The Ramanujan Machine: Using algorithms to discover new mathematics(ramanujanmachine.com)
ramanujanmachine.com
The Ramanujan Machine: Using algorithms to discover new mathematics
http://www.ramanujanmachine.com/
8 comments
The story of Eurisko doesn't add up. So advanced for its time, not replicated for 30 years (or at all) and Lenat lost the source code and was not interested in doing anything similar again, with the next version (Cyc, IIRC) not getting anywhere close?
I would guess there was a lot more manual handholding that was not documented.
I would guess there was a lot more manual handholding that was not documented.
Yes, I mostly agree here. But even with that manual handholding, it's no different from current, mostly-supervised AI approaches. Even if you think of Eurisko as a very clever optimiser when posed problems in a certain way, it clearly delivered some interesting results. Also, you _can_ find the source code for large parts of Eurisko if you look hard enough, I believe.
Cyc never seemed in the least bit interesting to me, tbh. Even today several "we taught our AI common sense!" articles have hit HN, and it's still not _really_ true.
Cyc never seemed in the least bit interesting to me, tbh. Even today several "we taught our AI common sense!" articles have hit HN, and it's still not _really_ true.
Well, 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.
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.
There are various continued fractions for e that were well-known: https://en.wikipedia.org/w/index.php?title=List_of_represent...
But the one found by this machine seems to be new; see the paper: http://www.ramanujanmachine.com/paper (or with some comments at https://fermatslibrary.com/s/the-ramanujan-machine-automatic... )
But the one found by this machine seems to be new; see the paper: http://www.ramanujanmachine.com/paper (or with some comments at https://fermatslibrary.com/s/the-ramanujan-machine-automatic... )
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.
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.
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.
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.
> the continuous fraction for pi is irregular
True, but there are regular generalized continuous fractions for pi [1].
[1] https://en.wikipedia.org/wiki/William_Brouncker,_2nd_Viscoun...
True, but there are regular generalized continuous fractions for pi [1].
[1] https://en.wikipedia.org/wiki/William_Brouncker,_2nd_Viscoun...
It 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
[1] https://en.wikipedia.org/wiki/Ramanujan_summation
Except he didn't. He created a way of assigning surrogate values to forms that otherwise wouldn't have one. To quote from the Wikipedia article you linked:
> assigning a value to divergent infinite series
Every time someone claims that series is EQUAL to -1/12, someone loses faith in math.
> assigning a value to divergent infinite series
Every time someone claims that series is EQUAL to -1/12, someone loses faith in math.
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.
If anything, it seems like dirty hacking, but this time done by a mathematician.
I read somewhere that the result is confirmed in applied physics where it somehow relates to the casimir force, but I'm not a physicist.
It's a theorem of analytic number theory. It's just a special value of the ζ-function. People misunderstand how analytic continuation works and therefore (wrongly) interpret ζ(-1) as the divergent series 1+2+3+... (The correct interpretation is as the unique analytic continuation of the holomorphic function $\sum_{n\geq 1}\frac{1}{n^s}$ defined on the half-plane $\Re(s)>1$ to $\mathbb{C}\setminus\{1\}$.)
It's actually a pretty simple consequence of the functional equation for ζ and a few special values of Γ. That in turn comes from a theta-function identity which can be proven using Poisson's summation formula.
What I'm trying to say is that it's legit maths, that has been distorted due to the shock value of writing the equation "1+2+3+... = -1/12".
If you're looking for a reference, go to Davenport's "Multiplicative Number Theory". It's short, self-contained, and extremely well-written. Serre's "A Course in Arithmetic" should also work.
It's actually a pretty simple consequence of the functional equation for ζ and a few special values of Γ. That in turn comes from a theta-function identity which can be proven using Poisson's summation formula.
What I'm trying to say is that it's legit maths, that has been distorted due to the shock value of writing the equation "1+2+3+... = -1/12".
If you're looking for a reference, go to Davenport's "Multiplicative Number Theory". It's short, self-contained, and extremely well-written. Serre's "A Course in Arithmetic" should also work.
