What is the most efficient numerical base system?

Solution 1:

First, let's specify the sources. From Wikipedia:

The base $e$ is the most economical choice of radix $\beta > 1$ (Hayes 2001), where the radix economy is measured as the product of the radix and the length of the string of symbols needed to express a given range of values.

Then we have a longish article by Hayes in American Scientist, which I don't feel like reading. The matter boils down to: the representation of number $n$ in base $\beta$ (integer or not) takes $\approx \log n/\log \beta$ digits. If your idea of "economy" is the product of this length with $\beta$ then of course, you are going to minimize $\beta /\log \beta$ and find that the minimum is at $\beta=e$. For example, with Wolfram Alpha, which can plot this function and compute its derivative.

Solution 2:

Assume there are $V$ independent states of information. Then we can represent approximately $\frac{V}{N}$ digits in base $N$.

The amount of information we can represent is $I=N^{\frac{V}{N}}$.

The value of $N$ that maximizes $I$ (either where the derivative is $0$ (if the second derivative is negative) or at infinity (if the second derivative is positive)) is the most "efficient" base.

So we take the natural log: $\ln(I)=\frac{V}{N}\ln(N)$

And the derivative: $(\ln(I))'=V\frac{(1-\ln(N))}{N^2}$.

We then set $(\ln(I))'=0$. Solving, $N=e$.

Take the second derivative: $(\ln(I))''=V\frac{(2\ln(N)-3)}{N^3}$

When $N=e$, $(\ln(I))''=-\frac{V}{e^3}$ which is negative (recall that $V$ is positive), so $I$ reaches its maximum when $N=e$.