Approximating e using the digits 1–9

The number e (approximately 2.718281828459045) is one of the most important irrational numbers in mathematics, being the base of the natural logarithm. Remarkably, it is possible to get a stunningly accurate approximation to e using the digits 1 up to 9, exactly once each, together with the standard operations of addition, negation, multiplication and exponentiation. According to the formula in the picture, e is close to
(1 + 9^{–4^{7x6}})^{3^{2^{85}}}.

So why does this work? The key to understanding what is going on is the number N=3^{2^{85}}, an astronomically huge number. The number N appears more subtly in another place in the formula, because 4^{7x6}, or 4^{42}, is equal to 2^{84}. Since 9 is 3^2, it follows that 9^{4^{7x6}} is equal to {3^2}^{2^{84}}, which is equal to 3^{2x(2^{84})}, or 3^{2^{85}}. This means that 9^{–4^{7x6}} is equal to 1/N.

In summary, the formula in the picture is equal to (1+1/N)^N for a really large value of N. The limit of this formula, as N tends to infinity, is well known to be exactly equal to e.

I'd be interested to know where this formula originally came from. I found it on twitter via Colin Beveridge and Chris Smith. Chris Smith says that the approximation is accurate to over 1.8x10^{25} decimal places. I haven't checked this, but it is certainly a very accurate approximation.

Wikipedia has much more about the number e here: http://en.wikipedia.org/wiki/E_(mathematical_constant)

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