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The Ramanujan Machine: Using algorithms to discover new mathematics

ramanujanmachine.com

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Re: The Ramanujan Machine: Using algorithms to discover new mathematics

#11

Post by mci »

Grue3 wrote: post #7

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 exactl…

> 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...

Re: The Ramanujan Machine: Using algorithms to discover new mathematics

#13

Post by m3kw9 »

Zenst wrote: post #12

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.

Can primes be estimated?

Re: The Ramanujan Machine: Using algorithms to discover new mathematics

#14

Post by drinfinity »

m3kw9 wrote: post #13
Zenst wrote: post #12

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.

Can primes be estimated?

Through my genius I've come to the conclusion that primes are related to entropy. There is no hope.

Re: The Ramanujan Machine: Using algorithms to discover new mathematics

#15

Post by klipt »

Zenst wrote: post #12

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.

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%."

Re: The Ramanujan Machine: Using algorithms to discover new mathematics

#16

Post by sweezyjeezy »

Zenst wrote: post #12

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.

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: The Ramanujan Machine: Using algorithms to discover new mathematics

#17

Post by svat »

Grue3 wrote: post #7

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 exactl…

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... )

Re: The Ramanujan Machine: Using algorithms to discover new mathematics

#18

Post by bryanrasmussen »

thaumasiotes wrote: post #4
mometsi wrote: post #3

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?

Re: The Ramanujan Machine: Using algorithms to discover new mathematics

#20

Post by eesmith »

klipt wrote: post #15
Zenst wrote: post #12

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.

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…

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)
With that in place, and once I realized that 2^1017 mean 2E17 not 2 to the power of 1017:

   >>> 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-06
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