An approximation to e

Take the following pandigital formula:

(1 + 9^{− 46 × 7})³²⁸⁵

Or, in a slightly more readable format:

+-------------------------+
|                      85 |
|           (6*7)    2^   |
|        -4^       3^     |
| (1 + 9^        )^       |
+------------------------ +

What do you get if you evaluate this number?

Let N = 3²⁸⁵. It is a very large number, with approximately 1.8 × 10²⁵ digits! It can be rewritten like so:

3²⁸⁵ = 9²⁸⁴ = 9⁴⁴² = 9^{4(6 × 7)}

Then, the following is true:

1/N = 9^{− 46 × 7}

Then, the formula listed above can be expressed like so:

(1 + (1/N))^N

Does it remind you of something?

It is known that the constant known as e can be defined as:

e = lim_{n → ∞}(1+(1/n))ⁿ

What happens if n is not infinity, but a very large number? Well, you get an approximation to e; its accuracy will depend entirely on how large n is.

So, the expression above must be a good approximation, right? Indeed it is, accurate to over 10²⁵ digits (error calculated to be –2.01 × 10^{18457734525360901453873570}). This can be seen when trying to evaluate the natural logarithm of the above expression (See this blog post for the whole enchilada):

3²⁸⁵log ((1 + 9^{− 46 × 7})) ≈ 1?

Neat, isn’t it? This little formula uses all nonzero digits and evaluates to one of the most famous mathematical constants. It was discovered by Richard Sabey and was submitted to Erich Friedman’s page of mathematical problems and recreations «Math Magic» as the solution to the monthly problem presented in August 2004:

This month’s problem is to approximate famous mathematical constants using only the first n digits (each used exactly once) and the mathematical symbols + – × / ( ) . and ^.

To the best of my knowledge, there is no real world application that needs this much accuracy, but then again, when has that stopped recreational mathematics? See the linked page for some truly amazing approaches to this and related problems.

Gilgamesh ← Andy’s Brevity Quest 2019 (297 words) → Grigori Perelman