I wonder if it somehow can be integrated with Coq[1] and Univalent Foundations[2]. Probably these can be used as a more substantial "base" for this machine.
[1] https://github.com/coq/coq
[2] https://github.com/UniMath
[1] https://github.com/coq/coq
[2] https://github.com/UniMath
Why 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".
Didn't Good Will Hunting teach the proper pronunciation?
Related: Robert Munafo's RIES [1] tries to synthesize increasingly complex formula which solution is a given number.
[1] https://mrob.com/pub/ries/
[1] https://mrob.com/pub/ries/
I was pondering a few weeks back about the prospects of using ML/AI to find new ways to factor primes and if their is any sequence in primes and how to calculate them in a way that you input N and it will produce the Nth prime.
That would be something.
That would be something.
This is not really ML for what it's worth, it's gradient descent which is just basic optimisation, there is no training data or anything like that.
re: using AI to learn to factor integers / find primes - probably not doable yet. There are neural networks that could model an algorithm that does it (memory networks, neural turing machine etc.). But any target algorithm would surely be too complicated for the neural net to converge towards it simply based on binary signals and gradient descent.
re: using AI to learn to factor integers / find primes - probably not doable yet. There are neural networks that could model an algorithm that does it (memory networks, neural turing machine etc.). But any target algorithm would surely be too complicated for the neural net to converge towards it simply based on binary signals and gradient descent.
We can already do this approximately (asymptotically): https://en.wikipedia.org/wiki/Prime_number_theorem
"The prime number theorem is equivalent to the statement that the nth prime number p_n satisfies
p_n ~ nlog(n)
the asymptotic notation meaning, again, that the relative error of this approximation approaches 0 as n increases without bound. For example, the 2^1017th prime number is 8512677386048191063, and (2^1017)log(2^1017) rounds to 7967418752291744388, a relative error of about 6.4%."
"The prime number theorem is equivalent to the statement that the nth prime number p_n satisfies
p_n ~ nlog(n)
the asymptotic notation meaning, again, that the relative error of this approximation approaches 0 as n increases without bound. For example, the 2^1017th prime number is 8512677386048191063, and (2^1017)log(2^1017) rounds to 7967418752291744388, a relative error of about 6.4%."
https://www.johndcook.com/blog/2019/06/20/bounds-on-the-nth-... is a recent blog posted titled "Bounds on the nth prime". It lists tighter upper and lower bounds:
from math import log
def f(n, k):
ln = log(n); lln = log(log(n))
return ln + lln - 1 + (lln-2)/ln - ((lln**2) - 6*lln+k)/(2*ln*ln)
>>> n = 2E17
>>> lo = n * f(n, 11.847)
>>> hi = n * f(n, 10.273)
>>> lo
8.512627944213742e+18
>>> hi
8.512727125430618e+18
>>> exact = 8512677386048191063
>>> (exact - lo)/exact
5.80802398678381e-06
>>> (exact - hi)/exact
-5.842977499434444e-06There's something called "Deep Algebra" (no links handy sorry) where they are trying to parse LaTeX source of math papers to extract equations and figure out new stuff using ML.
> you input N and it will produce the Nth prime
Look for Willans' formula (there's a nice explanation in the book "Hacker's Delight").
Look for Willans' formula (there's a nice explanation in the book "Hacker's Delight").
Can primes be estimated?
Through my genius I've come to the conclusion that primes are related to entropy. There is no hope.
The nth prime is about nlog(n).
"New mathematics" is a bit of an overstatement. There is a lot more to maths than continued fractions.
It bugs me a bit that the authors write (I would love for anyone affiliated with the project to talk to me about this) "Any new conjecture, proof, or algorithm suggested will be named after you.". No offense, but there are very few mathematicians out there with that kind of a world view.
Seems a bit like a high school project without proper guidance from a mathematician.
It bugs me a bit that the authors write (I would love for anyone affiliated with the project to talk to me about this) "Any new conjecture, proof, or algorithm suggested will be named after you.". No offense, but there are very few mathematicians out there with that kind of a world view.
Seems a bit like a high school project without proper guidance from a mathematician.
https://en.wikipedia.org/wiki/Automated_Mathematician