Consider the following polynomial expression.

Numerical_Precision =200
xt= Symbol("xt" , real=True)
xt_b=-0.121
test_exp = xt**2+xt +1

The question is how to expand this polynomial expression with high accuracy.

method_1= test_exp.subs(xt, xt+xt_b).expand().evalf(Numerical_Precision )
#display(method_1)
method_2= test_exp.subs(xt, xt+xt_b).evalf(Numerical_Precision ).expand()  # in each case, more preciece
#display(method_2)
#display(method_1-method_2)

At xt_b=0.121, the correct constant term ought to be 1.1356411, and

#float 1.135641
#method_1 1.1356409999999999005382278483011759817600250244140625
#method_2 1.135640999999999995587529610929777858362092781876032636587657044524579674771302961744368076324462890625

At xt_b=-0.121, the correct constant term ought to be 0.893641, and

#float 0.8936409999999999
#method_1 0.89364100000000001866595766841783188283443450927734375
#method_2 0.893641000000000002692956968530779717073335457657282636587657044524579674771302961744368076324462890625

It appeared that there’s no control over the precision when using the .expand() method, and in each case, the .evalf(Numerical_Precision ).expand() allowed high precision computation.

However, the former implied a potential reduction of precision at .expand() method, and I wonder if there’s a way to increase the precision.

In a further investigation, at the Numerical_Precision =200 and xt_b=-0.121, the accuracy for the more accurate method_2 had already deviate at 0.893641000000000002692 the 19th digit. The increase of Numerical_Precision for method_2 seemed to have been limited by the .expand(), where at Numerical_Precision =20, the constant term for method_2had already been 0.89364100000000000269.

Is there a way to increase the precision of .expand() method in sympy? I tried .expand(Numerical_Precision) but it didn’t affect the result.