Sympy polynomials with `mpfr` coefficients?

Question

I want to use Sympy's polynomials, but I also want to use higher-precision coefficients.

Just Doing It seems to give me polynomials with sympy.core.numbers.float coefficients.

import sympy
from sympy import Poly
from sympy.abc import x
from gmpy2 import mpfr, get_context

get_context().precision = 150

#float64 can't tell this from 1.0
one_and_change = mpfr('1.0000000000000000000000000000000000001')
#mpfr('1.0000000000000000000000000000000000001000000005',150)

p = [one_and_change]
px = Poly(p, x)

print(px)
# Poly(1.0, x, domain='RR')
print(px.is_one)
# True
print(type(px.all_coeffs()[0]))
# <class 'sympy.core.numbers.Float'>

I've also tried sympy.mpmath.mpf , with the same results.

This also didn't work:[ 1 ]

domain = sympy.polys.domains.realfield.RealField(150)
px = Poly(p, x, domain=domain)
print(type(px.all_coeffs()[0]))
# <class 'sympy.core.numbers.Float'>

Answer 1

There are a few obstacles:

• gmpy.mpfr has no ._sympy_ method, so it will be converted to float in an intermediate step.
• sympy.Poly , by default, uses sympy.polys.domain.RR for floating point coefficients, and will use it to convert.
• RR , when it loads, uses 53 as its precision, and ignores mpmath.mp.precision .
• When converting, RR uses RR.precision and ignores the argument's precision.

Solution:

1. Coefficients must be a Sympy type (eg sympy.Float , which has extended precision).
2. For the domain, one of the following:
   • Set sympy.polys.domain.RR._context.prec = my_precision .
   • Pass domain='EX' , which doesn't do conversion.
   • Pass in a custom domain (such as sympy.polys.domains.realfield.RealField(150) ).