The Ramanujan Machine: Using algorithms to discover new mathematics
ramanujanmachine.com
The Ramanujan Machine: Using algorithms to discover new mathematics
1–10 of 29 posts
Re: The Ramanujan Machine: Using algorithms to discover new mathematics
#2Seems slightly less rich than Doug Lenat's Automated Mathematician (which in turn led to Eurisko which was one of the earliest genuinely interesting AI systems):
Re: The Ramanujan Machine: Using algorithms to discover new mathematics
#3Why on earth did they not call it the Ramanutron??
Re: The Ramanujan Machine: Using algorithms to discover new mathematics
#4Why on earth did they not call it the Ramanutron??
Possibly because that only really works if you know the correct stress pattern of "Ramanujan". Americans, at least, are likely to want to stress it on the "nu".
Re: The Ramanujan Machine: Using algorithms to discover new mathematics
#5It seems aptly named because Ramanujan also dealt with infinite series. One of his most famous equations is the ramanujan summation [1] where he derives that the sum of infinite series 1 + 2 + 3 + 4 + ... = - 1 / 12
Re: The Ramanujan Machine: Using algorithms to discover new mathematics
#6Related: Robert Munafo's RIES [1] tries to synthesize increasingly complex formula which solution is a given number.
Re: The Ramanujan Machine: Using algorithms to discover new mathematics
#7Well, the continuous fraction for e is pretty well known. Doubt they discovered anything new here that can't be obtained from the original formula.
On the other hand, the continuous fraction for pi is irregular, so it's interesting to see what they discovered... but I can't really find any pattern in the "conjectures" for pi. Take the first one:
pi/−4 = 1/(−1 + 1/(−4 + −2 /(−7 + −9/(−10 + −20/(−13+...))))
What exactly am supposed to prove here? The denominators are an arithmetic progression but numerators (1, 1, -2, -9, -20, ...) are just some bizarre sequence without an obvious pattern. The thing with continous fractions is that every number has one, so the fact that pi is presented as continous fraction is not impressive in itself.
Re: The Ramanujan Machine: Using algorithms to discover new mathematics
#8Well, the continuous fraction for e is pretty well known. Doubt they discovered anything new here that can't be obtained from the original formula. On the other hand, the continuous fraction for pi is irregular, so it's interesting to see what they discovered... but I can't really find any pattern in the "conjectures" for pi. Take the first one: pi/−4 = 1/(−1 + 1/(−4 + −2 /(−7 + −9/(−10 + −20/(−13+...)))) What exactl…
All their conjectures have low-degree polynomial formulae for both numerators and denominators, with exceptions allowed in the first couple. So I guess the nth numerator, counting from 0, is -n(2n-3) which goes 0, 1, -2, -9, -20, ..., so n=0 is a special case.
It would seem more natural to rewrite it like this
-4/pi = −1 + 1/(−4 + −2 /(−7 + −9/(−10 + −20/(−13+...))))
which avoids the gratuitously different first numerator -- but I guess they wanted to reproduce results exactly as they happened to emerge from their program.
I think this is, further, equivalent to the following which avoids some gratuitous-looking minus signs:
4/pi = 1 + 1/(4 + −2 /(7 + −9/(10 + −20/(13+...))))
Continuing the fraction using the quadratic polynomial I gave above does indeed seem to make it converge to 4/pi, though not very quickly.
Re: The Ramanujan Machine: Using algorithms to discover new mathematics
#9It seems aptly named because Ramanujan also dealt with infinite series. One of his most famous equations is the ramanujan summation [1] where he derives that the sum of infinite series 1 + 2 + 3 + 4 + ... = - 1 / 12 [1] https://en.wikipedia.org/wiki/Ramanujan_summation
I still don't understand why this derivation is such an achievement.
If anything, it seems like dirty hacking, but this time done by a mathematician.
Re: The Ramanujan Machine: Using algorithms to discover new mathematics
#10Well, the continuous fraction for e is pretty well known. Doubt they discovered anything new here that can't be obtained from the original formula. On the other hand, the continuous fraction for pi is irregular, so it's interesting to see what they discovered... but I can't really find any pattern in the "conjectures" for pi. Take the first one: pi/−4 = 1/(−1 + 1/(−4 + −2 /(−7 + −9/(−10 + −20/(−13+...)))) What exactl…
All their conjectures have low-degree polynomial formulae for both numerators and denominators, with exceptions allowed in the first couple. So I guess the nth numerator, counting from 0, is -n(2n-3) which goes 0, 1, -2, -9, -20, ..., so n=0 is a special case. It would seem more natural to rewrite it like this -4/pi = −1 + 1/(−4 + −2 /(−7 + −9/(−10 + −20/(−13+...)))) which avoids the gratuitously different first nume…
I wonder whether this one might be equivalent to a known infinite product for pi or pi/4 or something. The ratios between successive convergents are all of the form (a_n+k_n)/a_n where k_n is 1, 1, 3, 3, 15, 15, 105, 105, 315, 315, ... -- i.e., the least common multiples of odd numbers up to n. [EDITED to add:] This should make you think of the Wallis product formula for pi/4, though this doesn't seem to be the Wallis product in disguise.
Clearly these k_n divide n!, so maybe the right way to say this is that the conjecture seems to be equivalent to 4/pi = product (a_n+n!)/a_n where (a_n) = (4,130,2464,45448,882528,18410640,...) ... though I don't know what that sequence _is_, haven't shown that it has a nice form, etc. None of (a_n), (a_n+n!), (a_n+n!/2) seems to occur in OEIS or to be a subsequence of anything in OEIS.