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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?

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    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.2012-11-10

5 Answers 5

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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 $\frac{1}{n!}\left(1+\frac{1}{n}\right).$

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    @Steven, [I'm not so sure](http://www.cs.tau.ac.il/~nachumd/horror.html). But I sure hope no one is computing the $n$'th digit of $e$ by expanding the Taylor series for $e^x$ around $0$ at $x=1$ to some point determined by Stirling's formula.2012-11-10
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You can use the remainder term in Taylor's expansion

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    @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.2012-11-10
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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.

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    @PeterTamaroff: nah... it still costs 1 rep to downvote an answer (afaik).2012-11-10
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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.

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    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.2012-11-10
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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 $c$.

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    @glebovg: Since $f$ was not speci$f$ied in the question, it would be good to mention it in the answer.2012-11-09