Numerical Polynomials

If the input is wrong, then the output will be wrong as well.

symbols

The solver computes with complex numbers and the imaginary unit can be denoted by i or I. For example:

p = 'x - i;'

which represents the polynomial x - i. While i and I are symbols, they should not be used as the names of variables, because i and I are interpreted as numbers.

The polynomials can be written in factored form. For example:

p = '(y - 2)*(x + y - 1)^2;'

To enter this polynomial, first, the number of variables must be set.

from phcpy.dimension import set_double_dimension, get_double_dimension
set_double_dimension(2)
get_double_dimension()

which returns 2, confirming the number of variables.

from phcpy.polynomials import set_double_polynomial, get_double_polynomial
set_double_polynomial(1, 2, p)
get_double_polynomial(1)

shows

'y^3 + 2*y^2*x + y*x^2 - 4*y^2 - 6*y*x - 2*x^2 + 5*y + 4*x - 2;'

Observe that the variable y appears first in the polynomial.

from phcpy.polynomials import string_of_symbols
string_of_symbols()

confirms the list of symbols used as variable names:

['y', 'x']

If that is inconvenient, the simple trick is to add the null polynomial x - x in front of the string representation to ensure x comes first.

q = 'x - x + ' + p
q

defines the string 'x - x + (y - 2)*(x + y - 1)^2;'

To continue, the polynomial that has been set must be cleared.

from phcpy.polynomials import clear_double_system
clear_double_system()

set_double_dimension(2)
set_double_polynomial(1, 2, q)
get_double_polynomial(1)

which shows

'x^2*y + 2*x*y^2 + y^3 - 2*x^2 - 6*x*y - 4*y^2 + 4*x + 5*y - 2;'

Polynomials added to the system will follow the same order of symbols.

r = 'y**2 + 4*x + y - 1;'
set_double_polynomial(2, 2, r)
get_double_polynomial(2)

confirms

'y^2 + 4*x + y - 1;'

Consider the entire system of polynomials:

from phcpy.polynomials import get_double_system
get_double_system()

returns the list

['x^2*y + 2*x*y^2 + y^3 - 2*x^2 - 6*x*y - 4*y^2 + 4*x + 5*y - 2;',
 'y^2 + 4*x + y - 1;']

numbers

The letters e and E appear in the scientific notation of floating-point numbers. Therefore, these letters e and E should not be used as the names of variables.

Even as coefficients can be entered as exact rational numbers, they are evaluated to floating-point numbers.

clear_double_system() # restart

p = 'x - 1/3;'

Entering p goes then as follows:

set_double_dimension(1)
set_double_polynomial(1, 1, p)
get_double_polynomial(1)

which then shows the decimal floating-point expansion of 1/3:

' + x - 3.33333333333333E-01;'

Let us double the precision, first importing the corresponding double double functions:

from phcpy.dimension import set_double_double_dimension
from phcpy.polynomials import set_double_double_polynomial

The statements

from phcpy.polynomials import get_double_double_polynomial
set_double_double_dimension(1)
set_double_double_polynomial(1, 1, p)
get_double_double_polynomial(1)"

then show

' + x-3.33333333333333333333333333333324E-1;'

Laurent polynomials

Laurent polynomials have integer numbers as expeonents and therefore admit negative values. Some problems benefit from representations as Laurent polynomials. Consider for example the cyclic 5-roots problem:

from phcpy.families import cyclic
for pol in cyclic(5): print(pol)

which shows

x0 + x1 + x2 + x3 + x4;
x0*x1 + x1*x2 + x2*x3 + x3*x4 + x4*x0;
x0*x1*x2 + x1*x2*x3 + x2*x3*x4 + x3*x4*x0 + x4*x0*x1;
x0*x1*x2*x3 + x1*x2*x3*x4 + x2*x3*x4*x0 + x3*x4*x0*x1 + x4*x0*x1*x2;
x0*x1*x2*x3*x4 - 1;

The cyclic n-roots problem reformulated as a Laurent system, appears in a paper of Uffe Haagerup, available as

from phcpy.families import cyclic_reformulated
for pol in cyclic_reformulated(5): print(pol)

which shows

x1 + x1^-1*x2 + x2^-1*x3 + x3^-1*x4 + x4^-1;
x2 + x1^-1*x3 + x2^-1*x4 + x3^-1 + x4^-1*x1;
x3 + x1^-1*x4 + x2^-1 + x3^-1*x1 + x4^-1*x2;
x4 + x1^-1 + x2^-1*x1 + x3^-1*x2 + x4^-1*x3;

Compared to the original cyclic 5-root problem, the Laurent version has a more regular structure.