Jeroen Demeyer on Mon, 04 Feb 2008 12:30:02 +0100 |
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integer n-roots like sqrtint() |
Hello list,I am wondering if there is an analogous function to sqrtint, but for n-th roots? With this I mean to compute floor(sqrtn(x,n)) exactly for integers x and n.
Unfortunately the obvious "floor(sqrtn(x,n))" does not work: gp> \p100 realprecision = 105 significant digits (100 digits displayed) gp> floor(sqrtn(6^3,3)) %33 = 5Of course I could do "floor(sqrtn(x,n) + 1e-90)" or something but that's not very elegant and doesn't work for very large numbers.
Cheers, Jeroen.