error function for Euler's number

Question

By the error function for the sum $$\sum_{i = 0}^\infty \frac{1}{i!},$$ I mean the function $$f : \mathbb{R}_{> 0} \rightarrow \mathbb{N}$$ defined as follows.

For each $\varepsilon \in \mathbb{R}_{>0}$, $f(\varepsilon)$ is the least natural number $n$ such that $$\left|e - \sum_{i = 0}^n \frac{1}{i!}\right| < \varepsilon$$

More generally, any convergent series has a corresponding error function.

Question. Regarding the above series for Euler's number $e$, is it known how to compute $f(\varepsilon)$ when $\varepsilon$ is an explicitly known rational number written as a quotient of two coprime integers?

You souldn't say the "error function". This expression is already used for something quite different. — , Mar 25 '19 at 10:10
@YvesDaoust, I'm well aware of that, but I don't think there's much scope for confusion here. I'll change my mind if you can suggest a situation where there could plausibly be some confusion, however. — goblin GONE, Mar 25 '19 at 10:11
The rational quality of $\epsilon$ doesn't seem important: the function is piecewise constant. — , Mar 25 '19 at 10:13
I mean that you can obtain $n$ almost exactly for a given $\epsilon$. — Claude Leibovici, Mar 25 '19 at 11:39
@ClaudeLeibovici, oh. So basically you're just saying that having seen the linked formula, you can never unsee it, because it's cool. Have I interpreted you correctly? — goblin GONE, Mar 25 '19 at 11:50
Not quite what you are looking for but closely related. Specifically, a technique for estimating the tail of power series using geometric series, see "Bounding Power Series Remainders" by Mark Bridger and John Frampton. Available from JSTOR. — Carl Christian, Mar 25 '19 at 21:31

score 1 · Answer 1 · answered Mar 25 '19 at 10:18

From Taylor's theorem, you know that

$$ e - \sum_{i=1}^n \frac{1}{i!} = \dfrac{e^{\xi_n}}{(n+1)!}, \quad \xi_n \in [0,1], $$

so your condition would be equivalent to

$$ \dfrac{e^{\xi_n}}{(n+1)!} < \varepsilon $$

Now, is the best case scenario, setting $\xi_n = 0$ we would get $$ (n+1)! > \frac{1}{\varepsilon} $$

and in the worst case scenario, $\xi_n=1$ we would get $$ (n+1)! > \frac{e}{\varepsilon} $$

Solving these inequalities will provide bounds for $f(\varepsilon)$ but in general will not give you the best $n$ that you are looking for.

Claude Leibovici · Answer 2 · 2019-03-26T10:30:09.780

Starting from PierreCarre's answer, let us consider the worst case and look for the solution of the equation $$(n+1)! = \frac{e}{\epsilon}$$ If you look at this question of mine, you will see a magnificent approximation by @robjohn for the inverse of the factorial (have a look here).

Edit

To make it clearer, the approximate solution of $m!=a$ is given by $$m \sim \exp\left(1+W\left(\frac{\log \left(\frac{a^2}{2 \pi }\right)}{2 e}\right) \right)-\frac 12$$

Computing the error function for Euler's number

2 Answers2