* smaller than the relative error in z^n/n (the absolute error of thelatter is the
* absolute error in z)
*/
BigDecimal c = divideRound(z.pow(n, mcloc), n);
MathContext m = new MathContext(err2prec(n * z.ulp().doubleValue() / 2. / z.doubleValue()));
c = c.round(m);
/* At larger n, zeta(n)-1 is roughly 1/2^n. The product is c/2^n.
* The relative error in c is c.ulp/2/c . The error in the product should be small versus eps/10.
* Error from 1/2^n is c*err(sigma-1).
* We need a relative error of zeta-1 of the order of c.ulp/50/c. This is an absolute