# An amazing approximation of $e$

As we can read in Wolfram Mathworld's article on approximations of $e$, the base of natural logarithm,

An amazing pandigital approximation to e that is correct to $18457734525360901453873570$ decimal digits is given by $$\LARGE \left(1+9^{-4^{6 \cdot 7}}\right)^{3^{2^{85}}}$$ found by R. Sabey in 2004 (Friedman 2004).

The cited paragraph raises two natural questions.

1. How was it found? I guess that Sabey hasn't used the trial and error method.
2. Using which calculator can I verify its correctness "to $184\ldots570$ decimal digits"?
• "Using which calculator..." that is not math within the scope defined by the help center, but the first question is answerable. Sep 28, 2016 at 13:37
• It is not that astonishing, once we realize it is a small variation on $$e=\lim_{n\to +\infty}\left(1+\frac{1}{n}\right)^n$$ Sep 28, 2016 at 13:41
• @JackD'Aurizio The astonishing bit is the pan-digitality, I think. Sep 28, 2016 at 13:46
• The Friedman reference is www2.stetson.edu/~efriedma/mathmagic/0804.html See also quora.com/… and stackoverflow.com/questions/4721650/checking-approximation-of-e Sep 28, 2016 at 13:46
• An even better approximation comes from $$e-\left(1+\frac{1}{n}\right)^{n+1/2}=-\frac{e}{12 n^2}+O\left(\frac{1}{n^3}\right).$$ Sep 28, 2016 at 13:51

\begin{aligned} (1+9^{-4^{42}})^{3^{2^{85}}} &=(1+9^{-4^{42}})^{3^{2*2^{84}}}\\ &=(1+9^{-4^{42}})^{9^{2^{84}}} \\ &=(1+9^{-4^{42}})^{9^{4^{42}}}\\ &=\Bigl(1+\frac1{9^{4^{42}}}\Bigr)^{9^{4^{42}}}\qquad\text{where }=\left(1+\frac1n\right)^n. \end{aligned} This is just the limit definition of $e$ with a large number as an approximation for $\infty$.
Edit: Numberphile just did a video on this, which also gives a pandigital approximation for $\pi$, but it's only accurate up to ten digits.
• @Eric No, it's not easy at all to make it pandigital, all digits from $1$ to $9$ are used exactly once. Sep 28, 2016 at 14:57
• That's right, all you need is a large $n$ for that. Sep 28, 2016 at 14:59