# Calculating $e$

If I calculate $e$ using the following formula.

$$e = \sum_{k=0}^{\infty}{\frac{1}{k!}}$$

Is it possible to predict how many correct decimal places I get when I stop summing at $n$ terms?

-
There is a subtlety in the contrast between "correct decimal places" and accuracy. Most of us had interpreted "$n$ correct decimal places as within $10^{-n}$, but as Dan Brumleve points out, you could be very close. If the correct answer is $1.9999$, an error of $10^{-4}$ can change the ones digit. Having a string of $9$'s is rare, but if you care about it you need to check. – Ross Millikan Nov 10 '12 at 4:37

If we use $n$ terms, the last term used is $\dfrac{1}{(n-1)!}$. The missing "tail" is therefore $$\frac{1}{n!}+\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}\cdots.\tag{1}$$ Note that $(n+1)!=n!(n+1)$ and $(n+2)!\gt n!(n+1)^2$, and $(n+3)!\gt n!(n+1)^3$ and so on. So our tail $(1)$ is less than $$\frac{1}{n!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\cdots \right).$$ Summing the geometric series, we find that the approximation error is less than $$\frac{1}{n!}\left(1+\frac{1}{n}\right).$$

-
This is a good answer, why was it downvoted? – robjohn Nov 9 '12 at 19:45
To end the error vs correct digits discussion: If you calculate $e_n:=\sum_{k=0}^n\frac1{n!}$ and $e_n+\frac1{n!}(1+\frac1n)$ and both agree on $d$ decimals, then both estimates are correct to (at least) $d$ decimals. – Hagen von Eitzen Nov 9 '12 at 20:39
@DanBrumleve How about instead of just downvoting (something that can't be undone after a few minutes) 10 (a slight exaggeration) different answers without a single comment (until after someone asked why), leave a comment and clarify. Then, if it turns out you misunderstood, you haven't just downvoted 10 different answers incorrectly. – Graphth Nov 9 '12 at 22:41
@DanBrumleve Well, obviously many other people (everyone who answered?) believe otherwise. So, is it possible that at the very least it's just your preference? And, if so, is it really worth downvoting? – Graphth Nov 9 '12 at 23:31
@DanBrumleve While this is more of a meta issue, IMHO there is a difference between 'this is not how I would have answered the question' or even 'this is not what I consider a complete answer to the question' and 'this is an unhelpful answer to the question' - I would consider the latter to be a minimal standard for downvoting (as opposed to simply not voting), and the FAQ seems to agree: 'Use your downvotes whenever you encounter an egregiously sloppy, no-effort-expended post, or an answer that is clearly and perhaps dangerously incorrect.' This answer is hardly egregious or dangerous. – Steven Stadnicki Nov 10 '12 at 1:08

You can use the remainder term in Taylor's expansion

-
To what function are we applying Taylor's theorem? – robjohn Nov 9 '12 at 19:49
@robjohn the definition of $e$ given by OP is actually the Taylor series for $e^x$ centered at $x=0$, evaluated at $x=1$. – Logan Stokols Nov 9 '12 at 23:48
@Logan: As I mentioned in a comment to glebovg, since $f$ was not specified in the question, it would be good to mention it in the answer. – robjohn Nov 10 '12 at 0:04

In this answer, it is shown, by comparison to a geometric series, that $$0\le n!\left(e-\sum_{k=0}^n\frac1{k!}\right)\le\frac1n$$ Therefore, the error after $n+1$ terms is at most $\frac1{nn!}$ .

To $n$ decimal places:

When asking for a number to $n$ decimal places, there are two common meanings

1. the error is less than $\frac12\times10^{-n}$.

2. the value is correct when rounded to $n$ decimal places. As has been pointed out, if a number is very close to $10^{-n}\left(\mathbb{Z}+\frac12\right)$, rounding to $n$ decimal places might require computing more decimal places to know the actual $n^{\mathrm{th}}$ digit of the rounded number. This is not as easy to use as meaning 1, so it is not as commonly used.

-
That was a fast downvote. Care to comment? – robjohn Nov 9 '12 at 20:21
Is someone just downvoting for free? – Pedro Tamaroff Nov 10 '12 at 0:24
@PeterTamaroff: nah... it still costs 1 rep to downvote an answer (afaik). – robjohn Nov 10 '12 at 0:26

The series converges rapidly. If you stop at $\frac 1{ k!}$ you can bound the error by $\frac 1{k(k!)}$ by bounding the remaining terms with a geometric series.

-
Why was this downvoted? It seems a bit scant, but valid. – robjohn Nov 9 '12 at 19:43
I downvoted for the same reason as most of the others: the question asks about the number of correct leading digits in the partial sum and this upper bound on the error term doesn't lead to any obvious answer to that question. – Dan Brumleve Nov 10 '12 at 4:29

The $n$-th Taylor polynomial is $${P_n}(x) = f(0) + \frac{{f'(0)}}{{1!}}x + \frac{{f''(0)}}{{2!}}{x^2} + \cdots + \frac{{{f^{(n)}}(0)}}{{n!}}{x^n}$$ (in this case $f(x)$ is simply $e$) and the error we incur in approximating the value of $f(x)$ by $n$-th Taylor polynomial is exactly $$f(x) - {P_n}(x) + \frac{{{f^{(n + 1)}}(c)}}{{(n + 1)!}}{x^{n + 1}}$$ where $0 < c < x$. This form of the remainder can be used to find an upper bound on the error. If the difference above is positive, then the approximation is too low, and likewise if the error is negative, then the approximation is too high. We only need to find an appropriate $c$.

-
To what function are we applying Taylor's theorem? – robjohn Nov 9 '12 at 19:50
@robjohn In this case $e^x$ for $x = 1$, but it works in general. – glebovg Nov 9 '12 at 20:17
Why downvote? Please comment. – glebovg Nov 9 '12 at 20:17
I downvoted because it doesn't address the distinction between the error term and the number of correct digits in the partial sum. (Also consider simplifying or clarifying: the question asks only about $e$ not $e^x$.) – Dan Brumleve Nov 9 '12 at 20:21
@glebovg: Since $f$ was not specified in the question, it would be good to mention it in the answer. – robjohn Nov 9 '12 at 22:49