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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"?
Rodrigo de Azevedo
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Santiago
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    "Using which calculator..." that is not math within the scope defined by the help center, but the first question is answerable. – Simply Beautiful Art Sep 28 '16 at 13:37
  • More on 1. can be found here. – Dietrich Burde Sep 28 '16 at 13:40
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    It is not that astonishing, once we realize it is a small variation on $$e=\lim_{n\to +\infty}\left(1+\frac{1}{n}\right)^n $$ – Jack D'Aurizio Sep 28 '16 at 13:41
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    @JackD'Aurizio The astonishing bit is the pan-digitality, I think. – Daniel Fischer Sep 28 '16 at 13:46
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    The Friedman reference is http://www2.stetson.edu/~efriedma/mathmagic/0804.html See also https://www.quora.com/How-did-Richard-Sabey-come-up-with-the-below-equation-which-is-accurate-to-over-a-septillion-digits and http://stackoverflow.com/questions/4721650/checking-approximation-of-e – Gerry Myerson Sep 28 '16 at 13:46
  • Well, $$ e-\left(1+\frac{1}{n}\right)^n = \frac{e}{2n}+O\left(\frac{1}{n^2}\right), $$ hence if $n$ is a colossal number, the approximation is extremely good. Still no wonder. – Jack D'Aurizio Sep 28 '16 at 13:50
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    An even better approximation comes from $$ e-\left(1+\frac{1}{n}\right)^{n+1/2}=-\frac{e}{12 n^2}+O\left(\frac{1}{n^3}\right).$$ – Jack D'Aurizio Sep 28 '16 at 13:51
  • The second question is answerable too, and I believe that it can be within the community's scope. The main point would be to communicate that any decimal representation of an expression is the result of an algorithm. Typical calculators just cut off the execution of this algorithm to a few digits, but in principle we can go on. And there are plenty examples where we have done just that... continue the execution of the algorithm (think of the billions of digits of $\pi$ we know of). Next step would be to give examples of algorithms that would calculate any digit of the expression given above. – Thanassis Sep 28 '16 at 13:54
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    One more reference: http://math.stackexchange.com/questions/449877/pandigital-rational-approximations-to-the-golden-ratio-and-the-base-of-the-natur – Gerry Myerson Sep 28 '16 at 13:55
  • @GerryMyerson thank you very much for finding all the references. You can post the comments as an answer and I will accept it. – Santiago Sep 28 '16 at 14:57
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    The point is that you don't have to compute the number to know how many digits are accurate. You "just" compute the error term and take its base $10$ log. – Ross Millikan Sep 28 '16 at 15:02
  • WolframAlpha reproduces the cited error using the (first) approximation cited by Jack above: $$\frac{e}{2\cdot 9^{4^{42}}}=10^{\log_{10} e/2\cdot 9^{4^{42}}}=\frac{1}{10^{-\log_{10}(e/2)+4^{42} \log_{10} 9}}\approx \frac{1}{10^{18457734525360901453873569}}$$ – Semiclassical Sep 28 '16 at 15:15
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    @Jack, can you come up with a pandigital instance of the better approximation? – Gerry Myerson Sep 28 '16 at 23:20

2 Answers2

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$$\begin{aligned} (1+9^{-4^{42}})^{3^{2^{85}}} &=(1+9^{-4^{42}})^{3^{2*2^{84}}}\\ &=(1+9^{-4^{42}})^{9^{2^{84}}} \\ &=(1+9^{-4^{42}})^{9^{4^{42}}}\\ &=\Bigl(1+\frac1{9^{4^{42}}}\Bigr)^{9^{4^{42}}}\qquad\text{where }=\left(1+\frac1n\right)^n. \end{aligned}$$ 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.

AlgorithmsX
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This may be best answered by looking for an improvement to that formula, and explaining how to find it.

The basic idea for all those approximations for $e$ is to write some large number $N$ in two different ways, using exactly the digits from $2$ to $9$, and then take $(1+\frac{1}{N})^{N}$ as an approximation for $e$. In Sabey's old (2004) example, he writes $N_{old}=3^{2^{85}}=9^{4^{6*7}}$. However, using a small digit like $3$ as the base of the exponential tower isn't optimal. For example, if we could find a way to replace the $3$ by a $5$, that would make $N$ much larger, and the resulting approximation much better...
Finding such an $N$ that works actually isn't too hard: Using $N_{new}=5^{3^{84}}$ frees up the digit $2$, which conveniently can be used to write $\frac{1}{5}$ as either $.2$ or $0.2$, depending on how you like to write your decimal numbers. At the same time, we have lost access to the digit $4$, which we have to take care of: Using properties of the exponential function, we can write $N_{new}=5^{9^{42}}=5^{9^{6*7}}$. The digit $4$ is no longer needed, so $(1+.2^{9^{6*7}})^{5^{3^{84}}}$ or $(1+0.2^{9^{6*7}})^{5^{3^{84}}}$ is a valid pan-digital solution, which is equal to $e$ up to $8368428989068425943817590916445001887164$ decimal places.
I'd be surprised if even this new solution was optimal. I replaced the digit $3$ by a $5$ in the base of the exponential tower, but using any of the digits $6$ through $9$ could lead to an even better solution. Maybe give it a try yourself!

About your second question:

The formula $e=\lim_{N\rightarrow\infty}{(1+\frac{1}{N})^{N}}$ actually approaches $e$ in a very predictable way. If you have an approximation $(1+\frac{1}{10^{n}})^{10^{n}}$, it will differ from $e$ by exactly $1.359...\times10^{-n}$, regardless of the value of $n$ (which can be any sufficiently large number). As a consequence, the number of correct decimal digits will always be equal to either $n$ or $n-1$. In general, $(1+\frac{1}{N})^{N}$ will be correct up to $\log_{10}{N}$ digits (or sometimes one less).
The number $3^{2^{85}}$ has $18457734525360901453873570$ digits, and Sabey's approximation is correct to that number of decimal places. The new solution $5^{3^{84}}$ has $8368428989068425943817590916445001887165$ digits, and my approximation turns out to be correct to $8368428989068425943817590916445001887164$ decimal places.
Unfortunately, I am not aware of an online calculator that will tell you whether a particular approximation is correct to $n$ or $n-1$ digits, respectively, but maybe you're okay with knowing the answer to within $\pm 1$? You can check the size of $N_{new}$ using WolframAlpha.

Erich Friedman actually asks on his Math Magic to get in touch if improvements to his puzzle solutions are found, which I just did. I guess he'll update his page accordingly.

I hope this helps!

Daniel Bamberger
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