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Yes, it looks like you're right. But I don't understand the reasoning - is he actually saying you can use this formula to compute pi^n to n decimal places? So i
by dmurray 4y ago
Yes, it looks like you're right. But I don't understand the reasoning - is he actually saying you can use this formula to compute pi^n to n decimal places? So it doesn't produce arbitrary digits on their own, but excellent approximations to pi^n?
And this was by taking 4 terms of the zeta function infinite product expansion. 2xn digits would presumably be given by taking more terms (but quite a few, one for every prime under 100, so in some sense this method has bad convergence when you want the mth digit of pi^n and m >> n).
- mattkrause 4y agoYeah, my sense is that the author knows this is more of a neat trick than a "practical" way of calculating pi. There's a bit on page 5 about how the Chudnovsky method totally blows this approach out of the water.
- dmurray 4y agoSeems like the real breakthrough was computing large Bernoulli numbers without a very precise value for pi, and this is a relatively easy corollary for someone to spot and get a paper out of.