[math-fun] computing products like (1 - 2/p^2)
This is quite nice. For my particular products Prod[p prime] (1 - J/p^K) you can get faster convergence by dividing by Prod[p prime] (1 - 1/p^K)^J = Zeta(K)^J and then the new product converges faster, since the discrepancy from 1 no longer is dying like 1/p^K but rather like a higher power of p. Then at the end you divide out your Zeta power (which you already knew to high precision) to reconstitute the original. However, the papers by Moree, Nivasch, and Cohen go beyond that trick. As far as I can tell from not reading them (reader software: no deal), that was only the first trick in an infinite set of tricks. You can then do it again to cancel out the next-most-dominant term to get even faster convergence, and so on forever. Thus after the infinite number of tricks are applied, your product gets re-expressed as a certain product of integer powers of zeta functions at integer arguments. So, all these kinds of prime products are evaluable to ultrahigh precision.
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Warren Smith