Approximating e with 2s and 3s - MathOverflow

Approximating e with 2s and 3s

2011-03-01T05:28:12Z

How can I generate a series of 2s and 3s such that the average of the generated values (so far) is as close to e as possible?

For example:

3:       avg=3      |2-e|    =0.282
3,2:     avg=2.5    |2.5-e|  =0.218
3,2,3:   avg=2.667  |2.667-e|=0.052
3,2,3,3: avg=2.75   |2.75-e| =0.032

Also, how can I quickly check if the nth index in the sequence is a 2 or 3?

Answer by Gerry Myerson for Approximating e with 2s and 3s

2011-03-01T05:30:18Z

Isn't the $n$ term just the closest integer to $ne$, minus the closest integer to $(n-1)e$?

Answer by Kevin O'Bryant for Approximating e with 2s and 3s

2011-03-01T14:29:55Z

As Gerry points out, the sequence $$a_n = [n e] - [(n-1)e],$$ where $[x]$ is the integer closest to $x$, has the desired extremal property. Unfortunately, one needs to know the value of $e$ to calculate the sequence in this way.

Fortunately, this is a typical example of a Sturmian sequence (on the alphabet $\{2,3\}$), and they can be generated quickly from the continued fraction expansion (of $e$, in this case). If one uses the floor function in place of rounding, this has already been worked out by Ken Stolarsky and Tom Brown, and you can find a simple proof in this article, which was published in Integers. This gives you quickly a large initial segment of the sequence; you cannot jump directly to $a_{1000000}$.

I haven't seen any detailed exposition using the "round" function (or ceiling function), but presumably it follows from the same principles.

A putman-ish followup question is to find a combinatorial process that generates a sequence $b_n$ with $\frac 1n \sum_{i=1}^n b_i \to e$. I don't have an answer for that. Yet.

Answer by S. Carnahan for Approximating e with 2s and 3s

2011-03-02T17:43:38Z

To compute the $n$th term in this sequence, you really only need decent estimates on the fractional parts of $(n-1)e$ and $ne$ (following Gerry Myerson's solution) - you get 2 if and only if the fractional part of $(n-1)e$ lies in $[0.5,1)$ and the fractional part of $ne$ lies in $[0,0.5)$. To find the fractional parts, you typically need about $m$ large integer divisions, where $m$ is such that $m!$ is a bit larger than $n$. A modern computer can do this quite quickly: SAGE took about 1 second to find that the $10^{100000}$th term is 3, and about 55 seconds to find that the $10^{1000000}$th term is also 3.

Edit: I'm still quite confused about Kevin O'Bryant's comments to the effect that knowledge of $e$ affects the operation count. To direct the conversation, I'll include some SAGE code that computes which half of the unit interval contains the fractional part of $ne$. An output of 0 means the fractional part lies in the lower half, while an output of 1 means it lies in the upper half.

def fracpart(n):
    ipart = n
    fpart = 0
    acc = 0
    k = 1
    while ipart != 0 or ceil(2*acc)-2*acc < 2/k:
        (ipart,rem) = ipart.quo_rem(k)
        fpart = RDF(rem/k + fpart/k)
        acc = acc + fpart
        if acc >= 1: acc = acc - 1
        k = k + 1
    return floor(2*acc)

The large integer divisions occur in the function quo_rem, while the other divisions are small. This code will return the correct answer for all but less than one out of a billion of the reasonable inputs - the remaining cases (where floating point precision isn't good enough) can be dealt with by using high-precision reals, removing the letters "RDF" to switch to rationals, or using some modular arithmetic to work with remainders.

The code uses the fact that $e$ expands as a sum of reciprocals of factorials in an essential way, but there doesn't seem to be any point where it explicitly computes the number $e$ itself. I'm not sure if this quality exempts the program from the previous criticism.