Takes less than 1 sec here to do (10**100)**(1./13) a million times, and
only about half as long to do (1e100)**(1./13), or about 14 times as
long as to do .2**2. Doesn't look like one could hope for it to be that
much quicker as you need 9 sig figs of accuracy to get the integer part
of (10**100)**(1./13) (floats have about 7 and doubles about 16).
Dan
Tim wrote:
In PythonWin I'm running a program to find the 13th root (say) of
millions of hundred-digit numbers. I'm using
n = 13
root = base**(1.0/n)
which correctly computes the root to a large number of decimal
places, but therefore takes a long time. All I need is the integer
component. Is there a quicker way?
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