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
>
>
>
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