I have a rational function in $\mathbb{Q}(x)$ which can be computed by substitution of a polynomial in the polynomial ring with integer compositions as variables, with coefficients in $\mathbb{Q}(x)$. The compositions can have unbounded parts. For example, one monomial might be $\frac{1}{1-x}C_{[2,2]}C_{[1,3,1]}$, where $C_\lambda$ is a formal variable related to the integer composition $\lambda$.
I would be happy to have this "symbolic" representation as polynomial over the set of all integer compositions, and not just the polynomial after substituting all of the $C_\lambda$'s. I have tried the following two approaches which failed.
- Using the
InfinitePolynomialRingclass. The setup used:
sage: R = FractionField(QQ["x"])
sage: R.inject_variables()
Defining x
sage: RC.<c> = InfinitePolynomialRing(R)
sage: f = 1/(1-x)*c[3] + 3*c[2]
But then f.subs does not work:
sage: f.subs({c[3]:5})
...[snip]
TypeError: <class 'sage.rings.polynomial.infinite_polynomial_ring.InfinitePolynomialGen'> is not hashable
sage: f.subs(c[3]=5) # Expected not to work.
SyntaxError: keyword can't be an expression
There is also a method called f.specialization which seems to work with substitution of just one variable, but not with more than one, nor when using an element from InfinitePolynomialRing(R, "cd"). There is also the annoyance of using a bijection between the natural numbers and compositions.
- Using
CombinatorialFreeModule. I know this might seem as an abuse, but a module can also have the structure of a ring...
sage: R = FractionField(QQ["x"])
sage: FM = FreeAbelianMonoid(Compositions())
sage: CR = Rings().Commutative()
sage: CMR = ModulesWithBasis(R)
sage: M = CombinatorialFreeModule(R, FM, category=(CR,CMR)) # Does not work also with category=None
sage: a = M.an_element(); a*a
...[snip]
TypeError: 'NotImplementedType' object is not callable
I assume some magic with the category argument might help. Note that in my case only a free abelian semigroup indexed by compositions is needed, and not a monoid.
Any help is welcomed.
