The previous post mentioned that Martin Gardner announced that Ramanujan’s conjecture that exp(π√163) in an integer had been proven. This was an April Fool’s joke in 1975. Gardner said
Working by hand, he [Ramanujan] found the value to be 262537412640768743.999999999999… The calculations were tedious, and he was unable to verify the next decimal digits.
Calculating exp(π√163) without a computer—Ramanujan died in 1920—would indeed be tedious, but not insurmountable. Certainly it would not stop someone like Ramanujan from testing a conjecture.
How might you go about calculating exp(π√163) by hand?
Algorithm
One possibility is an algorithm in [1].
where r = t′ / 256, t′ = t − n log 2, and n is chosen to minimize |t′|.
We can choose n so that |t′| < log(2)/2 and so |r| < 0.014. This means the infinite series converges rapidly and not too many terms will be needed, depending on the desired precision.
The calculation x256 can be done by squaring 8 times.
Although the context of this post is hand calculations, this would also be a viable algorithm for a program doing extended precision calculations.
Example
In our case, t = π√163 = 40.1091… and we choose n = 58 so that t′ = −0.09336….
Then r = −0.000364….
So each term in the series will contribute 3 or 4 decimal places to the desired precision at first, more once the factorial denominators get large.
Related posts
[1] Jonathan Borwein and David Bailey. Mathematics by Experiment. Volume 1.
So one might ask the next obvious question: using that series and a reasonable known value of pi at the time (and a reasonable calculation of the square root), how many terms would one need to get those twelve 9s on the joke?