Here’s an approximation to *e* by Richard Sabey that uses the digits 1 through 9 and is accurate to over a septillion digits. (A septillion is 10^{24}.)

MathWorld says that this approximation is accurate to 18457734525360901453873570 decimal digits. How could you get an idea whether this claim is correct? We could show that the approximation is near *e* by showing that its logarithm is near 1. That is, we want to show

Define *k* to be 3^(2^85) and notice that *k* also equals 9^(4^42). From the power series for log(1 + *x*) and the fact that the series alternates, we have

where η is some number between 0 and 1/*k*. This tells that the error is extremely small because 1/*k* is extremely small. It also tells us that the approximation underestimates *e* because its logarithm is slightly less than 1.

Just how small is 1/*k*? Its log base 10 is around -1.8 × 10^25, so it’s plausible that the approximation is accurate to 10^25 decimal digits. You could tighten this argument up a little and get the exact number of correct digits.

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Well, it is not exactly VERY surprising. (1+1/n)^n converges to e. In this case, you have an extremely large n = 3^(2^85)

Any idea why WolframAlpha says indeterminate?

http://www.wolframalpha.com/input/?i=%281%2B9%5E-4%5E%287*6%29%29%5E3%5E2%5E85

@vonjd: Because 3^2^85 generates an overflow.

…so this result is more of theoretical interest only?

John, do you mean that η (eta) should be a little bigger than k? That would make the second term small. If I’m following, then that error term would be something like 1/k–which means that the size of the error has a linear relation to the size of the input k. What would be really fun is some other approximation where this relation is a square or a cube.

@vonjd For this specific result, probably. If you’re using it for numerics (except perhaps for arbitrary precision perhaps) you’re better off just loading a pre-defined bit pattern which is accurate to whatever precision you want. (And possibly also for arbitrary precision).

For analytic purposes, you just keep e symbolic.

However, the method of finding the number of digits something is valid to (though perhaps in bits instead of decimal) is more generally useful.

IMHO. 🙂

Even more amazing if you add (15-14)/(11+13-10-12) to the exponent (It will roughly double the number of digits)