Amazing approximation to e

The Endeavour 2014-03-31

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 10²⁴.)

$e \approx \left( 1 + 9^{{-4}^{7ḑot6}}\right)^{3^{2^{85}}}$

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

$3^{2^{85}} \log \left( 1 + 9^{{-4}^{42}\right) \approx 1.$

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

$3^{2^{85}} \log \left( 1 + 9^{{-4}^{42}\right) = k \left( \frac{1}{k} - \frac{1}{2\eta^2} \right)$

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.