[The best with four digits is 2721/1001, and the best with five 25946/9545, the latter being the closest to e among all fractions with digits less or equal to five in the numerator].

]]>The most memorable rational number by this measure might be 3331/2220, kicking off Beethoven’s 5th (with 0 representing 7 from the previous octave).

]]>I think that 2721/1001 for e is the closest analog to 355/133 for pi. Here’s why:

If you look for the most accurate 3-digit rational representation of pi you indeed

get 355/133. If you look for the most accurate 4-digit rational representation of pi

then *by luck*, when reduced to lowest terms, you again get 355/133. This is why

355/133 is so good. For e, 2721/1001 is the best 4-digit rational approximation. The approximation error is comparable (actually it’s about half of) the approximation error of 355/133. 2721/1001 is better than a single precision… so it’s pretty decent. ]]>

import math

const = math.e # or math.pi or (math.sqrt(5) + 1) / 2 for phi

minErr = 1

for num in range(2, 999):

den = round(num / const)

err = abs(num/den - const) / const

if err < minErr:

minErr = err

print(str(num) + "/" + str(den) + "tt" + str(err))

The most accurate 3-digit rational approximation to e is also arguably the most memorable: 878/323, although perhaps less so than 355/113 for pi. Unfortunately it only gives us 4 correct digits after the point, so you’re still better off remembering the decimal digits. Of course, the above code only produces the most accurate approximations at the expense of omitting less accurate but more memorable ones. Now if only someone could come up with an algorithm to determine how memorable a rational number is!

]]>Whereas for pi it just so happens that there are some large c.f. coefficients early on: 3,7,**15**,1,**292**,1,1,1,2, etc.. Truncating before the 15 gives you 22/7, and truncating before the 292 gives you 355/113.

Of course “memorable” isn’t the same as “good”, but those memorable approximations to pi are only worth remembering because they *are* good. (If you don’t care about accuracy, I hereby propose the following memorable approximation to e: it’s about 3.)

@John did you consider embedding the tweets in your blog? They are:

https://twitter.com/divbyzero/status/296606825078460417

https://twitter.com/divbyzero/status/296606880367796225

the decimal expansion: 2721/1001… it’s the closest rational number with at most a 4-digit numerator. The “closest” with at most a 4-digit denominator is 25946/9545, which is closer but less memorable (to me anyway).

I’ve played around to see if there are sequences that seem to have nice patterns

in the digits, even in different bases, but I found nothing really striking so far… More generally it would be interesting if there digits in some base were algorithmically simple, like the BBP formula for pi in base 16. AFAIK, there are no BBP type formulas for e, though…

Here’s a memorable sequence. Suppose you’ve memorized the decimal expansion of *e* to *n* decimal places. Then use floor(10^n e)/10^n as a rational approximation. 🙂 OK, well maybe we could do better.

I suppose you could at sequences of digits that are memorable in the sense that they have some sort of pattern, then do a brute force search to see which best approximate *e*.

(24*n^2 – 12*n + 11) * n^(n-2)

@John: Interesting again. Thanks for the reference. (I did realize at least that none of the sequences I gave are very practical for the simple reason that the numerators/denominators climb so rapidly.)

]]>Here’s the answer: for any pair of integers *p* and *q*, the error | *e* – *p*/*q*| is always at least 0.5 log log *q*/(*q*^{2} log *q*). On the other hand, if you replace 0.5 with any larger number, you can always find a pair *p* and *q* where the inequality is reversed.

Source: C. S. Davis (1978). Rational approximations to e. Journal of the Australian Mathematical Society, 25, pp 497502 doi:10.1017/S1446788700021480.

]]>doing successively better corrections for the error:

1) c[n] = (n+1)^n

d[n] = n^n

2) c[n] = 2* (n+1)^n

d[n] = (2*n – 1)* n^(n-1)

3) c[n] = 24* (n+1)^n

d[n] = (24*n^2 – 12*n + 11) * n^(n-1)

Here is a boring sequence, but it probably is worth mentioning: just truncate the

Taylor series for exp(1) after some number of terms. At step n you’ll find the

rational approximation c[n]/d[n] where you can compute the terms via

c[0] = 1

d[0] = 1

and

c[n+1] = (n+1) c[n] + 1

d[n+1] = (n+1)d[n]

So you get 1, 2, 5/2, 16/6, 65/24, 326/120. 326/120 isn’t too bad.

]]>