Oscar you mentioned below finding the # of digits using log and the code ; In [1]: from sympy import mpmath In [2]: mpmath.mp.dps = 950 In [3]: mpmath.mpf('1.4142') ** 6000 % 400
How accurate of a return do you feel this will give? Are we losing any data, will the return be very accurate? > 10**d = 1.4142**6000 > Which gives that > d = log10(1.4142**6000) = 6000*log10(1.4142) ~= 903 > > So if you want an answer that's good for m digits you'll need to use > about 900+m digits for the exponentiation: > > In [1]: from sympy import mpmath > In [2]: mpmath.mp.dps = 950 > In [3]: mpmath.mpf('1.4142') ** 6000 % 400 > Out[3]: mpf('271.048181008630921815939109488389 > > BTW it's also possible to do this particular calculation easily enough > just with Python stdlib (and without numpy or sympy): > > In [1]: from fractions import Fraction > > In [2]: float(Fraction('1.4142')**6000 % 400) > Out[2]: 271.04818100863093 > > > -- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to sympy+unsubscr...@googlegroups.com. To post to this group, send email to sympy@googlegroups.com. Visit this group at https://groups.google.com/group/sympy. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/6da7b2d0-ce6f-4eb5-886f-36af91ac7ebe%40googlegroups.com. For more options, visit https://groups.google.com/d/optout.