S =[2,1,2]
T =0
for P in range(200):
S += [1,1,(4 + 2*P)]
for x in range(len(S)-1):
T = 1 / (T + S[-(x+1)])
print(S[0] + T) # value of e(The article's original title -- "To compute a constant of calculus: A treatise on multiple ways" -- described its purpose better than the truncated one submitted here on HN. I see posters suggesting various solutions. Perhaps someone will create a public repo based on the article and then folk can add solutions.)
> did not see the extremely easy and quite efficient method of calculating e through continued fractions.
(That could be misinterpreted by readers of this thread who have not read the OP. The OP includes several continued fractions, including ones that seem simpler to me than the one you've shown, though that might be because I'm no expert.)
I'm curious what name is normally given to the fraction you showed? And are you saying it is more efficient than the Brother series?
Since e ≈ 2.7 1828 1828 459, it is very close to the rational number 271,801/99,990 = 2.7(1828) . A continued fraction expansion should pick up on that approximation quite quickly, which will make the expansion look efficient. Will the continued fraction continue converging as quickly if you keep going after you reach that particular rational approximation?
Moreover, they are uncommonly good if you truncate just before a large integer. So for π, 22/7 is a great approximation because the next number in the continued fraction is 15, but it is blown out of the water by 355/113 which is not specified to only ~0.3% accuracy as you might expect from the denominator, but rather 0.000008%. This comes from the splendidly large next integer 292.
e is more boring, the continued fraction is [2; 1, 2, 1, 1, 4, 1, 1, 6, ...], and that pattern indeed repeats forever. So you do keep getting these large integers, and they keep getting larger. That first one was 19/7 (0.1%), then 193/71 (0.001%), then 2721/1001 (0.000004%), then 49171/18089, which has about the same error as your approximation, but less than 1/5 of the denominator. So the continued fraction doesn't pick that one up because as far as the continued fraction is concerned it's not very good :D
Side note: It turns out that the above computational approach of "build continued fraction and then evaluate it" is not necessary and you can actually stream these. So you can take a + 1/(b + 1/(c + 1/x)) and rephrase it as sort of f_a(f_b(f_c(x)))). Simplifying by just looking at
> f_c(m/n) = c + n/m = (c m + 1 n)/(1 m + 0 n)
we can rewrite this as a matrix product,
[ a 1 ] * [ b 1 ] * [ c 1 ] * [ m ]
[ 1 0 ] [ 1 0 ] [ 1 0 ] [ n ]
Prelude> (a, b, c, d) .* (w, x, y, z) = (a*w + b*y, a*x + b*z, c*w + d*y, c*x + d*z) :: (Integer, Integer, Integer, Integer)
Prelude> (2, 1, 1, 0) .* (1, 1, 1, 0) .* (2, 1, 1, 0) .* (1, 1, 1, 0) .* (1, 1, 1, 0) -- starting point
(19,11,7,4)
Prelude> let matrices = scanl (\matrix n -> matrix .* (n,1,1,0) .* (1, 1, 1, 0) .* (1, 1, 1, 0)) (19, 11, 7, 4) [4,6..]
Prelude> take 5 matrices
[(19,11,7,4),(193,106,71,39),(2721,1457,1001,536),(49171,25946,18089,9545),(1084483,566827,398959,208524)]
Prelude>
Prelude> :m +Data.Ratio
Prelude Data.Ratio> take 8 $ map (\(a, b, c, d) -> a % c) matrices
[19 % 7,193 % 71,2721 % 1001,49171 % 18089,1084483 % 398959,28245729 % 10391023,848456353 % 312129649,28875761731 % 10622799089]
Side side note: it also follows that the most irrational number is 1 + 1/(1 + 1/(1 + 1/(1 + ...))) for this metric of distance, as it always has the smallest possible number in its continued fractions; its approximants are never "very good". Solving for φ = 1 + 1/φ and some quadratic formula reveals this as the golden ratio φ = (1 + √5)/2. If you do this process you will find that the matrix always contains Fibonacci numbers and this matrix [1, 1; 1 0] kind of encodes the Fibonacci recurrence directly.
x = 1 + 1/x
x - 1 - 1/x = 0
x^2 - x - 1 = 0
x = 1 ± √(1 + 4)
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